Introduction to Mechanics of Materials (재료역학개론)

445.204

Introduction to Mechanics of Materials

(재료역학개론)

Myoung-Gyu Lee, 이명규

Tel. 880-1711; Email: [email protected]

TA: Seong-Hwan Choi, 최성환 Lab: Materials Mechanics lab.

Office: 30-521

Email: [email protected]

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.

J

= 0.5pc

J

= 0.5p(c

– b

)

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:

g_max = cf/L

f/L: rate of twist Equilibrium condition:

t_max = Tc/J Material behavior – Hooke’s law:

g_max = t_max/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

d T dx f = GJ

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

t = r J

T t

= r

or

Elastic c

r_o plastic

t r/r0*t

4 3

2 2

y

T J c c

c c

t t p^{ } p t

= =   =

( 

2

0 0 2

3

c u

T =  ^p rt r r d d = pt c 4

3

u y

T

T =

( 

( ( 

0

0

2

0 0

0

3 3

2 3 3 0

0 0 3

2 1 4

3 2 6

c

T d d

d d c c

c

p r

p r

r r t r r  r

r

pt p

r t r r  r t

 

=     

 

 

 

=  = 

     

“Yield torque”

r0

“Ultimate torque”

(Option) Inelastic torsion of circular shaft

g g= y

0

Lg y

r ⁼ f

y y

c Lg

= f ^{0 y}

c

r f

= f

3 3

4 1

3 1 4

y

u y

T T f

f

 

=     

 

Elastic c

r_o plastic

“Radius of elastic core”

fy Angle of twist at the onset of yielding Where,

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

d dt 0

t x s t s x

ds ds

t ^t t ^{ }

         =

dt d 0

t s

ds ds t t

   =

 

 

( 

d t

ds t ⁼ t_t = _{cons tant}

“Shear flow” is constant

• Moment (torque) equilibrium

(  (  ( (



Ti =  r tt ds = tt  rds = tt A i 2 _m

T = ttA

2 _m T t = tA

A_m: total plane area enclosed by the midline s or, 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

T L tdsdx

G f _{=  } t t

d L dx f = f

2 2

2 2

1 1

4 _m

d T

T tds tds

dx G G t A

f t

  =  = 

 

 

4 _m2

d T ds

dx GA t

 f  = 

 

  4 _m²

d T S

dx GA t

 f  =

 

 

Constant thickness

4 _m2

T L ds

GA t

f = 

Action:

Solve Example 5.13, 5.14