Strength Of Materials: Mechanical Engineering Civil Engineering

STRENGTH OF MATERIALS For MECHANICAL ENGINEERING CIVIL ENGINEERING

STRENGTH OF MATERIALS SYLLABUS

Stress and strain, elastic constants, Poisson's ratio; Mohr’s circle for plane stress and plane strain; thin cylinders; shear force and bending moment diagrams; bending and shear stresses; deflection of beams; torsion of circular shafts; Euler’s theory of columns; energy methods; thermal stresses; strain gauges and rosettes; testing of materials with universal testing machine; testing of hardness and impact strength.

ANALYSIS OF GATE PAPERS MECHANICAL 1 Mark 2 Mark Exam Year Ques. Ques. Total 2003 6 4 14 2004 3 4 11 2005 2 6 14 2006 2 5 12 2007 2 6 14 2008 4 8 20 2009 2 2 6 2010 1 2 5 2011 4 3 10 2012 3 3 9 2013 2 1 4 2014 Set-1 2 3 8 2014 Set-2 3 2 7 2014 Set-3 2 2 6 2014 Set-4 2 2 6 2015 Set-1 3 2 5 2015 Set-2 2 5 12 2015 Set-3 1 1 3 2016 Set-1 3 4 11 2016 Set-2 2 4 10 2016 Set-3 2 3 8 2017 Set-1 5 4 13 2017 Set-2 3 4 7 2018 Set-1 4 5 14 2018 Set-2 2 4 10

Exam Year 2003 2004 2005 2006 2007 2008 2009 2010 2011 2012 2013 2014 Set-1 2014 Set-2 2015 Set-1 2015 Set-2 2016 Set-1 2016 Set-2 2017 Set-1 2017 Set-2 2018 Set-1 2018 Set-2

CIVIL 1 Mark 2 Mark Ques. Ques. Total 2 5 12 1 6 13 2 3 8 3 9 21 3 5 13 1 8 17 2 5 12 5 2 9 1 3 7 3 3 9 2 2 9 1 3 7 1 5 11 2 2 6 2 2 6 3 6 4 8 2 2 6 1 4 9 1 3 7 1 2 5

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CONTENTS Topics 1.

Page No

SIMPLE STRESS AND STRAIN 1.1 1.2 1.3 1.4 1.5 1.6 1.7 1.8 1.9 1.10 1.11 1.12

2.

STRAIN ENERGY AND IMPACT LOAD

1 1 2 5 5 6 7 7 7 8 9 9 11

3.

2.1 Strain Energy Gate Questions

29 31

3.1 3.2 3.3 3.4 3.5 3.6 3.7

BENDING STRESS AND SHEAR STRESS

34 35 37 37 37 37 39 46

4.1 Pure Bending Moment 4.2 Beam under Uniform Strength 4.3 Direct Shear Stress Gate Questions

49 50 51 54

4.

5.

Introduction Stresses and Strains Types of stress and strain Hook’s law and Elastic Property True Stress and True Strain Elastic Constants Types of Materials Strain Under Tri Axial Loading Axial Stress in Compound Bars Elongation of the Taper Beam under Axial Load Elongation Due to Self-Weight of the Body Thermal Stress Gate Questions

SHEAR FORCE AND BENDING MOMENT DIAGRAM Shear Force and Bending Moment Beam Free Body Diagram Method of Analysis Equations of Equilibrium Example Problems Special Case of Shear Force Diagram and Bending Moment Gate Questions

PURE TORSION

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5.1 5.2 5.3 5.4 6.

Pure Torsion Composite Shafts Spring Connection of Springs Gate Questions

70 71 73 76 77

Introduction Normal Stress and Shear Stress Principal Planes and Principal Stress Maximum Shear Stress Plane and Maximum Shear Stress Principal Strain and Shear strain Combined Bending & Torsion Mohr’s circle Gate Questions

88 88 88 88 89 89 90 91

Introduction Types of Column Slenderness ratio Load analysis of column Pressure vessel Gate Questions

SLOPE AND DEFLECTION

95 95 95 95 96 100

8.1 Introduction 8.2 General Expression 8.3 Methods for Slope &Deflection 8.4 Special Case of Slope and Deflection

107 107 107 109

COMBINED STRESS AND STRAIN 6.1 6.2 6.3 6.4 6.5 6.6 6.7

7.

8.

9.

COLUMN AND PRESSURE VESSEL 7.1 7.2 7.3 7.4 7.5

ASSIGNMENT QUESTIONS

10. CIVIL GATE QUESTIONS

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114 126

1

SIMPLE STRESS AND STRAIN

1.1 INTRODUCTION 1. ENGINEERING MECHANICS The engineering mechanics may be defined as a branch of applied mechanics that deals with behaviors of solid bodies subjected to various types of loadings. This is usually subdivided into further two streams i.e Mechanics of rigid bodies or simply Mechanics and Mechanics of deformable solids. The mechanics of deformable solids is known by several names i.e. strength of materials, mechanics of materials etc. 2. MECHANICS OF RIGID BODIES

The mechanics of rigid bodies is concerned with the static and dynamic behavior under external forces of engineering components and systems which are treated as infinitely strong and under formable primarily we deal here with the forces and motions associated with particles and rigid bodies. Mechanics of rigid bodies is further divided in to two subdivision i.e. kinematics of mechanics and dynamics of mechanics.

the stresses, strains, and produced by loads. 1.2 STRESSES AND STRAINS

deflections

1. STRESS As we know that in mechanics of deformable solids, externally applied forces acts on a body and body suffers a deformation. From equilibrium point of view, this action should be opposed or reacted by internal forces which are set up within the particles of material due to cohesion. These internal forces give a concept of stress. Therefore, let us define a stress. Let us consider a rectangular bar of some cross – sectional area and subjected to some load or force. Let us imagine that the same rectangular bar is assumed to be cut into two halves at section XX. The each portion of this rectangular bar is in equilibrium under the action of load P and the internal forces acting at the section XX has been shown

3. STRENGTH OF MATERIALS

Strength of material is branch of mechanics that concerns with study of the forces and its effect on the properties of the deformed body. It is concerned with the internal forces and associated changes in the geometry of the components. Particular objective of SOM is that whether the components fail by breaking in service, and the amount of deformation they suffer is acceptable. Therefore, the subject of mechanics of materials or strength of materials is central to the whole activity of engineering design. Usually the objectives in analysis here will be the determination of

The stress can be defined as total internal resistance per unit area of the component. But after removal of the load, body regains its original position, so total internal resistance should be equal to applied load. So stress can be given by, P Stress ( σ ) = A

The basic units of stress are N / m2 (or Pa), N/mm2, MPa, GPa.

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MPa = 106  Pa GPa = 109  Pa KPa = 103  Pa

compressive stresses are considered as negative. 2. STRAIN

If a bar is subjected to a direct load and body Some times N / mm2 units are also used, is deformed body there will be deformation because this is an equivalent to MPa. under action of load. If the bar has an For direct stresses, if area under original length L and changes by an amount consideration is original area, then it is δL, The strain produce is defined as follows known as Engineering stress or nominal Change of length δL = Strain ( ε ) = stress or simply stress. But, if the area Original length L taken is actual area at the time of considering the stress, then stress is known as true stress. P P Nominal Stress = ; True Stress = A Ao Where, Ao = Original area of specimen. A = Actual area of specimen. Proof Stress: When a material such as Aluminum does not have an obvious yield point and yet undergoes large strains after the proportional limit is exceeded, an arbitrary yield stress may be determined by the offset method. A line parallel to initial linear part is drawn, which is offset by some standard amount of

Unit of Strain: Strain is a measure of the deformation of the material and is a non dimensional Quantity. It is simply a ratio of two quantities with the same unit. So strain is dimension less quantity. Sign convention for strain

Tensile strains are positive whereas compressive strains are negative. 1.3. TYPES OF STRESS AND STRAIN Strain such as 0.2%. The intersection of the off set point (A) defines the yield stress or off set yield stress, which is slightly above the proportional limit and is called proof stress.

1. TYPES OF STRESS According to direction of loading, Stress can be divided into following way:

Sign convention for stress Tensile Stresses or pulling Stresses are considered as positive whereas

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Stress produced due to load in the form of bending moment or eccentric axial load is known as bending stress. Nature of bending stress is tensile and compressive. Shear stress (τ )

Normal stress ( σ ) We have defined stress as force per unit area. If the stresses are normal to the areas concerned, then these are termed as normal stresses. The normal stresses are generally denoted by a Greek letter ( σ ).

Tensile or compressive stresses The normal stresses can be either tensile or compressive whether the stresses act out of the area or into the area.

Let us consider now the situation, where the cross – sectional area of a block of material is subject to a distribution of forces which are parallel, rather than normal, to the area concerned.

Such forces are associated with a shearing of the material, and are referred to as shear forces. The resulting force per unit area is known as shear stress. The resulting force intensities are known as shear stresses, the mean shear stress is given by Shear Stress τ =

Axial Stress (σ a ) Stress produced due to load acting along the axis of the component is called as axial stress. It may be tensile or compressive in nature. P Axial Stress = Ao Where, Ao = Original area of specimen

P A

Where P is the total force and A the area over which it acts. 2. TYPES OF STRAIN

According to basis of the deformation, strain can be classified as

Bending Stress (σ b )

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Normal Strain ( ε ) Deformation produced under the action of the load in the direction of the load is called as linear or normal strain. In normal strain, dimension of the body changes but the shape of the body remains same. Longitudinal Strain

The linear deformation in the direction of the load is known as the longitudinal deformation. It may be tensile and compressive in nature depend upon the nature of the loading.

Longitudinal = Strain ( ε x )

Change of length δL = Original length L

Lateral Strain The linear deformation in perpendicular direction of the load is known as the lateral deformation. It may be tensile and compressive in nature depend upon the nature of the longitudinal Strain. Change of diameter δD Lateral Strain ( ε y ) = = Original diameter D

Volumetric Strain It is the ratio of total change of the volume to the original volume of the component. Change of volume Volumetric Strain ( ε v ) = − Original volume

ε v =-

When a body is subjected to shearing stresses, the shape of the body gets distorted. The measurement of this distortion is done by angle of distortion. Under pure shear the shape of the body gets distorted but the volume remains same. shear stress Modulus of rigidity, ( G ) = shear strain τ G= γ As we know that the shear stresses acts along the surface. The action of the Stress is to produce or being about the deformation in the body considers the distortion produced shear strain on an element or rectangular block.

If under the shear, the shear strain is γ then the linear strain in the diagonal of the specimen is given by. γ εd = i.e. Linear strain of diagonal is half of 2 the shear strain in the body. 3. STRAIN SECTIONS

FOR

DIFFERENT

CROSS

a) Rectangular Co- ordinates Let us consider a rectangular block of length L, width D and thickness t. after application of load P in x direction change in dimension is dL, dD and dt respectively. The strain produced in x- direction is given by

dV V

Change of length dL = = (εx ) Original length L  Every longitudinal strain is associated The strain produced in y- direction is given with two lateral strains. by  Longitudinal strain and lateral strain are opposite in nature. Change of width dW = εy ) = ( Original width W Shear strain: The strain produced in z- direction is given by © Copyright Reserved by Gateflix.in No part of this material should be copied or reproduced without permission

Change of thickness dt = Original thickness t The volume of the rectangular body is given by V = Lwt Taking the log in both sides and differentiating the equation, volumetric strain dV dL dw dt εV = = + + V L w t ε V =ε x + ε y + ε z

(εz )

b) Cylindrical co-ordinates Let us consider a cylinder of length L, diameter. After application of load P in x direction change in dimension is dL, dD respectively. The strain produced in x and ydirection is given by Change of length dL (εx ) = Original length L Change of diameter dD (εr ) = Original diameter D Volume of the cylinder is given by π V = D2 L 4 Taking the log in both sides and differentiating, we get volumetric strain

εV =

dV dL dD = +2 = ε l + 2ε c V L D

c) Spherical co-ordinates Let us consider a sphere of diameter D. After application of load P in x direction change in diameter is dD respectively. Volumetric strain produced in spherical body is given by dV dD εV = = 3 = 3εc V D 1.4 HOOK’S LAW & ELASTIC PROPERTY

is removed. According to Hook’s law, Up to proportional limit stress is directly proportional to strain. Stress ∝ strain Stress = Const. = E Strain This constant is given by the symbol E and is termed as the modulus of elasticity or σ Young's modulus of elasticity E= ε The unit of modulus of elasticity is MPa.

 Curve 1 represents actual or true stress strain diagram.  Curve 2 represents theoretical stress strain diagram.  Hooke`s law is valid up to limit of proportionality. However for mild steel, Proportional limit and Elastic limit are almost equal, but for other metals & materials elastic limit may be higher than proportional limit, for exampleRubber.

1.5 True stress and True Strain

1. True stress: The true stress is defined

as the ratio of the load to the cross section area at any instant

(σ ) = T

load = σ (1 + ε ) Instantaneous area

Where σ and ε is the engineering stress and engineering strain respectively.

1. HOOKE’S LAW A material is said to be elastic if it returns to its original, unloaded dimensions when load © Copyright Reserved by Gateflix.in No part of this material should be copied or reproduced without permission

2. True Strain: L

( ε ) = ∫ dll = T

Lo



 L  ln   = ln (1 + ε )  L0 











( ε ) = ln  AA  = 2ln  dd  T



0

0



OR engineering strain ( ε ) = eεT - 1

The volume of the specimen is assumed to be constant during plastic deformation

Q A 0 L0 =AL  .It is valid till the neck

formation.

Where, Ao & A are original & final area respectively. Lo & L are original & final length respectively.

Note: Comparison of engineering and the true stress-strain curves shown below • The true stress-strain curve is also knows as the flow curve. • True stress-strain curve gives a true indication of deformation characteristics because it is based on the instantaneous dimension of the specimen.

• In engineering stress-strain curve, stress drops down after necking since it is based on the original area

• In true stress-strain curve, the stress however increases after necking since the cross sectional area of the specimen decreases rapidly after necking.

The flow curve of many metals in the region of uniform plastic deformation can be expressed by the simple power law.

σ f =K ( ε T )

n

Where K is the strength coefficient

n is the strain hardening exponent n = 0 for perfectly plastic solid n = 1 for Elastic solid For most metals, 0.1< n< 0.5

1.6 ELASTIC CONSTANTS: 1. Poisson’s ratio (μ) It has been observed that for elastic materials, the lateral strain is directly proportional to the longitudinal strain. The ratio of the lateral strain to longitudinal strain is known as the poison's ratio.

Poisson ratio ( μ ) =-

Lateral Strain longitudinal Strain

2. Modulus of rigidity(G) Up to proportion limit, shear stress is directly proportional to shear strain. Shear stress (τ)αshear strain(γ) Shear stress ( τ )= G × shear strain(γ)

G=

shear strain(γ) γ = Shear stress ( τ ) τ

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G is known as shear modulus and its unit is same as the unit of stress i.e. MPa. 3. Bulk Modulus(K)

Up to proportional limit, the hydrostatic stress is directly proportional to volumetric strain. dV dP ∝ − V dP K= dV V K is known as bulk modulus and its unit is MPa. Note: RELATIONSHIP BETWEEN ELASTIC CONSTANTS Linear stress Linear strain Shear stress Modulus of regidity, G = shear strain Direct stress Bulk mod ulus, K = Volumetric strain 1 Lateralstrain Poisson`s ratio, µ = = − m Linear strain Young`s mod ulus, E =

1  E = 2G  1 +  = 2G(1 + μ) m 

elastic properties are the same at every point in the body. However, the properties need not to be the same in all the direction for the material to be homogenous. Isotropic materials have the same elastic properties in all the directions. Therefore, the material must be both homogenous and isotropic in order to have the lateral strains to be same at every point in a particular component. 2. ISOTROPIC MATERIAL

If the response of the material is independent of the orientation of the load axis of the sample, then we say that the material is isotropic or in other words we can say that isotropy of a material is a characteristics, which gives us the information that the properties are the same in the different direction, on the other hand if the response is dependent on orientation it is known as anisotropic. 1.8 STRAIN UNDER TRI AXIAL LOADING

If a rectangular block is subjected to three normal stresses σx, σy & σz on all its faces, mutually perpendicular to each other in x, y & z directions respectively as shown in fig.

 2 E = 3K  1 -  = 3K(1 - 2μ)  m E=

9KG 3K + G

µ=

1 3K − 2G = m 6K + 2G

1.7 TYPES OF MATERIALS

Then volumetric strain of the specimen is given by εV = εx + εy + εZ

 σx + σ y + σz  = εv   [1 − 2µ ] E   A material is homogenous if it has the same composition throughout body. Hence the 1.9 AXIAL STRESS IN COMPOUND BARS 1. HOMOGENEOUS MATERIAL

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Case (a):Let the axial load is applied on the free end of the member as shown in fig. Then the load shared by each member P1 =P2 = P3 = P

Deformation produced in member 1 is given by Pl1 4Pl1 = δ1 = A1E1 πd12 E1 Deformation produced in member 2 is given by Pl2 4Pl2 = δ2 = A 2 E 2 πd 2 2 E 2 Deformation produced in member 1 is given by Pl3 4Pl3 = δ3 = A 3 E 3 πd 32 E 3 Total deformation produced in members is given by Pl3 Pl Pl2 δTotal = 1 + + A1E1 A 2 E 2 A 3 E 3 4Pl3 4Pl 4Pl2 δTotal = 2 1 + + 2 πd1 E1 πd 2 E 2 πd 32 E 3 Stress produced in member 1 is given by P1 4P = σ1 = A1 πd12 Stress produced in member 2 is given by P2 4P = σ2 = A 2 πd 2 2 Stress produced in member 3 is given by P3 4P = σ3 = A 3 πd 32 Case (B): Let both ends of the members are fixed as shown in the figure.

A beam is subjected under the load P at point B and it is fixed at both the ends, the deformation produced in the body is zero, so δ total = δ1 + δ1 = 0 Pl Pl or 1 1 + 2 2 = 0 A1E1 A 2 E 2 4P1l1 4P2 l2 or + = 0 2 πd1 E1 πd 2 2 E 2 Where P1 and P2 is the load on member 1 and member 2 respectively If the reaction at point A and B are RA and RB, then RA+ RB = P 1.10 ELONGATION OF THE TAPER BEAM UNDER AXIAL LOAD

A taper beam having circular cross section is subjected under axial load P at the free and as shown in the fig.

Maximum stress produced in the taper beam is given by P1 4P σ= = max A min πd min 2 Minimum stress produced in the taper beam is given by P1 4P = σ min = A max πd max 2

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Elongation of the taper beam at the free end 4Pl is given by δ total = πd1 d 2 E Where E is modulus of elasticity d1, d2 are the diameters of the smallest and biggest end of the beam and l is the length of the beam. 1.11. ELONGATION DUE TO SELF WEIGHT OF THE BODY

1. For bar of uniform section: Due to self weight of the component, the deformation produced in the component is given by

deformation in uniform cross section under its self weight.

1.12 THERMAL STRESS

If temperature of the component is increased or decreased, thermal expansion or contraction in the component takes place. If this thermal expansion or contraction is restricted either completely or partially, then thermal stress produces in the component. The stress produced due to thermal expansion or contraction is called as thermal stress.  If there is free thermal expansion and contraction or it is not resisted by any Resistance then thermal stress is zero. 1. FREE EXPANSION

λL2 WL = 2E 2AE Where , λ = ρg = Unit − weight of the material. L = Length of the specimen E = Modulus of Elasticity of the component W = Total self weight of specimen A = Cross-sectional area.  Thus, the deformation of the bar under its own weight is half the deformation, if the body is subjected to the direct load equal to the weight of the body. ∆L=

2. For bar of tapering section: due to self weight of the component, the deformation produced in the component is given by λL2 WL ∆L = = 6E 6AE  Thus, the deformation of taper bar under its own weight is one third of the

When beam is supported at one end and its other end is free, temperature increase ΔT, due to this temperature difference change in length of the beam is ΔL. If thermal expansion co-efficient of the material of the beam is α, Free expansion in length is given by

ΔL = α L ΔT Stress produced in the member is given by σ thermal = 0 2. WHEN ENDS OF THE COMPOSITE BAR ARE RESTRICTED If the free expansion or contraction is prevented then stresses will be developed

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Let us consider the separate effect of the temperature and the reaction at the fixed reactions.

Thermal expansion in the body is given by ΔL = α L ΔT Axial compression in the member due to reaction in the beam is given by ΔLa =-α L ΔT Thermal stress produced in the member is given by σ thermal = −α E ∆T  If temperature of the component is increased the thermal stress produced in the member is compressive and vice versa.

Since coefficient of expansion of copper ( α c ) is greater than coefficient of expansion of steel (α s ) hence free expansion of copper is more than that of steel. But due to combined effect of the component, final deformation will be same in both the metals. So steel bar expand and copper bar contracts and steel bar will be in tensile stress and copper bar will be in compressive stress.

3. WHEN ENDS OF THE COMPOSITE BAR ARE RESTRICTED PARTIALLY If partial elongation is allowed, thermal stress produced in the component  αLΔT − δ  σ thermal =  E L   Where δ is the allowed deformation under temperature change. 4. COMPOSITE BEAM

If the material consists of two metals and is subjected to temperature change, opposite kinds of stresses (tensile and compressive) will setup in the two materials. Let us consider Copper and steel composite section as shown in the fig. The temperature of the composite bar is increased. © Copyright Reserved by Gateflix.in No part of this material should be copied or reproduced without permission

GATE QUESTIONS Q.1

Q.2

Two identical circular rods of same diameter and same length are subjected to same magnitude of axial tensile force. One of the rod is made out of mild steel having the modulus of elasticity of 206 GPa. The other rod is made out of cast iron having the modulus of elasticity of 100 GPa. Assume both the materials to be homogeneous and isotropic and the axial force causes the same amount of uniform stress in both the rods. The stresses developed are within the proportional limit of the respective materials. Which of the following observations is correct? a) Both rods elongate by the same amount b) Mild steel rod elongates more than the cast iron rod c) Cast iron rod elongates more than the mild steel rods d) As the stresses are equal strains are also equal in both the rods [GATE-2003] The figure below shows a steel rod of 25 mm2 cross sectional area. It is loaded at four points, K, L, M and N. Assume Esteel =200 GPa. The total change in length of the rod due to loading is

a) 1 µm c) 16 µm Q.3

b) −10 µm d) −20 µm [GATE-2004]

A steel bar of 40mm×40 mm square cross-section is subjected to an

axial compressive load of 200 kN. If the length of the bar is 2m and E=200 GPa, the contraction of the bar will be a) 1.25 mm b) 2.70 mm c) 4.05 mm d) 5.40 mm [GATE-2006] Q.4

A bar having a cross-sectional area of 700 mm2 is subjected to axial loads at the positions indicated. The value of stress in the segment QR is

a) 40 MPa c) 70 MPa

b) 50 MPa d) 120 MPa [GATE-2006]

Q.5

A 200×100×50 mm steel block is subjected to a hydrostatic pressure of 15 MPa. The Young's modulus and Poisson's ratio of the material are 200 GPa and 0.3 respectively. The change in the volume of the block in mm3 is a) 85 b) 90 c) 100 d) 110 [GATE-2007]

Q.6

A rod of length L and diameter D is subjected to a tensile load P. Which of the following is sufficient to calculate the resulting change in diameter? a) Young’s modulus b) Shear modulus c) Poisson’s ratio d) Both Young’s modulus and shear modulus [GATE-2008]

Common Data For Q. 7 and 8:

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A cylindrical container of radius R = 1 m, wall thickness 1 mm is filled with water up to a depth of 2 m and suspended along its upper rim. The density of water is 1000 kg/m3 and acceleration due to gravity is 10 m/s2. The self-weight of the cylinder is negligible. The formula for hoop stress in a thin-walled cylinder can be used at all points along the height of the cylindrical container.

Q.7

The axial and circumference stress (σa,σc) experienced by the cylinder wall at mid-depth (1m as shown) are a) (10, 10)MPa b) (5,10)MPa c) (10, 5)MPa d) (5, 5)MPa [GATE-2008]

Q.8

If the Young's modulus and Poisson's ratio of the container material are 100 GPa and 0.3, respectively, the axial strain in the cylinder wall at mid-depth is a) 2 ×10−5 b) 6 ×10−5 c) 7 ×10−5 d) 1.2 ×10−4 [GATE-2008]

Q.9

A rod of length L having uniform cross-sectional area A is subjected to a tensile force P as shown in the figure below. If the Young’s modulus of the material varies linearly from E1 to E2 along the length of the rod, the normal stress developed at the section-SS is

P(E1 − E 2 ) A(E1 + E 2 ) PE1 d) PE 2 [GATE-2013]

P A PE 2 c) AE1 a)

b)

Q.10 A circular rod of length ‘L’ and area of cross-section ‘A’ has a modulus of elasticity ‘E’ and coefficient of thermal expansion ‘α’. One end of the rod is fixed and other end is free. If the temperature of the rod is increased by ∆T, then a) stress developed in the rod is E α ∆T and strain developed in the rod is α ∆T b) both stress and strain developed in the rod are zero. c) stress developed in the rod is zero and strain developed in the rod is α ∆T d) stress developed in the rod is EαΔT and strain developed in the rod is zero [GATE-2014 (1)] Q.11 The stress-strain curve for mild steel is shown in figure given below. Choose the correct option referring to both figure and table.

Point on the graph P. Q. R. S. T. U.

Description of the point

1. Upper yield point 2. Ultimate Tensile strength 3. Proportionality limit 4. Elastic limit 5. Lower yield point 6. Failure

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a) P-1, Q-2, R-3, S-4, T-5, U-6 b) P-3,Q-1,R-4,S-2,T-6,U-5 c) P-3, Q-4, R-1, S-5, T-2, U-6 d) P-4, Q-1, R-5, S-2,T-3,U-6 [GATE-2014(3)] Q.12 Which one of the following types of stress- strain relationship best describes the behavior of brittle materials, such as ceramics and thermosetting plastics, (σ = stress and ε = strain)? a) b)

c)

d)

developed in the rod is __________ MPa. Young’s modulus of the material of the rod is 200 GPa and the coefficient of thermal expansion is 10−5 per oC.

