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A rod, which tapers uniformly from 5cm diameter to 3cm diameter in a length of 50cm, is subjected to an axial load of 6000N. If E=2 x 105N/mm2, find the extension of the rod.​

1 Answer

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To find the extension of the rod, we can use the formula:

delta = PL / (AE)

where delta is the extension, P is the axial load, L is the length of the rod, A is the cross-sectional area of the rod, and E is the modulus of elasticity.

The cross-sectional area of the rod varies along its length, so we need to find the average area. The average diameter is:

(5 + 3) / 2 = 4 cm

The average area is:

A = pi/4 x (4)^2 = 12.57 cm^2

Substituting the given values, we get:

delta = 6000 x 50 / (12.57 x 2 x 10^5) = 0.015 cm

Therefore, the extension of the rod is 0.015 cm.
User Raathigesh
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