Question

Q.1 A solid shaft can resist a bending moment of 3.0 kNm and a twisting moment of 4.0
kNm together then what is the maximum torque that can be applied.
​

Answer 1

Explanation:

ft =  

M

Z

=  

3 × 103

{(π / 64 ) × d4 }

d / 2

ft =  

30.558

kPa

d3

Maximum shear stress due to torsion, fs =  

TR

J

=  

4 × 103 × (d / 2)

π

× d4

32

fs =  

20.7

kPa

d3

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

fs max. =  

1

[√(ft)² + 4(fs)² ]

2

Maximum torque that can be applied,

T =  

πd3

fs max.

16

=  

πd3

×  

1

[ √(30.558 / d3)² + 4(20.37 / d3)² ]

16 2

= 4 kNm

Answer 2

4kNm is your answer mate

Mark me as brainliest