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All questions of Truss Analysis for Civil Engineering (CE) Exam
A simply supported truss shown in the given figure carries a load of 20 kN at F the forces in the members EFand BE are respectively:
- a)zero and 10 kN (compression)
- b)zero and 10 kN (tension)
- c)10 kN (tension) and 10 kN (compression)
- d)10 kN (compression) and 10 kN (tension)
Correct answer is option 'B'. Can you explain this answer?
A simply supported truss shown in the given figure carries a load of 20 kN at F the forces in the members EFand BE are respectively:
a)
zero and 10 kN (compression)
b)
zero and 10 kN (tension)
c)
10 kN (tension) and 10 kN (compression)
d)
10 kN (compression) and 10 kN (tension)
|
Rahul Chauhan answered |
Force in EF is zero if joint equilibrium at E is considered force in SE will be tensile, and its value will be 10 kN by considering joint equilibrium at B.
Consider the following Assumptions in the analysis of a plane truss:
1. The individual members are straight.
2. The individual members are connected by frictionless hinges.
3. The loads and reactions acts only at the joints.Which of the following assumptions are valid?- a)1 and 2
- b)1 and 3
- c)2 and3
- d)1,2 and 3
Correct answer is option 'D'. Can you explain this answer?
Consider the following Assumptions in the analysis of a plane truss:
1. The individual members are straight.
2. The individual members are connected by frictionless hinges.
3. The loads and reactions acts only at the joints.
1. The individual members are straight.
2. The individual members are connected by frictionless hinges.
3. The loads and reactions acts only at the joints.
Which of the following assumptions are valid?
a)
1 and 2
b)
1 and 3
c)
2 and3
d)
1,2 and 3
|
Divya Kulkarni answered |
For plane truss assumptions are:
(i) Weight of all members are neglected. All members are straight.
(ii) All connections are smooth, frictionless pins.
(iii) External loads are applied directly to the pin joints.
(iv) All frames are perfect, hence statically determinate.
(i) Weight of all members are neglected. All members are straight.
(ii) All connections are smooth, frictionless pins.
(iii) External loads are applied directly to the pin joints.
(iv) All frames are perfect, hence statically determinate.
A truss ABC carries two horizontal and two vertical loads, as shown in the figure. The horizontal and vertical components of reactions at A will be
- a)HA = 1000 kN; VA = 1706.25 kN
- b)HA = 2000 kN; VA = 2550 kN
- c)HA = 1000 kN; VA = 2350 kN
- d)HA = 2000 kN; VA = 1706.25 kN
Correct answer is option 'D'. Can you explain this answer?
A truss ABC carries two horizontal and two vertical loads, as shown in the figure. The horizontal and vertical components of reactions at A will be
a)
HA = 1000 kN; VA = 1706.25 kN
b)
HA = 2000 kN; VA = 2550 kN
c)
HA = 1000 kN; VA = 2350 kN
d)
HA = 2000 kN; VA = 1706.25 kN
|
Muthu Selvi answered |
B
Axial forces in the members 1-2 and 1 - 5 of the truss shown in the given figure .are respectively
- a) 50 kN (compressive) and 25 kN (tensile)
- b)25 kN (tensile) and 50/√2 kN (compressive)
- c)100 kN (tensile) and 50√2 (compressive)
- d)25 kN (compressive) and 50/√2 k (tensile)
Correct answer is option 'C'. Can you explain this answer?
Axial forces in the members 1-2 and 1 - 5 of the truss shown in the given figure .are respectively
a)
50 kN (compressive) and 25 kN (tensile)
b)
25 kN (tensile) and 50/√2 kN (compressive)
c)
100 kN (tensile) and 50√2 (compressive)
d)
25 kN (compressive) and 50/√2 k (tensile)
|
Rohan Kumar answered |
Cutting a section through 1-2; 2-5; and 5-6. Taking moment of left part about 5,
Cut a section through 1-2, 1-5 and 4-5 and balance vertical force for left part only,
Cut a section through 1-2, 1-5 and 4-5 and balance vertical force for left part only,
The force induced in the vertical member CD of the symmetrical plane truss shown in the figure is
- a)50 kN (tension)
- b)100 kN (tension)
- c)50 kN (compression)
- d)zero
Correct answer is option 'A'. Can you explain this answer?
The force induced in the vertical member CD of the symmetrical plane truss shown in the figure is
a)
50 kN (tension)
b)
100 kN (tension)
c)
50 kN (compression)
d)
zero
|
|
Abhay Kapoor answered |
At join D, the equilibrium shows the force in member CD is 50 kN tensile.
Which one of the following is the correct statements regarding the force and deflection at point B in trusses I and II shown in the figure?
- a)I will have less member force and less deflection at B compared to II
- b)I will have less member force and more deflection at B compared to II
- c)I will have more member force and deflection at B compared to II
- d)I will have more member force and less deflection at B compared to II
Correct answer is option 'A'. Can you explain this answer?
