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veedub1955
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Mechanical principles - Stress and modulus of elasticity
A 2.2m long steel towing bar of solid circular section diameter of 45mm is expected to carry a maximum load of 210 kN. The safety factor is 4; and for the steel the UTS is 540 MNm-2 and the modulus of elasticity is 200 GNm-2
i) Calculate the actual towing capacity of the bar and state whether the proposed limit is acceptable
ii) Determine the extension of the bar under the maximum proposed load
i) σ = Load / Area
Proposed limit acceptable= SF = UTS / σ
ii) Change in length = (σ / Modulus of elasticity) / Length
Hi I have worked out what I think is right but I have trouble writing the correct units, I will put what i have worked out could someone check if I am right please, thank you very much:-
i) σ = Load / Area = 210kN / ((∏(45 x 10-3) / 4)
=
840x10-3 / 6.362x10-3 = 132.04 (now I think this is MNm2)
Proposed limit acceptable = SF = 540 / 132.04 = 4.09
So the proposed limit in this case is acceptable as it is over the required Safety Factor of 4
ii) Change in length = (132.04 / 200) x 2.2
= 0.66 x 2.2 = 1.45mm (??)
Homework Statement
A 2.2m long steel towing bar of solid circular section diameter of 45mm is expected to carry a maximum load of 210 kN. The safety factor is 4; and for the steel the UTS is 540 MNm-2 and the modulus of elasticity is 200 GNm-2
i) Calculate the actual towing capacity of the bar and state whether the proposed limit is acceptable
ii) Determine the extension of the bar under the maximum proposed load
Homework Equations
i) σ = Load / Area
Proposed limit acceptable= SF = UTS / σ
ii) Change in length = (σ / Modulus of elasticity) / Length
The Attempt at a Solution
Hi I have worked out what I think is right but I have trouble writing the correct units, I will put what i have worked out could someone check if I am right please, thank you very much:-
i) σ = Load / Area = 210kN / ((∏(45 x 10-3) / 4)
=
840x10-3 / 6.362x10-3 = 132.04 (now I think this is MNm2)
Proposed limit acceptable = SF = 540 / 132.04 = 4.09
So the proposed limit in this case is acceptable as it is over the required Safety Factor of 4
ii) Change in length = (132.04 / 200) x 2.2
= 0.66 x 2.2 = 1.45mm (??)
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