modulus of rupture

MODULUS OF RUPTURE

The maximum shear stress calculated by the torsion formula by using the experimentally found maximum torque required to rupture a shaft.

SOLVED PROBLEMS 

Example 3.48

Evaluate Poisson's ratio and Young's modulus, Bulk modulus, Rigidity modulus for a material, in a tensile test, a test piece of 25 mm diameter, 200 mm gauge length, stretched 0.0950 mm under a pull of kN. In a torsion test, the same rod twisted 0.03 radian over a length of 150 mm when a torque of 0.5 kN.m was applied.

Given:

To find:

Solution:

Result:

Example 3.49

Find the length of a 5 mm diameter aluminium wire, if it can be twisted through one complete revolution and no exceeding the shear stress of 45 MN/m². Take C = 30 GN/m².

Given:

To find:

l = ?

Solution:

Result:

l = 10.47 m

Example 3.50 

For a solid steel shaft, taking shear stress as 75 MN/m² and it has to transmit power 80 kW at 170 rpm. If the maximum torque transmitted on each revolution exceeds the mean by 30%, find the suitable diameter for the shaft. 80 kW = 80 × 103 W

Given:

To find:

D = ?

Solution:

Result:

D = 73.47 mm