steif chapter 4 excerpt

Torsion

Ch

ap

ter 4

4.1 Rotation (pg. 138)

4.2 Shear Strain in Circular Shafts (pg. 140)

4.3 Application and Transmission of Torque (pg. 148)

4.4 Shear Stress in Circular Shafts (pg. 150)

4.5 Strength and Stiffness (pg. 162)

4.6 Dependence of Stiffness and Strengthon Shaft Properties (pg. 164)

4.7 General Guidelines for TorsionalStiffness of Non-Circular Cross-Sections (pg. 166)

4.8 Torsion of Shafts with RectangularCross-Sections (pg. 176)

4.9 Torsion of Shafts with Thin-Walled Cross-Sections (pg. 178)

4.10 Shafts with Non-Uniform Twisting AlongTheir Lengths (pg. 186)

4.11 Internal Torque and the Relation to Twist and Stress (pg. 188)

4.12 Relation Between Senses and Signs of Internal Torque,Twist, and Stress (pg. 190)

4.13 Shafts with Varying Cross-Sections (pg. 192)

4.14 Statically Indeterminate StructuresSubjected to Torsion (pg. 202)

4.15 Power-Torque-Speed Relations for Rotating Shafts (pg. 210)

C O N T E N T S

ELEM

ENT

DEFORMATIONS

SEG

MEN

TSY

STEM Rotation

Shear strain

Twist

TwistrotationRelation

TwiststrainRelation

M04_STEI3340_01_SE_C04.QXD 7/13/11 1:30 AM Page 136

2012 Pearson Education, Inc., Upper Saddle River, NJ 07458. All Rights Reserved

Common Deformation

Modes

Design AgainstBody Composed of Elements

Chapter 2. Force and

Deformation in an Element

Chapter 3. Axial

Chapter 4. Torsion

Chapter 5. Bending

Chapter 6. Excessive

Deformations

Chapter 7. Material Failure

Chapter 8. Buckling

Unit 1 Unit 2 Unit 3

Chapter OutlineA shaft of a torsion bar suspension deforming due toup and down displacement of a tire is an example oftwisting or torsional deformation. A shaft subjectedto torsion can be viewed as composed of segments orslices along the length of the member. Each cross-section rotates through some angle (4.1). When endfaces of a segment rotate through different angles,the segment twists, giving rise to shear strains thatvary in the radial direction (4.2). External loads onthe bar result in an internal twisting moment ortorque that acts oppositely on the two ends of a seg-ment (4.3). The shear stresses acting on each face ofa segment add up to the internal twisting moment(4.4). For a shaft of circular cross-section loadedonly by opposite torques at its ends, the twist or rela-tive rotation depends on the bars length, shear mod-ulus, and its polar moment of inertia, a property ofthe cross-section (4.54.6). The relation betweentorque and twist and maximum shear stress can alsobe determined for some non-circular cross-sections(4.74.9). A shaft can be subjected to several exter-nal torques along its length, with the internal torquevarying in different segments (4.104.14). When amotor drives a shaft causing it to rotate at constantspeed, the power, rotation speed, and torque areinterrelated, but stress and deformation are related tointernal torque just as for a stationary shaft (4.15).

137

FORCES

SYSTEMSEG

MEN

TELEM

ENT

Externaltorque

Shear stress

Internal torque

Externalinternaltorque Relation

TorquetwistRelation

TorquestressRelation

StressstrainRelation

M04_STEI3340_01_SE_C04.QXD 7/13/11 1:31 AM Page 137


138 C H A P T E R 4 | Torsion

The pattern of deflection is a characteristic feature of a shaft in torsion.Successive cross-sections of a twisting shaft rotate by different amounts.Here, we describe this deflection pattern and how it is quantified.

4.1 Rotation

The pivoting leg rotates by f. If f is small(in radians), then the end deflects by .wf

Now, one leg is held down and the other ispivoted. We consider the center portion, whichstays straight and twists, to be the shaft.

1. Each cross-section of ashaft in torsion rotates aboutthe shaft axis.

This wire was lying on the flat surface.

