Products

## Holooly Rewards

We are determined to provide the latest solutions related to all subjects FREE of charge!

Enjoy Limited offers, deals & Discounts by signing up to Holooly Rewards Program

## Holooly Tables

All the data tables that you may search for.

## Holooly Arabia

For Arabic Users, find a teacher/tutor in your City or country in the Middle East.

## Holooly Sources

Find the Source, Textbook, Solution Manual that you are looking for in 1 click.

## Holooly Help Desk

Need Help? We got you covered.

## Q. 4.5

An initially unstressed short steel cylinder, internal radius 0.2 m and external radius 0.3 m, is subjected to a temperature distribution of the form $T = a + b log_{e} r$ to ensure constant heat flow through the cylinder walls. With this form of distribution the radial and circumferential stresses at any radius r , where the temperature is T. are given by

$\sigma _{r}=A-\frac{B}{r^{2}} -\frac{\alpha ET}{2(1-\nu )}$

$\sigma _{H}=A+\frac{B}{r^{2}} -\frac{\alpha ET}{2(1-\nu )}-\frac{ E\alpha b}{2(1-\nu )}$

If the temperatures at the inside and outside surfaces are maintained at 200°C and 100°C respectively, determine the maximum circumferential stress set up in the cylinder walls. For steel, E = 207 GN/m², ν = 0.3 and $\alpha = 11 × 10^{-6}$ per °C

## Verified Solution

$T=a+b\log _{e}r$
$200=a+b\log _{e}0.2=a+b(0.6931-2.3026)$
∴     $200=a-1.6095 b$                 (1)
also            $100=a+b\log _{e}0.3=a+b(1.0986-2.3026)$
$100=a-1.204 b$                       (2)
(2)-(1),               $100=-0.4055 b$
$b=-246.5=-247$
Also               $\frac{E\alpha }{2(1-\nu )} =\frac{207\times 10^{9}\times 11\times 10^{-6}}{2(1-0.29)}$
$=1.6\times 10^{6}$

Therefore substituting in the given expression for radial stress

$\sigma _{r}=A-\frac{B}{r^{2}} -1.6\times 10^{6}T$

At r = 0.3, $\sigma _{r} = 0$ and T = 100

$0=A-\frac{B}{0.09}-1.6 \times 10^{6} \times 100$      (3)

At r = 0.2, $\sigma _{r} = 0$ and T = 200

$0=A-\frac{B}{0.04}-1.6\times 10^{6}\times 200$    (4)

(4)-(3)    $0=B(11.1-25)-1.6\times 10^{8}$
$B=-11.5\times 10^{6}$

and from (4),

$A=25B+3.2\times 10^{8}$
$=(-2.88+3.2)10^{8}=0.32\times 10^{8}$

substituting in the given expression for hoop stress,

$\sigma _{H}=0.32\times 10^{8}-\frac{11.5\times 10^{6}}{r^{2}} -1.6\times 10^{6}T+1.6\times 10^{6}\times 247$

At  r = 0.2, $\sigma _{H}=(0.32-2.88-3.2+3.96)10^{8}=-180 MN/m^{2}$

At r = 0.3,    $\sigma _{H}=(0.32-1.28-1.6+3.96)10^{8}=+140 MN/m^{2}$

The maximum tensile circumferential stress therefore occurs at the outside radius and has a value of 140 MN/m². The maximum compressive stress is 180 MN/m² at the inside radius