# Thin Cylinder

 Question 1
A thin-walled cylindrical pressure vessel has mean wall thickness of $t$ and nominal radius of $r$. The Poisson's ratio of the wall material is 1/3. When it was subjected to some internal pressure, its nominal perimeter in the cylindrical portion increased by 0.1% and the corresponding wall thickness became $\bar{t}$. The corresponding change in the wall thickness of the cylindrical portion, i.e. $100\times (\bar{t}-t)/t$, is ________%(round off to 3 decimal places).
 A 0.06 B -0.06 C 0.12 D -0.12
GATE ME 2022 SET-1   Strength of Materials
Question 1 Explanation: $\pi(d+\delta d)=1.001 \pi d$
$\varepsilon _1=\frac{\delta d}{d}=0.001=\frac{Pd}{4tE}(2-v)$
$\frac{Pd}{4tE}=\frac{0.001}{(2-v)}=6 \times 10^{-4}$
$\varepsilon _2=\frac{\delta t}{t}=\frac{1}{E}\left [ \sigma _r-v(\sigma _h-\sigma _L) \right ]$
$\frac{\delta t}{t}=\frac{1}{E}\left [ -v\frac{Pd}{4t}(2+1) \right ]$
$\frac{\delta t}{t}=-\frac{Pd}{4tE}v(3)=-6 \times 10^{-4} \times \frac{1}{3}(3)=-6 \times 10^{-4}$
Therefore, The corresponding change in the wall thickness of the cylindrical portion
$=100 \times \left [ \frac{\bar{t}-t}{t} \right ]=100 \times \frac{\delta t}{t}=100 \times (-6) \times 10^{-4}=-0.06 \%$
 Question 2
Consider two concentric circular cylinders of different materials M and N in contact with each other at r=b, as shown below. The interface at r=b is frictionless. The composite cylinder system is subjected to internal pressure P. Let $(u_r^M,u_\theta ^M)$ and $(\sigma_{rr}^M, \sigma_{\theta \theta}^M)$ denote the radial and tangential displacement and stress components, respectively, in material M. Similarly, $(u_r^N,u_\theta ^N)$ and $(\sigma_{rr}^N,\sigma_{\theta \theta}^N)$ denote the radial and tangential displacement and stress components, respectively, in material N. The boundary conditions that need to be satisfied at the frictionless interface between the two cylinders are: A $u_r^M=u_r^N$ and $\sigma _{rr }^M=\sigma _{rr}^N$ B $u_r^M=u_r^N$ and $\sigma _{rr }^M=\sigma _{rr}^N$and $u_{\theta }^M=u_{\theta }^N$ and $\sigma _{\theta \theta }^M=\sigma _{\theta \theta }^N$ C $u_{\theta }^M=u_{\theta }^N$ and $\sigma _{\theta \theta }^M=\sigma _{\theta \theta }^N$ D $\sigma _{rr }^M=\sigma _{rr}^N$ and $\sigma _{\theta \theta }^M=\sigma _{\theta \theta }^N$
GATE ME 2019 SET-2   Strength of Materials
Question 2 Explanation:
As the contact is frictionless, one cylinder can rotate freely with respect other. The displacement in tangential directions need not be same at a point contact for two cylinders (i.e., $u_{\theta}^{M} \neq u_{\theta}^{N}$ ). Similarly
the Hoop stress at point of contact need not be same (i.e., $\sigma_{\theta \theta}^{M} \neq \sigma_{\theta \theta}^{N}$ ). As the interface will be always in contact the displacement in radial direction and stress in radial directions must be same for two cylinders. i.e., $u_{r}^{M}=u_{r}^{N}$ and $\sigma_{r}^{M}=\sigma_{r r}^{N}$
 Question 3
A thin cylindrical pressure vessel with closed-ends is subjected to internal pressure. The ratio of circumferential (hoop) stress to the longitudinal stress is
 A 0.25 B 0.5 C 1 D 2
GATE ME 2016 SET-2   Strength of Materials
Question 3 Explanation:
For thin cylinder
Circumferential stress
$\sigma_{h}=\frac{p d}{2 t}$
Longitudinal stress,
\begin{aligned} \sigma_{L} &=\frac{p d}{4 t} \\ \therefore \quad \frac{\sigma_{h} }{\sigma_{L}} &=2 \end{aligned}