[GATE-2016 (2)] Q.15 A hypothetical engineering stressstrain curve shown in the figure has three straight lines PQ, QR, RS with coordinates P(0,0), Q(0.2,100), R(0.6,140) and S(0.8,130). 'Q' is the yield point, 'R' is the UTS point and 'S' is the fracture point.

[GATE-2015 (1)] Q.13 A horizontal bar with a constant cross-section is subjected to loading as shown in the figure. The Young’s modul i for the sections AB and BC are 3E and E, respectively.

Q.14

The toughness of the material (in MJ/m3) is __________ [GATE-2016 (1)]

For the deflection at C to be zero, the ratio P/F is ____________ [GATE-2016 (1)]

Q.16 A square plate of dimension L × L is subjected to a uniform pressure load p = 250 MPa on its edges as shown in the figure. Assume plane stress conditions. The Young’s modulus E = 200 GPa.

A circular metallic rod of length 250 mm is placed between two rigid immovable walls as shown in the figure. The rod is in perfect contact with the wall on the left side and there is a gap of 0.2 mm between the rod and the wall on the right side. If the temperature of the rod is increased by 200℃, the axial stress

The deformed shape is a square of dimension 𝐿𝐿 − 2𝛿𝛿. If 𝐿𝐿 = 2 m and 𝛿𝛿 =

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0.001 m, the Poisson’s ratio of the plate material is __________ [GATE-2016 (3)] Q.17 The state of stress at a point 0 σ= σ= σ= τ= τ= τ= τ= x y z xz zx yz zy and

τ= τ= 50 MPa . xy yx

The

maximum normal stress (in MPa) at that point is _________ [GATE-2017 (2)] Q.18 In the engineering stress-strain curve for mild steel, the Ultimate Tensile Strength (UTS) refers to [GATE-2017 (1)] (A) Yield stress (B) Proportional limit (C) Maximum stress (D) Fracture stress.

Q.19 An initially stress-free massless elastic beam of length L and circular cross-section with diameter d (d << L) is held fixed between two walls as shown. The beam material has Young’s modulus E and coefficient of thermal expansion α .

If the beam is slowly and uniformly heated, the temperature rise required to cause the beam to buckle is proportional to (A) d (B) d2 (C) d3

(D) d4

[GATE-2017 (1)] Q.20 A horizontal bar, fixed at one end (x = 0), has a length of 1 m, and crosssectional area of 100mm2. Its elastic modulus varies along its length as

given by E ( x ) = 100 e − x GPa, Where

x is the length coordinate (in m) along the axis of the bar. An axial tensile load of 10 kN is applied at the free end (x=1). The axial displacement of the free end is _______ mm. [GATE-2017 (1)]

Q.21 A point mass of 100 kg is dropped onto a massless elastic bar (crosssectional area = 100 mm2, length = 1m, Young’s modulus = 100 GPa) from a height H of 10 mm as shown (Figure is not to scale). If g = 10m/s2, the maximum compression of the elastic bar is _______ mm. [GATE-2017 (1)]

Q.22 A steel bar is held by two fixed supports as shown in the figure and is subjected to an increases of temperature T =100 °C . If the coefficient of thermal expansion and Young’s Modulus of Elasticity of steel are 11×10−6 / °C and 200GPa, respectively, the magnitude of thermal stress (in MPa) induced in the bar is __________. [GATE-2017 (2)] Q.23 The state of stress at a point, for a body in plane stress, is shown in the

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figure below. If the minimum principal stress is 10 kPa, then the normal stress σ y (in kPa) is

[GATE 2018 (1)]

(A) 9.45

(C) 37.78

[GATE-2018 (1)] Q.25 The true stress (σ) - true strain (∈)

diagram of a strain hardening material is shown in figure. First, there is loading up to point A, i.e., up to stress of 500 MPa and strain of 0.5. Then from point A, there is unloading up to point B, i.e., to stress of 100 MPa. Given that the Young’s modulus E = 200 GPa, the natural strain at point B ( ∈B ) is _________ (correct to three decimal places). [GATE 2018 (1)]

(B) 18.88

(D) 75.50

Q.24 A carpenter glues a pair of cylindrical wooden logs by bonding their end faces at an angle of θ =300 as shown in the figure. The glue used at the interface fails if Criterion 1: the maximum normal stress exceeds 2.5 MPa. Criterion 2: the maximum shear stress exceeds 1.5 MPa. Assume that the interface fails before the logs fail. When a uniform tensile stress of 4 MPa is applied, the interface

[GATE-2018 (1)] Q.26 A bimetallic cylindrical bar of cross sectional area 1 m2 is made by bonding Steel (Young’s modulus = 210 GPa) and Aluminium (Young’s modulus = 70 GPa) as shown in the figure. To maintain tensile axial strain of magnitude 10 −6 in Steel bar and compressive axial strain of magnitude 10 −6 in Aluminum bar, the magnitude of the required force P (in kN) along the indicated direction is [GATE 2018 (2)]

(A) fails only because of criterion 1 (B) fails only because of criterion 2 (C) fails because of both criteria 1 and 2 (D) does not fail

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(A) 70 (C) 210

(B) 140 (D) 280 [GATE-2018 (2)]

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ANSWER KEY: 1 (c) 15 0.85

2 (b) 16 0.2

3 (a) 17 50

4 (a) 18 (c)

5 (b) 19 (b)

6 7 8 (d) (a) (c) 20 21 22 1.71 1.51 220

9 (a) 23 (c)

10 (c) 24 (c)

11 12 (c) (d) 25 26 0.498 280

13 4

14 240

EXPLANATIONS Q.1

Q.2

(c) Given : Ls = LI , Es = 206 GPa , EI = 100 GPa Ps = PI , Ds = DI. Where subscript S is for steel and I is for cast iron rod. We know that elongation is given by, PL ∆L= AE PL Elongation in mild steel, ∆LS = AES PL Elongation in cast iron, ∆L I = AE I ∆LS E I 100 Therefore, = = <1 ∆L I ES 206 (b) Given A = 25 mm2 ESteel =200 GPa =200×103 MPa First of all we have to make the F.B.D. of the section KL, LM and NM separately.

Now, From the F.B.D, PKL = 100 N (Tensile) PLM = 150 N (Compressive) PMN = 50 N (Tensile) LKL=500mm,LLM=800mm,LLM=400 mm

Total change in length = ∆ KL + ∆ LM + ∆ MN

PL AE ∴total change in length P L P L P L = KL KL + LM LM + MN MN AE AE AE [100 × 500 − 150 × 800 + 50 × 400] = 25 × 200 ×103 = −10µm We know that ∆L=

Q.3

Q.4

(a) Given: A= (40)2 = 1600 mm2, P= − 200 kN (Compressive) L = 2m = 2000 mm, = E 200GPa = 200 ×103 MPa PL Elongation of the bar, ∆L= AE 3 200 ×10 × 2000 ∆L = =−1.25mm 1600 × 200 ×103 Since the load is Compressive in nature.

(a) The FBD of segment QR is shown below

Given: A = 700 mm2 From the free body diagram of the segment QR.

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Force acting on QR=28 kN (Tensile)Stress in segment QR is given by 28 ×103 = σ = 40MPa 700 Q.5 (b) We know that, ∆V 3σ = (1 − 2μ ) V E ∆V 3 ×15 = [1 − (2 × 0.3)] 200 ×100 × 50 200 ×103 ∴ΔV = 90 mm3 Q.6

(d)

Q.7

(a) Pressure(P)=hρg=1×1000×10=10 kPa Axial Stress ( σ a )

⇒ σ a × 2πRt = ρg × πR 2 L Or ρgRL 1000 ×10 ×1×1 = σa = = 10MPa t 1×10−3 Circumferential Stress PR 10 ×1 = = 10MPa ( σ= c) t 1×10−3 Q.8

(c)

σa σ −μ c E E 10 10 = − 0.3 × = 7 ×10−5 −3 −3 100 ×10 100 ×10 ε= a

Q.9

(a)

P A We see that normal stress only depends on force and area and it does not depends on E. The normal stress is given by σ =

Q.10 (c) For thermal stress to be developed there must be constraint in the system. So, although strain develops but there is no thermal stress.

P:Proportional limit Q: Elastic limit R: Upper Yield Point S: Lower Yield Point T: Ultimate Tensile Strength U: Failure /Rupture

Q.12 (d) Q.13 (4)

δC = 0

⇒ δ AB + δ BC = 0 − [ P − F] L

+

A(3E) F−P +F= 0 3 4F = P P ∴ =4 F

FL = 0 AE

Q.14 (240)

Lα∆T −

PL = AE

( 0.2 ×10 ) −3

Multiplying above equation by

E L

E L 9 200 × 10   −5 −3 9 σ= ( 200 × 10 ) × 10 × 20 −  0.2 × 10 × 0.25   = σ 240 ×106 Pa σ = 240MPa Eα∆T − σ= (0.2 ×10−3 )

Q.11 (c)

Q.15 (0.85)

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Toughness is given by area under the curve up to the fracture point.

∴ Toughness = Area PQA + Area AQRB + Area BRSC 1  0.2  1  0.4  1  0.2  [100 + 140] + [140 + 130] = × 100 +     2  100  2  100  2  100 

= 0.85 MJ/m3

Q.16 (0.2) ∈v =∈x + ∈y [Q∈z = 0, planes tress ]

∆V  σ μσ   σ μσ  = + V  E E   E E  1.9982 -22 -P = ×2[1-μ] 22 E -250 -2×10-3 = ×2[1-μ] 200×103 200 = [1 − μ] 250 ⇒μ= 0.2

Q.19 (b) Fth = Reaction offered by support or Buckling load. π2 EI α∆TE A = 4 2 l π π2 E × D 4 π 2 64 α∆TE D = 4 4 l2 ∆T ∝ D 2

Q.20

Q.17 By Mohr’s circle

Maximum normal stress given at the point = 50 MPa Q.18 (c)

Pdx Elemental elongation (dδ) = AE x 1

∴ total elomgation (δ)= ∫ 0

1

=

P dx −x ∫ A 0 100e ×1000

10 ×103 x 1 e  δ= 100 ×105   0

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Pdx AE x

Ans. (c)

1 [e − 1] 103 δ =1.718mm

= δ

kPa, τ xy 50 kPa = = x 100 Minimum principal strees =

Q.21 Let the compression is x meter For the conservation of energy P.E=Elastic energy 1 mg[h + x] =kx 2 2 EA Where, k = l 1 EA 2 mg[h + x] = x 2 l

1 100 ×103 ×100x 2 100 ×10[0.01 + x] = × 2 1 2 0.01 + x = 5000x 2 5000x − x − 0.01 = 0 x1,2 =

−b ± b − 4ac 2a 2

1 ± 1 + 4 × 5000 × 0.01 1 ± 14.17 = 2 × 5000 10000 If take positive then 15.17 = x1 = meter 1.517mm 10000 x1 = 1.517mm x1,2

Q.22 Sol: (220) Since σ th = α∆t.E

=11×10−6 ×100 × 200 ×103 σ th = 220MPa

Q.23

x +y 2

 100 −y 100 +y 10 = −   2 2  ∴

 y  50 − 2 

2

 −y −  x  2 

 2  + τ xy 

2

 2  + 50 

2

 y 2 − 10  + 50 =50 + 2  = 40 +

y 2

By squaring 2y

2y

+ 40y − 50y + 2500 = 1600 + 4 4 ∴ 90 y = 3400 2500+

y = 37.78 MPa

Q.24(c)

Normal stress on inclined plane T ' x cos 2 θ = = 4  cos 2 30 = 3 MPa x sin 2θ 2 = 2  sin 60o = 1.73 MPa Since both the stress exceeds the given limits,

Shear stress on inclined plane τ ' =

So, Answer is option (c) Q.25

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Ans. ( 0.498 ) We know that Slope of AB line is E AC E= BC ( 500 − 100 ) MPa 200  103 MPa = BC 400 BC = = 0.002 200 103 ∴ ∈ E = 0.5 − 0.002= 0.498 Q.26

= ∈1 P1 = ∈2

P1 = 10−6 A1 E1 −6 = 1  210    210 kN 10

P2 = 10−6 A2E2

= ⇒ P2 10−6  = 1 70 109 70 kN P = P1 + P2 = 210 + 70 = 280 kN

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GATE QUESTIONS Q.1

Q.2

Q.3

A shaft subjected to torsion experiences pure shear stress τ on the surface. The maximum principal stress on the surface which is at 45° to the axis will have a value a)τ cos 45° b) 2τ cos 45° 2 c) τ cos 45° d) 2τ sin 45° cos 45° [GATE-2003]

In terms of Poisson's ratio (υ) the ratio of Young's Modulus (E) to Shear Modulus (G) of elastic materials is a) 2 (1 + υ) b) 2 (1 − υ) c) 1/2 (1 + υ) d) 1 /2 (1 − υ) [GATE-2004]

According to Von-Mises distortion energy theory, the distortion energy under three dimensional stress stage is represented by 1 2 2 2 σ 1 + σ 2 + σ 3 − 2υ (σ 1σ 2 + σ 3σ 2 + σ 1σ 3 )  2E  1 − 2υ  2 b) σ + σ 22 + σ 32 + 2 (σ1σ 2 + σ 3σ 2 + σ1σ 3 )   1

a) c)

d) Q.4

Q.5

6E 1+υ

σ 2 + σ 2 + σ 2 − (σ σ + σ σ + σ σ ) 2 3 1 2 3 2 1 3  3E  1

1 σ 12 + σ 22 + σ 32 − υ (σ 1σ 2 + σ 3σ 2 + σ 1σ 3 )  3E

[GATE-2006]

If the principal stresses in a plane stress problem are σ1 = 100 MPa, σ 2 = 40 MPa, the magnitude of the maximum shear stress (in MPa) will be a) 60 b) 50 c) 30 d) 20 [GATE-2010]

A solid circular shaft of diameter d is subjected to a combined bending moment M and torque, T. The material property to be used for

designing

Q.6

Q.7

the shaft using the 16 relation M 2 + T 2 is 3 πd a) Ultimate tensile strength (Su) b) Tensile yield strength (Sy) c) Torsional yield strength (Ssy) d) Endurance strength (Se) [GATE-2009] The state of plane-stress at a point is given by σx = −200MPa, σy=100 MPa τxy = 100 MPa. The maximum shear stress (in MPa) is a) 111.8 b) 150.1 c) 180.3 d) 223.6 [GATE-2010]

Match the following criteria of material failure, under biaxial stresses σ1 and σ2 and yield stress σy, with their corresponding graphic representations. List I P. Maximum-normal-stress criterion Q. Maximum-distortion-energy criterion R. Maximum-shear-stress criterion List II L.

M.

N.

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Q.8

Q.9

a) P-M, Q-L, R-N c) P-M, Q-N, R-L

b) P-N, Q-M, R-L d) P-N, Q-L, R-M [GATE-2011] The homogeneous state of stress for a metal part undergoing plastic 10 5 0    0 deformation is T=  5 20  0 0 -10    Where the stress component values are in MPa. Using Von Mises Yield criterion, the value of estimated shear yield stress, in MPa is a) 9.50 b) 16.07 c) 28.52 d) 49.41 [GATE-2012]

A metallic rod of 500 mm length and 50 mm diameter, when subjected to a tensile force of 100 kN at the ends, experiences an increase in its length by 0.5 mm and a reduction in its diameter by 0.015mm. The Poission’s ratio of the rod material is ___. [GATE-2014 (1)]

Q.10 A steel cube, with all faces free to deform has Young’s modulus, E, Poisson’s ratio, υ and coefficient of thermal expansion, α. The pressure (hydrostatic stress) developed within the cube, when it is subjected to a uniform increase in temperature, ∆T, is given by α ( ∆T ) E a) 0 b) 1 − 2υ α ( ∆T ) E α ( ∆T ) E c) − d) − 1 − 2υ 3 (1 − 2υ ) [GATE-2014 (2)] Q.11 A thin plate of uniform thickness is subject to pressure as shown in the figure below

Under the assumption of plane stress, which one of the following is correct? a) Normal stress is zero in the zdirection b) Normal stress is tensile in the z-direction c) Normal stress is compressive in the z-direction d) Normal stress varies in the zdirection [GATE-2014 (2)] Q.12 If the Poisson’s ratio of an elastic material is 0.4, the ratio of modulus of rigidity to Young’s modulus is____. [GATE-2014 (4)] Q.13 The number of independent elastic constants required to define the stress-strain relationship for an isotropic, elastic solid is ____. [GATE-2014 (4)] Q.14 The state of stress at a point is given by σ x = −6MPa , σ y = 4MPa , and

τ xy = −8MPa . The maximum tensile stress (in MPa) at the point is ____. [GATE-2014 (1)]

Q.15 A 200 mm long, stress free rod at room temperature is held between two immovable rigid walls. The temperature of the rod is uniformly raised by 250℃. If the Young’s modulus and coefficient of thermal expansion are 200GPa and 1× 10−5 / ℃, respectively, the magnitude of the longitudinal stress (in MPa) developed in the rod is ____. [GATE-2014 (1)]

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Q.16 A rod is subjected to a uni-axial load within linear elastic limit. When the change in the stress is 200 MPa, the change in the strain is 0.001, If the Possion’s ratio of the rod is 0.3, the modulus of rigidity (in GPa) is ____ [GATE-2015 (2)] Q.17 A shaft with a circular cross-section is subjected to pure twisting moment. The ratio of the maximum shear stress to the largest principal stress is a) 2.0 b) 1.0 c) 0.5 d) 0 [GATE-2016 (2)] Q.18 The state of stress at a point on an element is shown in figure (a). The same state of stress is shown in another coordinate system in figure (b).

under deformed rectangle are given by P(0,0), Q (4,0), S (0,3). The rectangle is subjected to uniform strains,

= ε xx 0.001, = ε yy 0.002, = γ xy 0.003 . The deformed length of the elongated diagonal, up to three decimal places, is _________ units. [GATE-2017 (1)] Q.21 If σ1 and σ3 are the algebraically largest and smallest principal stresses respectively, the value of the maximum shear stress is  +3  −3 (A) 1 (B) 1 2 2 (C)

1 +3 2

The components (τ xx , τ yy , τ xy ) are given by a) (P / 2, −P / 2, 0) b) (0 , 0 , P) c) (P, −P, P / 2)  

d) (0, 0, P / 2) [GATE-2016 (3)]

Q.19 The Poisson’s ratio for a perfectly incompressible linear elastic material is (A) 1 (C) 0

(B) 0.5

(D) infinity

[GATE-2017 (1)]

Q.20 A rectangular region in a solid is in a state of plane strain. The (x,y) coordinates of the corners of the © Copyright Reserved by Gateflix.in No part of this material should be copied or reproduced without permission

(D)

1 −3 2

[GATE-2018 (1)]

ANSWER KEY: 1 (d) 15 500

2 (a) 16 77

3 (c) 17 (b)

4 (c) 18 (b)

5 (c) 19 (b)

6 (c) 20 5.014

7 (c) 21 (b)

8 (b)

9 0.3

10 (a)

11 (a)

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12 0.35

13 2

14 0.43

EXPLANATIONS Q.1

(d)

We know that, for a shaft subjected to combined bending moment M and torque T, the equivalent Torque is,

σx + σy  σx − σy  2 σ1/2 = ±   + τ xy 2  2  For pure shear, σ x = 0 and σ y = 0 and τ xy = τ 2

= Te M2 + T2 Induced shear stress is,

⇒ σ1/2 =± 0 τ2

σ1/2 = ± τ ∴ maximum principal stress = τ It can be written as 2τ sin 45° cos 45° 1 1 = 2 × τ× × = τ 2 2 Q.2

Q.3 Q.4

Q.5

(a) Relation between, E G and u is given by, = E 2G (1 + υ ) Where E = Young’s modulus G = Shear Modulus υ = Poisson’s ratio E Now, = 2(1 + υ) G (c)

(c) Given = σ1 100MPa, = = σ 2 40MPa We know, the maximum shear stress for the plane complex stress is given by σ1 − σ 2 100 − 40 τ max == = 30 MPa 2 2 (c)

Q.6 Q.7 Q.8

16 M 2 + T 2 τ= πd 3 Now, for safe design, should be S less then SY N Where Ssy = Torsional yield strength and N = Factor of safety. (c)

(c) So correct pairs are, P-M, Q-N, R-L (b) σ x = 10, σ y = 20, τ xy = 5, σ z = −10

σx + σy  σx − σy  2 = ±   + τ xy 2 2   2

σ1,2

σ1 = 15 + 25 + 25 σ1 = 22.07MPa σ 2 = 7.928MPa σ3 = σ z = −10 σ yt = yield stress in tension σ ys = yield stress in shear

According to Von mises theory 2  2 1 ( σ1 − σ 2 )   ( σ yt ) ≥ 2  2 2   + ( σ 2 − σ3 ) + ( σ3 − σ1 ) 

(σ ) yt

2

≥ 772.435

σ yt ≥ 27.79

Now σ ys =

σ yt 3

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∴ σ ys = 16.046MPa

Q.9

(0.3)

Lateral strain Longitudinal strain  ∆D    D   = −  ∆L     L   L   ∆D  = µ   −   D   ∆L  Given , L = 500mm, D = 50mm, ∆D = - 0.015, ∆L = +0.5mm 500 0.015 ∴µ= × = 0.3 50 0.500 µ= −

Q.10 (a) As all faces are free to deform, there are no thermal stresses.

Q.11 (a) Thin plate of uniform thickness pertains to plane stress condition. So, stress out of plane would be zero. Q.12 (0.357) E = 2G (1 + µ) G 1 1 1 = = = E 2(1 + µ) 2 ×1.4 2.8 = 0.357

Q.13 (2) Either E or G, 2 independent constant Either G or K, 2 independent constant Either E or K, 2 independent constant Q.14

(8.43)

σx + σy  σx − σy  2 σ1 = +   + τ xy 2 2   2

−6 + 4  −6 + 4  2 +   + (−8) 2  2  2

=

=−1 + 25 + 64 =−1 + 89 =8.4339MPa Q.15 (500) Given, L = 200 mm ∆T = 250°C α = 1×10−5 / ℃ E = 200 GPa σ=? σ = α ∆TE = 10−5 × 250 × 200 ×103 = 500 MPa

Q.16 (77) Modulus of rigidity (G) E G= 2(1 + μ ) It has given change in stress=200 MPa Change in strain = 0.001 Here, 200 = E × 0.001 200MPa ⇒E= = = 200 ×103 MPa 0.001 = 200 GPa 200 200 100 1000 = G = = = 2(1 + 0.3) 2 ×1.3 1.3 13 = 77GPa Q.17 (b)

A state of pure shear can be depicted as shown in the above figure. σ ∴ 1 = 1 τ max Q.18 (b) The given state of stress is pure shear ∴ τ xx = τ yy = 0

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And τ xy = P

Q.19 (b) Volumetric strain for linear elastic material, ∆v (1 − 2µ) ∈ =v = ( σx + σ y + σz ) v E For incompressible flow ∆v = 0 ∴1 − 2µ = 0 ⇒ µ = 0.5

= 5.014 mm

Q.21 (b)

Q.20

PR =

42 + 32 = 5, & PQ=4 4 ∴ cos θ = 5 3 sin θ = 5

εθ =∈xx cos 2 θ+ ∈yy sin 2 θ + γ xy sin θcosθ 2

2

4 3  4  3  = εθ 0.001×   + 0.002   + 0.003     5 5  5  5 

εθ= 2.8 ×10−3 ∆PR εθ = PR ∆PR = εθ .PR Deformed length of diagonal

= PR + ∆PR = PR + εθ PR = PR[1 + εθ ]

= 5[1 + 2.8 ×10−3 ]

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2

STRAIN ENERGY AND IMPACT LOAD

2.1 STRAIN ENERGY Strain Energy is defined as the energy absorption capacity of a material of the component when it is strained under the action of the load. Let us consider case of bar of cross section area A length l and subjected under the load W. Deformation produced in the bar under the action of W is δl. The strain energy stored in member is equal to the work done by internal resisting force F developed due to deformation. Strain Energy (U) = Work done by internal resisting force (F) = Work done by Load W. 1 U = Wδl 2 If W vs. δl diagram is drawn, then it can be given as

Fig. load vs. deformation Strain Energy U = Area of W vs. δl 1 Wl W 2 l U= W × = 2 AE 2AE

U =

W 2 l σ 2 lA σ 2 = = × Volume 2AE 2E 2E

1. STRAIN ENERGY UNDER THE TORSION IS GIVEN AS

1 T 2l = Tθ 2 2JG Where T=Torsion applied on rotating beam J = Polar modulus of cross section

= U

G = Modulus of rigidity

2. STRAIN ENERGY UNDER THE BENDING MXX IS GIVEN AS

b

U=∫ a

M xx 2 dx 2EI

a and b are initial and final point of the variation of the moment 3. STRAIN ENERGY UNDER THE PURE SHEAR STRESS IS GIVEN BY τ2 = U × volume 2G τ = direct shear stress 4. RESILIENCE Resilience is defined as the energy absorption capacity of the material of the component within the elastic limit when the component is deformed under the action of the load.

Let at point A load is given by W1 and deflection is given by δ1, then resilience is given by 1 Resilience = W1δ1 2 5. PROOF RESILIENCE

Proof Resilience is defined as the energy absorption capacity of the material of the component up to the elastic limit when the

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component is deformed under the action of the load. Let at point E, load is given by WEL and deflection is given by δEL, then proof resilience is given by 1 Proof Resilience = WEL δ EL 2 6. MODULUS OF RESILIENCE

Modulus of Resilience is defined as the energy absorption capacity of the material of the component up to the elastic limit per unit volume when the component is deformed under the action of the load. Let at point E, load is given by WEL and deflection is given by δEL, then modulus of resilience is given by

Modulus of Resilience =

9. IMPACT LOAD The loads which apply in a very fast way are known as impact or falling load. The duration of the application of these loads is less but these type of loads are more dangerous than static load. Let consider a weight W falling from a height of h on a collar fitted on a rod of length l and cross section area A. Deformation and stress in impact loading is σimp and δimp respectively. Work done by weight = Strain Energy of the bar

Proof Resilience volume

1 WEL δ EL σ EL ε EL σ EL 2 2 = = = A×l 2 2E

7. TOUGHNESS Toughness is defined as the energy absorption capacity of the material of the component just before fracture under the action of the load.

