Team 1 : Report 3
PB-10.1, Sec. 10
Find the normal and shear stresses
(
σ
,
τ
)
{\displaystyle \left(\sigma ,\tau \right)}
t , angle
θ
{\displaystyle \theta }
dy , and given normal stress
σ
m
a
x
{\displaystyle \sigma _{max}}
τ
m
a
x
{\displaystyle \tau _{max}}
Are the stresses depending t and dy ? Equilibrium of triangle, normal and shear stresses on inclined plane in bar under tension and shaft under torsion. Contents taken from the notes of Dr. Loc Vu-Quoc
θ
1
=
π
6
{\displaystyle \displaystyle \theta _{1}={\frac {\pi }{6}}}
θ
2
=
π
4
{\displaystyle \displaystyle \theta _{2}={\frac {\pi }{4}}}
Triangle 1
σ
m
a
x
{\displaystyle \displaystyle \sigma _{max}}
σ
{\displaystyle \displaystyle \sigma }
σ
m
a
x
=
P
o
A
o
{\displaystyle \displaystyle \sigma _{max}={\frac {P_{o}}{A_{o}}}}
σ
=
P
A
=
P
o
c
o
s
(
θ
)
A
o
c
o
s
(
θ
)
=
P
o
c
o
s
(
θ
)
A
o
{\displaystyle \displaystyle \sigma ={\frac {P}{A}}={\frac {P_{o}cos(\theta )}{\frac {A_{o}}{cos(\theta )}}}={\frac {P_{o}cos(\theta )}{A_{o}}}}
σ
=
σ
m
a
x
c
o
s
2
(
π
6
)
=
3
2
σ
m
a
x
{\displaystyle \displaystyle \sigma =\sigma _{max}cos^{2}({\frac {\pi }{6}})={\frac {\sqrt {3}}{2}}\sigma _{max}}
σ
m
a
x
{\displaystyle \displaystyle \sigma _{max}}
τ
{\displaystyle \displaystyle \tau }
τ
=
V
A
=
P
o
s
i
n
(
θ
)
A
o
c
o
s
(
θ
)
=
P
o
s
i
n
(
θ
)
c
o
s
(
θ
)
A
o
{\displaystyle \displaystyle \tau ={\frac {V}{A}}={\frac {P_{o}sin(\theta )}{\frac {A_{o}}{cos(\theta )}}}={\frac {P_{o}sin(\theta )cos(\theta )}{A_{o}}}}
τ
=
σ
m
a
x
s
i
n
(
π
6
)
c
o
s
(
π
6
)
=
3
4
σ
m
a
x
{\displaystyle \displaystyle \tau =\sigma _{max}sin({\frac {\pi }{6}})cos({\frac {\pi }{6}})={\frac {\sqrt {3}}{4}}\sigma _{max}}
Triangle 2
τ
=
0
{\displaystyle \displaystyle \tau =0}
Problem Number 3.2, p.154
(a ) Determine the torque T that causes a maximum shearing stress of 45 MPa in the hollow cylindrical steel shaft shown. (b ) Determine the maximum shearing stress caused by the same torque T in solid cylindrical shaft of the same cross-sectional area. Contents taken from Mechanics of Materials: 6th Edition textbook. Authors: F.P BEER, E.R. JOHNSTON, J.T. DEWOLF AND D.F. MAZUREK ISBN:9780077565664
Given(s):
τ
m
a
x
h
=
45
M
P
a
{\displaystyle \displaystyle \tau _{max_{h}}=45MPa}
c
1
{\displaystyle \displaystyle c_{1}}
=
45
m
m
{\displaystyle \displaystyle =45mm}
c
2
{\displaystyle c_{2}}
=
30
m
m
{\displaystyle \displaystyle =30mm}
c
3
{\displaystyle \displaystyle c_{3}}
A
s
{\displaystyle \displaystyle A_{s}}
A
h
{\displaystyle \displaystyle A_{h}}
subscripts h and s identifies which cylinders are under consideration, whether hollow or solid, respectively
(a) The relation between torque and shearing stress is
τ
m
a
x
=
T
c
J
{\displaystyle \displaystyle \tau _{max}={\frac {Tc}{J}}}
