38883053 structural analysis ds doc09 10

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Statically determinate structures


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D e p a r t m e n t o f C i v i l E n g i n e e r i n g , U n i v e r s i t y o f I l o r i n + 2 3 4 8 0 3 3 7 7 4 6 1 6 a a a d e j i @ u n i l o r i n . e d u .n g a a d e o l a d e j i @ y a h o o .c o m ,

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.

A A Adedeji Statically Determinate Structures

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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) Lecturer: A A Adedeji

M.Sc. (Prague), Ph.D. (ABU, Zaria), MNIEM., MNICE, R. Engr. MIAENG

E-mail: [email protected]; [email protected]

Office Location: Block 8, F29, Main Campus, University of Ilorin, Ilorin

© 2010

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A A Adedeji

DEPARTMENT OF CIVIL ENGINEERING UNIVERSITY OF ILORIN

2010


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Contents Page PREFACE 6 CHAPTER ONE 7 INTRODUCTION 7 CHAPTER TWO 10 STATICALLY DETERMINATE STRUCTURES 10 2. ESTABLISHING DEGREES OF STATIC

DETERMINACY OF STRUCTURES 10 2.1 Plane system 10 2.2 Joint for component systems 13 2.3 Degree of statistical indeterminate

(or determinate) in a plane 14 2.4 Statical and form-determinate system of joint and rigid member 16 2.5 Degree of statical indeterminacy/

determinacy in space 22 2.6 Statical and form-kinematical

determinacy of trusses with hinge joints 24 2.7 Graphical method of analysis 28

2.7.1 Characteristics of statically indeterminate structure 28

CHAPTER THREE 34 REACTIONS AND SUPPORTS OF ELEMENTS 34

3.1 Reactions by calculation 34 3.2 Graphical solution for obtaining

system reactions 34 3.3 Examples 3.1 to 3.3 35 3.4 Calculation of reactions of a rigid body 43 3.5 Reactions of a three-hinged frame 49

3.5.1 Three-hinged frame 50 3.5.2 Continuous (Gerber’s) beam 58


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3.6 Calculation of reactions for complex systems with 3 or more rigid members and joints 61

CHAPTER FOUR 67 BEAM ELEMENTS 67

4.1 Straight beam 67 4.2 Simple beam 69

4.2.1 Examples 4.1 to 4.2 69 4.3 Cantilever beam 70

4.3.1 Examples 4.3 – 4.6 71 4.4 Inclined load and beam 72

4.4.1 Examples 4.7 to 4.10 72 Annex A 77

Course Programme 77 Annex B 79

Lectures 79 Annex C 84

Practicals 84 Bibliography 85


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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, free-polygon 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


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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


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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 Determi-nacy

Condition Support Determinate Condition

S= m - r

> 0

= 0

< 0

Form Statical Indeterminate Determinate Over- Determinate

Over-determinate Determinate Indeterminate

___

∆≠ 0

___

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


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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.


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CHAPTER TWO STATICALLY DETERMINATE STRUCTURES 2. ESTABLISHING DEGREES OF STATIC DETERMINACY OF STRUCTURES 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.

( c) Schematic representation

(a) Column on balls (b) Beam-column

Fig. 2.1 Swinging members


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

(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

(c ) Diagramatical Representation

(a) A member (b) A member on a roller on rollers

Fig. 2.2 Rollers (Movable hinges)


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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. 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

- a

- b - b

- a - a β

Α b -

(a) A closed member (b) a cut in point b (c) cuts in two-point

Fig. 2.4 A closed structure


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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).


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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 (2

1

na ),

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


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(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 fixed-ends (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)


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c) No part of a system must be form-kinematically indeterminate – must not have deficiency bonding, even if a part was unnecessarily (abundantly) bonded

d) 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. 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.

S = 2β + 3 – 33 + 22 + 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


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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.

OI,V OIII,V II d III c e i ii iv vi f g I iii v


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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 (I-XII) 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.

c a

b d

(i)

(ii(ii

(x (ix(iv (v) (vi

(vii) (viii)

(xi) (xiiI II

Fig.2.8 Truss system

a

(a)

III, IV

δ II, IV

b

c

b

(b) IV

III

II, III

I

Fig. 2.9 Frame object and its analysis


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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. 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

(a)

I II III IV

(b) Fig. 2.10 Arch structure


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

a b

Fig. 2.12 A Closed structure

q1 (KN/m)

q2 (KN/m1)

X3 X2

X1 XI1

XI2

XI3


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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 S = 6δ = 6 x 2 = 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.

