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Civil Engineering (CE) Exam  >  Civil Engineering (CE) Tests  >  Test: Method Of Joints And Sections - Civil Engineering (CE) MCQ

Test: Method Of Joints And Sections - Civil Engineering (CE) MCQ


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17 Questions MCQ Test - Test: Method Of Joints And Sections

Test: Method Of Joints And Sections for Civil Engineering (CE) 2024 is part of Civil Engineering (CE) preparation. The Test: Method Of Joints And Sections questions and answers have been prepared according to the Civil Engineering (CE) exam syllabus.The Test: Method Of Joints And Sections MCQs are made for Civil Engineering (CE) 2024 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for Test: Method Of Joints And Sections below.
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Test: Method Of Joints And Sections - Question 1
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 How many equilibrium equations do we need to solve generally on each joint of a truss?

Detailed Solution for Test: Method Of Joints And Sections - Question 1

Answer: b
Explanation: Summation of forces in x and y direction should be equated to 0. Since there is no bending moments in trusses, we don’t need to solve the third equation.

Test: Method Of Joints And Sections - Question 2
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If a member of a truss is in compression, then what will be the direction of force that it will apply to the joints?

Detailed Solution for Test: Method Of Joints And Sections - Question 2

Answer: a
Explanation: Member will apply outward force. Joint will in turn apply inward force resulting in compression of the member.

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Test: Method Of Joints And Sections - Question 3
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 If a member of a truss is in tension, then what will be the direction of force that it will apply to the joints?

Detailed Solution for Test: Method Of Joints And Sections - Question 3

Answer: b
Explanation: Member will apply inward force. Joint will in turn apply outward force resulting in compression of the member.

Test: Method Of Joints And Sections - Question 4
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What should be ideally the first step to approach to a problem using method of joints?
Detailed Solution for Test: Method Of Joints And Sections - Question 4

Answer: c
Explanation: Identifying zero force members should always be the first step to approach any truss problem as it eliminate a lot of variables and is fairly easy.

Test: Method Of Joints And Sections - Question 5
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What should be the angle (in degrees) in the given system (part of a bigger system) if both of the members have to be a zero force member?
Detailed Solution for Test: Method Of Joints And Sections - Question 5

Answer: d
Explanation: 90o would mean that without any external force, each one would carry no force to satisfy equations of equilibrium.

Test: Method Of Joints And Sections - Question 6
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In the above figure, force is applied at joint c and its magnitude is 10N with downward direction.

Q.  Which of the following are 0 force members?
Detailed Solution for Test: Method Of Joints And Sections - Question 6

Answer: a
Explanation: FH, HE and AE are non-zero force member as there are directly transmitting load from the external support. So, by option elimination we can say that the answer is (a).

Test: Method Of Joints And Sections - Question 7
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In the above figure, force is applied at joint c and its magnitude is 10N with downward direction.  .

Q.  What will the magnitude of force (in N) transmitted by FI?
Detailed Solution for Test: Method Of Joints And Sections - Question 7

Answer: a
Explanation: GF is a zero force member as stated in earlier question. Now, in joint F, BF and FH are in a line. This means that the only remaining member FI which is not in line will transmit zero force.

Test: Method Of Joints And Sections - Question 8
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In the above figure, force is applied at joint c and its magnitude is 10N with downward direction.  .

Q.  What will the magnitude of force (in N) transmitted by IC?
Detailed Solution for Test: Method Of Joints And Sections - Question 8

Answer: a
Explanation: IH is a zero member force as is FI. So, IC too will be zero force members.

Test: Method Of Joints And Sections - Question 9
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In the above figure, force is applied at joint c and its magnitude is 10N with downward direction.  .

Q.   What is total no. of zero force members in the above given system?
Detailed Solution for Test: Method Of Joints And Sections - Question 9

Answer: c
Explanation: Following are the zero force member based on the logics explained above: – GF, HI, HJ, ED, FI, IC, CH, JE and JC.

Test: Method Of Joints And Sections - Question 10
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How many equilibrium equations are used in method of sections?
Detailed Solution for Test: Method Of Joints And Sections - Question 10

Answer: c
Explanation: Moments too can be conserved along with forces in both directions. So, total no. of equations are three.

Test: Method Of Joints And Sections - Question 11
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 In trusses, a member in the state of tension is subjected to:-
Detailed Solution for Test: Method Of Joints And Sections - Question 11

Answer: b
Explanation: Pull is for tension, while push is for compression.

Test: Method Of Joints And Sections - Question 12
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In method of sections, what is the maximum no. of unknown members through which the imaginary section can pass?
Detailed Solution for Test: Method Of Joints And Sections - Question 12

Answer: c
Explanation: Since we have three equilibrium equations, so we can have maximum 3 unknown forces/members through which imaginary section can pass.

Test: Method Of Joints And Sections - Question 13
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Method of substitute members is use for which type of trusses?
Detailed Solution for Test: Method Of Joints And Sections - Question 13

Answer: a
Explanation: Method of substitute members is used to solve problems involving complex trusses.

Test: Method Of Joints And Sections - Question 14
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First step to solve complex truss using Method of substitute members is to convert it into unstable simple truss.
State whether the above statement is true or false.
Detailed Solution for Test: Method Of Joints And Sections - Question 14

Answer: b
Explanation: First step is to convert it to stable simple truss.
Shear force is represented by V
Bending moment is represented by M
Distance along the truss is represented by X
W is the uniform load applied.

Test: Method Of Joints And Sections - Question 15
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 On differentiating V wrt X we will get:-
Detailed Solution for Test: Method Of Joints And Sections - Question 15

Answer: b
Explanation: On applying equilibrium equation, V – W(x)Δx – V – ΔV = 0.

Test: Method Of Joints And Sections - Question 16
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On differentiating M wrt X we will get:-
Detailed Solution for Test: Method Of Joints And Sections - Question 16

Answer: c
Explanation: On applying equilibrium equation, M + VΔx – M – ΔM = 0.

Test: Method Of Joints And Sections - Question 17
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 If a member of a truss is in compression, then what will be the direction of force that it will apply to the joints?
Detailed Solution for Test: Method Of Joints And Sections - Question 17

Answer: a
Explanation: Member will apply outward force. Joint will in turn apply inward force resulting in compression of the member.

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