Question

A solid circular shaft is subjected to a bending moment of 3000 N-m and a torque of 10,000Nm. The shaft is made of steel having ultimate tensile stress of 700Mpa and a ultimate shear stress of 500Mpa. Assuming factor of safety as 6. Determine the diameter of the shaft.​

Answer 1

Answer:

ok

Explanation:

plz check it again incorrect question

Answer 2

The factor of safety as the dimeter of the shaft $0.636*106$ or $86mm.$

Explanation:

M=3000N-m; T=10,000Nm; σtu=700Mpa;τu=500Mpa

σ$t$orσ$b$=σ$tu$

$FOS=\frac{700}{6}$

=$116.7\frac{N}{mm_{2}}$

allowable shear stress,

τ=τ$u$

$FOS=\frac{500}{6}$

$=83.3\frac{N}{mm_{2}}$

$T_{e}=\sqrt{M_{2}+T^{2}$

$=\sqrt{(3*106)^{2}+(10*106)^{2}}$

$=10.44*106N_{mm}$

we also know that equivalent twisting moment($T_{e}$)

$10.44*106=\pi *$τ*$d3$

$d_{3}=\frac{10.44*106}{16.36}$

=$0.636*106$ or

$=86mm$.