We define twist as the difference in rotation:the rotation angle of one end minus the other.In this case, since one leg doesnt rotate, thetwist is equal to . The fingers apply forces to the legs that result in opposite torques on the central portion equal to Fw.

f

2. The rotation of one end ofthe shaft relative to the othercaptures the deformation in twisting and it is related to the torque. L

L

FF

F

T Fw

T Fw

F

ww

Twist, , is defined as the differencein rotation angle of one cross-section of a shaft relative to another, due to torsional deformation.

f

Here is a torsion bar suspension. Engineersdesign the torsion bar to give the rightresistance to up and down motion of thetire over the road. When driving overa bump, the tire moves up relative tothe rest of the vehicle, due to twistingof the torsion bar. The torsion bar isanalogous to the twisting centerportion of the wire. The upwardmotion of the tire is analogous to the end of the wire leg that lifts. If the torsion bar does not twistenough, the vehicle moves up toomuch (with the tire). If the twist istoo great, the tire deflects too much relative to the vehicle and could contact it.

Fixed to car frame

Pivot axis

Control arm

Torsion bar

Tire moves up and down with road surface, causing control arm to pivot.

w

w

M04_STEI3340_01_SE_C04.QXD 7/5/11 5:32 PM Page 138


4 . 1 | Rotation 139

>>End 4.1

A foam tube can illustrate the variation inrotation along a twisted shaft. A straight linewas drawn on the side.

4. For a uniform shaft twistedby torques at its ends, therotation angle varies linearlywith distance along the shaft.

Every 1 mm of the shaft feels the same twisting torque. So the additional rotation that accumulatesover every 1 mm segment is the same. Every segment of this shaft has the same twist per length(e.g., the same degrees of rotation per millimeter length).

5. The twist per length, thatis, the change in rotation angleper unit distance along thebody, captures the intensity of the twisting deformation.

3. Successive cross-sectionsalong the length of a twistedshaft rotate by differentamounts.

A twisted shaft not only rotates at its ends. Every cross-section rotates, usually through a different angle.

The pegs on this shaft were initially aligned. The shaft is then twisted by rotating the right endthrough some angle.

From the new orientations of the pegs on the shaft, you can see that each cross-section rotatesthrough a different angle.

The tube was then twisted slightly, movingthe line.

As viewed from the side, the line stays approximately straight. The vertical displacement of the linevaries linearly along its length. So the rotations of cross-sections vary linearly.

The twist per length captures the intensity of the twist. Twist per length captures the intensityof the torsional deformation, and it isdefined as the difference in rotationangle of one cross-section of a shaftrelative to another, divided by thedistance between the cross-sections.

1

2

L

x

T

>>End 4.1

We express the linear variation of rotation f from at to at as follows:

f = f1 + (f2 - f1)x

L

x = Lf2x = 0f1

where x is the distance alongthe shaft and L is its length.For example, halfway alongthe shaft ,

, that isequal to the mean of therotations at the ends.

f = (f1 + f2)>2(at x = L>2)

Rotation is measured relative to the orientation before twisting.A cross-section can rotate in either direction, which wedistinguish with a plus or minus sign. We use the right handrule to define positive rotation. The rotations and here,and at all cross-sections in between, are positive.

f2f1

Twist per length = f>L = (f2 - f1)>L

positive x positive




Rotation of successive cross-sections through different anglesdistorts or strains elements in a shaft. Here we look at the typeof strain, its variation from point to point in the shaft, and howthe overall twisting of the shaft affects the strain.

4.2 Shear Strain in Circular Shafts

1. Rotations of successivecross-sections throughdifferent angles cause shearstrain.

Rectangles were drawn on the side ofthis shaft.

The cross-section at x1 rotatesby f(x1). The cross-section at x2rotates by a greater angle f(x2).

x1

R

x2

(x1) (x2)

To see only the change in shape (strain), move the deformed element up by Rf(x1).

R(x2)R(x1)

R[(x2) (x1)]

R[(x2) (x1)]x2 x1

This element extends from x1 tox2. Follow its deformation whenthe shaft ends rotate.

2. The shear strain, which is equal to the relative sheardisplacement over an elementdivided by its length, can berelated to rotations at the endsof the shaft.

When the shaft is twisted, the angles areno longer 90. The rectangles have dis-torted or sheared into parallelograms.

Shear strain is due to the different rotations of successive cross-sections.

Here is the element, originally and as deformed

Compute the shear strain from the angle change asthe rectangle distorts into a parallelogram.

