lecture4 torsion

7/27/2019 Lecture4 Torsion

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2005 Pearson Education South Asia Pte Ltd

5. Torsion

1

CHAPTER OBJECTIVES

Discuss effectsof applyingtorsional loading to a long straight

member

Determine stress distribution

within the member under torsionalload

Determine angle of twist when material behaves

in a linear-elastic and inelastic manner

Discuss statically indeterminate analysis of shafts

and tubes


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5. Torsion

2

CHAPTER OUTLINE

1. Torsional Deformation of a Circular Shaft

2. The Torsion Formula

3. Absolute Maximum Torsional Stress

4. Power Transmission

5. Angle of Twist


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5. Torsion

3

Torsion is a moment that twists/deforms a

member about its longitudinal axis

Effect??Its effect is of primary concern in the

design of axles or drive shafts used in vehicles

and machinery

For example: When torque is applied to a circular

shaft (eg: rubber material), the circles and

longitudinal grid lines originally marked on theshaft tend to distort with circles remain circles

- grid lines deforms a helix

(Fig5.1a-b)

5.1 TORSIONAL DEFORMATION OF A CIRCULAR SHAFT


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5. Torsion

4

Torsion is a moment that twists/deforms a

member about its longitudinal axis

By observation, if angle of rotation is small, length

of shaftand its radius remain unchanged

5.1 TORSIONAL DEFORMATION OF A CIRCULAR SHAFT


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5. Torsion

5

5.1 Torsional Deformation

If the shaft is fixed at one and a torque is applied to its

other end, it will distorted with the cross-section at a

distance x from the fixed end of the shaft will rotate

through an angle (angle of twist).

x

y

zf(x)

x

f(x)


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5. Torsion

6

5.1 Torsional Deformation

HOW THIS

DISTORTIONSTRAINS THE

MATERIAL??


7/522005 Pearson Education South Asia Pte Ltd

5. Torsion

7

Geometry of Deformation

g

r

qo

r

T

Q

R

P

L

Assumption: Cross-sections which are plane

before twisting remain plane

and undistorted after twisting.

PQ Longitudinal

generator

gshear strain at radial

distance r(rad.)

qangle of twist (radians)

grq LQR arcCircular (1)



5. Torsion

8

DEFORMATION OF A CIRCULAR SHAFT

By definition, shear strain is

Let x dx andf= df

BD=rdf= dx g

g= (/2) lim qCA along CA

BA along BA

g =rdf

dx

Sincedf/ dx = g/r= gmax/c

g = gmaxr

c( )Equation 5-2



5. Torsion

9

Stress Strain Relation

G

P.L.t

g

Assumption: Shear stress does not

exceed the proportional

limit.

Gshear modulus (N/m)

Torsion Formulae

gt G

Substituting Eq. (3) into Eq. (2) results in

(2)

L

G

LG

q

r

trqt

or (3)



5. Torsion

10

5.2 THE TORSION FORMULA

For solid shaft, shear

stress varies from zero at

shafts longitudinal axis to

maximum value at its outer

surface. Due to proportionality of

triangles, or using Hookes

law and Eqn 5-2,

t = tmaxrc( ).

.

.

T=tmax

cAr2dA



5. Torsion

11

dAc

T

cwhere

dAdMT

dMT

M

A

A A

A

z

2max

max

0

0

rt

tr

t

rt

5.2 Torsion Formulas

TZ

r

J: Polar Moment of Inertia

Solid Shaft:

Tubular Shaft: ( )44

4

2

2

io ccJ

cJ

c

dM



5. Torsion

12


The integral in the equation can be represented asthe polar moment of inertiaJ, of shafts x-sectionalarea computed about its longitudinal axis

tmax

=Tc

J

tmax= max. shear stress in shaft, at the outer surface

T= resultant internal torque acting at x-section, from

method of sections & equation of momentequilibrium applied about longitudinal axis

J= polar moment of inertia at x-sectional area

c= outer radius of the shaft



5. Torsion

13


Shear stress at intermediate distance,r

t=Tr

J

The above two equations are referred to as the

torsion formula

Used only if shaft is circular, its material

homogenous, and it behaves in an linear-elastic

manner



5. Torsion

14


Solid shaft

Jcan be determined using area element in the formof a differential ring or annulus having thickness drand circumference 2r.

