Past Year Questions: Trusses Notes | EduRev

Topic wise GATE Past Year Papers for Civil Engineering

GATE : Past Year Questions: Trusses Notes | EduRev

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Q.1 Consider the pin-jointed plane truss shown in figure (not drawn to scale). Let RP, RQ, and RR denote the vertical reactions (upward positive) applied by the supports at P, Q and R, respectively, on the truss. The correct combination of (RP, RQ, RR) is represented by    [2019 : 1 Mark, Set-I]
Past Year Questions: Trusses Notes | EduRev
(a) (10, 30,-10) kN
(b) (30,-30, 30) kN
(c) (20, 0,10) kN
(d) (0,60,-30) kN
Ans.
(B)
Solution:
Past Year Questions: Trusses Notes | EduRev
Using, Past Year Questions: Trusses Notes | EduRev
Past Year Questions: Trusses Notes | EduRev ...(i)
Taking moment about P,
Past Year Questions: Trusses Notes | EduRev
Past Year Questions: Trusses Notes | EduRev
Again considering the equilibrium of RHS of section X-X
Past Year Questions: Trusses Notes | EduRev
∴ RR=-RQ ...(iii)
Past Year Questions: Trusses Notes | EduRev

Q.2 All the members of the planar truss (see figure), have the same properties in terms of area of cross-section (A) and modulus of elasticity (E),
Past Year Questions: Trusses Notes | EduRev
For the loads shown on the truss, the statement that correctly represents the nature of forces in the members of the truss is:    [2018 : 1 Mark, Set-II]
(a) There are 3 members in tension, and 2 members in compression
(b) There are 2 members in tension, 2 members in compression, and 1 zero-force member
(c) There are 2 members in tension, 1 member in compression, and 2 zero-force members
(d) There are 2 members in tension, and 3 zero- force Members
Ans. 
(D)
Solution:
Past Year Questions: Trusses Notes | EduRev
Since member BD neither elongate nor contract.
Hence, FBD 0
So there are 2 tension members (AB and DC) and 3 zero force members (AD, BD, BC).

Q.3 Consider the deformable pin-jointed truss with loading, geometry and section properties as shown in figure.
Past Year Questions: Trusses Notes | EduRev
Given that E = 2 x 1011 N/m2, A = 10 mm2, L = 1 m and P= 1 kN. the horizontal displacement of Joint C (in mm, up to one decimal place) is_____    [2018 : 2 Marks, Set-I]
Solution: 
Force is each member due to applied loading.
Past Year Questions: Trusses Notes | EduRev
Past Year Questions: Trusses Notes | EduRev
Force in each member due to unit load.

Q.4 Consider the structural system shown in the figure under the action of weight W. All the joints are hinged. The properties of the members in terms of length (L), are (A) and the modulus of elasticity (E) are also given in the figure. Let L, A and E be 1 m, 0.05 m2 and 30 x 106 N/m2, respectively, and W be 100 kN.
Past Year Questions: Trusses Notes | EduRev
Which one of the following sets gives the correct values of the force, stress and change in length of the horizontal member QR?    [2016 : 2 Marks, Set-II]
(a) Compressive force = 25 kN;
Stress = 250 kN/m2 ; Shortening = 0.0118 m
(b) Compressive force = 14.14 kN;
Stress = 141.4 kN/m2; Extension = 0.0118 m
(c) Compressive force = 100 kN;
Stress = 1000 kN/m2; Shortening = 0.0417 m
(d) Compressive force = 100 kN;
Stress = 1000 kN/m2; Extension = 0.0417 m
Ans.
(C)
Solution:
Past Year Questions: Trusses Notes | EduRev
Given data:
Past Year Questions: Trusses Notes | EduRev
Consider joint 'S’,
Past Year Questions: Trusses Notes | EduRev
Past Year Questions: Trusses Notes | EduRev
Past Year Questions: Trusses Notes | EduRev
Past Year Questions: Trusses Notes | EduRev
Past Year Questions: Trusses Notes | EduRev
Past Year Questions: Trusses Notes | EduRev
As the truss is symmetrical,
Past Year Questions: Trusses Notes | EduRev  (Tensile)
Now consider joint 'Q,
Past Year Questions: Trusses Notes | EduRev
Past Year Questions: Trusses Notes | EduRev
Past Year Questions: Trusses Notes | EduRev
Past Year Questions: Trusses Notes | EduRev
Past Year Questions: Trusses Notes | EduRev (Compressive)
Stress in member OR,
Past Year Questions: Trusses Notes | EduRev
Past Year Questions: Trusses Notes | EduRev
Past Year Questions: Trusses Notes | EduRev
Past Year Questions: Trusses Notes | EduRev
As the member QR is subjecterd to compression, it will go under shortening.
∴ Shortening,
Past Year Questions: Trusses Notes | EduRev

Past Year Questions: Trusses Notes | EduRev
∴ Δ = 0.0471 m

Q.5 A plane truss with applied loads is shown in the figure.
Past Year Questions: Trusses Notes | EduRev
The members which do not carry any force are    [2016 : 2 Marks, Set-I]
(a) FT, TG, HU, MP, PL
(b) ET, GS, UR, VR, QL
(c) FT, GS, HU, MP, QL
(d) MP, PL, HU, FT, UR
Ans. 
(A)
Solution:
If there members meet at a joint and out of them are collinear, then non collinear member will carry zero force provided that there is no external load at the joint.
Use this statement to check the members with zero force.

