# Shear-force and Bending Moment Diagrams

 Question 1
A beam of negligible mass is hinged at support P and has a roller support Q as shown in the figure.

A point load of 1200 N is applied at point R. The magnitude of the reaction force at support Q is __________ N.
 A 3000 B 1500 C 300 D 750
GATE ME 2020 SET-2   Strength of Materials
Question 1 Explanation:

\begin{aligned} \Sigma \vec{M}_{P} &=0 \\ 1200 \times 5-R_{Q} \times 4 &=0 \\ R_{Q} &=\frac{1200 \times 5}{4}=1500 \mathrm{N} \end{aligned}
 Question 2
The magnitude of reaction force at joint C of the hinge-beam shown in the figure is _______ kN (round off to 2 decimal places).
 A 40 B 10 C 50 D 20
GATE ME 2020 SET-1   Strength of Materials
Question 2 Explanation:

\begin{aligned} \Sigma M_{B_{\text {Right }}} &=0 \\ 4 R_{C} &=10 \times 4 \times 2 \\ R_{C} &=20 \mathrm{kN} \end{aligned}
 Question 3
The barrier shown between two water tanks of unit width (1 m) into the plane of the screen is modeled as a cantilever.

Taking the density of water as 1000 $kg/m^3$, and the acceleration due to gravity as 10 $m/s^2$, the maximum absolute bending moment developed in the cantilever is ______ kNm (round off to the nearest integer).
 A 85 B 105 C 128 D 146
GATE ME 2020 SET-1   Strength of Materials
Question 3 Explanation:

\begin{aligned} F_{1} &=\rho g A \cdot \bar{x}_{1} \\ &=1000 \times 10 \times(1 \times 4) \times 2=80 \mathrm{kN}\\ F_{2} &=\rho g A \cdot \bar{x}_{2} \\ &=1000 \times 10 \times(1 \times 1) \times 0.5=5 \mathrm{kN} \\ C P_{1} &=4 \times \frac{1}{3}=\frac{4}{3} \mathrm{m} \\ C P_{2} &=1 \times \frac{1}{3}=\frac{1}{3} \mathrm{m} \\ M_{A} &=80 \times \frac{4}{3}-5 \times \frac{1}{3}=105 \mathrm{kN}-\mathrm{m} \end{aligned}
 Question 4
A simply supported beam of width 100 mm, height 200 mm and length 4 m is carrying a uniformly distributed load of intensity 10 kN/m. The maximum bending stress (in MPa) in the beam is __________ (correct to one decimal place).
 A 25 B 30 C 35 D 50
GATE ME 2018 SET-1   Strength of Materials
Question 4 Explanation:
$\text { Maximum B.M.. } M=\frac{w L^{2}}{8}=\frac{10 \times 16}{8}=20 \mathrm{kNm} \quad (L=4)$
Maximum Bending Stress
\begin{aligned} \sigma_{\max } &=\frac{M}{I} y_{\max }=\frac{20 \times 10^{3}}{\left(\frac{0.1 \times 0.2^{3}}{12}\right)} \times 0.1 \\ &=30 \times 10^{6} \mathrm{N} / \mathrm{m}^{2}=30 \mathrm{MPa} \end{aligned}
 Question 5
For a loaded cantilever beam of uniform cross-section, the bending moment (in N.mm) along the length is $M(x)=5x^{2}+10x$ , where x is the distance (in mm) measured from the free end of the beam. The magnitude of shear force (in N) in the cross-section at x=10 mm is
 A 100 B 105 C 110 D 115
GATE ME 2017 SET-2   Strength of Materials
Question 5 Explanation:
$(\mathrm{BM})_{x-x}=M_{x-x}=5 x^{2}+10 x$

