Test: Trusses -1 - Civil Engineering (CE) MCQ

# Test: Trusses -1 - Civil Engineering (CE) MCQ

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## 10 Questions MCQ Test GATE Civil Engineering (CE) 2025 Mock Test Series - Test: Trusses -1

Test: Trusses -1 for Civil Engineering (CE) 2024 is part of GATE Civil Engineering (CE) 2025 Mock Test Series preparation. The Test: Trusses -1 questions and answers have been prepared according to the Civil Engineering (CE) exam syllabus.The Test: Trusses -1 MCQs are made for Civil Engineering (CE) 2024 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for Test: Trusses -1 below.
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Test: Trusses -1 - Question 1

### A simply supported truss shown in the given figure carries load as shown. The force in member BE is

Detailed Solution for Test: Trusses -1 - Question 1

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)

Test: Trusses -1 - Question 2

### A cantilever pin-jointed truss carries one load P at the point Z as shown in the given figure. The hinges in the vertical wall are at A and D. The truss has only three horizontal members AC, CG and GZ. The nature of force in member AB

Test: Trusses -1 - Question 3

### The force in member FD in the given figure is

Detailed Solution for Test: Trusses -1 - Question 3

Considering joint equilibrium at D. The force equilibrium equation in vertical direction gives FD = 50 kN tensile

Test: Trusses -1 - Question 4

In the cantilever truss shown in the given figure. the reaction at A is

Detailed Solution for Test: Trusses -1 - Question 4

Only a horizontal reaction can exist at A.

HA x 5 = 10 x 15
HA = 30t

Test: Trusses -1 - Question 5

Force in the member BC of the truss shown in the given figure is

Detailed Solution for Test: Trusses -1 - Question 5

By considering joint equilibrium at B,

Test: Trusses -1 - Question 6

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:

Detailed Solution for Test: Trusses -1 - Question 6

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.

Test: Trusses -1 - Question 7

Axial force in the member BC of the truss shown in the given figure is (where α = 30°)

Detailed Solution for Test: Trusses -1 - Question 7

The given truss is symmetrical. So force in member BO and EO will be same in magnitude and nature both. Joint equilibrium at O.

Test: Trusses -1 - Question 8

Axial forces in the members 1-2 and 1 - 5 of the truss shown in the given figure .are respectively

Detailed Solution for Test: Trusses -1 - Question 8

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,

Test: Trusses -1 - Question 9

A pin-jointed tower truss is loaded as shown in the below figure. The force induced in the member DF is

Detailed Solution for Test: Trusses -1 - Question 9

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,

Test: Trusses -1 - Question 10

For the truss shown in the figure, which one of the following members has zero force induced in it?

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## GATE Civil Engineering (CE) 2025 Mock Test Series

31 docs|280 tests