Question

A 2.5-m-long tension member (E = 70 GPa) has cross-sectional dimensions of 25 mm x 35 mm. Determine the maximum load P that may be supported by this member if the allowable normal stress may not exceed 90 MPa and the total elongation must not exceed 3 mm. State your answer in kN.

Answer 1

Force = 7.3 kN

Step-by-step explanation:

Elastic Modulus = Stress / Strain

Since the maximum stress applied can be 90 MPa, lets check the corresponding maximum strain using the above equation.

70 * 10^9 = 90 * 10^6 / Strain

Strain = 0.0012857

Strain = Change in length/total length

Change in Length = 0.0012857 * 2.5

Change in Length = 0.003214 m

Converted to mm this is: 0.003214 * 1000 = 3.214 mm

Thus we can see that strain is the limiting factor, and not stress. This is because with the maximum stress possible applied, we go over the strain limit.

Doing another quick calculation, we can take out the maximum stress with strain equal to 3mm/2.5m

Strain = (3/1000)/2.5

Strain = 0.0012

Modulus of elasticity = Stress/ Strain

Stress = 70 * 10^9 * 0.0012

Stress = 84 MPa

This stress is under both the strain and stress limit.

Taking out the force due to this stress:

Stress = force / area

Force = 84 * 10^6 * Area

Force = 84 * 10^6 * (25/1000) * (35/1000)

Force = 7350 N or 7.3 kN