# Mechanics of materials/Problem set 4

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## 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 ${\displaystyle T_{F}}$  =1200 lb*in. at F. ${\displaystyle r_{D}}$  =8in., ${\displaystyle r_{G}}$  =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.

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

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

FBD

#### Step 3

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

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

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

(a)

FBD

#### Step 4

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

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

#### Step 5

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

#### Step 6

Solve for the diameter of shaft CDE

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

(b)

FBD

#### Step 8

Solve ${\displaystyle \tau _{max}}$

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

#### Step 9

Solve for the diameter of shaft FGH

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

## 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 ${\displaystyle T_{C}}$  =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

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

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

Solve for F

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

#### Step 2

Sum the Forces

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

#### Step 3

Solve for ${\displaystyle T_{F}}$

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

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

#### Step 4

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

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

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

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

#### Step 5

Input the Given values into the equation for the radius

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

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