Structural Analysis (ds).doc09_10

Statically determinate structures

Statically determinate structures

Department of Civil Engineering, University of Ilorin +2348033774616 [email protected]. ngaadeoladeji@yahoo. com,

Adeola A Adedeji This course deals with statically determinate structures. In this volume theories are stated and problems are solved. In general, the analyses involved beams, trusses, frames and polygons element are discussed here.

f F

F D

d D

a A

A A Adedeji

Statically Determinate Structures

Department of Civil Engineering Faculty of Engineering and Technology University of Ilorin, Ilorin Kwara State, Nigeria www.unilorin.edu.ng Course:

CVE 365 -Structural Analysis I ( 2 Credits/Compulsory) A A Adedeji M.Sc. (Prague), Ph.D. (ABU, Zaria), MNIEM., MNICE, R. Engr. MIAENG [email protected]; [email protected] Block 8, F29, Main Campus, University of Ilorin, Ilorin

Lecturer:

E-mail: Office Location:

f F d

F

D

D

a A © 2010

2

A

A A Adedeji

Statically Determinate Structures

Statically determinate structures

A A Adedeji DEPARTMENT OF CIVIL ENGINEERING UNIVERSITY OF ILORIN

2010 3

A A Adedeji

Statically Determinate Structures

Contents

PREFACE CHAPTER ONE INTRODUCTION CHAPTER TWO STATICALLY DETERMINATE STRUCTURES 2. ESTABLISHING DEGREES OF STATIC DETERMINACY OF STRUCTURES 2.1 Plane system 2.2 Joint for component systems 2.3 Degree of statistical indeterminate (or determinate) in a plane 2.4 Statical and form-determinate system of joint and rigid member 2.5 Degree of statical indeterminacy/ determinacy in space 2.6 Statical and form-kinematical determinacy of trusses with hinge joints 2.7 Graphical method of analysis 2.7.1 Characteristics of statically indeterminate structure

CHAPTER THREE REACTIONS AND SUPPORTS OF ELEMENTS 3.1 Reactions by calculation 3.2 Graphical solution for obtaining system reactions 3.3 Examples 3.1 to 3.3 3.4 Calculation of reactions of a rigid body 3.5 Reactions of a three-hinged frame 3.5.1 Three-hinged frame 3.5.2 Continuous (Gerber’s) beam

4

Page 6 7 7 10 10 10 10 13 14 16 22 24 28 28 34 34 34 34 35 43 49 50 58

A A Adedeji

Statically Determinate Structures

3.6 Calculation of reactions for complex systems with 3 or more rigid members and joints

CHAPTER FOUR BEAM ELEMENTS 4.1 Straight beam 4.2 Simple beam 4.2.1 Examples 4.1 to 4.2 4.3 Cantilever beam 4.3.1 Examples 4.3 – 4.6 4.4 Inclined load and beam 4.4.1 Examples 4.7 to 4.10

Annex A Course Programme

Annex B Lectures

Annex C Practicals

Bibliography

5

61 67 67 67 69 69 70 71 72 72 77 77 79 79 84 84 85

A A Adedeji

Statically Determinate Structures

PREFACE (CVE 365 Structural Analysis I) This book is an instructional text relating to the structural analysis of statically determinate structures for 30hrs period. The course is prerequisite to engineering mechanics, for the students of civil and related engineering courses in the Faculty of Engineering and Technology, University of Ilorin, Ilorin, Nigeria. The book has been planned to serve as a basic and thorough knowledge in structural mechanics. The course is sectioned to be followed sequently. The structural analysis (I) is a course of study that deals with statically determinate structures of plane, special, truss, freepolygon and centenary. In this student’s edition, theories are taught and problems are solved for the statically determinate structures. In general, the analyses apply to beams, trusses, frames and polygons are discussed here. Graphical solutions to problems are presented to enhance student’s understanding of the application of technical drawing knowledge.

A.A. ADEDEJI ©2010 First produced 2006

6

A A Adedeji

Statically Determinate Structures

CHAPTER ONE INTRODUCTION In a statically determinate structure it is sufficient to compute, correctly, required reactions, forces and moments by simply applying the conditions of equilibrium. In this analysis we do not have to know the sectional dimensions and materials properties of the elements under consideration. Such quantities can be computed until only that time when we have to calculate the deformations of the structural elements. If, in a considered structure with joints, more degrees of freedom are restrained than what its individual free elements of the structure have, we talk of statically indeterminate member systems, and we should further engage in their calculation. In the statically indeterminate structures, the internal forces and reactions cannot be calculated only by the conditions of equilibrium, therefore further equations for deformation of structures have to be involved. An indeterminate or redundant structure is one that possesses more unknown member forces or reactions than the available equations of equilibrium. These additional forces or reactions are termed redundants. To determine the redundants, additional equations must be obtained from conditions of geometrical compatibility. The redundants may be removed from the structure, and a stable, determinate structure remains, which is known as the cut-back structure. External redundants are redundants that exist among the external reactions. Internal redundants are redundants that exist among the member forces. Under the concept of a building structure it is understood that a structural system is composed of basic elements - bar, beam, plate (slab), wall, shell etc and are joined together by bonds (rigid joint, movable placement, hinges etc) so as to

7

A A Adedeji

Statically Determinate Structures

withstand, safely, the presumed or applied loads. In this analysis we discussed only the structures having members with straight, constant or variable sectional elements. Such structures are referred to as elemental systems. In order to stabilize the support of an object, number of joints and its arrangement must be used so that it can eliminate all its degree of freedom. If a structural object has m degrees of freedom and its reactions eliminates r degrees of freedom then the difference: S= m - r

(1.1)

This is the degree of shape variability of the object supports. A reaction forming in a determinate joint occurs by the number of the unknown parameters, i.e. how much is the degree of freedom of the joint? The unknowns are the values of the component reactions, their sum is equivalent to degree of freedom r, which can prevent a joint from free movement. The values of statically determinacy of elements with respect to the support and determinate conditions are shown in Table 1.1. Table 1.1 Determination of statically determinacy of structural elements Determinacy

S= m - r

Condition

Support

>0

Form Indeterminate

= 0

Determinate

Statical Overdeterminate Determinate

<0

OverDeterminate

Indeterminate

Determinate Condition

___ ∆≠ 0 ___

Statically and form-kinematically determinacy are characterized by the relationship S = O, which is the very

8

A A Adedeji

Statically Determinate Structures

condition required for equilibrium, so that determinant of the system ∆ ≠ 0. And If ∆ = O an exceptional case occurs. An exception case of S support for a load bearing structure is a disadvantage.

9

A A Adedeji

Statically Determinate Structures

CHAPTER TWO STATICALLY DETERMINATE STRUCTURES 2. ESTABLISHING DEGREES STRUCTURES

OF

STATIC DETERMINACY

OF

Structural systems comprising of members (elements) can have its members joined thoroughly (mostly by the number of external or internal joints) at their ends. This may not necessarily be important for the stability of the support (stable structures). 2.1 Plane system Placing a joint material in a plane can be expressed as two components (projection to x, y-axis). It has two degrees of freedom, that must be eliminated or restrained in order to get a stable support or joint by removing one degree of freedom and drawing-in one component reaction. These are: (a) Swinging member:– member with reactions acting along the axis of the member Fig. 2.1.

(a) Column on balls (b) Beam-column ( c) Schematic representation Fig. 2.1 Swinging members

10

A A Adedeji

(b)

Statically Determinate Structures

Member supported on rollers (or movable hinge) – reaction acting perpendicular to the placement. Fig. 2.2

(a) A member on a roller

(b) A member on rollers

(c ) Diagramatical Representation

Fig. 2.2 Rollers (Movable hinges) (c)

Member placed in a movable guide (along a line or curve) – its reaction is acting perpendicular to guide. - Double – removes two degrees of freedom and draws in two component reactions. This involves a movable fixed-end support. Fig. 2.3.