For this ring,dA = 2rdr

J= c4

2

Jis a geometric property of the circular area andis always positive. Common units used for itsmeasurement are mm4and m4.



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15


Tubular shaft

J= (co4ci

4)2



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Absolute maximum torsional stress

Need to find location where ratio Tc/Jis maximum

Draw a torque diagram (internal torque tvs.xalongshaft)

Sign Convention: Tis positive, by right-hand rule, isdirected outward from the shaft

Once internal torque throughout shaft is determined,maximum ratio of Tc/Jcan be identified



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Procedure for analysis

Internal loading

Section shaft perpendicular to its axis at pointwhere shear stress is to be determined

Use free-body diagram and equations ofequilibrium to obtain internal torque at section

Section property

Compute polar moment of inertia and x-sectional

area For solid section,J= c4/2

For tube,J= (co4 ci

4)/2

5 T i



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Shear stress

Specify radial distancer, measured from centre

of x-section to point where shear stress is to be

found Apply torsion formula, t= Tr/Jor tmax= Tc/J

Shear stress acts on x-section in direction that is

always perpendicular tor

5 T i



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19

EXAMPLE 5.3

Shaft shown supported by two bearings and

subjected to three torques.

Determine shear stress developed at pointsAandB,

located at section a-aof the shaft.

5 T i



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20

EXAMPLE 5.3 (SOLN)

Internal torque

Bearing reactions on shaft = 0, if shaft weightassumed to be negligible. Applied torques satisfy

moment equilibrium about shafts axis.

Internal torque at section a-adetermined from free-body diagram of left segment.

5 T i



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EXAMPLE 5.3 (SOLN)

Internal torque

Mx= 0; 4250 kNmm 3000 kNmm T= 0

T= 1250 kNmm

Section property

J= /2(75 mm)4= 4.97107mm4

Shear stress

Since pointAis atr= c= 75 mmtA= Tc/J= ... = 1.89 MPa

5 T i



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EXAMPLE 5.3 (SOLN)

Shear stress

Likewise for pointB,atr= 15 mm

tB= Tr/J= ... = 0.377 MPa

Directions of the stresses

on elementsAandB

established from

direction of resultantinternal torque T.

5 T i



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5.2 Torsion Formulas

EXERCISE F5-1 & F5-2

5 Torsion


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EXERCISE F5-5 & F5-6

5 Torsion


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Problem 1 (5-2)

750Nm

c=100mm

a=75mm

25mm

Torque carried by the shaded section ?

( )MPa4793.0

025.01.02

)1.0(75044

max

t

JTc

( )

Nm515

075.01.02

)10(4793.0 446

max

cT

J

aT

c

a

a

a

a

a

tt

Nm5152

2

1.0

075.0

3max

1.0

075.0 max

rrt

rrr

rtrt

dc

d

c

dATa

(b)

(a)

5 Torsion


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5. Torsion

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Problem 2 (5-56)

300Nm500Nm

200Nm

400Nm

300Nm

500Nm200Nm

400Nm

( )

( )MPa7.75

015.02

015.0400

4

maxmax

tJ

cT

68Nm

20Nm

49Nm35Nm

68Nm

20Nm

49Nm

35Nm

132Nm

( )

( )MPa38.5

025.02

025.0132

4

maxmax

tJ

cT

5 Torsion


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5. Torsion

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Poweris defined as work performed per unit of

time

Instantaneous power is

Since shafts angular velocity = dq/dt, we can

also express power as

5.3 POWER TRANSMISSION

P= T (dq/dt)

P= T

Frequencyfof a shafts rotation is often reported.