P

X3 X2

X5 X4 X6

X1 X1 X3

X6 X4 X5

P

Fig. 2.13 Frame structure


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Fig. 2.14 Complex structures (Examples 2.8-2.11) 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

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

(a) 1.8

(b) 1.9

© 1.10 (d) 1.11


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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

(fixed-end)

Hinge

In plane 30 20 30 20 In space 60 30 60 30 - spherical

40 -cylindrical Symbols Δ Β α3 α2


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Table 2.1 continued Structure

Mov

able

hi

nge

Swin

ging

m

embe

rs

Mov

able

hi

nge

Mov

able

bu

ilt-in

In plane 10 10 - 20 In space 20 10 10 50 Symbols α1 α1 α1 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


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Example 2.14 Find the number of statical indeterminate/determinate of the structure in Fig. 2.16 Example 2.15 Obtain degrees of statical indeterminacy (determinacy) at the member system in Fig. 2.17 Solution: Member systems with hinge bonds has: α2 = 1, α1 = 1 and π =10

Fig. 2.17 Truss system

b e

a c d f

f

a b

c d

e

g

Fig. 2.16 Frame structure

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.


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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.

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.

a b c d e

f g h

i j k

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


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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 form-kinematically 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

Fig. 2.20 Truss system framework

b a c d e

a b

c

d e

f g

h

i

j

k

l

m

n

o

p

q

r


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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

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)

1m 2m 1m

Fig. 2.21 A solid element

P2 = 5kN C

P1 = 4kN

600 P3= 2kN 6m

1m

2

m

2m

P1

300

B A


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Fig. 2.21a Graphical solution Example 2.20 a) Use Cullman’s Method to find reactions in Fig.2.22 Fig. 2.22 An L-section plane structure

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

E

Solution

P

P

C

P(3

(4

(2

(1RR

PR

B A

A

C RR

B P

300 A (kN)

P3 = 5kN F

2

m

2m

P2 = 3 kN

P1 = 4 kN 2m 3m

C (kN)

D

B (kN)

600

(a


30

Solution

P3

PR

R

6 3

4

5

C

1

2 B

A P2

Fig. 2.23 Vectoring example 2.20


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Fig. 2.24 Solution to example 2.20 Example 2.21 Use resultant line method, find the reactions of the beam element.

3 kN

5kN 5

PR

2

3

8

4

C O

1

P2

P1

β 10 A

2..9 kN

7 9


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Fig.2.25 A simply supported beam Solution to resultant (Graphical) method:

1m Ax b

B P3 = 2kN

Ay

2m 8m 2m 2m 600

a 450

P2 = 10 kN P1 = 5 kN

2

Fig. 2.26 A simply supported beam Fig. 2.6a Solution to example 2.21

A

b

3 P3

P1 P2

B

1

3

B

4

1

A y

A P1

6.0 kN

8.2 kN

5.6 kN

P3

P2

1

2

AX


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Fig.2.27 A beam with loads Fig. 2.27a Solution to example 2.22

Example 2.22

C) Use Cullman’s Method to Solve Example 2.21

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

(2R

A

(1)

Ay P1

Ax

B

P2

P3

B (3)

P3

Ay

P2 P1 Ax

(2) (1)


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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 x-

axis 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

n r y : ∑ Fiy + Σ Rjy = 0 (3.2)

i=1, j=1 c ) Moment condition of equilibrium, for instance to the

origin of the coordinate system (0) is written as n r 0 : ∑ (Xi F iy - y i F iy) + ∑ (Xj Rjy – Yj Rjx) = 0 (3.3)

i=1 j=1

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


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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). Fig. 3.1 A plane object

P2 = 3kN

A 300 P3 = 5kN1

600

a A 2m 3m

C

C

P4= 4kN B

d

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.


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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

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

Fig. 3.2 A beam element

Solution Using condition of equilibrium, the reactions are:

A=2.03KN, B=3.32KN and C = 5.78 kN Control: d : 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

Ax

b a

B P3 =2kN

2m 2m 3m Ay

2m

45o

P1 =5kN P2 =10kN

1m


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

Fig. 3.3 An element inclined at 300 to the horizontal

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

– B x 9 = 0 y : Ay – P1 – P2 sin 450 – P3 sin600 + B = 0 x : - P2 cos 450

+ P3 sin 600 + Ax = 0 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

Ax

P1 = 5 kN

Q b q = 2KN/m’

Ay

2 2m 1 1 1

B 300

P2 = 1kN


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Fig. 3.4 A structural frame

3m

a Bx

P3 = 4KN

P1 = 10 kN

b B y

1m

P2 = 3 kN 1 m 2m 1m

2m

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


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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

1m 2m 1m

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

C P3 = 2 kN

600 1m 1m

A

B

300

P1 = 4 kN

P2 = 5kN

2m


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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.

