Question

7. A 6 ft long steel rod has a cross-sectional area of 0.08 in2. How much does the rod stretch under an axi- al tensile load of 400 lb? Report the answer in inches.

Answer 1

From, Young’s modulus:

Y = Stress/Strain

Stress = Y*strain

Strain = dL/L0

Stress = Force/Area

Y = F*L0/(A*dL)

dL = Stretch in rod = F*L0/(A*Y)

A = cross-sectional area of rod = 0.08 in^2 = 5.1613*10^-5 m^2 (Since 1 in = 0.0254 m)

given, F = tension in rod = 400 lb = 1779.2886 N (since 1lb = 4.448 N)

L0 = initial length = 6 ft = 1.8288 m (since 1 ft = 0.3048 m)

Y = Young modulus of steel = 21*10^10 N/m^2

So, Using these values:

dL = 1779.2886*1.8288/[(21*10^10)*5.1613*10^-5] = 0.000300 m

Since 1 in = 0.0254 m, So

dL = 3.00*10^-4 m = 3.00*10^-4/(0.0254) = 0.0118 in

Stretch in rod = 0.0118 inch

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