Design Of Curved Beams

DESIGN OF CURVED BEAMS Introduction. Straight Beam: - A beam is a straight structural member subjected to a system of external forces acting at right angles to its axis. They are classified by the type of supports. 1. If a member is fixed or built in one end while its other end is free, the member is called cantilever beam. 2. If the ends of the beam are made to freely rest on supports the beam is called a freely or simply supported beam. 3. If abeam is fixed at both its, it is called as built-in or fixed beam.

4. A beam which is provided with more than two supports is called as continuous beam.

A beam is said to be statically determinate beam, if its reaction components can be determined by using equations of static equilibrium only. Commonly encountered statically determinate beams are cantilever beams, SS beam and over hanging beams. Beams are subjected to transverse loads such as concentrated load, UDL, UVL & applied moments. Beam transfer applied load to supports, the beam develop resistance to moments & transverse shear forces at all its cross-sections. CURVED BEAMS A beam in which the neutral axis in the unloaded condition is curved instead of straight. Or If the beam is originally curved before applying the bending moment, are termed as “Curved Beams”. Applications of Curved Beams. Curved beams find applications in many machine members such as c – clampers , crane hooks, frames of presses, chains, links, rings, etc..,

Differences between Straight Beams & Curved Beams. Sl.no Straight Beams Curved Beams 1. Neutral axis of the Neutral axis does not co cross-section passes – incide with the crossthrough the centroid of section, but is shifted the section. towards the centre of curvature of the beam. 2. The variation of The distribution of the bending stress is linear, stress in the case of magnitude being curved beam is nonproportional to the linear (Hyper- bolic) distance of a fiber from because of the neutral the neutral axis. axis is initially curved. Note:- In all the machine members, the line of action the load does not pass through the centroid of the section over which the stress is to be found out. Due to this, the total stress over section of the member will be the algebraic sum of direct stress & bending stress.

Q) Discuss the stress distribution pattern in curved beams when compared to straight beam with sketches. [VTU /Feb 2002](4) Curved beam Hyperbolic stress Distribution [nonlinear] ►The neutral axis of the cross section does not coincide with the centroidal axis but is shifted towards the center of

curvature of the beam. ►Bending stress in curved beams donot follow linear variation because of the variation of arc length. Straight beam Linear variation of stress ►Neutral axis coincides with the centroidal axis. ►Variation of normal stress due to bending across its section is linear. ►The stress is proportional to the distance of the fiber from the neutral axis.

Q) Derive the expression for the normal stress due to bending at the extreme fibers of a curved beam. [VTU/Jan-Feb/2002](8) OR State the assumptions & derive the stress equation in case of a curved beam. [VTU/Jan/Feb/2003] (8) →

Assumptions:1. The beam is subjected to pure bending. 2. Material of the beam is isotropic & homogeneous & obeys hook’s law. 3. Plane sections perpendicular to the axis of the beam remain plane even after bending. 4. The stress induced do not exceed elastic limit. 5. Each layer of the beam is free to expand OR contract, independent of the layer above or below it. 6. Young’s modulus is same in tension &compression. Let F =Load M =Applied bending moment, N –mm e = distance from the centroidal axis to the neutral axis, measured towards centre of curvature, mm Ci = distance from neutral axis to inner fiber (radius) mm Co = distance from neutral axis to outer fiber (radius) mm. Ri = inner radius of curvature, mm Ro = outer radius of curvature, mm Rn = Radius of neutral axis, mm R (Rc) = Radius of centroidal axis, mm A = area of section, mm2 σi = Stress in the inner fiber, N/mm2 σo = Stress in the outer fiber, N/mm2 Consider a part of the curved beam between two radial planes ab & cd subtending an angle ‘θ’ at the centre of

curvature when the beam is subjected to bending moment ‘M’ as shown in fig, the plane ‘cd’ rotates with respect to ‘ab’ through an angle ‘θ’ & takes new position ‘fg’. The outer fiber are shantered due to compression and inner fibre are elongated due to tension. Now, consider a fiber of depth ‘dy’ & cross- sectional area ‘dA’ at a distance ‘y’ from the neutral axis. The original length of strip at a distance ‘y’ from the neutral axis is (Rn + y)θ. It is shortened by an amount ydθ. The strain across cross section dA would be Є= +yθ . (Rn+ y )θ Stress in the fibre is given by σ α Є within elastic limit or σ=EЄ E- Young’s modulus σ = + E y dθ (1) (Rn + y)θ Load on the strip having the thickness ‘dy’ & the crosssection dA is given by σ = dF or dF = σ dA dA i.e., dF = + E y dθ . dA (Rn + y)θ Now applying the condition of equilibrium (i.e., summation of forces over the whole cross-section is zero). i.e., ∫dF = 0 ∫ +Ed θ y

dA=0 Or

+Ed θ ydA

=0

(RN+y) θ But

θ

(RN+y)