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

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

7m

4m 2m 1m

8m 1m

B A

P1 = 4kN q = 3kN/m P2 = 6kN


41

(a) Cullman’s Method Solution

Fig.3.8b Resultant line solution

b Bx

P

By

A Fig. 3.8 (a) Cullman’ method

a

R (7)

(R 4 Bx

By

(5 (3)

P

A

A R Bx

By

P=10KN

4 m

By

3

m

2

m

3

m

B x

a

A

b

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

A = 2.5kN

Bx = 10kN

By= 2.5kN


42

Example 3.11 Find the reactions of the structure shown in Fig. 3 .9

4m

2m

560 450

2m 2m 2m 1 4m

a b

C

Q7 = 6kN P1 = 4kN

P2= 2kN

c

Fig. 3.9 A framed object

1

P Bx

2 A

P Bx

By

A

Fig. 3.8 (b) resultant line method

a 2

1 R

By 0


43

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

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.

2m 3m 4m

300

A

a

P7=70kN c

b

7m

7.5m

5m


44

x : n r ∑ Fix + ∑ Rjx = 0 (3.4)

i=1 j=1 y : n r

∑ Fiy + ∑ Rjy = 0 (3.5) i=1 j=1

z : n r ∑ Fiz + ∑ Rjz = 0 (3.6)

i=1 j=1

Fig. 3.11 Analysis of a point lying in space

Fiz

Xi

Xi

+X

+Z

Fix

Fi Fi y

+y

Zi

αi

i

yi


45

Moment Condition of equilibrium to X-axis n r X Σ (Yi Fiz - Zi Fiy) + Σ (Yj Rjz - Zj Rjy) = 0 (3.7) i=1 j=1 Moment Condition of equilibrium to y-axis

n r Y Σ (Zi Fix - Xi Fiz) + Σ (Zj Rjx - Xj Rjz) = 0 (3.8) i=1 j=1 Moment Condition of equilibrium to z-axis

n r Z Σ (αi Fiy - Yi Fiz) + Σ (Xj Rjy - Yj Rjz) = 0 (3.9) i=1 j=1 where: Fi = axial forces, Rj = reactions

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

Fig.3.12 A structural element loaded by horizontal load

+y x1 D

C

+x

B

P2=8kN 3m Ay Az

+Z Z

Ax a= 0

6

P1=4kN

1.5

m

2m

2m

2m


46

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 Az B C D 1 0 0 0 0 1 0 1 0 -1 0 0 = 0 0 -1 0 -1 0 = +12 (3.10) 0 0 0 2 0 0 0 0 0 0 -3 4 0 0 0 3 0 2 From the above the reactions: Ax= -24kN; Ay 8kN; Az = -30kN B = -16kN; C =30kN; D=20kN Control x1: 4Ay - 2Az - 2B - 2C = 0 y1: 2D + 0.5P1 - 2Az + 3Az = 0 Results OK

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


47

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

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 D x 2 = 0 y C x 2 + P2 x 2 sin 300 = 0

P2=5KN

2m 2m

a A

P7 =70KN 3m

2m

C c

F +x b

E

D

30

+y

Fig. 3.13 A structural body with horizontal and inclined loads supported by swinging members


48

Z 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

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

2m 3m

Az a

Ay P7=4kN 600

b

B

P2=3kN

2m

E C

D e

+y -z

+x

x1

3m

3m

5m


49

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. 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

C

a b


50

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 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.