Solve the definite integral for the centroidal polar moment of inertia
J
=
∫
c
2
d
A
,
A
=
π
c
2
,
d
A
=
2
π
c
{\displaystyle {\begin{aligned}J=\int c^{2}dA,&&A=\pi c^{2},&&dA=2\pi c\end{aligned}}}
d
c
{\displaystyle dc}
J
h
=
2
π
∫
c
2
c
1
c
3
d
c
=
π
2
(
c
1
4
−
c
2
4
)
=
5.1689
∗
10
−
6
m
4
{\displaystyle J_{h}=2\pi \int _{c_{2}}^{c_{1}}c^{3}dc={\frac {\pi }{2}}(c_{1}^{4}-c_{2}^{4})=5.1689*10^{-6}m^{4}}
The torque experienced by the hollow cylindrical steel shaft is
T
=
J
h
τ
m
a
x
h
c
1
=
5.169
K
N
∗
m
{\displaystyle \displaystyle T={\frac {J_{h}\tau _{max_{h}}}{c_{1}}}=5.169KN*m}
(b) Given that both the hollow and solid cylinders have the same cross-sectional area
A
h
=
A
s
{\displaystyle \displaystyle A_{h}=A_{s}}
A
h
=
π
(
c
1
2
−
c
2
2
)
{\displaystyle \displaystyle A_{h}=\pi (c_{1}^{2}-c_{2}^{2})}
A
s
=
π
c
3
2
{\displaystyle \displaystyle A_{s}=\pi c_{3}^{2}}
c
3
=
c
1
2
−
c
2
2
=
0.0335
m
{\displaystyle \displaystyle c_{3}={\sqrt {c_{1}^{2}-c_{2}^{2}}}=0.0335m}
Solve for the centroidal polar moment of inertia for the solid cylinder
J
s
=
2
π
∫
0
c
3
c
3
d
c
=
π
2
c
3
4
=
1.988
∗
10
−
6
m
4
{\displaystyle J_{s}=2\pi \int _{0}^{c_{3}}c^{3}dc={\frac {\pi }{2}}c_{3}^{4}=1.988*10^{-6}m^{4}}
Using
J
s
{\displaystyle \displaystyle J_{s}}
c
3
{\displaystyle \displaystyle c_{3}}
τ
m
a
x
s
{\displaystyle \displaystyle \tau _{max_{s}}}
τ
m
a
x
s
=
T
c
3
J
s
=
87.21
M
P
a
{\displaystyle \displaystyle \tau _{max_{s}}={\frac {Tc_{3}}{J_{s}}}=87.21MPa}
Problem Number 3.4, p.154
Knowing that the internal diameter of the hollow shaft shown d = 1.2 in., determine the maximum shearing stress caused by a torque of magnitude T = 7.5 kip*in. Contents taken from Mechanics of Materials: 6th Edition textbook. Authors: F.P BEER, E.R. JOHNSTON, J.T. DEWOLF AND D.F. MAZUREK ISBN:9780077565664
Given(s):
Internal diameter =
1.2
i
n
{\displaystyle \displaystyle 1.2in}
1.6
i
n
{\displaystyle \displaystyle 1.6in}
τ
m
a
x
=
7.5
k
s
i
{\displaystyle \displaystyle \tau _{max}=7.5ksi}
Use equation 3.9 from the text book to find maximum torque in a hollow shaft:
τ
m
a
x
=
T
c
2
J
{\displaystyle \displaystyle \tau _{max}={\frac {Tc_{2}}{J}}}
Where
T
:
{\displaystyle \displaystyle T:}
J
=
π
2
(
c
2
4
−
c
1
4
)
{\displaystyle \displaystyle J={\frac {\pi }{2}}(c_{2}^{4}-c_{1}^{4})}
c
1
{\displaystyle \displaystyle c_{1}}
=
1.2
2
=
0.6
i
n
{\displaystyle ={\frac {1.2}{2}}=0.6in}
c
2
{\displaystyle \displaystyle c_{2}}
=
1.6
2
=
0.8
i
n
{\displaystyle ={\frac {1.6}{2}}=0.8in}
τ
m
a
x
=
T
c
2
J
=
T
c
2
π
2
(
c
2
4
−
c
1
4
)
{\displaystyle \displaystyle \tau _{max}={\frac {Tc_{2}}{J}}={\frac {Tc_{2}}{{\frac {\pi }{2}}(c_{2}^{4}-c_{1}^{4})}}}
Solving for
T
{\displaystyle \displaystyle T}