Pin supports

Schematic representation Fig.2.3 Pin supports

11

A A Adedeji

Statically Determinate Structures

Triple – removes three degrees of freedom. It is rigidly fixed with rigid plate having its support built-in (encased or fixed) or rigidly fixed end. It has three component reactions, two forces and one moment. Note If a joint material will be supported as statically determinate bonds must restrain, rightly, two degrees of freedom. In supporting a material point in plane, two simple bonds are used. A plate in plane (straight element, curve bar, etc) must not depend on the calculation as a statically indeterminate element. If we consider such a plate as an independent object from a freely chosen point a to whichever of the points (such as point b in Fig. 2.4) we can attain, along the centerline, more than one path. From a free plate we can form a simple plate or more sections for some simple plates.

b

- b - a

(a) A closed member

- a

Α - a

(b) a cut in point b

bβ

(c) cuts in two-point

Fig. 2.4 A closed structure If supported, a rigid plate could be statically determinate where bonds must restrain three degrees of freedom and must be arranged in such a way that it will not form exceptional cases as when the directions of rays of reaction

12

-

A A Adedeji

Statically Determinate Structures

intersect in one point whereby rotating such a plate. See Fig. 2.5.

Fig. 2.5 Rigid bodies Simple plate, in plane, has three degrees of freedom :  Supports of fixed end (encastre’) removes three degrees of freedom (internal built-in),  Bond by a moving hinge (along a line or curve) restrains possible transfer in the direction of normal to the line or curve. It removes one degree of freedom.  Swinging member restrains a possible transfer in the direction of joint of the hinges of the member and one degree of freedom could be removed.  Movable fixed end will restrain a plate rotating and moving in the direction of normal to the direction of such movement and two degrees of freedom could be removed. 2.2 Joint for component systems Bonds are used to stiffen supports or rigid plate and joint material (rigid hinge, movable hinge placement, swinging members).

13

A A Adedeji

Statically Determinate Structures

Bonds can be either external or internal. External bonds are referred to as ones that join parts of system rigidly with supports while the internal bonds join one part to another. Component systems can form a movable mechanism. From movable mechanism for a building structure and it is better to use only the fibre polygons and chains in this respect. Building structures should be immovable (static) systems that is why we must not go about systems that can exert virtual movement or movable and dimensions even if it is in accordance with the calculations of form – kinematical determinate or over-determinate. Systems can be either free without external bond or supported with external bonds. Free rigid plate has in plane three degrees of freedom, material point two. If the component system is established of β (material point) and δ (rigid plate), then degree of freedom m = 2β + 3δ

(2.1)

Note Bond, which restrains only one degree of freedom is a simple bond. This includes swinging members, movable hinge, and movable guide along line or curve. Stiff hinge if it joins two parts (2 rigid plates or a rigid plate and a rigid support) is double bond and so restrains two degrees of freedom we say it is a simple hinge. If one hinge replaces two or more simple hinges, lying infinitely near each other. It is better to replace it (n-1 ) by a simple hinge ( a  n  1 ),

2

where n is the number of joined rigid plates, a is the built–in joining two parts (elements, rigid plates) is of three bonds which restrains three degrees of freedom.

2.3 Degree of statistical determinacy in a plane If parts of a component system are in between each other and are placed on supports 1 (simple or one bond), 1

14

A A Adedeji

Statically Determinate Structures

(double bond/joint) and 3 (triple bond), then the restraint is expressed as, r = 3α 3 + 2α 2 + 1α1 + 2α

(2.2)

The system has degrees of freedom: δ = 2β + 3α 2 – ( 3α 3 + 2α

2+

1α 1+ 2α ) = m + r

(2.3)

where β = number of points (material points ) of an ideal form, δ = number of plates(bodies), α3 = number of fixedends (built–in ) bond, α2 = number of simple hinge joint, α1 = number of movable hinges, pin joint or swinging members, and α = number of movable fixed-ends. Conditions  If s > 0, then the system forms movable mechanism and then it is referred to as form– kinematically indeterminate and statistically over-determinate.







If S= 0, then the system is immovable, then it is form- kinematically and statically determinate if only their e is no conditional case (if Δ = 0) If s < 0, then the system is unsuitable, structure is deficiently bonded, in general, acting force can set the structure in motion Conditions that S = 0 > are not enough to say that the system is stable.

Further conditions are: a) External and internal bonds are included in the computation of bond. b) Minimum of three units of bonds must be applied to external rigid formation (1 built-in; 1 hinge + 1 movable bond; 3 movable bonds)

15

A A Adedeji

c)

d)

Statically Determinate Structures

No part of a system must be form-kinematically indeterminate – must not have deficiency bonding, even if a part was unnecessarily (abundantly) bonded External and internal bonds must not form conditional cases (identified by formulation of large forces in bonds).

Equation for obtaining static and form-kinematic indeterminacy can also be expressed as shown in Fig. 2.6.

S = 2β + 3

–

33 + 22 + 1 + 2

(2.4)

No. of degree of freedom - No. of restraints for degrees of freedom by bond No. of condition of equilibrium - No. of unknown component reaction Fig. 2.6 Determination of degree of freedom in structures

2.4

Statical and form-determinate system of joint and rigid member

Example 2.1 Obtain the statical and form-kinematical determinacy of the following component system. See Fig.2.7.

16

A A Adedeji

Statically Determinate Structures

OI,V II

OIII,V d

III

c

e i

ii f

iv

vi g

I iii

v

Fig. 2.7 A component system Solution 1 a) The system has: δ = 4 (plates I to IV), β= 2 ( material points, f, g), α2= 5 (stiff hinges a, b, c, d, e,) and α1= 6 (swinging members, i to vi), therefore degree of freedom s = 3 x 4 + 2 x 2 – ( 2 x 5 + 1 x6) = 16- 16 = 0. Therefore the hinge at 0II,III virtual hinges at 0III,V and 0III,V did not lie on one line. The system is statistically and form-kinematically determinate. Solution 2 b) The unloaded rigid plates I and IV can be used as swinging members (i to vi + I and IV) and 1 stiff hinge so that the degree of freedom s = 3 x 2 + 2 x 2 – (2 x 1 + 1 x 8) = 10 –10 = 0. The system is statistically and form-kinematically determinate. Example 2.2 Obtain statical and form kinematical determinacy of the system shown in Fig.2.8.

17

A A Adedeji

Statically Determinate Structures

b a

(iv

(ii

(ii (x (i)

c

(ix

d (v) (viii) (vii) (vi

I

II

(xi)

(xii Fig.2.8 Truss system

Solution The system has: 2 member decks (i.e. I, II) = δ, 4 material points β (a-d), 1 stiff hinge α2, 12 swinging members α1 (IXII) so that, δ= 3x2 + 2x4 – (2x1 + 1x1) =14 – 14 = 0. Therefore the system is statically and form-kinematically determinate. Example 2.3 Obtain statical and form-kinematical determinacy of the system in Fig. 2.9a.

δ II, IV

c

a (a)

I

II, III

III

III, IV b

b (b)

IV

Fig. 2.9 Frame object and its analysis

18

A A Adedeji

Statically Determinate Structures

Solution See Fig.2.9b The system has: 3 plates (5 labs) or members - δ (I, II, III), 3 stiff hinges (α2) and 3 swinging members (α1) so that S

= 3 x 3 - (2 x 3 + 1 x 3) = 9–9=0

The Virtual hinges 0II,IV; 0II,III; 0III,IV do not lie on a line. Therefore the system is statically and form-kinematically determinate. Example 2.4 Establish degree of statical and form-kinematical determinacy in the structure. See Fig. 2.10.