Shear strain on the outer surface of shaft is

g = R3f(x2) - f(x1)4

x2 - x1

If the element extends over the length of theshaft and has a twist , thenf(x1 = 0, x2 = L) g =

RfL



4 . 2 | Shear Strain in Circular Shafts 141

Shear strain depends on , which is the twist per length. This is the intensity of twisting, or how

rapidly the rotation changes over the length L of the shaft.

fL

3. Shear strain is proportionalto the twist per length.

0.5 m

1

3

So far, we have considered strain on the outside surface of the shaft. What is the strain of an elementinside the shaft, located at a distance from the center? Use the same reasoning as above, exceptnow all displacements are equal to , instead of .Rfrf

r

4. Shear strain increases withradial distance from the shaftcenter.

(x)(x)

Shear strain varies with radial position : g = rf

Lr

We just considered how the strain magnitude varies with radial position in the shaft. The directionof the strain varies at different points around the shaft circumference.

r

The strain varies because rotation aboutthe shaft axis producesdisplacements in the

circumferential direction.

Because the shear strain depends on the relative displacements of neighboring cross-sections,the direction and magnitude of the strain do not vary along the length of the shaft.

>>End 4.2

As an example, for this shaft

fL

=3 - 1

0.5= 4

deg m

= 6.98 * 10-2 radm

5. The direction of the shearstrain depends on the positionof the element around thecircumference of the shaft,but not along the length.




EXAMPLE PROBLEM 4.1

(a) The shaft BC twists and arm AB pivots rigidly about B. As a result, endA of the arm displaces by 0.1 in. perpendicularly to AB in the y-direction.Determine the axis about which the rotation occurs and the rotation in degrees.

(b) The arm CD is initially oriented at 35 relative to the y-axis. BC twists,causing CD to rotate by 0.1 about the positive x-axis. Determine the displacement of the point D along each of the coordinate axes.

z

y 35

0.1

0.1

u20 in.

C

0.1

20 in.

u

A

B

C

D

15 in.

20 in.

z

xy

35Solution

(a) Here are 3-D and 2-D views of AB pivoting rigidly about B.B stays in position, and A displaces by 0.1 in. Pivoting rigidly means that ABdoes not deform, but keeps its original shape. With the right hand rule, the rotation is about the negative x-axis.

Displacement 0.1 in. 15 in. tan(f). Displacement 0.1 in. is small comparedto 15 in. 0.1 in. (15 in.) f, (angle f in radians). Therefore,

(b) Here are 3-D and 2-D views of CD pivoting rigidly about C. The rotation is 0.1about the positive x-axis. So as rotated, BC is 35.1 from the y-axis.

10-2 rad = 0.382.f = 0.667 *=Q

=

C stays in position, and D displaces by displacement is perpendicular to the

initial direction of CD.

Displacement is in. along y.Displacement is along z.(3.49 * 10-2 in.)(sin 55) = 2.86 * 10-2 in.

- (3.49 * 10-2 in.)(cos 55) = -2.00 * 10-2f = 3.49 * 10-2 in.u = (20 in.)

10-3 radians V 1 Q0.1 = 1.745 *u = (20 in.) tan (f).

A

B

x

15 in.

0.1 in.

0.1 in.

15 in.

0.1 in.

>>End Example Problem 4.1

90 35 55

u 3.49 102 in.z

y



| Example Problem 4.2 143

EXAMPLE PROBLEM 4.2

400 mm

15 mm

300 mm

A

C

B

C

z

z

y

y

x

Strainedelement

>>End Example Problem 4.2

Solution

Since equal and opposite torques act at its ends, the shaft twists uniformly. The strain isuniform along the length and varies in the radial direction (from the shaft centerline).Here are the rotation directions. The relative rotation is .If the rotations were in the same direction, instead we would have

.= 0.5- 1.5f = 2

2 = 3.5f = 1.5 +

In the formula for the shear strain must be in radians, so .

On the outer surface, so

Consider how the faces of a cubic element at point C, aligned with the coordinateaxes would move.

