Pb-10.1 in sec.10. )
edit
Find the normal and shear stresses (
σ
{\displaystyle \sigma }
τ
{\displaystyle \tau }
θ
{\displaystyle \theta }
σ
m
a
x
{\displaystyle \sigma _{max}}
τ
m
a
x
{\displaystyle \tau _{max}}
θ
=
30
∘
{\displaystyle \theta =30^{\circ }}
θ
=
45
∘
{\displaystyle \theta =45^{\circ }}
(a) Find the normal and shear stresses on the inclined facet in the triangles pictured below.
θ
=
30
∘
{\displaystyle \displaystyle \theta =30^{\circ }}
θ
=
45
∘
{\displaystyle \displaystyle \theta =45^{\circ }}
Problem 10.1
σ
m
a
x
{\displaystyle \displaystyle \sigma _{max}}
τ
m
a
x
{\displaystyle \displaystyle \tau _{max}}
Draw free body diagrams for top triangle.
Free Body Diagram
Using equations of equilibrium, we can obtain expressions for the shear force
F
S
{\displaystyle \displaystyle F_{S}}
F
N
{\displaystyle \displaystyle F_{N}}
∑
F
x
=
0
=
−
τ
m
a
x
A
0
+
F
S
c
o
s
(
θ
)
+
F
N
s
i
n
(
θ
)
{\displaystyle \sum F_{x}=0=-\tau _{max}A_{0}+F_{S}\,cos(\theta )+F_{N}\,sin(\theta )}
(3.1-1)
∑
F
y
=
0
=
−
τ
m
a
x
A
0
t
a
n
(
θ
)
−
F
S
s
i
n
(
θ
)
+
F
N
c
o
s
(
θ
)
{\displaystyle \sum F_{y}=0=-\tau _{max}A_{0}\,tan(\theta )-F_{S}\,sin(\theta )+F_{N}\,cos(\theta )}
(3.1-2)
Equation 3.1-1 can be rearranged as so,
F
S
=
τ
m
a
x
A
0
−
F
N
s
i
n
θ
c
o
s
θ
{\displaystyle F_{S}={\frac {\tau _{max}A_{0}-F_{N}\,sin\theta }{cos\,\theta }}}
(3.1-3)
Which can be substituted back into equation (3.1-2) and solved for
F
N
{\displaystyle \displaystyle F_{N}}
F
N
=
2
τ
m
a
x
A
0
s
i
n
θ
{\displaystyle F_{N}=2\tau _{max}A_{0}\,sin\theta }
(3.1-4)
And substituted back into equation (3.1-3)
F
S
=
τ
m
a
x
A
0
(
2
c
o
s
θ
−
1
c
o
s
θ
)
{\displaystyle F_{S}=\tau _{max}A_{0}(2\,cos\theta -{\frac {1}{cos\theta }})}
(3.1-5)
Now, using these forces we can solve for the normal stress,
σ
{\displaystyle \displaystyle \sigma }
τ
{\displaystyle \displaystyle \tau }
σ
=
F
N
A
{\displaystyle \sigma ={\frac {F_{N}}{A}}}
(3.1-6)
τ
=
F
S
A
{\displaystyle \tau ={\frac {F_{S}}{A}}}
(3.1-7)
Where
A
=
A
0
c
o
s
θ
{\displaystyle \displaystyle A={\frac {A_{0}}{cos\theta }}}
F
N
{\displaystyle \displaystyle F_{N}}
A
{\displaystyle \displaystyle A}
σ
=
2
τ
m
a
x
A
0
s
i
n
θ
A
0
/
c
o
s
θ
=
τ
m
a
x
s
i
n
(
2
θ
)
{\displaystyle \sigma ={\frac {2\tau _{max}A_{0}\,sin\theta }{A_{0}\,/\,cos\theta }}=\tau _{max}\,sin(2\theta )}
(3.1-8)
Substituting
F
S
{\displaystyle \displaystyle F_{S}}
A
