Ch3 Torsion Lecture
Transcript of Ch3 Torsion Lecture
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Topic 3: TORSION1. Introduction
2. Stresses in Elastic range3. Angle of twist
4. Design transmission shafts
5. Thin-wall hollow shafts
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1. Introduction
- Members subjected to loads along the longitudinal axes
(Axial deformation)
- Members subjected to torsional loads twisting the
members about their longitudinal centroid axesare
considered in this topic.
- Examples: a lug-wrench, power transmission shaft
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- A lug-wrench:
+ A lug-wrench shaft AB, arm CD
+ Apply equal and opposite forces (P) to ends of CD
AB is a torsion member
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Sign convention:
+ Torque T(x): right-hand-rule sense
+ Angle of rotation, (x)
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- Power transmission shaft:
+ Turbine exerts torque T on the shaft
+ The shaft transmits T to the generator+ Generator creates an equal & opposite torque T
Generator
Rotation
Turbine
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2. Stresses in Elastic range
Shear stress in shaft:
How is torque transferred through shaft?An element dA distance from the center of the shaft has
shear force dF.
For equilibrium: = 0 =
Since dF = dA = ()= T
Shear stress distribution?
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Shaft deformations:
Square shaft: wrapped under tension, cross section do not
remain plane.
Circular shaft: Every cross section remains plane and
undistorted.
- Circumferential lines remain in a plane after deformation- Longitudinal lines: Parallel to the axis become helical
- Right angleABC : Shear deformation
- : Torsional deformation
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Consider a circular shaft attached to a fixed support.
- Apply torque T: + Shaft will twist
+ Free end will rotate through angle
- By observation: + T
+ L
Relationship of , T, L?
Shear stress distribution?
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Shearing strain in circular shaft:
Consider a shaft of length L, radius c,
angle of twist .
AA = L= =
, [rad]
is proportional to and
varies linearly with the distance
from the axis of the shaft.
max
=
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Stresses in the elastic range:
We have: =
and max=
=
From Topic 2: = G
G= G
=
As long as the yield strength is not exceeded, varies
linearly to the distance from the axis of the shaft.
min
=
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Recall that ()= T
T =
T =
.
J = : the polar moment of inertia of the crosssection with respect to its center O.
max=
and =
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+ A solid cross section of radius r:
dJ = dA = 2udu
J =
2
= 2
3
J =
4 =
3
+ A tubular shaft of outer radius r0and inner radius ri:
=
4
4
Example 1:
a/ Tmax=? If max= 120 Mpa
b/ min= ?
Solution: T =
min=
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Example 2:
Determine maxin shaft AB, BC.
Solution:
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Axial shear components:
A torque applied to a shaft shearing stresses on the face
perpendicular to the axis.
Equilibrium equal stresses on planes containing the axis.
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Normal stresses due to Torsion:
- Element a: pure shear
- Element b:
Normal stress, shear stress or
combination of both.
- Consider an element at 45to the
shaft axis:
F = 2(maxA0) cos45= maxA0 2
45 =
=
= max
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Torsion failure modes:
When subjected to torsion:
+ Ductile specimen breaks along a
plane of maximum shear.
+ Brittle specimen breaks along planes perpendicular to the
direction in which tension is a maximum.
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- Shaft deformation: + Shear deformation
+ Torsional deformation
- Strain: Shearing strain max=cL
- Stress: max=Tc
J and =
TJ
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3. Angle of twist
max=
max=
In the elastic range: = G
= G
=
- For shaft with cross-section changes:=
Each segment has constant cross section an torque.
- If cross section changes continuously:
d=
=
()
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Gear assembly:
E/B= E- B=
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Example 3:
rA= 2rB Apply T at E
Determine E?
TAD= 2T
A=
=
CC = CC rAA= rBB
B= (rA/rB) A= 2A =4
E= B + E/B=4
+
= 5
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Example 4:
Shaft ABC with d = 60 mm
is supported by 2 journal
bearings.
