Question

A mild steel bar of 2.0cm diameter is subjected to a tensile test. The bar yields under a load of 80kN. It reaches a maximum load of 150kN and breaks finally at a load of 70kN. Estimate the.(a) Average tensile stress at the yield point. (b) Ultimate tensile stress. (c) Average stress at the bearing point if the diameter of the fractured neck is 1cm.

Answer 1

a) An average tensile stress at the yield point $σ_y=\frac{F_y}{A_1}=\frac{4\cdot80 \cdot10^3}{πd_1^2}=\frac{320 \cdot10^3}{(2\cdot10^{-2})^2π}=25.5\cdot10^7=255$ MPa.

b) An ultimate tensile stress $σ_{max}=\frac{F_{max}}{A_1}=\frac{4\cdot150 \cdot10^3}{πd_1^2}=\frac{4\cdot150 \cdot10^3}{(2\cdot10^{-2})^2π}=47.7\cdot10^7=477$ MPa.

c) An average tensile stress at the bearing point $σ_{d}=\frac{F_{d}}{A_2}=\frac{4\cdot70 \cdot10^3}{πd_2^2}=\frac{280 \cdot10^3}{(10^{-2})^2π}=89.1\cdot10^7=891$ MPa