(a)

III I

II

(b) Fig. 2.10 Arch structure Solution The system has: δ = 4, β = 0, α3 = 4, α2 = 3 :. S = 3 x 4 + 2 x 0 - (3 x 4 + 2 x 3) = 12 - 18 = -6

19

IV

A A Adedeji

Statically Determinate Structures

The structure is 6x statically indeterminate and formkinematically Over-determinate. Example 2.5 Obtain statical indeterminacy of the frame shown in Fig. 2.11

b

a Fig.2.11 A frame

Solution From Fig. 2.11, δ = 1, α3 = 1, α2 = 1 :. S

= 3 x 1 - (3 x 1 + 2 x 1) =3-5 =-2 The system is 2x statically indeterminate. Example 2.6 Fig. 2.12 shows a closed frame, find its degree as statical indeterminacy.

q1 (KN/m)

X2

X3 q2 (KN/m1)

X1

Fig. 2.12 A Closed structure

20

XI1

XI3 XI2

A A Adedeji

Statically Determinate Structures

Solution From Fig. 2.12 S

= - (2 x 1 + 1 x 1) =-3

The structure is 3 times (3 x) statically indeterminate Example 2.7 Find the degrees of statically indeterminate of the frame shown in Fig. 2.13

P

P

X3 X2 X5

X1 X6 X4

X3 X6

X1

X5 X4

Fig. 2.13 Frame structure S

= 6δ =6x2 = 12 x statically indeterminate

Exercises 2.8 to 2.13 Obtain degree of statical and form-kinematical determinate or indeterminate structures Figs 2.14 to 2.15 respectively.

21

A A Adedeji

Statically Determinate Structures

(a) 1.8 (b) 1.9

© 1.10

(d) 1.11 Fig. 2.14 Complex structures (Examples 2.8-2.11)

(e) 1.12 (f) 1.13 Fig. 2.15 Frame structures (Examples 2.12 – 2.13) 2.5

Degree of statical indeterminacy/determinacy in space Material point has, in space, three degrees of freedom; rigid body is degrees of freedom. In order that the material object

22

A A Adedeji

Statically Determinate Structures

becomes rigidly or freely supported, the degrees of freedom must be restrained (removed) by bonding. The restraints are: 









Swinging member restrains one degree of freedom and draws in one component reaction in the direction of axis of the member. Movable hinge (along area of placement restrains one degree of freedom and draws in one component reaction in the direction perpendicular to the bond (guide). Spherical hinge restrains three degrees of freedom, draws in three component reactions (such as in the direction of coordinate axis) or reaction of unknown size and direction. Prismatic hinge restrains four degrees of freedom. If the cylindrical hinge, it depends on restraints of three degrees of freedom along a line of action. Fixed-end (Built-in) restrains all the six degrees of freedom draws in six component reactions (three forces and three moments).

Table 2.1 shows, at a glance, degrees of statical indeterminacy S of some structures and bonds in plane and special structures.

Table 2.1 Degree of statically indeterminacy Structure Plate Point Built-in Hinge (fixedend) In plane In space

30 60

20 30

30 60

Symbols

Δ

Β

α3

23

20 30 - spherical 40 -cylindrical α2

A A Adedeji

Statically Determinate Structures

Structure

Movable hinge

Swinging members

Movable hinge

Movable built-in

Table 2.1 continued

In plane In space Symbols

10 20 α1

10 10 α1

10 α1

20 50 Α

For degrees of statical indeterminate for a special system S = 6 + 3 – (63 + 32 + 21 + 11 + 11 + 5)

(2.4)

2.6

Statical and form-kinematical determinacy of trusses with hinge joints The actual case of component system is a member (truss) system with hinge bond. Member system with hinge bonds (connections) is a component of material points (joints) mutually joined together by swinging elements. We can use, for obtaining statical and form-kinematical determinacy members of rigid plates mutually as joint hinges. The degrees of freedom (Equation 2.4a): S = 2β – (2α2 + α1)

(2.4a)

In the truss (member) systems with hinge bonds sometime in place of α1 (number of simple external and internal bonds) we use π + α1, where π denotes the number of members (of internal bonds-swinging members) and movable number of simple bonds (swinging members placement). Then: S = 2β – (π + α1 + 2α2)

(2.5)

Places where one element crosses and didn’t join with another can be represented by arch

24

A A Adedeji

Statically Determinate Structures

Example 2.14 Find the number of statical indeterminate/determinate of the structure in Fig. 2.16

e f c

d

g

a

b

β = 5(c, d, e, f, g) π = 10 (no. of members) :. S = 2 x 5 – (1 x 10) =0 The system is statically determinate.

Fig. 2.16 Frame structure Example 2.15 Obtain degrees of statical indeterminacy (determinacy) at the member system in Fig. 2.17

c

a

d

b

f

e Fig. 2.17 Truss system

Solution: Member systems with hinge bonds has: α2 = 1, α1 = 1 and π =10

25

A A Adedeji

Statically Determinate Structures

S = 2 x 6 – (10 + 1 + 2 x 1) = 12 –13 =-1 The system is 1x statically indeterminate

Example 2.16 Find the degrees of statical determinacy of the system shown in Fig. 2.18.

h a

g

c

d

i

j

f e

b

Solution In the system: β = 11 α2 = 1 α1 = 1 π = 20

k

Fig. 2.18 Truss system S

= 2 x 11 – (20 + 1 + 2 x 1) = 22 – 23 = - 1 (i.e 1x statically indeterminate).

Example 2.17 Obtain statical and form-kinematical determinacy of the component system in Fig. 2.19.

26

A A Adedeji

Statically Determinate Structures

b

a

c

e

d

Fig. 2.19 A component structure Solution: The system has: β = 5(a, b, c, d, e), α = 2(a, b) and π = 4 S

= 2β – (π+2α2) = 2 x 5 – (4 + 2 x 2) = 10 - 8 =+2 0 x = the system is statically overdeterminate and formkinematically indeterminate. It is a movable system (fibre polygon) Example 2.18 Obtain the statical and form-kinematical determinacy of the system in Fig. 2.20

g

f

d

i

k

n e

h

j

p

l o

c a

m

b Fig. 2.20 Truss system framework

27

q

r

A A Adedeji

Statically Determinate Structures

P2 = 5kN C 2m

P1 = 4kN 0

P1

P3= 2kN

6m

2m

60

1m

B A 300

1m

2m

1m

Fig. 2.21 A solid element Solution: β = 18= 18 α2 = 2(b, j) α1 = 1 π = 30 = 32 :. S = 2 x 16 – (30 + 1 + 2 x 2) =-3 The system is 3x statically indeterminate and form-kinematically over determinate ( - 3 < 0) 2.7 Graphical method of analysis 2.7.1

Characteristics of statically indeterminate structure

Example 2.19 Use graphical (Cullman’s) method, find the reactions in Fig. 2.21

28

A A Adedeji

Statically Determinate Structures

A = 2.7 kN B = 4.2 kN C = 2.0 kN PR = Resultant load RR = Resultant reaction

Solution

P C (2

P

P

RR

(1 PR (4 A

A

(3 P

R R

C B

B

Fig. 2.21a Graphical solution Example 2.20 a) Use Cullman’s Method to find reactions in Fig.2.22 0 A (kN) 30

(a

F

P2 = 3 kN 600

D

2m

2m

P3 = 5kN

E B (kN)

2m

P1 = 4 kN 3m

Fig. 2.22 An L-section plane structure

29

C (kN)

A A Adedeji

Statically Determinate Structures

Solution P3

P2

A R

3

6

B

2 C

4

1 5

Fig. 2.23 Vectoring example 2.20

30

PR

A A Adedeji

Statically Determinate Structures

2..9 kN

A

10

β P1

1

3 kN

9

7 O

2C 3

P2

8

5

4

5kN

PR

Fig. 2.24 Solution to example 2.20

Example 2.21 Use resultant line method, find the reactions of the beam element.

