MechanicsMechanics && MaterialsMaterials 11

ChapterChapter 1111 TorsionTorsion

FAMU-FSU College of Engineering Department of Mechanical Engineering TorsionTorsion

• Torsion refers to the twisting of a structural member when it is loaded by couples that produce rotation about the longitudinal axis

• The couples that cause the tension are called Torques, Twisting Couples or Twisting Moments TorsionTorsion ofof CircularCircular BarBar • From consideration of symmetry: – Cross sections of the circular bar rotates as rigid bodies about the longitudinal axis – Cross sections remain straight and circular TorsionTorsion ofof CircularCircular BarsBars

• If the right hand of the bar rotates through a small angle φ – φ: angle of twist • Line will rotate to a new position

• The element ABCD after applying torque b → b′ → Element in state of “pure d → d′ shear” ShearShear StrainStrain

• During torsion, the right-hand cross section of the original configuration of the element (abdc) rotates with respect to the opposite face and points b and c move to b' and c'. • The lengths of the sides of the element do not change during this rotation, but the angles at the corners are no longer 90°. Thus, the element is undergoing pure shear and the magnitude of the shear strain is equal to the decrease in the angle bab'. This angle is bb′ tan γ = ab ShearShear StrainStrain γγ tan γ ≈ γ because under pure torsion the angle γ is small. So ′ φ γ = bb = rd ab dx Under pure torsion, the rate of change dφ/dx of the angle of twist is constant along the length of the bar. This constant is equal to the angle of twist per unit length θ. dφ → rate of change of the angle of twist φ dx φ d = Θ dx → Angle of twist per unit length ShearShear StrainStrain

ρ φ γ = d = ρΘ dx •Since every cross section is subjected to the same torque so dφ/dx is constant, the shear strain varies along the radial line

γ = 0 at ρ = 0 (center of shaft)

γ = maximum at ρ = C (outer surface of shaft) φ γ d = dx ρ ShearShear StrainStrain since The shear strain within the shaft varies d φ = const linearly along any radial line, from dx zero at the axis of the shaft to a φ γ maximum to at its outer surface d = max dx C hence φ γ γ d = = max dx ρ C  ρ  γ =  γ  C  max PurePure TorsionTorsion In the case of pure torsion the rate of change d φ dx is constant along the length of the bar, because every cross section is subjected to the same torque. Therefore, we obtain

φ θ = , where L is the length of the shaft L Equation γ = r θ becomes r φ γ = r θ = L THEORYTHEORY OFOF TORSIONTORSION FORMULAFORMULA • The following conditions are used in the torsion of the circular shaft:

1. Sectional planes perpendicular to the axis of the shaft remain plane during torque application. 2. The shear strain varies linearly from a value of zero at the axis of the shaft to a maximum at the extreme radius . 3. For linearly elastic materials, Hook's Law is applicable and shear stress is linearly proportional to shear strain. TorsionTorsion FormulaFormula

•If the external torque applied to shaft in equilibrium requires internal torque inside the shaft Hook’s Law of Torsion

τ = Gγ = Gρθ τ : shear stresss G : shear modulus ( modulus of rigidity) γ : Shear strain TorsionTorsion Formula:Formula: ShearShear StressStress

• From Hook’s law for shear, if the material behavior is linear-elastic then a linear variation in shear strain leads to linear variation in shear stress. So the shear stress for a solid shaft will vary from zero at the shafts longitudinal axis to a τ maximum value max at its  ρ  outer surface such that: τ =  τ  c  max TorsionTorsion Formula:Formula: ShearShear StressStress

• The shear stress depends on both the Torque and the cross section Moment of inertia; for circular shafts that moment of inertia is referred to as polar Moment of Inertia, J ρ T.c τ = T . τ = J max J T : The resultant internal torque acting at the cross section J : Polar moment of Inertia of cross section ρ : The radius measured from the center of the shaft :c The outer radius of the shaft PolarPolar MomentMoment ofof InertiaInertia :: JJ

• For Solid Shaft π J = c 4 2

• For Tubular Shaft π J = (c4 −c4 ) 2 o i • J is always positive • Units for J (length) 4: m 4, mm 4, in 4, etc. PowerPower TransmissionTransmission

• Shafts mainly used to transmit mechanical power from one machine to another • Power (P) : work performed per unit time. • During instant of time dt an applied torque T will cause the shaft to rotate d θ so dθ P = T = T.ω dt • ω: shafts angular velocity( rad/s) PowerPower TransmissionTransmission

