Question 1 |
In the frame shown in the figure (not to scale), all four members (A B, B C, C D, and A D) have the same length and same constant flexural rigidity. All the joints A, B, C, and D are rigid joints. The midpoints of A B, B C, C D, and A D, are denoted by E, F, G, and H, respectively. The frame is in unstable equilibrium under the shown forces of magnitude P acting at E and G. Which of the following statements is/are TRUE?
Shear forces at \mathrm{H} and \mathrm{F} are zero | |
Horizontal displacement at \mathrm{H} and \mathrm{F} are zero | |
Vertical displacement at \mathrm{H} and \mathrm{F} are zero | |
Slopes at E, F, G, and \mathrm{H} are zero |
Question 1 Explanation:
Due to symmetry horizontal displacement of \mathrm{H} \& F= zero
\Rightarrow option (B) correct
Also due to symmetry slopes at E, F, G, H =0 \Rightarrow option (D) is correct
Since A D and B C are subjected to pure bending Hence shear force at \mathrm{H} \& \mathrm{F} are zero
\Rightarrow option (A) correct
Vertical displacement at \mathrm{H} \& \mathrm{~F} \neq 0
\Rightarrow option (C) is incorrect.
Question 2 |
The linearly elastic planar structure shown in the figure is acted upon by two
vertical concentrated forces. The horizontal beams UV and WX are connected
with the help of the vertical linear spring with spring constant k = 20 kN/m. The
fixed supports are provided at U and X. It is given that flexural rigidity
EI = 10^5 kN-m^2, P = 100 kN, and a = 5 m. Force Q is applied at the center of beam WX such that the force in the spring VW becomes zero.
The magnitude of force Q (in kN) is ________. (round off to the nearest integer)
The magnitude of force Q (in kN) is ________. (round off to the nearest integer)
640 | |
458 | |
251 | |
325 |
Question 2 Explanation:
If force in spring is zero, there will be no
deformation in spring i.e., deflection of point V
will be equal to deflection of point W
\begin{aligned} \delta _v&=\delta _w\\ \frac{P(2a)^3}{3EI}&=\frac{Q(a)^3}{3(2EI)}+\frac{Q(a)^2}{2(2EI)} +a\\ \frac{8}{3}P&=\left ( \frac{1}{6}+\frac{1}{4} \right )Q\\ Q&=\frac{32}{5}P=640kN \end{aligned}
\begin{aligned} \delta _v&=\delta _w\\ \frac{P(2a)^3}{3EI}&=\frac{Q(a)^3}{3(2EI)}+\frac{Q(a)^2}{2(2EI)} +a\\ \frac{8}{3}P&=\left ( \frac{1}{6}+\frac{1}{4} \right )Q\\ Q&=\frac{32}{5}P=640kN \end{aligned}
Question 3 |
Which of the following statement(s) is/are correct?
If a linearly elastic structure is subjected to a set of loads, the partial derivative
of the total strain energy with respect to the deflection at any point is equal to
the load applied at that point. | |
If a linearly elastic structure is subjected to a set of loads, the partial derivative
of the total strain energy with respect to the load at any point is equal to the
deflection at that point. | |
If a structure is acted upon by two force system Pa and Pb , in equilibrium
separately, the external virtual work done by a system of forces Pb during the
deformations caused by another system of forces Pa is equal to the external
virtual work done by the Pa system during the deformation caused by the Pb
system. | |
The shear force in a conjugate beam loaded by the M/EI diagram of the real
beam is equal to the corresponding deflection of the real beam. |
Question 4 |
Consider the linearly elastic plane frame shown in the figure. Members HF, FK
and FG are welded together at joint F. Joints K, G and H are fixed supports. A
counter-clockwise moment M is applied at joint F. Consider flexural rigidity
EI=10^5kN-m^2
for each member and neglect axial deformations.
If the magnitude (absolute value) of the support moment at H is 10 kN-m, the magnitude (absolute value) of the applied moment M (in kN-m) to maintain static equilibrium is ___________. (round off to the nearest integer)
If the magnitude (absolute value) of the support moment at H is 10 kN-m, the magnitude (absolute value) of the applied moment M (in kN-m) to maintain static equilibrium is ___________. (round off to the nearest integer)
60 | |
25 | |
85 | |
45 |
Question 4 Explanation:
\frac{M}{6}=10kNm\Rightarrow M=60kNm
Question 5 |
A frame EFG is shown in the figure. All members are prismatic and have equal flexural rigidity. The member FG carries a uniformly distributed load w per unit length. Axial deformation of any member is neglected.
