Completion status
Problem 4.1
edit
P3.23, Beer 2012
Problem Statement
edit
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}}
Objective
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Determine the required diameter of (a) shaft CDE, (b) shaft FGH.
Solution
edit
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}
Problem 4.2
edit
P3.25, Beer 2012
Problem Statement
edit
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}}
Objective
edit
Determine the required diameter of (a) shaft BC, (b) shaft EF.
Solution
edit
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)
