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25 October, 02:39

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.

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  1. 25 October, 02:44
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    Force = 7.3 kN

    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
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