# Mechanics of materials/Problem set 4 Completion status: this resource is a stub, which means that pretty much nothing has been done yet.

## Problem 4.1

P3.23, Beer 2012
On our honor, we did this problem on our own, without looking at the solutions in previous semesters or other online solutions.

### Problem Statement

Under normal operating conditions a motor exerts a torque of magnitude $T_{F}$  =1200 lb*in. at F. $r_{D}$  =8in., $r_{G}$  =3in., and the allowable shearing stress is 10.5 ksi in each shaft.

### Objective

Determine the required diameter of (a) shaft CDE, (b) shaft FGH.

### Solution

FBD

#### Step 2

We can write an equilibrium equation: The moment about the center point D.

$\sum M_{D}=0=T_{E}-F(r_{D})$

$\Rightarrow F={\frac {T_{E}}{r_{D}}}$

#### Step 3

We can write an equilibrium equation: The moment about the center point G.

$\sum M_{G}=0=T_{F}-F(r_{G})$

$\Rightarrow F={\frac {T_{F}}{r_{G}}}$

(a)

#### Step 4

Now we set the two Forces (F) equal to each other and solve.

${\frac {T_{E}}{r_{D}}}={\frac {T_{F}}{r_{G}}}\Rightarrow T_{E}={\frac {T_{F}*r_{D}}{r_{G}}}={\frac {1200lb*in*8in}{3in}}=3200lb*in$

#### Step 5

Solve $\tau _{max}$
$\tau _{max}={\frac {T_{E}*C_{D}}{J}}\Rightarrow C_{D}={\sqrt[{3}]{\frac {2T_{E}}{\pi \tau _{max}}}}={\sqrt[{3}]{\frac {2*3200lb*in}{\pi 10.5*10^{3}{\frac {lb}{in^{2}}}}}}=0.58in$

#### Step 6

Solve for the diameter of shaft CDE

$d_{D}=2*C_{D}=2*0.58in=1.158in$

(b)

#### Step 8

Solve $\tau _{max}$

$\tau _{max}={\frac {T_{F}*C_{F}}{J}}\Rightarrow C_{F}={\sqrt[{3}]{\frac {2T_{F}}{\pi \tau _{max}}}}={\sqrt[{3}]{\frac {2*1200lb*in}{\pi 10.5*10^{3}{\frac {lb}{in^{2}}}}}}=0.417in$

#### Step 9

Solve for the diameter of shaft FGH

$d_{F}=2*C_{F}=2*0.42in=0.835in$

## Problem 4.2

P3.25, Beer 2012
On our honor, we did this problem on our own, without looking at the solutions in previous semesters or other online solutions.

### Problem Statement

The two solid shafts are connected by gears as shown and are made of steel for which the allowable shearing stress is 8500 psi. A torque of magnitude $T_{C}$  =5 kip*in. is applied at C and the assembly is in equilibrium.

### Objective

Determine the required diameter of (a) shaft BC, (b) shaft EF.

### Solution

#### Step 1

First sum the moments around gears A and E

$\sum M_{A}=F_{A}*R_{C}+T_{C}$

$\sum M_{D}=F_{D}*R_{F}+T_{F}$

Solve for F

$F={\frac {T}{R}}$

#### Step 2

Sum the Forces

$\sum F=0=F_{C}+F_{F}={\frac {T_{C}}{R_{A}}}+{\frac {T_{F}}{R_{E}}}$

#### Step 3

Solve for $T_{F}$

$T_{F}={\frac {-T_{C}*R_{E}}{R_{A}}}={\frac {-5kipin*2.5in}{4in}}=-3.125kip*in$

The magnitude of $T_{F}$  = 3.125 kip*in

#### Step 4

Use the formula for $\tau _{max}$  to solve for the radius

$\tau _{max}={\frac {T*r}{J}}={\frac {2*T}{\pi *r^{3}}}$

$r^{3}={\frac {2*T}{\pi *\tau _{max}}}$

$r=({\frac {2*T}{\pi *\tau _{max}}})^{1/3}$

#### Step 5

Input the Given values into the equation for the radius

$d_{C}=2*r_{C}=2*({\frac {2*T_{C}}{\pi *\tau _{max}}})^{1/3}=2*({\frac {2*5000lb*in}{\pi *8500psi}})^{1/3}=1.44in$

$d_{F}=2*r_{F}=2*({\frac {2*T_{F}}{\pi *\tau _{max}}})^{1/3}=2*({\frac {2*3125lb*in}{\pi *8500psi}})^{1/3}=1.233in$