• Power units: SI: Watt (N.m/s), fps: (ft.lb/s) • 1horsepower=550 ft.lb/s • The shafts angular speed is given in terms of frequency f which is a measure of the number of revolutions or cycles the shaft makes per second ω = 2π f ⇒ P = 2π fT • In terms of the number of revolutions per minute(rpm) n = 60 f 2πnT P = (n = rpm ) 60 ShaftShaft DesignDesign

• From the power transmitted by the P shaft and its frequency of rotation we T = can find the torque , 2 π f

• Knowing T and the allowable shear τ stress for the material; allow we can determine the size of the shaft’s cross J = T τ section using the torsion formula, c allow assuming linear elastic behavior

π • The design geometric parameter is solid shaft → J = c 4 J/c, J: polar moment of inertia, for 2 solid shaft and c is the radius, for π → = 4 − 4 hollow shaft J (co ci ) hollow shaft we need c in and c out , 2 so we assume one and find the other AngleAngle ofof TwistTwist

• The formula for angle of twist is derived by examining an element of the shaft which is dx (infinitesimal) in length. It is known that an arc length equals the subtended angle times the radius of the arc: d = φr • A point on the radius on one end of the element will travel a relative distance, i.e. create an arc length, equal to the angle γ max dx. φ = T.L JG AngleAngle ofof TwistTwist

• As shown in the figure, this distance is also equal to the radius time the angle φ, where φ measures the angle of twist. • The angle of twist φ = T . L JG • G is the modulus of rigidity (shear modulus), J is Polar moment of inertia NonuniformNonuniform TorsionTorsion

A. Discrete Shaft

φ =φ +φ + = T1L1 + T2L2 + 1 2 ...... G1J1 G2J2

n TL
= ∑ i i = i 1 Gi Ji The signs of the torques should be observed B.Varying cross section shaft T dx L L T dx dφ = x ⇒ϕ = ∫ dφ = ∫ x GJ 0 0 GJ x StaticallyStatically IndeterminateIndeterminate TorsionalTorsional Members:Members: UniformUniform ShaftShaft

• Statically Indeterminate Torsional Members: When the moment equation of equilibrium applied about the axis of the shaft is not adequate to determine the unknown torques acting on the shaft. = → − − = ∑ M x 0 T TA TB 0 One Equilibriu m Eq   → Statically Indetermin ate Two Unknowns :TA &TB  StaticallyStatically IndeterminateIndeterminate TorsionalTorsional Members:Members: UniformUniform ShaftShaft

To solve The problem we transform the compatibility equation into an equation which involves the torques • Compatibility Condition: angle of twist of one end of the shaft w.r.t the the other end=0, since the supports are fixed φ = → φ − φ = A / B 0 A / C B / C 0

T L T L A AC − B BC = 0 JG JG T L = B BC substitute T A into the moment equation L AC L L T = T BC = T BC A + L BC L AC L L L T = T Ac = T AC B + L BC L AC L SuperpositionSuperposition ofof TorqueTorque

• We can solve the earlier statically indeterminate problem utilizing the superposition principle, using one torque at each time, while considering the other torque as redundant SuperpositionSuperposition ofof TorqueTorque

First we load the shaft by appling the external torque T then, the redundant torque T B is applied to unlaod the shaft ′ = φ − φ 0 B B TL T L  L  0 = AC − B ⇒ T = T  AC  JG JG B  L  From the moment equation = − − = ∑ M x o ⇒ T T A TB 0 L − L L T = T AC = T BC A L L L T = T AC B L StaticallyStatically IndeterminateIndeterminate TorsionalTorsional Members:Members: NonuniformNonuniform ShaftShaft • From static's equilibrium + = Ta Tb T0 • Using the superposition method, selecting the torque Tb as redundant, so that the released structure is obtained by removing support B StaticallyStatically IndeterminateIndeterminate TorsionalTorsional Members:Members: NonuniformNonuniform ShaftShaft • Compatibility condition φ = φ − φ = → b ac cb 0 because end B is fixed

φ = T0a − Tba − Tbb = b 0 GI pa GI pa GI pb = I p : polar moment of inertia J using the compatabil ity to write T b as a function of To , then substitute in the equilibriu m equation t o solve for Ta bI aI T = T pa and T = T pb a 0 + b 0 + aI pb bI pa aI pb bI pa StaticallyStatically IndeterminateIndeterminate TorsionalTorsional Members:Members: CompositeComposite ShaftShaft