Toughness = Area under W vs. δl up to fracture point. 8. Modulus of Toughness

Modulus of Toughness is defined as the energy absorption capacity of the material of the component per unit volume just before fracture under the action of the load.

W ( h + δi )= Wh +

Wσimp

σimp 2 2E l=

× Volume σimp 2

×A×l E 2E  A×l   Wl  σimp 2 ×  0  − σimp   − Wh =  2E   E  2

Wl  Wl   Al  +   + 4  Wh E  E   2E  σimp =  Al  2   2E  W  hAE   σimp= 1 + 1 + 2   A  Wl    2h  + + σ= σ 1 1   imp Static  δstatic   2h Impact Factor =+ 1 1+ δstatic If h = 0, there is a sudden applied load

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Impact factor for sudden applied load = 2 So, Impact factor ≥ 2 σimp ≥ 2σStatic

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GATE QUESTIONS Q.1

Q.2

Q.3

A uniform, slender cylindrical rod is made of a homogeneous and isotropic material. The rod rests on a frictionless surface. The rod is heated uniformly. If the radial and longitudinal thermal stresses are represented by σr and σz , respectively, then a) σr = 0, σz = 0 b) σr ≠ 0, σz = 0 c) σr = 0, σz ≠ 0 d) σr ≠ 0, σz ≠ 0 [GATE-2005]

a) 4.12 c) 1.73 Q.4

A steel rod of length L and diameter D , fixed at both ends, is uniformly heated to a temperature rise of ∆T . The Young's modulus is E and the co-efficient of linear expansion is α. The thermal stress in the rod is a) 0 b) α∆T c) E α∆T d) E α∆TL [GATE-2007]

A solid steel cube constrained on all six faces is heated so that the temperature rises uniformly by ΔT . If the thermal coefficient of the material is α, Young’s modulus is E and the Poisson’s ratio is υ, the thermal stress developed in the cube due to heating is α ( ∆T ) E 2α ( ∆T ) E a) − b) − (1 − 2υ) (1 − 2υ) c) −

A stepped steel shaft shown below is subjected to 10 Nm torque. If the modulus of rigidity is 80 GPa, the strain energy in the shaft in N-mm is

b) 3.46 d) 0.86 [GATE-2007]

3α ( ∆T ) E

d) −

(1 − 2υ )

3 (1 − 2υ ) [GATE-2012]

Q.5 A cantilever beam of length L and flexural modulus EI is subjected to a point load P at the free end. The elastic strain energy stored in the beam due to bending (neglecting transverse shear)

[GATE-2017(2)]

(A)

P 2 L3 6EI

PL3 (C) 3EI

(B)

2 (c)

3 (c)

4 (a)

P 2 L3 3EI

PL3 (D) 6EI

ANSWER KEY: 1 (a)

α ( ∆T ) E

5 (a)

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EXPLANATIONS Q.1

Q.2

(a) We know that due to temperature changes, dimensions of the material change. If these changes in the dimensions are prevented partially or fully, stresses are generated in the material and if the changes in the dimensions are not prevented, there will be no stress set up. (Zero stresses). So, σρ = 0 ανδ σζ = 0

Now since the cube is uniformly constrained to expand, the stress produced in all the three directions will be same ∴ Strain in x direction υσ y υσ x σ −α ( ∆T ) = x − − E E E σ= σ = σ = σ x y z

(c)

Q.3

(c) Given : T = 10 N m = 104 N mm G = 80 GPa = 80 ×103 N/mm2 L1 = L2 = 100 mm, d1 = 50 mm, d2 = 25 mm We know that for a shaft of length l and polar moment of inertia J, subjected to a torque T with an angle of twist θ. The expression of strain energy, TL 1 . U = × T × θ and θ = GJ 2 So total strain energy T2L  1 1  = U  +  2G  J1 J 2 

∴σ = − Q.5

α ( ∆T ) E (1 − 2υ)

(a) There can be more than one Possibility. So identities P,Q cannot be determined.

π 4 (d) 32 Substituting the values, we get U = 1.73 N-mm

Where J = Q.4

(a) Let the side of cube be ’a’

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3

SHEAR FORCE AND BENDING MOMENT DIAGRAM

3.1 SHEAR FORCE & BENDING MOMENT 1. SHEAR FORCE Shear force is defined at the vertical force at the section of a beam. The Shear force is given by summation of all vertical forces acting at the section of the beam either left or right of the section.

clockwise bending moment will be considered as negative. From the left of the Section anticlockwise bending moment will be considered as negative, while clockwise bending moment will be considered as positive. Sagging is the condition of positive bending moment. Hogging is the condition of the negative bending moment.

Sign convention of shear force:

The sign convention of the shear force is shown in the fig. From the right of the section at which shear force calculated downward shear force will be considered as positive while upward shear force will be considered as negative. From the left of the section downward shear force will be considered as negative, while upward shear force will be considered as positive.

3. SHEAR FORCE & BENDING MOMENT Shear force at any given section is dF = w dx  dF =w dx dF Where is the slop of shear force diagram dx w is distribution load. Shear force between two points 1 and 2 is given as 2

F2 − F1  wdx =∫ 1

2. BENDING MOMENT Bending moment at a given section is defined as the algebraic summation of all the bending moment and couples either left or right of the section of the beam. Sign convention of bending moment

The sign convention of the bending moment is shown in the fig. From the right of the section at which bending moment calculated anticlockwise bending moment will be considered as positive while

Shear force at any given section is dM = f dx dM =f dx Where

dM is the slop of bending moment diagram dx

f is distributed shear force. Bending moment between two points 1 and 2 is given as 2

M 2 − M1  = ∫f dx

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1

Note: Variation of shear force and bending moment diagram S. No. Load 1 Point Load 2 UDL 3 IVL 4 Pure Bending

3.2 BEAMS

SFD Horizontal Line Inclined Line Parabolic Curve Zero Value

BMD Inclined Line Parabolic Curve Cubic Parabolic Curve Horizontal Line

1. BEAM Beam is defined as any structural member which is subjected under transverse loads. Due to the transverse loading, there is a variation of shear force and bending moment over the entire span of the beam. So it is better to draw SFD and BMD to check the maximum value of shear force and bending moment. 2. SUPPORTS:

(i) ROLLER SUPPORTS:

Roller supports are free to rotate and translate along the surface upon which the roller rests. The surface can be horizontal, vertical, or sloped at any angle. The resulting reaction force is always a single force that is perpendicular to, and away from, the surface. Roller supports are commonly located at one end of long bridges. This allows the bridge structure to expand and contract with temperature changes. The expansion forces could fracture the supports at the banks if the bridge structure was "locked" in place. Roller supports can also take the form of rubber bearings, rockers, or a set of gears which are designed to allow a limited amount of lateral movement. A roller support cannot provide resistance to lateral forces. Imagine a structure (perhaps a person) on roller skates. It would remain in place as long as the structure must only support itself and perhaps a perfectly vertical load. As soon as a lateral load of any kind pushes on the structure it will roll away in response to

the force. The lateral load could be a shove, a gust of wind or an earthquake. Since most structures are subjected to lateral loads it follows that a building must have other types of support in addition to roller supports. (ii) PINNED SUPPORTS

A pinned support can resist both vertical and horizontal forces but not a moment. They will allow the structural member to rotate, but not to translate in any direction. Many connections are assumed to be pinned connections even though they might resist a small amount of moment in reality. It is also true that a pinned connection could allow rotation in only one direction; providing resistance to rotation in any other direction. The knee can be idealized as a connection which allows rotation in only one direction and provides resistance to lateral movement. The design of a pinned connection is a good example of the idealization of the reality. A single pinned connection is usually not sufficient to make a structure stable. Another support must be provided at some point to prevent rotation of the structure. The representation of a pinned support includes both horizontal and vertical forces. (iii) FIXED SUPPORTS

Fixed supports can resist vertical and horizontal forces as well as a moment. Since they restrain both rotation and translation, they are also known as rigid supports. This means that a structure only needs one fixed support in order to be stable. All three equations of equilibrium can be satisfied. A flagpole set into a concrete base is a good example of this kind of support. The representation of fixed supports always includes two forces (horizontal and vertical) and a moment. (iv) SIMPLE SUPPORTS

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Simple supports are idealized by some to be frictionless surface supports. This is correct in as much as the resulting reaction is always a single force that is perpendicular to, and away from, the surface. However, are also similar to roller supports in this. They are dissimilar in that a simple support cannot resist lateral loads of any magnitude. The built reality often depends upon gravity and friction to develop a minimal amount of frictional resistance to moderate lateral loading.

3) Fixed beam - fixed at both ends.

4) Overhanging beam:-

A simple beam extending beyond its support on one end.

3. TYPES OF BEAMS (a)Statically determinate or indeterminate Statically determinate - Equilibrium conditions sufficient to compute reactions. Statically indeterminate - Deflections (Compatibility conditions) along with equilibrium equations should be used to find out reactions. (b) Cross sectional Shapes - I,T,C or other cross sections. (c) Depending on the supports used

5) Double overhanging beam :A simple beam with both ends extending beyond its support on both ends.

1) Simply supported beam - pinned at one end and roller at the other.

6) Continuous beam :- A beam extending over more than 2 supports.

2) Cantilever beam - fixed at one end and the other end free.

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4. Continue using the Equilibrium equations to solve the structure.

3.3 Free-Body Diagrams Before calculating for the support reactions on a body, a free-body diagram should be drawn. A free-body diagram is a representation of a body that is free from any support or external object; however, the forces and moments from the supports attached to the body are to be considered when drawing a free-body diagram, along with any other external forces or moments. If there is a force that acts in both the x and y direction (i.e. the force is on an angle), it must be broken up into its respective x and y components. Once all the known and unknown forces are drawn, the free-body diagram is complete. From here, all the forces that need to be solved can be clearly seen, and their respective directions clearly labeled, therefore all forces and moments are able to be separated into their respective equations of equilibrium. All forces in the x-direction are to be placed into the x-direction equation of equilibrium; the same procedure is used for the y-direction as well as all the moments that act on the body.

5. Redraw the Free-Body Diagram with forces and reactions in the correct directions. A negative answer shows that the direction drawn on the initial FreeBody Diagram was assumed incorrectly. 3.5 Equations of Equilibrium The equations of equilibrium are as follows: 1. ∑ Fx = 0 , Equation 1 represents the sum of all the forces acting on a body in the x direction, and states that this sum must be equal to zero. 2. ∑ Fy = 0  , Equation 2 represents the sum of all the forces acting on a body in the y direction, and states that this sum must be equal to zero. 3. ∑ Mo = 0 , Equation 3 represents the sum of all the couple moments and the moments caused by all force components acting on the member around the z axis, which is perpendicular to both the x and y axis and passes through an arbitrary point o. 3.6 EXAMPLE PROBLEMS Problem 1 Solve the following support reactions.

3.4 Method of Analysis. 1. Draw the Free-Body Diagram including all applied forces and support reactions.

2. Using the Equations of Equilibrium, solve reactions if possible. 3. For more complicated structures, it may be cut into sections. Sections should be cut between every time the loading condition changes. It is also useful to cut at hinges where no moments occur.

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Step1: To determine the appropriate reactions, draw the free-body diagram. Include all reactions and external loads.

Step 2: Solve the reactions using the Equations of Equilibrium.

∑ MA = 0 −100kN × 5m + C y ×10m = 0

C y = 50kN   ∑ Fy = 0

Step1: To determine the appropriate reactions, draw the free-body diagram. Include all reactions and external loads.

Step 2: Split at the hinge. The hinge has no moments.

A y + C y − 100kN = 0 A y = 50kN   ∑ Fx = 0 Ax = 0 Step 3: Redraw FBD.

Step 3: Starting with the right side as it is simpler, solve for reactions.

∑ MB = 0

C y ×1m − 30kN × 3m = 0 Problem 2

C y = 90kN

Solve the following support reactions.

∑ Fy  = 0 −VB + C y − 30kN = 0 VB = 60kN  ∑ Fx = 0 NB = 0

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Step 4: Transfer the internal reactions to the other side of the hinge. Solve for remaining unknowns. ∑ MA = 0

−M A − 20kN × 2m + 60kN × 3m = 0

i) Cantilever of length l carrying a concentrated load W at the free end Sx = - w Mx = - W.x Mmax = - Wl

M A = 140kNm ∑ Fy = 0 A y + VB − 20kN = 0 A y = −40kN  ∑ Fx = 0 Ax = 0

Step 5: Redraw FBD. ii) Cantilever of length l carrying a uniformly distributed load of w per unit run over the whole length Sx = - wx wx 2 Mx = − 2 Step 6: Check the answers by checking the wl2 equilibrium of the entire system. M max = − 2 ∑ MA = 0 Smax = − wl −140kNm − 20kN × 2m + 90kN × 4m − 30kN × 6m = 0

0=0 ∑ Fy = 0 −40kN − 20kN + 90kN − 30kN = 0 0=0 ∑ Fx = 0 0=0

3.7 SPECIAL CASE OF SHEAR FORCE DIAGRAM AND BENDING MOMENT DIAGRAM Case (A) Cantilevers Beam

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iii) Cantilever of length l carrying a uniformly distributed load of w per unit run over the whole length and a concentrated load W at the free end. S= wx + W x  wx 2  − + Wx  Mx =  2  Smax= wl + W M max

iv) Cantilever of length l carrying a uniformly distributed load of w per unit for a distance “a” from free end.

 wl2  = − + WI   2 

From D to B Sx = + wx

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wx 2 2 from D to A Sx = + wa Mx = −

a  Mx = − wa  x −  2  v) Cantilever of length l carrying a load whose intensity varies uniformly from zero at free end to w per unit run at the fixed end.

wl wx 2 S= − x 2 2l wlx wx 3 wl2 Mx = − − 2 6l 3

Case (B) Simply Supported beams

wx 2 = area of load diagram 2l between X and B Mx = Moment of load acting on XB about X = area of the load diagram between X and B x distance of centroid of this diagram from X wx 3 wx 2 x = − . = − 6l 2l 3 2 wl M max = 6 Sx = +

i) Simply supported beam of span l carrying a concentrated load at mid span.

vi) Cantilever beam carrying a load whose intensity varies uniformly from zero at the fixed end to w per unit run at the free end.

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Sx = + = −

W (betweenAC) 2

W (between CB) 2

Mx = +

W x (from A to C) (at a dis tan ce`x`from A) 2

Wl 4 ii) Simply supported beam carrying a concentrated load eccentrically the span. The load W is subjected at the length ‘a’ from one end and at ‘b’ from the other end. Shear stress between A and D can be given as Wb Sx = + (from A to D) l Wa = − (from D to B) l at a dis tan ce `x`from A Wab M= M = max D l Note: Maximum B.M. occurs where S.F. changes its sign.

wl − wx 2 wl wx 2 = Mx .x − 2 2 wl2 M max = M = C 8 S= x

M max = M = C

iv) Simply supported beam carrying a load whose intensity varies uniformly from zero at each end to w per unit run at the mid span.

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wl wx 2 − 6 2l wl wx 3 = Mx .x − 6 6l 1 Max B.at x = from end A 3 wl2 M max = 9 3 S= x

v) Simply supported beam carrying a load whose intensity varies uniformly from zero at one end to w per unit run at the other end:

wl w Sx = + − .x 2 4 l wl Smax = + 4 wl w = Mx .x − .x 3 4 3l wl2 M max = M = C 12 Case (C) Overhanging Beam Simply supported beam with equal overhangs & carrying a uniformly distributed load of w per unit run over the whole length. S.F. at any section in EA at a distance x fromE, Sx = - wx At any section in AB, w Sx = (l + 2a) − wx 2 B.M.at any selection in EA, wx 2 2 at any sec tion in AB, Mx = −

w wx 2 (l1 + 2a)(x − a) − 2 2 = at x `a`and `a + 1`i.e., at A & B, M= x

wa 2 Mx = − 2 1 at x= a + i.e., at C 2

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= Mc

w 2 (l − 4a 2 ) 2

Case (a) l2 > 4a 2 ⇒ l > 2a l2 − 4a 2 > o ⇒ M c > o (Positive) ∴ B.M.D. will be an shown in figure and to find point of contra flexure, put Mx = O ⇒ point of contra flexure O1 & O2 are at a 1 2 distance = l − 4a 2 2

Case (b) l2= 4a 2 ⇒ =l 2a , The bending moment at C ⇒ MC = O The beam will be subjected to only hogging moments. Points of contra flexure O1 & O2 will coincide with C. B.M.D. will be as shown in figure

Case (c) l2 < 4a 2 ⇒ l < 2a w 2 = MC (l − 4a 2 ) 8 MC is negative, since l2 < 4a 2 W ∴M C = − (4a 2 − l2 ) 8 B.M. will be zero only at ends A and D and at all other sections B.M. will be of hogging type

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Case (D) Shear Force Diagram and Bending Moment Diagram due to a couple

(i): Cantilever Beam Let a cantilever beam is subjected under the constant bending moment at the free end as shown in the fig. There will be no shear force. So shear force diagram and bending moment diagram can be represented as shown in the fig.

Sx = −

M l

B.M., M x = − = +

Ma (left of C) l

Mb (right of C) l

(ii): Simply supported Let a simply supported beam is subjected under the constant bending moment a point C as shown in the fig. The shear force diagram and bending moment diagram can be represented as shown in the fig.

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GATE QUESTIONS Q.1

A cantilever beam carries the antisymmetric load shown, where w0 is the peak intensity of the distributed load. Qualitatively, the correct bending moment diagram for this beam is

magnitude of the vertical reaction force in N at the left support is L a) Zero b) 3π L 2L c) d) π π [GATE-2013] Q.4

a)

b)

c)

d)

A cantilever beam OP is connected to another beam PQ with a pin joint as shown in the figure. A load of 10 kN is applied at midpoint of PQ. The magnitude of bending moment (in kN-m) at fixed end O is______.

[GATE-2005] Q.2

A cantilever beam has the square cross section of 10 mm × 10 mm. It carries a transverse load of 10 N. Consider only the bottom fibers of the beam, the correct representation of the longitudinal variation of the bending stress is a) b) c)

d) [GATE-2005]

Q.3

a) 2.5 c) 10 Q.5

b) 5 d) 25 [GATE-2015(2)]

For the overhanging beam shown in figure, the magnitude of maximum bending moment (in kN-m) is _____

A simply supported beam of length L is subjected to a varying 3πx distributed load sin( ) Nm−1 L where the distance x is measured from the left support. The

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[GATE-2015(3)]

Q.6

The value of moment of inertia of the section shown in the figure about the X-axis is

All dimensions are in ‘mm’ a) 8.5050 ×106 mm 4 b) 6.8850 ×106 mm 4 c) 7.7625 ×106 mm 4 d) 8.5725 ×106 mm 4 [GATE-2015(3)] Q.7

For a loaded cantilever beam of uniform cross-section, the bending moment (in N.mm) along the length is M ( x ) = 5x 2 +10x , where x is the

distance (in mm) measured from the free end of the beam. The magnitude of shear force (in N) in the crosssection at x =10 mm is ________ [GATE-2017(2)]

ANSWER KEY: 1 (c)

2 (a)

3 (b)

4 (c)

5 40

6 (b)

7 110

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EXPLANATIONS Q.1

(c)

Q.4

(c)

Q.2

(a) The bending moment varies from zero to 10 N-m along the length of the beam from the centre of the beam.

Q.5

(40) BMD:

M × y max I 10 ×103 10 = 4 × = 60MPa 10 /12 2

∴σ =

Q.3

(b) Total load on the beam L  3πx  F = ∫ sin   dx 0  L  L

  3πx  L  = -cos  ×   L  3π  0 

−L L 2L = − − = 3π 3π 3π Now taking the moment about point B. we have ∑ M B = 0

RA × L = RA =

2L L × 3π 2

Q.6 Q.7

R A = 10kN R B = 50kN Maximum bending momentum occurs at reaction B and has a magnitude of 40 kN-m. (b)

∵ S.F =

dM dx

∴ S.F=10x+10

⇒ [S.F]at x =10 = 10 ×10 + 10 = 110N

L 3π

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4

BENDING STRESS AND SHEAR STRESS

4.1 PURE BENDING MOMENT

•

A member is said to be under pure bending moment when it subjected to two equal and opposite couple in the plane passing through the axis of the beam.

• • • • •

Fig. Pure bending 1. EULER BERNOULLI’S EQUATION OR

(BENDING STRESS BENDING EQUATION

M = I

σ = y

FORMULA)

OR

E R

Where σ =Bending Stress M = Bending Moment I = Moment of Inertia E = Modulus of elasticity

The material is homogenous and isotropic i.e. it has a uniform composition and its mechanical properties are the same in all directions The stress-strain relationship is linear and elastic Young’s Modulus is the same in tension as in compression Sections are symmetrical about the plane of bending Sections which are plane before bending remain plane after bending

3. BENDING STRESS DISTRIBUTION

As per bending stress equation, bending stress is directly proportional to y, extreme fiber of the component from the neutral axis is under the maximum bending stress. For the symmetrical cross section, extreme fiber will be under same amount of stress but for unsymmetrical cross section extreme fiber will not be under same amount of bending stress.

R = Radius of curvature y = Distance of the fibre from NA ( Neutral axis ) 2. Assumptions in Simple Bending Theory All of the foregoing theory has been developed for the case of pure bending i.e. constant B.M along the length of the beam. In such case: • The shear force at each c/s is zero. • Normal stress due to bending is only produced. • Beams are initially straight • •

Fig. Bending stress distribution So the bending strain can be given as y εy = ± R  It means if at the top side from neutral axis is tensile bending stress then bottom side from neutral axis will have compressive bending stress As per Hook’s law, up to elastic limit σ by ∝ ε y ∝ y

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 y σ= Eε= E±  by y  R σ by E = y R So the ratio of bending stress and strain at the top of the fiber and bottom of the fiber is given as ε ytop y top = ε ybottom y bottom σ bytop = σ bybottom

ε ytop = ε ybottom

y top y bottom

4. Section Modulus(Z) I Z = y • Z is a function of beam cross section only • It is also called the strength of the beam • Section modulus is the first moment of area about the axis of bending for a beam cross-section • The strength of the beam sections depends mainly on the section modulus

The flexural formula may be written as. The flexural formula may be written as.  =

M Z

Rectangular cross section of width is ‘‘b’’ & depth ‘’h’’ with sides horizontal: Z =

bh 2 6

Square beam with sides horizontal: Z =

a3 6

Square cross section with diagonal horizontal :

Z =

3

a 6 2

Circular cross section of diameter ‘‘d’’

Z=

πd 3 32

5. Flexural Rigidity (EI) Reflects both • Stiffness of the material (measured by E) • Proportions of the cross section area(measured by I ) 4.2 BEAM UNDER UNIFORM STRENGTH A beam is said to be under uniform bending strength if the bending stress developed is same on every cross section along the length of the beam.

= σb

M 6M 6Px = = Zxx bd 2 bd 2

For uniform strength M ∝ bd 2 So uniform bending strength can be achieved by varying width of the cross section or depth of the cross section. There are following methods:

a) By changing the width and keeping depth as constant Let d and b are depth and width of the cross section at the fixed end as shown in the fig. The length of the beam is l, then for uniform strength σ b2− 2 = σ b3−3 6Px 6Pl = b x d 2 bd 2

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x bx = b   l

b) By changing the depth and keeping width as constant Let d and b are depth and width of the cross section at the fixed end as shown in the fig. The length of the beam is l, then for uniform strength σ b2− 2 = σ b3−3 6Px 6Pl = bd x 2 bd 2

x dx = d   l 4.3 DIRECT SHEAR STRESS Let us consider a beam of a rectangular cross section having a load of P at its free end as shown in the fig. The applied shear force will be distributed as a shear stress across transverse sections of the beam. At each point on a section the transverse shear stress will produce a complementary horizontal shear stress.

Shear stress at distance y is given by PAY τ= I NA .b Where Y is distance between centroid of the section and centre of the cross section. I NA is the moment of inertia about NA

d  Area of shaded portion A = b  − y  2  1d  Y =  + y 22  3 bd I NA = 12 So shear stress d 1d  Pb  − y   + y  PAY 2 2 2  = τ = bd 3 I NA .b .b 12   d 2  6P    − y 2   2    τ=  3 bd When y =0, shear stress is maximum, It can be given as 3P τ max = 2bd The average shear stress is given by P P τ avg= = A bd 3 τ max = τ avg 2

Let b is the width and d is depth of the rectangular cross section. τ is the shear stress in a layer at the distance y from neutral axis. Section is subjected to P.

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Shear stress distribution for L section is given as 1 .SHEAR STRESS DISTRIBUTION Shear stress is a parabolic function of y for the rectangular cross section. So the shear stress distribution can be given by Shear stress distribution circular section is given as

Shear stress distribution for I section is given by

Shear stress distribution for T section is given as

for

Hollow

Shear stress distribution for Cross section is given as

2. SHEAR CENTRE Shear centre of a section can be defined as a point about which the applied force is © Copyright Reserved by Gateflix.in No part of this material should be copied or reproduced without permission

balanced by the shear forces obtained by summing the shear stresses over the section. It is also known as centre of twist. When the load passes through the shear centre then there will be only bending in the cross section and no twisting. Shear centre for channel can be given by

b 2 d 2 t web 4I NA Shear centre for circular tube can be given as e=

e = 2r

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GATE QUESTIONS Q.1

The second moment of a circular area about the diameter is given by (D is the diameter). a) πd 4 / 4 b) πd 4 /16 c) πd 4 / 32 d) πd 4 / 64 [GATE-2003]

Q.2

A concentrated load of P acts on a simply supported beam of span L at a distance L/3 from the left support. The bending moment at the point of application of the load is given by a) PL/3 b) 2PL/3 c) PL/9 d) 2PL/9 [GATE-2003]

Q.3

The beams, one having square cross section and another circular crosssection, are subjected to the same amount of bending moment. If the cross sectional area as well as the material of both the beams are same then a) maximum bending stress developed in both the beams is same b) the circular beam experience more bending stress than the square one c) the square beam experience more bending stress than the circular one d) as the material is same, both the beams will experience same deformation. [GATE-2003]

Q.4

A simply supported laterally loaded beam was found to deflect more than a specified value. Which of the following measures will reduce the deflection? a) Increase the area moment of inertia

b) Increase the span of the beam c) Select a different material having lesser modulus of elasticity d) Magnitude of the load to be increased [GATE-2003] Common Data For Q.5 and 6 are given below. A steel beam of breadth 120 mm and height 750 mm is loaded as shown in the figure. Assume Esteel = 200 GPa.

