Thermal Stress in a Rod fixed between two rigid supports

 

 

In this article we are going to discuss about thermal stress produced in a rod which is fixed between two rigid supports.

Consider a rod of length L₀ with Young’s modulus Y is fixed between two rigid supports. Let it is heated for some time and its temperature increases by ΔT.

Now the rod tries to expend in length(due to thermal expansion) but since it is fixed between rigid supports, so a stress is produced in the rod.

This stress is called Thermal Stress.

Now stress is given by = F/A

Also from the concepts of elasticity Young’s modulus :-

Y = Thermal Stress/Thermal Strain

Hence,

Thermal Stress(F/A) = YαΔT

And the force which is responsible for this stress :-

Force (F) = YAαΔT

 

Example 1 :- A steel rod 50 cm long has cross-sectional area of 0.8 cm². What force would be required to stretch this rod by the same amount as the expansion produced by heating it through 10ºC. For steel it is given that α = 10^-5 K^-1 and Y = 2×10¹¹ Nm^-2.

Solution. Thermal Stress (F/A) = YαΔT

⇒ F = YAαΔT

⇒ F = (2×10¹¹)×(0.8×10^-4)×(10^-5)×(10)

⇒ F = 1.6×10³ N

Example 2 :- In the given figure a rod is free at one end and other end is fixed. When we change the temperature of rod by Δθ, then strain produced in the rod, will be

(A) αΔθ      (B) αΔθ/2      (C) zero      (D) information incomplete

Solution. Here rod is free to expand from one side. So by changing temperature, no strain will be produced in the rod. Hence option (C) is correct.

Example 3 :- A steel rod is clamped at its two ends and rests on a fixed horizontal base. The rod is in natural length at 20°C. Find the longitudinal strain developed in the rod if the temperature rises to 50°C. Coefficient of linear expansion of steel = 1.2 × 10^–5/°C.

Solution. We know that strain is given by

∴ Strain = αΔT

Strain = 1.2 × 10^–5 × (50 – 20) = 3.6 × 10^–4

Here strain is compressive strain.

Example 4 :- A steel rod of length 1m rests on a smooth horizontal base. If it is heated from 0°C to 100°C, what is the longitudinal strain developed?

Solution. A longitudinal strain develops if and only if, there is an opposition to the expansion. Since there is no opposition for expansion in this case, hence the longitudinal stain here = Zero.

Note :- In absence of external force no strain or stress is created.

Example 5 :- A steel wire of cross-sectional area 0.5 mm² is held between two fixed supports. If the wire is just taut at 20°C, determine the tension when the temperature falls to 0°C. Coefficient of linear expansion of steel is 1.2 × 10^–5/°C and its Young’s modulus is 2.0 × 10¹¹ N/m².

Solution. With fall in temperature a tensile force is developed in the rod, which is given by :-

Force (F) = YAαΔT

∴ F = 0.5 × 10^–6 × 2 × 10¹¹ × 1.2 × 10^–5 × 20 = 24 N

 

Example 6 :- An iron ring measuring 15.00 cm in diameter is to be shrunk on a pulley which is 15.05 cm in diameter. All measurements refer to the room temperature 20°C. To what minimum temperature should the ring be heated to make the job possible? Calculate the strain developed in the ring when it comes to the room temperature. Coefficient of linear expansion of iron = 12 × 10^–6/°C.

Solution. The ring should be heated to increase its diameter from 15.00 cm to 15.05 cm.

Using l₂ = l₁ (1 + αΔT)

So minimum temperature the ring be heated = 20°C + 278°C = 298°C.

The strain developed in the rod = (l₂ – l₁)/l₁ = 3.33 × 10^–3

 

 

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