• A composite bar is made of concentric, circular torsional bars that are firmly bonded together to act as a single member. • If the tube and the core have different materials properties then the bar is statically indeterminate StaticallyStatically IndeterminateIndeterminate TorsionalTorsional Members:Members: CompositeComposite ShaftShaft Assuming that the composite bar is acted upon by a total torque T , which is resisted by torques T a and T b developed in the core and tube respectively. = + Equilibriu m :T Ta Tb Compatibil ity : Angle of twist φ,must be the same for both T L T L parts so they htold together φ = a = b Ga J a Gb Jb G J solving → T = T a a a + Ga J a Gb Jb G J T = T b b b + Ga J a Gb Jb The angle of twist φ T L T L TL φ = φ = a = b ⇒= + Ga J a Gb Jb Ga J a Gb Jb ExampleExample :: CompositeComposite ShaftsShafts

• A composite aluminum and steel shaft is used to transmit a torque T=6000 π in.-lb as shown. The two materials are assumed to act as a unit, meaning no relative motion occurs between the aluminum and steel portions at their common interface. • Determine (a) the resisting torque in the aluminum and in the steel (b) the angle of twist of the free end relative to the fixed end, and (c) the maximum stress in the aluminum and in the steel SolutionSolution

•Geometric compatibility: ϕ = ϕ ST AL T l T l ST ST = AL AL J ST G ST J AL G AL so

= J ST G ST T ST T AL J AL G AL

π 15 π J = in 4 and J = in 4 •Now ST 32 AL 32 1 So that J / J = , Also ST AL 15 G 12 × 10 6 ST = = 3 × 6 G AL 4 10 Consequent ly 1 T = T ST 5 AL SolutionSolution M = :0 T +T = 6000 π •Equilibrium: ∑ Z AL ST

T = 5000 πin. lb-  Solution (a) AL = π  TST 1000 in. lb-   ϕ = T AL l  J G  AL AL  Solution (b): The angle of twist ϕ = T ST l   J ST G ST 

5000 π )1( ϕ = = 0 .267 rad 15 π ( 4 × 10 6 ) 32 Solution,Solution, cont..cont..

Solution c: max Shearing stress

5000 π )1( τ = = 10 ,667 psi (aluminum) max 15 π 32 π τ = 1000 )5.0( = max π 16000 psi s( teel ) 32 Example:Example: ShaftShaft DesignDesign

• A solid alloy shaft of 50mm in diameter is to be friction welded concentrically to the end of a hollow steel shaft of the same external diameter. Find the internal diameter of the shaft if the angle of twist per unit length is to be 75% that of the alloy shaft. • What is the maximum torque that can be transmitted if the limiting shear stresses in the alloy and the steel are 50MN/m and 75MN/m respectively.

•Gsteel = 2.2G alloy SolutionSolution •Equilibrium:

Talloy = T steel = T θ θ Ts = Ta •Geometry of deformation: s = 75.0 a 75.0 JsGs JaGa Ls La Since = = T s T a and G s 2.2 G a

= J a 2.2 * 75.0 J s π d 4 π a = 2.2 * 75.0 ( D 4 − d 4 ) 32 32 s s 4 = 4 − 4 50 2.2( * 75.0 * 50 ) 2.2( * 75.0 * d s ) 65.0 * 50 4 d 4 = s 2.2 * 75.0 = d s 39 6. mm SolutionSolution The torque that can be carried by the alloy is: πd 3 π × 50 3 T = τ = *50 *10 6 = 1227 Nm 16 16 *10 9

The torque that can be carried by the steel is: π 50( 4 * 39 6. 4 ) T = * 75 *10 6 = 1120 Nm 16 50 *10 9

The maximum torque allowable is 1120 N.m Example:Example: StaticallyStatically IndeterminateIndeterminate TorsionalTorsional ElementElement • The composite shaft consists of a steel section 25mm in diameter and an aluminum section 50mm in diameter. The ends of the shaft are fixed so that rotation cannot occur there. • Determine (a) the resisting torques exerted by the supports on the shaft and (b) the maximum stress in the aluminum and the max. stress in the steel. SolutionSolution