E.dθis not equal to zero. θ ∫ y dA =0 (Rn + y)

Now talking moments about the neutral axis, we get Mb = +∫y dF substitute the value of dF Mb = Ed θ ∫ y2 .dA θ (y + RN) 0

= Edθ ∫ y – y RN dA θ y + RN

from (2)

Mb = Edθ ∫ y dA θ Since ∫y.dA represents the moment of area, it may be replaced by A.e, i.e., the product it may be reduced by A.e , i.e., the product of total area A and the distance ‘e’ from the centroidal axis to the neutral axis. Mb = E dθ .A.e (3) θ Or E dθ/θ = Mb/Ae from eqn (1) Edθ = +σ (RN+y) θ On comparing we get, Mb = +σ (RN+y) Ae y Or σ = +Mb.y Eqn 10.1/pg 132 (RN+y)Ae

This is the general equation for the stress in a fiber at a distance ‘y’ from the neutral axis. At the outer fiber, y=+Co; Bending stress σb = - MbCo (RN+Co)Ae Or

(compressive)

σbo = - MbCo [RN + Co = Ro] A e Ro At inner fibre, y = + Ci Bending stress σbi = MbCi (RN - Ci)Ae σi = + MbCi A e Ri

[RN – Ci = Ri]

Problems:The section of a crane hook is rectangular in shape whose width is 30mm & depth is 60mm. The centre of curvature of the section is at a distance of 125mm from the inside section & the load line is 100mm from the same point. Find the capacity of the hook if the allowable stress in tension is 75 N/mm2. Soln:1)

Given data :- b= 30mm; h=60mm; Ri= 125mm; σRi = 75N/mm2

To find the capacity of the the hook, use eqn. σRi = F + MbCi A A e Ri Inner radius = Ri = 125mm

Outer radius = Ro = 125 + 60 = 185mm Radius of = Rc = 125 + 60/2 = 155mm Centroidal axis Radius of = RN = h --- table 10.1/pg 134 Neutral axis ln((R + c)/(R - c)) But R = Rc, R + C2 = Ro, R - C 1 = Ri RN = h ln(Ro/Ri) RN = 60 = 153.045mm Ln(185/125) Distance of centroidal axis from neutral axis = e = Rc – RN = 155- 153.045 e = 1.955mm Distance of neutral axis to inner radius = Ci = RN - Ri = 153.045-125 = 28.045mm Distance of neutral axis to outer radius = Co = Ro - RN = 185-153.045 = 31.955mm Area of cross-section = A = b × h = 30 × 60 = 1800mm2 Distance from centroidal axis to force = l = Rc – 25 = 155-25 = 130mm

Bending moment about the centroidal axis = Mb = F × l = 130.F N-mm Combined stress at the inner fiber σRi = F + Mb Ci A AeRi 75 = F/1800 + (130.F × 28.045)/ (1800 × 1.955 × 125) Solving for ‘F’ we get F = 8480.4N -------- Capacity for the hook. A crane hook of trapezoidal cross-section has an inner fiber width = 120mm, depth = 100mm & inner radius = 120mm. Calculate the width if stresses are numerically equal at inner & outer fibres. Also determine the capacity of the hook, if the permissinble stress is 100MPa. soln:2)

Given data:- for stress to be numerically equal, MCi = MCO AeRi AeRO ⇒ Ci = CO Ri RO But Ci = RN – Ri & CO = RO – RN R N – R i = RO - R N Ri RN RN -1 = 1 - RN Ri RO RN [ 1/Ri + 1/RO ] = 2 ⇒

R N = 2 Ri R O R i + RO

But Ri = 120; RO = 220mm; RN = 2 × 120 × 220 = 155.29mm 120 + 220 But RN = A B1(R + c2) – b(R – C1) ln (R + C2) - (b1 - b) h (R – C1) table 10.1/pg 135 But R + C2 = RO & R – C1 = Ri But RN =

A B1RO – bRi ln RO - (b1 - b) h Ri Substituting the values,

155.29 = 50(120 + b) 120 × 220 – 120b ln 220 - (120 - b) 100 120 On solving, we get b = 27.6mm ≈ 28mm (ii)