Q.5

The beam is subjected to a maximum bending moment of a) 3375 kN-m b) 4750 kN-m c) 6750 kN-m d) 8750 kN-m [GATE-2004]

Q.6

The value of maximum deflection of the beam is a) 93.75 mm b) 83.75 mm c) 73.75 mm d) 63.75 mm [GATE-2004]

Q.7

Two identical cantilever beams are supported as shown, with their free ends in contact through a rigid roller. After the load P is applied, the free ends will have

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Q.8

a) equal deflections but not equal slopes b) equal slopes but not equal deflections c) equal slopes as well as equal deflections d) neither equal slopes nor equal deflections [GATE-2005]

A beam is made up of two identical bars AB and BC, by hinging them together at B. The end A is built-in (cantilevered) and the end C is simply-supported. With the load P acting as shown, the bending moment at A is a) zero c) 3PL/2

b) PL/2 d) indeterminate [GATE-2005]

Common Data for Q.9 and Q.10: A simply supported beam of span length 6 m and 75 mm diameter carries a uniformly distributed load of 1.5 kN/m Q.9

What is the maximum value of bending moment? a) 9 kN-m b) 13.5 kN-m c) 81 kN-m d) 125 kN-m [GATE-2006]

Q.10 What is the maximum value of bending stress? a) 162.98 MPa b) 325.95 MPa c) 625.95 MPa d) 651.90 MPa [GATE-2006]

a) 25 c) 35

b) 30 d) 60 [GATE-2007]

Q.12 A uniformly loaded propped cantilever beam and its free body diagram are shown below. The reactions are

5qL = ,R 2 8 3qL = = ,R 2 b) R1 8 3qL c) R1 = = ,R 2 8 3qL d) R1 = = ,R 2 8 = a) R1

3qL qL2 = ,M 8 8 3qL qL2 = ,M 8 8 3qL = ,M 0 8 3qL = ,M 0 8 [GATE-2007]

Common Data for Q.13 and 14: A machine frame shown in the figure below is subjected to a horizontal force of 600 N parallel to Z -direction.

Q.11 In a simply-supported beam loaded as shown below, the maximum bending moment in Nm is Q.13 The normal and shear stresses in © Copyright Reserved by Gateflix.in No part of this material should be copied or reproduced without permission

MPa at point P are respectively a) 67.9 and 56.6 b)56.6 and 67.9 c) 67.9 and 0.0 d) 0.0 and 56.6 [GATE-2007] Q.14 The maximum principal stress in MPa and the orientation of the corresponding principal plane in degrees are respectively a)- 32.0 and-29.52 b) 100.0 and 60.48 c)- 32.0 and 60.48 d) 100.0 and -29.52 [GATE-2007] Q.15 The transverse shear stress acting in a beam of rectangular cross-section, subjected to a transverse shear load, is a) variable with maximum at the bottom of the beam b) Variable with maximum at the top of the beam c) Uniform d) Variable with maximum on the neutral axis [GATE-2008] Q.16 An axial residual compressive stress due to a manufacturing process is present on the outer surface of a rotating shaft subjected to bending. Under a given bending load, the fatigue life of the shaft in the presence of the residual compressive stress is a) decreased b) increased or decreased, depending on the external bending load c) neither decreased nor increased d) increased [GATE-2008] Q.17 For the component loaded with a force F as shown in the figure, the axial stress at the corner point P is

F(3L – b) 4b3 F(3L – 4b) c) 4b3 a)

F(3L + b) 4b3 F(3L – 2b) d) 4b3 [GATE-2008] b)

Q.18 A solid circular shaft of diameter 100 mm is subjected to an axial stress of 50 MPa. It is further subjected to a torque of 10 kNm. The maximum principal stress experienced on the shaft is closest to a) 41 MPa b) 82 MPa c) 164 MPa d) 204 MPa [GATE-2008]

Q.19 The strain energy stored in the beam with flexural rigidity EI and loaded as shown in the figure is

P 2 L3 3EI 4P 2 L3 c) 3EI a)

2P 2 L3 3EI 8P 2 L3 d) 3EI [GATE-2008] b)

Q.20 A frame of two arms of equal length L is shown in the adjacent figure. The flexural rigidity of each arm of the frame is EI. The vertical deflection at the point of application of load P is

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PL3 3EI PL3 c) EI a)

2PL3 3EI 4PL3 d) 3EI [GATE-2009] b)

Common Data For Q.21 and Q.22 A mass less beam has a loading pattern as shown in the figure. The beam is of rectangular cross-section with a width of 30 mm and height of 100 mm

(The length of the beam shown in the figure is in mm)

Q.21 The maximum bending moment occurs at a) Location B b) 2675 mm to the right of A c) 2500 mm to the right of A d) 3225 mm to the right of A [GATE-2010] Q.22 The maximum magnitude of bending stress (in MPa) is given by a) 60.0 b) 67.5 c) 200.0 d) 225.0 [GATE-2010] Q.23 A simply supported beam PQ is loaded by a moment of 1 kN-m at the mid-span of the beam as shown in the figure the reaction forces RP and RQ at supports P and Q respectively are

a) 1 kN downward, 1 kN upward b) 0.5 kN upward, 0.5 kN downward c) 0.5 kN downward, 0.5 kN upward d) 1 kN upward, 1 kN upward [GATE-2011] Common Data For Q.24 and 25 : A triangular-shaped cantilever beam of uniform-thickness is shown in the figure The Young’s modulus of the material of the beam is E. A concentrated load P is applied at the free end of the beam.

Q.24 The area moment of inertia about the neutral axis of a cross-section at a distance x measured from the free end is bxt 3 bxt 3 a) b) 12l 6l 3 bxt xt 3 c) d) 24l 12l [GATE-2011] Q.25 The maximum deflection of the beam is 24Pl3 12Pl3 a) b) Ebt 3 Ebt 3 8Pl3 6Pl3 c) d) Ebt 3 Ebt 3 [GATE-2011] Q.26 A cantilever beam of length L is subjected to a moment M at the free

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end. The moment of inertia of the beam cross section about the neutral axis is I and the Young's modulus is E . The magnitude of the maximum deflection is ML2 ML2 a) b) 2EI EI 2 2ML 4ML2 c) d) EI EI [GATE-2012] Q.27 Consider a cantilever beam, having negligible mass and uniform flexural rigidity, with length 0.01 m. The frequency of vibration of the beam, with a 0.5 kg mass attached at the free tip, is 100 Hz. The flexural rigidity (in Nm 2 ) of the beam is __________. [GATE-2014 (1)] Q.28 The flexural rigidity (EI) of a cantilever beam is assumed to be constant over the length of the beam shown in figure. If a load P and bending moment PL/2 are applied at the free end of the beam then the value of the slope at the free end is

1 PL2 2 EI 3 PL2 c) 2 EI a)

PL2 EI 5 PL2 d) 2 EI [GATE-2014 (2)] b)

Q.29 A cantilever beam of length, L with uniform cross section and flexural rigidity, EI, is loaded uniformly by a veridical load, w per unit length. The maximum vertical deflection of the beam is given by wL4 wL4 a) b) 8EI 16EI

c)

wL4 4EI

wL4 24EI [GATE-2014 (2)] d)

Q.30 Consider a simply supported beam of length, 50h, with a rectangular cross –section of depth, h, and width, 2h The beam carries a vertical point load, P, at its midpoint. The ratio of the maximum shear stress to the maximum bending stress in the beam is a) 0.02 b) 0.10 c) 0.05 d) 0.01 [GATE-2014 (3)] Q.31 A force P is applied at a distance x from the end of the beam as shown in the figure. What would be the value of x so that the displacement at ‘A’ is equal to zero?

a) 0.5L c) 0.33L

b) 0.25L d) 0.66L [GATE-2014 (3)]

1 PL3 3 EI PL4 c) EI

2 PL3 3 EI 4 PL3 d) 3 EI [GATE-2014(4)]

Q.32 A frame is subjected to a load P as shown in the figure. The frame has a constant flexural rigidity EI. The effect of axial load is neglected. The deflection at point A due to the applied load P is

a)

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b)

Q.33 A cantilever beam with flexural rigidity of 200 Nm2 is loaded as shown in the figure. The deflection (in mm) at the tip of the beam is ____

located at the top surface of the beam at a distance of 1.5L from the left end is _____________ (Indicate compressive stress by a negative sign and tensile stress by a positive sign.) [GATE-2016 (1)] Q.37

[GATE-2015 (1)] Q.34 A cantilever beam with square cross section of 6 mm side is subjected to a load of 2 kN normal to the top surface as shown in the figure. The young’s modulus of elasticity of the material of the beam is 210 GPa. The magnitude of slope (in radian) at Q (20 mm from the fixed end) is _____.

[GATE-2015 (2)] Q.35

A cantilever beam having square cross-section of side a is subjected to an end load. If a is increased by 19%, the tip deflection decreases approximately by a) 19% b) 29% c) 41% d) 50% [GATE-2016 (1)]

Q.36

A simply-supported beam of length 3L is subjected to the loading shown in the figure. It is given that P = 1 N, L = 1 m and Young’s modulus E = 200 GPa. The cross-section is a square with dimension 10 mm×10 mm. The bending stress (in Pa) at the point A

A beam of length L is carrying a uniformly distributed load w per unit length. The flexural rigidity of the beam is EI. The reaction at the simple support at the right end is

wL 2 wL c) 4 a)

3wL 8 wL d) 8 [GATE-2016 (3)] b)

Q.38 The figure shows cross-section of a beam subjected to bending. The area moment of inertia (in mm4) of this cross-section about its base is _______

[GATE-2016 (1)] Q.39 A simply supported beam of width 100 mm, height 200 mm and length 4 m is carrying a uniformly distributed load of intensity 10 kN/m. The maximum bending stress (in MPa) in the beam is __________ (correct to one decimal place).

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[GATE-2018 (1)]

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ANSWER KEY: 1 (d) 15 (d) 29 (a)

2 (d) 16 (d) 30 (a)

3 (b) 17 (d) 31 (c)

4 (a) 18 (b) 32 (d)

5 6 (a) (a) 19 20 (c) (d) 33 34 0.25 0.16

7 (a) 21 (c) 35 (d)

8 (b) 22 (b) 36 0

9 6.75 23 (a) 37 (b)

10 (a) 24 (b) 38 1875

11 (b) 25 (d) 39 30

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12 (a) 26 (a)

13 (a) 27

14 (b) 28 (b)

EXPLANATIONS Q.1

(d)

Q.2

(d) We know that, the simplest form of the simply supported beams is the beam supported on rollers at ends. The simply supported beam and the FBD are shown in the figure.

Where, RA and RB are the reactions acting at the ends of the beam. In equilibrium condition of forces, P = RA + RB ...(i) Taking the moment about point A L RB × L = P× 3 P ∴RB = 3 From equation (i) 2P RA = 3 Now bending moment at the point of application of the load 2P L 2PL = M    = × 3 3 9 Q.3

(b) Let, d = diameter of circular crosssection a = side of square cross-section Since cross sectional area of square and circular cross-section are equal π ∴ d2 = a2 …(i) 4 For circular cross-section,

d M 2 32M = πd 4 πd 3 64 Square cross-section M.y σs = I a M 2 6M = = 4 a a3 12 32M 3 σc Now, = πd 6M σs a3 32a 3 = 6πd 3 My = σc = I

2

32  a   a  =     6π  d   d  32  π   π  = = ×   ×   6π  4   4 

32 π 32 π = >1 24 4 48 ∴ σ c > σs

=

Q.4

(a) We know, differential equation of flexure for the beam is, d2 y d2 y M EI 2 =M ⇒ 2 = dx dx El Integrating both side dy l l = = mdx Mx + Ci dx El ∫ El Again integrating, y = 2 l  Mx  ...(i)   + C1x + C2 El  2  where, y gives the deflection at the given point. It is easily shown from the equation (i), If we increase the

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Q.5

value of E and I, then deflection reduces.

Q.6

(a)

(a) Given : b = 120 mm , h = 750 mm , E steel = 200 GPa = 200 ×103 N / mm 2 W = 120 kN/m, L = 15 m For a simply supported beam subjected to uniformly distributed load, maximum bending moment occurs at the center and is given by wL2 120 × 152 B.M = = = 3375 kN − m 8 8

For a simply supported beam subjected to uniformly distributed load, maximum deflection occurs at Q.9 the center and is given by 5 wL4 δ= × max 384 EI For rectangular cross section bh 3 120 × 7503 I = = = 42.1875 ×108 mm4 12 12

(15 ×10 ) 5 120 ×103 δ max =× × 3 384 10 200 ×103 × 42.1875 ×108 3 4

δ max = 93.75mm

Q.7

(a)

Q.8

(b) First of all we have to make a free body diagram of the given beam.

Where, RA and RC are the reactions acting at points A and C. The point B is a point of contra flexure or point of inflexion or a virtual hinge. The characteristic of

the point of contra flexure is that, about this point moment equal to zero. For span BC, MB = 0 L Rc × L = P × 2 P Rc = 2 Also, RA + RC = P P P RA =P − = 2 2 Now for the bending moment about point A take the moment about point A. L  M a + R c × 2L − P ×  L= = 0 2  P 3L M a + × 2L − P × = 0 2 2 M a = PL / 2 None of these options is correct

Given: L=6m, W=1.5kN/m, d= 75mm We know that for a uniformly distributed load, maximum bending moment at the centre is given by, WL2 15 ×108 × (6) 2 B.M. = = 8 8 B.M = 6750 N-m = 6.75 kN–m

Q.10 (a)

σb M = y I Where M = Bending moment acting at the given section π 4 I = Moment of inertia = ( d ) 64 Y = Distance from the neutral axis to d the external fibre = 2

From the bending equation.

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σ b = Bending stress M So, σ b = × y I Substitute the values, we get 6.75 ×106 75 σb 162.98 MPa = = × π 4 2 ( 75) 64 Q.11 (b) Resultant beam diagram

RA + RB = 100 ...(1) Taking moment A, −100 × 0.5 − 10 + RB × 1 = 0 ∴ RB = 60 N And RA= 40 N Bending moment diagram

So, deflection occurs at B due to the UDL alone, qL4 δ UDL = 8EI Also deflection at B due to point load R L3 δ PL = 2 3EI Deflections are equal at B. δ UDL = δ PL qL4 R 2 L3 = 8EI 3EI 3qL ∴R2 = 8 And from equation (i), we have 5qL R1 = 8 Taking the moment about B, we get qL2 M= 8 ∴

Q.13 (a) Normal stress, σ = 32 × 600 × 300 π × (30)3 = 67.9 MPa

32M πd 3

=

Maximum bending moment = 40 × 0.5 + 10 = 30 Nm

Q.12 (a) First of all, we have to make a FBD of the beam. We know that a UDL acting at the mid-point of the beam and its magnitude is equal to (qL) So. In equilibrium of forces R1 + R2 = qL ….(i) This cantilever beam is subjected to two types of load. First load is due to UDL and second load is due to point load at B. Due to this deflection occurs at B, which is equal in amount.

Shear stress, τ = 16 × 600 × 500 π × (30)3 = 56.6 MPa

16T πd 3

=

Q.14 (b) The stress element

Centre of Mohr circle σx + σy = = 33.95MPa 2 Radius of mohr circle

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1

 σ x − σ y  2 2 2 =  + τ  xy   2   1

 0 − 67.9  2 2 2 =  66MPa  + (56.6)  = 2   

Principal stress = 66 + 33.95 = 99.95 = 100Mpa 2θ p = 120.95 θ p = 60.47°

Q.15 (d)

For a rectangle cross- section.  FAY 6F  d 2 = τv = − y2  3  lb bd  4  F = Transverse shear load Maximum value of τ v occurs at the neutral axis where, y = 0 3F τmax =

Q.16 (d)

2bd

compressive stress. The fatigue life of the material is effectively increased by the introduction of a compressive mean stress, whether applied or residual.

Q.17 (d) M σ = l y F(L − b) ×12 σ b ⇒ = (2b) 4 b 3F(L − b) ⇒ σb = 3 (due to bending) 4b F ⇒ ⇒ σ a =2 (due to axial force) 4b ∴ Total axial stress 3F(L − b) F = + 2 4b3 4b 3F(L − b) F = + 3 4b3 4b 3FL − 3Fb + Fb = 4b3 F(3L − 2b) = 4b3 Q.18 (b) T τ = lp r

10 ×106 τ = π 100 (100) 4 32 2 ⇒  τ = 50.95 N / mm 2 Maximum principal stress ⇒

σ σ = +   + τ2 2 2 2

2

50  50  = +   + (50.95) 2 2  2 

The figure shows the Gerber’s parabola. It is the characteristic curve of the fatigue life of the shaft in the presence of the residual

= 25 + 252 + (50.95) 2 = 82 MPa

Q.19 (c)

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Total strain energy stored is given by L (Px) 2 dx (PL) 2 .2L 2∫ + 2EI 2EI 0 L

2P 2 2 P 2 L3 x dx + 2EI ∫0 EI

=

P 2 L3 P 2 L3 × + EI 3 EI 2 3 2 3 P L P L 4 P 2 L3 = + = 3EI EI 3 EI

=

At any section Y-Y at a distance y from B, M y = PL M y 2 dy ∴ U BC = ∫ 2EI L 2 (PL) dy =∫ 2EI 0

P 2 L2 P 2 L3 L = (y) 0 2EI 2EI Total strain energy stored in the frame of two arm

= U BC

= U U AB + U BC

Q.20 (d)

For Arm AB: At any section X-X, at a distance x from A M x = −Px M x 2 dx ∴ U AB = ∫ 2EI L (−Px) 2 =∫ dx 2EI 0

L

P2  x3  P 2 L3 = U AB =   2EI  3 0 6EI For Arm BC:

P 2 L3 P 2 L3 = U + 6EI 2EI 4P 2 L3 = 6EI Therefore, vertical deflection of A is ∂U ∂  4P 2 L3  = δA = ∂P ∂P  6EI  4L3 ∂ ⇒ δA = ( P2 ) 6EI ∂P 4L3 = (2P) 6EI 4PL3 δA = 3EI

Q.21 (c)

RA + RC = 6000N Taking moment about A © Copyright Reserved by Gateflix.in No part of this material should be copied or reproduced without permission

6000 × 3 − R C × 4 = 0

Let, b=width of the base of triangle ABD BD=t= thickness of cantilever beam

⇒ RC = 4500N R A = 1500N Taking any section at x m from A Shear force =1500-3000(x-2) And maximum bending moment occurs where shear force = 0 ⇒ 1500 – 3000(x-2) = 0 x =2.5m or 2500 mm from A

Q.22 (b) Maximum bending moment 3000 (BM)= 1500 × − (x − 2) 2 max 2 at x = 2.5 = 1500 × 2.5 − 1500(0.5) 2 =3375 N.m Bending stress, M σ= z 1 Z = bd 2 6 6 × 3375 σ= 30 ×1002 ×10−9 = 67.5 ×106 N / m 2 = 67.5MPa

From similarity of triangles b/2 h = let OE = h l x bx h= ...(i) 2l Now from figure (ii), For rectangular cross section. (2h)t 3 bx t 3 bxt 2 l= =2 × × = 12 2l 12 12l From equation (i)

Q.25 (d) Q.26 (a) Q.27

Q.23 (a) First of all we have to make a free body diagram of the given beam

(0.064 to 0.067)

P =mg =0.5 × 9.81 =4.905N L = 0.01 m f n = 100Hz

= ωn 2πf = 628rad / s n Here RP and RQ are the reaction forces acting at P and Q For equilibrium of forces on the beam, RP + PQ = 0 Taking the moment about the point P, R Q Q ×1 = 1 R ⇒ = 1kNm From equation RP = −R Q = −1kNm

Q.24 (b)

(i),

k m Or k = mωn 2 = 0.5 × (628) 2 = 197192 N/m PL3 δ= 3EI P 3EI ⇒ =3 δ L P L3 Or EI= × δ 3 ωn =

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a

(0.01)3 3 EI = 0.06573 Nm2 =197192 ×

Q.31 (c)

Q.28 (b)

PL3 (− ve) 3EI Deflection due to moment 2 ML2 P ( L − x ) L = = (+ ve) 2EI 2EI δp + δM = 0 Deflection due to load =

θ1 =

PL2 2EI

P ( L − x ) L2 PL3 = 2EI 3EI L−x L = Or 2 3 Or 3L – 3x = 2L L Or x =   or 0.33L 3 ∴

ML PL2 = EI 2EI By superposition, PL2 θ = θ1 + θ 2 = EI

= θ2

Q.29 (a)

wx 2 2 (x + 6L2 − 4Lx) 24EI wL4 and δ max = 8EI

= y

Q.30 (a)

3 P 3P P = = 0.75 2 2 2 A 4h h PL P × 50h M max = = = 12.5Ph 4 4 bh 2 h 3 = Z = 6 3 M max 12.5Ph × 3 = σ max = Z h3 P = 37.5 2 h τ max 0.75 ∴ = = 0.02 σ max 37.5 τ= max

Q.32 (d)

Bending moment in section AB M = - Px ∂M = −x ∂P Bending moment in section BC: M = −PL ∂M = −L ∂P L   ∫ ( −Px )( − x ) dx +  1 0  δA =  L  EI   ∫ ( −PL )( −L ) dx   0  L L  P  2 2 = δA x dx + L dx  ∫ ∫ EI  0 0  3 3 P L 4PL 3 =  +L=  EI  3  3EI

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−P M x = ×1.5L + P[1.5L − L] 3 Mx = 0 i.e. MA = 0

Q.33 (0.24 to 0.28) Q.34 (0.15 to 0.17) Q.35 (d) We know that δ ∝ ∴

δ 2 I1 = δ1 I 2

⇒ σ A= 0[Q σ ∝ M]

1 I

Q.37 (b) The deflection at right support is zero because of reaction at the right support wL4 R A L3 ∴ = 8EI 3EI 3wL ∴RA = 8

a4 δ2 = 12 4 δ1 [1.19a ] 12 δ2 1 = = 0.5 δ1 1.194 δ −δ % decrease =   1 2 ×100 δ1 δ − 0.5δ1 = 1 ×100 δ1 = 50%

Q.38 (1873 to 1879) Q.39 Ans. (30)

wL2 10  16 = 20 kNm = Maximum B.M. ( M ) = = ( L 4) 8 8 Maximum Bending Strees

Q.36 (-1 to 1)

M 20  103 ymax =  0.1 I  0.1 0.23     12  6 2  10 N/m = 30 = 30 MPa max =

∑F

y

=0

⇒ Rp + RQ = 0

∑M

p

=0

−PL + P ( 2L ) − R Q ( 3L ) = 0

PL − R Q ( 3L ) = 0

P 3 and R P = −P / 3 RQ =

Considering section X-X M x = R p x + P[x − L] at x = 1.5L

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5

PURE TORSION

5.1 PURE TORSION A component is said to be under pure torsion when two equal and opposite couple are subjected in a plane perpendicular to axis of the component.

Fig. pure torsion 1. TORSION EQUATION Let us consider a shaft under the pure torsion T as shown in the fig. the angle of twist at the free end is θ and maximum shear angle is ɸ. Radius of the shaft is R and length of the shaft is l.

change with radius and it change with respect to length of the shaft. So  Shear angle is directly proportional to radius of the shaft but independent from the length of the shaft.  Angle of twist is directly proportional to length of the shaft but independent from the radius of the shaft. In triangle ABB’, Shear angle at maximum radius is given by BB' Rθ ∅ = tan ∅ = = L L Rθ Shear strain γ = L So shear strain is directly proportional to radius of the shaft. As per Hook’s law, up to elastic limit Shear stress

Figure: Twist deformation of a torque circular shaft. Deformations and rotations greatly exaggerated for visualization convenience. From the figure it is cleared under pure torsion shear angle does not change with length of the shaft and it changes with the radius of the shaft under the same value of the torsion. But angle of twist does not

τ max = Gγ =

GRθ L

τ max Gθ = R L Where G = Modulus of rigidity in MPa. So τ max ∝ γ ∝ r τ max τ = R r Where, r is radius of the same shaft other than the extreme fiber of the shaft. The shear stress equation (Torsion equation) can be given by T τ max Gθ = = J R L

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Where J = Polar moment of inertia τ max = Shear stress induced due to torsion T. G = Modulus of rigidity θ=Angular deflection of shaft R, L = Shaft radius & length respectively TR T 16T = = J ZP πD3 ZP = polar modulus of cross section  2. Shear stress is inversely proportional to polar modulus of cross section. For the circular solid shaft maximum shear Stress is given as τ max =

τ max =

16T πD3

For the circular hollow shaft maximum shear stress is given as having a diameter ratio k d  k =  D  16T τ max = πD3 (1 − k 4 ) d = inner diameter of the shaft D = Outer diameter of the shaft 3.TORSIONAL SHEAR STRESS DISTRIBUTION As per equation of the shear stress distribution it is cleared that Torsion shear stress is directly proportional to the radius of the shaft. The shear stress distribution can be given as

Fig. Torsion Shear Stress Distribution 4 TORSIONAL STIFFNESS Torsion stiffness is defined as the applied torsion per unit angle of twist. It is represented by q and its unit is N-m/rad. T GJ q= = θ l 5.2 COMPOSITE SHAFTS: 1. Series Connection If two or more shafts of different materials, lengths and diameters are connected in such a way that each carries the same torque, then a composite shaft connected in series is produced. Fig. shows such a shaft, in which diameter, length and modulus of rigidity of each component is described. According to the definition of series connected shaft and making use of torsion equation, one can write,

G1 J1 θ1 G 2 J 2θ 2 G 3 J 3θ3 T= = = L1 L2 L3

……. (1)

where, θ1 =θΒΑ , θ2 = θCB , θ3 = θDC And θΒΑ ( means angle of twist of section B with respect to section A, etc. Therefore,

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angle of twist of section C with respect to A. i.e. θΑD = θ1 + θ2 + θ3 The ratio of the torque to the angle of twist, ( T/ θ ) is defined as spring constant. Thus the spring constant of the series connected system will be T/θAC , whereas respective spring constants of components will be T/θ1 , T/θ2 and T/θ3 . If the spring constants are denoted by k1. k2, and k3 respectively, then from Equation. T T T + + θ AD = k1 k 2 k 3 or

relationship between the deformations or angles of twist will also have to be used. Figure shows a composite shaft made up of parts AB and BC. The shaft is rigidly held at sections A and C and a torque T is applied at section B when elements 1 and 2 are rigidly connected together. If the torques carried by components 1 and 2 are T1 and T2 respectively, then from equation of equilibrium.

