Torsion - RC. Hibbeler

12005 Pearson Education South Asia Pte Ltd

5. Torsion

CHAPTER OBJECTIVES

• Discuss effects of applying torsional loading to a long straight member

• Determine stress distribution within the member under torsional load

• Determine angle of twist when material behaves in a linear-elastic and inelastic manner

• Discuss statically indeterminate analysis of shafts and tubes

• Discuss stress distributions and residual stress caused by torsional loadings


5. Torsion

CHAPTER OUTLINE

1. Torsional Deformation of a Circular Shaft2. The Torsion Formula3. Power Transmission4. Angle of Twist5. Statically Indeterminate Torque-Loaded Members6. *Solid Noncircular Shafts7. *Thin-Walled Tubes Having Closed Cross

Sections8. Stress Concentration9. *Inelastic Torsion10. *Residual Stress


5. Torsion

• Torsion is a moment that twists/deforms a member about its longitudinal axis

• By observation, if angle of rotation is small, length of shaft and its radius remain unchanged

5.1 TORSIONAL DEFORMATION OF A CIRCULAR SHAFT


5. Torsion

5.1 TORSIONAL DEFORMATION OF A CIRCULAR SHAFT

• By definition, shear strain is

Let x dx and = d

BD = d = dx

= (/2) lim ’CA along CA

BA along BA

= ddx

• Since d / dx = / = max /c

= max

c( )Equation 5-2


5. Torsion

5.2 THE TORSION FORMULA

• For solid shaft, shear stress varies from zero at shaft’s longitudinal axis to maximum value at its outer surface.

• Due to proportionality of triangles, or using Hooke’s law and Eqn 5-2,

= max

c( ) ...

= max

c∫A 2 dA


5. Torsion


• The integral in the equation can be represented as the polar moment of inertia J, of shaft’s x-sectional area computed about its longitudinal axis

max =Tc

J

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

T = resultant internal torque acting at x-section, from method of sections & equation of moment equilibrium applied about longitudinal axis

J = polar moment of inertia at x-sectional area

c = outer radius pf the shaft


5. Torsion


• Shear stress at intermediate distance,

=TJ

• 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


Solid shaft• J can be determined using area element in the form

of a differential ring or annulus having thickness d and circumference 2 .

• For this ring, dA = 2 d

J = c42

• J is a geometric property of the circular area and is always positive. Common units used for its measurement are mm4 and m4.


5. Torsion


Tubular shaftJ = (co

4 ci4)

2


5. Torsion


Absolute maximum torsional stress• Need to find location where ratio Tc/J is maximum• Draw a torque diagram (internal torque vs. x along

shaft)• Sign Convention: T is positive, by right-hand rule, is

directed outward from the shaft• Once internal torque throughout shaft is determined,

maximum ratio of Tc/J can be identified


5. Torsion


Procedure for analysisInternal loading• Section shaft perpendicular to its axis at point

where shear stress is to be determined• Use free-body diagram and equations of

equilibrium to obtain internal torque at sectionSection property• Compute polar moment of inertia and x-sectional

area• For solid section, J = c4/2• For tube, J = (co

4 ci2)/2


5. Torsion


Procedure for analysis

Shear stress• Specify radial distance , measured from centre

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

• Apply torsion formula, = T /J or max = Tc/J

• Shear stress acts on x-section in direction that is always perpendicular to


5. Torsion

EXAMPLE 5.3

Shaft shown supported by two bearings and subjected to three torques.

Determine shear stress developed at points A and B, located at section a-a of the shaft.


5. Torsion

EXAMPLE 5.3 (SOLN)

Internal torque

Bearing reactions on shaft = 0, if shaft weight assumed to be negligible. Applied torques satisfy moment equilibrium about shaft’s axis.

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


5. Torsion

EXAMPLE 5.3 (SOLN)

Internal torque

Mx = 0; 4250 kN·mm 3000 kN·mm T = 0

T = 1250 kN·mm

Section property

J = /2(75 mm)4 = 4.97 107 mm4

Shear stress

Since point A is at = c = 75 mm

B = Tc/J = ... = 1.89 MPa


5. Torsion

EXAMPLE 5.3 (SOLN)

Shear stress

Likewise for point B, at = 15 mm

B = T /J = ... = 0.377 MPa

Directions of the stresses on elements A and B established from direction of resultant internal torque T.


