# Bending of Beams

 Question 1
A cantilever beam with a uniform flexural rigidity $(EI = 200 \times 10^6 N.m^2)$ is loaded with a concentrated force at its free end. The area of the bending moment diagram corresponding to the full length of the beam is $10000 \;N.m^2$. The magnitude of the slope of the beam at its free end is ________micro radian (round off to the nearest integer).
 A 42 B 50 C 65 D 84
GATE ME 2021 SET-2   Strength of Materials
Question 1 Explanation:

Assume:
\begin{aligned} &A=10000 \mathrm{~N}-\mathrm{m}^{2}\\ &\mathrm{El}=200 \times 10^{6} \mathrm{~N}-\mathrm{m}^{2} \end{aligned}

As per moment area first theorem.
\begin{aligned} \theta_{\mathrm{B}}-\theta_{A} &=\left(\frac{A}{E I}\right) A B \\ \theta_{\mathrm{B}}-0 &=\frac{10000}{200 \times 10^{6}}=0.5 \times 10^{-4} \mathrm{radian} \\ \theta_{\mathrm{B}} &=50 \mu \text { radians } \end{aligned}
 Question 2
A plane frame PQR (fixed at P and free at R) is shown in the figure. Both members (PQ and QR) have length, L, and flexural rigidity, EI. Neglecting the effect of axial stress and transverse shear, the horizontal deflection at free end, R, is
 A $\frac{5FL^3}{3EI}$ B $\frac{4FL^3}{3EI}$ C $\frac{2FL^3}{3EI}$ D $\frac{FL^3}{3EI}$
GATE ME 2021 SET-2   Strength of Materials
Question 2 Explanation:

\begin{aligned} U&=U_{P Q}+U_{Q P} \\ U&=\frac{M^{2} L}{2 E I}+\int_{0}^{L} \frac{\left(M_{x-x}\right)^{2}(d x)}{2 E I} \\ U&=\frac{(F L)^{2} L}{2 E I}+\int_{0}^{L}\left(\frac{(F x)^{2}(d x)}{2 E I}\right) \\ U&=\frac{F^{2} L^{3}}{2 E I}+\frac{F^{2} L^{3}}{6 E I}=\frac{2 F^{2} L^{3}}{3 E I} \end{aligned}
By Castigliano's theorem:
$\left(\delta_{H}\right)_{R}=\frac{\partial U}{\partial F}=\frac{4 F L^{3}}{3 E I}$
 Question 3
An overhanging beam PQR is subjected to uniformly distributed load 20 kN/m as shown in the figure.

The maximum bending stress developed in the beam is ________MPa (round off to one decimal place).
 A 125 B 250 C 325 D 450
GATE ME 2021 SET-1   Strength of Materials
Question 3 Explanation:

\begin{aligned} M_{D}-M_{A} &=\frac{1}{2} \times 15 \times 0.75 \\ M_{D} &=5.625 \mathrm{kN}-\mathrm{m}(\mathrm{s}) \\ \Sigma M_{A}&=60 \times 1.5-R_{B} \times 2=0 \\ R_{B} &=45 \mathrm{kN}(\uparrow) ; \quad R_{A}=15 \mathrm{kN}(\uparrow) \end{aligned}
Location of D:
\begin{aligned} \frac{15}{x} &=\frac{25}{2-x} \\ 6-3 x &=5 x \\ &=\frac{3}{4} m \end{aligned}
\begin{aligned} M_C-M_B &=\frac{1}{2} \times 20 \times 1 \\ M_B&=-10 kN- m \\ M_B &= 10kN - m \;(+1)\\ \text{Max B.M.}&=\text{Larger of }(M_B \text{ and } M_D) \\ M_B&=10kN- m\; (+1) \\ (\sigma _b)_{max} &=\frac{M_{max}}{Z_{N.A.}}\\&=\frac{6 \times 10 \times 10^6}{224 \times 100^2}=250MPa \end{aligned}
 Question 4
A cantilever beam of length, $L$, and flexural rigidity, $EI$, is subjected to an end moment, $M$, as shown in the figure. The deflection of the beam at $x=\frac{L}{2}$ is

 A $\frac{ML^2}{2EI}$ B $\frac{ML^2}{4EI}$ C $\frac{ML^2}{8EI}$ D $\frac{ML^2}{16EI}$
GATE ME 2021 SET-1   Strength of Materials
Question 4 Explanation:

\begin{aligned} Y_{C}-Y_{A}&=\left(\frac{A \bar{X}}{E I}\right)_{A C} \\ Y_{C}-0&=\frac{1}{E I}\left[\frac{-ML}{2} \times \frac{L}{4}\right]\\ Y_{C}&=\frac{ML^{2}}{8 E I} \text { (downward) } \end{aligned}
 Question 5
A cantilever of length $l$, and flexural rigidity EI, stiffened by a spring of stiffness k, is loaded by transverse force P, as shown