θΑD 1 1 1 = + + T k1 k 2 k 3

1 1 1 1 = + + k k1 k 2 k 3 where k is the spring constant of composite shaft, or the equivalent spring constant. The units of k are N-m/radian or N-mm/radian. Torsional shear Stress produced in member 1 is given by T1 16T = τ1 = Z1 πd13 Torsional shear Stress produced in member 2 is given by T2 16T = τ2 = Z2 πd 23 Torsional Shear Stress produced in member 3 is given by T3 16T = τ3 = Z3 πd 33 i.e.

2. Statically Indeterminate Torsion Members Or Parallel Connection A parallel connected composite shaft system is obtained if two or more shafts are connected together such that the applied torque is changed between them. Such a system is statically indeterminate because in addition to equation of equilibrium,

T1 + T2 = T

---------

(4)

This equation by itself will not give the values of T1 and T2 and hence stresses or angles of twist in two parts cannot be calculated. Since section 8 is common to both parts AB and BC, ∴ θ BA = θ BC or θ1 = θ2 = 0 ------(5) OR

T1 L1 TI = 2 2 G1 J1 G 2 J 2

------- (6)

Simultaneous solution of Equation(4 ) and (6) will yield the values of T1 and T2. If Equation(4) is divided by θ .

T1 T2 T = + θ θ θ or k = k1 + k 2

----------

(7)

Where, k is the spring constant of parallel connected shaft, and k1 and k2, are the spring constants of respective components. Note : One obvious fact from Equation( 3 ) and ( 7 ) may now be noted that the spring constant of a series connected system is less than the spring constant of each component shaft, whereas the spring constant of a

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parallel connected system is the sum of the individual spring constants. 5.3 SPRING Spring is the elastic member which is used to store the energy and release the energy. It deforms under the action of the load and regains its original position after removal of the load. The functions of the spring are following: a) To store the energy & release the energy b) To absorb the shocks and reduce the vibrations. c) To control the motion and apply the forces d) In spring balance, to measure the weight of the body etc. 1. HELICAL SPRING It is made of wire coiled in the form of helix having circular, square or rectangular cross section

d = wire diameter of spring (mm) Di = inside diameter of spring coil (mm) Do = outside diameter of spring coil (mm) D = mean coil diameter (mm) Therefore

D=

Di + DO 2

There is an important parameter in spring design called spring index. It is denoted by letter C. The spring index is defined as the ratio of mean coil diameter to wire diameter. Or C = D/d In design of helical springs, the designer should use good judgment in assuming the value of the spring index C. The spring index indicates the relative sharpness of the curvature of the coil. A low spring index means high sharpness of curvature. NOTE: When the spring index is low(C < 3), the actual stresses in the wire are excessive due to curvature effect. Such a spring is difficult to manufacture and special care in coiling is required to avoid cracking in some wires. When the spring index is high (C >15), it results in large variation in coil diameter. Such a spring is prone to buckling and also tangles easily during handling. Spring index from 4 to 12 is considered better from manufacturing considerations.

2. TERMINOLOGY OF HELICAL SPRING: The main dimensions of a helical spring subjected to compressive force are shown in the figure. They are as follows:

There are three terms - free length, compressed length and solid length related to helical compression spring. These lengths are determined by following way

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1) Solid length: solid length is defined as the axial length of the spring which is so compressed, that the adjacent coils touch each other. In this case, the spring is completely compressed and no further compression is possible. The solid length is given by,

Pitch :The pitch of the coil is defined as the axial distance between adjacent coils in uncompressed state of spring. It is denoted by p. It is given by, p=

Full length Nt −1

Solid length = Nt d Where Nt = total number of coils 2) Compressed length: Compressed length is defined as the axial length of the spring that is subjected to maximum compressive force. In this case, the spring is subjected to maximum deflection. When the spring is subjected to maximum force, there should be some gap or clearance between the adjacent coils. The gap is essential to prevent clashing of the coils. The clashing allowance or the total axial gap is usually taken as 15% of the maximum deflection. Sometimes, an arbitrary decision is taken and it is assumed that there is a gap of 1 or 2 mm between adjacent coils under maximum load condition. In this case, the total axial gap is given by, Total gap = (Nt-1) x gap between adjacent coils 3) Free length: Free length is defined as the axial length of an unloaded helical compression spring. In this case, no external force acts on the spring. Free length is an important dimension in spring design and manufacture. It is the length of the spring in free condition prior to assembly. Free length is given by, Free length = compressed length + y = solid length + total axial gap + y

3. Stresses in Helical Springs The flexing of a helical spring create torsion in the wire and the force applied induces a direct stress. The maximum, stress in the wire may be computed be super position. The result is :

Tr F + J A

τ max = +

Replacing the terms, FD d πd 4 πd 4 ,r = ,J= and A = 2 2 32 4 And re-arranging, 8FD 8FC τ = K S 3 or τ = K S πd 2 πd T=

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Where KS is the shear-stress correction factor and is defined by the equation: 2C+1 KS = 2C

curved beams. Consequently the stress on the inside surface of the wire of the spring, increases but decreases it only slightly on the outside. The curvature stress is highly localized that it is very important only fatigue if is present. This effect can be neglected for static loading, because local yielding with the first application of the load will relieve it.

To take care of the curvature effect, the earlier equation for maximum shear stress in the spring wire is modified as. 

8FD 

τ max =  K s 3  πd   Where. KW is Wahl correction factor, which takes care of both curvature effect and shear stress correction factor and is expressed as. = KW

4C − 1 0.615 + 4C − 4 C

4. STRAIN ENERGY When Subjected to Axial load Curvature Effect The curvature of the wire increases the stress on the inside of the spring, an effect very similar to stress concentration but due to shifting of the neutral axis away from the geometric center, as could be observed in

Neglecting strain energy due to direct shear W, strain energy stored,

U=

τ2 4G

 ( Volume of the spring )

where G = modulus of rigidity.

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2

∴ ∴

 8FD   3   πd 2  πd   U= ( πDn )  4G  4  2 3 4F D n U= Gd 4

5. DEFLECTION OF SPRINGS

 If spring is cut into number of the coils, the stiffness of the spring increases n times. 5.4 CONNECTION OF SPRINGS 1. Spring in Parallel If two spring having stiffness K1 and K2 are connected in parallel,

Let δ is the axial deflection.

Then, work done by the load =

1 ( Fδ ) 2

Equating the work done to the strain energy stored in the spring,

1 4F2 D3 n ( Fδ ) = 2 Gd 4 8FD3 n 64FR 3 n ∴ δ= = Gd 4 Gd 4 6. STIFFNESS OF SPRINGS The stiffness of the spring is defined as the

then equivalent spring stiffness KEq = K1 + K2 2. Spring in series If two spring having stiffness K1 and K2 are connected in series, then equivalent spring stiffness

load required to produce unit deflection. .

F Gd 4 ∴ Stiffness of the spring = = δ 64R 3 n This is also termed as spring Constant k.

Stiffness of the spring is inversely proportional to the number of coils (n) of the 1 spring. k ∝ n

KEq =

K1.K 2 K1 + K 2

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GATE QUESTIONS Q.1

Maximum shear stress developed on the surface of a solid circular shaft under pure torsion is 240 MPa. If the shaft diameter is doubled then the maximum shear stress developed corresponding to the same torque will be a) 120 MPa b) 60 MPa c) 30 MPa d) 15 MPa [GATE-2003]

are built-in (cantilevered). A twisting moment T is applied to the coupling. If TA and TC represent the twisting moments at the ends A and C , respectively, then

Q.2

A torque of 10 Nm is transmitted through a stepped shaft as shown in figure. The torsional stiffness of individual sections of length MN, NO and OP are 20 Nm/rad, 30 Nm/rad and 60 Nm/rad respectively. The angular deflection between the ends M and P of the shaft is

a) TC = TA c)TC = 16TA

a) 0.5 rad c) 5.0 rad Q.3

Q.4

b) 1.0 rad d)10.0 rad [GATE-2004]

A solid circular shaft of 60 mm diameter transmits a torque of 1600 Nm. The value of maximum shear stress developed is a) 37.72 MPa b) 47.72 MPa c) 57.72 MPa d) 67.72 MPa [GATE-2004] Two shafts AB and BC, of equal length and diameters d and 2d, are made of the same material. They are joined at B through a shaft coupling, while the ends A and C

b) TC = 8TA d) TA = 16TC [GATE-2005]

Q.5

For a circular shaft of diameter d subjected to torque T, the maximum value of the shear stress is 32T 64T a) b) 3 πd 3 πd 8T 16T c) 3 d) πd 3 πd [GATE-2006]

Q.6

A solid shaft of diameter d and length L is fixed at both the ends. A torque, T0 is applied at a distance L/4 from the left end as shown in the figure given below

The maximum shear stress in the shaft is 12To 16To a) b) 3 πd 3 πd 8T 4T c) o3 d) o3 πd πd [GATE-2009]

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Q.7

A torque T is applied at the free end of a stepped rod of circular crosssection as shown in the figure. The shear modulus of material of the rod is G. The expression for d to produce an angular twist θ at the free end is

1

 32TL  4 a)    πθG   16TL  c)    πθG  Q.8

1 4

1 4

 2TL  d)    πθG  [GATE-2011]

Two solid circular shafts of radii R1 and R2 are subjected to same torque. The maximum shear stresses developed in the two shafts are τ1 and τ 2 . If R1/ R2=2, then

τ 2 / τ1 is ___

Q.9

1

 18TL  4 b)    πθG 

a) 11.29 mm c) 33.87 mm

b) 22.58 mm d) 45.16 mm [GATE-2015 (2)] Q.11 A hollow shaft of 1 m length is designed to transmit a power of 30 kW at 700 rpm. The maximum permissible angle of twist in the shaft is 1°. The inner diameter of the shaft is 0.7 times the outer diameter. The modulus of rigidity is 80GPa. The outside diameter (in mm) of the shaft is _____. [GATE-2015(2)] Q.12 The cross sections of two hollow bars made of the same material are concentric circles as shown in the figure. It is given that r3 > 𝑟𝑟1 and 𝑟𝑟4 > 𝑟𝑟2, and that the areas of the crosssections are the same. J1 and J2 are the torsional rigidities of the bars on the left and right, respectively. The ratio J2/J1 is

[GATE-2014 (3)]

Consider a stepped shaft subjected to a twisting moment applied at B as shown in the figure. Assume shear modulus, G=77 GPa. The angle of twist at C (in degrees) is ____

[GATE-2015 (1)] Q.10 A hollow shaft ( d o = 2d i , where d o

& d i are outer and inner diameters respectively) needs to transmit 20 kW power at 3000 rpm. If the maximum permissible shear stress is 30 MPa, d o is _____.

a) > 1 c) =1

b) < 0.5 d) between 0.5 and 1 [GATE-2016 (1)]

Q.13 A machine element XY, fixed at end X, is subjected to an axial load P, transverse load F and a twisting moment T at its free end Y. The most critical point from the strength point of view is

a) a point on the circumference at location Y b) a point at the center at location Y

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c) a point on the circumference at location X d) a point at the center at location X [GATE-2016 (2)] Q.14 A simply supported beam of length 2L is subjected to a moment M at the mid-point x=0 as shown in the figure. The deflection in the domain 0 ≤ x ≤ L is given by −Mx = W ( L − x ) (x + c) 12EIL where E is the Young’s modulus, I is the area moment of inertia and c is a constant (to be determined) .

The slope at the center x = 0 is a) ML/(2EI) b)ML/(3) c) ML/(6EI) d) ML/(12EI) [GATE-2016 (2)] Q.15 The cross-sections of two solid bars made of the same material are shown in the figure. The square cross-section has flexural (bending) rigidity I1, while the circular cross-section has flexural rigidity I2. Both sections have the same cross-sectional area. The ratio I1/I2 is

a) 1/𝜋𝜋 c) 𝜋𝜋/3

b) 2/𝜋𝜋 d) 𝜋𝜋/6 [GATE-2016 (3)]

Q.16 Two circular shafts made of same material, one solid (S) and one hollow (H), have the same length and polar moment of inertia. Both are subjected to same torque. Here,

θS is the twist and τS is the maximum shear stress in the solid shaft, whereas θH is the twist and τH is the maximum shear stress in the hollow shaft. Which one of the following is TRUE?

a) θS θ= = τH H andτS

b) θS > θ H andτS > τ H c) θS < θ H andτS < τ H

d) θS θ H andτS < τ H = [GATE-2016 (3)] Q.17 A rigid horizontal rod of length 2L is fixed to a circular cylinder of radius R as shown in the figure. Vertical forces of magnitude P are applied at the two ends as shown in the figure. The shear modulus for the cylinder is G and the Young’s modulus is E.

The vertical deflection at point A is a) 𝑃𝑃𝑃𝑃3/(𝜋𝜋𝜋𝜋4𝐺𝐺) b) 𝑃𝑃𝑃𝑃3/(𝜋𝜋𝜋𝜋4𝐸𝐸) 3 4 c) 2𝑃𝑃𝑃𝑃 /(𝜋𝜋𝜋𝜋 𝐸𝐸) d) 4𝑃𝑃𝑃𝑃3/(𝜋𝜋𝜋𝜋4𝐺𝐺) [GATE-2016 (2)] Q.18 A motor driving a solid circular steel shaft transmits 40kW of power at 500 rpm. If the diameter of the shaft is 40 mm, the maximum shear stress in the shaft is ________MPa. [GATE-2017 (1)]

Q.19 Consider a beam with circular crosssection of diameter d. The ratio of the second moment of area about the neutral axis to the section modulus of the area is. [GATE-2017 (1)]

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(A)

d 2

(C) d

(B)

πd 2

(D) π d

Q.20 A bar of circular cross section is clamped at ends P and Q as shown in the figure. A torsional moment 𝑇𝑇 = 150 Nm is applied at a distance of 100 mm from end P. The torsional reactions (𝑇𝑇𝑃𝑃 , 𝑇𝑇𝑄𝑄 ) in Nm at the ends P and Q respectively are

(All dimensions are in mm)

(A) (50, 100) (B) (75, 75) (C) (100, 50) (D) (120, 30)

[GATE-2017 (2)]

Q.21 A hollow circular shaft of inner radius 10 mm, outer radius 20 mm and length 1 m is to beused as a torsional spring. If the shear modulus of the material of the shaft is 150 GPa, the torsional stiffness of the shaft (in kN-m/rad) is ________ (correct to two decimal places). [GATE-2018 (2)]

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ANSWER KEY: 1 (c) 15 (c)

2 (b) 16 (d)

3 (a) 17 (d)

4 5 (c) (c) 18 19 60.79 (a)

6 7 (b) (b) 20 21 (c) 35.543

8 8

9 0.23

10 (b)

11 44.52

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12 (a)

13 (c)

14 (c)

EXPLANATIONS Q.1

(c)

16T = 240 πd 3 When shaft diameter is doubled then 16T 16T = τ 'max = 3 π(2d) 8πd 3 τ max 240 = = 8 8 ' τ max = 30MPa τ= max

`

(b) Given : T = 10 N m KMN =20 Nm/rad, KNO=30Nm/rad & KOP=60 Nm/rad We know that, torsional stiffness K T = θ T ∴ Angular deflection θ = K For section MN, NO or OP 10 10 10 = θ MN = rad,θ NO rad = and θ OP 20 30 60 Since M, NO and OP are connected in series combination. So angular deflection between the ends M and P of the shaft is, θ MP = θ MN + θ NO + θ OP 10 10 10 = + + = 1rad 20 30 60

Q.5

Here both the shafts AB and BC are in parallel connection. So, deflection in both the shafts is equal. …..(i) θ AB = θ BC T Gθ We know that = J L From equation (i) TC × L TA × L = π π 4 G × d 4 G × ( 2d ) 32 32 TC TA = 16 ∴ TC = 16TA

Q.6

(b)

Q.2

Q.3

(a) Given : d = 60 mm = T = 1600 Nm = 1600 × 103 Nmm 16T We know that τ max = 3 πd 3 16 ×1600 ×10 τ max = = 37.72 MPa π603

Q.4

(c)

(c)

First, the shaft is divided in two parts (1) and (2). By the nature of these twisting moments, we can say that shafts are in parallel combination. So, T0 = T1 + T2 …. (i) From the torsional equation, T Gθ = J L But, here G1 = G2 θ1 = θ2 For parallel connection J1 = J2 Diameter is same

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So, T1L1 =T2L2 T1 = 3T2 Now From equation (i), TO = 3T2 + T2 = 4T2 T T2 = o 4 3T And T1 = o 4 Here T1   > T2 So, maximum shear stress developed due to T1, Substitute the values, we get 3 To d 12T 4 τ max = × = 3o π 4 2 πd d 32

TL GJ 10 × 0.5 × 32 4.133 ×10−3 rad ⇒ = 9 4 77 ×10 × π × 0.02 ⇒ θ =0.2360 ⇒θ=

is

Q.7

(b) Here we see that shafts are in series combination. For series combination Total angular twist, … (i) θ= θ1 + θ 2 From the torsion equation τ T Gθ = = r J L TL ∴θ = GJ Now from equation (i) TL TL / 2 θ = = π π 4 4 (d) G ( 2d ) G 32 32 On simplifying, we get 1

 18TL  4 d=   πθG 

Q.8

(8)

2T πR 3 3 τ 2  R1  3 =  = = 8  (2) τ1  R 2  (Q T1 = T2 = T) τ max =

Q.9

(0.23) Angle of twist at(C)= Angle of twist at(B)

Q.10 (b) P = Tω

2π× 3000 60 ∴ T = 63.662 N-m T τ = Now J R 63.662 30 ×106 So, = π di (15d i4 ) 32 ∴ di = 11.295mm ∴ d 0 = 2d i = 22.59mm 20 ×103 =T ×

Q.11 (44.52) P = Tω

2π × 700 60 ∴ T = 409.256 N-m T Gθ = J L 409.256 80 ×109 π = × π 1 180 (1 − 0.7 4 )d 04 32 Solving, we get, d 0 = 44.5213mm 30 ×1000 = T×

Q.12 (a) Given A1 = A 2

⇒ π  r22 − r12  = π  r42 − r32  π G ×  r44 − r34  J2 2 Now = π J1 G ×  r 4 − r 4  2 1  2 π 2 2 r − r  r 2 + r 2  J2 2  4 3   4 3  = J1 π  r 2 − r 2   r 2 + r 2  2 1  2 1  2 Q A1 = A 2 & r4 > r2 & r3 > r1

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16T 16 P = . πd 3 πd 3 ω 16T P Q= τmax . × 60 πd 3 2πN 16 × 40 ×103 × 60 = π(0.04)3 × 2π× 500 τmax = 60.79Mpa

J2 >1 J1 Q.13 (c)

Q τmax =

∴

At the center, shear and torsional stresses are zero.

Q.14 (c)

Q.19 (a)

Q sec tion mod ulus =

Q.15 (c) Given: a 2 =

πd 2 4

------- (1)

4 I1 a 12 ------ (2) = π 4 I2 d 64 From equation (1) and (2) I1 π = I2 3

I N.A Ymax

&

d for circular cross sec tion. 2 I d ∴ ratio = N.A .Y= max I N.A 2 Ymax =

Q.20 (c) Angular twist at torque must be same. TL T1 L1 T2 L 2 = = GJ GJ GJ T1 ×100 = T2 ×200

θ=

Q.16 (d)

∴= T1 2 T2 − − − − − (1)

Q.17 (d)

T1 + T2 = T= 150 − − − − − ( 2 ) Solving 2T2 + T2 = 150 ∴ T2 = 50 N .m.

= T P= [ 2L] 2PL

( 2PL ) L TL = GJ G π [2R]4 32 2 4PL = πGR 4 4PL3 y = Rθ = πGR 4

θ =

Q.18 2πN ω= 60 P= T.ω ⇒ T=

∴ T1 = 100 N.m T = (100, 50 )

Q.21 Ans. ( 35.343)

Torsionalstiffness =

GI P L

π 0.044 − 0.024  32 = 1 = 35343 Nm/rad = 35.343 kNm/rad 150  109 

P ω

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GATE QUESTIONS Q.1

Q.2

A weighing machine consists of a 2 kg pan resting on a spring. In this condition, with the pan resting on the spring, the length of the spring is 200 mm. When a mass of 20 kg is placed on the pan, the length of the spring becomes 100 mm. For the spring, the un-deformed length L and the spring constant k (stiffness) are a) L = 220 mm, k = 1862 N/m b) L = 210 mm, k = 1960 N/m c) L = 200 mm, k = 1960 N/m d) L = 200 mm, k = 2156 N/m [GATE-2005]

Q.4

A helical compression spring made of wire of circular cross-section is subjected to a compressive load. The maximum shear stress induced in the cross-section of the wire is 24 MPa. For the same compressive load, if both the wire diameter and the mean coil diameter are doubled, the maximum shear stress (in MPa) induced in the cross-section of the wire is _____.

A compression spring is made of music wire of 2 mm diameter having a shear strength and shear modulus of 800 MPa and 80 GPa respectively. The mean coil diameter is 20 mm, free length is 40 mm a the number of active coils is 10. If the mean coil diameter is reduced to 10 mm, the stiffness of the spring is approximately a) decreased by 8 times b) decreased by 2 times c) Increased by 2 times d) increased by 8 times [GATE-2008]

Q.3

The spring constant of a helical compression spring DOES NOT depend on a) coil diameter b) material strength c) number of active turns d) wire diameter [GATE-2016(1)]

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[GATE 2017 (2)]

ANSWER KEY: 1 (b)

2 (d)

3 (b)

4 6

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EXPLANATIONS Q.1

(b) F = k∆x ∴ 2g= k ( Lo − 0.2 )

…. (i)

When mass of 20 kg is placed on the pan = 22g k ( Lo − 0.1) ….. (ii) Q.2

Q.3 Q.4

Solving equation (i) and (ii), we get L o = 210 mm and k = 1960N / m (d) Gd 4 K= 8D3 N k 2 D13 = k1 D32

k 2 203 = = 8 k1 103 So, stiffness is increased by 8 times. (b) (6) Maximum shear stress induced in spring wired of helical compression spring is given by 8WD τMax = kw πd 3 D τMax α 3 d

τ2 D 2  d1  =   τ1 D1  d 2 

3

D 2 = 2D1 & d 2 = 2d1 3

1 = τ2 2   .τ1 2 1 τ2 = × 24 = 6MPa 4

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6

COMBINED STRESS AND STRAIN

6.1 INTRODUCTION

The principal plane is the plane where the normal stress has maximum and minimum value. The shear stress is zero at maximum normal stress plane. So it is also known as zero shear stress plane or pure normal stress plane. The plane having the maximum value of the normal stress is known as major principal plane and the plane having the 6.2 NORMAL STRESS AND SHEAR STRESS minimum normal stress is known as minor principal stress plane. Let a point is subjected under normal stress Normal stress at any plane is given as σx and σy and the shear stress about the x-y  σx + σy   σx − σy  = σ  +  cos 2θ + τxy sin 2θ n plane is τxy. the normal and shear stress at  2   2  the plane having an angle θ is σn and τ . For the maximum and minimum value of the respectively as shown in the fig. dσ normal stress, n = 0 dθ  σ − σy  0= 0 − 2  x  sin 2θ + 2τxy cos 2θ  2  A point is said to be under the combined stress when it is subjected under normal and shear stress. Example: A rotating shaft is subjected under the bending moment and shear stress. So it is said to subject under the combined load.

Considering the triangle BCP and converting the stress in the term of the forces and then resolving all the forces in the direction of the normal stress and shear stress, we get Normal stress at an angle θ

 σ +σ y   σ x −σ y   +   cos 2θ + τ xy sin 2θ σ n =  x  2   2 

Shear stress at the same plane can be given by

 σ −σ y τ =  x  2

  sin 2θ − τ xy cos 2θ 

6.3 PRINCIPAL PLANES AND PRINCIPAL STRESS PRINCIPAL PLANES

 2τ xy  tan 2θp =  σ − σ  y   x Where, θp is plane of principal stress.

So θp is angle of major principal stress, the minor principal stress plane is θ P2 = θp ± 90

Substituting the value of sin2θ and cos2θ for principal stress plane condition, we get

σ1 = σmax =

σ2 = σmin =

σx + σy 2

σx + σy 2

 σx − σy  2 +   + τ xy  2  2

 σx − σy  2 −   + τxy  2  2

6.4 MAXIMUM SHEAR STRESS PLANE AND MAXIMUM SHEAR STRESS The Maximum Shear stress plane is the plane where the shear stress has maximum value.