31

A A Adedeji

Statically Determinate Structures

P1

= 5 kN

P2 = 10 kN 450

a

b 0

60

Ax 2m

8m

2m

1m

2m B

Ay

P3 = 2kN Fig.2.25 A simply supported beam

Solution to resultant (Graphical) method:

P1

6.0 kN AX

P2 B b B

A

A

P1

8.2 kN

1 1

1

P2 3

2

Ay 5.6 kN

P3

P3

3 4

2 Fig. 2.26 A simply supported beam

Fig. 2.6a Solution to example 2.21

32

A A Adedeji

Statically Determinate Structures

Example 2.22 C) Use Cullman’s Method to Solve Example 2.21

P2

P1 Ax (3)

Ay

B (2)

(1) P3

Fig.2.27 A beam with loads

Ax

Reactions

Ay Ax = 6.0kN Ay = 8.9kN B = 5.6kN

P1 A (1) (2

R P2

B

P3

Fig. 2.27a Solution to example 2.22

33

A A Adedeji

Statically Determinate Structures

CHAPTER THREE REACTIONS AND SUPPORTS OF ELEMENTS 3.1 Reactions by calculation Primary external forces (loads) and secondary external forces (reactions) act on a rigid plate must be in equilibrium system of force. a)

Total sum (directional) condition of equilibrium along xaxis is written as: n r X : ∑ Fix + Σ Rjx = 0 (3.1) i=1 i=1

b)

Total sum (directional) condition of equilibrium along y-axis

y

:

c)

Moment condition of equilibrium, for instance to the origin of the coordinate system (0) is written as

0 :

n ∑ Fiy i=1,

n ∑ (Xi F i=1

iy

+

r Σ Rjy = 0 j=1

r - y i F iy) + ∑ (Xj Rjy – Yj Rjx) = 0 j=1

(3.2)

(3.3)

3.2 Graphical solution for obtaining system reactions Graphical solution to obtain reaction of a system must be a component of drawings and resultants closed line. This can be

34

A A Adedeji

Statically Determinate Structures

obtained using two methods 1) Cullman and 2) Resultant line methods. 3.3 Examples 3.1 to 3.3 In examples 3.1 to 3.3 use conditions of equilibrium calculate the reactions. Example 3.1 (Fig. 3.1).

P3 = 5kN1 300

A

P2 = 3kN 600

d

a

A

C

2m

3m C

B

P4= 4kN

Fig. 3.1 A plane object Solution (See Fig. 3.1) From the condition of equilibrium X:

– A – P2 Cos 600 + P3 Cos 300 = 0

y:

B +C – P1 – P2 sin 600 – P3 sin 300 = 0

a

P7 x 2 + P3 sin 300 x 2 + P2 sin 600 x 5 + P2 cos 600 x 2 – C x 5 = 0

Solving simultaneously: A = 2.03 KN, B = 3.32KN, C = 5.78KN.

35

A A Adedeji

Statically Determinate Structures

Example 3.2 Find the control for the results in Example 3.1. Refer to Fig.3.1 and the solution to Example 3.1 Solution Using condition of equilibrium, the reactions are: A=2.03KN, B=3.32KN and C = 5.78 kN Control: 3.32x2 = 5700 x 3 – 2 - 3 x 2 + 5x 0.86 x 2 + 3 x 0.866 x 3 = 0 0.09  0 Computed reactions are OK d

:

Example 3.3 Refer to Fig. 2. 26 (as in Fig. 3.2) in Example 2.2.

P2 =10kN

P1 =5kN

45o 1m

Ax a P3 =2kN Ay

b B

2m

3m

2m

Fig. 3.2 A beam element

36

2m

A A Adedeji

Statically Determinate Structures

Solution : P1 x 2 + P2 x 5 sin 450 – P2 x 1 cos 45 + P8 x 7 sin 600 –Bx9=0

a

y:

Ay – P1 – P2 sin 450 – P3 sin600 + B = 0 - P2 cos 450 + P3 sin 600 + Ax = 0

x:

The results of the reactions: Ax = 6.07 kN, Ay = 8.19kN and B = 5.6 kN Control : b :

8.19 x 9 – 6.07 x 1 – 5 x 7 - 7.07 x 4 – 7.732 x 2 1.1 = 0 73.71 – 73.724  0

Example 3.4 Find the reactions and control the results of the element inclined at angle 30 to the horizontal. See Fig. 3.3

Q

P1 = 5 kN b

q = 2KN/m’

P2 = 1kN

300 Ax

B

Ay 2

2m

1

1

1 0

Fig. 3.3 An element inclined at 30 to the horizontal

37

A A Adedeji

Statically Determinate Structures

Solution y = Ay – Q – P1 cos 300 + P2 + B = 0 x = Ax + P1 sin 300 a

G x 2 + P1

5 - B x 6 - Px2 x 8 = 0 Cos 300

The reactions are: Ax = -3.0kN, Ay = 5.1. KN, B = 7 – 1 kN Control b

= 5.1 x 6 – 8 x 4 – 2 x – 6 (1/ Cos 300 ) + 6x 3 tan 300 = 0 40. 96 – 40.94 = 0.06  0

P2 = 3 kN

P1 = 10 kN

1m

2m P3 = 4KN Bx1m

b

a

By 2m

1m

3m

Fig. 3.4 A structural frame

38

A A Adedeji

Statically Determinate Structures

Example 3.5 Find the reactions and control the results of the element. See Fig. 3.4. Solution Condition of equilibrium Y: A+ By –P1 = 0 x - Bx + P3 – P2 = 0 b:

A x 6 – p1 x 4 + p3 x 1 = 0

Reactions : A = 8k kN, By = 2 kN and Bx = 1 kN Examples 3.6 Solve (see section 3 – 4 – 1). Also solve for the reaction by calculation

P2 = 5kN P1 = 4 kN C

600

1m 1m

P3 = 2 kN

2m A 300 B 1m 2m

1m

Fig. 3.5 A solid structure under vertical, inclined and horizontal loads

39

A A Adedeji

Statically Determinate Structures

Example 3.7 (See section 3 .4 . 1 ) Also calculate the reactions See Fig. 3.6

P1 = 4kN

q = 3kN/m

P2 = 6kN

8m 7m B 1m A 4m

2m

1m

Fig.3. 6 A framed structure with an evenly distributed load (q) and two concentrated loads

Example 3.8 to 3.10 In examples 3.8 to 3.10 and in Fig. 3.7 to 3.9, using Cullman’s resultant line methods for each problem, find the reactions of the structures.

40

Statically Determinate Structures

3m

3m

A A Adedeji

P=10KN

Bx

b 2m

a By A

4m

Fig. 3.7 A frame with one leg longer than the other (a)

Cullman’s Method Solution

4 (5 b

Bx

By

Bx R (

A (3)

P

Fig.3.8b Resultant line solution

R (7) P

A a A

A = 2.5kN Bx = 10kN By= 2.5kN

By

Fig. 3.8 (a) Cullman’ method

41

R

Bx By

A A Adedeji

Statically Determinate Structures

Bx P

1 2 a

P

R Bx

By A

2

1

By

A

0 Fig. 3.8 (b) resultant line method Example 3.11 Find the reactions of the structure shown in Fig. 3 .9

Q7 = 6kN

c 2m

P1 = 4kN

C 4m

P2= 2kN 450

a

2m

b

2m

2m

1

Fig. 3.9 A framed object

42

560

4m

A A Adedeji

Statically Determinate Structures

Example 3.12 Find the reactions of the structure shown in figure (Fig.) 3.10

P7=70kN

7.5m

c

A

7m

a

30

5m

0

b 2m

3m

4m

Fig. 3.10 Y-shape frame supported by a tensile member 3.4 Calculation of reactions of a rigid body primary external force (load) a secondary external force (reaction) acting on a solid body must, togetehr, form equilibrium system of forces. For solution, six conditions of equilibrium must be fulfilled. See equations (3.4) t0 (3,6) and Fig.3.12.

