Completion status
P3.23, Beer 2012
Under normal operating conditions a motor exerts a torque of magnitude
T
F
{\displaystyle T_{F}}
r
D
{\displaystyle r_{D}}
r
G
{\displaystyle r_{G}}
Determine the required diameter of (a) shaft CDE, (b) shaft FGH.
FBD
We can write an equilibrium equation: The moment about the center point D.
∑
M
D
=
0
=
T
E
−
F
(
r
D
)
{\displaystyle \sum M_{D}=0=T_{E}-F(r_{D})}
⇒
F
=
T
E
r
D
{\displaystyle \Rightarrow F={\frac {T_{E}}{r_{D}}}}
We can write an equilibrium equation: The moment about the center point G.
∑
M
G
=
0
=
T
F
−
F
(
r
G
)
{\displaystyle \sum M_{G}=0=T_{F}-F(r_{G})}
⇒
F
=
T
F
r
G
{\displaystyle \Rightarrow F={\frac {T_{F}}{r_{G}}}}
Now we set the two Forces (F) equal to each other and solve.
T
E
r
D
=
T
F
r
G
⇒
T
E
=
T
F
∗
r
D
r
G
=
1200
l
b
∗
i
n
∗
8
i
n
3
i
n
=
3200
l
b
∗
i
n
{\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}
Solve
τ
m
a
x
{\displaystyle \tau _{max}}
τ
m
a
x
=
T
E
∗
C
D
J
⇒
C
D
=
2
T
E
π
τ
m
a
x
3
=
2
∗
3200
l
b
∗
i
n
π
10.5
∗
10
3
l
b
i
n
2
3
=
0.58
i
n
{\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}
Solve for the diameter of shaft CDE
d
D
=
2
∗
C
D
=
2
∗
0.58
i
n
=
1.158
i
n
{\displaystyle d_{D}=2*C_{D}=2*0.58in=1.158in}
(b)
Solve
τ
m
a
x
{\displaystyle \tau _{max}}
τ
m
a
x
=
T
F
∗
C
F
J
⇒
C
F
=
2
T
F
π
τ
m
a
x
3
=
2
∗
1200
l
b
∗
i
n
π
10.5
∗
10
3
l
b
i
n
2
3
=
0.417
i
n
{\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}
Solve for the diameter of shaft FGH
d
F
=
2
∗
C
F
=
2
∗
0.42
i
n
=
0.835
i
n
{\displaystyle d_{F}=2*C_{F}=2*0.42in=0.835in}
P3.25, Beer 2012
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
{\displaystyle T_{C}}
Determine the required diameter of (a) shaft BC, (b) shaft EF.
First sum the moments around gears A and E
∑
M
A
=
F
A
∗
R
C
+
T
C
{\displaystyle \sum M_{A}=F_{A}*R_{C}+T_{C}}
∑
M
D
=
F
D
∗
R
F
+
T
F
{\displaystyle \sum M_{D}=F_{D}*R_{F}+T_{F}}
Solve for F
F
=
T
R
{\displaystyle F={\frac {T}{R}}}
Sum the Forces
∑
F
=
0
=
F
C
+
F
F
=
T
C
R
A
+
T
F
R
E
{\displaystyle \sum F=0=F_{C}+F_{F}={\frac {T_{C}}{R_{A}}}+{\frac {T_{F}}{R_{E}}}}
Solve for
T
F
{\displaystyle T_{F}}
T
F
=
−
T
C
∗
R
E
R
A
=
−
5
k
i
p
i
n
∗
2.5
i
n
4
i
n
=
−
3.125
k
i
p
∗
i
n
{\displaystyle T_{F}={\frac {-T_{C}*R_{E}}{R_{A}}}={\frac {-5kipin*2.5in}{4in}}=-3.125kip*in}
T
F
{\displaystyle T_{F}}
Use the formula for
τ
m
a
x
{\displaystyle \tau _{max}}
τ
m
a
x
=
T
∗
r
J
=
2
∗
T
π
∗
r
3
{\displaystyle \tau _{max}={\frac {T*r}{J}}={\frac {2*T}{\pi *r^{3}}}}
r
3
=
2
∗
T
π
∗
τ
m
a
x
{\displaystyle r^{3}={\frac {2*T}{\pi *\tau _{max}}}}
r
=
(
2
∗
T
π
∗
τ
m
a
x
)
1
/
3
{\displaystyle r=({\frac {2*T}{\pi *\tau _{max}}})^{1/3}}
Input the Given values into the equation for the radius
d
C
=
2
∗
r
C
=
2
∗
(
2
∗
T
C
π
∗
τ
m
a
x
)
1
/
3
=
2
∗
(
2
∗
5000
l
b
∗
i
n
π
∗
8500
p
s
i
)
1
/
3
=
1.44
i
n
{\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}
d
F
=
2
∗
r
F
=
2
∗
(
2
∗
T
F
π
∗
τ
m
a
x
)
1
/
3
=
2
∗
(
2
∗
3125
l
b
∗
i
n
π
∗
8500
p
s
i
)
1
/
3
=
1.233
i
n
{\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}
![{\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}](https://wikimedia.org/api/rest_v1/media/math/render/svg/e3e4726a2c9121633370b66c856baac74d46c1a8)
![{\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}](https://wikimedia.org/api/rest_v1/media/math/render/svg/25d5e114e0ce6d64b616fec8a7b701ec8c447f4b)