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The shear stress equation can be given as  σx − σy  τ   sin 2θ − τxy cos 2θ  2  For the maximum shear stress

dτ =0 dθ

 σx − σy  0   2cos 2θ + 2τ xysin2θ  2   σ x −σ y  tan 2θ s = −   2τ xy    θs2 =θs1 + 90 Maximum shear stress can be given as  σx − σy  2 =+   + τ xy 2   2

τmax

 σx − σy  2 τ 'max = −   + τ xy  2  τmax and τ’max are complementary shear stress. τmax is given by 2

τ max =

σ1 − σ 2 2

6.5 PRINCIPAL STRAIN & SHEAR STRAIN

The maximum and minimum strain at the principal plane can be given as εx + εy  ε x − ε y   γ xy  ε1,2 = ±   +  2  2   2  The maximum shear strain is given as γ max = − ( ε1 − ε 2 ) 2

2

6.6 COMBINED BENDING & TORSION Let a shaft of diameter `d` be subjected to bending moment `M` and a twisting moment `T` at a section. At any point in the section at radius `r` and at a distance y from the neutral axis, the bending stress is given by M σ = .y I and shear stress is given by T τ = .r Ip Where, I = Moment of Inertia about its NA. π I = d 4 and 64 Polar moment of inertia π Ip = d 4 32 σ σ σ1 ,2= ±   + τ2 2 2 16  = σ1 ,2 M ± M2 + T2  3   πd The maximum shear stresses is given by σ − σ2 τmax =1 or 2 16 = τmax M2 + T2 3 πd 2

The normal strain at an angle θ is given by εx + εy  εx − εy  γ xy sin 2θ ε n =+   cos 2θ + 2 2 2   The shear strain at an angle is given by

γ xy  εx − εy  γθ = − cos 2θ  sin 2θ + 2 2  2 

1. Equivalent Bending Equivalent Torque:

Moment

&

Let `Me` be the equivalent bending moment which acting alone produces the maximum tensile stress equal to σ1, as produced by M & T.

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Me Me 16 = = (M + M 2 + T 2 ) 3 3 πd Z πd 32 Therefore 1 Me = M + M2 + T2 2 Let `Te` be the equivalent torque, which acting alone produces the same maximum shearing stress τmax as produced by M & T. Te 16 τmax = = M2 + T2 3 3 πd πd 16 σ= 1

(

= Te

)

M2 + T2

6.7 MOHR’S CIRCLE Mohr’s circle is a graphical representation of the plane stress state at a point. Instead of using the methods of equations, a circle is drawn on the {σ, τ } plane. The normal stress σ(θ) and the shear stress τ(θ) are plotted along the horizontal and vertical axes, respectively, with θ as a parameter. All stress states obtained as the angle θ is varied fall on a circle called Mohr’s circle. This representation was more important for engineers before computers and calculators appeared. But it still retains some appealing features, notably the clear visualization of principal stresses and maximum shear. It also remains important in theories of damage, fracture and plasticity that have a “failure surface”. Procedure for making Mohr’s circle

Let consider the problem of bi axial stress and shear stress is acted on the point. Normal stress in x and y direction isσx and σy and the shear stress at x − y plane is τxy . The normal stress and shear stress at an angle of θ areσn and τ. The procedure for making Mohr’s circle is as follows:

1. Draw a set of coordinate axes with σx1 as positive to the right and τxy as positive downward. 2. Locate point A, representing the stress conditions on the x face of the element by plotting its coordinates σx1 = σx and - τxy. 3. Locate point B, representing the stress conditions on the y face of the element by plotting its coordinates σx2= σy and τxy = τxy. 4. Draw a line from point A to point B, a diameter of the circle passing through point c (center of circle). Points A and B are at opposite ends of the diameter (and therefore 180° apart on the circle). 5. Using point c as the center, draw Mohr’s circle through points A and B. This circle has radius R. The center of the circle c at the point having coordinates σx1 = σavg and τxy = 0. 6. On Mohr’s circle, point A corresponds to θ = 0 represents maximum or minimum value ot normal stress. Thus it’s the reference point from which angles are measured. The angle 2θ locates the point D on the circle, which has coordinates σx1and τxy. D represents the co –ordinates of normal stresses on the x face of the inclined element and shear stress. 7. The maximum shear stress is radius of the Mohr’s circle. This is represented by CE.

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Substituting the value of the radius of the Mohr’s circle τmax

 σx − σy  2 =+   + τ xy  2 

τmax

 σx − σy  2 =−   + τxy 2  

2

2

Case (1) A point is subjected under pure torsion due to which shear stress developed at that point is τ.

Normal stress on the plane of θ is given as σ n = OP = OC + CP =

 σx + σy   σx − σy   +  cos 2θ + τxy sin 2θ  2   2  The shear stress on the plane of the θ is given by  σ − σy  = τ DP =  x  sin 2θ − τxy cos 2θ  2 

Case (2) A point is subjected under equal and same nature of the normal stress and shear stress is zero.

Major Principal stress and minor principal stress

is given by = σ1 σ max = OF = OC + CF = σ1 σ max = OG = OC + CG Substituting the value of OC, CF, CG, we get σ1 = σmax =

σx + σy 2

σx + σy

 σx − σy  2 +   + τ xy  2  2

 σx − σy  2 σ2 = σmin = −   + τ xy 2 2   The maximum shear stress is given by τmax = CE = Radius of the Mohr′ scircle 2

 So shear stress in this condition is zero.  In all the direction, normal stress is same and of same nature. Mohr’s circle in this case will be a point.

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GATE QUESTIONS Data for Q.1 and 2 is given below. The state of stress at a point "P" in a two dimensional loading is such that the Mohr's circle is a point located at 175 MPa on the positive normal stress axis.

Q.1 The maximum and minimum principal stresses respectively from the Mohr's circle are a) +175 MPa, −175 MPa b) +175 MPa, +175 MPa c) 0, −175 MPa d) 0, 0 [GATE-2003] Q.2

The directions of maximum and minimum principal stresses at the point "P " from the Mohr's circle are a) 0, 90° b) 90°, 0 c) 45c, 135° d) all directions [GATE-2003]

Q.3

The figure shows the state of stress at a certain point in a stressed body. The magnitudes of normal stresses in x and y directions are 100 MPa and 20 MPa respectively. The radius of Mohr's stress circle representing this state of stress is

begin if the designer chooses a ductile material whose yield strength is

Q.5

Q.6

a) 45 MPa c) 90 MPa

b) 50 MPa d) 100 MPa [GATE-2005]

A two dimensional fluid element rotates like a rigid body. At a point within the element, the pressure is 1 unit. Radius of the Mohr's circle, characterizing the state of stress at that point, is a) 0.5 unit b) 0 unit c) 1 unit d) 2 unit [GATE-2008] The state of stress at a point under plane stress condition is σ xx =40 MPa, σ yy = 100 MPa and τxy = 40 MPa. The radius of the Mohr's circle representing the given state of stress in MPa is a) 40 b) 50 c) 60 d)100 [GATE-2012]

Q.7 a) 120 MPa c) 60 MPa Q.4

b) 80 MPa d) 40 MPa [GATE-2004]

The Mohr's circle of plane stress for a point in a body is shown. The design is to be done on the basis of the maximum shear stress theory for yielding. Then, yielding will just

In a plane stress condition, the components of stress at a point are = σ x 20MPa;σ = 80MPa; = τ xy 40MPa y The maximum shear stress (in MPa) at the point is ____ a) 20 b)25 c) 50 d) 100 [GATE-2015(2)]

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ANSWER KEY: 1 (b)

2 (d)

3 (c)

4 (c)

5 (b)

6 (b)

7 (c)

EXPLANATIONS Q.1

Q.2

Q.3

(b)

Given, Mohr’s circle is a point located at 175 MPa on the positive Normal stress (at point P) So, σ= σ= 175MPa 1 2 So, both maximum and minimum principal stresses are equal.

(d) Mohr’s circle is a point. So, maximum and minimum principal stresses at point P are same in all directions.

  100 – (−20) 2  Radius   = =  + 0  60MPa 2       Q.4

(c) As per maximum shear stress theory, σ x − σ y σ yp = τ max = ( σ yp =yield point) 2 2 −100 − (−10) σ yp = = 2 2 ∴σ    yp = −90 MPa

Q.5

(b) Since the fluid will be subjected to hydrostatic loading therefore Mohr circle reduce to a point on σ- axis. Therefore, radius of Mohr circle is zero.

Q.6

(b)

(c)

σ x = 100MPa (Tensile) σ y = 20MPa (Compressive),

We know that,   σ − σ 2  y 2  Radius =   x + τ  XY   2    

Radius of Mohr’s circle

= OR

( AR ) + ( AO ) 2

2

AB BN − AN 100 − 40 = = = 30 2 2 2 Therefore, AO =

OR =

( 40 ) + ( 30 ) 2

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2

= 50MPa

Q.7

(c)

 σx − σy  2   + τ xy  2  2

= τ max

 20 − 80  2 50MPa   + 40 =  2  2

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7

COLUMN AND PRESSURE VESSEL

7.1 INTRODUCTION A member is said to be a strut if it is subjected under axial compressive load. The strut may be vertical, horizontal and inclined. The examples of the struts are piston rods, connecting rods etc. A member is said to column if it is subjected under axial compressive load in vertical direction.  All columns are the struts but all the struts are not columns. Buckling Load A maximum limiting load at which the column tends to have lateral displacement or tends to buckle is called as buckling or crippling load. Buckling of the column takes place about the axis having least moment of inertia. 7.2 TYPE OF COLUMN

There are three types of columns: a) Short Column The column is said to be short column if slenderness ratio of the column is less than 30. When short column are subjected under axially compressive load, the buckling produced in the column is negligible. This type of column mainly fails under compressive stress. So the short columns are stronger in buckling but weaker in crushing. b) Long Column The column is said to be long column if slenderness ratio of the column is higher than 120. When long columns are subjected under axially compressive load, the crushing produced in the column is negligible. This type of column mainly fails under buckling of the column. So the short

columns are stronger in crushing but weaker in buckling. c) Medium Column The column is said to be medium column if slenderness ratio of the column lies between 30 and 120. The medium column fails under the combined effect of crushing and buckling of the column. 7.3 SLENDERNESS RATIO

Slenderness ratio is defined as the ratio of the effective length to the minimum radius of gyration of the cross section of the member. It is represented by S and it has no unit. If the effective length of the column is Le and minimum radius of gyration is k, then the slenderness ratio L S= e k I k = min A Imin = minimum moment of inertia A = Cross section area of column 7.4 LOAD ANALYSIS OF COLUMN

1. LOAD ANALYSIS OF LONG COLUMN Euler theory is valid for the long column which is based on following assumptions i) Column fails only due to buckling. ii) The material of the column is perfect homogeneous and isotropic iii) The weight of the column is neglected. iv) The Column deformation is up to elastic limit. Critical load for a column is given by

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π 2 EI min nπ 2 EI min = le 2 l2 Where PEuler = Critical load E= modulus of elasticity I min = Minimum moment of inertia of cross section of the column le = Effective length of the column n = End fixity constant l = Actual lengthof the column

PRankine =

= PEuler

End Condition

Both end are fixed

One end is fixed and one end is hinged Both ends are hinged One end is fixed and one end is free

Effective Length Le L 2 L √2 L

2L

End fixity constant n 4 2 1

0.25

The short column is subjected under crushing load Pc and yield strength of the component of the column is σyt, the critical load for the short column can be given as Pc = σyt. A 3. LOAD ANALYSIS OF MEDIUM COLUMN

Rankine’s theory of column is applicable for short, medium and long column. This theory is valid for all the dimension of the columns and struts. The critical load for medium column can be given as 1 1 1 = + PRankine PC PEuler PEuler =

in

1 + b ( S)

2

σc = Rankine's Constant π2E The Critical load of medium column from Rankin’s formula is inversely proportional to slenderness ratio of the cross section of the column. Where b =

2. LOAD ANALYSIS OF SHORT COLUMN

π 2 EI min , Pc = σ yt .A le 2 Substituting the values formula, we get σ c .A PRankine = 2 σ l  1 + 2c  e  π E k 

σ c .A

Rankine’s

7.5 PRESSURE VESSEL

The pressure vessel is the closed container used to store the liquid or gas at the pressure other than the pressure of the atmosphere. Generally the pressure vessels are made of cast iron, steels and alloys. There are two types of pressure vessels: a) On the basis of the shape i) Cylindrical Pressure Vessel ii) Spherical Pressure Vessel b) On the basis of d/t ratio i) Thin pressure vessel ii) Thick pressure vessel

1. THIN PRESSURE VESSEL A pressure vessel is called as the thin d pressure vessel if the ≥ 20. Where d is the t inner diameter of the pressure vessel and t is the thickness of the pressure vessel. 2. THIN CYLINDRICAL PRESSURE VESSEL

Let consider a thick cylindrical pressure vessel is subjected under uniform pressure P. The stress distribution in thin cylinder is uniform because the thickness of the material is very less as compare to diameter of the cylinder. Three type of normal stress develops in the cylinder. (I) STRESSES

(a) Hoop of Circumferential Stress

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It acts in the tangential direction of the circumference of the cylinder. It is represented by σh.

(b) Longitudinal Stress It acts in the direction of the longitudinal axis of the cylinder. It is represented by σl. (c) Radial Stress It acts in the radial direction of the cylinder. It is represented by σr. The value the radial stress is very less so it can be neglected for the thin cylinder. The radial stress is compressive stress in nature.

(a) Hoop of Circumferential Stress Let a thin cylinder of inner diameter d and thickness t is subjected under the pressure P.

The pressure force on the material of the cylinder in circumferential direction is given as FL = Pdl Where, l = length of the cylinder Resisting force developed in the material of the cylinder is given by FR = 2σ h lt By equating the pressure force and the resisting force, we get Pd σh = 2t (b) Longitudinal Stress

The pressure force on the material of the cylinder in circumferential direction is given as π FL  = P × d 2 4 Where l = length of the cylinder Resisting force developed in the material of the cylinder is given by FR = σ l πdt By equating the pressure force and the resisting force, we get Pd σl = 4t (c) Radial Stress The radial stress in thin cylinder can be given as σ r = −P

Note: 1. Maximum & Minimum Principal Stress Pd Major Principal stress σ= σ= 1 h 2t Minor Principal Stress σ= σ=l 2

Pd 4t

2. Maximum Shear Stress Maximum shear stress under Pressure P can be given as σ1 − σ 2 σ h − σ l Pd = τ max = = 2 2 8t (II) Strains in Thin Cylinder Due to pressure, longitudinal strain and hoop strain produces and it increases the length and other dimension of the cylinder and volume of the cylinder. (a) Circumferential Strain

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ε= h

δd σ h − μσ l Pd = = (2 − μ) d E 4tE

(b) Longitudinal Strain δl σ − μσ h Pd ε=l = l = (1 − 2μ ) l E 4tE (c) Volumetric Strain

δV δl δd = +2 V l d ε v = ε v + 2ε c ε= v

= εv

Pd ( 5 − 4μ ) 4tE

3. THIN SPHERICAL PRESSURE VESSEL a) Hoop of Circumferential Stress Let a thin sphere of inner diameter d and thickness t is subjected under the pressure P.

b) Longitudinal Stress The pressure force on the material of the sphere in circumferential direction is given π as FL  = P × d 2 4 Where l = length of the cylinder Resisting force developed in the material of the sphere is given by FR = σ l πdt By equating the pressure force and the resisting force, we get Pd σl = 4t c) Radial Stress The radial stress in thin sphere can be given as σ r = −P Maximum & Minimum Principal Stress Pd = σ= Major Principal stress σ1,2 h 4t

Maximum Shear Stress Maximum shear stress under Pressure P can be given as σ1 − σ 2 σh − σl = τ max = = 0 2 2

The pressure force on the material of the sphere in circumferential direction is given as π FL  = P × d 2 4 Resisting force developed in the material of the Sphere is given by FR = σ h πdt By equating the pressure force and the resisting force, we get Pd σh = 4t

Strain in Thin Sphere Due to pressure, longitudinal strain and hoop strain produces and it increases the length and other dimension of the cylinder and volume of the cylinder. Circumferential Strain δd σ h − μσ l Pd ε= = = (1 − μ ) h d E 4tE Volumetric Strain δV δd 3Pd = εh = 3 = (1 − μ ) V d 4tE 4. THICK PRESSURE VESSEL

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A pressure vessel is called as the thick d pressure vessel if the < 20 . Where d is the t inner diameter of the pressure vessel and t is the thickness of the pressure vessel.Examples of thick pressure vessel are gun barrels, container under high pressure.

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GATE QUESTIONS Q.1

Q.2

A pin-ended column of length L, modulus of elasticity E and second moment of the cross-sectional area I is loaded eccentrically by a compressive load P. The critical buckling load (Pcr) is given by π 2 EI EI a) Pcr = 2 2 b) Pcr = 3L2 πL π 2 EI πEI c) Pcr = 2 d) Pcr = 2 L L [GATE-2006]

Q.4

Q.5

Q.6

The rod PQ of length L and with flexural rigidity EI is hinged at both ends. For what minimum force F is it expected to buckle?

π2E I a) L2 π2E I c) 2L2 Q.3

a) 1 c) 4

2π 2 E I b) L2 π2E I d) 2L2 [GATE-2008]

A column has a rectangular crosssection of 10×20 mm and a length of 1m. The slenderness ratio of the column is close to a) 200 b) 346 c) 477 d) 1000 [GATE-2011] For a long slender column of uniform cross section, the ratio of critical buckling load for the case with both ends clamped to the case with both the ends hinged is

b) 2 d) 8 [GATE-2012]

Consider a steel (Young’s modulus E = 200 GPa) column hinged on both sides. Its height is 1.0 m and cross – section is 10mm×20mm. The lowest Euler critical buckling load (in N) is _____. [GATE-2015] An initially stress-free massless elastic beam of length L and circular cross-section with diameter d (d << L) is held fixed between two walls as shown. The beam material has Young’s modulus E and coefficient of thermal expansion α .

If the beam is slowly and uniformly heated, the temperature rise required to cause the beam to buckle is proportional to [GATE 2017 (1)] (A) d Q.7

(C) d3

(B) d2

(D) d4

A steel column of rectangular section (15 mm × 10 mm) and length 1.5 m is simply supported at both ends. Assuming modulus of elasticity, E = 200 GPa for steel, the critical axial load (in kN) is ____ (correct to two decimal places). [GATE-2018(1)]

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ANSWER KEY: 1 (d)

2 (b)

3 (b)

4 (c)

5 3290

6 (b)

7 1.079

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EXPLANATIONS Q.1

(d)

According to Euler’s theory, the crippling or buckling load (Pcr) under various end conditions is represented by a general equation, π 2 EI min Pcr = L2e Where E = Modulus of elasticity I=Minimum mass-moment of inertia Le = Effective length of column For both end pinned Le = L

Q.4

π 2 EI min ∴ Pcr = 2 g L

Q.2

(b) For both ends hinged, π 2 EI P= 2 L π 2 EI F cos 45° = 2 (∵ F cos 45°=P) L 2 F π EI ⇒ = 2 L 2

2π 2 EI ∴F = 2 L Q.3

(b) Given : l =1 meter, b = 20 mm, h = 10 mm

Q.5

We know that, Slenderness ratio l = k I Where k = min A Where, Substitute the value, we get k 2.88 ×10−3 = 1 Slenderness ration = = 2.88 ×10−3 347.22

(c) According to Euler’s theory, the crippling or buckling load (PEuler ) under various end conditions is given by, π 2 EI min PEuler = L2e L For (i). Both ends fixed L e = 2 (ii).Both ends hinged  L e = L ∴ Ratio is 4 (3285 to 3295)

Q.6 (b) Fth = Reaction offered by support or Buckling load. π2 EI α∆TE A = 4 2 l π π2 E × D 4 π 2 64 α∆TE D = 4 4 l2 ∆T ∝ D 2 Q.7

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Ans. (1.097 ) π2 EI min L2 15103 π2  200 103  12 = 15002 = 1096.62 N =1.097 kN Buckling load =

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GATE QUESTIONS Q.1

Q.2

Q.3

A thin cylinder of inner radius 500 mm and thickness 10 mm is subjected to an internal pressure of 5 MPa. The average circumferential (hoop) stress in MPa is a) 100 b) 250 c) 500 d) 1000 [GATE-2011] A thin walled spherical shell is subjected to an internal pressure. If the radius of the shell is increased by 1% and the thickness is reduced by 1%, with the internal pressure remaining the same, the percentage change in the circumferential (hoop) stress is a) 0 b) 1 c) 1.08 d) 2.02 [GATE-2012]

A long thin walled cylindrical shell, closed at both the ends, is subjected to an internal pressure. The ratio of the hoop stress (circumferential stress) to longitudinal stress developed in the shell is a) 0.5 b) 1.0 c) 2.0 d) 4.0 [GATE-2013]

radius of the tank is 2 m, and it has wall thickness of 10 mm. The magnitude of maximum in-plane shear stress (in MPa) is ____ [GATE-2015 (3)]

Q.6

A thin cylindrical pressure vessel with closed-ends is subjected to internal pressure. The ratio of circumferential (hoop) stress to the longitudinal stress is a) 0.25 b) 0.50 c) 1.0 d) 2.0 [GATE-2016 (2)]

Q.7 A thin-walled cylindrical can with rigid end caps has a mean radius 100R mm and a wall thickness of 5 t mm. The can is pressurized and an additional tensile stress of50 MPa is imposed along the axial direction as shown in the figure. Assume that the state of stress in the wall is uniform along its length. If the magnitudes of axial and circumferential components of stress in the can are equal, the pressure (in MPa) inside the can is ___________ (correct to two decimal places).

Q.4 A thin gas cylinder with an internal radius of 100 mm is subject to an internal pressure of 10 MPa. The maximum permissible working stress is restricted to 100 MPa. The minimum cylinder wall thickness (in mm) for safe design must be___ [GATE-2014 (4)] Q.5

A cylindrical tank with closed ends is filled with compressed air at a pressure of 500 kPa. The inner

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[GATE 2018 (2)]

[GATE-2018 (2)]

ANSWER KEY: 1 (b)

2 (d)

3 10

4 (c)

5 25

6 (d)

7 5

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EXPLANATIONS Q.1

(b) Given : r = 500mm, t = 10 mm, P = 5MPa We know that average circumferential (hoop) stress is given by, Pr 5 × 500 σ= = 250 MPa c = t 10

Q.2

(d)

pd 4t p= internal pressure d=diameter of shell t=thickness of shell pd ' Now σ ' = ' 4t d’=1.01d t’=0.99t p ×1.01d = ∴σ ' = 1.0202σ 4 × 0.99t σ’ − σ % change = ×100 σ (1.0202 − 1)σ = = ×100 2.02% σ

Hoop stress =

Q.3

(10) pd σ working =   ( d = 2r = 200m ) 2t pd 10 × 200 = t = 2σ w 2 × 100 = 10 mm

Q.4

Q.5

Q.6

(c) Hoop streets of circumferential stress is Pr σ1 = t and longitudinal or axial stress is Pr σ2 = 2t σ Pr 2t Ratio = 1 = × = 2 σ2 t pr (25) Maximum in-plane shear stress pd 500 × 4 τ max = = = 25MPa 8t 8 × 10 (d)

Pr σc t 2 = = P σl r 2t Q.7

Sol:

Circumferential stress, n = Axialstress,  = L Now,

PR t

PR + 50 MPa 2t

n =L

PR PR = + 50 MPa t 2t PR = 50 MPa 2t 50  2  5 P= = 5 MPa 100

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8

SLOPE AND DEFLECTION

8.1 INTRODUCTION For modern design there are two governing criteria i) Strength ii) Stiffness As per the strength criteria, the beam should be strong enough to resist bending moment and shear force or in other words beam should be strong enough to resist the bending stresses and shear stresses. And as per the stiffness criterion of the beam, the beam should be stiff enough not to deflect more than the permissible limit. It means serviceability condition must be satisfied. The deflection may be linear or angular deflection

8.3 METHODS FOR SLOPE & DEFLECTION

8.2 General expression

d2 y = Mx dx 2 we may easily find out the following relations.

From the equation EI

d4 y = − ω ⇒ Shear force density (Load) dx 4 d3 y EI = Vx ⇒ Shear force dx 3 d2 y EI = M x ⇒ Bending moment dx 2 dy = θ⇒ slope dx y = δ Defliction, Displacement Flexural rigidity = EI

• EI • • • • •

1. DOUBLE INTEGRATION METHOD: This method is suitable for simple loading in simply supported and cantilever beams with uniformly distributed loads and triangular loadings. As per the relation of the deflection an bending moment, equation can be given as d2 y EI 2 = M dx M  = +ive if sagging   = −ive if hogging    After integrating the equation, we get dy EI = ∫Mdx + C1 dx

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By applying the boundary condition we can find the value of C1. = EIθ ∫Mdx + C1 θ = Slop of the beam at the given point. Again integrating the equation of slope , we get EIy = ∫∫Mdx + C1x + C2

y = deflection of the beam at the given section EI = Flexural Rigidity of the beam M = Bending moment at the given section

2. MACAULAY`S METHOD:

This is the convenient method for the beams subjected to point loadings or in general discontinuous loads or beams subjected to couples (concentrated moments). This method is similar to the double integration method but specialty of this method lies in the manner in which the bending moment at any section is expressed and in the manner in which integration is carried out. In Macaulay’s method a single equation is formed for all the loadings on a beam the equation is constructed in such a way that constants of the integration apply to all portions of the beam. 3. MOMENT AREA METHOD: (MOHR`S THEOREM)

This method is suitable for cantilevers and simply supported beams carrying symmetrical loadings and beams fixed at both ends i.e. those beams for which the area and Centre of gravity of area of B.M.D. can be found easily. It means this method is not suitable for triangular loading and irregular loading. This can also be used for non prismatic bars. This method can be used easily for i) Cantilever beams because slope at the fixed end is zero) ii) Simply supported beams carrying symmetrical loading because slope at mid span is zero.

iii) Beams fixed at both ends because the slope at each end is zero. (a) Slop of beam As per Mohr’s theorem, total change in slope between B and A equals the area of B.M. diagram between B and A divided by the flexural rigidity EI. 1 θB − θA = (Area of B.M.D between A & B) EI  In case of point A is not of zero slope and zero deflection, then the deflection of B with respect to the tangent at A equals the first moment about (b) Deflection of beam

As per Mohr’s theorem, total change in deflection between B and A equals the moment of area of B.M. diagram between B and A divided by the flexural rigidity EI. 1 y B − y A =(Moment.of Area of EI B.M.D between A & B) 4. CONJUGATE BEAM METHOD: MOHR`S THEORIES This method is convenient if flexural rigidity of the beam is not uniform throughout the length of the beam. This method is suitable for beams carrying internal hinge. An Imaginary beam for which the load diagram M is diagram, of the given beam is called EI the conjugate beam. If the slope of point A is 1 δB = (Moment.of Area of B.M.D zero, EI between A & B about B) If the slope of point A is not zero, then slope of point B is given as δB = δ1 + δ2

= θA .Iab +

1 (Moment.of Area of B.M.D EI between A & B about B)

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In order to calculate the deflection under the point of application of the load P, the ∂U deflection can be given as y = ∂P In order to calculate the slope under the point of application of the moment M, the ∂U Slope can be given as θ = ∂M 6. CASTIGLIANO`S THEOREM

Let U is total strain energy of any structural member due to application of loads P1, P2, P3 at the points A1, A2,A3 and M1, M2, M3 at the points B1, B2,B3. Then differentiate of the strain energy with respect to any load gives the deflection at that point and with respect to Moment gives the slope at that point. ∂U ∂U ∂U = y= y= y A3 A1 , A2 ∂P1 ∂P2 ∂P3 ∂U ∂U ∂U = θ= θ= θ B3 B1 , B2 , ∂M1 ∂M 2 ∂M 3 Points to be remembers:  Treat all the load as a variables and carry out partial differential  To find the deflection or slope of a point here there is no load or bending moment given assume a dummy load P or dummy moment respectively at that point and give a value zero at the end.  ∂U   ∂U  = θ A,   = y A, and    ∂P  P =0  ∂M M =0 5. STRAIN ENERGY METHOD: It is Suitable for cantilevers and beams having varying EI or varying depth of beam. This method is very useful in case of determinate frames and arches. This can also be used when internal hinge is provided. Let U is total strain energy of the structure M 2 dx given by U = ∫ x 2EI

8.4 SPECIAL DEFLECTION

CASE

OF

SLOPE

AND

Case (A) Cantilever Beam 1) A cantilever of length l is subjected to a couple M at its free end. Find the slope and deflection of the free end.

