- #1

- 122

- 0

## Homework Statement

A locked door handle is composed of a solid circular shaft AB with a diameter of b = 108mm and a flat plate BC with a force P = 69N applied at point C as shown. Let c = 532mm , d = 126mm , and e = 153mm .

## Homework Equations

[itex]\tau[/itex]

_{max}= [itex]\frac{Tc}{J}[/itex]

T = internal torque acting at cross section

c = outer radius

J = polar moment of inertia → J = [itex]\frac{\pi}{2}[/itex]c

^{4}

## The Attempt at a Solution

I've converted all mm to m.

I just wanted to see if someone could verify that I am on the right track.

radius, c = 0.054 m

1. The torque caused by the P force:

69 N × 0.532 m = 36.71 N*m → acting along the length of the solid shaft.

2. Internal torque, T:

T = 36.71 N*m *0. 126 m = 4.62546 N*m → acting at section m-m

3. J = [itex]\frac{\pi}{2}[/itex]0.054 m

^{4}= 13.36×10

^{-6}m

^{4}

[itex]\tau[/itex]

_{max}= [itex]\frac{(4.62546 N*m)*(0.054 m)}{13.36×10^{-6} m^4}[/itex]