T
=
τ
m
a
x
∗
π
2
(
c
2
4
−
c
1
4
)
c
2
{\displaystyle \displaystyle T={\frac {\tau _{max}*{\frac {\pi }{2}}(c_{2}^{4}-c_{1}^{4})}{c_{2}}}}
=
7.5
∗
π
2
(
0.8
4
−
0.6
4
)
0.8
{\displaystyle \displaystyle ={\frac {7.5*{\frac {\pi }{2}}(0.8^{4}-0.6^{4})}{0.8}}}
Unit conversion:
K
s
i
∗
i
n
4
i
n
=
k
i
p
∗
i
n
{\displaystyle \displaystyle {\frac {Ksi*in^{4}}{in}}=kip*in}
T
=
4.12
k
i
p
∗
i
n
{\displaystyle \displaystyle T=4.12kip*in}
Problem Number 3.7, p.155
The solid spindle AB has a diameter ds = 1.5 in. and is made of a steel with an allowable shearing stress of 12 ksi, while sleeve CD is made of a brass with an allowable shearing stress of 7 ksi. Determine the largest torque T that can be applied at A . Figure P3.7 Contents taken from Mechanics of Materials: 6th Edition textbook. Authors: F.P BEER, E.R. JOHNSTON, J.T. DEWOLF AND D.F. MAZUREK ISBN:9780077565664
For solid spindle AB
c
=
0.5
d
s
=
0.5
∗
1.5
=
0.75
i
n
{\displaystyle c=0.5d_{s}=0.5*1.5=0.75in}
J
=
π
2
c
4
=
π
4
(
0.75
)
4
=
0.49701
i
n
4
{\displaystyle J={\frac {\pi }{2}}c^{4}={\frac {\pi }{4}}(0.75)^{4}=0.49701in^{4}}
τ
m
a
x
=
T
c
J
{\displaystyle \tau _{max}={\frac {Tc}{J}}}
T
A
B
=
J
∗
T
d
C
=
(
0.49701
i
n
4
)
(
12
k
s
i
)
0.75
i
n
=
7.952
k
i
p
−
i
n
{\displaystyle T_{AB}={\frac {J*T_{d}}{C}}={\frac {(0.49701in^{4})(12ksi)}{0.75in}}=7.952kip-in}
For sleeve CD
c
2
=
(
0.5
)
d
2
=
(
0.5
)
(
3
)
=
1.5
i
n
{\displaystyle c_{2}=(0.5)d_{2}=(0.5)(3)=1.5in}
c
1
=
c
2
−
t
⟹
1.5
i
n
−
0.25
i
n
=
1.25
i
n
{\displaystyle c_{1}=c_{2}-t\implies 1.5in-0.25in=1.25in}
J
=
π
2
(
c
2
4
−
c
1
4
)
=
π
2
(
1.5
4
−
1.25
4
)
=
4.1172
i
n
4
{\displaystyle J={\frac {\pi }{2}}(c_{2}^{4}-c_{1}^{4})={\frac {\pi }{2}}(1.5^{4}-1.25^{4})=4.1172in^{4}}
T
C
D
⟹
J
T
d
C
d
⟹
(
4.1172
i
n
4
)
(
7
k
s
i
)
1.5
i
n
=
19.213
k
i
p
−
i
n
{\displaystyle T_{CD}\implies {\frac {JT_{d}}{C_{d}}}\implies {\frac {(4.1172in^{4})(7ksi)}{1.5in}}=19.213kip-in}
Problem Number 3.9, p.155
The torques shown are exerted on pulleys A and B . Knowing that both shafts are solid, determine the maximum shearing stress in (a ) in shaft AB , (b ) in shaft BC . Contents taken from Mechanics of Materials: 6th Edition textbook. Authors: F.P BEER, E.R. JOHNSTON, J.T. DEWOLF AND D.F. MAZUREK ISBN:9780077565664
We begin this problem by drawing a free body diagram of the system.
Using the right hand rule we can figure out the direction of the torque on the elements.
By analyzing disk A:
T
A
=
T
A
B
=
300
N
∗
m
{\displaystyle \displaystyle T_{A}=T_{AB}=300N*m}
and from disk B
T
A
B
+
T
B
=
T
B
C
=
700
N
∗
m
{\displaystyle \displaystyle T_{AB}+T_{B}=T_{BC}=700N*m}
Now that all the torques have been calculated we must calculate the maximum sheer stress in each of the shafts.