Max load:σRi = F/A + MbCi/AeRi 100 = F + F × 220 ×

(3) A crane hook shown in figure below is made of 30mm diameter steel rod. The distance between the centroidal axis of the rod & the centre of curvature of the hook is 50mm. Determine the load ‘F’ so that the maximum stress in the rod is not to exceed 40 N/mm2.] soln:-

Given data:- d = 30mm; Rc = 50mm; σRi = 40 N/mm2 From the figure; Ri = Rc – (d/2) = 50 - 15 =35mm RO = Rc + (d/2) = 50 + 15 =65mm C1 = C2 = d/2 = 15 mm. From DDHB; table 10.1/pg 134; for circular c/s RN = ½ C2 = ½ (152) R – √ R2 – C 2 50 – √ 502 – 152 RN = 48.85mm e = Rc – RN = 50 – 48.85 = 1.15mm Ci = C1 – e = 15 – 1.15 = 13.85mm CO = C2 + e = 15 + 1.15 = 16.15mm Area of cross-section: A = π d2/4 = π302/4 = 707.14mm2 Bending moment : M = F × Rc = f × 50 = 50F N-mm Total stress = Direct stress + Bending stress at inner fiber = F/A + MbCi/AeRi 40 = F/707.14 + 50F × 13.85/707.14 × 1.15× 35 On solving for ‘F’ we get F = 1556.42 N 4)

A section of frame for a punch press is shown in figure below. Determine the capacity of the press if the maximum tensile stresses in the frame is not to exceed 60MPa.

Soln:Given data:σRi = 60 N/mm2 ; Ri = 80 mm ; RO = 80 + 40+ 100 = 220mm. To find RN;From table 10.1 /pg 136 RN = A B ln R + d – C1 + a ln R + C2 R – C1 R + d + C2 To find C1 & C2 from table 1.3/pg 8 C1 = aH2 + bd2 2(aH + bd)

& C2 = H – C1

where a = 40mm, H= 140 mm, b = 40mm, d = 40mm. C1 = 40(140)2 + 40(40)2 = 58.88mm 2(40 × 140 + 40 × 40) IIIly

C2 = 140 – 58.88 = 81.12mm.

Again RN =

A

B ln Ri– C1 + a ln RO Ri RO + d Because R – C1 = Ri & R + C2 = RO A = area of cross-section = 80 × 40 + 100 × 40 = 7200mm2 RN =

7200 80 ln 80 + 40 + 40 ln 220 80 80 + 40

RN = 127.02mm e = RC – RN = 138.88 – 127.02 = 11.85mm Bending moment : Mb = F × (Rc + l) = F(138.88 + 200) Mb = 338.88F Total stress = σRi =F/A + MbCi/AeRi Ci = C1 – e = 58.88 – 11.85 = 47.03mm 60 = F + 338.88 × 47.03 7200 7200 × 11.85 × 80 F = 24253.32N

capacity of the frame.

STRESS IN CLOSED RINGS:-

A closed ring is an example of a Curved beam with restrained ends. Load is acting at section B-B & sectiona A-A is 90o away from the point of application of load. Let the ring be subjected to a central load shown in the figure. The bending moment at any cross-section of the ring is given by BM = Mb = F.R [ 2/π - sinθ]/2 At section A –A:θ = 0; MbA = 0.318FR eqn 10.5/pg 133 At section B-B :MbB = -0.182FR eqn 10.6/pg 133 here -ve sign is considered as tensile. Direct stress at any cross section ‘DD’ at an angle θ with vertical is given. σd = F.sinθ/2A

Problem :1) Detrermine the stress induced in a circular ring of circular c/s of 25mm diameter subjected to a tensile load of 6500N. The inner ring diameter is 60mm. Soln:-

Given data:Ri = 30mm RO = 30 + 25 = 55mm RC = 30 + 25/2 = 42.5mm RN = ( √RO + √Ri ) 2 /4 ---- for circular c/s = ( √55 + √30 )2 /4 = 41.51mm e = Rc – RN = 42.5 – 41.51 = 0.99mm Ci = RN – Ri = 41.51 – 30 = 11.51mm CO = RO + RN = 55 – 41.51 = 13.49mm A = π × d2/4 = 491.07mm2 Consider section A-A; θ = 0o Direct stress = σd = F sinθ/2A = 0 Bending moment: Mb = 0.318 F RC

(because sin 0 = 0)

