Torsion

©2005 Pearson Education South Asia Pte Ltd

5. Torsion

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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


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CHAPTER OUTLINE1. 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


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• 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


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5.1 TORSIONAL DEFORMATION OF A CIRCULAR SHAFT

• By definition, shear strain is

Let Δx → dx and Δ φ = dφBD = ρ dφ = dx γ

γ = (π/2) − lim θ’C→A along CAB→A along BA

γ = ρdφdx

• Since dφ / dx = γ /ρ = γmax /c

γ = γmaxρc( )Equation 5-2


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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,

τ = τ maxρc( )

...

τ = τ max

c ∫A ρ2 dA


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• 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

τ max =TcJ

τ 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 areac = outer radius pf the shaft


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• Shear stress at intermediate distance, ρ

τ =TρJ

• 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


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

of a differential ring or annulus having thickness dρand circumference 2πρ .

• For this ring, dA = 2πρ dρ

J = c4π2

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


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Tubular shaftJ = (co

4 − ci4)

π2


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


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5.2 THE TORSION FORMULAProcedure 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 = πc4/2• For tube, J = π(co

4 − ci2)/2


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5.2 THE TORSION FORMULAProcedure for analysisShear stress• Specify radial distance ρ, measured from centre

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

• Apply torsion formula, τ = Tρ /J or τmax = Tc/J• Shear stress acts on x-section in direction that is

always perpendicular to ρ


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EXAMPLE 5.3Shaft 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.


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EXAMPLE 5.3 (SOLN)Internal torqueBearing 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.


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EXAMPLE 5.3 (SOLN)Internal torqueΣ Mx = 0; 4250 kN·mm − 3000 kN·mm − T = 0

T = 1250 kN·mmSection property

J = π/2(75 mm)4 = 4.97× 107 mm4

Shear stressSince point A is at ρ = c = 75 mm

τB = Tc/J = ... = 1.89 MPa


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EXAMPLE 5.3 (SOLN)Shear stressLikewise for point B, at ρ = 15 mm

τB = Tρ /J = ... = 0.377 MPa

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


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• Power is defined as work performed per unit of time

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

also express power as

5.3 POWER TRANSMISSION

P = T (dθ/dt)

P = Tω

• 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 ω = 2πf T, then power

P = 2πf TEquation 5-11


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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 T and allowable shear stress for material, τallow and applying torsion formula,


Jc

Tτallow

=


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

2 − ci2) to

determine co and ci



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EXAMPLE 5.5Solid steel shaft shown used to transmit 3750 W from attached motor M. Shaft rotates at ω = 175 rpm and the steel τallow = 100 MPa.Determine required diameter of shaft to nearest mm.


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EXAMPLE 5.5 (SOLN)Torque on shaft determined from P = Tω, Thus, P = 3750 N·m/s

Thus, P = Tω, T = 204.6 N·m

( )ω = = 18.33 rad/s175 rev

min2π rad1 rev

1 min60 s( )

= =Jc

π c4

2 c2

Tτallow. . .

c = 10.92 mmSince 2c = 21.84 mm, select shaft with diameter of d = 22 mm


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5.4 ANGLE OF TWIST

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

φ =T(x) dxJ(x) G∫0

L

φ = angle of twist, in radiansT(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 xG = shear modulus of elasticity for material


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5.4 ANGLE OF TWIST

Constant torque and x-sectional area

φ =TLJG

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:

φ =TLJGΣ


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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


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5.4 ANGLE OF TWISTProcedure 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


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5.4 ANGLE OF TWISTProcedure 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, φ = ∫ (T(x)/J(x)G) dx or φ = TL/JGmust be applied to each segment for which J, Tand G are continuous or constant

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


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EXAMPLE 5.950-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


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EXAMPLE 5.9 (SOLN)Internal torqueFrom free-body diagramΣ Mz = 0; TAB = 100 N(300 mm) = 30 × 103 N·mm


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EXAMPLE 5.9 (SOLN)Internal torqueMagnitude of the uniform distribution of torque along buried segment BC can be determined from equilibrium of the entire post.