400 mm = 1.145 * 10-3.g = rf>L = (7.5 mm)(6.11 * 10-2)>r = R = 7.5 mm,= 6.11 * 10-2 radf = 3.5

g = rf>L. f

Line originally parallelto shaft axis

Same line withshaft twisted

1.5

2.0

relative to this face

This face moves down(toward negative z)

The shaft twists due to equal and opposite torques acting at its ends.End A rotates by 1.5 about the positive x-axis, and end B rotates by 2 about the negative x-axis.

Determine the shear strain of an element located 300 mm from Aat a point C on the side of the cylinder. Find the magnitude of the shear strain, and draw a strained element.

The element deforms as shown under this strain.The direction of straining would change as one considers different elements aroundthe circumference.



PROBLEMS


200 mm

C

B

A

z

yx

25

Prob. 4.2

4.2 Arm AB is initially oriented at 25 from the x-axis. The shaft BCtwists, while AB moves rigidly. A displaces by 0.5 mm in the y-direction. (A also displaces in the x-direction.) What is therotation angle of the shaft at B, and about what axis is the rotation?

4.4 The shaft twists due to two equal and opposite torques applied toits ends. The rotations are 0.7 about the -z-axis at B, and 0.1about the -z-axis at D. What is the direction and magnitude of the rotation (a) at C and (b) at E?

40 in.

20 in.

20 in.

30 in.

A

B

C

D

E

z

yx

Prob. 4.4

y

z x

60 mm

100 mm

150 mm

300 mm

200 mm

A

BC

Prob. 4.5

A

C

B

z

x y

80 mm C

z

x30

Prob. 4.3

A

B

Cy

z x15 in.

Prob. 4.1

4.1 The shaft AB twists, and BC moves rigidly. At B the shaft rotatesby 1.5 about the -axis. How much does the end C displaceand in what direction?

+x

4.5 The shaft twists due to two equal and opposite torques applied to its ends. Point C displaces 0.2 mm in the -x-direction. Point Bdisplaces 0.1 mm in the y-direction. Determine the direction and magnitude of the displacement of point A.

4.3 The shaft at B and the attached disk rotate by 2about the -y-axis. By how much does the point Cdisplace? (Indicate direction and magnitude alongthe x- and y-axes.)

Additional data on material properties needed to solve problems can be found in Appendix D or inside back cover.



Cz

y

300 mm

| Problems 145

4.6 The shaft twists due to equal and opposite torques applied to itsends. The cross-section A rotates by 0.3 about the -z-axis, andB rotates by 0.1 about the z-axis. What is the magnitude ofthe shear strain at point C? Also, draw the deformed shape of anelemental cube at C.

4.7 The shaft twists due to equal and opposite torquesapplied to its ends. The cross-section B rotates by 1about the -x-axis. At point C, the shear strain is

(the deformed element is shown in thediagram). What is the direction and magnitude of the rotation at A?

2 * 10-4

4.9 The 16 mm diameter shaft twists due to two equal and opposite torques applied to its ends. Points A, B, and D are 40 mm, 30 mm, and 50 mm from the center of the shaft. The displacement at B is 0.3 mm in the x-direction. The shear strain at C is (deformed element is shown in the diagram). Determinethe displacement of points D and A.

3 * 10-4

B

0.75 in.

20 in.

A

Cz

yx

C

x

y

Prob. 4.6

A

B

C

C

z

y

z

yx

300 mm12 mm

Prob. 4.7

z

x y

3 in.

15 in.

A

C

z

yB

Prob. 4.8

D

C

B

400 mm250 mm

200 mm

A y

xz

C

z

x

Prob. 4.9

4.8 A thin disk is bonded to each end of a shaft, whichhas outer and inner diameters of 1 in. and 0.75 in.Twisting is due to equal and opposite torquesapplied to the ends. Point A, located along the z-axis,displaces by in. in the -direction, andC, located along the y-axis, displaces by in. in the -direction. Determine the magnitude of the shear strain at point B, which is located at the shaft inner diameter. Also, draw the deformedshape of an elemental cube at B.

+z6 * 10-2

-y4 * 10-2

4.10 The shaft ABC, 40 mm in diameter, twists under equaland opposite torques applied to its ends. Rotation at Ais 1 about the -y-axis. Due to the rotation at C, point Dof the rigid disk moves 0.2 mm in the -z-direction(D also displaces in the x-direction). Determine the magnitude of the shear strain at point B. Also, drawthe deformed shape of an elemental cube at B.