= 0.318 × 6500 × 42.5 = 87847.5 N – mm Max stress at inner fiber ; σRi = σd + σb = 0 + Mb.Ci/(A.e.Ri) = 87847.5 × 11.51 491.07 × 0.99 × 30 = 69.32 N/mm2 (tensile) Max stress at outer fiber ; σRo = σd - σb = 0 - Mb.CO/(A.e.RO) = 87847.5 × 13.49 491.07 × 0.99 × 55 σRo = -44.32 N/mm2 Consider section B-B; θ = 90o ( w.r.t horizontal) Direct stress = σd = F sinθ/2A = 6500 × sin90 2 × 491.07 = 6.618 N/mm2 Bending moment: Mb = -0.182 F RC = -0.182 × 6500 × 42.5 = - 50277.5 N – mm Max stress at inner fiber ; σRi = σd + σb = 6.618 + Mb.Ci/(A.e.Ri) = 6.618 + (- 50277.5) × 11.51 491.07 × 0.99 × 30 σRi = -33.052 N/mm2 (compressive) Max stress at outer fiber ; σRo = σd - σb

= 6.618 - Mb.Ci/(A.e.Ri) = 6.618 - (- 50277.5) × 11.51 491.07 × 0.99 × 30 σRo = 31.98 N/mm2

SPRINGS Introduction A spring is defined as an elastic body, whose function is to distort when loaded and to recover its original shape when the load is removed. Applications 1) To absorb or control energy due to either shock or vibratgion as in automotives, railways, aircrafts, landing gears and vibration dampers etc. 2) To apply forces, as in brakes, clutches and spring loaded valves. 3) To control motion by maintaining control between two elements as in CAMS & followers. 4) To measure forces as in spring balances and engine indicators. 5) To store energy as in watches, toys etc. TYPES OF SPRINGS 1).HELICALSPRINGS

Helical springs are made of wire coiled in the form of helix and are primarily intended for compressive or tensile loads. The cross-section of the wire from which the spring is made may be

circular, square or rectangular. The two forms of helical springs are compression helical spring and tension helical spring as shown in figure. o Helical springs are said to be closely coiled, when the helix angle is very small (< 10 ), where as in open coil helical spring the helix angle is large. Advantages. a) These springs are easy to manufacture. b) They are available in wide range. c) They are highly reliable. d) They have constant spring rates. e) Their performance can be predicted more accurately. f)

There characteristics can be varied by changing dimensions.

2). Conical and Volute springs.

The conical and volute spring shown in the figure are used in special applications where the spring rate increases in increase in load. Another feature of these types of springs is the decreasing number of coils results in an increasing spring rate. This characteristic is some times utilized in vibrations problems where springs are used to support to body that have varying mass. 3). Torsion springs.

These springs may be of helical or spiral type as shown in figure. Helical types of springs are used where the load tends to wind up the springs and are used in electrical mechanisms. Spiral type is used where the loads tends to increase the number of coils and are used in watches and clocks. 4). Laminated or Leaf springs.

The laminated or leaf spring (also known as flat spring) consists of a number of flat plates (known as leaves) of varying lengths held together by means of clamps and bolts. These types of springs are most used in automobiles.

5). Disc springs.

These springs consists of a number of conical discs held together by a central bolt or tube as shown in figure. These springs are used in applications where high spring rates and compact spring units are required. Terms used in compression springs 1. Solid length: -

When the springs are compressed until the coils come in contact with each other, then the spring is said to be solid. The solid length of a spring is the product of total number of coils & the diameter of the wire. Mathematically, l Solid length- Ls = n .d l Where n - no of coils d - dia of the coils

2. Free length:-

Free length of a compression spring is the length of the spring in the free or unloaded condition & is equal to the solid length plus the maximum deflection or compression of the spring & the clearance between the adjacent coils. Mathematically, Free length - Lf = solid length + max. def + clearance between adjacent coils. l Lf = n .d + δmax + 0.15 δmax 3. Spring index: - It is defined as the ratio of the man diameter of the coil to

the diameter of the

wire. Spring index = C = D/d 4. Spring Rate: - spring rate (stiffness/spring constant) is the defined as the load required per unit deflection of the spring. Spring Rate, K= F/ δ

F- load, N δ - Defection, mm

5. Pitch: - Pitch of the coils is defined as the axial distance between adjacent coils in un compressed state. l Pitch of the coil, p = ( LF – Ls)/n + d where, LF - Free length Ls - Solid length l n - Total number of coils d - diameter of coil. Stress in helical spring and circular wire.