43

A A Adedeji

x:

y:

z:

Statically Determinate Structures

n ∑ Fix i=1

+

r ∑ Rjx = 0 j=1

(3.4)

(3.5)

n ∑ Fiy i=1

+

r ∑ Rjy = 0 j=1

n ∑ Fiz i=1

+

r ∑ Rjz j=1

=0

Fi y

(3.6)

Fi

+y

αi Fix i

Fiz yi

+X Xi Zi Xi

+Z Fig. 3.11 Analysis of a point lying in space

44

A A Adedeji

Statically Determinate Structures

Moment Condition of equilibrium to X-axis n r X Σ (Yi Fiz - Zi Fiy) + Σ (Yj Rjz - Zj Rjy) = 0 i=1 j=1

(3.7)

Moment Condition of equilibrium to y-axis n r Y Σ (Zi Fix - Xi Fiz) + Σ (Zj Rjx - Xj Rjz) = 0 i=1 j=1

(3.8)

Moment Condition of equilibrium to z-axis n r Z Σ (αi Fiy - Yi Fiz) + Σ (Xj Rjy - Yj Rjz) = 0 i=1 j=1 where: Fi = axial forces, Rj = reactions

(3.9)

Example 3.12 Find the reactions of the body shown in Fig.3.12.

+y 1

D

1.5m

x

P1=4kN

2m

6 B

Ax a= 0 +Z Z

Az

Ay 3m

2m

2m

C

P2=8kN

+x

Fig.3.12 A structural element loaded by horizontal load

45

A A Adedeji

Statically Determinate Structures

Solution Condition of Equilibrium

x: Ax + D + P1 = 0 y Ay- B - P2 = 0 z: - Az - C = 0 x: P2 x 4 + B x 2 = 0 y: P1 x 2.5 + D x 4 – C x 3 = 0 z: P1 x 2 + D x 2 + B x 3 = 0 In order that the system equation has one result, the determinant of such system must not be equal to zero. The system determinant is expressed as:

=

Ax Ay 1 0 0 1 0 0 (3.10) 0 0 0 0 0 0

Az 0 0 -1

B 0 -1 0

C 0 0 -1

D 1 0 0

0 0 0

2 0 3

0 -3 0

0 4 2

From the above the reactions: Ax= -24kN; B = -16kN;

Ay 8kN; C =30kN;

Az = -30kN D=20kN

Control x1: 4Ay - 2Az - 2B - 2C = 0 y1: 2D + 0.5P1 - 2Az + 3Az = 0 Results OK

46

= +12

A A Adedeji

Statically Determinate Structures

Example 3.11 Find the reaction of the body shown in Fig. 3.13 P2 lies in plane x

+y

P7 =70KN E c

C

3m 2m

D b

a

F

+x

30 P2=5KN

A

2m

2m

Fig. 3.13 A structural body with horizontal and inclined loads supported by swinging members Solution From the condition of equilibrium: X: F+C+P1+P2cos 300 = 0 y: A-B-D = 0 Z 0 = -E+ P2 sin 300 X Dx2=0 y C x 2 + P2 x 2 sin 300 = 0

47

A A Adedeji

Z

Statically Determinate Structures

B x 4 +P1 x 3 = 0

Determinant : = + 16 Results of the reactions A = -7.5KN, B =7.5KN, C = -2.5KN D = 0KN, E =2.5KN, F = -77.83KN Example 3.12 Calculate the reaction of the special structure supported by rigid body and loaded as shown in Fig.14

+y -z e

+x

D 3m

E

3m

C

2m

P2=3kN

a

b 0

60 P7=4kN Ay 2m 3m

B 5m

Az

x1

Fig. 3.14 A body loaded by an inclined load supported by swinging members

48

A A Adedeji

Statically Determinate Structures

3.5 Reactions of a three-Hinged Arch In a compound system of two members, the members are joined together mutually with the support of three rigid hinges (even virtually). This is unknown shape, three-hinged arch, even, in a case when the center-line is not curved (see Fig. 2.15 where hinge b is virtual). Hinge c joining mutually the two members is the “Tip hinge” and hinges a, b joining embers to the supports is the base hinge”. But if hinges a, b and c lie on a line, the system is treated as special.

C

a

b

Fig. 3.15 A structure supported by an hinge, a swinging and a roller In order to obtain components of reactions e can write for each member, three (3) conditions of equilibrium, or each from both members must be in equilibrium. From such system 6 equations must be obtained b 6 component reactions (2-internal and 4-external). Apart from that, we can write only 3 conditions of equilibrium as a whole which is linearly dependent on the previous 6 which can be used as control. If we want to obtain only the component of external reactions, we use the three conditions of equilibrium as a system of the whole moment conditions of equilibrium. From

49

A A Adedeji

Statically Determinate Structures

such, 4 equations are obtained with 4 unknown components of external reactions. In graphical method, we divide it into two cases, when it is loaded only by one element and again by the two elements. 3.5.1 Three-hinged frame

q

C

4.5m

2m

Q

3m

Examples 3.12 to 3.13 Find the reactions in each of the following examples (3.12 to 3.13) of the structures as loaded. Use Cullman’s and Resultant line methods. See Fig. 3.16.

3m

Bx

b

0

30 P=6kN 4x

By

4y 3m

4m

Fig. 3.16 (a) A frame with a hinge joint in the roof

50

A A Adedeji

Statically Determinate Structures

Cullman’s Method Refer to Fig.3.17 ( c)

(0) PRI (2)

(8) Q2 =PRI

Q1 (1)

C b

P

BI BIV

AII

a

Fig. 3.16(b) Cullman’s method

AI

Procedure 1) Superposition from 2 leading systems for the member I, we obtain reactions AI, BI, CI, then AII, BII, CII 2) Total sum of reactions I and II. 3) See Fig.3.17(b)

51

A A Adedeji

Statically Determinate Structures

P

(1) Q1 (3)

PRI

(2) A

AI (4) (5) AII

(O)

B

BI (C) AII

Fig. 3.16© Cullman’s solution

(O) BII

52

Q2 (II)

A A Adedeji

Statically Determinate Structures

Example 3.13 Using Resultant Line method and given: P=6kN, Q=18kN, q=2kN/m1

Q1

Q2

C

(7)

(8)

P b (5)

a

B

(6)

P (10) (7)

A

(2) 12

A=7.7kN, B=8.5kN, C=4.6kN

(7)

(6)

(5)

Q1 O2

O1

(8)

(3) (13) O

(b)

(9) Q2

(4)

11 Fig. 3.17 Resultant line solution

53

B

A A Adedeji

Statically Determinate Structures

Example 3.14 (By calculation) Refer to Fig. 3.17a. Solution Q=2kN/m1, …..Q=10kN, P=6kN Statical determinacy: δ=2, α2=3, so that s=3x2-(2x3)=0 From the conditions of equilibrium as a whole: x:

Ax-Bx+Pcos 300=0

y:

Ay+By+Pcos 300-Q=0

b:

7Ay-2Ax+7Psin 300+7Pcos 300-4.5Q=0

Taking left hand side of the structure (Refer to Fig. 3.17a).

x:

Ax+Pcos300-Cx=0

y:

Ay+Psin 300- 5q+ cy=0

c

3Ay –5Ax-2Psin 300+3Psin 300-5xqx2.5=0

Also for the right part: A=-0.693kN,

Ay=7.65kN,

BX=4.50kN

By=7.36kN

54

A A Adedeji

Statically Determinate Structures

Example 3.14 (use graph and find the reactions). See fig. 3.18.