{\displaystyle \displaystyle A}
τ
=
τ
m
a
x
A
0
(
2
c
o
s
θ
A
0
/
c
o
s
θ
−
1
/
c
o
s
θ
A
0
/
c
o
s
θ
)
=
τ
m
a
x
c
o
s
(
2
θ
)
{\displaystyle \tau =\tau _{max}A_{0}({\frac {2\,cos\theta }{A_{0}\,/\,cos\theta }}-{\frac {1\,/\,cos\theta }{A_{0}\,/\,cos\theta }})=\tau _{max}\,cos(2\theta )}
(3.1-9)
F
S
=
σ
m
a
x
A
0
c
o
s
θ
{\displaystyle F_{S}=\sigma _{max}A_{0}\,cos\theta }
(3.1-10)
F
N
=
−
σ
m
a
x
A
0
c
o
s
θ
{\displaystyle F_{N}=-\sigma _{max}A_{0}\,cos\theta }
(3.1-11)
And using the definition of stress, the shear stress and normal stress are
τ
=
F
S
A
=
σ
m
a
x
s
i
n
θ
c
o
s
θ
{\displaystyle \tau ={\frac {F_{S}}{A}}=\sigma _{max}\,sin\theta \,cos\theta }
(3.1-12)
σ
=
F
N
A
=
−
σ
m
a
x
c
o
s
2
θ
{\displaystyle \sigma ={\frac {F_{N}}{A}}=-\sigma _{max}\,cos^{2}\theta }
(3.1-13)
σ
{\displaystyle \displaystyle \sigma }
τ
{\displaystyle \displaystyle \tau }
τ
m
a
x
{\displaystyle \displaystyle \tau _{max}}
θ
{\displaystyle \displaystyle \theta }
σ
m
a
x
{\displaystyle \displaystyle \sigma _{max}}
θ
{\displaystyle \theta }
d
y
{\displaystyle dy}
t
{\displaystyle t}
Using equation (3.1-8) and
θ
=
30
∘
{\displaystyle \theta =30^{\circ }}
σ
=
τ
m
a
x
s
i
n
(
60
)
=
.866
τ
m
a
x
{\displaystyle \sigma =\tau _{max}\,sin(60)=.866\,\tau _{max}}
(3.1-14)
τ
=
τ
m
a
x
c
o
s
(
60
)
=
.5
τ
m
a
x
{\displaystyle \tau =\tau _{max}\,cos(60)=.5\,\tau _{max}}
(3.1-15)
When
θ
=
45
∘
{\displaystyle \theta =45^{\circ }}
σ
=
τ
m
a
x
s
i
n
(
90
)
=
τ
m
a
x
{\displaystyle \sigma =\tau _{max}\,sin(90)=\tau _{max}}
(3.1-16)
τ
=
τ
m
a
x
c
o
s
(
90
)
=
0
{\displaystyle \tau =\tau _{max}\,cos(90)=0}
(3.1-17)
Using equation (3.1-12) and (3.1-13) when
θ
=
30
∘
{\displaystyle \theta =30^{\circ }}
σ
=
σ
m
a
x
c
o
s
2
(
30
)
=
.75
σ
m
a
x
{\displaystyle \sigma =\sigma _{max}\,cos^{2}(30)=.75\,\sigma _{max}}
(3.1-18)
τ
=
σ
m
a
x
s
i
n
(
30
)
c
o
s
(
30
)
=
.433
σ
m
a
x
{\displaystyle \tau =\sigma _{max}\,sin(30)\,cos(30)=.433\,\sigma _{max}}
(3.1-19)
When
θ
=
45
∘
{\displaystyle \theta =45^{\circ }}
σ
=
σ
m
a
x
c
o
s
2
(
45
)
=
.5
σ
m
a
x
{\displaystyle \sigma =\sigma _{max}\,cos^{2}(45)=.5\,\sigma _{max}}
(3.1-20)
τ
=
σ
m
a
x
s
i
n
(
45
)
c
o
s
(
45
)
=
.5
σ
m
a
x
{\displaystyle \tau =\sigma _{max}\,sin(45)\,cos(45)=.5\,\sigma _{max}}
(3.1-21)
P3.2, Beer 2012 )
edit
(a) Determine the torque T that causes a maximum shearing stress of 45 MPa in the hollow cylindrical stell shaft shown.
