Determine size of a standard square key made of 1045 hot rolled steel for a 2 inch DIA shaft transmitting 100 HP at 500 rpm with factor of safety 2.5 for yield in direct shear. Winch Linch

Width = Length = 1.148 inches

Explanation:

We have been given the following data:

D = diameter = 2 inch = 0.0508m

P = Power = 100 HP = 74570 W

N = 500 rpm

Safety Factor = 2.5

We need to find yield strength which is represented by σ(y).

σ(y) for 1045 hot rolled steel = 330MPa

Find Shear Strength. Formula is given:

τ(y) = σ(y) / 2

τ(y) = 330 / 2

τ(y) = 165 MPa

τ(y) = 165 × 10⁶ Pa

τ(y) = 165 × 10⁶ kg.m⁻¹.s⁻²

Find Torque. Formula is given:

T = 60P / 2πN

T= (60)(74570) / 2π(500)

T = 1424.9 Nm

Find Shear Force. Formula is given:

F = 2T/d

F = 2(1424.9)/0.0508

F = 56098.43 N

F = 56098.43 kg.m.s⁻²

Find length by the given formula:

F/L² = τ(y)/Safety factor

Rearrange for L:

L = √(F· Safety factor / τ(y))

Substitute the values found in previous steps to calculate L.

L = 0.02915 meters

Convert it into inches:

L = 1.148 in

As it si a square key:

L = W

Width = 1.148 in

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