A shaft of 50 mm diameter transmits a torque of 800 N-m. The width of the rectangular key used is 10 mm. The allowable shear stress of the material of the key being 40 MPa, the required length of the key would be

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ESE Mechanical 2016 Paper 2: Official Paper
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  1. 60 mm
  2. 70 mm
  3. 80 mm
  4. 90 mm

Answer (Detailed Solution Below)

Option 3 : 80 mm
Free
ST 1: UPSC ESE (IES) Civil - Building Materials
20 Qs. 40 Marks 24 Mins

Detailed Solution

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Concept:

For a rectangular key Allowable shear stress 

The torque transmitted by the shaft is given by, 

Calculation:

Given, T = 800 N-m, d = 50 mm, b = 10 mm, τper = 40 MPa

∴ Force on key F = 800/25 = 32 kN

Allowable shear stress 

⇒ l = 3200/40 = 80 mm

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