Question 1 |
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 1 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 2 |
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 2 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 3 |
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 3 Explanation:
\theta _R=\frac{(PL/2)L}{6EI}
\theta _R=\frac{PL^2}{12EI}
Question 4 |
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 4 Explanation:
(P_1+P_2)^2 \gt P_1^2+P_2^2
U \gt U_1+U_2
Question 5 |
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 5 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 6 |
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 6 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 7 |
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 7 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}
Question 8 |
A vertical load of 10 kN acts on a hinge located at a distance of L/4 from the roller support Q of a beam of length L (see figure).
The vertical reaction at support Q is
The vertical reaction at support Q is
0.0 kN | |
2.5 kN | |
7.5 kN | |
10.0 kN |
Question 8 Explanation:
Bending moment about hinge point A=0
(consider the right hand side of A )
\begin{aligned} R_{0} \times \frac{L}{4} &=0 \\ R_{0} &=0 \mathrm{kN} \end{aligned}
Question 9 |
The figure shows a simply supported beam PQ of uniform flexural rigidity EI carrying two moments M and 2M.
The slope at P will be
The slope at P will be
0 | |
ML/(9EI) | |
ML/(6EI) | |
ML/(3EI) |
Question 9 Explanation:
\begin{array}{c} \text { Now } \quad R_{1}+R_{2}=\frac{1}{2} \times \frac{L}{3} \times \frac{M}{E I}=\frac{M L}{6 E I} \\ \Sigma M_{O}=0 \\ R_{1} L-\frac{1}{2} \times \frac{L}{3} \times \frac{M}{E I}\left(\frac{2 L}{3}+\frac{L}{9}\right)-\frac{1}{2} \times \frac{L}{3} \\ \times \frac{M}{E I}\left(\frac{L}{3}+\frac{L}{9}\right)+\frac{1}{2} \times \frac{L}{3} \times \frac{M}{E I} \times \frac{2 L}{9}=0\\ \quad R_{1}=\frac{7 M L}{54 E I}+\frac{4 M L}{54 E I}-\frac{M L}{27 E I}=\frac{M L}{6 E I} \\ \therefore \text{Slope at }P, \theta_{P}=R_{1}=\frac{M L}{6 E I} \end{array}
Question 10 |
Consider the portal frame shown in the figure and assume the modulus of elasticity E=2.5\times 10^{4} MPa and the moment of inertial, I=8\times 10^{8}\; mm^{4} for all the members of the frame.
The rotations (in degrees, up to decimal place) at the rigid joint Q would be ___
The rotations (in degrees, up to decimal place) at the rigid joint Q would be ___
0.017 | |
1.201 | |
0.034 | |
1.003 |
Question 10 Explanation:
\begin{aligned} \text { Moment at } Q &=\frac{1650 \times 2^{2}}{2}-2000 \times 2 \\ &=-700 \mathrm{kNm}(\mathrm{ACW}) \\ &=2.5 \times 10^{4} \times 8 \times 10^{8} \\ &=20 \times 10^{12} \mathrm{Nmm}^{2} \\ &=20000 \mathrm{kNm}^{2} \\ \theta_{Q} &=\frac{M l}{4 E I}=\frac{350 \times 4}{4 \times 20000} \\ &=0.0175 \mathrm{rad} \\ &=\frac{0.0175 \times 180^{\circ}}{\pi} \\ &=1.0027^{\circ} \end{aligned}
There are 10 questions to complete.