Inner radius of cylinder:
c
1
=
30
m
m
=
0.030
m
{\displaystyle c_{1}=30mm=0.030m}
c
2
=
45
m
m
=
0.045
m
{\displaystyle c_{2}=45mm=0.045m}
edit
Consider a hollow cylindrical shaft having torque T, with inner radius
c
1
{\displaystyle c_{1}}
c
2
{\displaystyle c_{2}}
(
τ
m
a
x
)
{\displaystyle (\tau _{max})}
From the torsion equation,
τ
m
a
x
=
T
c
2
J
{\displaystyle \tau _{max}={\frac {Tc_{2}}{J}}}
(3.2-1)
Calculate polar moment of inertia fro the hollow cylindrical shaft.
c
2
{\displaystyle c_{2}}
c
1
{\displaystyle c_{1}}
J
=
1
2
π
(
0.045
4
−
0.030
4
)
{\displaystyle J={\frac {1}{2}}\pi (0.045^{4}-0.030^{4})}
(3.2-2)
J
=
5.168
∗
10
−
6
m
4
{\displaystyle J=5.168*10^{-6}m^{4}}
(3.2-3)
Calculate the torque for the hollow cylindrical shaft
T
=
τ
m
a
x
J
c
2
{\displaystyle T={\frac {\tau _{max}J}{c_{2}}}}
(3.2-4)
Substitute in values.
T
=
(
45
∗
10
6
)
(
5.168
∗
10
−
6
)
0.045
{\displaystyle T={\frac {(45*10^{6})(5.168*10^{-}6)}{0.045}}}
(3.2-4)
Therefore, the torque is
T
=
5.168
k
N
{\displaystyle \displaystyle T=5.168kN}
(3.2-5)
edit
Consider a solid cylindrical shaft having torque T, with radius c, and polar moment of inertia J.
τ
m
a
x
=
T
c
J
{\displaystyle \tau _{max}={\frac {Tc}{J}}}
(3.2-6)
Calculate polar moment of inertia for the solid cylindrical shaft.
J
=
1
2
π
(
c
)
4
{\displaystyle J={\frac {1}{2}}\pi (c)^{4}}
(3.2-7)
J
=
1
2
π
(
0.045
)
4
{\displaystyle J={\frac {1}{2}}\pi (0.045)^{4}}
(3.2-8)
J
=
6.4412
∗
10
−
6
m
4
{\displaystyle J=6.4412*10^{-}6m^{4}}
(3.2-9)
Insert all values into equations 3.2-6
τ
m
a
x
=
(
5.168
∗
10
3
)
(
0.045
)
6.4412
∗
10
−
6
{\displaystyle \tau _{max}={\frac {(5.168*10^{3})(0.045)}{6.4412*10^{-}6}}}
(3.2-10)
Therefore, maximum shearing stress required is
τ
m
a
x
=
36.1050
M
P
a
{\displaystyle \displaystyle \tau _{max}=36.1050MPa}
(3.2-11)
P3.4, Beer 2012 )
edit
Knowing that the internal diameter of the hollow shaft shown is
d
=
0.9
i
n
{\displaystyle d=0.9in}
T
=
9
k
i
p
∗
i
n
{\displaystyle T=9kip*in}
Torque acting on shaft Internal Diameter
D
1
=
.9
i
n
{\displaystyle D_{1}=.9in}
(3.4-1)
External Diameter
D
2
=
1.6
i
n
{\displaystyle D_{2}=1.6in}
(3.4-2)
Torque
T
=
9
k
i
p
∗
i
n
{\displaystyle T=9kip*in}
(3.4-3)
Inner Radius
C
1
=
0.45
{\displaystyle C_{1}=0.45}
(3.4-4)
Outer Radius
C
2
=
0.8
i
n
{\displaystyle C_{2}=0.8in}
(3.4-5)
The Torsional formula allows the ability ti find the maximum shearing stress:
(
τ
)
=
T
C
J
{\displaystyle (\tau )={\frac {TC}{J}}}
(3.4-6)
In this case
J
{\displaystyle J}
C
1
{\displaystyle C_{1}}
C
2
{\displaystyle C_{2}}
(
τ
)
=
T
C
π
2
[
c
2
4
−
C
1
4
]
{\displaystyle (\tau )={\frac {TC}{{\frac {\pi }{2}}[c_{2}^{4}-C_{1}^{4}]}}}
(3.4-7)
Substituting values into the equation:
(
τ
)
=
9
∗
0.8
π
2
[
0.8
4
−
.45
4
]
{\displaystyle (\tau )={\frac {9*0.8}{{\frac {\pi }{2}}[0.8^{4}-.45^{4}]}}}
(3.4-8)
(
τ
)
=
12.44
k
s
i
{\displaystyle \displaystyle (\tau )=12.44ksi}
(3.4-9)
P3.7, Beer 2012 )
edit
A solid spindle AB made of steel has a diameter of 1.5 in. and an allowable shear stress of 12 ksi. The sleeve CD around it is made of brass and has an allowable shear stress of 7 ksi.
What is the largest torque that can be applied at point A?