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Ml Slope at the end B, θB = EI

Ml2 2EI 2) A cantilever of length `l` carrying point load `W` at the free end Deflection at the free end δ =

wl3 The slope at B, θB = 6EI Wl2 Slope at the free end, θB = 2EI

Wl3 Deflection of B, δb = 3EI 3) Cantilever of length `l` carrying a concent rated load W at a distance a from the fixed end.

W.a 2 Slope at B = slope at C θB =θC = 2EI Wa 3 Deflection at B, δb = 3EI Deflection at C, = δc

Wa 3 Wa 2 + (l − a) 3EI 2EI

4) Cantilever of length `l` carrying a uniformly distributed load w per unit length over the whole span

The deflection at B,

wl4 δB = 6EI

5) Cantilever of length `l` carrying a udl of w per unit run over a distance `a` from the fixed end.

The slope at B = the slope at C w.a 3 θB =θC = 6EI

w.a 4 Deflection at B, δb = 8EI

w.a 4 wa 3 Deflection at C, δ C = + (l − a ) 8 EI 6 EI 6) Cantilever carrying moment `Mo` at a distance a from fixed end

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M .a Slope at free end, θb = o EI Deflection at free end, M o .a = δb (2L − a) 2EI 7) Cantilever of length ‘l’ carrying a distributed load whose intensity varies uniformly from zero at the fixed end to w per unit run at the free end.

Slope at the end A = slope at the end B wl3 θA =θB = 24EI Maximum deflection at the center of 5 wl4 beam δc = 384 EI

3) A simply supported beam of span `l` carries a UDL for a distance `l ` from left support A.

wl3 Slope at B, θb = 8EI Deflection at the end δ b =

11 wl 4 120 EI

2

The deflection at the center is given by 1 5 wl4 δC = × × 2 384 EI

Case (B) Simply supported beam 1) Simply supported beam of span l carrying a point load W at mid span.

wl2 Slope at A = Slope at B θA =θB = 16EI Deflection at the center of span C Wl3 δC = 48EI 2) Simply supported beam AB of span `l` carrying a uniformly distributed load of w per unit length over the whole span.

4) A beam of length `l` simply supported at the ends carries a point load W at a distance `a` from the left end. Find the deflection under the load the maximum deflection.

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M (2l2 − 6al + 3a 2 ) 6EI l M = θB (3a 2 − l2 ) EI l The deflection at C below the moment applied is given by. M.a (1 − a)(1 − 2a) δC = 3EI l 7) Simply supported beam, carrying two concentrated loads at a distance `a` from each support. = θA

The deflection below the load (C) is Wa 2 b 2 δc = 3EI l The maximum deflection occurs at x

l2 − b 2 from left end, and is given by 3 W.b.(a 2 + 2ab)3/2 δmax = 9 3 EI l It has been found that maximum deflection always occurs within the l l  range of position  ±   2 13  5) A beam of length L is supported on two supports allowing equal over hangs. The distance between the support being `l`. The beam carries a point load W at its mid span. =

The ratio between the l/L, if the down ward defection at the center is equal to the upward deflection at the ends is 1 given by. = 0.6 L 6) A been of length `l` with supports at the ends is subjected to a couple M at a distance a from the left end.

W.a(L − a) Slope at each end θa =θb = 2EI Deflection at the center W.a(3L2 − 4a 2 ) δc =δmax = 24EI 8) A simply supported beam carries a uniformly distributed load of varying intensity 0 to w from left end to right end B. 7 wl3 Slope at A = 360 EI wl3 Slope at B = 45EI 5wl4 Deflection at the center, δc = 768 EI Maximum deflection wl4 δ max = 0.00652 EI

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9) Simply supported beam, carrying triangular loading, of maximum intensity w at the center.

The maximum deflection occurs at 0.422l from the prop end B. wl4 δmax = 0.005415 EI Case (D) A quadrant ring AB shown in figure is of radius r. It supports a concentrated load P at the free end.

5 wl3 Slope at A = slope at B θA =θB = 192 EI Maximum deflection at the center wl4 δc =δmax = 120EI Case (C) Propped Beam 1) A propped cantilever of length `l` carrying a ud1 of w per unit run. If the propped beam carries UDL on the entire span of beam.

P.πr 3 Vertical deflection, δ v = 4EI P.r 3 Horizontal deflection, δh = 2EI

3 i) The reaction at B, R b = wl 8 5 ii) The reaction at A, R a = wl 8 © Copyright Reserved by Gateflix.in No part of this material should be copied or reproduced without permission

ASSIGNMENT QUESTIONS Q.1

Q.2

Q.3

A material has identical properties in all directions, it is said to be a) homogeneous b) isotropic c) elastic d) orthotropic

The energy absorbed in a body, when it is strained within the elastic limits, is known as a) strain energy b) resilience c) proof resilience d) modulus of resilience

Q.5)

Q.6)

Toughness for mild steel under uniaxial tensile loading is given by the shaded portion of the stressstrain diagram as shown in Q.7)

a) Kinematic viscosity b) Surface tension c) Bulk modulus d) Strain

Hooke's law holds good up to a) yield point b) limit of proportionality c) breaking point d) elastic limit

The linear relation between the stress and strain of a material is valid until a) fracture stress b) elastic limit c) ultimate stress d) proportional limit

Match List-I (Materials) with List-II (Stress Strain curves) and select the correct Answer using the codes given below the lists: List-I List-II A. Mild steel 1.

Q.4

B. Pure copper

2.

C. Cast iron

3.

D. Pure aluminum

4.

Which of the following has no unit?

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Q.8

Q.9

Codes: A a) 3 b) 3 c) 2 d) 4

B 2 1 4 1

C 4 4 3 3

D 1 2 1 2

Match List-1 (Types of tests and Materials) with List-11 (Types of Fractures) and Select the correct answer using the codes given below the lists: List-1 A. Tensile test on Cl B. Torsion test on MS C. Tensile test on MS D. Torsion test on Cl List-11 1. Plain fracture on a transverse plane 2. Granular helicoidal fracture 3. Plain granular at 45° to the axis 4. Cup and cone 5. Granular fracture on a transverse plane Codes: A B C D a) 4 2 3 1 b) 5 1 4 2 c) 4 1 3 2 d) 5 2 4 1

Poisson's ratio is defined as the ratio of a) longitudinal stress and longitudinal strain b) longitudinal stress and lateral stress c) lateral strain and longitudinal strain d) lateral stress and lateral strain

Q.10 Which one of the following properties cannot be evaluated by static tension test? a) Shear strength b) Modulus of elasticity c) Ductility d) Poisson's ratio

Q.11 The percentage elongation of a material as obtained from static tension test depends upon the a) diameter of the test specimen b) gauge length of the specimen c) nature of end-grips of the testing machine d) geometry of the test specimen Q.12 Modulus of resilience is a) a property to resist shocks b) a property to withstand heavy pressure c) a property to store energy without undergoing permanent deformation d) an index of elasticity

Q.13 What is the relationship between the linear elastic properties Young's modulus (E), rigidity modulus (G). and bulk modulus (K)? 1 9 3 3 9 1 a) = b) = + + E K G E K G 9 3 1 9 1 3 c) = d) = + + E K G E K G

Q.14 Which one of the following is the correct expression for maximum shear stress induced in the wire of a closed-coiled helical spring of wire diameter d and mean coil radius R which carries an axial load W ? 16WR W a) b) 3 πd  πd 2     4  (W / 2) 16WR c) d) 2 πd / 4 πd Q.15 The ratio of the area under the bending moment diagram to the flexural rigidity between any two points along a beam gives the change in a) deflection b) slope c) shear force d) bending moment

Q.16 The shear force diagram is shown above for a loaded beam. The

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corresponding bending moment diagram is represented by a) b) c) d) Q.17 A beam carrying a uniformly distributed load rests on two supports 'b' apart with equal overhangs 'a' at each end. The ratio b/a for zero bending moment at mid-span is 1 a) b) 1 2 2 c) d) 2 3 Q.18 For a simply supported beam of length 'L', subjected to downward load of uniform intensity w, Match List-I with List-II and select the correct answer using the codes given below the lists : List-I A. Slope of shear force diagram B. Maximum shear force C. Maximum deflection D. Magnitude of maximum bending moment List-II 5wL4 1. 384EI

2. w wL4 3. 8 wL 4. 2 Codes: A a) 1 b) 3 c) 3 d) 2

B 2 1 2 4

C 3 2 1 1

D 4 4 4 3

Q.19 The distribution of shear stress of a beam is shown in the given figure.

The cross-section of the beam is a) I b) T b) d)

Q.20 Select the correct shear stress distribution diagram for a square beam with a diagonal in a vertical position. a) b)

c)

d)

Q.21 A structural member subjected to an axial compressive force is called a) beam b) column c) frame d) strut

Q.22 Euler's formula holds good only for a) short columns b) long columns c) both short and long column d) weak columns

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Q.23 What causes a short steel strut to fail? a) Fracture b) Shearing c) Buckling d) Yielding

Q.24 A column of rectangular section (Ixx>Iyy) is subjected to an axial load. What is the axis about which the column will have a tendency to buckle? a) x-x axis b) y-y axis c) The diagonal of the section d) x-x or y-y axis without any preference Q.25 Given that PE = the crippling load given by Euler Pc = the load at failure due to direct compression PR = the load in accordance with the Rankine's criterion of failure Then PR is given by (P + P ) a) E C b) PE × PC 2 PP c) E E d)None of these PC + PE

Q.26 Principal stresses on the outside surface element of a thin cylindrical shell subjected to internal fluid pressure as shown in the figure, are represented by a)

b)

c)

d)

Q.27 The stiffness of the beam shown in the figure below is = ( I 375 ×10−6 m 4 ; L=0.5 m and E=200 GPa) a) 12 ×108 N / m c) 4 ×108 N / m

b) 10 ×108 N / m d) 8 ×108 N / m

Q.28) A mild steel specimen is tested in tension up to fracture in a Universal Testing machine. Which of the following mechanical properties of the material can be evaluated from such a test? 1. Modulus of elasticity 2. Yield stress 3. Ductility 4. Tensile strength 5. Hardness 6. Modulus of rigidity Select the correct answer using the code given below: a) 1, 3, 5 and 6 b) 2, 3, 4 and 6 c) 1, 2, 5 and 6 d) 1, 2, 3 and 4

Q.29 A steel rod of length 300 mm is held between two fixed supports so that the rod cannot elongate or contract in the axial direction. If the temperature of the rod is raised by 20°C, the axial stress induced in the rod due to this rise in temperature is (Take E=200GPa,= α 11.5 ×10−6 / 0 C ) a) 46 MPa (Tension) b) 46 MPa (Compression) c) 23 MPa (Tension) d) 23 MPa (Compression) Q.30 A steel cube of 1m ×1m ×1m is subjected to hydrostatic pressure of 1 MPa. If for steel, Young's modulus is 200 GPa, and Poisson's ratio is 0.3, the approximate change in volume of the cube due to the hydrostatic pressure will be a) −6 ×10−6 m3 b) −2 ×10−6 m3 c) −1×10−3 m3 d) −0.5 ×10−6 m3

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Q.31 A bar of 20 mm diameter is tested in tension. It is observed that when a load of 30 kN is applied the extension measured over a gauge length of 200 mm is 0.12 mm and contraction in diameter is 0.0036 mm. The Poisson's ratio is a) 0.2 b) 0.3 c) 0.25 d) 0.33 Q.32 The bending moment for a loaded beam is shown below:

The loading on the beam is represented by which one of the following diagram? a) b) c) d) Q.33

Shear force and bending moment diagrams for a beam ABCD are shown in figure. It can be concluded that

a) b) c) d)

The beam has three supports End A is fixed A couple 2000 Nm acts at C A uniformly distributed load is confined to portion BC only

Q.34 For a specific application Young's modulus of a material is 125 GPa and Poisson's ratio is 0.25, the modulus of rigidity is a) 50 GPa b) 60 GPa c) 80 GPa d) 90 GPa Q.35

Match List-I (Property) with List-II (Testing Machine) and select the correct answer using the codes given below the lists: List-I A. Tensile strength B. Impact strength C. Bending strength D. Fatigue strength List-II 1. Rotating bending machine 2. Three-point loading machine 3. Universal testing machine 4. lzod testing machine Codes: A B C D a) 4 3 2 1 b) 3 2 1 4 c) 2 1 4 3 d) 3 4 2 1

Q.36 A uniform bar lying in the xdirection is subjected to pure bending. Which one of the following tensors represents the strain variation when bending moment is about the z-axis (p, q and r constants)?  py 0 0   py 0 0    a)  0 q y 0  b)  0 q y 0   0 0 0 0 0 r  y   

 py  c)  0 0 

0 py 0

0  0 p y 

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 py  d)  0 0 

0 qy 0

0  0 q y 

Q.37 Assuming E = 160 GPa and G = 100 GPa for a material a strain tensor is given as  0.002 0.004 0.006    0   0.004 0.003  0.006 0 0   The shear stress, τ xy is a) 400 MPa c) 8OO MPa

b) 500 MPa d) 1000 MPa

Q.38 The principal stresses at a point in an elastic material are 60 N/mm2 (tensile), 20 N/mm2 (tensile) and 50 N/mm2 (compressive). If the material properties are: μ= 0.35 and E = 105 N/mm2 then the volumetric strain of the material is a) 9 × 10-5 b) 3 × 10-4 -5 c) 10.5 × 10 d) 21 × 10-5 Q.39 Consider the following statements: 1. There are only two independent elastic constants 2. Elastic constants are different in orthogonal directions 3. Material properties are same everywhere 4. Elastic constants are same in all loading directions 5. The material has ability to withstand shock loading Which of the above statements are true for a linearly elastic, homogeneous and isotropic material? a) 1, 3, 4 and 5 b) 2, 3 and 4 c) 1, 3 and 4 d) 2 and 5 Q.40 If at a point in a body σ x =70MPa , −5MPa , then σ y =60MPa and σ xy = the radius of Mohr's circle is equal to a) 5 5 MPa b) 2 5 MPa c) 5 2 MPa

d) 25 MPa

Q.41 Match List-I (Stress induced) with List-II (Situation/Location) and select

the correct answer using the codes given below the lists: List-I A) Membrane stress B) Tensional shear stress C) Double shear stress D) Maximum shear stress List-II 1. Neutral axis of beam 2. Closed coil helical spring under axial load 3. Cylindrical shell subject to fluid pressure 4. Rivets of double strap butt joint Codes: Codes: A B C D a) 3 1 4 2 b) 4 2 3 1 c) 3 2 4 1 d) 4 1 3 2

Q.42 State of stress at a point of a loaded component is given by: σ x = 30 MPa; σ y = 18 MPa and τxy = 8 MPa. If the

larger principal stress at the point is 34 MPa. What is the value of smaller principal stress? a) 12 MPa b) 14 MPa c) 16 MPa d) 18 MPa

Q.43 The bar of a boring machine is 45 mm in diameter. During operation, the bar may be twisted through O.01 radians and subjected to a shear stress of 45 MPa. What is the required length of the bar? (Take= G 0.84 ×105 N / mm 2 ) a) 380 mm b) 550 mm c) 420 mm d) 395 mm

Q.44 A close-coiled helical spring has wire diameter 10 mm and spring index 5. If the spring contains 10 turns, then the length of the spring wire would be a) 100mm b) 157mm c) 500 mm d) 1570 mm

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Q.45 A pipe of external diameter 3 cm, internal diameter 2 cm and of length 4 m is supported at its center. The sectional modulus of the pipe will be a) 65 π /64 cm3 b) 65 π /32 cm3 3 c) 65 π /96 cm d)65 π /128cm3

Q.46 For the beam loaded as shown in figure we use below. The magnitude of bending moment at the roller support is

a) 100 N-m c) 0

b) 1000 N-m d) 2000 N-m

Q.47 Match List-l (Cantilever loading) with List-II (Shear force diagram) and select the correct answer using the code given below the lists: List I List II A. 1. B.

2.

C.

3.

D.

4. 5. Codes: A a) 1 b) 4 c) 1 d) 4

B 5 5 3 2

C 2 2 4 5

D 4 3 5 3

figure. What is the nature of the beam?

a) Simply supported with a concentrated load at its mid length b) Simply supported and is subject to a couple M at its midpoint c) Simply supported and carries a uniformly varying load from zero at the support to a maximum at its mid length d) A cantilever subjected to end moment M

Q.49 For a particular load distribution and support conditions in a beam of length L, Bending moment at any section x (0<x
Q.48 The bending moment diagram for a beam AB is shown in the below © Copyright Reserved by Gateflix.in No part of this material should be copied or reproduced without permission

ANSWER KEY: 1 (b) 15 (b) 29 (b) 43 (c)

2 (b) 16 (a) 30 (a) 44 (d)

3 (d) 17 (d) 31 (b) 45 (c)

4 (d) 18 (d) 32 (d) 46 (a)

5 (b) 19 (b) 33 (c) 47 (b)

6 (d) 20 (d) 34 (a) 48 (b)

7 (a) 21 (d) 35 (d) 49 (a)

8 (b) 22 (b) 36 (d) 50 (a)

9 (c) 23 (a) 37 (c)

10 (a) 24 (b) 38 (a)

11 (b) 25 (c) 39 (a)

12 (c) 26 (d) 40 (c)

13 (d) 27 (c) 41 (c)

14 (a) 28 (d) 42 (b)

EXPLANATIONS Q.1

Q.2

Q.3

Q.4 Q.5

Q.6

Q.7

(b) Isotropy is the property in which material possess same properties in all directions.

Q.8

(b) Resilience can be defined as the ability of the material to absorb energy when deformed elastically and to release this energy when unloaded

(d) Toughness is the ability of material to absorb the energy upto fracture point. i.e. toughness of material is the total area under stress-strain curve.

In Torsion test of brittle material, granular helicoidal fracture takes place and fracture plane is 45° from the longitudinal axis while in case of ductile material, plane fracture on a transverse plane takes place.

(d)

(b) Hooke’s law is proportional limit.

valid

up

to

(d) According to Hooke’s law stress is directly proportional to strain up to proportional limit. (a)

(b) In Tensile test of ductile materials like mild steel, cup and cone failure occurs and failure plane is at 45° from the longitudinal axis. Where as in tensile test of brittle materials like cast iron, granular fracture occurs and failure plane is at 90° with longitudinal axis.

Q.9

(c) Poisson’s ratio is the ratio of lateral strain to the longitudinal strain.

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Q.10 (a) Q.11 (b) Q.12 (c) Q.13 (d)

9KG 3K+G 3K+G 1 or = 9KG E 1 1 1 + = 3G 9K E 3 1 9 or + = G K E E=

Q.14 (a)

= τ

8WD 16WR = πd 3 πd 3

The beam is loaded symmetrically; therefore the reactions at the supports will be equal w(b + 2a) ∴ VA = VB = 2 Bending moment in BD, Mx (x from A) − wx 2 w(b + 2a) = + × (x − a) 2 2 W 2 ⇒ M= b − 4a 2 ) ( c 8 But M c = 0 (given) W ∴ ( b 2 − 4a 2 ) = 0 8 2 b b ⇒   =4 ⇒ =2 a a Q.18 (d) Q.19 (b)

Q.15 (b) Q.16 (a)

Q.20 (d)

Q.21 (d) Q.22 (b) Q.23 (a) Q.17 (d)

Q.24 (b) Q.25 (c) 1 = PR 1 = PR

1 1 + PC PE PC + PE PC PE

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σ = 11.5 ×10−6 × 20 × 200 ×103 σ = 46MPa(Compression)

PC PE PC + PE

PR = Q.26 (d)

Q.30 (a)

Q.27 (c)

Q.31 (b)

Lateral Strain Longitudinal Strain −0.036 = − 20 0.12 200 μ = 0.3 μ= −

M x = Px l

= U

2l

(Px) 2 (Px) 2 dx + ∫0 2El ∫1 2E(I) dx l

Q.32 (d) 2l

 P2 x3   P2 x3  =   +   3 × 2EI  0  3 × 2E × 2I  l 2l

P 2 l3  P 2 x 3  = + 6EI  3 × 2E × 2I  l P 2 l3 7P 2 l3 = + 6EI 12EI 9 P 2 l3 3 P 2 l3 = = 12 EI 4 4EI 3 P 2 l3 1 P 2 = U = 4 EI 2 K 2 EI K= 3 l3 2 200 ×109 × 375 ×10−6 K= × 3 (0.5)3

= 4 ×105 ×103 = 4 ×108 N / m Q.28 (d) Modulus of elasticity, yield stress, ductility and tensile strength can be determined from the tensile testing on universal testing machine. But hardness and modulus of rigidity cannot be determined. Modulus of rigidity is determined by torsion test. Q.29 (b) σ = αΔT E

Q.33 (c) Q.34 (a) E = 2G (1+μ) 125 = 2G (+ 0.25) G = 50 GPa

Q.35 (d) Tensile Strength Universal Testing machine Impact Strength Izod/charpy test Bending Strength - Three point loading machine Fatigue Strength - Rotating bending machine Q.36 (d) Stress in x- direction = σ x σ σ Therefore ε x = x ,ε y =-μ x E E σx and ε z =-μ E Q.37 (c)

 ∈xx  φ Strain Tensor =  yx 2   φzx  2 

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φxy 2 ∈yy φzy 2

φxz   2  φ yz  2    ∈zz  

on comparison, we get φxy = 0.004 ⇒= φxy 0.008 2 τ xy τ xy ⇒ 100 ×103 = G= φxy 0.008 τ xy = 800MPa

Q.38 (a) Volumetric strain, ( σx + σy + σz ) (1 − 2µ) = ∈v E 60 + 20 − 50 = (1 − ( 2 × 0.35 )) 105 30 × 0.3 = = 9 ×10−5 5 10 Q.39 (a) Q.40 (c) Radius of Mohr’s circle

 σx − σy  2   + τxy  2  2

=

 70 − 60  2   + (−5) 2   2

=

= 25 + 25 = 5 2 MPa

Alternatively, Summation of normal stresses on two mutually perpendicular planes is always constant i.e. σ x + σ y = σ1 + σ2

⇒ 30 + 18= 34 + σ2 ⇒ σ2 =14MPa

Q.43 (c) From the torsion equation τ T Gθ = = R J L GθR ∴ L= τ On substituting the values, we get L = 420 mm Q.44 (d) Wire diameter d = 10 mm D Spring index= 5= 2 D = 5d D = 50 mm Number of turns n = 10 Length of wire = πDn =  π× 50 ×1 0 = 1570 mm Q.45 (c) Section modulus 1 Z= y max

Q.41 (c) Q.42 (b) σ x =30 MPa σ y =18 MPa τ xy =8 MPa

 σ x +σ y = σ2   2

  σ x +σ y  2 -   +τxy 2    2

2

=

Z=

π  Do4 − D14    64  Do / 2 

Z=

π  Do4 − D14    32  Do 

π  81 − 16  π  65  65π =  =   32  3  32  3  96

Q.46 (a)

 30+18   30+18  2 = σ2  -   +8  2   2 

σ 2 =24 − 36 + 64 σ2 = 24 − 10 = 14 MPa

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Bending, moment at roller support = Moment about B in Beam portion AB = 100 N-m

Q.47 (b)

Q.48 (d) Q.49 (a) Given M(x) = Ax- Bx2 dM Shear force= = A − 2Bx dx For Shear force to be zero, ⇒ A − 2Bx =0 ⇒ A =2Bx A ∴x = 2B Q.50 (a) d = 1 m = 1000 m m= 2 N/mm2 pd Hoop stress σn = 2t 2 ×1000 = 80 = 12.5mm 80

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CIVIL GATE QUESTIONS

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GATE SOLVED PAPER - CE STRENGTH OF MATERIALS

YEAR 2013 Q. 1

Q. 2

ONE MARK

Creep strain is (A) caused due to dead load only

(B) caused due to live load only

(C) caused due to cyclic load only

(D) independent of load

The “Plane section remain plane” assumption in bending theory implies (A) strain profile is linear (B) stress profile is linear (C) both profiles are linear

(D) shear deformation is neglected

Q. 3

A symmetric I-section (with width of each flange = 50 mm , thickness of web = 10 mm ) of steel is subjected to a shear force of 100 kN. Find the magnitude of the shear stress (in N/mm2) in the web at its junction with the top flange.

