# Bending and Shear Stresses

 Question 1
The cross-section of a built-up wooden beam as shown in the figure is subjected to a vertical shear force of 8 kN. The beam is symmetrical about the neutral axis (N.A.) shown, and the moment of inertia about N.A. is $1.5 \times 10^9 mm^4$. Considering that the nails at the location P are spaced longitudinally (along the length of the beam) at 60 mm, each of the nails at P will be subjected to the shear force of A 60N B 120N C 240N D 480N
GATE CE 2019 SET-1   Solid Mechanics
Question 1 Explanation:
Shear stress acting at the joint at which the nail 'P' is resisting the shear.
\begin{aligned} \tau &=\frac{FA\bar{y}}{Ib}\\ &=\frac{(8 \times 10^3)(50 \times 100)(150)}{1.5 \times 10^9 \times 50}\\ &=0.08N/mm^2 \end{aligned}
The shear force acting on nail at 'P'.
$F=\tau (60\times 50)=240N$
 Question 2
A cantilever beam of length 2 m with a square section of side length 0.1 m is loaded vertically at the free end. The vertical displacement at the free end is 5 mm. The beam is made of steel with Young's modulus of $2.0 \times 10^{11} N/m^{2}$. The maximum bending stress at the fixed end of the cantilever is
 A 20.0 MPa B 37.5 MPa C 60.0 MPa D 75.0 MPa
GATE CE 2018 SET-1   Solid Mechanics
Question 2 Explanation: \begin{aligned} I&=\frac{(0.1)^4}{12} \\ \text{Deflection }\delta &=\frac{Pl^2}{3eI} \\ 5 \times 10^{-3}&=\frac{P(2)^3}{3 \times 2 \times 10^11 \times \frac{(0.1)^4}{12}} \\ P&=3125N \\ \text{Now, } M&= Pl=3125 \times 2\\ &= 6250Nm\\ \text{As, }\frac{M}{I} &=\frac{\sigma }{y}=\frac{E}{R} \\ \sigma _{max}&=\frac{M}{Z}=\frac{6250}{\frac{(0.1)^3}{6}}\\ &=37.5 \times 10^6 N/m^2=37.5MPa \end{aligned}
 Question 3
A 450 mm long plain concrete prism is subjected to the concentrated vertical loads as shown in the figure. Cross section of the prism is given as 150 mm x 150 mm. Considering linear stress distribution across the cross-section, the modulus of rupture (expressed in MPa) is _______ . A 1.7 B 2.7 C 2 D 3
GATE CE 2016 SET-2   Solid Mechanics
Question 3 Explanation: \begin{aligned} \mathrm{BM}_{\mathrm{Q}} &=11.25 \times 150 \\ &=1.6875 \times 10^{6} \mathrm{N}-\mathrm{mm} \\ \Rightarrow\quad \frac{\sigma}{y} &=\frac{M_{Q}}{I} \\ \text{where}\quad y &=\frac{150}{2}=75 \mathrm{mm} \text { and } 1=\frac{(150)^{4}}{12} \\ \Rightarrow\quad \sigma &=\frac{1.6875 \times 10^{6} \times 75}{(150)^{4} / 12}=3 \mathrm{MPa} \end{aligned}
 Question 4
A symmetric I-section (with width of each flange = 50 mm, thickness of each flange = 10 mm, depth of web = 100 mm, and thickness of web = 10 mm) of steel is subjected to a shear force of 100 kN. Find the magnitude of the shear stress(in $N/mm^{2}$) in the web at its junction with the top flange. __________
 A 50.45 B 71.12 C 85.25 D 96.36
GATE CE 2013   Solid Mechanics
Question 4 Explanation: \begin{aligned} 1 &=\frac{50 \times 120^{3}}{12}-\frac{40 \times 100^{3}}{12} \\ &=3.866 \times 10^{6} \mathrm{mm}^{4} \\ q &=\frac{S A \bar{y}}{16}=\frac{100 \times 10^{3} \times 50 \times 10 \times 55}{3.866 \times 10^{6} \times 10} \\ &=71.12 \mathrm{N} / \mathrm{mm}^{2} \end{aligned}
 Question 5
The 'plane section remains plane' assumption in bending theory implies:
 A strain profile is linear B stress profile is linear C both strain and stress profiles are linear D shear deformations are neglected
GATE CE 2013   Solid Mechanics
 Question 6
The maximum tensile stress at the section X-X shown in the figure below is A $\frac{8P}{bd}$ B $\frac{6P}{bd}$ C $\frac{4P}{bd}$ D $\frac{2P}{bd}$
GATE CE 2008   Solid Mechanics
Question 6 Explanation:
The section at X-X may be shown as in the figure below: The maximum tensile stress at the section X-X is