Q

q c

3m

P2 M

7m

P1 d

W

e

W

2m

S

Given that: P1 = 3kN P2 = 6kN M = 7kNm q = 2kN/m1 Q = 6kN w = 7kN/m1 W = 4kN

a

b

Ax Ay

3m

B 1m 1m

Fig. 3.18 A frame a tensile force and an inclined support system Example 3.15 Calculate the reactions for both Figs 3.18 and 3.19. Use all methods of analyses. 1

2m

c P =4kN

I

II M=6kNm1

2m 1m

3m

q=7kN/m

d 3m

III b 2m

3m a

1m

2m

Fig. 3.19 A structural frame with a tensile force and a hinge in55 the roof

2m

A A Adedeji

Statically Determinate Structures

Example 3.16 Use conditions of equilibrium to calculate the reactions of the structures shown in Figs 3.20. 1

q=7kN/m

1

q=7kN/m

D

600 II

2m

I

4kN

c1 1m

3m

d

C

f

3m

e1 P=

c

2m 300 600 B

a

A

b Note: Virtual hinges:

6m

2m

d1, e1, c1

Fig.3.20 supported with multiple supports

q Cy Cy

300 P Ay Ay Fig.3.21 (a)

56

A A Adedeji

Statically Determinate Structures

Solution From Fig. 19. δ = 2 (separate plates) α = 6 (supports of one restraint) Then: s= 3x2 - (6x7) = 0 The system is statically and form-kinematically determinate. The virtual hinges (d1, e1, and c1) do not lie in the same line. From the conditions of equilibrium for the whole system (Fig.3.21a): x:

-D – P cos600 + B cos600 = 0

y:

A- 4q + C + B sin600 – P sin600 - 2q = 0

a:

2 x 4q +7x 2q - 8C - 6B sin600-3B cos600 + 2P sin600

=0 Condition of equilibrium for the plate II

F E d D Fig. 3.21 (b)

x: y:

-F cos450 - P cos600 + B cos600=0 C + B sin600 – F sin450 + E – P sin600 -qx2=0

e: -5C - 3B sin600 - 5B cos600 + 4q x2 + 2P cos6007P sin600=0

57

A A Adedeji

3.5.2

Statically Determinate Structures

Continuous (Gerber’s) beam

Examples 3.17 Find the reactions in Fig 3.22 of the structures as loaded. Use both the graphical and computation methods.

2m 1 q =7kN/m/

P= 3kN

1m

Ax a

b

I

Ay

e

1m

4m

q=1kN/m1

P=3kN

II

P=3kN

B

1

III d M=4kNm

f

D

C

4m

1

4m

1m

Fig. 3.22 A continous beam Solution δ = 3, α1 = 3, α2 = 3, then: s = 3x3 - (3 + 2x3) = 0 The system is statically and kinematically determinate without a movable mechanism. We can then start to find the reactions of the plate III or any part with three component reactions, so as to form 3 conditions of equilibrium

P (kN)

q (kN/m)

Fx

M f

III

d

Fy 4m

7m

Fig. 3.22 (a) Left hand portion

58

A A Adedeji

Statically Determinate Structures

q

P

Fy

Fx

II Fy

Fy 1m

2m 1m 1m

Fig 3.22 (b) Right hand portion

See Fig. 3.22 (a) x:

-Fx=0,  Fx=0

f:

-4D + M + 5 x 2q.5 = 0, D = 4.725kN

a:

4Fy - 4P – 5 x 1.5q5 + M =0,



Fy = 3.875kN

If we know the internal reaction of the hinge (f) in the part II members, we are left with 3 unknowns. See Fig. 3.22(b). x:

Ex+ Fx=0,  Ex = Fx = 0kN

y:

Ey + C - Fy + P- P -1q=0,  C=4.444kN

e: -4C + 5Fy - 3P + 1P + 1 x 4 x 5q = 0,  Ey = 0.431kN

59

A A Adedeji

Statically Determinate Structures

Examples 3.18 to 3.20 In examples 3.18 to 3.20, use conditions of equilibrium, calculate the reactions of the structures as shown in Figs. 3.23, 3.24 and 3.25 respectively.

q (kN/m)

Ax I

B

Ay 3m

Ex

e

b

a

Ey

1

1

Fig. 3.23 A simply placed beam

x: Ax + Ex=0,

 Ax=0kN

y: Ay + BFq – Ey =0 a: -4B + 5Ey + 3 x 1 x q.5 = 0  B =1.664kN b: 4Ay + 1Ey + 3 x 2q.5 =0  Ay=1.76kN To control the result, we can employ the whole system. y: Ay + Bc + D -3q - 6q – P = 0= 12 – 12 =0 Example 3.19

a

q = 4kN/m e

3

b 3

c

f

2

2

2

d

g 2

Fig. 3.24 A beam hinged in three places

60

(all in m)

A A Adedeji

Statically Determinate Structures

Example 3.20

q = 1kN/m

q = 1kN/m e

d

b

a

c 2m

3m

2m

3m

Fig. 3.25 A beam hinged at e 3.6 Calculation of reactions for complex systems with three or more rigid members and joints When we have 3 or more rigid members and joints, we can solve for the reactions using conditions of equilibrium by part. From each part we should solve for the reactions from the condition of equilibrium directly. In other words, we can reduce the number of the unknowns. The unknowns which otherwise must be solved by system equations. The following examples are self-explanatory: Example 3.21 Calculate the reactions of the structure in Fig. 3.26

f

f1

I

1m 1m



Ay

300

e P = 100 kN

d II

2m

Ax

g

e1 a

c

b III

3m

2m

2m

Fig.3.26 A Complex structure

61

A A Adedeji

Statically Determinate Structures

Solution Statical determinacy δ =3,

α1=3,

α2=3

Then: S=3x3-(3+2x3)=0 The system is statically & kinimatically determinate. For solution in external and internal reaction is distributed into individual part of the whole system. Let us take the system by parts.

f1

Dx

F

f

g

d Dy Fig. 3.26(a) Analytical solution

62

A A Adedeji

Statically Determinate Structures

Example 3.22 Calculation the reactions of the Structures in Fig. 3.27.

0

7

1m

2

P= 8kN 30

3m

q = 2kN/m q

5m

9

a A

Bx b

3m

Cx

c Cy

By 4m

4m

Fig. 3.27 A frame Solution If δ = 3, α1 = 3, α2 = 3 so that:

(I)

P

f F

d

D

s=3x3-(3+2x3)=0 the structure is stable and statically determinate.

a A Fig. 3.27(a) Analytical solution

63

A A Adedeji

Statically Determinate Structures

From the condition of equilibrium (I) x: D+P cs30=0 y: A- F – P sin300 - 7q = 0,

 D= - 6.93kN  A = 9.02kN

d: 7F+2x7x3.5 + 2Pcos300=0,

F

Gx D e

g (II)

Bx

Fx

Gy

 F = - 8.78kN

q

f Fy

b By

(III)

c Cy

Cx

Fig. 3.27(b) Analytical solution Solving as a whole in Fig. 3.27b y:

By+ Cy + F- q x 4=0

x:

Bx – Cx - D=0

c:

By x 8 D x 5 + F x 4 q x 4 x 2=0

Moment condition of part II: g:

4By - 5Bx = 0

So that: By=2.16 kN, Bx=7.73 kN, Cx=8.66 kN and Cy=74.82 kN. From the sum total condition of part II, we can obtain the internal reaction:

64

A A Adedeji

Statically Determinate Structures

x: y:

 Gx =8.66kN  Gy = -2.76kN

Bx – D – Gk = 0, By + Gy = 0,

Control: y: A + By + C - Psin300 - 77.q = 0, i.e., 0=0 Example 3.23 Find the reactions in Fig. 3.28 and 29.

q = 7kN/m

P = 5kN

e g

(II)

b

2q

i

2q 1m

(I) b a 4m

c 2m

4m

w

2m

j (III) (IV)

3m

f

d

1 1 2m 2m

1m

1 1m

(a)

Fig. 3.28 A complex frame structure

3m q

I

a

e

b F

A

III

c II

g

B 2m

2m

D

Cy 1m

3m

1m

2m

1m

(b) Fig. 3.29 A structure with multiple rigid joints

65

A A Adedeji

Statically Determinate Structures

Example 3. 24 Find the reactions of the system shown in Fig. 3.30.

q= 2kN/m

Ma I b

Ax Ay

II 2m

1m

2m

1m

1m

2m

Fig. 3.30 A truss system Example 3.25 Find the reactions of the system shown in Fig. 3.31.

