Luqman Nul Hakim Bin Juwara

Jabatan Kejuruteraan Mekanikal

Politeknik Muadzam Shah

luqman@pms.edu.my

DJJ3053

ENGINEERING MECHANICS

CHAPTER 4 :

STRUCTURES

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

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.

Structures

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.

Trusses

Trusses

June 2015 Session : 2(a)

Describe a plane truss. (3 marks)

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

Simple Trusses

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.

Assumption For Design

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 SENSE of an unknown member force

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, ∑Fx = 0

and ∑Fy = 0.

Method of JOINTS

Truss

Simple Beam

Two Methods To Analyze Force In Simple Truss

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

NA

NA

NC

NF

NF

NF

y

x

y

CA

BA

BC

Solution:

+ve ∑Fx = 0; 500 N + FBC sin 45o = 0

FBC = -707 N = 707 N (C)

+ve ↑ ∑Fy = 0; -FBC cos 45o – FBA = 0

-(-707 N) cos 45o = FBA

FBA = 500 N (T)

Joint B :

Solution: Joint C :

+ve ∑Fx = 0; -FCA + 707 N cos 45o = 0

FCA = 500 N (T)

+ve ↑ ∑Fy = 0; -707 sin 45o + Cy = 0

Cy = 500 N

Solution: Joint A :

+ve ∑Fx = 0; 500 N – Ax = 0

Ax = 500 N (T)

+ve ↑ ∑Fy = 0; 500 N – Ay = 0

Ay = 500 N (T)

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

NF

NF

NF

F

NF

NF

DE

DC

BC

BD

AB

AD

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:

NC

NF

NF

NF

y

CA

BC

BA

43.271

(C) 271

(T) 86.383

(T) 71.285

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

NC

NF

NF

NF

NA

NF

NF

NC

NA

NC

y

CB

DC

DB

y

AD

AB

y

y

x

Solution: Support Reaction :

+ve ∑Fx = 0; 600N – Cx = 0 Cx = 600 N (C)

+ve ∑Mc = 0; Ay(6m) – 400N(3m) – 600N(4m) = 0

Ay = 600 N (C)

+ve ↑ ∑Fy = 0; Ay – 400N – Cy = 0 Cy = 200 N (C)

Solution:

+ve ↑ ∑Fy = 0; 600N + (⅘)FAB = 0 FAB = -750 N

FAB = 750 N (C)

+ve ∑Fx = 0; FAD + (⅗)FAB = 0 FAD = 450 N (T)

Joint A :

Ay = 600N

FAD

FAB

3

4 5

Solution:

+ve ∑Fx = 0; -450N – (⅗)FDB + 600N = 0

FDB = 250 N (T)

+ve ↑ ∑Fy = 0; (⅘)FDB + FDC = 0 FDC = -200 N

FDC = 200 N (C)

Joint D :

FDC

600 N

FDB

FAD = 450N 3

4 5

Solution:

+ve ∑Fx = 0; -600N – FCB = 0 FCB = -600 N

FCB = 600 N (C)

+ve ↑ ∑Fy = 0; FDC– Cy = 0 FDC = Cy = 200 N (C)

Joint C :

Cx = 600N

Cy = 200N

FDC =

200N

FC

B

Solution:

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

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

Method of SECTIONS

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

NF

NF

NF

GC

GE

BC

Solution:

+ve ∑Fx = 0; 400N – Ax = 0

Ax = 400 N

+ve ∑MA = 0; 1200N (8m) + 400N (3m) - Dy (12m) = 0

Dy = 900 N

+ve ↑ ∑Fy = 0; Ay – 1200N + 900= 0

Ay = 300 N

Solution:

+ve ∑MG = 0; 300N(4m) + 400N (3m) - FBC (3m) = 0

FBC = 800 N (T)

+ve ∑Mc = 0; 300N (8m) - FGE (3m) = 0

FGE = 800 N (C)

+ve ↑ ∑Fy = 0; 300N - (3/5)FGC = 0

FGC = 500 N (T)

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

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

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

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

QUESTION

&

ANSWER SESSION

Thank You

For Listening