Consider a helical compression spring made of circular wire & subjected to an axial load F, as shown in figure. Let, D = Mean diameter of the coil d = Diameter of the spring wire, n = number of active coils, G = Modulus of Rigidity for the spring material, F = Axial load on the spring, τ = Max. Shear stress induced in the wire, C = spring index = D/d p = pitch of the coils

&

δ = deflection of the spring. Consider a point of the spring shown in fig (b). The load ‘F’ tends to rotate the wire & as a result twisting moment (T) is developed in the wire, & thus torsional shear stress is induced in the wire. Let us consider that part of spring is in equilibrium under the action of two forces ‘F’ & twisting ‘T’. Therefore,

3 T = F × D/2 = π × τ1 × d

n eq 3.1/pg 142

16 Therefore

τ1 = 8.F.D 3 π.d

In addition to the torsional shear stress (τ1) induced in the wire, the following stresses also act on the wire. 1. Direct stress due to the load. F, & 2. Stress due to curvature of wire. Direct stress due to load F τ2 =

Load =

F

= 4F 2

c/s Area π × d /4

2 πd

τ2 = 4F 2 πd Now, the resultant shear stress induced in the wire τ = τ1 ± τ2 = 8FD ± 4F 3 2 πd πd Positive sign is used for the inner edge of the wire & Negative sign is used for the outer edge of the wire. Since the stress is maximum at the inner edge of the wire, Therefore, Max shear stress

= Torsional shear stress + Direct shear stress

induced in the wire = 8FD + 4F 3 2 πd πd = 8FD 1 + d 3 πd 2D

(but C = D/d)

= 8FD 1 + 1 3 πd 2C Therefore

τ = 8FD .K 3 πd

Where K = 4C - 1 + 0.615 4C - 4

C

Stress distribution diagram.

n eq 11.1a/pg 139

Wahl stress Conc. Factor n eq 11.26/pg 139

Deflection of helical springs of circular wire. Let l = total active length of wire = πD × n θ = Angular deflection of the wire due to Torque t. Therefore, Axial deflection of the spring δ = θ × D/2 Also,

(1)

T = τ = Gθ J

OR

-----------

R

l

T = τ = Gθ

⇒

θ = T .l

J

But

D/2 l J.G 4 J = πd /32 & G – Modulus of Rigidity.

Substituting the values of l & J, we have, 2 = (F × D/2) (πDn ) = 16FD n 4 4 J.G π × d × G/32 G×d n Substituting eq (2) in (1) for ‘θ’ 2 2 3 δ = 16.F.D .n × D = 8.F.D .n = 8.F.C .n 4 4 G.d 2 G.d G.d θ = T .l

n eq 11.3/pg139

(b’coz C = D/d)

3 3 Therefore δ = 8FD n = 8FC n 4 Gd G.d

n eq 11.5a/pg139

Where n – number of active coils. Design Procedure for Helical Springs. 1. Diameter of the wire: d = ((8FDK)/πτ)

1/3

2. Mean diameter of the coil: D = cd (a) Outer diameter of the coil: Do = D + d (b) Inner diameter of the coil: Di = D – d 4 3. Number of coils: I = yGd 3 8FD

n eq 11.6/ pg139 n eq 11.20a/ pg142 n eq 11.7a/ pg139

4. Free length: lo ≥ (i + 2) d + y + a 5. Stiffness or Spring Rate: Fo = F/y 6. Pitch: p = (lo - 2d)/i

------------ table 11.7 / pg 152

PROBLEMS (1) Design a helical compression spring to support an axial load of N. The deflection under load is limited to 60 mm. The spring index is 6. The spring is made of chrome-vanadium steel & FOS = 2. n Sol : - Given data; - F = 3000N y = 60mm c=6 FOS = 2 Mat - Chrome - Vanadium Steel From table 11.8/pg 153, for Chrome-Vanadium steel, 3 2 τy = 690N/mm & G = 0.07485 × 10 N/mm = 78.45 × 10 N/mm 2 τ = τy/ FOS = 690/2 = 345 N/mm 1. Diameter of wire n --------- eq 11.1a/ pg139

τ = 8FDK 3 πd d=

8FDK

1/3

n --------- eq 11.1b/ pg139

πτ But K =4C - 1 + 0.615 4C - 4

C

C = 6 therefore, k = 4×6 – 1

+ 0.615

= 1.2525

4×6 – 4 Also c = D/d

6

⇒ 6 = D/d or D =6d.

Therefore 345 = 8×3000×6d×1.2525 3 π×d Solving, d = 12.89mm Select standard diameter of the wire from table 11.3a/pg 150. Standard diameter size = d = 13mm

2.

Diameter of the coil:C =D/d 6 = D/13 D = 78mm --- Mean diameter of the coil. Outer dia: Do = D + d = 78 + 13 = 91mm Inner dia : Di = D - d = 78 - 13 = 65mm

3.