P = 4kN q=2kN/m W = 1kN/m

e I

II

2

3

4 2m

1

2m

d

a

b 4m

4m

4m

Fig. 3.31 Triangular frame

66

4m

A A Adedeji

Statically Determinate Structures

CHAPTER FOUR BEAM ELEMENTS 4.1 Straight beam Straight beams are the real cases of the so called rigid members. Their reactions are obtained as in the previours chapter. Beam can be sectioned perpendicular to the axis of the member into 2 mutual independent parts and should be analysed for the removing part of the internal forces reactoins at the internal fixed end. The internal forces in each section are 3: Bending Moment (M), Shearing Force (V) and Normal Force (N). Bending Moment, in a section of a beam, is equal to the algebraic sum of statial moments of all forces acting in one side of the beam section, with respect to its centre of gravity. This is true also for V and N. See Fig. 4.1.

V

M

M

N

N

V Fig.4.1 Bar analysis From the condition of equilibrium of the element which is loaded by elementary part of the load intnesity it can obtian the relatioship between the load intensity q, shear force, normal force and bending moment. See Fig. 4.2.

67

A A Adedeji

Statically Determinate Structures

V

dx 2

+y q(x) q + dx

M + dM +x

qndx

N

N + dN

dx V + dv Fig. 4.2 Condition of equilibrium

Refer to Fig. 4.2. From the conditions of equilibrium along x-axis, we have: qn

dN dX From the total conditions along y-axis qt

=

=

-dV dX

(4.1)

(4.2)

From the moment condition, for instance, to point ‘O’ neglecting small addition in 2nd series. V

=

dM dX

(4.3)

This is the Schwedler Theorem The first derivation (differentiation) of bending monemnt, with respection X-axis is equal to the shear force.

68

A A Adedeji

Statically Determinate Structures

Note that maximum moment is in the posiiton where 1st derivation function is equal to zero. dM dX

=

0 (4.3a)

or rather, in a position (point) where the shear force changes its sign (under point load). The point of maximum moment is called “Transitive Section”. 4.2 Simple Beam Simple beam is a straight beam supported on one end by a rigid hinge and at the other end by a sliding support along the axis of the beam (also gliding support perpendicular to the beam axis). 4.2.1 Examples 4.1 Find the bending moments shear forces for the strucutre loaded as shown in Figs. 4.3.

P=5kN

Ax

x

b

Ay

B 2m x

3m X

Fig. 4.3 A simply supported beam

(2)

Forces/Moments (a,1) Mx = Ax Vx = A

Solution: (a) Reactions: x:

Ax = 0  Ax=0kN a: 2P - 5B = 0  B=2.0kN y: Ay + B – P = 0  Ay=3kN

i.e. Ma1 = 0, M1a = 6kN i.e. Va1 = V1a= 3kN

69

A A Adedeji

(1,b)

Statically Determinate Structures

Nx = 0

i.e. Nx = 0

Mx = Ax - P(x-2);

M1b = 6kNm Mb1 = 0kNm V1b = Vb1 - 3kN

Vx = A - P Nx=0 Let’s take the calculation at b-1 (b,1)

Mx1 = B x1 V x1 = -B N x1 = 0

i.e. Mb1 = 0, M1b = 6kNm i.e. V1b = Vb1 = -3kN

Solving example 4.1 using graphical solution (Fig. 4.4)

P = 5kN Ax a

b

Ay f=4

B A

(1)

V P

B (2) P

(2) 3 (2)

O2 O1

M Mmax Fig. 4.4 A simple straight beam 4.3 Cantilever Beam This is a beam with a fixed end support. The internal forces MiN and N due to the external load lie infinitely near the support (fixed end). This brings the beam to equilibrium.

70

A A Adedeji

4.3.1

Statically Determinate Structures

Examples 4.2

Find and skech the bending moments, shearing and axial forces of the structures shown in Fig. 4.5. Example 4.2

P=3kN Solution

a

The internal forces

2 3m x

1m

(1,2)

Mx1 =0, Nx1 = 0, Vx1 = 0

x1

(2,a)

Mx1 = -P(x1-1) Nx1 = 0 Vx1 = P

9.0

(2,a)

M

Ma2 = -9kNm M 2a = 0

3.0

V a2 = V 2a = 3kN

V

Fig.4.5 A camtilever ant its bending moment and shear force diagrams Example 4.3 Find and sketch the bending moments, shearing and axial forces of the structures shown in Fig. 4.6.

71

A A Adedeji

Statically Determinate Structures

q=3kN/m

Solution Internal forces Mx1 Vx1 Nx

2

1

2m

3 2m

(1, 2) M x =-2x  M1,2 = 0, 2 M21=-6kN Vx=-qx  V12=0, V21=-6kN Nx=0

1

x x

6.0

6.0

18.0

M

(2, 3) Mx=-q x 2(x-1)

V M23=-6kNm,

M32=-18.0kNm Vx=-2q V23=V32=-6kN Nx=0 Q=3 x 2 = 6kN

Fig. 4.6 Graphical Solution

f=6

0

V

M

M =3 x 6 =18 y=3 Fig. 4.7 Grafical solution 4.4 Inclined load and beam These are better explained by the following examples 4.4.1 Examples 4.4 to 4.6 Find and skecth the bending moments, shearing forces and normal forces, as required, in the structures shown in examples 4.4 to 4.6

72

A A Adedeji

Statically Determinate Structures

Example 4.4

P1= 4kN 450

A

P2= 3kN 3m

50 2 a

1

1.5m

P3= 8kN

0

b 3

P4 By = 2kN 2m 1m

1m x’ 2.0 3.66 Fig. 4.8 V, M and V diagrams

73

3.66

2.0

3.42

2.0

3.47

0.09 5.29

2.0

5.2

3.47

2.97

x

4 Bx

A A Adedeji

Solution Reactions A=338 kN,

Statically Determinate Structures

Bx=-3.66 kN

and By =4.77 kN

Internal foxes (1,a) Mx = - P1 sin 600, M1a=0, Ma1 = - 5.2kNm Vx =- P1 sin 600 V1a Va1=-3.47kN Nx= P1 cos 600, N1a=Na1=2kN Mx – P1 sin 600 + A (x - 1.5), Ma2 = - 5.2kNm, M2a = -5.29kN Vx = -P1 sin 600 + A Va2 = V2a = 0.09kN Nx = + P1 sin 600 Na2 = N2a = 2kN (a,2)

(2,3)

Mx = - P1x sin 600 + A (x – 1.5) + P2(x –2 M23 = - 5.29kNm, M32 =3.42kNm Vx = -P1 sin 600 + A +P2 V23 = V32 = 2.97kN 0 Nx = P1 cos 60 , N23 =N32 = 2kN

(b,3)

Mx1 = - P4 x 1 + By (x1-1) Mb3 = - 2kNm, M3b = 3.42kNm Vx1 = P4 – By Vb3 = V3b = - 2.77kN Nx1 = Bx Nb3 = N3b = - 3.66kN

(4b)

Mx1 = - Px1 Vx1 = - P4 Nx1 = 0

M4b =0, Mb4 = -2kNm V4b = Vb4 = 2kN

74

A A Adedeji

Statically Determinate Structures

Example 4.5 See 4.4.1

P =6kN 300 0 60

Q =8kN q = 2 kN/m a

2

A

By

2m

2m

Bx 3 M= 4 kNm

b

2m

4m X

7.3 5.2

x = 5.03m

Fig. 4.9, M, V N diagrams

75

4.0 2.73

2.47

2.47

4.0

4.6

1.8

10.