•Equilibrium: = + = π ∑ M Z :0 TA TB 200 ϕ •Geometric compatibility: A/B = 0 π ϕ = T B )3.0( + T B )3.0( − 200 )2.0( B / A π π π .0( 025 ) 4 84( × 10 9 ) 05.0( ) 4 28( × 10 9 ) 05.0( ) 4 28( × 10 9 ) 32 32 32 = π Fulfilling the compatibility TB 21 05. N * m = π condition we find T B, then T A 178 95. N * m from equilibrium we find T A (178 95. π )( .0 025 ) (τ ) = = 22 90. MPa AL max π •Finally; using the Torsion 05.0( ) 4 32 formula, we find the max 21( 05. π )( .0 0125 ) (τ ) = = 21 55. MPa ST max π shearing stresses .0( 025 ) 4 32 Example:Example: TorqueTorque

• A circular shaft rigidly clamped at its left and free to rotate in a frictionless bearing at its right end.A 16in diameter pulley and a 12in. Diameter pulley are keyed to the shaft as shown. Cables attached to the pulleys exert the forces shown. • The internal or resisting torque

TR is required for the intervals (0

Find the reactions:

For 0 ≤ z ≤ 2 = − − = ∑ M z :0 TR 1000 16( ) 1500 12( ) 0 or = TR 34000 in lb-.

For 2 ≤ z ≤ 6 = − = ∑ M z :0 TR 1500 12( ) 0 or = TR 18000 in. lb- ExampleExample

• Determine the maximum shear stress and rate of twist of the given shaft if a 10 kN.m torque is applied to it. • If the length of the shaft is 15 m, how much would it rotate by? Let G = 81 GPa, D = 75 mm AnswerAnswer

4 4 πD π .0( 075 ) − J = = = .3 106 × 10 6 m 4 32 32 T .r 10 × 10 3 × .00375 τ = = = 120 7. MPa J .3 106 × 10 − 6 dθ T 10 × 10 3 = = = .0 03974 rad / m dx GJ 81 × 10 9 × .3 106 × 10 − 6 Example:Example: PowerPower TransmissionTransmission

• A pump is connected to an electric motor through steel shafting as shown. An offset necessary to connect the pump to its power source is provided by the gear arrangement as shown. • If the motor delivers 100 πHP at 330rpm at its shaft, determine (a) the maximum shearing stress in the shaft and (b) the relative rotation of sections A and C. • G=12*10 6psi for steel SolutionSolution

550 12( 60) T = 100 π = 60000 in. lb- AB 2π (330 ) T 60000 )5.1( τ = C = = 11318 psi max J 7.952 Calculate R: = = ∑ M Z :0 5R 60000 R =12000lb

From FBD: = = = − ∑ M Z :0 TBC 12000 )3( 36000 in . lb 36000 )5.1( τ = = 6791 psi max .7 952 SolutionSolution

ϕ = ϕ +ϕ A/ C A/ B

T l 60000 (120 ) ϕ = AB AB = = .0 07545 rad A/ B JG .7 952 12( ×10 6 ) T l and 5 ϕ = 3ϕ , where ϕ = BC BC bC BC JG 3 36000 (144 ) ϕ = = .0 03259 rad 5 .7 952 12( *10 6 )

ϕ = + = = AC .0 074545 .0 03259 .0 108 rad 19.6 deg Example:ShaftExample:Shaft DesignDesign

• The steel shaft shown is required to transmit 20 πHP at 5.5Hz. If the allowable angle of twist per meter of the shaft is not to exceed 4.5 degrees, and if the allowable shearing stress is not to exceed 84MPa, calculate the minimum permissible diameter. SolutionSolution

P 745 20(7. π ) The torque that the shaft T = = = 1355 8. N .m must transmit is 2πf 2π )5.5( Tc 1355 (8. d / )2 Strength requirement|: τ = ⇒ 84 × 10 6 = allow π from the design criteria J d 4 32 Stiffness requirement:We calculate the diameter such that the allowable angle of twist is not to be exceeded allowable angle of twist per unit length = 5.4 deg = .0 07854 rad φ = × allow 5 .07854 rad TL 1355 8. ×5 φ = ⇒ × = ⇒ = .0 07854 5 π d 38 mm Hence, the minimum JG d 4 ×10 6 permissible diameter is 43.5 32 mm Example:Example: NonuniformNonuniform TorsionTorsion

• A solid steel shaft ABCD having diameter d=3in. turns freely in a bearing at D and is loaded at B and C by torques T 1 =20in.-kips and T 2 =12in.-kips. The shaft is connected in the gear box at A to gears that are temporarily locked in position. • Determine the maximum shear stress in each part of the shaft and the angle of twist at and D.