Using equation 3.9 from the text book we see that
τ
m
a
x
=
T
c
J
{\displaystyle \displaystyle \tau _{max}={\frac {Tc}{J}}}
Where T = torque, c = the radius of the cross section, J = centroidal polar moment of inertia (
1
2
π
c
4
{\displaystyle {\frac {1}{2}}\pi c^{4}}
By analyzing shaft AB we find:
τ
(
m
a
x
)
A
B
=
T
A
B
c
A
B
J
A
B
=
2
T
A
B
π
c
A
B
3
=
56.6
M
P
a
{\displaystyle \displaystyle \tau _{(max)AB}={\frac {T_{AB}c_{AB}}{J_{AB}}}={\frac {2T_{AB}}{\pi c_{AB}^{3}}}=56.6MPa}
By analyzing shaft BC we find:
τ
(
m
a
x
)
B
C
=
T
B
C
c
B
C
J
B
C
=
2
T
B
C
π
c
B
C
3
=
36.6
M
P
a
{\displaystyle \displaystyle \tau _{(max)BC}={\frac {T_{BC}c_{BC}}{J_{BC}}}={\frac {2T_{BC}}{\pi c_{BC}^{3}}}=36.6MPa}
Problem Number 3.17, p.156
The allowable stress is 50 MPa in the brass rod AB and 25 MPa in the aluminum rod BC . Knowing that a torque of magnitude T = 1250 N*m is applied at A , determine the required diameter of (a ) rod AB , (b ) rod BC . Contents taken from Mechanics of Materials: 6th Edition textbook. Authors: F.P BEER, E.R. JOHNSTON, J.T. DEWOLF AND D.F. MAZUREK ISBN:9780077565664
Egm3520.s13.team1.scheppegrell.jas (discuss • contribs ) 16:58, 22 February 2013 (UTC)
Known Values:
τ
A
B
m
a
x
=
50
M
P
a
{\displaystyle \tau _{ABmax}=50MPa}
τ
B
C
m
a
x
=
25
M
P
a
{\displaystyle \tau _{BCmax}=25MPa}
T
A
=
1250
N
m
{\displaystyle T_{A}=1250Nm}
T
A
=
T
A
B
=
T
B
C
{\displaystyle T_{A}=T_{AB}=T_{BC}}
Formulas:
τ
m
a
x
=
T
C
J
{\displaystyle \tau _{max}={\frac {TC}{J}}}
J
=
π
2
C
4
⟹
C
=
(
2
T
π
τ
m
a
x
)
1
3
{\displaystyle J={\frac {\pi }{2}}C^{4}\implies C=({\frac {2T}{\pi \tau _{max}}})^{\frac {1}{3}}}
(a)
Maximum Allowable Shear Stress in the Rod:
τ
A
B
m
a
x
=
50
M
P
a
=
50000000
N
m
2
{\displaystyle \tau _{ABmax}=50MPa=50000000{\frac {N}{m^{2}}}}
Torque Applied to the Rod:
T
A
B
=
T
A
=
1250
N
m
{\displaystyle T_{AB}=T_{A}=1250Nm}
Radius of the Rod AB:
C
A
B
=
(
2
T
A
B
π
τ
A
B
m
a
x
)
1
3
{\displaystyle C_{AB}=({\frac {2T_{AB}}{\pi \tau _{ABmax}}})^{\frac {1}{3}}}
Substitute Known Values into the Radius Equation:
C
A
B
=
(
2
(
1250
N
m
)
π
(
50000000
N
m
2
)
)
1
3
{\displaystyle C_{AB}=({\frac {2(1250Nm)}{\pi (50000000{\frac {N}{m^{2}}})}})^{\frac {1}{3}}}
Simplify and Solve for Radius:
C
A
B
=
(
1
m
3
20000
π
)
1
3
=
.025
m
{\displaystyle C_{AB}=({\frac {1m^{3}}{20000\pi }})^{\frac {1}{3}}=~.025m}
Find Diameter of AB:
D
A
B
=
2
C
A
B
⟹
D
A
B
=
.050
m
{\displaystyle D_{AB}=2C_{AB}\implies D_{AB}=~.050m}
τ
B
C
m
a
x
=
25
M
P
a
=
25000000
N
m
2
{\displaystyle \tau _{BCmax}=25MPa=25000000{\frac {N}{m^{2}}}}
Torque Applied to the Rod:
T
B
C
=
T
A
=
1250
N
m
{\displaystyle T_{BC}=T_{A}=1250Nm}
Radius of the Rod BC:
