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A tensile test is carried out on a bar of mild steel of diameter 3 cm. The bar

yields under a load of 90 KN. It reaches a maximum load of 160 KN and

breaks finally at a load of 80KN.

Compute:

I. the tensile stress at the yield point

II. the ultimate tensile stress

III. the average stress at the breaking point, if the diameter of the fractured

neck is 2 cm.

Respuesta :

abiolataiwo2015

The tensile stress at the yield point is 286 MPa, the ultimate tensile stress is 509 MPa and the average stress is 1,018 MPa.

Tensile stress

I. Tensile stress at the yield point

Tensile stress=Fy/A1=4×90×10^3÷πd1²

=360×10^3÷(2×10^-2)²π

=286 MPa

II. Ultimate tensile stress

Fmax/A1=4×160×10^3÷πd1²

=640×10^3÷(2×10^-2)²π

=509 MPa

III. Average stress at the breaking point

Fd/A2=4×80×10^3÷πd2²

=320×10^3÷(10^-2)²π

=1,018 MPa


Therefore the tensile stress at the yield point is 286 MPa, the ultimate tensile stress is 509 MPa and the average stress is 1,018 MPa.

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