Ch3 Torsion Lecture

7/24/2019 Ch3 Torsion Lecture

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Topic 3: TORSION1. Introduction

2. Stresses in Elastic range3. Angle of twist

4. Design transmission shafts

5. Thin-wall hollow shafts


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1. Introduction

- Members subjected to loads along the longitudinal axes

(Axial deformation)

- Members subjected to torsional loads twisting the

members about their longitudinal centroid axesare

considered in this topic.

- Examples: a lug-wrench, power transmission shaft


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- A lug-wrench:

+ A lug-wrench shaft AB, arm CD

+ Apply equal and opposite forces (P) to ends of CD

AB is a torsion member


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Sign convention:

+ Torque T(x): right-hand-rule sense

+ Angle of rotation, (x)


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- Power transmission shaft:

+ Turbine exerts torque T on the shaft

+ The shaft transmits T to the generator+ Generator creates an equal & opposite torque T

Generator

Rotation

Turbine


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2. Stresses in Elastic range

Shear stress in shaft:

How is torque transferred through shaft?An element dA distance from the center of the shaft has

shear force dF.

For equilibrium: = 0 =

Since dF = dA = ()= T

Shear stress distribution?


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Shaft deformations:

Square shaft: wrapped under tension, cross section do not

remain plane.

Circular shaft: Every cross section remains plane and

undistorted.

- Circumferential lines remain in a plane after deformation- Longitudinal lines: Parallel to the axis become helical

- Right angleABC : Shear deformation

- : Torsional deformation


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Consider a circular shaft attached to a fixed support.

- Apply torque T: + Shaft will twist

+ Free end will rotate through angle

- By observation: + T

+ L

Relationship of , T, L?

Shear stress distribution?


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Shearing strain in circular shaft:

Consider a shaft of length L, radius c,

angle of twist .

AA = L= =

, [rad]

is proportional to and

varies linearly with the distance

from the axis of the shaft.

max

=


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Stresses in the elastic range:

We have: =

and max=

=

From Topic 2: = G

G= G

=

As long as the yield strength is not exceeded, varies

linearly to the distance from the axis of the shaft.

min

=


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Recall that ()= T

T =

T =

.

J = : the polar moment of inertia of the crosssection with respect to its center O.

max=

and =


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+ A solid cross section of radius r:

dJ = dA = 2udu

J =

2

= 2

3

J =

4 =

3

+ A tubular shaft of outer radius r0and inner radius ri:

=

4

4

Example 1:

a/ Tmax=? If max= 120 Mpa

b/ min= ?

Solution: T =

min=


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Example 2:

Determine maxin shaft AB, BC.

Solution:


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Axial shear components:

A torque applied to a shaft shearing stresses on the face

perpendicular to the axis.

Equilibrium equal stresses on planes containing the axis.


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Normal stresses due to Torsion:

- Element a: pure shear

- Element b:

Normal stress, shear stress or

combination of both.

- Consider an element at 45to the

shaft axis:

F = 2(maxA0) cos45= maxA0 2

45 =

=

= max


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Torsion failure modes:

When subjected to torsion:

+ Ductile specimen breaks along a

plane of maximum shear.

+ Brittle specimen breaks along planes perpendicular to the

direction in which tension is a maximum.


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- Shaft deformation: + Shear deformation

+ Torsional deformation

- Strain: Shearing strain max=cL

- Stress: max=Tc

J and =

TJ


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3. Angle of twist

max=

max=

In the elastic range: = G

= G

=

- For shaft with cross-section changes:=

Each segment has constant cross section an torque.

- If cross section changes continuously:

d=

=

()


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Gear assembly:

E/B= E- B=


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Example 3:

rA= 2rB Apply T at E

Determine E?

TAD= 2T

A=

=

CC = CC rAA= rBB

B= (rA/rB) A= 2A =4

E= B + E/B=4

+

= 5


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Example 4:

Shaft ABC with d = 60 mm

is supported by 2 journal

bearings.