5. Torsion

• Power is defined as work performed per unit of time

• Instantaneous power is• Since shaft’s angular velocity = d/dt, we can

also express power as

5.3 POWER TRANSMISSION

P = T (d/dt)

P = T

• Frequency f of a shaft’s rotation is often reported. It measures the number of cycles per second and since 1 cycle = 2 radians, and = 2f T, then power

P = 2f TEquation 5-11


5. Torsion

Shaft Design• If power transmitted by shaft and its frequency of

rotation is known, torque is determined from Eqn 5-11

• Knowing T and allowable shear stress for material, allow and applying torsion formula,


J

cT

allow

=


5. Torsion

Shaft Design• For solid shaft, substitute J = (/2)c4 to determine c

• For tubular shaft, substitute J = (/2)(co2 ci

2) to determine co and ci



5. Torsion

EXAMPLE 5.5

Solid steel shaft shown used to transmit 3750 W from attached motor M. Shaft rotates at = 175 rpm and the steel allow = 100 MPa.

Determine required diameter of shaft to nearest mm.


5. Torsion

EXAMPLE 5.5 (SOLN)

Torque on shaft determined from P = T, Thus, P = 3750 N·m/s

Thus, P = T, T = 204.6 N·m

( ) = = 18.33 rad/s175 rev

min

2 rad

1 rev

1 min

60 s( )

= =J

c

c4

2 c2

T

allow. . .

c = 10.92 mm

Since 2c = 21.84 mm, select shaft with diameter of d = 22 mm


5. Torsion

5.4 ANGLE OF TWIST

• Angle of twist is important when analyzing reactions on statically indeterminate shafts

=T(x) dx

J(x) G∫0

L

= angle of twist, in radians

T(x) = internal torque at arbitrary position x, found from method of sections and equation of moment equilibrium applied about shaft’s axis

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

G = shear modulus of elasticity for material


5. Torsion

5.4 ANGLE OF TWIST

Constant torque and x-sectional area

=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 adding each segment’s angle of twist:

=TL

JG


5. Torsion

5.4 ANGLE OF TWIST

Sign convention

• Use right-hand rule: torque and angle of twist are positive when thumb is directed outward from the shaft


5. Torsion

5.4 ANGLE OF TWIST

Procedure for analysisInternal torque• Use method of sections and equation of moment

equilibrium applied along shaft’s axis• If torque varies along shaft’s length, section made

at arbitrary position x along 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

5.4 ANGLE OF TWIST

Procedure for analysisAngle of twist• When circular x-sectional area varies along

shaft’s axis, polar moment of inertia expressed as a function of its position x along its axis, J(x)

• If J or internal torque suddenly changes between ends of shaft, = ∫ (T(x)/J(x)G) dx or = TL/JG must be applied to each segment for which J, T and G are continuous or constant

• Use consistent sign convention for internal torque and also the set of units


5. Torsion

EXAMPLE 5.9

50-mm-diameter solid cast-iron post shown is buried 600 mm in 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 t N·mm/mm along its 600 mm buried length. G = 40(103) GPa


5. Torsion

EXAMPLE 5.9 (SOLN)

Internal torque

From free-body diagram Mz = 0; TAB = 100 N(300 mm) = 30 103 N·mm


5. Torsion

EXAMPLE 5.9 (SOLN)

Internal torque

Magnitude of the uniform distribution of torque along buried segment BC can be determined from equilibrium of the entire post.

Mz = 0;

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

t = 50 N·mm


5. Torsion

EXAMPLE 5.9 (SOLN)

Internal torque

Hence, from free-body diagram of a section of the post located at position x within region BC, we have

Mz = 0;

TBC 50x = 0

TBC = 50x


5. Torsion

EXAMPLE 5.9 (SOLN)

Maximum shear stress

Largest shear stress occurs in region AB, since torque largest there and J is constant for the post. Applying torsion formula

max = = ... = 1.22 N/mm2TAB c

J


5. Torsion

EXAMPLE 5.9 (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 segments AB and BC twist, so

A = +TAB LAB

JG

TBC dx

JG∫0

LBC. . .

A = 0.00147 rad


5. Torsion

5.5 STATICALLY INDETERMINATE TORQUE-LOADED MEMBERS

• A torsionally loaded shaft is statically indeterminate if moment equation of equilibrium, applied about axis of shaft, is not enough to determine unknown torques acting on the shaft


5. Torsion


• From free-body diagram, reactive torques at supports A and B are unknown, Thus,

Mx = 0; T TA TB = 0

• Since problem is statically indeterminate, formulate the condition of compatibility; end supports are fixed, thus angle of twist of both ends should sum to zero

A/B = 0


5. Torsion


• Assume linear-elastic behavior, and using load-displacement relationship, = TL/JG, thus compatibility equation can be written as

TA LAC

JG

TB LBC

JG = 0

• Solving the equations simultaneously, and realizing thatL = LAC + LBC, we get

TA = TLBC

L( ) TB = T

LAC

L( )


5. Torsion


Procedure for analysisEquilibrium• Draw a free-body diagram• Write equations of equilibrium about axis of shaftCompatibility• Express compatibility conditions in terms of

rotational displacement caused by reactive torques

• Use torque-displacement relationship, such as = TL/JG

• Solve equilibrium and compatibility equations for unknown torques


5. Torsion

EXAMPLE 5.11

Solid steel shaft shown has a diameter of 20 mm. If it is subjected to two torques, determine reactions at fixed supports A and B.