$\text { (S.F) }_{x-x}=F_{x-x}=?$
We know that at section X-X
\begin{aligned} F_{X-X} &=\frac{d}{d x}\left[5 x^{2}+10 x\right] \\ F_{X-X} &=10 x+10 \\ \left(F_{X-X}\right)_{x=10 \mathrm{mm}} &=10(10)+10=110 \mathrm{N} \end{aligned}
 Question 6
For the overhanging beam shown in figure, the magnitude of maximum bending moment (in kN-m) is ________
 A 54 B 36 C 40 D 14
GATE ME 2015 SET-3   Strength of Materials
Question 6 Explanation:

$\therefore \quad R_{A}+R_{B}=60 \mathrm{kN}$
Taking moment from point B
\begin{aligned} R_{A} \times 4+20+2 &=40 \times 2 \\ 4 R_{A} &=10 \mathrm{kN} \\ R_{B} &=50 \mathrm{kN} \end{aligned}
Taking x from A
\begin{aligned} M x &=R_{A} \times x-W \frac{x^{2}}{2} \\ &=10 x-5 x^{2}\\ \text{at}\;x&=1,M_{1}=5 \end{aligned}

\begin{aligned} \therefore \text{at}\; B, M_{B}&=-40\\ |M|&=40 \end{aligned}
 Question 7
The value of moment of inertia of the section shown in the figure about the axis-XX is
 A $8.5050\times 10^{6} mm^{4}$ B $6.8850\times 10^{6} mm^{4}$ C $7.7625\times 10^{6} mm^{4}$ D $8.5725\times 10^{6} mm^{4}$
GATE ME 2015 SET-3   Strength of Materials
Question 7 Explanation:

\begin{aligned} I_{tot}&=I_{A1}-2I_{A2} \\ I_{A1}&=60 \times \frac{120^3}{12} \\ &= 8640000\\ I_{A2}&=30 \times \frac{30^3}{12} +30^2 \times 30^2 \\ I_{A2}&= 877500\\ I&=8640000-2 \times 877500 \\ &= 68850000 \; mm^4\\ &=6.885 \times 10^6 \; mm^4 \end{aligned}
 Question 8
A cantilever beam OP is connected to another beam PQ with a pin joint as shown in the figure. A load of 10 kN is applied at the mid-point of PQ. The magnitude of bending moment (in kN-m) at fixed end O is
 A 2.5 B 5 C 10 D 25
GATE ME 2015 SET-2   Strength of Materials
Question 8 Explanation:

At hinge moment releases
Taking summation of all forces

$P_{V}+R_{Q}=10$
Taking moment of all forces about Q
$M=\underset{clockwise}{\left(P_{V} \times 1\right)}-\underset{anticlockwise}{(10 \times 0.5)}$
At equilibrium M=0
$\begin{array}{rl} \therefore \quad 0 & =\left(P_{V} \times 1\right)-5 \\ P_{V} & =5 \mathrm{kN} \end{array}$

Summation of all vertical forces is zero.
$\therefore \quad O_{V}+P_{V}=0$
Summation of Moments of all forces at equilibrium is zero.
$\therefore \quad\underset{clockwise}{+M}+\underset{clockwise}{P_{v}\times 2}=0$
$\therefore M=-P_{V} \times 2$
$\text{where}\quad P_{V}=-5 \mathrm{kN}$
$\therefore M=10 \mathrm{kN}$
 Question 9
A simply supported beam of length L is subjected to a varying distributed load $sin(3\pi x/L)Nm^{-1}$ , where the distance x is measured from the left support. The magnitude of the vertical reaction force in N at the left support is
 A zero B $L/3\pi$ C $L/\pi$ D $2L/\pi$
GATE ME 2013   Strength of Materials
Question 9 Explanation:
$W=\int_{0}^{L} \sin \left(\frac{3 \pi x}{L}\right) d x=\frac{2 L}{3 \pi}$
Vertical reaction,
$R_{1}=R_{2}=\frac{L}{3 \pi}$
\begin{aligned} \therefore \quad \sigma &=\frac{M}{I} \times y_{\max } \\ &=\frac{10 \times 10^{3}}{10^{4} / 12} \times \frac{10}{2}=60 \mathrm{MPa} \end{aligned}