Σ Mz = 0;

100 N(300 mm) − t(600 mm) = 0t = 50 N·mm


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EXAMPLE 5.9 (SOLN)Internal torqueHence, from free-body diagram of a section of the post located at position x within region BC, we haveΣ Mz = 0;

TBC − 50x = 0TBC = 50x


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EXAMPLE 5.9 (SOLN)Maximum shear stressLargest shear stress occurs in region AB, since torque largest there and J is constant for the post. Applying torsion formula

τmax = = ... = 1.22 N/mm2TAB c

J


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EXAMPLE 5.9 (SOLN)Angle of twistAngle 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

φA = +TAB LAB

JGTBC dx

JG∫0

LBC

. . .

φA = 0.00147 rad


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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


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

Σ 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

φA/B = 0


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

TA LAC

JGTB LBC

JG− = 0

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

TA = TLBC

L( ) TB = TLAC

L( )


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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 φ = TL/JG

• Solve equilibrium and compatibility equations for unknown torques


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EXAMPLE 5.11Solid steel shaft shown has a diameter of 20 mm. If it is subjected to two torques, determine reactions at fixed supports A and B.


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EXAMPLE 5.11 (SOLN)EquilibriumFrom free-body diagram, problem is statically indeterminate.

Σ Mx = 0;

− TB + 800 N·m − 500 N·m − TA = 0

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

φA/B = 0


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EXAMPLE 5.11 (SOLN)

1.8TA − 0.2TB = −750

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

. . .

Solving simultaneously, we get

TA = −345 N·m TB = 645 N·m


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*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.


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*5.6 SOLID NONCIRCULAR SHAFTS

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


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EXAMPLE 5.136061-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 τallow = 56 MPa, φallow = 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?


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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,

τallow = 20T/a3; ... T = 179.2 N·m

φallow = 46TL/a3Gal; ... T = 24.12 N·m

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


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EXAMPLE 5.13 (SOLN)Circular x-sectionWe need to calculate radius of the x-section.Acircle = Atriangle; ... c = 14.850 mm

Limitations of stress and angle of twist require

τallow = Tc/J; ... T = 288.06 N·m

φallow = 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


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*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


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Shear flow• Force equilibrium requires the

forces shown to be of equal magnitude but opposite direction, thus τAtA = τBtB

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

q = τavgt

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


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Average shear stress

τavg = average shear stress acting over thickness of tube

T = resultant internal torque at x-sectiont = thickness of tube where τavg is to be

determinedAm = mean area enclosed within

boundary of centerline of tube’s thickness

τavg = T

2tAm


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

q = T

2Am

Angle of twistCan be determined using energy methods

φ = ∫TL4Am

2Gdst

O


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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, largestaverage 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


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EXAMPLE 5.16Square 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.


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EXAMPLE 5.16 (SOLN)Average shear stressAm = (50 mm)(50 mm) = 2500 mm2

τavg = = ... = 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 τavg acts upward on darker-shaded face, since it contributes to internal resultant torque T at the section


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EXAMPLE 5.16 (SOLN)Angle of twist

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

φ = ∫TL4Am

2Gdst

O = ... = 0.196(10-4) mm-1 ∫ dsO

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


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5.8 STRESS CONCENTRATION• Three common discontinuities

of the x-section are:a) is a coupling, for connecting

2 collinear shafts togetherb) is a keyway used to connect

gears or pulleys to a shaftc) is a shoulder fillet used to

fabricate a single collinear shaft from 2 shafts with different diameters


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5.8 STRESS CONCENTRATION• Dots on x-section indicate

where maximum shear stress will occur

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


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5.8 STRESS CONCENTRATION• K, can be obtained from

a graph as shown• Find geometric ratio D/d

for appropriate curve• Once abscissa r/d

calculated, value of Kfound along ordinate

• Maximum shear stress is then determined from

τmax = K(Tc/J)


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5.8 STRESS CONCENTRATIONIMPORTANT• 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 fatigueloadings, then stress concentrations need to be considered in design/analysis.