5.2

4.2

4.2

X

A A Adedeji

Statically Determinate Structures

Example 4.6 See 4.4.1

P =10 kN

q = 2 kN/m

1

a

2

o

b Bx 3

45

A 1.5m

By 1.5m

2m

x

x’

Fig. 4.10 A simple beam

76

1.5m

M= 4 kNm

A A Adedeji

Statically Determinate Structures

ANNEX A COURSE PROGRAMME A1 Course Content: Structural analysis of statically determinate elements

structural

A2 Course Description This course is an instructional piece, relating to the structural analysis of statically determinate structural elements (30hrs period: pr to GET 252) for the students of civil and related engineering courses in the faculty of engineering and technology of the University of Ilorin, Ilorin-Nigeria. This volume is strictly for internal circulation. A3 Course Justification This book, Structural Analysis, has been planned to serve as a basic and thorough knowledge in mechanics and structures. The course is sectioned to be followed sequent. This course of study deals with statically determinate structures (plane, special, truss and polygon and centenary). In this volume theories and problems are solved for the statically determinate structures. In general, the analyses apply to beams, trusses, frames and polygons are discussed here. A4 At the   

Course Objectives end of the course, the students will be able to: Determine the statically determinate structure Determine the statically indeterminate structure Determinate the form-Kinematical Determinacy of Truss  Graphical method of analysis  Solve problem on Cantilever beam  Draw bending moment and shear force diagram

77

A A Adedeji

A5 

  

Statically Determinate Structures

Course Requirements: This is a compulsory course for all students studying civil engineering. In view of this, students are expected to participate in all the course activities and have minimum of 75% attendance to be able to write the final examination. Practical analysis will be conducted by the students They will also be expected to treat the study questions and assignments. Students are also expected to have e-mail

No

Methods of Grading: Item

Score %

1.

Class/Assignment/test

40

2.

Comprehensive final examination

60

Total

100

A6 Course Delivery Strategies: The lecture will be delivered through face-to-face method, theoretical material (lecturer note) provided during lecture. Students will be encouraged and required to read around the topics and follow current issues in the media. Web – interactions will be employed by requesting each student to have yahoo e-mail address to enable them participate in the yahoo discussion group that had been created for the course (Unilorin/CVE365). Additional materials and links will be provided on the board. The delivery strategies will also be supported by tutorial sessions and review of study questions.

78

A A Adedeji

Statically Determinate Structures

ANNEX B LECTURES B1

Week 1: Introduction Objective: the student will be able to explain the basic knowledge in mechanics and structures. Description: the course outline will be introduced with emphasis on the objective and delivery strategies, the definition and scope of structural analysis, career opportunities and expectation for the study of the programme. Study questions  What is structural analysis?  Explain by what is meant by statically determinate structure.

B2

Week 2: Establishing degrees of statically determinacy of structures Objective: the objective is for the student to be able to establish degrees of statically determinacy of structures. Description: structural systems comprising of members (elements) can have its members joined thorough (mostly by the number of external or internal joints) at their ends. This is may not necessarily be important for the stability of the support (stables structures). Joint for component system Study question: Learn on how to establish the degree of determinacy of the following member; Supports of fixed end, Bond by a moving hinge, Swinging member and movable fixed end. Reading list: Visit the school library and internet

B3

Week 3: Degree of statistical indeterminate (or determinate) in a plane arrangement

79

A A Adedeji

Statically Determinate Structures

Objective: the objective is for the student to be able to establish degrees of statically indeterminacy of structures. Statically and Form-Kinematical Determinacy of truss (member) systems with hinge joints. Description: statically and form-determinate system of joint and rigid member, Degree of Statically Indeterminacy /Determinacy in special arrangement. Study question  Study how to obtain the statically and form kinematically determinacy in a structure, frame, closed frame  Shows the degrees of statically indeterminate S of some forms and bonds in plane and special structures Reading lists: Visit the school library and internet B4

Week 4: Graphical Method of Analysis Objective: the objective of the week lecture is for the student to be able to explain the graphical method of analysis Description: Characteristics of statically indeterminate structure Study question: Use graphical (Cullman’s) method finds the reactions of several structures Reading List: Visit the school library and internet

B5

Week 5: Reactions of material objects and compound system Objective: the main objective is for the student to be able to find the reactions of material objects and compound system. Description: Reactions By calculation, Graphical Solution for Obtaining System Reactions. Study question: known how to calculate the reactions.

80

A A Adedeji

Statically Determinate Structures

Reading List: Visit the school library and internet

B6

Week 6: A structural element loaded by horizontal load Objective: the objective of the lecture is to describe the structural element loaded by horizontal load Description: In order that the system equation has one result, the determinant of such system must not be equal to zero Study question: study on structural element load by horizontal load Reading List: Visit the school library and internet

B7

Week 7: Reactions of Three –Hinged Arch Objective: The student to be able to describe the react ions of three- hinged arch. Description: Compound system of two members, the member are joined together mutually with the support of three rigid hinges. Cullman’s method Study question: study on Reactions of Three-Hinged Arch Reading List: Visit the school library and internet

B8

Week 8: Continuous Beam (Gerber’s Beam) Objective: the student to be able to find the reaction in a continuous beam Description: Calculation of reactions for continuous beams and compound systems with 3 or more rigid members and joints. Study question: study on continuous beams Reading Lists: Visit the school library and internet B9

Week 9: Straight Beam Objective: the student to be able to find the reaction in a straight beam.

81

A A Adedeji

Statically Determinate Structures

Description: Simple beam; simple beam is a straight beam supported on one end by a rigid hinge and at the other end by a sliding support along axis of the beam. Study question: study on straight beam Reading List: Visit the school library and internet B10

Week 10: Cantilever beam Objectives: the student to be able to find the reaction at cantilever beams Description: this is a beam with a fixed end support. The internal forces MN and N due to the external load lie infinitely near the support (fixed end). Study question: know more about cantilever beam Study List: Visit the school library and internet

B11

Week 11: Bending moment diagrams Objectives: the student to be able to identify and understand the bending moment diagrams Description: to know the deflection of beams. Study question: find the bending moments of structures. Study List: Visit the school library and internet

B12

Week 12: Shear Force Diagrams Objectives: the student to be able to identify and understand the shear force diagrams Description: to know the force acting on beams Study question: find the shear force of structures. Study List: Visit the school library and internet

B12

Week 12: Class Test Description: the students will be assessed on all the treated topics for 45 minutes Week 13: Revision Description: the students will be reminded of what the course is all about.

82

A A Adedeji

Statically Determinate Structures

Course Content: Use of knowledge of leveling and compass traversing for detailed topographical mapping of areas (2 week duration in harmattan semester break) 90h(P): PR : CVE 351: C Course Description: The course is designed to expose students in the department of civil engineering to practical surveying using various surveying instruments. Course Justification: The course is majorly practical and report is written at the end and submitted to the department of civil engineering. Course Objectives:  Determination of vertical height btw two stations,  Contouring,  Planning of roads, and similar services,  Fixing site levels, Course Requirements:  This is a compulsory course for all students studying Civil Engineering. In view of this, students are expected to participate in all the course activities and have minimum of 60% attendance to be able to write the final examination.  They will also be expected to treat the study questions and assignments. Course Delivery Strategies: The lecturer introduces the students to the course and teaches them how to use survey instruments. After then, the students go the field and carry out the expected work as required and directed by the lecturer.

83

A A Adedeji

Statically Determinate Structures

ANNEX C PRACTICALS C1

Week 1: Pacing Objective: The students are to adopt a moderately equal length of pace under various circumstances. Equipment: Tape Procedure: 1. Measure long distance with a tape preferably 50 – 100m on a fairly level ground 2. Pace the same length and know the number of your pacings. 3. Compute pacing by dividing the known distance by the number of pacing. Requirement: each student to calculate his or her own pacing on 1. A fairly level ground 2. Upward a slope 3. Downwards a slope and compare the result of the three items.

C2

Week 2: Use of levels Objective: to know the difference between two points i.e. two benchmarks (BM) and to draw the profile along the benchmarks Equipments: Level, two leveling staffs, umbrella, chain or tape. Requirement: the students are to level btw two given points and do the reduction. Each student is to draw the profile btw two points.

84

A A Adedeji

Statically Determinate Structures

BIBLIOGRAPHY Hoit , M. I. (1995), Computer assisted structural analysis and modeling, Upper Saddle River, NJ . Prentice Hall. Murthy, V. S. and Sood, R. (2009), Question bank in civil engineering, S. K. Kataria and Sons. www.katariabooks.com Structural Analysis in theory and practice, (2009) ButterworthHeinemann-imprint of Elsevier, www.elsevierdirect.com

85