Shaft EH with d = 80 mm,
is fixed at E and supported
by a bearing at H.
If A/C= 0.04 rad, determine T1and T2?
Solution:
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Example 4:
d = 40 mm.
Determine B/A(rad) =?
Solution:
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Comparison of axial and torsion:
4 D i f T i i h ft
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4. Design of Transmission shaft:
- Select the shaft material (G) and cross section (J) to meet
performance specifications (power - P and speed - ) without
exceeding the allowable and allowable .Step 1: Determine T? From dynamics: P = T= T(2f) T =
SI units: T: torque [N.m]
P: power [N.m/s] [W]: angular velocity (rad/s)
f: frequency of rotation in Hz
US units: f [rpm], 1 rpm =
1
6HzP [hphorsepower]
1 hp = 550 ft.lb/s (6600 in.lb/s) = 746 W
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Step 2: Select a material (set all)
Design requirement: max=
all
Step 3: Find shaft cross section?
If considering the allowable angle of twist: all
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Example 5:
P = 16 hp from motor A machine tool D
Motor A: f = 1260 rpm
all= 8 ksi
Determine dABand dCD?
Solution:
all= 8 ksi = 8 x 103
(lb/in2
)P = 16 hp = 105600 in.lb/s
f = 1260 rpm =16
6= 21 Hz
T = = 156..1 = 800.72 (lb.in)
max=.
alldAB0.798 [in.]
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3
=
5TCD= 1334.53 (lb.in)
max=
.
alldCD0.947 [in.]
Example 6:
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Example 6:
dAB= dCD;
max80 Mpa,
D1.5; G = 70 Gpa
Determine d AB= dCD= ?
Solution:
Design based on allowable stress:
=.
=
max =.
=
max
TAB = rB.F
TCD= rC.F TAB=.
= 2500 (N.m)
dAB= 59.6 mm
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Design based on allowable angle of twist:
D= D/C+ C =.
+ C1.5
C rC= B rBC
dAB= 62.9 mm
Design must use the lager value for d: d = 62.9 mm
St ti ll d t i t h ft
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Statically determinate shaft:
Example 7:
a/ Determine T1, T2: = 0
TCT2= 0 TC+ TBT1= 0
b/ Determine C, B
- Torque-twist behavior: 1= 11 =
1=
1 =
- Geometry of deformation: = + ;
= 1+ =1
St ti ll i d t i t h ft
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Statically indeterminate shaft:
The unknown torques exceeds the number of applicable
equilibrium equations.
- Step 1: Write equil. eqs.
- Step 2: Formulate compatibility eqs.
- Step 3: Use torque-displacement
eqs. to relate and T.
Example 8:
E ilib i
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- Equilibrium eqs:
TA= -T1
T1T2= TB
TC= T2
- Element torque-twist behavior:
1
= 1
1
=
1
=
=
T
2
- Geometry of deformation:
=1+ = 0
Results: T1,
T2are determined
1=
3 =
3
- Determine rotation angle B
=1 =
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1/
a/ Determine expressions of
max1
, max2,
max3
b/ B= ?
2/
Problem 3.53 in Mechanics of Materials, 6thedition, Beer
and Johnston.
5 Thi ll d h ll h ft
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5. Thin-walled hollow shafts
Consider a thin-walled circular hollow shaft with inner
radius r1, outer radius r2, thickness t = r2r1, and median
radius rm= 1(r1+ r2)
The stress distribution for a circular shaft increases linearly
as we move from the center axis to the surface of the
member.
=
=
where J =
(
4 1
4)
Write r1and r2in terms of rm:
r2= rm+ t/2
r1= r
mt/2
4 4
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J =
+
4
4
t2
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The average force per unit length on the member cross-
section is: q = mt
where q is the shear flow
The total torque can be estimate as:
T = shear flow x total length x moment arm
T = (mt)(2rm)(rm)
Therefore: m=