5. Torsion

EXAMPLE 5.11 (SOLN)

Equilibrium

From free-body diagram, problem is statically indeterminate.

Mx = 0;

TB + 800 N·m 500 N·m TA = 0

Compatibility

Since ends of shaft are fixed, sum of angles of twist for both ends equal to zero. Hence,

A/B = 0


5. Torsion

EXAMPLE 5.11 (SOLN)

1.8TA 0.2TB = 750

Compatibility

The condition is expressed using the load-displacement relationship, = TL/JG.

. . .

Solving simultaneously, we get

TA = 345 N·m TB = 645 N·m


5. Torsion

*5.6 SOLID NONCIRCULAR SHAFTS

• Shafts with noncircular x-sections are not axisymmetric, as such, their x-sections will bulge or warp when it is twisted

• Torsional analysis is complicated and thus is not considered for this text.


5. Torsion

*5.6 SOLID NONCIRCULAR SHAFTS

• Results of analysis for square, triangular and elliptical x-sections are shown in table


5. Torsion

EXAMPLE 5.13

6061-T6 aluminum shaft shown has x-sectional area in the shape of equilateral triangle. Determine largest torque T that can be applied to end of shaft if allow = 56 MPa, allow = 0.02 rad, Gal = 26 GPa.

How much torque can be applied to a shaft of circular x-section made from same amount of material?


5. Torsion

EXAMPLE 5.13 (SOLN)

By inspection, resultant internal torque at any x-section along shaft’s axis is also T. Using formulas from Table 5-1,

allow = 20T/a3; ... T = 179.2 N·m

allow = 46TL/a3Gal; ... T = 24.12 N·m

By comparison, torque is limited due to angle of twist.


5. Torsion

EXAMPLE 5.13 (SOLN)

Circular x-section

We need to calculate radius of the x-section.Acircle = Atriangle; ... c = 14.850 mm

Limitations of stress and angle of twist require

allow = Tc/J; ... T = 288.06 N·m

allow = TL/JGal; ... T = 33.10 N·m

Again, torque is limited by angle of twist.

Comparing both results, we can see that a shaft of circular x-section can support 37% more torque than a triangular one


5. Torsion

*5.7 THIN-WALLED TUBES HAVING CLOSED CROSS SECTIONS

• Thin-walled tubes of noncircular shape are used to construct lightweight frameworks such as those in aircraft

• This section will analyze such shafts with a closed x-section

• As walls are thin, we assume stress is uniformly distributed across the thickness of the tube


5. Torsion


Shear flow• Force equilibrium requires the

forces shown to be of equal magnitude but opposite direction, thus AtA = BtB

• This product is called shear flow q, and can be expressed as

q = avgt

• Shear flow measures force per unit length along tube’s x-sectional area


5. Torsion


Average shear stress

avg = average shear stress acting over thickness of tube

T = resultant internal torque at x-section

t = thickness of tube where avg is to be determined

Am = mean area enclosed within boundary of centerline of tube’s thickness

avg = T

2tAm


5. Torsion


Average shear stressSince q = avgt, the shear flow throughout the x-section is

q = T

2Am

Angle of twist

Can be determined using energy methods

= ∫ TL

4Am2G

ds

tO


5. Torsion


IMPORTANT• Shear flow q is a product of tube’s thickness and

average shear stress. This value is constant at all points along tube’s x-section. Thus, largest average shear stress occurs where tube’s thickness is smallest

• Both shear flow and average shear stress act tangent to wall of tube at all points in a direction to contribute to resultant torque


5. Torsion

EXAMPLE 5.16

Square aluminum tube as shown.

Determine average shear stress in the tube at point A if it is subjected to a torque of 85 N·m. Also, compute angle of twist due to this loading.

Take Gal = 26 GPa.


5. Torsion

EXAMPLE 5.16 (SOLN)

Average shear stress

Am = (50 mm)(50 mm) = 2500 mm2

avg = = ... = 1.7 N/mm2T

2tAm

Since t is constant except at corners, average shear stress is same at all points on x-section.