Number of coils or turns: 4 i = yGd 3 8FD

n eq 11.6 / pg 139

3 4 = 60×78.45×10 ×13 = 11.8 3 8×3000×78 i = 12 4. Free length:lo ≥ (i + 2)d + y +a

n eq 11.20a / pg 142

a = 25% of max. deflection = 25×60/100 = 15mm lo ≥(12 + 2)13 + 60 +15 lo ≥257mm 5. Pitch:- Assuming square and ground end P = (lo – 2d)i = (257 – 2×13)/12 = 19.25mm 6. Stiffness/spring rate:Fo = F/y = 3000/60 = 50 N/mm Spring specifications (i) Wire dia - 13mm (ii) Mean dia - 78mm (iii) Free length - 257mm

l (iv) Total no of coils – i = i+2 = 12 + 2 = 14coils (v) Style of ends = squared & ground (vi) Pitch - p – 19.25mm (vii)Spring rate – Fo = 50N/mm (viii)Material – Chrome- vanadium (2) Design a helical compression spring for a max. load of 1000N for a deflection of 25mm using 2 the spring index as 5. the max permissible shear stress for spring wire is 420 N/mm & G = 3 2 84×10 N/mm n Sol : - Given data; - F = 1000N y = 25mm c=5 3 G =84 × 10 N/mm 2 τ = 420N/mm 1. Diameter of wire n --------- eq 11.1a/ pg139

τ = 8FDK 3 πd d=

8FDK

1/3

n --------- eq 11.1b/ pg139

πτ But K =4C - 1 + 0.615 4C - 4

C

C = 5 therefore, k = 4×5 – 1 4×5 – 4 Also c = D/d

+ 0.615

= 1.31

5

⇒ 5 = D/d or D =5d.

Therefore 420 = 8×3000×5d×131 3 π×d Solving, d = 6.3mm Select standard diameter of the wire from table 11.3a/pg 150.

4.

Diameter of the coil:C =D/d 5 = D/6.3 D = 31.5mm --- Mean diameter of the coil. Outer dia: Do = D + d = 37.8mm Inner dia: Di = D - d = 25.2mm

5.

Number of coils or turns: 4 i = yGd 3 8FD

n eq 11.6 / pg 139

3 4 = 25×84×10 ×6.3 ≈ 14 3 8×1000×31.5 i = 14 4. Free length:lo ≥ (i + 2)d + y +a

n eq 11.20a / pg 142

a = 25% of max. Deflection = 25×25/100 = 6.25mm lo ≥ (14 + 2)6.3 + 25 +6.25 lo ≥ 131.2mm 5. Pitch:- Assuming square and ground end P = (lo – 2d)i = (131.2 – 2×6.3)/14 = 8.75mm 6. Stiffness/spring rate:Fo = F/y = 1000/25 = 40 N/mm Spring specifications (i) Wire dia – 6.3mm (ii) Mean dia – 31.5mm (iii) Free length – 131.2mm l (iv) Total no of coils – i = i+2 = 14 + 2 = 16coils (v) Style of ends = squared & ground (vi) Pitch - p – 8.75mm (vii)Spring rate – Fo = 40N/mm (3) A railway carriage weighing 40KN & moving at 8km/hr is to be brought to rest by 2 buffer springs. The compression between the coils must be twice the wire diameter. Assume spring index as 8. And allowable shear stress for the spring material = 450N/mm. Take G = 0.8 * 10 N/mm. Design the spring? n Sol : Given data:w- Weight of the carriage - 40 * 10 N v - Velocity - 8km/hr = /3600) m/s = 2.22m/sec n - Number of springs = 2 y = 500mm = 0.5 m Clearance = a = 2d

τ = 450N/mm; G = 0.80 * 10 N/mm c=8 K.E imported on the two springs due to the impact 2 2 U = 0.5mv = 0.5wv /g 3 2 = 0.5 ×40 × 10 × 2.22 /9.81 = 10047.7N-m

K.E on the each spring = 10047.70/2 = 5023.85 N-m If 'F' is the gradually applied force which would defect the spring by 0.5m, then energy stored U = F × y/2 i.e., 5023.85 = F × 0.5/2 F = 20095.4N ≈ 20.09KN (i). Wire diameter C =8 (given) Therefore K = 4C – 1 + 0.615 = 4 × 8 - 1 + 0.615 = 1.184 4C – 4

C

4 ×8 - 4

8

w.k.t. τ = 8FDK = 8FCK 3 2 πd πd 550 =8×2489×6×1.12525 2 πd ⇒ d = 9.306mm ≈ 9.5mm (standard value) (ii) Mean diameter D = c.d D = 6 × 9.5 = 57mm Outside dia = Do = D +d = 57 + 6 = 66.5mm Inside dia = Di = D – d = 57 – 6 = 47.5mm (iii) Number of coils

