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445.204
Introduction to Mechanics of Materials
(재료역학개론)
Myoung-Gyu Lee, 이명규
Tel. 880-1711; Email: myounglee@snu.ac.kr
TA: Seong-Hwan Choi, 최성환
Lab: Materials Mechanics lab.
Office: 30-521
Email: cgr1986@snu.ac.kr
Homework #3
Page 140-143: #4.2, 4.5, 4.17
(Option) Page 144: Example 4.5
(Option) Page 150: Example 4.8
Page 151-155: #4.27, 4.32, 4.34, 4.37, 4.44, 4.47
Page 158: #4.60
(Option) Page 162: Example 4.10
Due by Mid-night on May 6 (Wed.)
Through ETL !
Chapter 5
Torsion
Outline
• Deformation of a Circular Shaft
• Axial and Transverse Shear Stresses
• Stresses on Inclined Planes
• Angle of Twist
• Statically Indeterminate Shafts
• Design of Circular Shafts
• Stress Concentrations
• (Option) Inelastic Torsion of Circular Shafts
• (Option) Torsion of Noncircular Solid Bars
• Thin-Walled Hollow Members
What we learn ...
• Chapters 3 and 4 introduced the concepts of axial stress and deformation
• In this chapter, angular deflection (or twist) analysis and (shear) stress analysis of circular bars are considered under torsional loads
• Linear strain (= small strain) theory is used = elasticity under torsional loading
Geometry and deformation
Assumptions
• The plane cross sections perpendicular to the axis of the bar remain plane after the application of a torque
• That is, points in a given plane remain in that plane after twisting.
• Furthermore, expansion or contraction of a cross section does not occur, nor does a shortening or lengthening of the bar.
• In othere words, only shear deformation is involved without normal strains
• The material is homogeneous, and under linear isotropic elasticity
The maximum shear strain at the outer radius c is given by,
Shear strain at any arbitrary radius r is given by:
The shear strain in a circular shaft varies linearly with the distance from the axis of the shaft.
Therefore, in the linear elasticity theory, the shear stress also varies linearly from the axis of a circular shaft.
FIGURE 5.3 Distribution of shearing stress: (a) a circular shaft in torsion; (b) cross section of a hollow circular tube in torsion.
Stress by a torque
Unit: In SI units, T - N-m, c – m, J - m4, t - Pa.
In English units, T - in-lb, c – inch, J - inch4, t - psi.
J : Polar moment of inertia.
Jsolid = 0.5pc4
Jhollow = 0.5p(c4 – b4)
Stepped bar with multiple torque loadings
Q: Find Max. shearing stress in the shaft when a shaft consisting of two prismatic
circular parts is in equilibrium under the torques ….
Example of an application
Q: An electric motor exerts a constant torque T. And, coupling connects the two
shafts together by bolts. At two gears, driving torques equal to T_B and T_C.
The shaft is hollow with inner and outer diameter d and D. Coupling has N bolts with
a bolt circle radius of R.
Find (a) max. permissible torque in the shaft with a safety factor=1.4. (b) required bolt
diameter.
Stresses on inclined planes
Stresses on inclined planes
Torsion failures
Example 5.3
Q: Maximum permissible torque if allowable tensile stress of weld is given!!
Deformation of circular memebers under torsion - Angle of twist
Angle of twist
Geometry of deformation – Max. shear stress:
gmax = cf/L
f/L: rate of twist
Equilibrium condition:
tmax = Tc/J
Material behavior – Hooke’s law:
gmax = tmax/G
Angle of twist
Combining the previous equations yield:
f= TL/GJ
“Positive torque produces positive angle of twist”
GJ : Torsional rigidity
Torsional spring stiffness:
c.f., 1/k = Torsional flexibility
Torsion tests
𝜙
𝑇
𝐺𝐽/𝐿
Linear elastic material
Stepped bar with multiple loads
Sign convention
• Apply right-hand screw rule for internal torque and angle of
twist.
Positive
Non-prismatic bars
x
x
T dxd
GJf =
Example 5.4: Hollow shaft with various torques
Statically indeterminate stepped bar example
Design of circular shafts – transmission of power
What are the optimum material and cross-section of shaft?
In SI units:
In English units:
Procedure for shaft design
Case study
Cross section of a gearbox.
From Novagear AG
Stress Concentrations
(Option) Inelastic torsion of circular shaft
From elastic circular shaft T
J
rt =
JT
t
r=or
Elastic
c
ro
plastic
t
r/r0*t
4 3
2 2y
JT c c
c c
t t p pt
= = =
( 3
2
0 02
3
c
u
cT d d
prt r r pt= =
4
3
u
y
T
T=
(
( (
0
0
2
0 0
0
332 3 3 0
00 3
12 4
3 2 6
c
T d d
cd d c
c
p r
p
r
rr t r r
r
rpt pr t r r r t
=
= =
“Yield torque”
r0
“Ultimate torque”
(Option) Inelastic torsion of circular shaft
yg g=
0
yLgr
f=
y
y
Lc
g
f= 0 y
c
fr
f=
3
3
4 11
3 4
y
u yT Tf
f
=
Elastic
c
ro
plastic
“Radius of elastic core”
yf Angle of twist at the onset of yieldingWhere,
Action:Solve Example 5.11
(Option) Torsion of non-circular shafts
Action:Solve Example 5.12
(Optional) Thin-walled hollowed members
Assumptions
- Tube is prismatic (or constant cross section)
- Wall thickness can be variable
- Define middle surface or midline (along this coordinate s is defined)
(Optional) Thin-walled hollowed members
7
• Applying equilibrium into the infinitesimal element
0d dt
t x s t s xds ds
tt t
=
0dt d
t sds ds
tt
=
( 0d
tds
t = tant cons tt =
“Shear flow” is constant
• Moment (torque) equilibrium
( ( ( ( 2 mT t d t d t At t t= = = i r s r s i
2 mT tAt=2 m
T
tAt =
Am: total plane area enclosed by the midline sor, area enclosed by the mean perimeter
(Optional) Thin-walled hollowed members
• Applying conservation of energy law,For the net rotation
0
1 1
2 2
LT tdsdx
G
tf t=
dL
dx
ff =
22
2 2
1 1
4 m
d TT tds tds
dx G G t A
ft
= =
24 m
d T ds
dx GA t
f =
24 m
d T S
dx GA t
f =
Constant thickness
24 m
T L ds
GA tf =
Action:Solve Example 5.13, 5.14
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