Shaft EH with d = 80 mm,

is fixed at E and supported

by a bearing at H.

If A/C= 0.04 rad, determine T1and T2?

Solution:


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Example 4:

d = 40 mm.

Determine B/A(rad) =?

Solution:


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Comparison of axial and torsion:

4 D i f T i i h ft


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4. Design of Transmission shaft:

- Select the shaft material (G) and cross section (J) to meet

performance specifications (power - P and speed - ) without

exceeding the allowable and allowable .Step 1: Determine T? From dynamics: P = T= T(2f) T =

SI units: T: torque [N.m]

P: power [N.m/s] [W]: angular velocity (rad/s)

f: frequency of rotation in Hz

US units: f [rpm], 1 rpm =

1

6HzP [hphorsepower]

1 hp = 550 ft.lb/s (6600 in.lb/s) = 746 W


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Step 2: Select a material (set all)

Design requirement: max=

all

Step 3: Find shaft cross section?

If considering the allowable angle of twist: all


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Example 5:

P = 16 hp from motor A machine tool D

Motor A: f = 1260 rpm

all= 8 ksi

Determine dABand dCD?

Solution:

all= 8 ksi = 8 x 103

(lb/in2

)P = 16 hp = 105600 in.lb/s

f = 1260 rpm =16

6= 21 Hz

T = = 156..1 = 800.72 (lb.in)

max=.

alldAB0.798 [in.]


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3

=

5TCD= 1334.53 (lb.in)

max=

.

alldCD0.947 [in.]

Example 6:


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Example 6:

dAB= dCD;

max80 Mpa,

D1.5; G = 70 Gpa

Determine d AB= dCD= ?

Solution:

Design based on allowable stress:

=.

=

max =.

=

max

TAB = rB.F

TCD= rC.F TAB=.

= 2500 (N.m)

dAB= 59.6 mm


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Design based on allowable angle of twist:

D= D/C+ C =.

+ C1.5

C rC= B rBC

dAB= 62.9 mm

Design must use the lager value for d: d = 62.9 mm

St ti ll d t i t h ft


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Statically determinate shaft:

Example 7:

a/ Determine T1, T2: = 0

TCT2= 0 TC+ TBT1= 0

b/ Determine C, B

- Torque-twist behavior: 1= 11 =

1=

1 =

- Geometry of deformation: = + ;

= 1+ =1

St ti ll i d t i t h ft


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Statically indeterminate shaft:

The unknown torques exceeds the number of applicable

equilibrium equations.

- Step 1: Write equil. eqs.

- Step 2: Formulate compatibility eqs.

- Step 3: Use torque-displacement

eqs. to relate and T.

Example 8:

E ilib i


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- Equilibrium eqs:

TA= -T1

T1T2= TB

TC= T2

- Element torque-twist behavior:

1

= 1

1

=

1

=

=

T

2

- Geometry of deformation:

=1+ = 0

Results: T1,

T2are determined

1=

3 =

3

- Determine rotation angle B

=1 =


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1/

a/ Determine expressions of

max1

, max2,

max3

b/ B= ?

2/

Problem 3.53 in Mechanics of Materials, 6thedition, Beer

and Johnston.

5 Thi ll d h ll h ft


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5. Thin-walled hollow shafts

Consider a thin-walled circular hollow shaft with inner

radius r1, outer radius r2, thickness t = r2r1, and median

radius rm= 1(r1+ r2)

The stress distribution for a circular shaft increases linearly

as we move from the center axis to the surface of the

member.

=

=

where J =

(

4 1

4)

Write r1and r2in terms of rm:

r2= rm+ t/2

r1= r

mt/2

4 4


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J =

+

4

4

t2


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The average force per unit length on the member cross-

section is: q = mt

where q is the shear flow

The total torque can be estimate as:

T = shear flow x total length x moment arm

T = (mt)(2rm)(rm)

Therefore: m=

Ch3 Torsion Lecture

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