)=0 So, σz=µ(σx+σy)=µ(−40+σv) Maximum shear stress = larger of |[
σx−σy
2
],[
σy−σz
2
],[
σx−σx
2
]| Maximum shear stress = larger of
|[
−40−σy
2
],[
σy−µ(σy−40)
2
],[
µ(−40+σy)+40
2
]|
From the above equation, [
−40−σy
2
] will be maximum So, |
−40−σy
2
|≤
sjt
2FOS
or −40−σy=300 ⇒σy=−340MPa=340( compression ) Or −40−σy=−300 ⇒σy=260MPa=260 (compression) But in forging, force cannot be tensile, hence the correct answer must be 340MPa (compressive)