Stress Analysis Lecture 3 ME 276 Spring 2017-2018
Transcript of Stress Analysis Lecture 3 ME 276 Spring 2017-2018
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Stress Analysis
Lecture 3
ME 276
Spring 2017-2018
Dr./ Ahmed Mohamed Nagib Elmekawy
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Axial Stress
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Beam under the action of two tensile forces
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Beam under the action of two tensile forces
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Shear Stress
Average Shear Stress
is the average shear stress at the section
V is internal resultant shear force at the section determined
from the equations of equilibrium
A is the area at the section
A
V
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Stress
A
F
A
P
Single Shear
A
F
2
Double Shear
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Stress
Example 1
Determine the average shear stress in the 20-mm-diameter pin at A and the 30-mm
diameter pin at B that support the beam in the attached figure.
kNAAF
kNAAF
kNFFM
yyy
xxx
BBA
200305
45.120
5.705
35.120
5.1202*3065
40
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Stress
The resultant force acting on pin A is
kNAAF yxA 36.21205.72222
MPa
A
F
A
AA 34
204
*2
1000*36.21
2 2
MPa
A
F
B
BB 7.17
304
1000*5.21
2
double shear
single shear
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Stress
Allowable Stress
• An engineer on charge of the design of a structural or mechanical
element must restrict the stress in the material to a level that will be
safe.
• So it becomes necessary to perform the calculations using a safe or
allowable stress.
• To ensure safety, it is necessary to choose an allowable stress that
restrict the applied load to one that is less than the load the member
can fully support.
• One method of specifying the allowable load for the design or
analysis of a member is to use a number called the factor of safety.
allow
failsf
..
allow
failsf
..
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Stress
Example 1
The control arm is subjected to the loading shown in the figure. Determine to the
nearest ¼ in. the required diameter of the steel pin at C if the allowable shear stress
for the steel is allow = 8 ksi.
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Stress
kipF
FM
AB
ABC
3
05*5
353*38*0
kipC
CF
x
xx
1
05
4530
kipC
CF
y
yy
6
05
3530
kipCCF yxC 082.6612222
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Stress
in6956.0
in6956.03801.0*44
4
in3801.082
082.6
2
2
2
d
AddA
AAA
Fallow
CC
Use a pin having a diameter of
in75.04
3d
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Torsion Stress
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Torsion Stress
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Torsion Stress
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Torsion Stress
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Torsional Deformation of a Circular Shaft
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Angle of Twist
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The Torsion Formula
If the material is linear-elastic, then Hook’s law
applies, = G, and consequently a linear variation
in shear strain, leads to a linear variation in shear
stress along any radial line on the cross section.
J
Tcand
J
T max
Where
T the resultant internal torque acting at the cross section.
J the polar moment of inertia of the cross-sectional area.
c the outer radius of the shaft.
max the maximum shear stress in the shaft, which occurs at the outer surface.
the radius
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The Torsion Formula
4
4
32
2
d
cJ
44
44
32
2
io
io
dd
ccJ
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Shearing Strain
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Torque stress summary
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Shear Stress
𝜏 = 𝐺𝛾= 𝑇𝑟
𝐽=
16𝑇
𝜋 𝑑3
𝜏 : Shear Stress 𝑟 : shaft radius𝐺: Modulus of rigidity 𝑑 : Shaft diameter𝛾: Shear Strain𝐽 : Polar moment of inertiaFor a hollow circular shaft of inner radius c1 and outer radius c2,the polar moment of inertia is
𝐽 =𝜋
32𝑑𝑜4 − 𝑑𝑖
4
∅: Angle of Twist
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Torsion of a Shaft with Circular Cross-SectionFinite Element Analysis
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Torsion of a Shaft with Circular Cross-SectionFinite Element Analysis
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Torsion of a Beam with the Square Cross-SectionFinite Element Analysis
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Torsion of a Beam with the Square Cross-SectionFinite Element Analysis
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The Torsion Formula
Example 1
The shaft shown in the attached figure 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. The shaft diameter is 75 mm.
4.25 kN.m
3 kN.m
1.25 kN.m
Torque diagram
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The Torsion Formula
MPa89.1
75032
2/750*1000*1000*25.1
4
J
TrA
MPa337.0
75032
2/150*1000*1000*25.1
4
J
TrB
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The Torsion Formula
Example 2
The pipe shown in the attached figure has an inner diameter of 80 mm and an outer
diameter of 100 mm. If its end is tightened against the support at A using a torque
wrench at B, determine the shear stress developed in the material at the inner and
outer walls along the central portion of the pipe when the 80-N forces are applied
to the wrench.
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The Torsion Formula
N.m40
02.0*803.0*80;0
T
TM y
MPa345.0
40502
50*00040
44
J
Tcoo
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The Power Transmission
MPa276.0
40502
40*00040
44
J
Tcii
The Power Transmission
Shafts and tubes having circular cross sections
are often used to transmit power developed by
a machine. When used for this purpose, they
are subjected to a torque that depends on the
power generated by the machine and the
angular speed of the shaft.
TP
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Example 1
A solid steel shaft AB shown in the figure is to be used to transmit 3750 W from
the motor M to which it is attached. If the shaft rotates at = 175 rpm and the steel
has an allowable shear stress of allow = 100 MPa determine the required diameter
of the shaft to the nearest mm.
N.mm8.204627
60
175*2
1000*3750
T
TP
allowd
T
d
dT
J
Tc
3
4max
16
32
2
The Power Transmission
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The Angle of Twist
mm22mm843.211008.20462716
3 dd
dAngle of Twist
Occasionally the design of a shaft depends on restricting the amount of rotation or
twist that may occur when the shaft is subjected to a torque. Furthermore, being able
to compute the angle of twist for a shaft is important when analyzing the reactions on
statically indeterminate shafts.
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The Angle of Twist
J
Tand
Gdxd
;
L
dxGxJ
xTdx
GxJ
xTd
0
Constant Torque and Cross-Sectional Area
JG
TL
The similarities between the above equations
and those for an axially loaded members
should be noted.
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The Angle of Twist
The equation of angle of twist is often
used to determine the shear modulus of
elasticity G of a material. To do so, a
specimen of known length and diameter is
placed in a testing machine like shown in
the attached figure. The applied torque T
and angle of twist are then measured
along the length L.
Multiple Torques
JG
TL
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The Angle of Twist
Sign Convention
JG
L
JG
L
JG
L
JG
TL CDBCABDA
000100007000080/
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Example 1
The gears attached to the fixed-end steel shaft are subjected to the torques shown in
the figure. If the shear modulus of elasticity is 80 GPa and the shaft has a diameter
of 14 mm, determine the displacement of tooth P on gear A. The shaft turns freely
within the bearing at B.
The Angle of Twist
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The Angle of Twist
rad2121.0
500*000170300*000130400*000150
0008072
1
4
JG
TLA
Since the answer is negative, by the right–
hand rule the thumb is directed toward the
end E of the shaft and therefore gear A will
rotate as shown in the attached figure.
The displacement of tooth P on gear A is
mm2.21100*2121.0 rs AP