Question

Consider a cast iron machine part with tensile strength S ut =30 Sut=30 kpsi, compressive strength S uc =90 Suc=90 kpsi, subjected to a state of plane stress ( σ 3 =0 σ3=0 ) with σ x =−7, σ y =−8, τ xy =−15 σx=−7,σy=−8,τxy=−15 kpsi. ∙ ∙ Calculate the safety factor using the Brittle Mohr-Coulomb (BMC) criterion.

Answer 1

The safety factor using the Brittle Mohr-Coulomb (BMC) criterion is approximately 3.46.

Step-by-step explanation:

To calculate the safety factor using the Brittle Mohr-Coulomb (BMC) criterion, we need to determine the shear strength (τ_ult). For cast iron, the shear strength is typically given as a fraction of the compressive strength, commonly assumed to be 0.577 times the compressive strength (τ_ult = 0.577 * S_uc). In this case, the shear strength is 0.577 * 90 kpsi = 52.03 kpsi. The maximum shear stress (τ_max) in the given state of plane stress can be calculated using the equations: τ_max = (σ_x - σ_y) / 2 ± sqrt[((σ_x - σ_y) / 2)^2 + τ_xy^2]. Substituting the given values, we have τ_max = (-7 - (-8)) / 2 ± sqrt[((-7 - (-8)) / 2)^2 + (-15)^2] = 1 / 2 ± sqrt[1 / 4 + 225] = 1 / 2 ± sqrt[901 / 4] = 1 / 2 ± sqrt[225.25] ≈ 15.034 kpsi. The safety factor (SF) is then calculated as SF = τ_ult / τ_max = 52.03 / 15.034 ≈ 3.46.