Question

A mild steel rod of diameter 25 mm and 400 mm long is incased centrally inside a hollow copper tube
of external diameter 35 mm and inside diameter 30 mm. The ends of the rod and tube are rigidly attached,
and the composite bar is subjected to an axial pull of 40kN. If E for steel and copper is 200GN/m² and
100GN/m² respectively, find the stress developed in the rod and the tube. Find also the extension of the rod.
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Answer 1

Answer:

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Explanation:

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Answer 2

Concept:

As stress is the force operating on a material's unit area, it can be find out by dividing Force with Area on a material.

Given:

Diameter of rod = 25 mm

Length of rod = 400 mm

External diameter of tube = 35 mm

Inside diameter of tube = 30 mm

E for steel = 200GN/m²

E for copper = 100GN/m²

Find:

Stress in the rod and tube

Extension in the rod

Solution:

Let as use 1 for the steel rod are 2 for the upper tube

A₁ = π/4 × (25)²  

A₂ = 490.87 mm²

A₂ = π/4 (35² - 30²)

A₂ = 255.25 mm²

from the equilibrium -->

P₁A₁ + P₂A₂ = P = 40 × 10³

therefore,

P₁²/E₁ = P₂²/E₂

P₁ = P₂ × (200/100)

P₁ = 2P₂

2P₂A₂ + P₂A₂ = 40 × 10³

P₂ = 32.34 N/mm³

P₁ = 2P₂ = 64.68 N/mm³

P₁ = 64.68 N/mm³

also extension of rod = (64.88 × 4000) / (2×10³)

extension of rod = 1.294 mm

Hence the stress developed in steel rod is P₁ which is 64.68 N/mm³ and stress developed in outer tube is P₂ which is 32.34 N/mm³, The rod also extended by 1.294 mm.

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