Question 1 |
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 1 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 2 |
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 3 |
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 3 Explanation:
\frac{M}{6}=10kNm\Rightarrow M=60kNm
Question 4 |
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 4 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}
Question 5 |
A propped cantilever beam XY, with an internal hinge at the middle, is carrying a uniformly distributed load of 10 kN/m, as shown in the figure.
The vertical reaction at support X ( in kN, in integer) is _____
The vertical reaction at support X ( in kN, in integer) is _____
25 | |
30 | |
35 | |
40 |
Question 5 Explanation:
\begin{aligned} B M &=0 \text { at hinge } \\ R_{Y} \times 2-10 \times 2 \times 1 &=0 \\ R_{Y} &=10 \mathrm{kN} \\ R_{X}+R_{Y} &=10 \times 4 \\ R_{X}+10 &=40 \\ R_{X} &=30 \mathrm{kN} \end{aligned}
Question 6 |
The planar structure RST shown in the figure is roller-supported at S and pin-supported
at R. Members RS and ST have uniform flexural rigidity (EI) and S is a rigid joint. Consider
only bending deformation and neglect effects of self-weight and axial stiffening
When the structure is subjected to a concentrated horizontal load P at the end T, the magnitude of rotation at the support R, is
When the structure is subjected to a concentrated horizontal load P at the end T, the magnitude of rotation at the support R, is
\frac{PL^3}{12EI} | |
\frac{PL^2}{12EI} | |
\frac{PL^2}{6EI} | |
\frac{PL}{6EI} |
Question 6 Explanation:
\theta _R=\frac{(PL/2)L}{6EI}
\theta _R=\frac{PL^2}{12EI}
Question 7 |
A prismatic linearly elastic bar of length, L, cross-sectional area A, and made up of a
material with Young's modulus E, is subjected to axial tensile force as shown in the
figures. When the bar is subjected to axial tensile force P_1\; and\; P_2, the strain energies
stored in the bar are U_1\; and\; U_2, respectively.
If U is the strain energy stored in the same bar when subjected to an axial tensile force (P_1 + P_2), the correct relationship is
If U is the strain energy stored in the same bar when subjected to an axial tensile force (P_1 + P_2), the correct relationship is
U=U_1+ U_2 | |
U=U_1 - U_2 | |
U \lt U_1+ U_2 | |
U \gt U_1+ U_2 |
Question 7 Explanation:
(P_1+P_2)^2 \gt P_1^2+P_2^2
U \gt U_1+U_2
Question 8 |
A portal frame shown in figure 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, EI, of each member is 10 kNm^2. Under the applied load, the horizontal displacement (in mm, round off to 1 decimal place) of joint R would be _______
12.2 | |
25.0 | |
32.6 | |
38.8 |
Question 8 Explanation:
\begin{aligned} \Sigma M_p&=0\;\;[\text{Take clockwise as positive}]\\ R_R \times 5-50 \times 10&=0\\ R_R&=-100kN(\downarrow )\\ R_P&=100kN(\uparrow )\\ EI&=10^6kN-m^2\\ \end{aligned}
Using unit load method:
The horizontal displacement at 'R'
\delta _{HR}=\int_{0}^{L}\frac{M_xm_x}{EI}dx
where
M_x= BM at X-X due to real loads
m_x= BM at X-X due to vertical unit load applied where we want to find the deflection.
\begin{aligned} \Sigma M_R &=0\;\;[\text{Take clockwise as positive}]\\ R_P \times 5-1 \times 10&=0\\ R_P&=2kN(\uparrow )\end{aligned}
Sign conventions:
Sagging +ve
Hogging -ve
\delta _{HR}=\int_{0}^{5}\frac{200x^2}{10^6}dx+\int_{0}^{10}\frac{50x^2}{10^6}dx=\frac{1}{40}=0.025m=25mm
Question 9 |
The rigid-jointed plane frame QRS shown in the figure is subjected to a load P at the joint R. Let the axial deformations in the frame be neglected. If the support S undergoes a settlement of \Delta =\frac{PL^3}{\beta EI}, the vertical reaction at the support S will become zero when \beta is equal to
0.1 | |
3 | |
7.5 | |
48 |
Question 9 Explanation:
Adopting moment distribution method:
Thus, fnal vertical reaction at S (both due f_0 sinking and P)
\begin{aligned} R_S&=P-V \\ &= P-\frac{7.5EI}{L^3} \times \frac{PL^3}{\beta EI}\\ R_S&=\frac{P-7.5T}{\beta}\;\;\text{when}\; \beta =\text{ reaction at S = 0} \end{aligned}
Thus, fnal vertical reaction at S (both due f_0 sinking and P)
\begin{aligned} R_S&=P-V \\ &= P-\frac{7.5EI}{L^3} \times \frac{PL^3}{\beta EI}\\ R_S&=\frac{P-7.5T}{\beta}\;\;\text{when}\; \beta =\text{ reaction at S = 0} \end{aligned}
Question 10 |
A prismatic beam P-Q-R of flexural rigidity EI=1\times 10^{4}kNm^{2} 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 ______
The rotation at Q (in rad, up to two decimal places) is ______
0.01 | |
0.02 | |
0.06 | |
0.08 |
Question 10 Explanation:
\begin{aligned} K_{Q} &=K_{Q P}+K_{Q R} \\ &=\frac{4 E I}{5}+\frac{4 E I}{4}=1.8 \mathrm{EI} \\ \theta_{Q} &=\frac{M_{Q}}{K_{Q}}=\frac{180}{1.8 \times 10^{4}}=0.01 \mathrm{rad} \end{aligned}
Alternate solution
\begin{aligned} &\begin{aligned} M_{Q P} &=M_{f Q P}+\frac{2 E I}{l}\left[2 \theta_{Q}+\theta_{P}\right] \\ &=\frac{4 \times 10^{4}}{5} \times \theta_{Q}=8000 \theta_{Q} \\ M_{Q R}=& M_{f Q R}+\frac{2 E I}{l}\left[2 \theta_{Q}+\theta_{R}\right] \\ & \quad\left\{M_{f Q R}=0, \theta_{R}=0\right\} \\ &=\frac{2 \times 10^{4}}{4}\left[2 \theta_{Q}\right]=10000 Q \\ \Sigma M_{Q} &=0 \end{aligned} \\ &\Rightarrow \quad M_{Q P}+M_{Q R}+180=0 \\ & \Rightarrow \quad 18000 \cdot \theta_{Q}=-180 \\ & \Rightarrow \quad \theta_{Q}=0.01 \;\text{radian (anticlockwise)} \end{aligned}
There are 10 questions to complete.