(b’coz D/d = c)

4 i = Gd y 3 8FD

n eq 11.5a/ pg139

3 4 i = 3.5×84×10 ×9.5 3 8×166×57

= 9.73 ≈ 10

(iv) Free length: lo n eq 11.20a/ pg152

lo ≥ (i + 2)d +y + a a = 0.25y = 0.25×3.5 = 0.875mm lo ≥ (10 + 2)9.5 + 3.5 + 0.875 lo ≥ 118.375mm (v)Pitch: - Assume Square & Ground ends P = (lo – 2d)/i P = (118.375 – 2 × 9.5)/10 P = 9.9375mm

n eq 11.7a/ pg139

(vi)Spring Rate: - Fo =F/y Fo = 166 /3.5 = 47.42N/mm Spring specifications (i) Wire dia - 33mm (ii) Mean dia - 264mm (iii) Free length - 1160mm l (iv) Total no of coils – i = i+2 = 16 + 2 = 18coils (v) Style of ends = squared & ground (vi) Pitch - p - 68.375mm (vii)Spring rate - F = 40.18N/mm

(4)A load of 2KN is dropped axially on a helical spring from a height of 250mm. the spring has 20 turns, & it is made of 25mm diameter wire. The spring index is 8. Find the max. Shear stress induced in the spring 7 the amount of compression produced. Take G = 82.7GN/mm n Sol : - Given data: W = 2KN = 2000N; h= 250mm i = 20 turns; d = 25mm; c = 8 τ = ? y =? G = 82.7GN/m 3 = 82.7 × 10 N/mm Potential energy of falling weight = U = W (h + y) = 2000(250 + y) Energy absorbed by spring = U = Fy/2

Equating the above two equations: Fy/2 = 2000(250 + y) Fy = 4000(250 + y) 6 Fy = 4000y + 10 ------------ (1) 3 3 Now, Deflection: y = 8FD i = 8FC i 4 Gd Gd 3 y = 8×F×8 ×20 3 82.7×10 ×25 y =0.03962 F

------------- (2)

Substituting eqn (2) in (1) for ‘y’ 2 6 0.03962F – 4000(0.03962F)- 10 = 0 2 6 0.03962F – 158.48F – 10 = 0 F = 7407.37N Now, K = 4C – 1 + 0.615 4C – 4

but c = 8

C

K = 4×8 – 1 + 0.615 4×8 – 4

8

K = 1.184 Shear stress = τ = 8FDK = 8FCK 3 2 πd πd τ = 8×7407.37×8×1.184 2 π × 25 2 τ = 285.876N/mm Therefore y = 0.03962F = 0.03962×7407.37 y = 293.47 ≈ 293.5mm 5). Design a helical spring for a spring loaded safety valve for the conditions: (i) Diameter of the valve = 65mm (ii)Operating pressure

2 = 0.7N/mm

(iii) Max. Pressure on

2 = 0.75N/mm

the valve (iv) Max. lift of the valve when pressure = 3.5mm

following

rises from 0.7 to 0.75 N/mm

2

(v) Max. Allowable stress = 550MPa (vi) Spring index

=6 ----------------- VTU/August-2005

n

Sol :Given data:Let D1 = dia of the valve = 65mm P1 = Operating pressure = 0.7N/mm 2 P2 = Max. Pressure = 0.75N/mm

2

y = Max. Deflection = 3.5mm τ = allowable shear stress = 550N/mm 3 2 G = 84×10 N/mm

2

C=6 Let ‘F1’ be the initial force due to operating pressure & ‘F2’ be the force at Max. Pressure. Therefore F1 = P1 × A 2 = 0.7 ×π × 65 /4

III

ly

F1 = 2322.8 ≈ 2323N 2 F2 = P2 = 0.75× π×65 /4 = 2489N

Therefore, Force which produces deflection of 3.5mmis F = F2 – F1 = 2489 – 2323 = 166N Note:- Always design the springs for maximum load & maximum deflection. Maximum force = F2 = 2489N (i). Wire diameter C =6 Therefore K = 4C – 1 + 0.615 = 4 × 6 - 1 + 0.615 = 1.2525 4C – 4

C

w.k.t. τ = 8FDK = 8FCK 3 2 πd πd 3 450 =8×20.09×10 ×8×1.184 2 πd ⇒ d = 32.81mm ≈ 33mm

4 ×6 - 4

6 ( b’coz D/d = c)