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

4m

Q q

By

Bx

3m

4y

4x

4.5m C

b P=6kN 300

3m

3m

2m


51

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

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)

b

a AI

AII

BIV

BI

Q2 =PRI

PRI

Q1 (2)

C

(8)

(0)

(1)

P

Fig. 3.16(b) Cullman’s method


52

A

B (C) (O)

AII

BII

Q1

(3)

(II) Q2

PRI

BI

(5)

(2) AI (4)

AII

(O)

P (1)

Fig. 3.16© Cullman’s solution


53

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

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

a

b (5)

P

Q1

C

Q2

(6)

(10)

B

(8) (7)

P

B

Q1

Q2

(7)

(9)

O (13)

(3)

(4)

11

12 (2)

(6)

(8)

(5)

(7)

O1 O2

A

(b)

Fig. 3.17 Resultant line solution


54

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


55

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

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. a

2m

Q q

P2

M W

W S d

P1

Ay

Ax b

B 1m 1m 3m

a

c

e 7m

3m Given that:

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

1m

P =4kN I II c

b

2m 2m 1m 3m 2m

d

3m

q=7kN/m1

3m

2m

2m

M=6kNm1

III

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


56

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

Fig.3.20 supported with multiple supports

Fig.3.21 (a)

P 300

Cy

Ay

q

Cy

Ay

6m 2m

300

C I

D

d c1

c q=7kN/m1

q=7kN/m1

4kN

II 600

1m 2m

600

P=

B

b a

e1 f

A

3m

3m

2m

Note:

Virtual hinges:

d1, e1, c1


57

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

Fig. 3.21 (b)

D

d

F

E

x: -F cos450 - P cos600 + B cos600=0

y: C + B sin600 – F sin450 + E – P sin600 -qx2=0

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

7P sin600=0


58

3.5.2 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.

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

Fig. 3.22 (a) Left hand portion

4m III

q (kN/m)

d

P (kN)

f Fx

Fy 7m

M

1 m 4m 1 4m 1m 4m

Ay

Ax

B

q =7kN/m/

C D

III II I e f d a b

q=1kN/m1

2m

P=3kN

P=3kN P= 3kN

M=4kNm

1 1

1m


59

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

P q

Fy

Fy Fy

Fx

2m 1m 1m 1m

II


60

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.

Fig. 3.23 A simply placed beam To control the result, we can employ the whole system. y: Ay + Bc + D -3q - 6q – P = 0= 12 – 12 =0

Example 3.19

Fig. 3.24 A beam hinged in three places

Ax

3m

Ay I

B Ey

Ex q (kN/m) b e

a

1 1

3 3 2 2 2 2

a

q = 4kN/m

b c d e f g

(all in m)

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


61

Example 3.20

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

Fig.3.26 A Complex structure

a

q = 1kN/m

3m 2m 2 m 3m

q = 1kN/m

b d e

c

P = 100 kN

f1

e

c

d

f g I

II

300

Ax e1 a b

Ay 3m 2m 2m III

2m

1m

1m


62

Fig. 3.26(a) Analytical solution

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 d

f g

Dy

F


63

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

Fig. 3.27(a) Analytical solution

4m 4m 3m Cy

Cx Bx b

By A

a c

P= 8kN q = 2kN/m

q 7

5m

3m

2

9

1m

300

Solution If δ = 3, α1 = 3, α2 = 3 so that: s=3x3-(3+2x3)=0 the structure is stable and statically determinate.

f

d F

a

A

(I) P

D


64

From the condition of equilibrium (I) x: D+P cs30=0 D= - 6.93kN y: A- F – P sin300 - 7q = 0, A = 9.02kN d: 7F+2x7x3.5 + 2Pcos300=0, F = - 8.78kN

Fig. 3.27(b) Analytical solution 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:

Fx

c Cy

(III)

q

f

Gy

b By

(II)

D

Bx

e g Gx

Fy

F

Cx

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


65

x: Bx – D – Gk = 0, Gx =8.66kN y: By + Gy = 0, Gy = -2.76kN 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.

(a)

Fig. 3.28 A complex frame structure

(b)

4m 2m

1 1 2m 2m 1m

d c b a

(I) (II)

g

P = 5kN

2q 2q

(III) (IV) j

f

w

4m

3m

2m

e

1 1m

q = 7kN/m

b i 1m

2m 1m 3m 1m 2m 1m Fig. 3.29 A structure with multiple rigid joints

A

a

2m

B F g III

Cy D

b e I II c

3m q


66

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

Fig. 3.30 A truss system

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

4m 4m 4m 4m

4 3 2

d e

a b

1

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

2m

2m

I II

P = 4kN

Fig. 3.31 Triangular frame

2m

1m 2m 1m 1m 2m

b I

Ax Ay

II

Ma q= 2kN/m


67

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.

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.

V

N N

V

M M

Fig.4.1 Bar analysis


68

Refer to Fig. 4.2.

From the conditions of equilibrium along x-axis, we have: qn = dN dX (4.1) From the total conditions along y-axis qt = -dV

dX (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.

