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Bài giảng Strength of materials - Chương 9: Torsion - Nguyễn Sỹ Lâm

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CHAPTER 9: TORSION  
9.0 Introduction  
9.1 Torsional loads on circular shafts  
9.2 Torsion of noncircular members  
9.3 Helical spring under axial load  
9.1 INTRODUCTION  
A bar subjected to loadings has internal torsion moment only on cross sections: Mz  
Mz > 0 (counter clockwise)  
Eye’s direction  
z
x
y
9.1 INTRODUCTION  
9.1 TORSIONAL LOADS ON CIRCULAR SHAFTS  
• Interested in stresses and strains of  
circular shafts subjected to twisting  
couples or torques  
• Turbine exerts torque T on the shaft  
• Shaft transmits the torque to the  
generator  
• Generator creates an equal and  
opposite torque T’  
9.1 TORSIONAL LOADS ON CIRCULAR SHAFTS  
NET TORQUE DUE TO INTERNAL STRESSES  
• Net of the internal shearing stresses is an  
internal torque, equal and opposite to the  
applied torque,  
T dF   
dA  
• Although the net torque due to the shearing  
stresses is known, the distribution of the stresses  
is not  
• Distribution of shearing stresses is statically  
indeterminate – must consider shaft  
deformations  
• Unlike the normal stress due to axial loads, the  
distribution of shearing stresses due to torsional  
loads can not be assumed uniform.  
9.1 TORSIONAL LOADS ON CIRCULAR SHAFTS  
AXIAL SHEAR COMPONENTS  
• Torque applied to shaft produces shearing  
stresses on the faces perpendicular to the  
axis.  
• Conditions of equilibrium require the  
existence of equal stresses on the faces of the  
two planes containing the axis of the shaft  
• The existence of the axial shear components is  
demonstrated by considering a shaft made up  
of axial slats.  
The slats slide with respect to each other when  
equal and opposite torques are applied to the  
ends of the shaft.  
9.1 TORSIONAL LOADS ON CIRCULAR SHAFTS  
SHAFT DEFORMATIONS  
• From observation, the angle of twist of the  
shaft is proportional to the applied torque and  
to the shaft length.  
  T  
  L  
• When subjected to torsion, every cross-section  
of a circular shaft remains plane and  
undistorted.  
• Cross-sections for hollow and solid circular  
shafts remain plain and undistorted because a  
circular shaft is axisymmetric.  
• Cross-sections of noncircular (non-  
axisymmetric) shafts are distorted when  
subjected to torsion.  
9.1 TORSIONAL LOADS ON CIRCULAR SHAFTS  
SHEARING STRAIN  
• Consider an interior section of the shaft. As a  
torsional load is applied, an element on the  
interior cylinder deforms into a rhombus.  
• Since the ends of the element remain planar,  
the shear strain is equal to angle of twist.  
• It follows that  
  
L or   
L
• Shear strain is proportional to twist and radius  
c  
L
and   
c
max  
max  
9.1 TORSIONAL LOADS ON CIRCULAR SHAFTS  
STRESSES IN ELASTIC RANGE  
• Multiplying the previous equation by the  
shear modulus,  
c
GG  
max  
G  
From Hooke’s Law,  
, so  
c
  
max  
The shearing stress varies linearly with the  
radial position in the section.  
4
1
J c  
2
• Recall that the sum of the moments from  
the internal stress distribution is equal to  
the torque on the shaft at the section,  
2
max  
c
max  
c
T  dA   
dA   
J
• The results are known as the elastic torsion  
formulas,  
4
4
1
J c c  
2
1
2
Tc  
J
T  
J
and   
max  
9.1 TORSIONAL LOADS ON CIRCULAR SHAFTS  
NORMAL STRESSES  
• Elements with faces parallel and perpendicular  
to the shaft axis are subjected to shear stresses  
only. Normal stresses, shearing stresses or a  
combination of both may be found for other  
orientations.  
• Consider an element at 45o to the shaft axis,  
F 2  
A
cos45   
A 2  
A 2  
max 0  
max 0  
F   
max 0  
   
  
max  
o
45  
A
A 2  
0
• Element a is in pure shear.  
• Element c is subjected to a tensile stress on  
two faces and compressive stress on the other  
two.  
• Note that all stresses for elements a and c have  
the same magnitude  
9.1 TORSIONAL LOADS ON CIRCULAR SHAFTS  
TORSIONAL FAILURE MODES  
• Ductile materials generally fail in  
shear. Brittle materials are weaker in  
tension than shear.  
• When subjected to torsion, a ductile  
specimen breaks along a plane of  
maximum shear, i.e., a plane  
perpendicular to the shaft axis.  
• When subjected to torsion, a brittle  
specimen breaks along planes  
perpendicular to the direction in  
which tension is a maximum, i.e.,  
along surfaces at 45o to the shaft  
axis.  
9.1 TORSIONAL LOADS ON CIRCULAR SHAFTS  
EXAMPLES 9.01  
SOLUTION:  
• Cut sections through shafts AB  
and BC and perform static  
equilibrium analysis to find  
torque loadings  
• Apply elastic torsion formulas to  
find minimum and maximum  
stress on shaft BC  
Shaft BC is hollow with inner and outer  
diameters of 90 mm and 120 mm,  
respectively. Shafts AB and CD are solid  
of diameter d. For the loading shown,  
determine (a) the minimum and maximum  
shearing stress in shaft BC, (b) the  
required diameter d of shafts AB and CD  
if the allowable shearing stress in these  
shafts is 65 MPa.  
• Given allowable shearing stress  
and applied torque, invert the  
elastic torsion formula to find the  
required diameter  
9.1 TORSIONAL LOADS ON CIRCULAR SHAFTS  
EXAMPLES 9.01  
SOLUTION:  
• Cut sections through shafts AB and BC  
and perform static equilibrium analysis  
to find torque loadings  
M 0   
6kNm  
T  
M 0   
6kNm  
14kNm T  
x
AB  
x
BC  
T 6kNm T  
T 20kNm  
AB  
CD  
BC  
9.1 TORSIONAL LOADS ON CIRCULAR SHAFTS  
EXAMPLES 9.01  
• Apply elastic torsion formulas to  
find minimum and maximum  
stress on shaft BC  
• Given allowable shearing stress and  
applied torque, invert the elastic torsion  
formula to find the required diameter  
2
2
4
4
4
4
Tc Tc  
6kNm  
J   
c c  
0.060  
0.045  
2
1
65MPa   
max  
4
3
2
2
J
c
c
6 4  
13.9210 m  
3  
c 38.910 m  
T c  
BC 2  
20kNm0.060m  
13.9210 m  
  