Figure for 3.4 From the torsion equation,
τ
A
B
=
T
A
B
c
J
{\displaystyle \tau _{AB}={\frac {T_{AB}c}{J}}}
(3.4-1)
Rearranging Equation 3.4-1 to solve for the the torque,
T
A
B
{\displaystyle T_{AB}}
T
A
B
=
τ
A
B
J
c
{\displaystyle T_{AB}={\frac {\tau _{AB}J}{c}}}
(3.4-2)
Now
T
A
B
{\displaystyle T_{AB}}
Step Two: Calculating c and J
edit
The allowable shearing stress of solid spindle AB is
τ
a
=
12
k
s
i
{\displaystyle \tau _{a}=12ksi}
d
s
=
1.5
i
n
{\displaystyle d_{s}=1.5in}
Free Body Diagram of Solid Spindle AB
Radius c is half the diameter
d
s
{\displaystyle d_{s}}
c
=
1
2
d
s
=
1
2
1.5
i
n
=
0.75
i
n
{\displaystyle c={\frac {1}{2}}d_{s}={\frac {1}{2}}1.5in=0.75in}
The polar moment of AB,
J
{\displaystyle J}
c
{\displaystyle c}
J
=
1
2
π
c
4
=
1
2
π
(
0.75
i
n
.
)
4
=
0.4970
i
n
.
4
{\displaystyle J={\frac {1}{2}}\pi c^{4}={\frac {1}{2}}\pi (0.75in.)^{4}=0.4970in.^{4}}
(3.4-4)
The maximum sheer stress is equal to the torque at the radius in this case
c
2
{\displaystyle c_{2}}
[ 1]
τ
m
a
x
=
T
c
J
{\displaystyle \tau _{max}={\frac {Tc}{J}}}
Solving for
T
A
B
{\displaystyle T_{AB}}
T
A
B
=
J
τ
a
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\tau _{a}}{c}}={\frac {(0.49701in^{4})(12ksi)}{0.75in}}=7.952kip*in}
The allowable shearing stress of the sleeve CD is is
τ
c
=
7
k
s
i
{\displaystyle \tau _{c}=7ksi}
The diameter of the sleeve CD is
d
2
=
3
i
n
{\displaystyle d_{2}=3in}
The thickness of the sleeve CD is
t
=
1
4
i
n
=
0.25
i
n
{\displaystyle t={\frac {1}{4}}in=0.25in}
Free Body Diagram of Sleeve CD
Radius
c
2
{\displaystyle c_{2}}
d
2
{\displaystyle d_{2}}
c
2
=
1
2
d
2
=
1
2
(
3
i
n
)
=
1.5
i
n
{\displaystyle c_{2}={\frac {1}{2}}d_{2}={\frac {1}{2}}(3in)=1.5in}
Radius
c
1
{\displaystyle c_{1}}
t
{\displaystyle t}
c
1
=
c
2
−
t
=
1.5
i
n
−
0.25
i
n
=
1.25
i
n
{\displaystyle c_{1}=c_{2}-t=1.5in-0.25in=1.25in}
The polar moment of inertia is
J
=
1
2
π
(
c
2
4
−
c
1
4
)
=
π
2
(
(
1.5
i
n
)
4
−
(
1.25
i
n
)
4
)
=
4.1172
i
n
4
{\displaystyle J={\frac {1}{2}}\pi (c_{2}^{4}-c_{1}^{4})={\frac {\pi }{2}}((1.5in)^{4}-(1.25in)^{4})=4.1172in^{4}}
The maximum sheer stress is equal to the torque at the radius in this case
c
2
{\displaystyle c_{2}}
τ
m
a
x
=
T
c
2
J
{\displaystyle \tau _{max}={\frac {Tc_{2}}{J}}}
Solving for
T
C
D
{\displaystyle T_{CD}}
T
C
D
=
J
τ
c
c
2
=
(
4.1172
i
n
4
)
(
7
k
s
i
)
1.5
i
n
=
19.213
k
i
p
∗
i
n
{\displaystyle T_{CD}={\frac {J\tau _{c}}{c_{2}}}={\frac {(4.1172in^{4})(7ksi)}{1.5in}}=19.213kip*in}
Allowable value of torque T is the smaller one of the two
Step Three: Substitute given and calculated values and solve for TAB
edit
Substituting
τ
A
B
=
12
k
s
i
{\displaystyle \tau _{AB}=12ksi}
J
=
0.4970
i
n
.
