Chapter 4 - Structures

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Luqman Nul Hakim Bin Juwara Jabatan Kejuruteraan Mekanikal

Politeknik Muadzam Shah [email protected]

DJJ3053

ENGINEERING MECHANICS

CHAPTER 4 :

STRUCTURES

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Structures

 This topic introduces the concept to analyze truss structures by using methods of joints

and methods of sections. Learning Outcomes (LO)

 Upon completion of this topic, students should be able to :

4.1 Explain plane trusses

4.1.1 Define a truss

4.1.2 Explain plane trusses

4.2 Analyze force in trusses by using related method

4.2.1 Analyze a truss by using the method of joints

4.2.2 Analyze the force in the members of a truss by using method of sections

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Consist of three categories of Engineering Structures.

1.

TRUSSES



Framework

composed of

members

joined at their ends

to form a rigid structures.

2.

FRAMES



Stationary

and can support external

loads, which contain at least one

multi-force member.

3.

MACHINES

 Structures containing moving parts

designed to

transmit

and

modify forces

.

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TRUSS



Framework

composed of

members

joined at their ends

to form a rigid structures.

PLANE TRUSS



Member of truss

lie in same plane.

- A planar truss lies in a single plane. In

plane truss, both the truss structure and

the applied loads lie in the same plane. The

analysis of forces will be in 2 dimension.

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June 2015 Session : 2(a)

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 3 bars joined with pins at end

form a triangular truss.

 Rigid Bars (non-collapsible).  Non-Rigid Bars can be made

rigid.

 SIMPLE TRUSS  Attaching 2

or more members &

connecting these members by expanding the basic triangular truss will form a larger truss

Non-Rigid Bars Rigid Bars Bar

Pin

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1.

All loadings are applied at the

joints

.

2.

Weight can be included.

3.

The members are joined together by

smooth pins

(bolting or welding).

4.

Two force members –

equilibrium

only in two forces

either

TENSION

or

COMPRESSION

.

5.

Two force are applied at the end; they are

equal

,

opposite

and

collinear

for equilibrium.

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SENSE

– unknown member force can be assumed.



POSITIVE

– indicates the sense is

correct.



NEGATIVE

– indicates the sense is

reversed.

Always assume the unknown member force acting on the

joint FBD as tension (pull) on the pin.



POSITIVE

– in

Tension (T)



NEGATIVE

– in

Compression (C)

Method to determine the correct

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 Satisfying the conditions of equilibrium for the forces acting

on the connecting pin of each joint.

 Each joint is subjected to a force system that is coplanar and

concurrent.

 Only 2 independent equilibrium equations are solved, ∑F_x = 0

and ∑F_y= 0.

Method of JOINTS

Truss Simple Beam

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Example 1

Determine the force in each member of the truss shown and indicate whether the members are in tension or compression. Ans: (T) 500 (T) 500 500 (T) 500 (T) 500 (C) 707 N A N A N C N F N F N F y x y CA BA BC      

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Solution:

+ve  ∑F_x= 0; 500 N + F_BC sin 45o_{= 0}  F_BC = -707 N = 707 N (C) +ve ↑ ∑F_y= 0; -F_BC cos 45o_{– F} BA = 0 -(-707 N) cos 45o_{= F} BA  F_BA = 500 N (T) Joint B :

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Solution:

Joint C : +ve  ∑F_x= 0; -F_CA + 707 N cos 45o_{= 0}  F_CA = 500 N (T) +ve ↑ ∑F_y= 0; -707 sin 45o_{+ C} y = 0  C_y = 500 N

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Solution:

Joint A : +ve  ∑F_x= 0; 500 N – A_x = 0  A_x = 500 N (T) +ve ↑ ∑F_y= 0; 500 N – A_y = 0  A_y = 500 N (T)

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Example 2

The truss used to support a balcony, is subjected to the loading shown. Approximate each joint as a pin and determine the force in each member. State whether the members are in tension or compression. Set Set P₁ = 800 N and P₂ = 0 N. Ans: (C) 1600 (T) 4 . 1131 (T) 800 0 (T) 800 (C) 4 . 1131 N F N F N F F N F N F DE DC BC BD AB AD      

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Example 3

Determine the force in each member of the truss and state if the members are in tension (T) or compression (C).

