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.

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

Step 1
FBD



Step 2
We can write an equilibrium equation: The moment about the center point D.

$$\sum M_D=0=T_E-F(r_D)$$ $$\Rightarrow F=\frac{T_E}{r_D}$$

FBD

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

Step 4
Now we set the two Forces (F) equal to each other and solve. $$\frac{T_E}{r_D}=\frac{T_F}{r_G}\Rightarrow T_E=\frac{T_F*r_D}{r_G}=\frac{1200lb*in*8in}{3in}=3200 lb*in$$

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 7
FBD

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.

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

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

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