Step-by-step solved example
Solid shaft d=40 mm, P=50 kN, T=200 N·m. Surface τ and σ?
1. σ
A=π×20²=1257 mm², σ=50000/1257=39.8 MPa.
2. τ
J=π/32×40⁴=0.251×10^6 mm⁴, τ=Tr/J=200e3×20/0.251e6=15.9 MPa.
Answer: σ ≈ 40 MPa, τ ≈ 16 MPa
Solid shaft d=40 mm, P=50 kN, T=200 N·m. Surface τ and σ?
1. σ
A=π×20²=1257 mm², σ=50000/1257=39.8 MPa.
2. τ
J=π/32×40⁴=0.251×10^6 mm⁴, τ=Tr/J=200e3×20/0.251e6=15.9 MPa.
Answer: σ ≈ 40 MPa, τ ≈ 16 MPa
1.J for a solid round is
2.I used in bending is
3.Hollow shaft uses J= π/32 (Do⁴−Di⁴)
4.If P=0, principals are ±τ
5.Angle of twist still θ=TL/GJ if
6.von Mises with σ and τ (σy=0) is
7.Power and torque: P_power =
8.Surface r is
9.Keyway / hole locally
10.Units: 200 N·m and mm⁴ require T in N·mm (×1000) to get MPa