Question

The cross-sectional area of a 7.3 m long mild steel bar is 968 sq. mm.
When subjected to a tensile force, it stretches by 2.286 mm. Calculate the
load, if E = 21000 n / sq mm

Answer 1

Given that,

Area = 968 mm²

Length = 7.3 m

Change in length = 2.286 mm

Young's modulus =21000 N/mm²

We need to calculate the load

Using formula of young modulus

$Y=\dfrac{F\times L}{A\times \Delta L}$

$F=\dfrac{A\times\Delta L\times Y}{L}$

Where, A = area

$\Delta L$ change in length

L = length

Y = young's modulus

F = load

Put the value into the formula

$F=\dfrac{968\times2.286\times10^{-3}\times21000}{7.3}$

$F=6365.7\ N$

Hence, The load is 6365.7 N.

Answer 2

Given that,

Area = 968 mm²

Length = 7.3 m

Change in length = 2.286 mm

Young's modulus =21000 N/mm²

We need to calculate the load

Using formula of young modulus

$Y=\dfrac{F\times L}{A\times \Delta L}$

$F=\dfrac{A\times\Delta L\times Y}{L}$

Where, A = area

$\Delta L$ change in length

L = length

Y = young's modulus

F = load

Put the value into the formula

$F=\dfrac{968\times2.286\times10^{-3}\times21000}{7.3}$

$F=6365.7\ N$

Hence, The load is 6365.7 N.

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