editionfourthmechanics of materials beer • johnston

MECHANICS OF MATERIALS

FourthEdition

Beer • Johnston • DeWolf

Sample Problem 3.4• Find the T0 for the maximum

allowable torque on each shaft –choose the smallest.

• Find the corresponding angle of twist for each shaft and the net angular rotation of end A.

choose the smallest.

( )( )in24inlb561 ⋅ABLT

( )

( )( )( ) ( )

o

642

/

2.22rad387.0

psi102.11in.375.0.in24in.lb561

==

×==

AB

ABBA GJ

LTφπ

( )( )

ilb663

in.375.0in.375.08000 4

2

0max ==

T

TpsiJ

cT

AB

ABπ

τ ( )( )( ) ( )64

2/

psi102.11in.5.0.in24in.lb5618.2

×

⋅==

CD

CDDC GJ

LTφπ

( )( )in.5.0

in.5.08.28000

in.lb663

42

0max

0

==

⋅=

TpsiJ

cT

T

CD

CDπ

τ ( ) oo

o

26.895.28.28.2

95.2rad514.0

===

==

CB φφ

© 2006 The McGraw-Hill Companies, Inc. All rights reserved. 3 - 1

( )

in.lb5610

2

⋅=T

CD

inlb5610 ⋅=Too

/ 2.2226.8 +=+= BABA φφφ o48.10=Aφ


FourthEdition


Design of Transmission Shafts

• Principal transmission shaft performance specifications are:

• Determine torque applied to shaft at specified power and speed,performance specifications are:

- power- speed

specified power and speed,

PPT

fTTP πω 2

==

==

fT

πω 2==

• Find shaft cross-section which will not d th i ll bl

• Designer must select shaft material and cross-section to

exceed the maximum allowable shearing stress,

τ Tc

meet performance specifications without exceeding allowable shearing stress.

( )shafts solid2

3

max

τπ

τ

TccJ

J

==

=g

( ) ( )shafts hollow2

2

max

41

42

22

max

τπ

τ

Tcccc

J

c

=−=



FourthEdition


Stress Concentrations

• The derivation of the torsion formula,Tc

=τ

assumed a circular shaft with uniform cross-section loaded through rigid end

J=maxτ

cross section loaded through rigid end plates.

• The use of flange couplings, gears and pulleys attached to shafts by keys in keyways, and cross-section discontinuities can cause stress concentrations

• Experimental or numerically determined concentration factors are applied as

JTcK=maxτ



FourthEdition


Plastic Deformations• With the assumption of a linearly elastic material,

Tc=maxτ

Jmax

• If the yield strength is exceeded or the material has a nonlinear shearing-stress-strain curve, this

• Shearing strain varies linearly regardless of material

a nonlinear shearing stress strain curve, this expression does not hold.

Th i t l f th t f th i t l t

properties. Application of shearing-stress-strain curve allows determination of stress distribution.

• The integral of the moments from the internal stress distribution is equal to the torque on the shaft at the section,

( ) ∫=∫=cc

ddT0

2

022 ρτρπρπρρτ



FourthEdition


Elastoplastic Materials• At the maximum elastic torque,

YYY cJT τπτ 321==

L YY

γφ =

• As the torque is increased, a plastic region ( ) develops around an elastic core ( )ττ = Yτ

ρτ =

YYY c 2 cYφ

( ) develops around an elastic core ( )Yττ = YYτ

ρτ =

⎞⎛⎞⎛ 33φγρ YL

Y =

⎟⎟⎠

⎞⎜⎜⎝

⎛−=⎟

⎟⎠

⎞⎜⎜⎝

⎛−= 3

3

41

34

3

3

413

32 11

cT

ccT Y

YY

Yρρτπ

⎞⎛ 3φ

• As the torque approaches a limiting value0→

⎟⎟⎠

⎞⎜⎜⎝

⎛−= 3

3

41

34 1

φφY

YTT

• As , the torque approaches a limiting value,0→Yρ

torque plastic TT YP == 34



FourthEdition


Residual Stresses

• Plastic region develops in a shaft when subjected to a large enough torque.

• When the torque is removed, the reduction of stress and strain at each point takes place along a straight line to a generall non ero resid al stress

• On a T-φ curve, the shaft unloads along a straight line to an angle greater than zero.

to a generally non-zero residual stress.

to a a g e g eate t a e o.

