User:Tshankwitz

Problem Statement

Aluminum and brass rods secured to a wall.

There is a 1250 N*m torque applied at point A. The maximum allowable stress in the brass rod, AB, is 50 MPa and the maximum allowable stress in the aluminum rod, BC, is 25 MPa. Determine the minimum diameter of (a) rod AB and (b) rod BC.

Given

$\displaystyle \tau _{max,brass}=50\,MPa$
$\displaystyle \tau _{max,alum}=25\,MPa$

Solution

Step One:

Using the elastic torsion formula, which relates the shearing stress to the torque and properties of the rod, solve for $\displaystyle c$, the radius of the rod.

$\displaystyle \tau _{max}={\frac {Tc}{J}}$

(3.6-1)

Where $\displaystyle J$ for a cylindrical rod can be expressed as

$\displaystyle J={\frac {1}{2}}\pi c^{4}$

(3.6-2)

Replacing equation (3.6-2) with $\displaystyle J$ from equation (3.6-1), we recieve

$\displaystyle \tau _{max}={\frac {Tc}{{\frac {1}{2}}\pi c^{4}}}$

(3.6-3)

$\displaystyle \tau _{max}={\frac {2T}{\pi c^{3}}}$

(3.6-4)

Solving for $\displaystyle c$ gives us

$\displaystyle c^{3}={\frac {2T}{\tau _{max}\pi }}$

(3.6-5)

$\displaystyle c={\sqrt[{3}]{\frac {2T}{\tau _{max}\pi }}}$

(3.6-6)

Since the diameter is twice the bar's radius,

$\displaystyle d=2*c$

(3.6-7)

This can now be used to solve for the diameter of each bar separately

Step 2:

Starting with the brass rod, AB, and using equation (3.6-6)

$\displaystyle c_{brass}={\sqrt[{3}]{\frac {2(1250\,N\cdot m)}{(50\,MPa)\pi }}}$

(3.6-8)

$\displaystyle d_{brass}=2*c_{brass}$

(3.6-9)

$\displaystyle d_{brass}=.0503\,m$

(3.6-10)

Now for the aluminum rod,

$\displaystyle c_{alum}={\sqrt[{3}]{\frac {2(1250\,N\cdot m)}{(25\,MPa)\pi }}}$

(3.6-11)

$\displaystyle d_{alum}=2*c_{alum}$

(3.6-12)

$\displaystyle d_{alum}=0.0634\,m$

(3.6-13)