Note that avg acts upward on darker-shaded face, since it contributes to internal resultant torque T at the section


5. Torsion

EXAMPLE 5.16 (SOLN)

Angle of twist

Here, integral represents length around centerline boundary of tube, thus

= ∫ TL

4Am2G

ds

tO = ... = 0.196(10-4) mm-1 ∫ dsO

= 0.196(10-4) mm-1[4(50 mm)] = 3.92 (10-3) rad


5. Torsion

5.8 STRESS CONCENTRATION

• Three common discontinuities of the x-section are:

a) is a coupling, for connecting 2 collinear shafts together

b) is a keyway used to connect gears or pulleys to a shaft

c) is a shoulder fillet used to fabricate a single collinear shaft from 2 shafts with different diameters


5. Torsion


• Dots on x-section indicate where maximum shear stress will occur

• This maximum shear stress can be determined from torsional stress-concentration factor, K


5. Torsion


• K, can be obtained from a graph as shown

• Find geometric ratio D/d for appropriate curve

• Once abscissa r/d calculated, value of K found along ordinate

• Maximum shear stress is then determined from

max = K(Tc/J)


5. Torsion


IMPORTANT• Stress concentrations in shafts occur at points of

sudden x-sectional change. The more severe the change, the larger the stress concentration

• For design/analysis, not necessary to know exact shear-stress distribution on x-section. Instead, obtain maximum shear stress using stress concentration factor K

• If material is brittle, or subjected to fatigue loadings, then stress concentrations need to be considered in design/analysis.


5. Torsion

EXAMPLE 5.18

Stepped shaft shown is supported at bearings at A and B. Determine maximum stress in the shaft due to applied torques. Fillet at junction of each shaft has radius r = 6 mm.


5. Torsion

EXAMPLE 5.18 (SOLN)

Internal torque

By inspection, moment equilibrium about axis of shaft is satisfied. Since maximum shear stress occurs at rooted ends of smaller diameter shafts, internal torque (30 N·m) can be found by applying method of sections


5. Torsion

EXAMPLE 5.18 (SOLN)


From shaft geometry, we have

D

d

r

d

2(40 mm)

2(20 mm)

6 mm)

2(20 mm)

= = 2

= = 0.15

Thus, from the graph, K = 1.3

max = K(Tc/J) = ... = 3.10 MPa


5. Torsion

EXAMPLE 5.18 (SOLN)


From experimental evidence, actual stress distribution along radial line of x-section at critical section looks similar to:


5. Torsion

*5.9 INELASTIC TORSION

• To perform a “plastic analysis” for a material that has yielded, the following conditions must be met:

1.Shear strains in material must vary linearly from zero at center of shaft to its maximum at outer boundary (geometry)

2.Resultant torque at section must be equivalent to torque caused by entire shear-stress distribution over the x-section (loading)


5. Torsion


• Expressing the loading condition mathematically, we get:

T = 2∫A 2 dEquation 5-23


5. Torsion


A. Maximum elastic torque• For maximum elastic shear strain Y,

at outer boundary of the shaft, shear-strain distribution along radial line will look like diagram (b)

• Based on Eqn 5-23,

TY = (/2) Yc3

• From Eqn 5-13,

d = (dx/)


5. Torsion


B. Elastic-plastic torque• Used when material starts yielding, and

the yield boundary moves inward toward the shaft’s centre, producing an elastic core.

• Also, outer portion of shaft forms a plastic annulus or ring

• General formula for elastic-plastic material behavior,

T = (Y /6) (4c3 Y3)


5. Torsion


B. Elastic-plastic torquePlastic torque• Further increases in T will shrink the radius of

elastic core till all the material has yielded• Thus, largest possible plastic torque is

TP = (2/3)Y c3

• Comparing with maximum elastic torque,

TP = 4TY / 3

• Angle of twist cannot be uniquely defined.


5. Torsion


C. Ultimate torque• Magnitude of Tu can be determined “graphically”

by integrating Eqn 5-23


5. Torsion


C. Ultimate torque• Segment shaft into finite

number of rings• Area of ring is multiplied

by shear stress to obtain force

• Determine torque with the product of the force and

• Addition of all torques for entire x-section results in the ultimate torque,

Tu ≈ 2 2


5. Torsion


IMPORTANT• Shear-strain distribution over radial line on shaft

based on geometric considerations and is always remain linear

• Shear-stress distribution must be determined from material behavior or shear stress-strain diagram

• Once shear-stress distribution established, the torque about the axis is equivalent to resultant torque acting on x-section

• Perfectly plastic behavior assumes shear-stress distribution is constant and the torque is called plastic torque


5. Torsion

EXAMPLE 5.19

Tubular shaft made of aluminum alloy with elastic - diagram as shown. Determine (a) maximum torque that can be applied without causing material to yield, (b) maximum torque or plastic torque that can be applied to the shaft. What should the minimum shear strain at outer radius be in order to develop a plastic torque?