The transverse deflection under the load is
 A $\frac{Pl^3}{3EI}\left [ \frac{3EI}{3EI+2kl^3} \right ]$ B $\frac{Pl^3}{3EI}\left [ \frac{6EI-kl^3}{6EI} \right ]$ C $\frac{Pl^3}{3EI}\left [ \frac{3EI-kl^3}{3EI} \right ]$ D $\frac{Pl^3}{3EI}\left [ \frac{3EI}{3EI+kl^3} \right ]$
GATE ME 2020 SET-2   Strength of Materials
Question 5 Explanation:

\begin{aligned} \Delta_{\text {beam }} &=\Delta_{\text {spring }} \\ \frac{(P-R) l^{3}}{3 E I} &=\frac{R}{k} \\ P-R &=\frac{3 E I}{K l^{3}} R \\ \therefore \quad R &=\frac{P}{1+\frac{3 E I}{K l^{3}}}=\frac{P l^{3} K}{K l^{3}+3 E I} \\ \therefore \quad \Delta_{\text {beam }} &=\frac{R}{K}=\frac{P l^{3}}{3 E I+K l^{3}} \end{aligned}
 Question 6
A horizontal cantilever beam of circular cross-section, length 1.0 m and flexural rigidity EI= 200 $N\cdot m^2$ is subjected to an applied moment MA= 1.0 $N\cdot m$ at the free end as shown in the figure. The magnitude of the vertical deflection of the free end is _______mm (round off to one decimal place).
 A 1.2 B 5.3 C 8.2 D 2.5
GATE ME 2019 SET-2   Strength of Materials
Question 6 Explanation:
Deflection at free and $y_{B}=\frac{M_{A} L^{2}}{2 E I}$
$\begin{array}{l} \quad=\frac{1 \times 1^{2}}{2 \times 200}=2.5 \times 10^{-3} \mathrm{m} \\ \therefore y_{B}=2.5 \times 10^{-3} \times 10^{3} \mathrm{mm}=2.5 \mathrm{mm} \\ \therefore \quad y_{B}=2.5 \mathrm{mm} \end{array}$
 Question 7
A prismatic, straight, elastic, cantilever beam is subjected to a linearly distributed transverse load as shown below. If the beam length is L, Young's modulus E, and area moment of inertia I, the magnitude of the maximum deflection is
 A $\frac{qL^4}{15EI}$ B $\frac{qL^4}{30EI}$ C $\frac{qL^4}{10EI}$ D $\frac{qL^4}{60EI}$
GATE ME 2019 SET-2   Strength of Materials
Question 7 Explanation:
Double Integration method:
Let x be distance from the free end
\begin{aligned} \mathrm{M}_{\mathrm{x}} &=-\left(\frac{1}{2} \times \mathrm{x} \times \frac{\mathrm{q} \mathrm{x}}{\mathrm{L}}\right)\left(\frac{\mathrm{x}}{3}\right) \\ &=-\frac{\mathrm{q} \mathrm{x}^{3}}{6 \mathrm{L}} \end{aligned}
$\mathrm{EI} \cdot \frac{\mathrm{d}^{2} \mathrm{y}}{\mathrm{dx}^{2}}=-\left(\frac{\mathrm{qx}^{3}}{6 \mathrm{L}}\right)$
$\mathrm{EI} \cdot \frac{\mathrm{d} \mathrm{y}}{\mathrm{d} \mathrm{x}}=-\frac{\mathrm{q} \mathrm{x}^{4}}{24 \mathrm{L}}+\mathrm{C}_{1} \rightarrow \text{Slope equation} \;\; ...(1)$
$\text {EI.y }=-\frac{\mathrm{q} \mathrm{x}^{5}}{120 \mathrm{L}}+\mathrm{C}_{1} \mathrm{x}+\mathrm{C}_{2} \rightarrow \text { Deflection equation }\; \; ...(2)$
Boundary Conditions:
(a) Fixed end, $\mathrm{x}=\mathrm{L}, \theta_{\mathrm{A}}=0 ; \mathrm{Y}_{\mathrm{A}}=0$
$\therefore \quad \text{from }(1)$
$0=-\frac{\mathrm{q} \mathrm{L}^{4}}{24 \mathrm{L}}+\mathrm{C}_{1} \Rightarrow \mathrm{C}_{1}=\frac{\mathrm{qL}^{3}}{24}$
$\therefore \quad \text{from(2) }$
$0=-\frac{\mathrm{qL}^{5}}{120 \mathrm{L}}+\frac{\mathrm{qL}^{3}}{2 \mathrm{L}}(\mathrm{L})+\mathrm{C}_{2}$
$\Rightarrow \mathrm{C}_{2}=\frac{\mathrm{qL}^{4}}{30}$