C
B
C
=
(
2
T
B
C
π
τ
B
C
m
a
x
)
1
3
{\displaystyle C_{BC}=({\frac {2T_{BC}}{\pi \tau _{BCmax}}})^{\frac {1}{3}}}
Substitute Known Values into the Radius Equation:
C
B
C
=
(
2
(
1250
N
m
)
π
(
25000000
N
m
2
)
)
1
3
{\displaystyle C_{BC}=({\frac {2(1250Nm)}{\pi (25000000{\frac {N}{m^{2}}})}})^{\frac {1}{3}}}
Simplify and Solve for Radius:
C
B
C
=
(
1
m
3
10000
π
)
1
3
=
.032
m
{\displaystyle C_{BC}=({\frac {1m^{3}}{10000\pi }})^{\frac {1}{3}}=~.032m}
Find Diameter of BC:
D
B
C
=
2
C
B
C
⟹
D
B
C
=
.063
m
{\displaystyle D_{BC}=2C_{BC}\implies D_{BC}=~.063m}
The solid spindle AB is made of a steel with an allowable shearing stress of 12 ksi, and sleeve CD is made of brass with an allowable shearing stress of 7 ksi. Determine (a) the largest torque T that can be applied at A if the allowable shearing stress is not to be exceeded in sleeve CD, (b) the corresponding required value of the diameter d, of spindle AB.
Contents taken from Mechanics of Materials: 6th Edition textbook. Authors: F.P BEER, E.R. JOHNSTON, J.T. DEWOLF AND D.F. MAZUREK ISBN:9780077565664
(a ) For the sleeve CD
c
2
=
1
2
∗
d
2
=
.5
∗
3
=
1.5
i
n
{\displaystyle \displaystyle c_{2}={\frac {1}{2}}*d_{2}=.5*3=1.5in}
c
1
=
c
2
−
t
{\displaystyle \displaystyle c_{1}=c_{2}-t}
=>
1.5
i
n
−
0.25
i
n
=
1.25
i
n
{\displaystyle \displaystyle =>1.5in-0.25in=1.25in}
c
2
=
d
2
2
=
.5
∗
3
=
1.5
i
n
{\displaystyle \displaystyle c_{2}={\frac {d_{2}}{2}}=.5*3=1.5in}
J
=
π
2
(
(
c
2
)
4
−
(
c
1
)
4
)
{\displaystyle \displaystyle J={\frac {\pi }{2}}((c_{2})^{4}-(c_{1})^{4})}
J
=
π
2
(
(
1.5
)
4
−
(
1.25
)
4
)
{\displaystyle \displaystyle J={\frac {\pi }{2}}((1.5)^{4}-(1.25)^{4})}
J
=
4.1172
i
n
4
{\displaystyle \displaystyle J=4.1172in^{4}}
T
C
D
=
J
T
d
C
2
{\displaystyle \displaystyle T_{C}D={\frac {JT_{d}}{C_{2}}}}
=>
(
4.1172
i
n
4
)
(
7
k
s
i
)
1.5
i
n
=
19.213
k
i
p
−
i
n
{\displaystyle \displaystyle =>{\frac {(4.1172in^{4})(7ksi)}{1.5in}}=19.213kip-in}
b ) At the solid spindle AB
τ
=
T
C
J
=
2
T
π
c
3
{\displaystyle \displaystyle \tau ={\frac {TC}{J}}={\frac {2T}{\pi c^{3}}}}
=>
(
2
T
π
τ
)
1
/
3
=
(
2
∗
19.213
k
i
p
−
i
n
12
k
s
i
∗
π
)
1
/
3
=
1.0064
i
n
{\displaystyle \displaystyle =>({\frac {2T}{\pi \tau }})^{1/3}=({\frac {2*19.213kip-in}{12ksi*\pi }})^{1/3}=1.0064in}
d
s
=
2
c
{\displaystyle \displaystyle d_{s}=2c}
=>
2
∗
1.0064
i
n
=
2.01
i
n
{\displaystyle \displaystyle =>2*1.0064in=2.01in}
d
s
=
2.01
i
n
{\displaystyle \displaystyle d_{s}=2.01in}
A
for the sleeve CD
=>
1.5
i
n
−
0.25
i
n
=
1.25
i
n
{\displaystyle \displaystyle =>1.5in-0.25in=1.25in}
B
at the solid spindle AB
τ
=
T
C
J
=
2
T
(
π
c
3
)
{\displaystyle \displaystyle \tau ={\frac {TC}{J}}={\frac {2T}{(\pi c^{3})}}}