Considering the joint F being rigid, the support reaction at G is
Considering the joint F being rigid, the support reaction at G is
0.375 wL | |
0.453 wL | |
0.482 wL | |
0.500 wL |
Question 5 Explanation:
Compatibility condition
\begin{array}{l} \qquad \qquad \qquad \qquad \qquad \frac{\partial U}{\partial R}=0 \\ \Rightarrow \qquad \qquad \frac{\partial}{\partial R}\left[\int \frac{M^{2}}{2 E \mid} d x\right]=0 \\ \Rightarrow \frac{\partial}{\partial R}\left[\int_{0}^{L} \frac{\left(R x-\frac{w x^{2}}{2}\right)^{2}}{2 E I} d x+\int_{0}^{2 L} \frac{\left(R L-\frac{w L^{2}}{2}\right)^{2}}{2 E I} d x\right]=0 \\ \Rightarrow \int_{0}^{L} \frac{2\left(R x-\frac{w x^{2}}{2}\right)}{2 E I}(x) d x+\int_{0}^{2 L} \frac{\left(R L-\frac{w L^{2}}{2}\right)}{E I}(L) d x=0 \\ \Rightarrow \frac{R L^{3}}{3}-\frac{W}{2} \times \frac{L^{4}}{4}+R L^{2}(2 L)-\frac{W^{3}}{2}(2 L)=0 \\ \Rightarrow \frac{R L^{3}}{3}-\frac{w^{4}}{8}+2 R L^{3}-w L^{4}=0 \\ \Rightarrow 2 R L^{3}+\frac{R L^{3}}{3}=\frac{w^{4}}{8}+w^{4}=\frac{9}{8} w^{4} \\ \Rightarrow \frac{7 R L^{3}}{3}=\frac{9}{8} w L^{4} \\ \Rightarrow R=\frac{27}{56} w L=0.482 w L \end{array}
Alternative
\begin{aligned} \frac{w L^{4}}{8 E I}+\frac{w L^{2}(2 L)}{2 E I} \times L &=\frac{R L^{3}}{3 E I}+\frac{R L \times 2 L \times L}{E I} \\ \frac{w^{4}}{8 E I}+\frac{w^{4}}{E I} &=\frac{R L^{3}}{3 E I}+\frac{2 R L^{3}}{E I} \\ \frac{9 W^{4}}{8 E I} &=\frac{7 R L^{3}}{3 E I} \\ R &=\frac{27 w L}{56}=0.482 \mathrm{wL} \end{aligned}
\begin{array}{l} \qquad \qquad \qquad \qquad \qquad \frac{\partial U}{\partial R}=0 \\ \Rightarrow \qquad \qquad \frac{\partial}{\partial R}\left[\int \frac{M^{2}}{2 E \mid} d x\right]=0 \\ \Rightarrow \frac{\partial}{\partial R}\left[\int_{0}^{L} \frac{\left(R x-\frac{w x^{2}}{2}\right)^{2}}{2 E I} d x+\int_{0}^{2 L} \frac{\left(R L-\frac{w L^{2}}{2}\right)^{2}}{2 E I} d x\right]=0 \\ \Rightarrow \int_{0}^{L} \frac{2\left(R x-\frac{w x^{2}}{2}\right)}{2 E I}(x) d x+\int_{0}^{2 L} \frac{\left(R L-\frac{w L^{2}}{2}\right)}{E I}(L) d x=0 \\ \Rightarrow \frac{R L^{3}}{3}-\frac{W}{2} \times \frac{L^{4}}{4}+R L^{2}(2 L)-\frac{W^{3}}{2}(2 L)=0 \\ \Rightarrow \frac{R L^{3}}{3}-\frac{w^{4}}{8}+2 R L^{3}-w L^{4}=0 \\ \Rightarrow 2 R L^{3}+\frac{R L^{3}}{3}=\frac{w^{4}}{8}+w^{4}=\frac{9}{8} w^{4} \\ \Rightarrow \frac{7 R L^{3}}{3}=\frac{9}{8} w L^{4} \\ \Rightarrow R=\frac{27}{56} w L=0.482 w L \end{array}
Alternative
\begin{aligned} \frac{w L^{4}}{8 E I}+\frac{w L^{2}(2 L)}{2 E I} \times L &=\frac{R L^{3}}{3 E I}+\frac{R L \times 2 L \times L}{E I} \\ \frac{w^{4}}{8 E I}+\frac{w^{4}}{E I} &=\frac{R L^{3}}{3 E I}+\frac{2 R L^{3}}{E I} \\ \frac{9 W^{4}}{8 E I} &=\frac{7 R L^{3}}{3 E I} \\ R &=\frac{27 w L}{56}=0.482 \mathrm{wL} \end{aligned}
There are 5 questions to complete.