$\therefore$ maximum deflection at free end i.e. at $\mathrm{x}=0$
$\therefore \mathrm{y}_{\max }=\frac{\mathrm{C}_{2}}{\mathrm{EI}}=\frac{\mathrm{qL}^{4}}{30 \mathrm{EI}}$
 Question 8
Consider a prismatic straight beam of length $L= \pi m$, pinned at the two ends as shown in the figure. The beam has a square cross-section of side p=6 mm. The Young's modulus E= 200 GPa, and the coefficient of thermal expansion $\alpha =3 \times 10^{-6}K^{-1}$. The minimum temperature rise required to cause Euler buckling of the beam is ________ K.
 A 0 B 0.5 C 1 D 2
GATE ME 2019 SET-1   Strength of Materials
Question 8 Explanation:
$\begin{array}{l} \mathrm{L}=\pi \mathrm{m} \\ \text { side }=6 \mathrm{mm} \\ \mathrm{E}=200 \mathrm{GPa} \\ \alpha=3 \times 10^{-6} \mathrm{K} \end{array}$
Thermal thrust due to restricted expansion,
$P=\left(\sigma_{t h}\right) A=(E \alpha t) A$
Euler buckling load, $P_{e}=\frac{\pi^{2} E I_{\min }}{L_{c}^{2}}$
For hinged ends, $\mathrm{L}_{\mathrm{c}}=\mathrm{L}$
$\therefore P e=\frac{\pi^{2} E \frac{a^{4}}{12}}{L^{2}}$
Condition for buckling, $\mathrm{P}=\mathrm{P}_{\mathrm{c}}$
$\begin{array}{c} (E \alpha t) A=\frac{\pi^{2} E}{L^{2}} \frac{a^{4}}{12} \\ (\alpha t) a^{2}=\frac{\pi^{2} a^{4}}{12 L^{2}} \\ t=\frac{1}{\alpha} \times \frac{\pi^{2} a^{2}}{12 L^{2}} \\ =\frac{1}{3 \times 10^{-6}} \times \frac{\pi^{2} \times 6^{2}}{12(\pi \times 1000)^{2}} \\ \mathrm{t}=1^{\circ} \mathrm{K} \end{array}$
 Question 9
Consider an elastic straight beam of length $L=10 \pi m$, with square cross-section of side a=5 mm, and Young's modulus E=200 GPa. This straight beam was bent in such a way that the two ends meet, to form a circle of mean radius R. Assuming that Euler-Bernoulli beam theory is applicable to this bending problem, the maximum tensile bending stress in the bent beam is __________ MPa.
 A 100 B 200 C 50 D 150
GATE ME 2019 SET-1   Strength of Materials
Question 9 Explanation:
\begin{aligned} \mathrm{L}&=10 \pi \mathrm{m} \\ \mathrm{a}&=5 \mathrm{mm} \\ \mathrm{E}&=200 \mathrm{GPa} \\ \text { Length of wire }&=\mathrm{L}=\pi \mathrm{D}=2 \pi \mathrm{R} \\ 10 \pi&=2 \pi \mathrm{R} \\ \mathrm{R}&=5 \mathrm{m}\\ &\text { From bending equation, } \\ \frac{\mathrm{M}}{\mathrm{I}}&=\frac{\mathrm{f}}{\mathrm{Y}}=\frac{\mathrm{E}}{\mathrm{R}}\\ \Rightarrow \; \frac{f}{y}&=\frac{E}{R}\\ \mathrm{f}_{\max }&=\mathrm{y}_{\max } \times \frac{\mathrm{E}}{\mathrm{R}} \\ &=2.5 \times \frac{200 \times 10^{3}}{5 \times 10^{3}}\\ &=100 \mathrm{MPa} \end{aligned}
 Question 10
The minimum axial compressive load, P, required to initiate buckling for a pinned-pinned slender column with bending stiffness EI and length L is
 A $P=\frac{\pi ^{2}EI}{4L^{2}}$ B $P=\frac{\pi ^{2}EI}{L^{2}}$ C $P=\frac{3\pi ^{2}EI}{4L^{2}}$ D $P=\frac{4\pi ^{2}EI}{L^{2}}$
GATE ME 2018 SET-2   Strength of Materials
Question 10 Explanation:
For both ends hinged buckling load,
$P=\frac{\pi^{2} E I}{L^{2}}$
There are 10 questions to complete.

### 1 thought on “Bending of Beams”

1. Question 6 has wrong figure. kindly correct.