Failed to parse (syntax error): {\displaystyle \displaystyle => \frac{2T}{(\pi \tau)^{1/3}} = (\frac{2*19.213 kip-in}{12 ksi *\pi)^1/3 = 1.0064 in }
d
s
=
2
c
{\displaystyle \displaystyle ds=2c}
Failed to parse (unknown function "\math"): {\displaystyle \displaystyle => 2*1.0064 in = 2.01 in <\math> <math> \displaystyle ds = 2.01 in \sigma }
Problem Number 3.10, p.155
Figure P3.8 In order to reduce the total mass of the assembly of Prob 3.9, a new design is being considered in which the diameter of shaft BC will be smaller. Determine the smallest diameter of shaft BC for which the maximum value of the shearing stress in the assembly will not increase. Contents taken from Mechanics of Materials: 6th Edition textbook. Authors: F.P BEER, E.R. JOHNSTON, J.T. DEWOLF AND D.F. MAZUREK ISBN:9780077565664
We begin this problem by drawing a free body diagram of the system.
Using the right hand rule we can figure out the direction of the torque on the elements.
By analyzing disk A:
T
A
=
T
A
B
=
300
N
∗
m
{\displaystyle \displaystyle T_{A}=T_{AB}=300N*m}
and from disk B
T
A
B
+
T
B
=
T
B
C
=
700
N
∗
m
{\displaystyle \displaystyle T_{AB}+T_{B}=T_{BC}=700N*m}
Now that all the torques have been calculated we must calculate the maximum sheer stress in each of the shafts.
Using equation 3.9 from the text book we see that
τ
m
a
x
=
T
c
J
{\displaystyle \displaystyle \tau _{max}={\frac {Tc}{J}}}
Where T = torque, c = the radius of the cross section, J = centroidal polar moment of inertia (
1
2
π
c
4
{\displaystyle {\frac {1}{2}}\pi c^{4}}
By analyzing shaft AB we find:
τ
(
m
a
x
)
A
B
=
T
A
B
c
A
B
J
A
B
=
2
T
A
B
π
c
A
B
3
=
56.6
M
P
a
{\displaystyle \displaystyle \tau _{(max)AB}={\frac {T_{AB}c_{AB}}{J_{AB}}}={\frac {2T_{AB}}{\pi c_{AB}^{3}}}=56.6MPa}
By analyzing shaft BC we find:
τ
(
m
a
x
)
B
C
=
T
B
C
c
B
C
J
B
C
=
2
T
B
C
π
c
B
C
3
=
36.6
M
P
a
{\displaystyle \displaystyle \tau _{(max)BC}={\frac {T_{BC}c_{BC}}{J_{BC}}}={\frac {2T_{BC}}{\pi c_{BC}^{3}}}=36.6MPa}
By comparing the maximum sheer stress in both AB and BC we see that the torque is greater in AB
No we create the variable d' and solve for it using the known values.
Let 2cunknown = dunknown
τ
m
a
x
=
2
T
B
C
π
c
u
n
k
n
o
w
n
3
{\displaystyle \displaystyle \tau _{max}={\frac {2T_{BC}}{\pi c_{unknown}^{3}}}}
c
u
n
k
n
o
w
n
3
=
2
T
B
C
π
τ
m
a
x
{\displaystyle \displaystyle c_{unknown}^{3}={\frac {2T_{BC}}{\pi \tau _{max}}}}
c
u
n
k
n
o
w
n
=
19.8
m
m
{\displaystyle \displaystyle c_{unknown}=19.8mm}
d
u
n
k
n
o
w
n
=
39.6
m
m
{\displaystyle \displaystyle d_{unknown}=39.6mm}
Therefore the smallest possible diameter for shaft BC is 39.6mm
