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# Past Year Questions: Methods of Structural Analysis Notes | EduRev

## GATE : Past Year Questions: Methods of Structural Analysis Notes | EduRev

The document Past Year Questions: Methods of Structural Analysis Notes | EduRev is a part of the GATE Course Structural Analysis.
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Q.1 A portal frame shown in figure (not drawn to scale) has a hinge support at joint P and a roller support at joint R. A point load of 50 kN is acting at joint R in the horizontal direction. The flexural rigidity. El, of each member is 106 kNm2. Under the applied load, the horizontal displacement (in mm, round off to 1 decimal place) of joint R would be______.    [2019 : 2 Marks, Set-I] Solution:    Q.2 The rigid-jointed plane frame QRS shown in figure is subjected to load P at the joint R. Let the axial deformations in the frame be neglected. If the support S undergoes a settlement of the vertical reaction at the support S will be become zero when β is equal to    [2019 : 2 Marks, Set-I] (a) 7.5
(b) 3.0
(c) 48.0
(d) 0.1
Ans. (A)

Solution:
Assume 'R sinks by Δ.
Sway analysis:  End moment distribution, Sway force  Alternate method:   Q.3 A prismatic beam P-Q-R of flexural rigidity El = 1 x 104 kNm2 is subjected to a moment of 180 kNm at Q as shown in the figure. The rotation at Q (in rad, up to two decimal places) is ________ .    [2018 : 2 Marks, Set-II]
Solution:

Method-1  Method-ll     Q.4 A vertical load of 10 kN acts on a hinge located at a distance of LI4 from the roller support Q of a beam of length L (see figure).    [2018 :1 Mark, Set-II] The vertical reaction at support Q is
(a) 0.0 kN
(b) 2.5 kN
(c) 7.5 kN
(d) 10.0 kN
Ans.
(A)
Solution: Bending moment about hinge point A = 0
(consider the right hand side of A) Q.5 Consider the portal frame shown in the figure and assume the modulus of elasticity, E = 2.5 x 104 MPa and the moment of inertia, I = 8 x 108 mm4 for all the members of the frame. The rotation (in degrees, up to one decimal place) at the rigid joint Q would b e _________ .    [2017 : 2 Marks, Set-II)
Solution:

Method-I  = -700 kNm (ACW)
EI = 2.5 x 104 x 8 x 108
= 20 x 1012 Nmm2
= 20000 kNm2
As the two members QR and QS are identical, moment will be equally divided i.e., 350 kNm each. Method-ll Fixed end moment: Slope deflection equation: Joint equilibrium equation at joint Q, Q.6 The value of M in the beam ABC shown in the figure is such that the joint B does not rotate. The value of support reaction (in kN) at B should be equal to ________.    [2017 : 2 Marks, Set-I]
Solution:

Method-I
Fixed end moment:  By super position, Note : The value of 'M' shown in the figure, Q.7 The portal frame shown in the figure is subjected to a uniformly distributed vertical load w(per unit length). The bending moment in the beam at the join ‘Q ’ is    [2016 : 2 Marks, Set-II]
(a) zero
(b) (c) (d) Ans.
(A)
Solution: As there is no horizontal force,
Hence,
HP = HS = 0 ∴ BM at Q = 0

Q.8 For the beam shown below, the value of the support moment M is ____ kN-m.    [2015 : 1 Mark, Set-I] Solution:

By symmetry we can consider two parts of beam separately with half the load. Moment at joint B = 10 kNm
Carry over moment at joint A due to 10 kNm moment at propped end = 10/2= 5kN-m

Q.9 Considering the symmetry of a rigid frame as shown below, the magnitude of the bending moment (in kNm) at P (preferably using the moment distribution method) is    [2014 : 2 Marks, Set-II] (a) 170
(b) 172
(c) 176
(d) 178
Ans.
(C)
Solution:
As the frame and loading is symmetrical, θP= 0. So, joint Pcan be replaced by a fixed support as shown. Fixed end moment: Distribution factor:  Q.10 All members in the rigid-jointed frame shown are prismatic and have the same flexural stiffness EI. Find the magnitude of BM at Q (in kN-m) due to given loading.    [2013 : 2 Marks] Solution:  Offer running on EduRev: Apply code STAYHOME200 to get INR 200 off on our premium plan EduRev Infinity!

## Structural Analysis

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