• Assume L 1 = 20in., L 2 = 30in., L3 = 20in., and G=11500ksi SolutionSolution First find the torque in T = T +T = 32 in. - kips T = T =12 in. - kips T = 0 each part of the shaft ab 1 2 bc 2 cd

•The corresponding maximum 16 T 16 (32000 in. - lb) τ = ab = = 6040 psi stresses are as follows: ab πd 3 π (3.0in.) 3 16 T 16 (12000 in. - lb) τ = bc = = 2260 psi bc πd 3 π (3.0in.) 3 τ = cd 0

n T L 1 •The angle φ = i i = + + ∑ (Tab L1 Tbc L2 Tcd L3 ) = G I GI of twist: i 1 i i p 1 = [( 32 in. - k)(20in.) + (12in. - k)(30in) + 0] (11500 ksi )( π )( 0.3 in .) 4 32 = 0.0109rad = 0.627 o Example:Example: NonuniformNonuniform TorsionTorsion

• A tapered bar AB of solid circular cross-section is twisted by torques T applied at the ends. The diameter of the bar caries uniformly from d a on the left end to d b on the right end. • Derive the formula for the angle of twist of the bar. SolutionSolution

•Diameter d at distance x from end A is: d − d x d = d + b a x x a L

πd 4 π  d − d 4 •The polar moment of inertia: I = x = d + b a x px 32 32  a L 

•With constant T x L Tdx the expression of φ = ∫ the angle of twist 0 GJ becomes 32 TL  1 1  =  −  π −  3 3  3 G(db da )  da db  Example:NonuniformExample:Nonuniform ShaftShaft

• A step-shaft constructed from aluminum(G=12*10 6psi)bar stock carries the torsional loads shown. • Determine the angle of twist of the section at A relative to the section at D. Assume that all materials behave in a linearly elastic manner. • Solve using (a)discrete element procedure and , (b) the superposition procedure. Solution:Solution: DiscreteDiscrete ElementElement ProcedureProcedure

ϕ =ϕ +ϕ +ϕ A/ D A/ B B/C C/ D π π − π = (1200 20) + (1200 15) + ( 2000 15) π π π 12( ×10 6 ) 4( ×10 6 ) 4( ×10 6 ) 32 2 2 = .0 0640 + .0 0090 − .0 0150 = .0 058 rad = .3 323 deg rees SolutionSolution :: SuperpositionSuperposition PrinciplePrinciple

π π ϕ′ = (1200 20) + (1200 30) A/ D π π 12( ×10 6 ) 4( ×10 6 ) 32 2 = .0 0640 + .0 0180 = 0.082rad

− π ϕ ′′ = + + ( 3200 15) A/D 0 0 π 4( ×10 6 ) 2 = -0.0240rad

ϕ′ +ϕ ′′ = − = A/D A/ D .0 082 .0 024 .0 058 rad ExampleExample TorsionTorsion FormulaFormula

• Calculate the maximum shearing stress that occurs at any point in each of the three segments AB, BC, and CD of the shaft in the previous Example. SolutionSolution

τ = T R r J  1200 π )5.0(  τ = = 19200 psi 0 ≤ z ≤ 20 max π   32  1200 π )0.1(  τ = = 2400 psi 20 ≤ z ≤ 35 max π   2  − 2000 π )0.1(  τ = = −4000 psi 35 ≤ z ≤ 50  max π  2  Example:Example: StaticallyStatically IndeterminateIndeterminate TorsionalTorsional ElementElement • The composite shaft consists of a steel section 25mm in diameter and an aluminum section 50mm in diameter. The ends of the shaft are fixed so that rotation cannot occur there. • Determine (a) the resisting torques exerted by the supports on the shaft and (b) the maximum stress in the aluminum and the max. stress in the steel. SolutionSolution

•Equilibrium: = + = π ∑ M Z :0 TA TB 200 ϕ •Geometric compatibility: A/B = 0 π ϕ = T B )3.0( + T B )3.0( − 200 )2.0( B / A π π π .0( 025 ) 4 84( × 10 9 ) 05.0( ) 4 28( × 10 9 ) 05.0( ) 4 28( × 10 9 ) 32 32 32 = π Fulfilling the compatibility TB 21 05. N * m = π condition we find T B, then T A 178 95. N * m from equilibrium we find T A (178 95. π )( .0 025 ) (τ ) = = 22 90. MPa AL max π •Finally; using the Torsion 05.0( ) 4 32 formula, we find the max 21( 05. π )( .0 0125 ) (τ ) = = 21 55. MPa ST max π shearing stresses .0( 025 ) 4 32