5. Torsion

EXAMPLE 5.19 (SOLN)

Maximum elastic torqueShear stress at outer fiber to be 20 MPa. Using torsion formula

Y = (TY c/J); TY = 3.42 kN·m

Values at tube’s inner wall are obtained by proportion.


5. Torsion

EXAMPLE 5.19 (SOLN)

Plastic torqueShear-stress distribution shown below. Applying = Y into Eqn 5-23:

TP = ... = 4.10 kN·m

For this tube, TP represents a 20% increase in torque capacity compared to elastic torque TY.


5. Torsion

EXAMPLE 5.19 (SOLN)

Outer radius shear strainTube becomes fully plastic when shear strain at inner wall becomes 0.286(10-3) rad. Since shear strain remains linear over x-section, plastic strain at outer fibers determined by proportion:

o = ... = 0.477(10-3) rad


5. Torsion

*5.10 RESIDUAL STRESS

• Residual stress distribution is calculated using principles of superposition and elastic recovery


5. Torsion

EXAMPLE 5.21

Tube made from brass alloy with length of 1.5 m and x-sectional area shown. Material has elastic-plastic - diagram shown. G = 42 GPa.

Determine plastic torque TP. What are the residual-shear-stress distribution and permanent twist of the tube that remain if TP is removed just after tube becomes fully plastic?


5. Torsion

EXAMPLE 5.21 (SOLN)

Plastic torqueApplying Eqn 5-23,

When tube is fully plastic, yielding started at inner radius, ci = 25 mm and Y = 0.002 rad, thus angle of twist for entire tube is

TP = ... = 19.24(106) N·mm

P = Y (L/ci) = ... = 0.120 rad


5. Torsion

EXAMPLE 5.21 (SOLN)

r = (TPco)/J = ... = 104.52 MPa

Plastic torque

Then TP is removed, then “fictitious” linear shear-stress distribution in figure (c) must be superimposed on figure (b). Thus, maximum shear stress or modulus of rupture computed from torsion formula,

i = (104.52 MPa)(25 mm/50 mm) = 52.26 MPa


5. Torsion

EXAMPLE 5.21 (SOLN)

’P = (TP L)/(JG) = ... = 0.0747 rad

Plastic torque

Angle of twist ’P upon removal of TP is

+ = 0.120 0.0747 = 0.0453 rad

Residual-shear-stress distribution is shown.

Permanent rotation of tube after TP is removed,


5. Torsion

CHAPTER REVIEW

• Torque causes a shaft with circular x-section to twist, such that shear strain in shaft is proportional to its radial distance from its centre

• Provided that material is homogeneous and Hooke’s law applies, shear stress determined from torsion formula, = (Tc)/J

• Design of shaft requires finding the geometric parameter, (J/C) = (T/allow)

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


5. Torsion

CHAPTER REVIEW

• Angle of twist of circular shaft determined from

• If torque and JG are constant, then

• For application, use a sign convention for internal torque and be sure material does not yield, but remains linear elastic

=TL

JG

=T(x) dx

JG∫0

L


5. Torsion

• If shaft is statically indeterminate, reactive torques determined from equilibrium, compatibility of twist, and torque-twist relationships, such as = TL/JG

• Solid noncircular shafts tend to warp out of plane when subjected to torque. Formulas are available to determine elastic shear stress and twist for these cases

• Shear stress in tubes determined by considering shear flow. Assumes that shear stress across each thickness of tube is constant

CHAPTER REVIEW


5. Torsion

CHAPTER REVIEW

• Shear stress in tubes determined from = T/2tAm

• Stress concentrations occur in shafts when x-section suddenly changes. Maximum shear stress determined using stress concentration factor, K (found by experiment and represented in graphical form). max = K(Tc/J)

• If applied torque causes material to exceed elastic limit, then stress distribution is not proportional to radial distance from centerline of shaft


5. Torsion

CHAPTER REVIEW

• Instead, such applied torque is related to stress distribution using the shear-stress-shear-strain diagram and equilibrium

• If a shaft is subjected to plastic torque, and then released, it will cause material to respond elastically, causing residual shear stress to be developed in the shaft

Date post:	27-Oct-2015
Category:	Documents
Upload:	fadlienst
View:	545 times
Download:	46 times

Torsion - RC. Hibbeler

Documents