Which one of the following is the correct statements regarding the force and deflection at point B in trusses I and II shown in the figure?
a)
I will have less member force and less deflection at B compared to II
b)
I will have less member force and more deflection at B compared to II
c)
I will have more member force and deflection at B compared to II
d)
I will have more member force and less deflection at B compared to II
|
|
Prerna Menon answered |
Force in AB or AC [Consider joint B],
So in case I force in members will be less than that in case II.
From symmetry, there will be only vertical deflection of joint B.
So in case I force in members will be less than that in case II.
From symmetry, there will be only vertical deflection of joint B.
Axial force in the member BC of the truss shown in the given figure is (where α = 30°)
- a)15 kN
- b)10√3 kN
- c)15√3 kN
- d)30 kN
Correct answer is option 'D'. Can you explain this answer?
Axial force in the member BC of the truss shown in the given figure is (where α = 30°)
a)
15 kN
b)
10√3 kN
c)
15√3 kN
d)
30 kN
|
Janhavi Sharma answered |
The given truss is symmetrical. So force in member BO and EO will be same in magnitude and nature both. Joint equilibrium at O.
Force in the member BC of the truss shown in the given figure is
- a)5t, tensile
- b)zero
- c)2.88t, compressive
- d)5t, compressive
Correct answer is option 'D'. Can you explain this answer?
Force in the member BC of the truss shown in the given figure is
a)
5t, tensile
b)
zero
c)
2.88t, compressive
d)
5t, compressive
|
Samridhi Khanna answered |
By considering joint equilibrium at B,
A simply supported truss shown in the given figure carries load as shown. The force in member BE is
- a)√2t (tensile)
- b)√2t (compressive)
- c)1t (tensile)
- d)1t (compressive)
Correct answer is option 'C'. Can you explain this answer?
A simply supported truss shown in the given figure carries load as shown. The force in member BE is
a)
√2t (tensile)
b)
√2t (compressive)
c)
1t (tensile)
d)
1t (compressive)
|
|
Naveen Sharma answered |
At joint D, there is no external force so force in FD and BD is zero.
Taking moment about A, the reaction at B,
Considering joint B, force in member BF,
force in member BE,
FBE=1t (tensile)
Taking moment about A, the reaction at B,
Considering joint B, force in member BF,
force in member BE,
FBE=1t (tensile)
In the pin-jointed plane truss shown below, what is the magnitude and nature of force in member BC?
- a)Zero
- b)50 kN (tensile)
- c)50 kN (compressive)
- d)
Correct answer is option 'A'. Can you explain this answer?
In the pin-jointed plane truss shown below, what is the magnitude and nature of force in member BC?
a)
Zero
b)
50 kN (tensile)
c)
50 kN (compressive)
d)
|
|
Pallabi Chavan answered |
At joint C there is no external force so force in member BC is zero.
The force in member FD in the given figure is
- a)50 kN compression
- b)50 kN tension
- c)150 kN tension
- d)100 kN compression
Correct answer is option 'B'. Can you explain this answer?
The force in member FD in the given figure is
a)
50 kN compression
b)
50 kN tension
c)
150 kN tension
d)
100 kN compression
|
Yash Verma answered |
Considering joint equilibrium at D. The force equilibrium equation in vertical direction gives FD = 50 kN tensile
A loaded pin-jointed truss is shown in the given figure. The force in member AC is
- a)
- b)
- c)zero
- d)10 kN (Tensile)
Correct answer is option 'C'. Can you explain this answer?
A loaded pin-jointed truss is shown in the given figure. The force in member AC is
a)
b)
c)
zero
d)
10 kN (Tensile)
|
|
Preethi Datta answered |
Cutting a section through members CA, CD, DEand EB and taking moment of the upper part about E.
A pin-jointed tower truss is loaded as shown in the below figure. The force induced in the member DF is
- a)1.5 kN (tension)
- b)4.5 kN (tension)
- c)1.5 kN (compression)
- d)4.5 kN (compression)
Correct answer is option 'C'. Can you explain this answer?
A pin-jointed tower truss is loaded as shown in the below figure. The force induced in the member DF is
a)
1.5 kN (tension)
b)
4.5 kN (tension)
c)
1.5 kN (compression)
d)
4.5 kN (compression)
|
Amar Mukherjee answered |
Cutting a section through AC, CD and DF and taking moment of upper portion about C.The forces in the members AC and CD meet at C so they do not produce any moment. The force in member DF,
In the cantilever truss shown in the given figure. the reaction at A is
- a)10t
- b)20t
- c)30t
- d)40t
Correct answer is option 'C'. Can you explain this answer?
In the cantilever truss shown in the given figure. the reaction at A is
a)
10t
b)
20t
c)
30t
d)
40t
|
Madhurima Mukherjee answered |
Only a horizontal reaction can exist at A.
Taking moments about hinge.
HA x 5 = 10 x 15
HA = 30t
HA x 5 = 10 x 15
HA = 30t
Chapter doubts & questions for Truss Analysis - Structural Analysis 2025 is part of Civil Engineering (CE) exam preparation. The chapters have been prepared according to the Civil Engineering (CE) exam syllabus. The Chapter doubts & questions, notes, tests & MCQs are made for Civil Engineering (CE) 2025 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests here.
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