It measures the number of cycles per second

and since 1 cycle = 2 radians, and = 2f T, thenpower

P= 2fTEquation 5-11

5 Torsion


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5.3 Power Transmission

fTP

TP

dt

dTP

q

2

Power:

1watts(W)=1Nm/s1hp=550ftlb/s

where = angular velocity(rad/s)

f = frequency (cycle/s)

fgmax

JG

TL

G

Lc

J

Tc

cL

GJ

Tc

f

ffg

gt

/

wheremax

max

max

5 Torsion


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5. Torsion

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Shaft Design If power transmitted by shaft and its frequency of

rotation is known, torque is determined from Eqn

5-11

Knowing Tand allowable shear stress formaterial, tallowand applying torsion formula,

5.3 POWER TRANSMISSION: Shaft Design

J

c

T

tallow= 5-12

5 Torsion


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5. Torsion

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Shaft Design

For solid shaft, substituteJ= (/2)c4to determine c

For tubular shaft, substituteJ= (/2)(co2 ci

2)to

determine coand ci

5.3 POWER TRANSMISSION

5 Torsion


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5. Torsion

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EXAMPLE 5.5

Solid steel shaft shown used to transmit 3750 Wfrom

attached motor M. Shaft rotates at = 175 rpmandthe steel tallow= 100 MPa.

Determine required diameter of shaft to nearest mm.

5 Torsion


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5. Torsion

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EXAMPLE 5.5 (SOLN)

Torque on shaft determined fromP= T,

Thus,P= 3750 Nm/s

Thus,P= T, T= 204.6 Nm

( ) = = 18.33 rad/s175 revmin

2rad

1 rev

1 min

60 s( )

= =J

c

c4

2 c

T

tallow..

.

c= 10.92 mm

Since 2c= 21.84 mm, select shaft with diameter of

d= 22 mm

5-12

5 Torsion


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5. Torsion

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EXERCISE

A steel tube having an outer diameter of 62.5mmis used to transmit 3kW when turning at 27

rev/min.

Determine the inner diameter d of the tube the

nearest multiples of 5mm if the allowable shear

stress is tallow= 70MPa

5 Torsion


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5. Torsion

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5.4 ANGLE OF TWIST

Angle of twistis important when analyzing reactions

on statically indeterminate shafts

f =T(x) dx

J(x) G0

L

f = angle of twist, in radians

T(x)= internal torque at arbitrary positionx, found

from method of sections and equation of

moment equilibrium applied about shafts axis

J(x)= polar moment of inertia as a function ofx

G= shear modulus of elasticity for material

5-14

5 Torsion


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5.4 ANGLE OF TWIST

Constant torque and x-sectional area

f =TL

JG

If shaft is subjected to several different torques, or x-

sectional area or shear modulus changes suddenly

from one region of the shaft to the next, then apply

Eqn 5-15 to each segment before vectorially addingeach segments angle of twist:

f =TL

JG

5-15

5 Torsion


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5. Torsion

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5.4 ANGLE OF TWIST

Sign convention

Use right-hand rule: torque and angle of twist arepositive when thumb is directed outward from the

shaft

5 Torsion


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5. Torsion

37

5.4 ANGLE OF TWIST


Internal torque

Use method of sections and equation of moment

equilibrium applied along shafts axis

If torque varies along shafts length, section madeat arbitrary positionxalong shaft is represented

as T(x)

If several constant external torques act on shaft

between its ends, internal torque in each segment

must be determined and shown as a torque

diagram

5. Torsion


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5. Torsion

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5.4 ANGLE OF TWIST


Angle of twist

When circular x-sectional area varies along

shafts axis, polar moment of inertia expressed as

a function of its positionxalong its axis,J(x) IfJor internal torque suddenly changes between

ends of shaft, f= (T(x)/J(x)G) dxor f= TL/JGmust be applied to each segment for whichJ, T

and Gare continuous or constant

Use consistent sign convention for internal torque

and also the set of units

5. Torsion


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5. Torsion

39

EXAMPLE 5.7

50-mm-diameter solid cast-

iron post shown is buried 600mmin soil. Determine

maximum shear stress in the

post and angle of twist at its

top. Assume torque about to

turn the post, and soil exerts

uniform torsional resistance of

tNmm/mm along its 600 mmburied length. G= 40(103) MPa

5. Torsion


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40

EXAMPLE 5.7 (SOLN)

Internal torque

From free-body diagram

Mz= 0; TAB= 100 N(300 mm) = 30 103Nmm

5. Torsion


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EXAMPLE 5.7 (SOLN)

Internal torque: Magnitude of the uniform

distribution of torque along buried segmentBCcanbe determined from equilibrium of the entire post.