Fig. 4.2 Condition of equilibrium

dx V + dv

N + dN +x

M + dM

N q + dx

qndx

q(x)

+y V dx 2


69

Note that maximum moment is in the posiiton where 1st derivation function is equal to zero. dM = 0 dX (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. (2) Forces/Moments (a,1) Mx = Ax i.e. Ma1 = 0, M1a = 6kN Vx = A i.e. Va1 = V1a= 3kN

P=5kN Ax

Ay 2m 3m

B

b

x X

x Solution: (a) Reactions: x: Ax = 0

Ax=0kN a: 2P - 5B = 0

B=2.0kN y: Ay + B – P = 0

Ay=3kN Fig. 4.3 A simply supported beam


70

Nx = 0 i.e. Nx = 0 (1,b) Mx = Ax - P(x-2); M1b = 6kNm Mb1 = 0kNm Vx = A - P V1b = Vb1 - 3kN Nx=0 Let’s take the calculation at b-1 (b,1) Mx

1 = B x1 i.e. Mb1 = 0, M1b = 6kNm

V x1 = -B i.e. V1b = Vb1 = -3kN

N x1 = 0

Solving example 4.1 using graphical solution (Fig. 4.4)

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.

Ax a b P = 5kN

Ay

B

Mmax M

V

A f = 4

(1)

B (2) P

O2

O1 P

3 (2)

(2)


71

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

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.

P=3kN

1m

x1 x

3m 2 a

M

9.0

V

3.0

Solution

The internal forces

(1,2) Mx1 =0, Nx1 = 0, Vx1 = 0

(2,a) Mx1 = -P(x1-1)

Nx1 = 0

Vx1 = P

(2,a) Ma2 = -9kNm

M 2a = 0

V a2 = V 2a = 3kN


72

Fig. 4.6 Graphical Solution

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

Solution Internal forces Mx1 Vx1 Nx

(1, 2) M x =-2x M1,2 = 0, 2

M21=-6kN Vx=-qx V12=0, V21=-6kN Nx=0 (2, 3) Mx=-q x 2(x-1) M23=-6kNm, M32=-18.0kNm Vx=-2q V23=V32=-6kN Nx=0 Q=3 x 2 = 6kN

q=3kN/m

1 2m 2m

2 3

x x1

M 18

.0

6.0 V 6.0

0

M

V

f = 6

M =3 x 6 =18 y = 3


73

Example 4.4

Fig. 4.8 V, M and V diagrams

P4 = 2kN P2= 3kN

500 P1= 4kN

2 450 P3= 8kN

4 Bx

By

b 3

2m 3m A

a 1

1.5m

1 m

x x’

1m 3.

47

5.2

3.42

2.0

5.29

3.66

3.66

2.0 2.0

2.0

2.

97

0.09

3.47


74

Solution Reactions A=338 kN, 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 (a,2) 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 (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 Nx = P1 cos 600,

N23 =N32 = 2kN

(b,3) Mx1 = - P4 x 1 + By (x1-1)

Mb3 = - 2kNm, M3b = 3.42kNm Vx

1 = P4 – By Vb3 = V3b = - 2.77kN

Nx1 = Bx

Nb3 = N3b = - 3.66kN (4b) Mx

1 = - Px1

M4b =0, Mb4 = -2kNm Vx

1 = - P4 V4b = Vb4 = 2kN

Nx1 = 0


75

Example 4.5 See 4.4.1

Fig. 4.9, M, V N diagrams

5.2

2.47

2.73

2.47

4.0

4.0

4.6

1

.8 10

.5.

2 4.

2

4.2

7.3

x = 5.03m

2m 2m 4m 2m

X X

A

300

a 2 b Bx 3 M= 4 kNm

q = 2 kN/m

By

Q =8kN P =6kN

600


76

Example 4.6 See 4.4.1

1.5m 1.5m 2 m 1.5m

x’ x Fig. 4.10 A simple beam

45o

A

1 a 2 b Bx 3 M= 4 kNm

q = 2 kN/m

By

P =10 kN


77

ANNEX A COURSE PROGRAMME

A1 Course Content: Structural analysis of statically determinate structural elements 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 Course Objectives At the 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


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A5 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

Methods of Grading:

No 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.


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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


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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.


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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.


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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.


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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.


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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.


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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) Butterworth-Heinemann-imprint of Elsevier, www.elsevierdirect.com

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