(ii) Mean diameter D = c.d D = 8 × 33 = 264mm Outside dia = Do = D +d = 264 + 8 = 272mm Inside dia = Di = D – d = 264 – 8 = 256mm (iii) Number of coils 3 y = 8FD i 4 Gd

n eq 11.5a/ pg139

3 3 500 = 8×20.09×10 ×264 ×i. 5 4 0.8×10 ×33 i = 16.04 ≈ 16 (iv) Free length: lo n eq 11.20a/ pg152

lo ≥ (i + 2)d +y + a lo ≥ (16 + 2)33 + 500 + 2 × 33

(given a = 2d)

lo ≥ 1160mm (v)Pitch: - Assume Square & Ground ends P = (lo – 2d)/i P = (1160 – 2 × 33)/16 P = 68.375mm (vi)Spring Rate: - Fo =F/y

n eq 11.7a/ pg139

3 Fo = 20.09 × 10 /500 = 40.18 N/mm Spring specifications (i) Wire dia – 9.5mm (ii) Mean dia - 57mm (iii) Free length – 118.375mm l (iv) Total no of coils – i = i+2 = 10 + 2 = 12coils (v) Style of ends = squared & ground (vi) Pitch - p – 9.9375mm (vii)Spring rate - F = 47.42N/mm 6). The valve spring of an I.C Engine is 40mm long, when the valve is open & 48mm long when the valve is closed . The spring loads are 250N when the valve is closed & 400n when the valve is open. The inside diameter of the spring is not to be less than 25mm & take FOS = 2. Assume 3 2 spring index to be 6 &G = 79.34×10 MPa & τy = 690N/mm . Design the spring.

n Sol :Given data:- F1 = 250N; F2 = 400N; Di = 25mm; FOS = 2; C = 6; 3 2 G = 79.34×10 N/mm ; 2 2 τy = 690N/mm ⇒ τall = 690/2 = 345N/mm l l Let y be the deflection of the spring between opening & closing of the valve. Therefore, y = 48 – 40 = 8mm Now Max. Deflection is given by l y2 = (F2y )/ (F2 – F1) = 21.33mm

----- refer fig 11.2a / pg156

Design the spring for max. Load & max. Deflection. (i). Wire diameter C =6 Therefore K = 4C – 1 + 0.615 = 4 × 6 - 1 + 0.615 = 1.2525 4C – 4

C

4 ×6 - 4

w.k.t. τ = 8FDK = 8FCK 3 2 πd πd

6 (b’coz D/d = c)

345 =8×400×6×1. 2525 2 πd ⇒ d = 4.710m ≈ 5.00mm (ii) Mean diameter D = c.d D = 6 × 5 = 30mm Outside dia = Do = D +d = 30 + 5 = 35mm Inside dia = Di = D – d = 30 – 5 = 25mm (iii) Number of coils 3 y = 8FD i 4 Gd

n eq 11.5a/ pg139

3 21.33 = 8×400×30 ×i. 5 4 0.79×10 ×5 i = 12.24 ≈ 13 (iv) Free length: lo lo ≥ (i + 2) d +y + a lo ≥ (13 + 2)5 + 21.33 + 0.25 × 21.33 lo ≥ 101.66mm

n eq 11.20a/ pg152

(v)Pitch: - Assume Square & Ground ends P = (lo – 2d)/i P = (101.66 – 2 × 5)/13 P = 7.05mm n eq 11.7a/ pg139

(vi)Spring Rate: - Fo =F/y Fo = 400/21.33 = 18.75 N/mm Spring specifications (i) Wire dia – 5mm (ii) Mean dia - 30mm (iii) Free length – 101.66mm l (iv) Total no of coils – i = i+2 = 13 + 2 = 15coils (v) Style of ends = squared & ground (vi) Pitch - p – 7.05mm (vii)Spring rate - F = 18.75N/mm (viii) Allowable shear stress = 345N/mm

2

Assignment:(1) A helical spring made from 6.3mm diameter steel wire has an outside diameter of 57.3mm with squared & ground ends and has 12 coils. The allowable shear stress is 827MPa. Determine the following (i) Spring rate (ii) Free length (iii) Pitch (2) The following data refers to the valve of a petrol engine Length of the spring when the valve is open – 40mm Length of the spring when the valve is closed – 48mm Spring load when the valve is closed – 350N Spring load when the valve is open – 220N Spring index – 6.8 (3) The maximum shear stress allowed is 150MPa & the modulus of rigidity is 84GPa. The ends are squared & ground and the gap between the adjacent coils is 0.1 times the wire diameter. Determine the following (i) Wire dia (ii)Mean dia

(iii) Number of coils (iv) Free length (v) Pitch