A

B

CD 160 mm

300 mm100 mm B

D

z

x

60z

x y

Prob. 4.10




4.11 The bottom bracket spindle undergoes torsion during pedaling. If the spindleis fixed against rotation at the chain ring and the rotation where the spindleconnects to the far crank is 0.1, by how much has the initially horizontalcrank and pedal moved downwards? Take the dimensions to be a 70 mm,b 25 mm, c 9.5 mm, d 80 mm, and L 170 mm.=

=

4.12 A large downward force on the right pedal results in a moment about theforward axis of a bike. This moment can be balanced by an oppositemoment of the hands pulling up on right handlebar and pushing down onthe left. Thus, two equal and opposite moments have to be absorbed by theframe. A portion of this load results in twisting of the down tube. If thetube material has allowable shear strain of 0.002, what is the maximumallowable relative angle of rotation of the head tube and the seat tube? Letthe down tube be 24 in. long and have a circular cross-section with outerdiameter of 1.75 in. and wall thickness of 0.125 in.

4.13 Consider a simplified drill string consisting of single sized drill pipe. Say thebit does not rotate, and the maximum shear strain that the pipe material cansafely take is known to be 0.0012. Determine how many turns of the string areallowable at the top of the L 400 m string without exceeding the maximumshear strain. Take the parameters to be Do 120 mm and Di 76 mm.

cPedal

Crank

Spindle

Chain ring

d

ab

L

Prob. 4.11 (Appendix A1)

ToptubeSeat tube

Downtube

Headtube


Do

Di

L


R

w

q

D2

s

Pivotbeam

Cord disk

Shaft


Focused Application Problems

4.14 Say that the pivot-beam is initially horizontal and that the legs lift so thatthe end of the pivot-beam displaces upward by 1 in. Determine the rotationof the shaft where the pivot-beam connects to it. Take the dimensions to beD2 1 in., q 6 in., R 9 in., s 3 in., and w 17 in.



ScrewL

Bone

Intramedullarynail

| Problems 147

>>End Problems

4.15 Say that the pivot-beam is initially horizontal and that the legs lift the paddedbeam by 1.5 in. If the maximum shear strain in the shaft to the cord disk is0.001, by how much has the plate stack been lifted by the cord? Neglect anyelongation in the cord. Take the dimensions to be D2 1 in., q 6 in., R 9 in., s 3 in., and w 17 in.

R

w

q

D2

s

Pivotbeam Shaft

Cord disk


4.16 A torsion analysis of an external fracture fixation system under a forward loadon the foot predicts that the carbon fiber rod twists about its axis by a 5 angleat the plane of the lower pin relative to the upper pin. Say that the otherdeflections of the rod are neglected, and the pins and bone are rigid. What isthe transverse deflection of one bone fragment relative to the other at thepoints in the bone nearest and farthest from the rod? Approximate the bone as circular with an outer diameter of 26 mm and an inner diameter of 20 mm.Let the diameter of the solid carbon fiber bar be 10.5 mm. Take the dimensionsto be Lr 300 mm and Lp 60 mm.

RodPin

Bone

F0

Lp

Lr

Ls

Fractureplane


4.17 An intramedullary nail is fixed by screws to the bone at its two ends a distanceL 300 mm apart across the fracture. Approximate the bone as circular withan outer diameter of 26 mm and an inner diameter of 20 mm. The nail itself hasan outer diameter of 10 mm and an inner diameter of 5.4 mm. Say the twoscrews rotate by a 10 angle with respect to each other due to twisting of theintermedullary nail. What is the maximum relative displacement between thetwo faces of the fractured bone fragments? Assume that no load is transmittedfrom the rod to the bone between the screws (the nail is fully within theintramedullary canal).

ScrewL

Bone

Intramedullarynail


4.18 An intramedullary nail is fixed by screws to the bone at its two ends a distanceL 300 mm apart across the fracture. Approximate the bone as circular withan outer diameter of 26 mm and an inner diameter of 20 mm. The nail itself has an outer diameter of 10 mm, and an inner diameter of 5.4 mm. If the shearstrain in the nail is not to exceed 0.005, what is the allowable rotation of thetwo bone fragments with respect to each other?




steif chapter 4 excerpt

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