4
{\displaystyle J=0.4970in.^{4}}
c
=
0.75
i
n
.
{\displaystyle c=0.75in.}
T
A
B
=
12
k
s
i
×
0.4970
i
n
.
4
0.75
i
n
.
=
7.95
×
k
i
p
s
i
n
.
{\displaystyle \displaystyle T_{AB}={\frac {12ksi\times 0.4970in.^{4}}{0.75in.}}=7.95\times {kips}{in.}}
P3.9, Beer 2012 )
edit
Figure for 3.5 The torques shown are exerted on pulleys A and B. Knowing that both shafts are solid, determine the maximum shearing stess in (a) in shaft AB, (b) in shaft BC.
Diameter of the shaft AB,
d
A
B
=
30
m
m
{\displaystyle d_{AB}=30mm}
d
B
C
=
46
m
m
{\displaystyle d_{BC}=46mm}
T
A
=
300
N
−
m
{\displaystyle T_{A}=300N-m}
T
B
=
400
N
−
m
{\displaystyle T_{B}=400N-m}
edit
To solve the problem, the radius for shaft AB and BC must be found. To calculate, divide the given diameter's by 2.
c
A
B
=
30
2
=
15
m
m
{\displaystyle c_{AB}={\frac {30}{2}}=15mm}
(3.9-1)
c
B
C
=
46
2
=
23
m
m
{\displaystyle c_{BC}={\frac {46}{2}}=23mm}
(3.9-2)
For shaft AB acting torque is
T
A
B
=
300
N
−
m
{\displaystyle T_{AB}=300N-m}
(
τ
m
a
x
)
A
B
=
T
A
B
c
A
B
J
A
B
{\displaystyle (\tau _{max})_{AB}={\frac {T_{AB}c_{AB}}{J_{AB}}}}
(3.9-3)
Substituting the cross sectional area for J
(
τ
m
a
x
)
A
B
=
T
A
B
∗
c
π
2
c
4
{\displaystyle (\tau _{max})_{AB}={\frac {T_{AB}*c}{{\frac {\pi }{2}}c^{4}}}}
(3.9-4)
Simplifying the equation, we get
(
τ
m
a
x
)
A
B
=
2
T
A
B
π
c
3
A
B
{\displaystyle (\tau _{max})_{AB}={\frac {2T_{AB}}{\pi {c^{3}}_{AB}}}}
(3.9-5)
Now, insert the values
(
τ
m
a
x
)
A
B
=
2
∗
300
π
∗
15
3
∗
10
−
9
{\displaystyle (\tau _{max})_{AB}={\frac {2*300}{\pi *15^{3}*10^{-9}}}}
(3.9-6)
The torque in shaft AB is,
(
τ
m
a
x
)
A
B
=
56.58
M
P
a
{\displaystyle \displaystyle (\tau _{max})_{AB}=56.58MPa}
(3.9-7)
edit
For shaft BC acting torque
T
B
C
=
T
A
+
T
B
{\displaystyle T_{BC}=T_{A}+T_{B}}
(3.9-8)
T
B
C
=
300
+
400
{\displaystyle T_{BC}=300+400}
(3.9-9)
T
B
C
=
700
N
−
m
{\displaystyle T_{BC}=700N-m}
(3.9-9)
Therefore, the maximum sheer stress for shaft BC can be given by
(
τ
m
a
x
)
B
C
=
T
B
C
L
C
B
J
B
C
{\displaystyle (\tau _{max})_{BC}={\frac {T_{BC}L_{CB}}{J_{BC}}}}
(3.9-10)
The equations then simplifies similarly to equation 3.9-5
(
τ
m
a
x
)
B
C
=
2
T
B
C
π
c
3
B
C
{\displaystyle (\tau _{max})_{BC}={\frac {2T_{BC}}{\pi {c^{3}}_{BC}}}}
(3.9-11)
Now, insert the values
(
τ
m
a
x
)
B
C
=
2
∗
700
π
∗
23
3
∗
10
−
9
{\displaystyle (\tau _{max})_{BC}={\frac {2*700}{\pi *23^{3}*10^{-9}}}}
(3.9-11)
Torque in shaft BC is,
(
τ
m
a
x
)
B
C
=
36.62
M
P
a
{\displaystyle \displaystyle (\tau _{max})_{BC}=36.62MPa}
(3.9-7)
P3.17, Beer2012 )
edit
Aluminum and brass rods secured to a wall.