\begin{aligned} \sigma &=\frac{P}{A}+\frac{M}{Z} \\ &=\frac{P}{b \times(d / 2)}+\frac{P \times(d / 4) \times 6}{b \times\left(d^{2} / 4\right)} \\ &=\frac{2 P}{b d}+\frac{6 P}{b d}=\frac{8 P}{b d} \end{aligned}
 Question 7
The shear stress at the neutral axis in a beam of triangular section with a base of 40 mm and height 20 mm, subjected to a shear force of 3 kN is
 A 3MPa B 6MPa C 10MPa D 20MPa
GATE CE 2007   Solid Mechanics
Question 7 Explanation:
$\text { Shear stress, } \tau=\frac{S A \bar{y}}{1 b}$ Width at a distance of $\frac{40}{3} \mathrm{mm}$ from the top
\begin{aligned} &=\frac{40}{20} \times \frac{40}{3}=\frac{80}{3} \mathrm{mm}\\ \therefore \tau &=\frac{3 \times 10^{3} \times\left(\frac{1}{2} \times \frac{80}{3} \times \frac{40}{3}\right) \times\left(\frac{1}{3} \times \frac{40}{3}\right)}{\left(\frac{40 \times 20^{3}}{36}\right) \times \frac{80}{3}} \\ &=\frac{3 \times 10^{3} \times 3200 \times 40 \times 36 \times 3}{162 \times 3200 \times 20^{3}} \\ &=10 \mathrm{MPa} \end{aligned}
 Question 8
If a beam of rectangular cross-section is subjected to a vertical shear force V , the shear force carried by the upper one-third of the cross-section is
 A zero B $\frac{7V}{27}$ C $\frac{8V}{27}$ D $\frac{V}{3}$
GATE CE 2006   Solid Mechanics
Question 8 Explanation: Shear stress at distance y from neutral axis,
\begin{aligned} \tau&=\frac{S A \bar{y}}{1 b} =\frac{V \times\left(\frac{d}{2}-y\right) \times b \times\left(\frac{(d / 2)+y}{2}\right)}{16} \\ \Rightarrow \quad \tau&= \frac{V \times\left(\frac{d^{2}}{4}-y^{2}\right)}{21}\\ \therefore \quad d F&=\tau \times b d y=\frac{V \times\left(\frac{d^{2}}{4}-y^{2}\right)}{21} \times b d y\\ \end{aligned}
Integrating both sides, we get
\begin{aligned} \Rightarrow \quad F &=\frac{V b^{d / 2}}{21} \int_{d / 6}^{d}\left(\frac{d^{2}}{4}-y^{2}\right) d y \\ &=\frac{V b}{21}\left[\frac{d^{2}}{4} y-\frac{y^{3}}{3}\right]_{d / 6}^{d / 2} \\ &=\frac{V b}{21}\left[\frac{d^{3}}{8}-\frac{d^{3}}{24}-\frac{d^{3}}{24}+\frac{d^{3}}{648}\right] \\ &=\frac{V b}{21} \times \frac{d^{3}}{8} \times \frac{28}{81} \\ &=\frac{V b}{2 b d^{3}} \times \frac{d^{3}}{8} \times \frac{28}{81} \times 12=\frac{7 V}{27} \end{aligned}
 Question 9
I-section of a beam is formed by gluing wooden planks as shown in the figure below. If this beam transmits a constant vertical shear force of 3000 N, the glue at any of the four joint will be subjected to a shear force (in kN per meter length) of A 3 B 4 C 8 D 10.7
GATE CE 2006   Solid Mechanics
Question 9 Explanation: Considering shaded area 'abcd' of plank.
$\text { Shear flow, } \quad a=\frac{V Q}{1}$
Shear flow is measured as force per unit length along the longitudinal axis of a beam and over the cross-section is as shown above.
\begin{aligned} Q &=A \bar{y} \\ Q &=(75 \times 50) \times 125 \\ &=468750 \mathrm{mm}^{3} \end{aligned}
Moment of inertia of I-section about NA.
\begin{aligned} 1&=350 \times 10^{6} \mathrm{mm}^{4}\\ V &=3000 N \\ q &=\frac{468750 \times 3000}{350 \times 10^{6}} \\ &=4.018 N / m m \end{aligned}
So, resistance offered by glue to keep I-section
intact.
$\mathrm{q}=4.018 \mathrm{N} / \mathrm{mm} \simeq 4 \mathrm{kN} / \mathrm{m}$
 Question 10
A beam with the cross-section given below is subjected to a positive bending moment (causing compression at the top) of 16 kN-m acting around the horizontal axis. The tensile force acting on the hatched area of the cross-section is A zero B 5.9kN C 8.9kN D 17.8kN
GATE CE 2006   Solid Mechanics
Question 10 Explanation: \begin{aligned} f &=\frac{M}{1} \times y \\ &=\frac{16 \times 10^{6} \times 12}{100 \times 150^{3}} \times 25=14.22 \mathrm{N} / \mathrm{mm}^{2} \end{aligned}
From similar triangles, we have
\begin{aligned} \therefore \text { Tensile force } &=\frac{1}{2} \times 25 \times 14.22 \times 50 \times 10^{3} \\ &=8.88=8.9 \mathrm{kN} \end{aligned}
There are 10 questions to complete.