max  
2
d 2c 77.8mm  
6 4  
J
86.2MPa  
c
1
45mm  
min  
min  
c
86.2MPa 60mm  
86.2MPa  
max  
2
max  
min  
64.7MPa  
64.7MPa  
min  
9.1 TORSIONAL LOADS ON CIRCULAR SHAFTS  
ANGLE OF TWIST IN ELASTIC RANGE  
• Recall that the angle of twist and maximum  
shearing strain are related,  
c  
L
max  
• In the elastic range, the shearing strain and shear  
are related by Hooke’s Law,  
Tc  
max  
G
max  
JG  
• Equating the expressions for shearing strain and  
solving for the angle of twist,  
TL  
  
JG  
• If the torsional loading or shaft cross-section  
changes along the length, the angle of rotation is  
found as the sum of segment rotations  
T L  
i i  
  
J G  
i
i i  
9.1 TORSIONAL LOADS ON CIRCULAR SHAFTS  
SUMMARY  
M z  
Mz = M0  
  
J
P
M0  
M z  
x
max c  
J
z
max  
exists on the  
perimeter of the  
circular section  
4
1
J c  
2
The relative angle of twist (góc xoắn tỷ  
đối),  
M z  
  
GJ  
The angle of twist (góc xoắn),  
M z L  
  
4
4
1
J c c  
GJ  
2
1
2
9.1 TORSIONAL LOADS ON CIRCULAR SHAFTS  
STATICALLY INDETERMINATE SHAFTS  
• Given the shaft dimensions and the applied  
torque, we would like to find the torque reactions  
at A and B.  
• From a free-body analysis of the shaft,  
T T 90lb ft  
A
B
which is not sufficient to find the end torques.  
The problem is statically indeterminate.  
• Divide the shaft into two components which  
must have compatible deformations,  
T L T L  
L J  
1 2  
L J  
2 1  
A 1  
B 2  
  
0  
T   
T
A
1
2
B
J G J G  
1
2
• Substitute into the original equilibrium equation,  
L J  
1 2  
T   
T 90lb ft  
A
A
L J  
2 1  
9.1 TORSIONAL LOADS ON CIRCULAR SHAFTS  
EXAMPLE 9.02  
SOLUTION:  
• Apply a static equilibrium analysis on  
the two shafts to find a relationship  
between TCD and T0  
• Apply a kinematic analysis to relate  
the angular rotations of the gears  
• Find the maximum allowable torque  
on each shaft – choose the smallest  
Two solid steel shafts are connected  
by gears. Knowing that for each shaft  
G = 11.2 x 106 psi and that the  
allowable shearing stress is 8 ksi,  
determine (a) the largest torque T0  
that may be applied to the end of shaft  
AB, (b) the corresponding angle  
through which end A of shaft AB  
rotates.  
• Find the corresponding angle of twist  
for each shaft and the net angular  
rotation of end A  
9.1 TORSIONAL LOADS ON CIRCULAR SHAFTS  
EXAMPLE 9.02  
SOLUTION:  
• Apply a static equilibrium analysis on  
the two shafts to find a relationship  
between TCD and T0  
• Apply a kinematic analysis to relate  
the angular rotations of the gears  
r r   
B B  
C C  
M 0 F  
0.875in.  
T  
0
B
r
2.45in.  
C
  
C
B
C
M 0 F  
2.45in. T  
r
0.875in.  
C
CD  
B
T 2.8T  
2.8  
CD  
0
B
C
9.1 TORSIONAL LOADS ON CIRCULAR SHAFTS  
EXAMPLE 9.02  
• Find the T0 for the maximum  
• Find the corresponding angle of twist for each  
allowable torque on each shaft – shaft and the net angular rotation of end A  
choose the smallest  
T L  
561lb in.24in.  
   
AB  
A/ B  
4
6
J G  
0.375in.  
11.210 psi  
AB  
2
o
0.387 rad 2.22  
T c  
T
2
0.375in.  
AB  
0
T L  
2.8  
561lb in.24in.  
   
8000psi   
8000psi   
CD  
max  
4
C / D  
J
0.375in.  
AB  
4
6
2
J G  
0.5in.  
11.210 psi  
CD  
T 663lb in.  
0
o
0.514rad 2.95  
T c  
2.8T  
0.5in.  
CD  
0
o
o
o
max  
4
2.82.8  
2.95  
8.26  
J
0.5in.  
B
C
CD  
2
o
o
10.48  
  
8.26 2.22  
T 561lb in.  
T 561lbin  
A
A
B
A/ B  
0
0
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