There is a 1250 N*m torque applied at point A. The maximum allowable stress in the brass rod, AB, is 50 MPa and the maximum allowable stress in the aluminum rod, BC, is 25 MPa. Determine the minimum diameter of (a) rod AB and (b) rod BC.
τ
m
a
x
,
b
r
a
s
s
=
50
M
P
a
{\displaystyle \displaystyle \tau _{max,brass}=50\,MPa}
τ
m
a
x
,
a
l
u
m
=
25
M
P
a
{\displaystyle \displaystyle \tau _{max,alum}=25\,MPa}
Using the elastic torsion formula, which relates the shearing stress to the torque and properties of the rod, solve for
c
{\displaystyle \displaystyle c}
τ
m
a
x
=
T
c
J
{\displaystyle \displaystyle \tau _{max}={\frac {Tc}{J}}}
(3.6-1)
Where
J
{\displaystyle \displaystyle J}
J
=
1
2
π
c
4
{\displaystyle \displaystyle J={\frac {1}{2}}\pi c^{4}}
(3.6-2)
Replacing equation (3.6-2) with
J
{\displaystyle \displaystyle J}
τ
m
a
x
=
T
c
1
2
π
c
4
{\displaystyle \displaystyle \tau _{max}={\frac {Tc}{{\frac {1}{2}}\pi c^{4}}}}
(3.6-3)
τ
m
a
x
=
2
T
π
c
3
{\displaystyle \displaystyle \tau _{max}={\frac {2T}{\pi c^{3}}}}
(3.6-4)
Solving for
c
{\displaystyle \displaystyle c}
c
3
=
2
T
τ
m
a
x
π
{\displaystyle \displaystyle c^{3}={\frac {2T}{\tau _{max}\pi }}}
(3.6-5)
c
=
2
T
τ
m
a
x
π
3
{\displaystyle \displaystyle c={\sqrt[{3}]{\frac {2T}{\tau _{max}\pi }}}}
(3.6-6)
Since the diameter is twice the bar's radius,
d
=
2
∗
c
{\displaystyle \displaystyle d=2*c}
(3.6-7)
This can now be used to solve for the diameter of each bar separately
Starting with the brass rod, AB, and using equation (3.6-6)
c
b
r
a
s
s
=
2
(
1250
N
⋅
m
)
(
50
M
P
a
)
π
3
{\displaystyle \displaystyle c_{brass}={\sqrt[{3}]{\frac {2(1250\,N\cdot m)}{(50\,MPa)\pi }}}}
(3.6-8)
d
b
r
a
s
s
=
2
∗
c
b
r
a
s
s
{\displaystyle \displaystyle d_{brass}=2*c_{brass}}
(3.6-9)
d
b
r
a
s
s
=
.0503
m
{\displaystyle \displaystyle d_{brass}=.0503\,m}
(3.6-10)
Now for the aluminum rod,
c
a
l
u
m
=
2
(
1250
N
⋅
m
)
(
25
M
P
a
)
π
3
{\displaystyle \displaystyle c_{alum}={\sqrt[{3}]{\frac {2(1250\,N\cdot m)}{(25\,MPa)\pi }}}}
(3.6-11)
d
a
l
u
m
=
2
∗
c
a
l
u
m
{\displaystyle \displaystyle d_{alum}=2*c_{alum}}
(3.6-12)
d
a
l
u
m
=
.0634
m
{\displaystyle \displaystyle d_{alum}=.0634\,m}
(3.6-13)
P3.8, Beer2012 )
edit
Figure for 3.7 The solid steel spindle AB has an allowable shearing
stress of 12 ksi and the brass sleeve CD has an allowable
shearing stress of 7 ksi.
Determine
Free Body Diagram of Sleeve CD. Credit: diaz .
