Luqman Nul Hakim Bin Juwara Jabatan Kejuruteraan Mekanikal Politeknik Muadzam Shah [email protected] DJJ3053 ENGINEERING MECHANICS CHAPTER 4 : STRUCTURES
Luqman Nul Hakim Bin Juwara
Jabatan Kejuruteraan Mekanikal
Politeknik Muadzam Shah
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