What is thermal Stress and Strain? Definition & Hoop stress

Thermal Stress & Thermal Strains

The Stresses produced due to the change of the materials are known as thermal stresses or temperature stresses. If the temperature of the body is lowered its dimensions will decrease and if the temperature of the body is raised its dimensions will increase. The corresponding strains caused by thermal stresses known as temperature strains or Thermal strains.

Let us assume a bar of uniform cross-section with initial length l and initial temperature $t_1$ . Due to the change of temperature, its final temperature is $t_2$ . Let the coefficient of linear expansion is α.

The expansion of the bar due to a rise in temperature will be:

$\boxed{ \mbox{Expansion of bar} = \alpha . \left( t_2 - t_1\right). l}$

If the elongation of the bar is prevented by some external force or by fixing the bar ends, the Thermal strain thus produced will be :

⇒ $= \frac{ \alpha . \left( t_2 - t_1 \right). l }{l}$

⇒ $\boxed{ \mbox{ Thermal strain } = \alpha . \left( t_2 - t_1 \right) }$ ( Compressive)

Note: “If the temperature of the bar is decreased , the thermal strain and stress will be tensile in nature.”

Example (1): A steel rod is 10 m long is at the temperature of $10 ^oC$ . Find the free expansion of the length when the temperature is raised to $50 ^oC$ . Find the thermal stress produced when:

The expansion of the rod is prevented.
The rod is permitted to expand by 3 mm.

Note: Use $\alpha = 12 \times 10^{-6} \mbox{ per ^oC and E} = 200 GN/m^3$

Solution :

The Free expansion of the rod :

⇒ $\alpha . \left( t_2 - t_1\right). l$

⇒ $12 \times 10^{-6} \left( 50 - 10\right) \times 10$

$= 0.0048 mm = \bold{4.8 mm }$ .Ans

1. Thermal stress When the expansion of the rod is fully prevented:

$= \alpha . \left( t_2 - t_1 \right) . E$

⇒ $12 \times 10^{-6} \left( 50 - 10\right) \times 200 \times 10^9$

⇒ $\bold{96 \times 10^{6} N/m^2}$ .Ans

2. Thermal stress when the rod is permitted to expand by 3 mm:

The amount of expansion prevented = 4.8 – 3 =1.8 mm.

∴ Thermal Strain, $e = \frac{\mbox{Expansion Prevented}}{\mbox{Original Length}} = \frac{1.8}{10 \times 1000}$

= 0.00018

∴ thermal stress:

$e \times E = 0.00018 \times 200 \times 10^9$

⇒ $\bold{36 \times 10^6 N/m^2}$

What is Hoop Stress?

If a thin tire of a metal is to be shrunk on to a wheel, the diameter of the tire is to be slightly smaller than the diameter of the wheel for tightly fits and does not come out easily.

Let the temperature of the tire is raised so that the diameter of the tire is increased and equals to the diameter of the wheel. When the tire is slipped on to the wheel and temperature decreases, the steel tire will try to come to its original diameter and the hoop stress or Circumferential Stress( tensile stress) will be set up.

Let us assume,

The diameter of tire = d ( smaller)

The diameter of the wheel = D (bigger)

Temperature difference = $t^o$