Set P₁ = 500 N and P₂ = 100 N. Ans: N C N F N F N F y CA BC BA 43 . 271 (C) 271 (T) 86 . 383 (T) 71 . 285    

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Example 4

Determine the force in each member of the truss shown. Indicate whether the members are in tension or compression. Ans: (C) 200 (C) 600 (C) 200 (T) 250 (T) 500 (T) 450 (C) 750 (C) 200 (C) 600 (C) 600 N C N F N F N F N A N F N F N C N A N C y CB DC DB y AD AB y y x          

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Solution:

Support Reaction : +ve  ∑F_x= 0; 600N – C_x= 0  C_x= 600 N (C) +ve ∑M_c= 0; A_y(6m) – 400N(3m) – 600N(4m) = 0  A_y = 600 N (C) +ve ↑ ∑F_y= 0; A_y– 400N – C_y = 0  C_y= 200 N (C)

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Solution:

+ve ↑ ∑F_y= 0; 600N + (⅘)F_AB = 0  F_AB= -750 N  F_AB= 750 N (C) +ve  ∑F_x= 0; F_AD + (⅗)F_AB= 0  F_AD= 450 N (T) Joint A : A_y = 600N F_AD F_AB 3 4 5

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Solution:

+ve  ∑F_x= 0; -450N – (⅗)F_DB+ 600N = 0  F_DB= 250 N (T) +ve ↑ ∑F_y= 0; (⅘)F_DB + F_DC = 0  F_DC= -200 N  F_DC= 200 N (C) Joint D : F_DC 600 N F_DB F_AD = 450N 3 4 5

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Solution:

+ve  ∑F_x= 0; -600N – F_CB= 0  F_CB= -600 N  F_CB= 600 N (C) +ve ↑ ∑F_y= 0; F_DC– C_y = 0  F_DC= C_y = 200 N (C) Joint C : C_x= 600N C_y = 200N F_DC = 200N F_C B

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June 2015 Session : 2(c)

Figure below shows a truss is subjected to a horizontal force of 500N.

i. Calculate the force in each member of the truss. (16 marks) ii. Identify whether the members are in tension or

compression form. (4 marks)

Ans: (T) N 553 . 353 N 0 (T) N 250 (T) N 250 (C) N 553 . 353 N 250 N 250 N 500         CD BD BC AB AD y y x F F F F F C A C

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1. A truss is divided into two parts by taking an imaginary “cut”

through the truss.

2. Decide how you need to cut the truss:

a. Where you need to determine forces

b. Where the total number of unknown does not exceed than 3.

3. Decide which side of the cut truss will be easier to work with

(minimize the number of force you have to find).

4. If required, determine the necessary support reactions by drawing

the FBD of the entire truss.

Two Methods To Analyze Force In Simple Truss

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Example 5

Determine the force in members GE, GC & BC of the truss shown. Indicate whether the members are in tension or compression. Ans: (T) 500 (C) 800 (T) 800 N F N F N F GC GE BC   

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Solution:

+ve  ∑F_x= 0; 400N – A_x= 0  A_x= 400 N +ve ∑M_A= 0; 1200N (8m) + 400N (3m) - D_y(12m) = 0  D_y = 900 N +ve ↑ ∑F_y= 0; A_y– 1200N + 900= 0  A_y= 300 N

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Solution:

+ve ∑M_G= 0; 300N(4m) + 400N (3m) - F_BC(3m) = 0  F_BC= 800 N (T) +ve ∑M_c= 0; 300N (8m) - F_GE(3m) = 0  F_GE = 800 N (C) +ve ↑ ∑F_y= 0; 300N - (3/5)F_GC = 0  F_GC= 500 N (T)

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Example 6

Determine the force in members BC, CG & GF of the Warren truss shown. Indicate whether the members are in tension or compression. Ans: (C) kN 770 . 0 (C) kN 70 . 7 (T) kN 08 . 8    CG BC GF F F F

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Example 7

Determine the force in members CD, CF & FG of the Warren truss shown. Indicate whether the members are in tension or compression. Ans: (T) kN 770 . 0 (C) kN 47 . 8 (T) kN 08 . 8    CF CD FG F F F

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Example 8

Determine the force in members BC, HC & HG of the bridge truss and indicate whether the members are in tension or compression. Ans: (T) kN 0 . 12 (T) kN 5 . 20 (C) kN 0 . 29    HC BC HG F F F

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Example 9

Determine the force in members CD, CF & GF of the bridge truss and indicate whether the members are in tension or compression. Ans: (T) kN 78 . 7 (T) kN 5 . 23 (C) kN 0 . 29    CF CD GF F F F