Q. 4

Two steel column P (length L and yield strength fy = 250 MPa ) have the same cross-section and end condition the ratio of bucking load of column P to that of column Q is (A) 0.5 (B) 1.0 (C) 2.0

(D) 4.0

YEAR 2013 Q. 5

TWO MARKS

2D stress at a point is given by a matrix s xx t xy 100 30 > H => H MPa t yx s yy 30 20 The maximum shear stress in MPa is (A) 50 (B) 75 (C) 100 (D) 110 YEAR 2012

Q. 6

Q. 7

The Poisson’s ratio is defined as (A) axial stress lateral stress (C) lateral stress axial stress

ONE MARK

(B) lateral strain axial strain (D) axial strian laterial strain

The following statement are related to bending of beams I. The slope of the bending moment diagram is equal to the shear force. II. The slope of the shear force diagram is equal to the load intensity. III. The slope of the curvature is equal to the flexural rotation IV. The second derivative of the deflection is equal to the curvature. The only FALSE statements is

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Q. 8

STRENGTH OF MATERIALS

(A) I

(B) II

(C) III

(D) IV

The ratio of the theoretical critical buckling load for a column with fixed ends to that of another column with the same dimensions and material, but with pinned ends, is equal to (A) 0.5 (B) 1.0 (C) 2.0 (D) 4.0 YEAR 2012

Q. 9

The simply supported beam is subjected to a uniformly distributed load of intensity w per unit length, on half of the span from one end. The length of the span and the flexural stiffness are denoted as l and El respectively. The deflection at mid-span of the beam is 4 4 (B) 5 wl (A) 5 wl 6144 El 768 El (C)

Q. 10

5 wl 4 384 El

4 (D) 5 wl 192 El

This sketch shows a column with a pin at the base and rollers at the top. It is subjected to an axial force P and a moment M at mid-height. The reaction(s) at R is/are.

(A) (B) (C) (D) Q. 11

TWO MARKS

a a a a

vertical vertical vertical vertical

force force force force

equal equal equal equal

to to to to

P P/2 P and a horizontal force equal to M/h P/2 and a horizontal force equal to M/h

A symmetric frame PQR consists of two inclined members PQ and QR connected at ‘Q ’ with a rigid joint and hinged at ‘P ’ and ‘R’. The horizontal length PR is l . If a weight W is suspended at Q , the bending moment at Q is

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STRENGTH OF MATERIALS

(A) Wl 2

(B) Wl 4

(C) Wl 8

(D) zero

YEAR 2011 Q. 12

ONE MARK

Consider a simply supported beam with a uniformly distributed load having a neutral axis (NA) as shown. For points P (on the neutral axis) and Q (at the bottom of the beam) the state of stress is best represented by which of the following pairs ?

(A)

(B)

(C)

(D)

YEAR 2011 Q. 13

TWO MARKS

For the cantilever bracket, PQRS , loaded as shown in the adjoining figure ( PQ = RS = L , and, QR = 2L ), which of the following statements is FALSE ?

(A) The portion RS (B) The portion QR WL. (C) The portion PQ WL. (D) The portion PQ

has a constant twisting moment with a value of 2WL . has a varying twisting moment with a maximum value of has a varying bending moment with a maximum value of has no twisting moment.

Statement for Linked Q. 14 and 15 : A rigid beam is hinged at one end and supported on linear elastic springs (both having a stiffness of ‘k ’) at point ‘1’ and ‘2’, and an inclined load acts at ‘2’, as shown.

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GATE SOLVED PAPER - CE

Q. 14

Which of the following options represents the deflections d1 and d2 at points ‘1’ and ‘2’ ? (A) d1 = 2 b 2P l and d2 = 4 b 2P l (B) d1 = 2 b P l and d2 = 4 b P l 5 k 5 k 5 k 5 k (C) d1 = 2 c P m and d2 = 4 c P m 5 2k 5 2k

Q. 15

STRENGTH OF MATERIALS

(D) d1 = 2 c 5

2 P and d = 4 2 5c k m

2P k m

If the load P equals 100 kN, which of the following options represents forces R1 and R2 in the springs at point ‘1’ and ‘2’ ? (B) R1 = 50 kN and R2 = 50 kN (A) R1 = 20 kN and R2 = 40 kN (C) R1 = 30 kN and R2 = 60 kN (D) R1 = 40 kN and R2 = 80 kN YEAR 2010

ONE MARK

Q. 16

Two people weighing W each are sitting on a park of length L floating on water at L/4 from either end. Neglecting the wright of the plank, the bending moment at the centre of the plank is (A) WL (B) WL 8 16 WL (C) (D) zero 32

Q. 17

The major and minor principal stresses at a point are 3 MPa and - 3 MPa respectively. The maximum shear stress at the point is (A) zero (B) 3 MPa (C) 6 MPa (D) 9 MPa

Q. 18

The number of independent elastic constants for a linear elastic isotropic and homogeneous material is (A) 4 (B) 3 (C) 2 (D)1

Q. 19

The effective length of a column of length L fixed against rotation and translation at one end is (A) 0.5 L (B) 0.7 L (C) 1.414 L (D) 2 L

Q. 20

A solid circular shaft of diameter d and length L is fixed at one end and free at the other end. A torque T is applied at the free end. The shear modulus of the material is G . The angle of twist at the free end is (A) 16TL (B) 32TL pd 4 G pd 4 G (C) 64TL (D) 1284TL pd 4 G pd G

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STRENGTH OF MATERIALS

YEAR 2010 Q. 21

For the simply supported beam of length L, subjected to a uniformly distributed moment M kN-m per unt length as shown in the figure, the bending moment (in kN-m) at the mid-span of beam is

(A) zero (C) ML Q. 22

TWO MARKS

(B) M (D) M/L

A disc of radius r has a hole of radius r/2 cut-out as shown. The centroid of the remaining disc (shaded portion) at a radial distance from the centre “O ” is

(A) r/2 (C) r/6

(B) r/3 (D) r/8

YEAR 2009

ONE MARK

Q. 23

A thin walled cylindrical pressure vessel having a radius of 0.5 m and wall thickness of 25 mm is subjected to an internal pressure of 700 kPa. The hoop stress developed is (A) 14 MPa (B) 1.4 MPa (C) 0.14 MPa (D) 0.014 MPa

Q. 24

The point within the cross sectional plane of a beam through which the resultant of the external loading on the beam has to pass through to ensure pure bending without twisting of the cross-section of the beam is called (A) moment centre (B) centroid (C) shear centre (D) elastic center YEAR 2009

Q. 25

Consider the following statements : 1. On a principal plane, only normal stress acts. 2. On a principal plane, both normal and shear stresses act. 3. On a principal plane, only shear stress acts 4. Isothermal state of stress is independent of frame of reference. Which of the above statements is/are correct ? (A) 1 and 4 (B) 2 only (C) 2 and 4 (D) 2 and 3

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TWO MARKS

GATE SOLVED PAPER - CE

STRENGTH OF MATERIALS

Q. 26

A hollow circular shaft has an outer diameter of 100 mm and a wall thickness of 25 mm. The allowable shear stress in the shaft is 125 MPa. The maximum torque the shaft can transmit is (A) 46 kNm (B) 24.5 kNm (C) 23 kNm (D) 11.5 kNm

Q. 27

Match List-I (Shear Force Diagrams) beams with List-II (Diagram of beams with supports and loading) and select the correct answer by using the codes given below the lists :

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GATE SOLVED PAPER - CE

Codes : a (A) 3 (B) 3 (C) 2 (D) 2

b 1 4 1 4

STRENGTH OF MATERIALS

c 2 2 4 3

d 4 1 3 1

Statement for Linked Q. 28 & 29 : In the cantilever beam PQR shown in figure below, the segment PQ has flexural rigidity EI and the segment QR has infinite flexural rigidity

Q. 28

Q. 29

The deflection and slope of the beam at Q are respectively 3 2 3 2 (B) WL and WL (A) 5 WL and 3 WL 3 EI 2 EI 6 EI 2 EI 3 2 (C) WL and WL 2 EI EI

3 2 (D) WL and 3 WL 3 EI 2 EI

The deflection of the beam at R is 3 (A) 8 WL EI

3 (B) 5 WL 6 EI

3 (C) 7 WL 3 EI

3 (D) 8 WL 6 EI

YEAR 2008 Q. 30

A mild steel specimen is under uniaxial tensile stress. Young’s modulus and yield stress for mild steel are 2 # 105 MPa and 250 MPa respectively. The maximum amount of strain energy per unit volume that can be stored in this specimen without permanent set is (A) 156 Nmm/mm3 (B) 15.6 Nmm/mm3 (C) 1.56 Nmm/mm3 (D) 0.156 Nmm/mm3 YEAR 2008

Q. 31

ONE MARK

TWO MARKS

Cross-section of a column consisting to two steel strips, each of thickness t and width b is shown in the figure below. The critical loads of the column with perfect bond and without bond between the strips are P and P0 respectively. The ratio P/P0 is

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GATE SOLVED PAPER - CE

Q. 32

STRENGTH OF MATERIALS

(A) 2

(B) 4

(C) 6

(D) 8

A rigid bar GH of length L is supported by a hinge and a spring of stiffness K as shown in the figure below. The buckling load, Pcr , for the bar will be

(A) 0.5 KL

(B) 0.8 KL

(C) 1.0 KL

(D) 1.2 KL

Q. 33

The maximum shear stress in a solid shaft of circular cross-section having diameter d subjected to a subject to a torque T is t. If the torque is increased by four times and the diameter of the shaft is increased by two times, the maximum shear stress in the shaft will be (A) 2t (B) t (C) t/2 (D) t/4

Q. 34

A vertical rod PQ of length L is fixed at its top end P and has a flange fixed to the bottom end Q . A weight W is dropped vertically from a height h (< L) on to the flange. The axial stress in the rod can be reduced by (A) increasing the length of the rod (B) decreasing the length of the rod (C) decreasing the area of cross-section of the rod (D) increasing the modulus of elasticity of the material

Q. 35

The maximum tensile stress at the section X -X shown in the figure below is

(A) 8P bd (C) 4P bd

(B) 6P bd (D) 2P bd

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Q. 36

STRENGTH OF MATERIALS

The stepped cantilever is subjected to moments, M as shown in the figure below. The vertical deflection at the free end (neglecting the self weight) is

2 (A) ML 8 EI

2 (B) ML 4 EI

2 (C) ML 4 EI

(D) Zero

Statement for Linked Q. 37 and 38 : Beam GHI is supported by three pontoons as shown in the figure below. The horizontal cross-sectional area of each pontoon is 8 m2 , the flexural rigidity of the beam is 10000 kNm2 and the unit weight of water is 10 kNm3

Q. 37

When the middle pontoon is removed, the deflection at H will be (A) 0.2 m (B) 0.4 m (C) 0.6 m (D) 0.8 m

Q. 38

When the middle pontoon is brought back to its position as shown in the figure above, the reaction at H will be (A) 8.6 kN (B) 15.7 kN (C) 19.2 kN (D) 24.2 kN YEAR 2007

ONE MARK

Q. 39

An axially loaded bar is subjected to a normal stress os 173 MPa. The stress in the bar is (A) 75 MPa (B) 86.5 MPa (C) 100 MPa (D) 122.3 MPa

Q. 40

A steel column, pinned at both end, has a buckling load of 200 kN. If the column is restrained against lateral movement at its mid-height, it buckling load will be (A) 200 kN (B) 283 kN (C) 400 kN (D) 800 kN

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GATE SOLVED PAPER - CE

Q. 41

STRENGTH OF MATERIALS

For an isotropic material, the relationship between the Young’s modulus (E), shear modulus (G) and Poisson’s ratio (m) is given by E E (A) G = (B) E = 2 (1 + m) 2 (1 + m) E E (C) G = (D) G = (1 + 2m) 2 (1 - 2m) YEAR 2007

TWO MARKS

Q. 42

A metal bar of length 100 mm is inserted between two rigid supports and its temperature is increased by 10cC . If the coefficient of thermal expansion is 12 # 10- 6 per cC and the Young’s modulus is 2 # 105 MPa, the stress in the bar is (A) zero (B) 12 MPa (C) 24 MPa (D) 2400 MPa

Q. 43

A rigid bar is suspended by three rods made of the same material as shown in the figure. The area and length of the central rod are 3 A and L, respectively while that of the two outer rods are 2A and 2L , respectively. If a downward force of 50 kN is applied to the rigid bar, the forces in the central and each of the outer rods will be (A) 16.67 kN each (B) 30 kN and 15 kN (C) 30 kN and 10 kN

Q. 44

(D) 21.4 kN and 14.3 kN

The maximum and minimum shear stresses in a hollow circular shaft of outer diameter 20 mm and thickness 2 mm, subjected to a torque of 92.7 Nm will be

(A) 59 MPa and 47.2 MPa (C) 118 MPa and 160 MPa

(B) 100 MPa and 80 MPa (D) 200 MPa and 160 MPa

Q. 45

The shear stress at the neutral axis in a beam of triangular section with a base of 40 mm and height 20 mm, subjected to a shear force of 3 kN is (A) 3 MPa (B) 6 MPa (C) 10 MPa (D) 20 MPa

Q. 46

U1 and U2 are the strain energies stored in a prismatic bar due to axial tensile forces P1 and P2 , respectively. The strain energy U stored in the same bar due to combined action of P1 and P2 will be (B) U = U1 U2 (A) U = U1 + U2 (D) U > U1 + U2 (C) U < U1 + U2

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GATE SOLVED PAPER - CE

STRENGTH OF MATERIALS

YEAR 2006 Q. 47

ONE MARK

30 0 Mohr’s circle for the state of stress defined by > MPa is a circle with 0 30H (A) centre at (0, 0) and radius 30 MPa (B) centre at (0, 0) and radius 60 MPa (C) centre at (30, 0) and radius 30 MPa (D) centre at (30, 0) and zero radius

Q. 48

A long shaft of diameter d is subjected to twisting moment T at its ends. The maximum normal stress acting at its cross-section is equal to (A) zero (B) 16T3 pd 32 T 64 (C) (D T3 pd 3 pd

Q. 49

The buckling load P = Pcr for the column AB in figure, as KT approaches infinity, 2 becomes a p EI L2

Where a is equal to (A) 0.25 (C) 2.05

(B) 1.00 (D) 4.00

YEAR 2006

TWO MARKS

Q. 50

A thin-walled long cylindrical tank is inside radius r is subjected simultaneously to internal gas pressure p and axial compressive force F at its ends. In order to produce ‘pure shear’ state of stress in the wall of the cylinder, F should be equal to (B) 2ppr2 (A) ppr2 (D) 4ppr2 (C) 3ppr2

Q. 51

Consider the beam AB shown in the figure below. Part AC of the beam is rigid while Part CB has the flexural rigidity EI . Identify the correct combination of deflection at end B and bending moment at end A, respectively

3 (A) PL , 2PL 3EI

3 (B) PL , PL 3EI

3 (C) 8PL , 2PL 3EI

3 (D) 8PL , PL 3EI

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Q. 52

STRENGTH OF MATERIALS

A simply supported beam AB has the bending moment diagram as shown in the following figure.

The beam is possibly under the action of following loads (A) Couples of M at C and 2M at D (B) Couples of 2M at C and M at D (C) Concentrated loads of M/L at C and 2M/L at D (D) Concentrated load of M/L at C and couple of 2M at D Q. 53

A beam with the cross-section given is subjected to a positive bending moment (causing compression at the top) of 16 kNm acting around the horizontal axis. The tensile force acting on the hatched area of the cross-section is

(A) zero (C) 8.9 kN

(B) 5.9 kN (D) 17.8 kN

Q. 54

If a beam of rectangular cross-section is subjected to a vertical shear force V , the shear force carried by the upper one-third of the cross-section is (A) zero (B) 7V 27 (C) 8V (D) V 3 27

Q. 55

For the section shown below, second moment of the area about an axis d/4 distance above the bottom of the area is

3 (A) bd 48

3 (B) bd 12

3 (C) 7bd 48

3 (D) bd 3

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Q. 56

STRENGTH OF MATERIALS

I -section of a beam is formed by gluing wooden planks as shown in the figure below. If this beam transmits a constant vertical shear force of 3000 N, the glue at any of the four joint will be subjected to a shear force (in kN per meter length) of

(A) 3.0

(B) 4.0

(C) 8.0

(D) 10.7

Common Data For Q. 57 and 58 : Consider a propped cantilever beam ABC under two leads of magnitude P each as shown in the figure below. Flexural rigidity of the beam is EI .

Q. 57

Q. 58

The reaction at C is (A) 9Pa (upwards) 16L 9 (C) Pa (upwards) 8L

(B) 9Pa (downwards) 16L (D) 9Pa (downwards) 8L

The rotation at B is (A) 5PLa (clockwise) 16EI (C) 59PLa (clockwise) 16EI

(B) 5PLa (anticlockwise) 16EI (D) 59PLa (anticlockwise) 16EI

YEAR 2005 Q. 59

ONE MARK

The symmetry of stress tensor at a point in the body under equilibrium is obtained from (A) conserved of mass (B) force equilibrium equations (C) moment equilibrium equations (D) conservation of energy

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Q. 60

STRENGTH OF MATERIALS

The components of strain tensor at a point in the plane strain case can be obtained by measuring longitudinal strain in following directions (A) along any two arbitrary directions (B) along any three arbitrary directions (C) along two mutually orthogonal directions (D) along any arbitrary direction YEAR 2005

Q. 61

If principal stresses in a two-dimensional case are - 10 MPa and 20 MPa respectively, then maximum shear stress at the point is (A) 10 MPa (B) 15 MPa (C) 20 MPa

Q. 62

TWO MARKS

(D) 30 MPa

The bending moment diagram for a beam is given below :

The shear force at sections aal and bbl respectively are of the magnitude (A) 100 kN, 150 kN (B) zero, 100 kN (C) zero, 50 kN (D) 100 kN, 100 kN Q. 63

A circular shaft shown in the figure is subjected to torsion T at two point A and B. The torsional rigidity of portions CA and BD is GJ1 and that of portion AB is GJ2 . The rotations of shaft at points A and B are q1 and q2 . The rotation q1 is

TL GJ1 + GJ2 (C) TL GJ2 (A)

(B) TL GJ1 TL (D) GJ1 - GJ2

YEAR 2004 Q. 64

ONE MARK

For linear elastic systems, the type of displacement function for the strain energy is (A) linear (B) quadratic (C) cubic (D) quartic

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GATE SOLVED PAPER - CE

STRENGTH OF MATERIALS

YEAR 2004 Q. 65

Q. 66

In a two dimensional stress analysis, the state of stress at a point is shown below. If s = 120 MPa and t = 70 MPa , sx and sy , are respectively,

(A) 26.7 MPa and 172.5 MPa

(B) 54 MPa and 128 MPa

(C) 67.5 MPa and 213.3 MPa

(D) 16 MPa and 138 MPa

For the linear elastic beam shown in the figure, the flexural rigidity, EI is 781250 kNm2 . When w = 10 kN/m , the vertical reaction RA at A is 50 kN. The value of RA for w = 100 kN/m is

(A) 500 kN (C) 250 kN Q. 67

Q. 68

TWO MARKS

(B) 425 kN (D) 75 kN

A homogeneous, simply supported prismatic beam of width B , depth D and span L is subjected to a concentrated load of magnitude P . The load can be placed anywhere along the span of the beam. The maximum flexural stress developed in beam is (A) 2 PL2 (B) 3 PL2 3 BD 4 BD (C) 4 PL2 (D) 3 PL2 3 BD 2 BD A circular solid shaft of span L = 5 m is fixed at one end and free at other end. A twisting moment T = 100 kNm is applied at the free end. The torsional rigidity GH is 50000 kNm2 /rad . Following statements are made for this shaft : 1. The maximum rotation is 0.01 rad 2. The torsional strain energy is 1 kNm With reference to the above statements, which of the following applies ? (A) Both statements are true (B) Statement 2 is true but 2 is false (C) Statement 2 is true but 1 is false (D) Both the statements are false

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STRENGTH OF MATERIALS

Common Data For Q. 69 and 70 : A three-span continuous beam has an internal hinge at B. Section B is at the mid-span of AC. Section E is at the mid-span of CG. The 20 kN load is applied at section B whereas 10 kN loads are applied at sections D and F as shown in the figure. Span GH is subjected to uniformly distributed load of magnitude 5 kN/m. For the loading shown, shear force immediate to the right of section E is 9.84 kN upwards and the hogging moment at section E is 10.31 kNm

Q. 69

The magnitude of the shear force immediate to the left and immediate to the right of section B are respectively (A) 0 and 20 kN (B) 10 kN and 10 kN (C) 20 kN and 0 (D) 9.84 kN and 10.16 kN

Q. 70

The vertical reaction at support H is (A) 15 kN upward (B) 9.84 kN upward (C) 15 kN downward (D) 9.84 kN downward YEAR 2003

Q. 71

ONE MARK

A bar of varying square cross-section is loaded symmetrically as shown in the figure. Loads shown are placed on one of the axes of symmetry of cross-section. Ignoring self weight, the maximum tensile stress in N/mm2 anywhere is

(A) 16.0 (C) 25.0

(B) 20.0 (D) 30.0

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Q. 72

STRENGTH OF MATERIALS

A curved member with a straight vertical leg is carrying a vertical load at Z , as shown in the figure. The stress resultants in the XY segment are

(A) bending moment, shear force and axial force (B) bending moment and axial force only (C) bending moment and shear force only (D) axial force only YEAR 2003 Q. 73

TWO MARKS

The state of two dimensional stresses acting on a concrete lamina consists of a direct tensile stress, sx = 1.5 N/mm2 , and shear stress, t = 1.20 N/mm2 , which cause cracking of concrete. Then the tensile strength of the concrete in N/mm2 is (A) 1.50 (B) 2.08 (C) 2.17 (D) 2.29

Q. 74

A "H" shaped frame of uniform flexural rigidity EI is located as shown in the figure. The relative outward displacement between points K and O is

2 (A) RLh EI 2 (B) RL h EI 2 (C) RLh 3EI 2 (D) RL h 3EI

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Q. 75

STRENGTH OF MATERIALS

A simply supported beam of uniform rectangular cross-section of width b and depth h is subjected to linear temperature gradient 0c at the top and Tc at the bottom, as shown in the figure. The coefficient of linear expansion of the beam material is a. The resulting vertical deflection at the mid-span of the beam is

2 (A) aTh upward 8L 2 (B) aTL upward 8h 2 (C) aTh downward 8L 2 (D) aTL downward 8h

Q. 76

List I shows different loads acting on a beam and List II shows different bending moment distributions. Match the load with the corresponding bending moment diagram.

Codes : A (A) 4 (B) 5 (C) 2 (D) 2

B 2 4 5 4

C 1 1 3 1

D 3 3 1 3

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Q. 77

STRENGTH OF MATERIALS

A long structural column (length = L ) with both ends hinged is acted upon by an axial compressive load, P . The differential equation governing the bending of column is given by : d2 y EI 2 =- Py dx Where y is the structural lateral deflection and EI is the flexural rigidity. The first critical load on column responsible for its buckling is given by 2 2 (A) p EI (B) 2 p2 EI 2 L L 2 2 (C) 2p EI (D) 4p 2EI L2 L YEAR 2002

Q. 78

Q. 79

ONE MARK

The shear modulus ^G h, modulus of elasticity ^E h and the Poisson’s ratio ^v h of a material are related as E G (A) G = (B) E = 62 ^1 + v h@ 62 ^1 + v h@ E E (C) G = (D) G = 62 ^1 - v h@ 62 ^v - 1h@ For the loading given in the figure below, two statements (I and II) are made

I. Memeber AB carries shear force and bending moment II. Member BC carries axial load and shear force Which of these statement is true? (A) Statement I True but II is False (B) Statement I is False but II is True (C) Both statement I and II are True (D) Both statement I and II are False YEAR 2002 Q. 80

TWO MARKS

In the propped cantilever beam carrying a uniformly distributed load of w N/m , shown in the following figure, the reaction at the support B is

(A) 5 wL 8 (C) 1 wL 2

(B) 3 wL 8 (D) 3 wL 4

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GATE SOLVED PAPER - CE

STRENGTH OF MATERIALS

YEAR 2001 Q. 81

ONE MARK

The bending moment (in kNm units) at the mid span location X in the beam with overhangs shown below is equal to

z (A) 0 (C) –15

(B) –10 (D) –20

**********

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STRENGTH OF MATERIALS

ANSWER KEY STRENGTH OF MATERIALS 1

2

3

4

5

6

7

8

9

10

(A)

(A)

71.12

(D)

(A)

(B)

(C)

(D)

(B)

(C)

11

12

13

14

15

16

17

18

19

20

(D)

(A)

(B)

(B)

(D)

(D)

(B)

(C)

(D)

(B)

21

22

23

24

25

26

27

28

29

30

(A)

(C)

(A)

(C)

(A)

(C)

(A)

(A)

(A)

(D)

31

32

33

34

35

36

37

38

39

40

(B)

(C)

(C)

(A)

(A)

(C)

(B)

(C)

(B)

(D)

41

42

43

44

45

46

47

48

49

50

(A)

(C)

(C)

(B)

(C)

(D)

(D)

(A)

(D)

(C)

51

52

53

54

55

56

57

58

59

60

(A)

(A)

(C)

(B)

(C)

(B)

(C)

(A)

(C)

(B)

61

62

63

64

65

66

67

68

69

70

(B)

(C)

(B)

(B)

(C)

(B)

(D)

(B)

(A)

(B)

71

72

73

74

75

76

77

78

79

80

(C)

(D)

(C)

(A)

(D)

(D)

(A)

(A)

(A)

(B)

81 (C)

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