The torque exerted on the shaft is calculated by
T
=
τ
m
a
x
J
c
{\displaystyle \displaystyle T={\frac {\tau _{max}J}{c}}}
(3.7-1)
where
t
m
a
x
{\displaystyle t_{max}}
J
{\displaystyle J}
c
{\displaystyle c}
τ
m
a
x
{\displaystyle \tau _{max}}
τ
C
D
,
m
a
x
{\displaystyle \tau _{CD,max}}
First, calculate the outer diameter of the sleeve CD, which is identified as
c
o
{\displaystyle c_{o}}
c
o
=
d
s
2
{\displaystyle \displaystyle c_{o}={\frac {d_{s}}{2}}}
(3.7-2)
=
3.0
i
n
2
=
1.5
i
n
{\displaystyle ={\frac {3.0in}{2}}=1.5in}
c
i
{\displaystyle c_{i}}
c
i
=
c
o
−
t
{\displaystyle \displaystyle c_{i}=c_{o}-t}
(3.7-3)
=
1.05
i
n
−
0.25
i
n
=
1.25
i
n
{\displaystyle =1.05in-0.25in=1.25in}
J
{\displaystyle J}
J
=
π
2
[
c
o
4
−
c
i
4
]
{\displaystyle \displaystyle J={\frac {\pi }{2}}[c_{o}^{4}-c_{i}^{4}]}
(3.7-4)
=
π
2
[
1.5
4
−
1.25
4
]
=
4.1172
i
n
4
{\displaystyle ={\frac {\pi }{2}}[1.5^{4}-1.25^{4}]=4.1172in^{4}}
T
=
7
k
s
i
×
4.1172
i
n
4
1.5
i
n
{\displaystyle \displaystyle T={\frac {7ksi\times 4.1172in^{4}}{1.5in}}}
(3.7-1)
T
=
19.213
k
i
p
∗
i
n
{\displaystyle \displaystyle T=19.213\,kip*in}
To calculate the diameter of the solid steel spindle AB simple rearrangements of Eqs. (3.7-1) and (3.7-2) will yield a solution.
T
=
τ
m
a
x
J
c
{\displaystyle \displaystyle T={\frac {\tau _{max}J}{c}}}
(3.7-1)
c
=
τ
m
a
x
J
T
{\displaystyle \displaystyle c={\frac {\tau _{max}J}{T}}}
c
3
=
2
T
π
τ
m
a
x
{\displaystyle \displaystyle c^{3}={\frac {2T}{\pi \tau _{max}}}}
Substituting the values for
T
{\displaystyle T}
τ
m
a
x
{\displaystyle \tau _{max}}
c
=
1.0064
i
n
{\displaystyle c=1.0064in}
Once again, rearranging Eq (3.7-2)will ultimately give the solution wanted, which is the diameter of spindle AB.
c
=
d
s
2
{\displaystyle \displaystyle c={\frac {d_{s}}{2}}}
(3.7-2)
d
s
=
2
c
{\displaystyle \displaystyle d_{s}=2c}
d
s
=
(
2
)
(
1.0064
)
=
2.01
i
n
{\displaystyle \displaystyle d_{s}=(2)(1.0064)=2.01\,in}
P3.10, Beer2012 )
edit
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.
Figure for 3.8 The torques shown are exerted on pulleys A and B. Determine the diameter of BC for which the maximum allowable shearing stress in the system will not increase.
Diameter of the shaft AB,
d
A
B
=
30
m
m
{\displaystyle d_{AB}=30mm}
d
B
C
=
?
{\displaystyle d_{BC}=?}
T
A
=
300
N
−
m
{\displaystyle T_{A}=300N-m}
T
B
=
400
N
−
m
{\displaystyle T_{B}=400N-m}
Step One: Determine maximum shearing stress in shaft AB:
edit
To solve the problem, the radius for shaft AB and BC must be found. To calculate, divide the given diameter's by 2.
c
A
B
=
30
2
=
15
m
m
{\displaystyle c_{AB}={\frac {30}{2}}=15mm}
(3.8-1)
For shaft AB acting torque is
T
A
B
=
300
N
−
m
{\displaystyle T_{AB}=300N-m}
(
τ
m
a
x
)
A
B
=
T
A
B
c
A
B
J
A
B
{\displaystyle (\tau _{max})_{AB}={\frac {T_{AB}c_{AB}}{J_{AB}}}}
(3.8-2)
Substituting the cross sectional area for J
(
τ
m
a
x
)
A
B
=
T
A
B
∗
c
π
2
c
4
{\displaystyle (\tau _{max})_{AB}={\frac {T_{AB}*c}{{\frac {\pi }{2}}c^{4}}}}
(3.8-3)
Simplifying the equation, we get
(
τ
m
a
x
)
A
B
=
2
T
A
B
π
c
3
A
B
{\displaystyle (\tau _{max})_{AB}={\frac {2T_{AB}}{\pi {c^{3}}_{AB}}}}
(3.8-4)
Now, insert the values
(
τ
m
a
x
)
A
B
=
2
∗
300
π
∗
15
3
∗
10
−
9