 Question 4
A cylindrical tank with closed ends is filled with compressed air at a pressure of 500 kPa. The inner radius of the tank is 2 m, and it has wall thickness of 10 mm. The magnitude of maximum in-plane shear stress (in MPa) is ________
 A 25MPa B 34MPa C 65MPa D 45MPa
GATE ME 2015 SET-3   Strength of Materials
Question 4 Explanation:
Maximum in plane shear stress (in MPa)
\begin{aligned} \tau &=\frac{\sigma_{1}-\sigma_{2}}{2}=\frac{\frac{p d}{2 t}-\frac{p d}{4 t}}{2} \\ &=\frac{500 \times 10^{3} \times 4000}{8 \times 10}=25 \times 10^{6} \mathrm{Pa}=25 \mathrm{MPa} \end{aligned}
 Question 5
A gas is stored in a cylindrical tank of inner radius 7 m and wall thickness 50 mm. The gage pressure of the gas is 2 MPa. The maximum shear stress (in MPa) in the wall is
 A 35 B 70 C 140 D 280
GATE ME 2015 SET-2   Strength of Materials
Question 5 Explanation:
Maximum shear stress in the wall
$=\frac{\sigma_{1}}{2}=\frac{p d}{4 t}=\frac{2 \times 14 \times 1000}{4 \times 50}=140 \mathrm{MPa}$
 Question 6
A thin gas cylinder with an internal radius of 100 mm is subject to an internal pressure of 10 MPa. The maximum permissible working stress is restricted to 100 MPa. The minimum cylinder wall thickness (in mm) for safe design must be _______
 A 10mm B 20mm C 30mm D 40mm
GATE ME 2014 SET-4   Strength of Materials
Question 6 Explanation:
\begin{aligned} \sigma_{\text {wothing }} &=\frac{p d}{2 t} \quad(d=2 r=200 \mathrm{mm}) \\ t &=\frac{p d}{2 \sigma_{w}}=\frac{10 \times 200}{2 \times 100}=10 \mathrm{mm} \end{aligned}
 Question 7
A long thin walled cylindrical shell, closed at both the ends, is subjected to an internal pressure. The ratio of the hoop stress (circumferential stress) to longitudinal stress developed in the shell is
 A 0.5 B 1 C 2 D 4
GATE ME 2013   Strength of Materials
Question 7 Explanation:
Hoop stress: $\sigma_{h}=\frac{p d}{2 t}$
Longitudinal stress: $\sigma_{L}=\frac{p d}{4 t}$
$\therefore \frac{\sigma_{h}}{\sigma_{L}}=2$
 Question 8
A thin walled spherical shell is subjected to an internal pressure. If the radius of the shell is increased by 1% and the thickness is reduced by 1%, with the internal pressure remaining the same, the percentage change in the circumferential (hoop) stress is
 A 0 B 1 C 1.08 D 2.02
GATE ME 2012   Strength of Materials
Question 8 Explanation:
Hoop stress $=\frac{p d}{4 t}$
p= internal pressure
d= diameter of shell
t= thickness of shell
\begin{array}{l} \text { Now } \sigma=\frac{p d^{\prime}}{4 t^{\prime}} \\ \begin{aligned} d^{\prime}=1.01 d &\; \text { and } t^{\prime}=0.99 t \\ \therefore \quad \sigma^{\prime} &=\frac{p \times 1.01 d}{4 \times 0.99 t}=1.0202 \sigma \\ \text { \% change } &=\frac{\sigma^{\prime}-\sigma}{\sigma} \times 100 \\ &=\frac{(1.0202-1) \sigma}{\sigma} \times 100=2.02 \% \end{aligned} \end{array}
 Question 9
A thin cylinder of inner radius 500mm and thickness 10mm is subjected to an internal pressure of 5MPa. The average circumferential (hoop) stress in MPa is
 A 100 B 250 C 500 D 1000
GATE ME 2011   Strength of Materials
Question 9 Explanation:
Circumferential or hoop stress.
\begin{aligned} \sigma_{h}&=\frac{p d}{2 t}\\ \text{where }p&=5 \mathrm{MPa} \\ d &=1000 \mathrm{mm} \\ t &=10 \mathrm{mm} \\ \therefore \quad \sigma_{H} &=\frac{5 \times 1000}{2 \times 10}=250 \mathrm{MPa} \end{aligned}
There are 9 questions to complete.