# Methods of Structural Analysis

 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
 A 0.375 wL B 0.453 wL C 0.482 wL D 0.500 wL
GATE CE 2021 SET-2   Structural Analysis
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}
 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 _____
 A 25 B 30 C 35 D 40
GATE CE 2021 SET-2   Structural Analysis
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
 A $\frac{PL^3}{12EI}$ B $\frac{PL^2}{12EI}$ C $\frac{PL^2}{6EI}$ D $\frac{PL}{6EI}$
GATE CE 2020 SET-2   Structural Analysis
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
 A $U=U_1+ U_2$ B $U=U_1 - U_2$ C $U \lt U_1+ U_2$ D $U \gt U_1+ U_2$
GATE CE 2020 SET-2   Structural Analysis
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 _______ A 12.2 B 25 C 32.6 D 38.8
GATE CE 2019 SET-1   Structural Analysis
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}
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 A 0.1 B 3 C 7.5 D 48
GATE CE 2019 SET-1   Structural Analysis
Question 6 Explanation: 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 ______
 A 0.01 B 0.02 C 0.06 D 0.08
GATE CE 2018 SET-2   Structural Analysis
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
 A 0.0 kN B 2.5 kN C 7.5 kN D 10.0 kN
GATE CE 2018 SET-2   Structural Analysis
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
 A 0 B ML/(9EI) C ML/(6EI) D ML/(3EI)
GATE CE 2018 SET-1   Structural Analysis
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 ___
 A 0.017 B 1.201 C 0.034 D 1.003
GATE CE 2017 SET-2   Structural Analysis
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. 