Question 5.1: A hot-rolled steel has a yield strength of Syt = Syc = 100 k...

Since ε_{f} > 0.05 and S_{yc} and S_{yt} are equal, the material is ductile and the distortionenergy (DE) theory applies. The maximum-shear-stress (MSS) theory will also be applied and compared to the DE results. Note that cases a to d are plane stress states.

(a) The ordered principal stresses are σ_{A} = σ_{1} = 70, σ_{B} = σ_{2} = 70, σ_{3} = 0 kpsi.

DE From Eq. (5–13),

σ′ =\left (σ^{2}_{A} − σ_{A}σ_{B} + σ^{2}_{B} \right)^{1/2}                 (5–13)

σ′ = [70^{2} − 70(70) + 70^{2}]^{1/2} = 70  kpsi

n =\frac {S_{y}}{σ′} =\frac {100}{70} = 1.43

MSS Case 1, using Eq. (5–4) with a factor of safety,

σ_{A} ≥ S_{y}                       (5–4)

n =\frac {S_{y}}{σ_{A}} =\frac {100}{70} = 1.43

(b) The ordered principal stresses are σ_{A} = σ_{1} = 70, σ_{B} = σ_{2} = 30, σ_{3} = 0  kpsi.

DE            σ′ = [70^{2} − 70(30) + 30^{2}]^{1/2} = 60.8  kpsi

n =\frac {S_{y}}{σ′} =\frac {100}{60.8} = 1.64

MSS Case 1, using Eq. (5–4),

n =\frac {S_{y}}{σ_{A}} =\frac {100}{70} = 1.43

(c) The ordered principal stresses are σ_{A} = σ_{1} = 70, σ_{2} = 0, σ_{B} = σ_{3} = −30  kpsi.

DE                 σ′ = [702 − 70(−30) + (−30)^{2}]^{1/2} = 88.9  kpsi

n =\frac {S_{y}}{σ′} =\frac {100}{88.9 }= 1.13

MSS             Case 2, using Eq. (5–5),

σ_{A} − σ_{B} ≥ S_{y}                     (5–5)

n =\frac {S_{y}}{σ_{A}-σ_{B}} =\frac {100}{70- (-30)} = 1.00

(d) The ordered principal stresses are σ_{1} = 0, σ_{A} = σ_{2} = −30, σ_{B} = σ_{3} = −70 kpsi.

DE      σ′ = [(−70)^{2} − (−70)(−30) + (−30)^{2}]^{1/2} = 60.8  kpsi

n =\frac {S_{y}}{σ′} =\frac {100}{60.8 }= 1.64

MSS       Case 3, using Eq. (5–6),

σ_{B} ≤ −S_{y}             (5–6)

n = −\frac {S_{y}}{σ_{B}} = −\frac {100}{−70} = 1.43

e) The ordered principal stresses are σ_{1} = 30, σ_{2} = 30, σ_{3} = 30  kpsi

DE       From Eq. (5–12),

σ′ =\left [ \frac {(σ_{1} − σ_{2})^{2} + (σ_{2} − σ_{3})^{2} + (σ_{3} − σ_{1})^{2}}{2}\right]^{1/2}      (5-12)

σ′ =\left [\frac {(30 − 30)^{2} + (30 − 30)^{2} + (30 − 30)^{2}}{2}\right]^{1/2}= 0  kpsi

n =\frac {S_{y}}{σ′} =\frac {100}{0}→∞

MSS         From Eq. (5–3),

τ_{max} =\frac {S_{y}}{2n}            or     σ_{1} − σ_{3} =\frac {S_{y}}{n}                       (5–3)

n =\frac {S_{y}}{σ_{1} − σ_{3}} =\frac {100}{30 − 30}→∞

A tabular summary of the factors of safety is included for comparisons.

Since the MSS theory is on or within the boundary of the DE theory, it will always predict a factor of safety equal to or less than the DE theory, as can be seen in the table. For each case, except case (e), the coordinates and load lines in the σ_{A}, σ_{B} plane are shown in Fig. 5–11. Case (e) is not plane stress. Note that the load line for case (a) is the only plane stress case given in which the two theories agree, thus giving the same factor of safety.