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EXAMPLE 5.18Stepped shaft shown is supported at bearings at Aand B. Determine maximum stress in the shaft due to applied torques. Fillet at junction of each shaft has radius r = 6 mm.


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EXAMPLE 5.18 (SOLN)Internal torqueBy 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


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EXAMPLE 5.18 (SOLN)Maximum shear stressFrom shaft geometry, we have

Ddrd

2(40 mm)2(20 mm)

6 mm)2(20 mm)

= = 2

= = 0.15

Thus, from the graph, K = 1.3

τmax = K(Tc/J) = ... = 3.10 MPa


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EXAMPLE 5.18 (SOLN)Maximum shear stressFrom experimental evidence, actual stress distribution along radial line of x-section at critical section looks similar to:


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*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)


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*5.9 INELASTIC TORSION• Expressing the loading

condition mathematically, we get:

T = 2π∫A τρ 2 dρEquation 5-23


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*5.9 INELASTIC TORSIONA. Maximum elastic torque• For maximum elastic shear strain γY,

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

• Based on Eqn 5-23,

TY = (π/2) τYc3

• From Eqn 5-13,

dφ = γ (dx/ρ)


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*5.9 INELASTIC TORSIONB. 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 = (πτY /6) (4c3 − ρY3)


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*5.9 INELASTIC TORSIONB. 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 = (2π/3)τY c3

• Comparing with maximum elastic torque,

TP = 4TY / 3

• Angle of twist cannot be uniquely defined.


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*5.9 INELASTIC TORSIONC. Ultimate torque• Magnitude of Tu can be determined “graphically”

by integrating Eqn 5-23


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*5.9 INELASTIC TORSIONC. 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 ρ

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

Tu ≈ 2πΣτρ 2Δρ


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*5.9 INELASTIC TORSIONIMPORTANT• 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


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EXAMPLE 5.19Tubular shaft made of aluminum alloy with elastic τ-γdiagram 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?


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EXAMPLE 5.19 (SOLN)Maximum elastic torqueShear stress at outer fiber to be 20 MPa. Using torsion formula

τY = (TY c/J); TY = 3.42 kN·m

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


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EXAMPLE 5.19 (SOLN)Plastic torqueShear-stress distribution shown below. Applying τ = τY 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.


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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:

γo = ... = 0.477(10-3) rad


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*5.10 RESIDUAL STRESS• Residual stress distribution is calculated using

principles of superposition and elastic recovery


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EXAMPLE 5.21Tube made from brass alloy with length of 1.5 m and x-sectional area shown. Material has elastic-plastic τ-γ 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?


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EXAMPLE 5.21 (SOLN)Plastic torqueApplying Eqn 5-23,

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

TP = ... = 19.24(106) N·mm

φP = γY (L/ci) = ... = 0.120 rad


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EXAMPLE 5.21 (SOLN)

τr = (TPco)/J = ... = 104.52 MPa

Plastic torqueThen 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,

τi = (104.52 MPa)×(25 mm/50 mm) = 52.26 MPa


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EXAMPLE 5.21 (SOLN)

φ’P = (TP L)/(JG) = ... = 0.0747 rad

Plastic torqueAngle of twist φ’P upon removal of TP is

+ φ = 0.120 − 0.0747 = 0.0453 rad

Residual-shear-stress distribution is shown.Permanent rotation of tube after TP is removed,


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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, τ = (Tc)/J

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

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


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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

φ =TLJGΣ

φ =T(x) dx

JG∫0L


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


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CHAPTER REVIEW• Shear stress in tubes determined from

τ = 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). τmax = 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


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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

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Torsion