Mz= 0;

100 N(300 mm) t(600 mm) = 0

t= 50 Nmm

5. Torsion


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EXAMPLE 5.7 (SOLN)

Internal torque

Hence, from free-body diagram of a section of the

post located at positionxwithin regionBC, we have

Mz= 0;

TBC50x= 0

TBC= 50x

5. Torsion


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2005 Pearson Education South Asia Pte Ltd 43

EXAMPLE 5.7 (SOLN)

Maximum shear stress

Largest shear stress occurs in regionAB, since

torque largest there andJis constant for the post.

Applying torsion formula

tmax = = 1.22 x106N/m2= 1.22 MPaT

AB cJ

5. Torsion


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EXAMPLE 5.7 (SOLN)

Angle of twist

Angle of twist at the top can be determined relative

to the bottom of the post, since it is fixed and yet is

about to turn. Both segmentsABandBCtwist, so

fA = +

TABLAB

JG

TBC dx

JG0LBC

.

.

.

fA= 0.00147 rad

5. Torsion


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EXERCISE

45

The assembly is made of A-36 steel and consists of aSolid rod 15 mm in diameter connected to the inside

of a tube using rigid disk at B. Determine the angle of

twist A. The tube has an outer diameter of 30 mmwall thickness of 3mm

5. Torsion


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Problem 5-61 and 62

A

D

B

C

50Nm

30Nm

15mm

30mm

3mm

D

B

C

A

rad04706.0

)10(75

)3.0(50

)10(75

)3.0(8099

///

ABBC

AB

ABAB

BC

BCBC

BACBACA

JJ

GJ

LT

GJ

LT

ffff

rad00681.0)10(75

)3.0(809

//

BC

BC

BCBCBCDCB

J

GJ

LTfff

( )( )4

44

0075.02

012.0015.0

2

AB

BC

J

J

5. Torsion


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EXERCISE F5-7/F5-8/F5-10/F5-11

5. Torsion


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CHAPTER REVIEW

Torque causes a shaft with circular x-section to

twist, such that shear strain in shaft isproportional to its radial distance from its centre

Provided that material is homogeneous and

Hookes law applies, shear stress determinedfrom torsion formula, t= (Tc)/J

Design of shaft requires finding the geometric

parameter, (J/C) = (T/tallow)

Power generated by rotating shaft is reported,from which torque is derived; P = T

5. Torsion


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CHAPTER REVIEW

Angle of twist of circular shaft determined from

If torque andJGare constant, then

For application, use a sign convention for

internal torque and be sure material does notyield, but remains linear elastic

f =TL

JG

f =T(x) dx

JG0L

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If shaft is statically indeterminate, reactive

torques determined from equilibrium,compatibility of twist, and torque-twist

relationships, such as f= TL/JG

Solid noncircular shafts tend to warp out ofplane when subjected to torque. Formulas are

available to determine elastic shear stress and

twist for these cases

Shear stress in tubes determined byconsidering shear flow. Assumes that shear

stress across each thickness of tube is

constant

CHAPTER REVIEW

5. Torsion


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CHAPTER REVIEW

Shear stress in tubes determined from

t= T/2tAm

Stress concentrations occur in shafts when x-

section suddenly changes. Maximum shear

stress determined using stress concentrationfactor,K(found by experiment and represented

in graphical form). tmax= K(Tc/J)

If applied torque causes material to exceed

elastic limit, then stress distribution is notproportional to radial distance from centerline

of shaft

5. Torsion


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CHAPTER REVIEW

Instead, such applied torque is related to stress

distribution using the shear-stress-shear-straindiagram and equilibrium

If a shaft is subjected to plastic torque, and

then released, it will cause material to respondelastically, causing residual shear stress to be

developed in the shaft

lecture4 torsion

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