strength of material chapter 5
TRANSCRIPT
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PPB 25403 Strength of
Materials
Lecture 5: Torsion
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Learning Outcomes
Torsion twisting of an object due to an applied torque
Torsional Deformation of Circular Shaft
Torsion Formula
Angle of Twist
Transmission of Power
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Torsional Deformation of a Circular Shaft
Torque is a moment that twists a member about its
longitudinal axis.
If the angle of rotation is small, the length of the shaft
and its radius will remain unchanged.
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The Torsion Formula
When material is linear-elastic, Hooke’s law applies.
A linear variation in shear strain leads to a
corresponding linear variation in shear stress
along any radial line on the cross section.
J
Tp
J
Tcor max
= maximum shear stress in the shaft
= shear stress
= resultant internal torque
= polar moment of inertia of cross-sectional area
= outer radius of the shaft
= intermediate distance
max
TJc
p
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The Torsion Formula
If the shaft has a solid circular cross section,
If a shaft has a tubular cross section,
4
2cJ
44
2io ccJ
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Example 5.2The solid shaft of radius c is subjected to a torque T. Find the fraction of T that is
resisted by the material contained within the outer region of the shaft, which has an
inner radius of c/2 and outer radius c.
Solution:
dcdAdT 2' max
maxc
For the entire lighter-shaded area the torque is
(1) 32
152' 3
max
2/
3max cdc
T
c
c
Stress in the shaft varies linearly, thus
The torque on the ring (area) located within
the lighter-shaded region is
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Using the torsion formula to determine the maximum stress in the shaft, we have
Solution:
(Ans) 16
15' TT
3max
4max
2
2
c
T
c
Tc
J
Tc
Substituting this into Eq. 1 yields
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Example 5.3The shaft is supported by two bearings and is subjected to three torques. Determine
the shear stress developed at points A and B, located at section a–a of the shaft.
Solution:From the free-body diagram of the left segment,
mm 1097.4752
74J
kNmm 1250030004250 ;0 TTM x
The polar moment of inertia for the shaft is
Since point A is at ρ = c = 75 mm,
(Ans) MPa 89.11097.4
7512507J
TcB
Likewise for point B, at ρ =15 mm, we have
(Ans) MPa 377.01097.4
1512507J
TcB
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Power Transmission
Power is defined as the work performed per unit of
time.
For a rotating shaft with a torque, the power is
Since , the power equation is
For shaft design, the design or geometric parameter
is
dtdTP / locity,angular veshaft where
f2rad 2cycle 1
fTP 2
allow
T
c
J
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Example 5.5A solid steel shaft AB is to be used to transmit 3750 W from the motor M to which it
is attached. If the shaft rotates at w =175 rpm and the steel has an allowable shear
stress of allow τallow =100 MPa, determine the required diameter of the shaft to the
nearest mm.
Solution:The torque on the shaft is
Nm 6.20460
21753750 TT
TP
Since
mm 92.10100
10006.20422
2
3/13/1
4
allow
allow
Tc
T
c
c
c
J
As 2c = 21.84 mm, select a shaft having a diameter of 22 mm.
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Angle of Twist
Integrating over the entire length L of the shaft, we have
Assume material is homogeneous, G is constant, thus
Sign convention is
determined by right hand rule,
L
GxJ
dxxT
0
Φ = angle of twist
T(x) = internal torque
J(x) = shaft’s polar moment of inertia
G = shear modulus of elasticity for the material
JG
TL
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Example 5.8The two solid steel shafts are coupled together using the meshed gears. Determine
the angle of twist of end A of shaft AB when the torque 45 Nm is applied. Take G to
be 80 GPa. Shaft AB is free to rotate within bearings E and F, whereas shaft DC is
fixed at D. Each shaft has a diameter of 20 mm.
Solution:From free body diagram,
Nm 5.22075.0300
N 30015.0/45
xDT
F
Angle of twist at C is
rad 0269.01080001.02
5.15.2294
JG
TLDCC
Since the gears at the end of the shaft are in mesh,
rad 0134.0075.00269.015.0B
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Solution:
Since the angle of twist of end A with respect to end B of shaft AB caused by the
torque 45 Nm,
rad 0716.01080010.02
24594/
JG
LT ABABBA
The rotation of end A is therefore
(Ans) rad 0850.00716.00134.0/ BABA
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Example 5.10The tapered shaft is made of a material having a shear modulus G. Determine the
angle of twist of its end B when subjected to the torque.
Solution:From free body diagram, the internal torque is T.
L
ccxcc
x
cc
L
cc 122
212
Thus, at x,
4
122
2 L
ccxcxJ
For angle of twist,
(Ans) 3
223
2
3
1
2
121
2
2
0
4
122
cc
cccc
G
TL
L
ccxc
dx
G
TL
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Example 5.11The solid steel shaft has a diameter of 20 mm. If it is subjected to the two torques,
determine the reactions at the fixed supports A and B.
