Mechanics of materials/Problem set 4

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.
File:3.23 prob.png

Objective

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

Solution

Step 1

FBD

File:FBD 3.23.png

Step 2

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

$\sum M_D=0=T_E-F(r_D)$

$\Rightarrow F=\frac{T_E}{r_D}$

FBD
File:FBD 2 3.23.png

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)

FBD
File:Fbd shaft1.png

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 ${\tau }_{max}$
${\tau }_{max}=\frac{T_E*C_D}{J}\Rightarrow C_D=\sqrt[3]{\frac{2T_E}{\pi {\tau }_{max}}}=\sqrt[3]{\frac{2*3200lb*in}{\pi 10.5*10^3\frac{lb}{i{n}^2}}}=0.58in$

Step 6

Solve for the diameter of shaft CDE

$d_D=2*C_D=2*0.58in=1.158in$

(b)

Step 7

FBD
File:FBD shaft 2.png

Step 8

Solve ${\tau }_{max}$

${\tau }_{max}=\frac{T_F*C_F}{J}\Rightarrow C_F=\sqrt[3]{\frac{2T_F}{\pi {\tau }_{max}}}=\sqrt[3]{\frac{2*1200lb*in}{\pi 10.5*10^3\frac{lb}{i{n}^2}}}=0.417in$

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.
File:3.25 prob.png

Objective

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

Solution

Step 1

File:Fbd 3.25 1.png File:Fbd 3.25 2.png

File:Fbd 3.25 shaft1.png File:Fbd 3.25 shaft2.png

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

$d_C=2*r_C=2*(\frac{2*T_C}{\pi *{\tau }_{max}}{)}^{1/3}=2*(\frac{2*5000lb*in}{\pi *8500psi}{)}^{1/3}=1.44in$

$d_F=2*r_F=2*(\frac{2*T_F}{\pi *{\tau }_{max}}{)}^{1/3}=2*(\frac{2*3125lb*in}{\pi *8500psi}{)}^{1/3}=1.233in$