{\displaystyle (\tau _{max})_{AB}={\frac {2*300}{\pi *15^{3}*10^{-9}}}}
(3.8-5)
The torque in shaft AB is,
(
τ
m
a
x
)
A
B
=
56.58
M
P
a
{\displaystyle (\tau _{max})_{AB}=56.58MPa}
(3.8-6)
Step Two: Determine maximum shearing stress in shaft BC
edit
For shaft BC acting torque
T
B
C
=
T
A
+
T
B
{\displaystyle T_{BC}=T_{A}+T_{B}}
(3.8-7)
T
B
C
=
300
+
400
{\displaystyle T_{BC}=300+400}
(3.8-)
T
B
C
=
700
N
−
m
{\displaystyle T_{BC}=700N-m}
(3.8-9)
Therefore, the maximum sheer stress for shaft BC can be given by
(
τ
m
a
x
)
B
C
=
T
B
C
L
C
B
J
B
C
{\displaystyle (\tau _{max})_{BC}={\frac {T_{BC}L_{CB}}{J_{BC}}}}
(3.8-10)
The equations then simplifies similarly to equation 3.9-5
(
τ
m
a
x
)
B
C
=
2
T
B
C
π
c
3
B
C
{\displaystyle (\tau _{max})_{BC}={\frac {2T_{BC}}{\pi {c^{3}}_{BC}}}}
(3.8-11)
Now, insert the values
(
τ
m
a
x
)
B
C
=
2
∗
700
π
∗
23
3
∗
10
−
9
{\displaystyle (\tau _{max})_{BC}={\frac {2*700}{\pi *23^{3}*10^{-9}}}}
(3.8-11)
(
τ
m
a
x
)
B
C
=
36.62
M
P
a
{\displaystyle (\tau _{max})_{BC}=36.62MPa}
(3.8-12)
Because the shear stress is highest in part AB (56.68MPa is greater than 36.62 MPa), this will be the limiting value used to calculate the minimum allowable diameter for member BC in Step Three.
Step Three: Solve for the minimum allowable diameter of BC
edit
The radius of CB can be determined by manipulating the maximum torque equation as shown,
c
B
C
4
=
2
T
B
C
π
τ
m
a
x
=
2
×
700
N
×
m
π
×
56.68
×
10
6
P
a
=
0.0198
m
{\displaystyle c_{BC}^{4}={\frac {2T_{BC}}{\pi \tau _{max}}}={\frac {2\times 700N\times m}{\pi \times 56.68\times 10^{6}Pa}}=0.0198m}
(3.8-13)
This mean that the minimum diameter is
d
b
c
=
0.0396
m
{\displaystyle \displaystyle d_{bc}=0.0396m}
Beer, F. P., Johnston, E. R., Jr., DeWolf, J. T., & Mazurek, D. F. (2012). Mechanics of materials (6th ed.). New York, NY: McGraw Hill.
↑ Jshminer46 (31 December 2015). "User:Jshminer46/Mom-s13-team4-R3, In: Wikiversity " . San Francisco, California USA: Wikimedia Foundation, Inc. Retrieved 2017-10-09 .
![{\displaystyle (\tau )={\frac {TC}{{\frac {\pi }{2}}[c_{2}^{4}-C_{1}^{4}]}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/0d9d288db5fdf657cebc941aa22f6a003c1312f8)
![{\displaystyle (\tau )={\frac {9*0.8}{{\frac {\pi }{2}}[0.8^{4}-.45^{4}]}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/39ea3ad8d5375b781a32eefa1c8483b503f3a014)
![{\displaystyle \displaystyle c={\sqrt[{3}]{\frac {2T}{\tau _{max}\pi }}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/f4bda915b1486ac076354fb418a8954d5316badb)
![{\displaystyle \displaystyle c_{brass}={\sqrt[{3}]{\frac {2(1250\,N\cdot m)}{(50\,MPa)\pi }}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/75b2005022e2852c6eb8c968d36e02d2bcf3768a)
![{\displaystyle \displaystyle c_{alum}={\sqrt[{3}]{\frac {2(1250\,N\cdot m)}{(25\,MPa)\pi }}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/8bdce87cd70b1935b2b2277eb52eeb16a90fc9e2)
![{\displaystyle \displaystyle J={\frac {\pi }{2}}[c_{o}^{4}-c_{i}^{4}]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/7efb1dc9a9af69f18f17492d7d8253f3828733d9)
![{\displaystyle ={\frac {\pi }{2}}[1.5^{4}-1.25^{4}]=4.1172in^{4}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/209784cd9e2c422727cf327356bc29d0025d7c6e)