Solution:By inspection of the free-body diagram,
(1) 0500800 ;0 Abx TTM
Since the ends of the shaft are fixed, 0/ BA
Using the sign convention,
(2) 7502.08.1
03.05.15002.0
BA
AAB
TT
JG
T
JG
T
JG
T
Solving Eqs. 1 and 2 yields TA = -345 Nm and TB = 645 Nm.
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Solid Noncircular Shafts
The maximum shear stress and the angle of twist for
solid noncircular shafts are tabulated as below:
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Example 5.13The 6061-T6 aluminum shaft has a cross-sectional area in the shape of an
equilateral triangle. Determine the largest torque T that can be applied to the end of
the shaft if the allowable shear stress is τallow = 56 MPa and the angle of twist at its
end is restricted to Φallow = 0.02 rad. How much torque can be applied to a shaft of
circular cross section made from the same amount of material? Gal = 26 GPa.
Solution:By inspection, the resultant internal torque at any cross section
along the shaft’s axis is also T.
(Ans) Nm 12.24102640
102.146.020 ;
46
Nm 2.177940
2056 ;
20
34
3
4
33
TT
Ga
T
TT
a
T
al
allow
allow
By comparison, the torque is limited due to the angle of twist.
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Solution:
(Ans) Nm 10.33102685.142/
102.102.0 ;
Nm 06.28885.142/
85.1456 ;
34
3
4
TT
JG
TL
TT
J
Tc
al
allow
allow
For circular cross section, we have
mm 85.1460sin40402
1c ; 2 cAA trianglecircle
The limitations of stress and angle of twist then require
Again, the angle of twist limits the applied torque.
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Thin-Walled Tubes Having Closed Cross Sections
Shear flow q is the product of the tube’s thickness and
the average shear stress.
Average shear stress for thin-walled tubes is
For angle of twist,
tq avg
m
avgtA
T
2
τavg = average shear stress
T = resultant internal torque at the cross section
t = thickness of the tube
Am = mean area enclosed boundary
t
ds
GA
TL
m
24
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Example 5.14Calculate the average shear stress in a thin-walled tube having a circular cross
section of mean radius rm and thickness t, which is subjected to a torque T. Also,
what is the relative angle of twist if the tube has a length L?
Solution:The mean area for the tube is
2
mm rA
For angle of twist,
(Ans) 22 2
mm
avgtr
T
tA
T
(Ans) 24 32 Gtr
TL
t
ds
GA
TL
mm
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Example 5.16A square aluminum tube has the dimensions. Determine the average shear stress in
the tube at point A if it is subjected to a torque of 85 Nm. Also compute the angle of
twist due to this loading. Take Gal = 26 GPa.
Solution:By inspection, the internal resultant torque is T = 85 Nm.
(Ans) N/mm 7.12500102
1085
2
23
m
avgtA
T
For average shear stress,
22 mm 250050mAThe shaded area is
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Solution:
dsds
t
ds
GA
TL
m
1-4
32
33
2mm 10196.0
10102625004
105.11085
4
For angle of twist,
Integral represents the length around the centreline boundary of the tube, thus
(Ans) rad 1092.350410196.0 34
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Stress Concentration
Torsional stress concentration factor, K, is used to
simplify complex stress analysis.
The maximum shear stress is then determined from the
equation
J
TcKmax
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Example 5.18The stepped shaft is supported by bearings at A and B. Determine the maximum
stress in the shaft due to the applied torques. The fillet at the junction of each shaft
has a radius of r = 6 mm.
Solution:By inspection, moment equilibrium about the axis
of the shaft is satisfied
15.0202
6 ;2
202
402
d
r
d
D
(Ans) MPa 10.3020.02
020.0303.1
4maxJ
TcK
The stress-concentration factor can be determined
by the graph using the geometry,
Thus, K = 1.3 and maximum shear stress is
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Inelastic Torsion
Considering the shear stress acting on an element of
area dA located a distance p from the center of the
shaft,
Shear–strain distribution over a radial line on a shaft is
always linear.
Perfectly plastic assumes the shaft will continue to twist
with no increase in torque.
It is called plastic torque.
dTA
2
2
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Example 5.20A solid circular shaft has a radius of 20 mm and length of 1.5 m. The material has an
elastic–plastic diagram as shown. Determine the torque needed to twist the shaft Φ
= 0.6 rad.
Solution:The maximum shear strain occurs at the surface of
the shaft,
rad 008.002.0
5.16.0 ; max
maxL
mm 4m 004.0008.0
02.0
0016.0Y
Y
Based on the shear–strain distribution, we have
The radius of the elastic core can be obtained by
(Ans) kNm 25.1004.002.046
10754
6
336
33
YY cT