Q.6 Consider the plane truss with load Pas shown in the figure. Let the horizontal and vertical reactions at the joint B be HB and VB, respectively and VC be vertical reaction at the joint C.
Past Year Questions: Trusses Notes | EduRev
Which one of the following sets gives the correct values of VB, Hand VC?   [2016 : 1 Mark, Set-I]
(a) VB = 0; HB = 0; VC = P
(b) VB = P/2; HB = 0; VC = P/2
(c) VB = P/2; HB = P (sin600); VC = P/2
(d) VB = P: HB = P (cos(600); VC = 0
Ans.
(A)
Solution:
Past Year Questions: Trusses Notes | EduRev
⇒ VB + VC = P ...(i)
Past Year Questions: Trusses Notes | EduRev
Past Year Questions: Trusses Notes | EduRev

Q.7 For the 2D truss with the applied loads shown below, the strain energy in the member XY is __________ kN-m. For member XY, assume AE = 30 kN, where A is cross-section area and E is the modulus of elasticity.    [2015 : 2 Marks, Set-I]
Past Year Questions: Trusses Notes | EduRev
Solution: 

First calculating reactions,
Past Year Questions: Trusses Notes | EduRev
Past Year Questions: Trusses Notes | EduRev
Past Year Questions: Trusses Notes | EduRev
Past Year Questions: Trusses Notes | EduRev
RA = -30 kN
Let us cut a section 1-1 as shown in figure and consider the lower part.
Past Year Questions: Trusses Notes | EduRev
Given AE = 30 kN and L = 3m,
Calculation for force Pin XY member.
Past Year Questions: Trusses Notes | EduRev
Past Year Questions: Trusses Notes | EduRev
Strain energy,
Past Year Questions: Trusses Notes | EduRev

Q.8 For the truss shown below, the member PQ is short by 3 mm. The magnitude of vertical displacement of joint R (in mm) is _______ .    [2014 : 2 Marks, Set-I]
Past Year Questions: Trusses Notes | EduRev
Solution:
PQ is short by 3 mm
We have to find out vertical displacement of joint R in mm,
Past Year Questions: Trusses Notes | EduRev
Apply unit load at R as shown below
Past Year Questions: Trusses Notes | EduRev
Due to symmetric loading, Past Year Questions: Trusses Notes | EduRev
Consider P:
∑V = 0,
Past Year Questions: Trusses Notes | EduRev
Past Year Questions: Trusses Notes | EduRev
Past Year Questions: Trusses Notes | EduRev 

Q.9 Mathematical idealization of a crane has three bars with their vertices arranged as shown in the figure with a load of 80 kN hanging vertically. The coordinates of the vertices are given in parentheses. The force in the member QR, FQR will be   [2014 : 2 Marks, Set-I]
Past Year Questions: Trusses Notes | EduRev
(a) 30 kN Compressive
(b) 30 kN Tensile
(c) 50 kN Compressive
(d) 50 kN Tensile
Ans.
(A)
Solution:
Past Year Questions: Trusses Notes | EduRev
Past Year Questions: Trusses Notes | EduRev
Consider joint Q,
Past Year Questions: Trusses Notes | EduRev
Past Year Questions: Trusses Notes | EduRev

Q.10 A uniform beam weighing 1800 N is supported at E & F by cable ABCD. Determine the tension force in segment AB at this cable (correct to 1 decimal place). Assume the cable ABCD, BE and CF are weightless    [2013 : 2 Marks]
Past Year Questions: Trusses Notes | EduRev
Solution:

Past Year Questions: Trusses Notes | EduRev 
Taking moment about A,
RD x 4 = 1800 x 1.5
⇒ RD = 675 N
RA = 1800 - RD = 1125 N
At joint D,
Past Year Questions: Trusses Notes | EduRev
Past Year Questions: Trusses Notes | EduRev
Taking free body diagram as shown in figure,
Past Year Questions: Trusses Notes | EduRev
Past Year Questions: Trusses Notes | EduRev
Past Year Questions: Trusses Notes | EduRev 

Q.11 The pin-jointed 2-D truss is loaded with a horizontal force of 15 kN at joint S and another 15 kN vertical force at joint U as shown in figure. Find the force in member RS (in kN) and report your answer taking tension as +ve and compression as -ve.    [2013 : 1 Mark]
Past Year Questions: Trusses Notes | EduRev
Solution:

Take moment about V,Mv = 0,
⇒ RH, x 4 = 0
∴ RH = 0
Past Year Questions: Trusses Notes | EduRev
Past Year Questions: Trusses Notes | EduRev
We know that if two members meet at a. joint which are not collinear and also there is no external forces acting on that joint, then both members will carry zero forces.
∴ FQV - FQR = 0
Now, consider joint P,
Past Year Questions: Trusses Notes | EduRev
Joint V,
Past Year Questions: Trusses Notes | EduRev
Past Year Questions: Trusses Notes | EduRev
Now, member RS and RU area non-collinear members meeting at a joint with no external force acting on the joint.
Past Year Questions: Trusses Notes | EduRev

Q.12 For the truss shown in the figure, the force in the member QR is    [2010 : 1 Mark]
Past Year Questions: Trusses Notes | EduRev
(a) zero
(b) P/√2
(c) P
(d) √2 P
Ans.
(C)
Solution:
Using method of joints and considering joint S, we get,
Past Year Questions: Trusses Notes | EduRev
Past Year Questions: Trusses Notes | EduRev
Considering joint R, we get
Past Year Questions: Trusses Notes | EduRev
Past Year Questions: Trusses Notes | EduRev 

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