A solid shaft can resist a bending moment of 3.0 kNm and

A solid shaft can resist a bending moment of 3.0 kNm and a twisting moment of 4.0 kNm together, then the maximum torque that can be applied is

Maximum tensile stress due to bending,

f_t =	M
	Z

f_t =	30.558	kPa
	d³

Maximum shear stress due to torsion, f_s =	TR
	J

f_s =	20.7	kPa
	d³

When both the stress acts simultaneously, maximum induced shear stress,

f_{s max.} =	1	[√(f_t)² + 4(f_s)² ]
	2

Maximum torque that can be applied,

T =	πd³	f_{s max.}
	16

=	πd³	×	1	[ √(30.558 / d³)² + 4(20.37 / d³)² ]
	16		2

= 4 kNm