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