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Two gage marks are placed exactly 250 mm apart on a 12-mm-diameter aluminum rod with E 5 73 GPa and an ultimate strength of 140 MPa. Knowing that the distance between the gage marks is 250.28 mm after a load is applied, determine (a) the stress in the rod, (b) the factor of safety

Respuesta :

Answer:

81.76 N/mm² ( MPa), 1.71233

Explanation:

Modulus of elasticity = stress / strain

stress = modulus of elastic × strain

strain = ΔL / L = 250.28 mm - 250 mm / 250 mm = 0.00112

Modulus of elasticity E = 73 GPa = 73 × 10³ MPa where 1 MPa = 1 N/mm²

E = 73 × 10³N/mm²

stress =  73 × 10³N/mm²× 0.00112 = 81.76 N/mm² ( MPa)

b) Factor of safety = maximum allowable stress / induced stress = 140 MPa / 81.76 MPa = 1.71233

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