Strength of Materials: Unit III: Torsion

replacing a shaft problems

Torsion - Strength of Materials

replacing a shaft problems: Torsion - Strength of Materials

REPLACING A SHAFT PROBLEMS

Example 3.17 

A solid steel shaft of 50 mm diameter is to be replaced by a hollow steel shaft whose internal diameter is 0.5 times outer diameter. Find the diameter of the hollow shaft and percentage saving in weight.

Given data: 

Solid shaft diameter, D = 50 mm


To find: d, D1 and % saving in weight

Solution: 

We know that, the torque transmitted by the solid shaft should be equal to that hollow shaft when solid shaft is replaced by hollow shaft.

Torque transmitted by solid shaft


Equating (1) and (2),


Result:

External diameter, D1 = 51.08 mm

Internal diameter, d = 25.5 mm

% of saving in weight = 21.7%

Example 3.18 

A solid shaft transmits 240 kW at 110 rpm. If the shear stress is not to exceed 70 N/mm2, find the diameter of the shaft. If this shaft is to be replaced by a hollow shaft whose internal diameter is 0.6 times outer diameter, determine the size and percentage of saving in material, the maximum shearing stress being the same.

Given data:


To find:

For solid shaft

1. Diameter of the shaft (D)

For hollow shaft

1. External diameter (D1)

2. Internal diameter (d)

3. % of saving in material

Solution:

We know that,



Solid shaft is replaced by hollow shaft. So torque transmitted by solid shaft is equal to hollow shaft

Equating (1) and (2)



Result:

For solid shaft

Diameter of the shaft, D = 114.86 mm

For hollow shaft

External diameter, D1 = 120.3 mm

Internal diameter, d = 72.1 mm

% of saving in material = 29.7%

Strength of Materials: Unit III: Torsion : Tag: : Torsion - Strength of Materials - replacing a shaft problems


Strength of Materials: Unit III: Torsion



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Strength of Materials

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