• Residual stresses found from principle of superposition

( ) 0∫ dA


( ) 0=∫ dAτρ

JTc

m =′τ


FourthEdition


Example 3.08/3.09

SOLUTION:

S l E (3 32) f / d l• Solve Eq. (3.32) for ρY/c and evaluate the elastic core radius

• Solve Eq (3 36) for the angle of twist

A solid circular shaft is subjected to a torque at each endmkN6.4 ⋅=T

• Evaluate Eq. (3.16) for the angle which the shaft untwists when the

• Solve Eq. (3.36) for the angle of twist

torque at each end. Assuming that the shaft is made of an elastoplastic material with and determine (a) the

MPa150=YτGPa77=G

mkN6.4T which the shaft untwists when the torque is removed. The permanent twist is the difference between the angles of twist and untwistand determine (a) the

radius of the elastic core, (b) the angle of twist of the shaft. When the t i d d t i ( ) th

GPa77=G

• Find the residual stress distribution by a superposition of the stress due to

angles of twist and untwist

torque is removed, determine (c) the permanent twist, (d) the distribution of residual stresses.

p ptwisting and untwisting the shaft



FourthEdition


Example 3.08/3.09SOLUTION:• Solve Eq. (3.32) for ρY/c and

evaluate the elastic core radius

• Solve Eq. (3.36) for the angle of twist

φφevaluate the elastic core radius3

1

341 3

3

41

34

⎟⎟⎠

⎞⎜⎜⎝

⎛−=⇒⎟

⎟⎠

⎞⎜⎜⎝

⎛−=

Y

YYY T

Tcc

TT ρρ

( )( )3 m21mN10683 ⋅×

=⇒=ρφφρ

φφ

Y

Y

YY

Y

LT

cc

⎠⎝

( )m10614

m1025

49

3214

21

×=

×==

−

−cJ ππ( )( )

( )( )3

49-

rad104.93

Pa1077m10614m2.1mN1068.3

×=

×××

==

−φ

φ

Y

YY JG

LT

( )( )

m10614

=⇒=

×

YY

YY c

JTJ

cT ττo3

38.50rad103.148

630.0rad104.93

=×=×

= −−

φ

( )( )

mkN683m1025

m10614Pa101503

496

×

××= −

−

YTo50.8=φ

mkN68.3 ⋅=

630.068.36.434

31

=⎟⎠⎞

⎜⎝⎛ −=

cYρ


⎠⎝

mm8.15=Yρ


FourthEdition


Example 3.08/3.09

• Evaluate Eq. (3.16) for the angle which the shaft untwists when

• Find the residual stress distribution by a superposition of the stress due to

the torque is removed. The permanent twist is the difference between the angles of twist and

p ptwisting and untwisting the shaft

( )( )m1025mN106.4 33 ×⋅×′−Tcbetween the angles of twist and

untwist

′φ TL

( )( )

MPa3.187m10614 49-max

=

×==′

Jτ

( )( )( )( )949

3

P107710146m2.1mN106.4 ⋅×

=

=′

−

φJG

( )( )3

949

6.69rad108.116

Pa1077m1014.6

′−=°=×=

××−

φφpφ

o1.81

69.650.8

=

°−°=

φφpφ


o81.1=pφ


FourthEdition


Torsion of Noncircular Members• Previous torsion formulas are valid for

axisymmetric or circular shafts

• Planar cross-sections of noncircular shafts do not remain planar and stress

d t i di t ib ti d t

• For uniform rectangular cross sections

and strain distribution do not vary linearly

GabcTL

abcT

32

21

max == φτ

• For uniform rectangular cross-sections,

• At large values of a/b, the maximum shear stress and angle of twist for other open sections are the same as a rectangular bar.



FourthEdition


Thin-Walled Hollow Shafts• Summing forces in the x-direction on AB,

( ) ( )0 ∆−∆==∑ xtxtF BBAAx ττ

shear stress varies inversely with thickness

flowshear ==== qttt BBAA τττ

shear stress varies inversely with thickness

• Compute the shaft torque from the integral of the moments due to shear stress

( ) ( )qAdAqdMT

dAqpdsqdstpdFpdM

22

2

0

0

===

====

∫∫

τof the moments due to shear stress

tAT

2=

∫∫

τ

A l f t i t (f Ch t 11)

∫=t

dsGA

TL24

φ

• Angle of twist (from Chapter 11)



FourthEdition


Example 3.10Extruded aluminum tubing with a rectangular cross-section has a torque loading of 24 kip-i D i h h i i h fin. Determine the shearing stress in each of the four walls with (a) uniform wall thickness of 0.160 in. and wall thicknesses of (b) 0.120 in. on AB and CD and 0.200 in. on CD and BD.

SOLUTIONSOLUTION:

• Determine the shear flow through the tubing wallstubing walls.

• Find the corresponding shearing stress with each wall thickness .with each wall thickness .



FourthEdition


Example 3.10SOLUTION:

• Determine the shear flow through the

• Find the corresponding shearing stress with each wall thickness.

gtubing walls.

With a uniform wall thickness,

in.160.0in.kip335.1

==tqτ

ksi34.8=τ

With a variable wall thicknessinkip3351

( )( )

( )kip3351in.-kip24

in.986.8in.34.2in.84.3 2

===

==

Tq

A in.120.0in.kip335.1

== ACAB ττ

ksi13.11== BCAB ττ( ) in.

335.1in.986.822 2 ===

Aq

in.200.0in.kip335.1

== CDBD ττ

BCAB


ksi68.6== CDBC ττ

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