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STRUCTURAL MECHANICS: CE203Chapter 5
Torsion
Notes are based on Mechanics of Materials: by R. C. Hibbeler,
7th Edition, Pearson
Dr B. Achour & Dr Eng. K. El-kashifCivil Engineering
Department, University of Hail, KSA(Spring 2011)
Chapter 5: Torsion
Chapter 5: Torsion
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Chapter 5: TorsionTorsional Deformation of a Circular
ShaftTorque is a moment that twists a member about its longitudinal
axis.If the angle of rotation is small, the length of the shaft and
its radius will remain unchanged.
Chapter 5: Torsion
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Chapter 5: TorsionThe Torsion FormulaWhen material is
linear-elastic, Hookes law applies. A linear variation in shear
strain leads to a corresponding linear variation in shear stress
along any radial line on the cross section.
Chapter 5: Torsion
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Chapter 5: TorsionThe Torsion FormulaIf the shaft has a solid
circular cross section,
If a shaft has a tubular cross section,
Chapter 5: Torsion
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Chapter 5: TorsionExample 5.2The solid shaft of radius c is
subjected to a torque T. Find the fraction of T that is resisted by
the material contained within the outer region of the shaft, which
has an inner radius of c/2 and outer radius c.Solution:Stress in
the shaft varies linearly, thusThe torque on the ring (area)
located within the lighter-shaded region isFor the entire
lighter-shaded area the torque is
Chapter 5: Torsion
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Chapter 5: TorsionSolution:Using the torsion formula to
determine the maximum stress in the shaft, we haveSubstituting this
into Eq. 1 yields
Chapter 5: Torsion
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Chapter 5: TorsionExample 5.3The shaft is supported by two
bearings and is subjected to three torques. Determine the shear
stress developed at points A and B, located at section aa of the
shaft.Solution:From the free-body diagram of the left segment,The
polar moment of inertia for the shaft isSince point A is at = c =
75 mm,Likewise for point B, at =15 mm, we have
Chapter 5: Torsion
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Chapter 5: TorsionPower TransmissionPower is defined as the work
performed per unit of time.For a rotating shaft with a torque, the
power is
Since , the power equation is
For shaft design, the design or geometric parameter is
Chapter 5: Torsion
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Chapter 5: TorsionExample 5.5A solid steel shaft AB is to be
used to transmit 3750 W from the motor M to which it is attached.
If the shaft rotates at w =175 rpm and the steel has an allowable
shear stress of allow allow =100 MPa, determine the required
diameter of the shaft to the nearest mm.Solution:The torque on the
shaft isSinceAs 2c = 21.84 mm, select a shaft having a diameter of
22 mm.
Chapter 5: Torsion
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Chapter 5: TorsionAngle of TwistIntegrating over the entire
length L of the shaft, we have
Assume material is homogeneous, G is constant, thus
Sign convention is determined by right hand rule, = angle of
twistT(x) = internal torqueJ(x) = shafts polar moment of inertia G
= shear modulus of elasticity for the material
Chapter 5: Torsion
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Chapter 5: TorsionExample 5.8The two solid steel shafts are
coupled together using the meshed gears. Determine the angle of
twist of end A of shaft AB when the torque 45 Nm is applied. Take G
to be 80 GPa. Shaft AB is free to rotate within bearings E and F,
whereas shaft DC is fixed at D. Each shaft has a diameter of 20
mm.Solution:From free body diagram,Angle of twist at C isSince the
gears at the end of the shaft are in mesh,
Chapter 5: Torsion
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Chapter 5: TorsionSolution:Since the angle of twist of end A
with respect to end B of shaft AB caused by the torque 45 Nm,The
rotation of end A is therefore
Chapter 5: Torsion
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Chapter 5: TorsionExample 5.10The tapered shaft is made of a
material having a shear modulus G. Determine the angle of twist of
its end B when subjected to the torque.Solution:From free body
diagram, the internal torque is T.Thus, at x,For angle of
twist,
Chapter 5: Torsion
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Chapter 5: TorsionExample 5.11The solid steel shaft has a
diameter of 20 mm. If it is subjected to the two torques, determine
the reactions at the fixed supports A and B.Solution:By inspection
of the free-body diagram,Since the ends of the shaft are
fixed,Using the sign convention,Solving Eqs. 1 and 2 yields TA =
-345 Nm and TB = 645 Nm.
Chapter 5: Torsion
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Chapter 5: TorsionSolid Noncircular ShaftsThe maximum shear
stress and the angle of twist for solid noncircular shafts are
tabulated as below:
Chapter 5: Torsion
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Chapter 5: TorsionExample 5.13The 6061-T6 aluminum shaft has a
cross-sectional area in the shape of an equilateral triangle.
Determine the largest torque T that can be applied to the end of
the shaft if the allowable shear stress is allow = 56 MPa and the
angle of twist at its end is restricted to allow = 0.02 rad. How
much torque can be applied to a shaft of circular cross section
made from the same amount of material? Gal = 26 GPa.Solution:By
inspection, the resultant internal torque at any cross section
along the shafts axis is also T.By comparison, the torque is
limited due to the angle of twist.
Chapter 5: Torsion
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Chapter 5: TorsionSolution:For circular cross section, we
haveThe limitations of stress and angle of twist then requireAgain,
the angle of twist limits the applied torque.
Chapter 5: Torsion
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Chapter 5: TorsionThin-Walled Tubes Having Closed Cross
SectionsShear flow q is the product of the tubes thickness and the
average shear stress.
Average shear stress for thin-walled tubes is
For angle of twist, = average shear stressT = resultant internal
torque at the cross sectiont = thickness of the tubeAm = mean area
enclosed boundary
Chapter 5: Torsion
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Chapter 5: TorsionExample 5.14Calculate the average shear stress
in a thin-walled tube having a circular cross section of mean
radius rm and thickness t, which is subjected to a torque T. Also,
what is the relative angle of twist if the tube has a length
L?Solution:The mean area for the tube isFor angle of twist,
Chapter 5: Torsion
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Chapter 5: TorsionExample 5.16A square aluminum tube has the
dimensions. Determine the average shear stress in the tube at point
A if it is subjected to a torque of 85 Nm. Also compute the angle
of twist due to this loading. Take Gal = 26 GPa.Solution:By
inspection, the internal resultant torque is T = 85 Nm.The shaded
area isFor average shear stress,
Chapter 5: Torsion
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Chapter 5: TorsionSolution:For angle of twist,Integral
represents the length around the centreline boundary of the tube,
thus
Chapter 5: Torsion
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Chapter 5: TorsionStress ConcentrationTorsional stress
concentration factor, K, is used to simplify complex stress
analysis.The maximum shear stress is then determined from the
equation
Chapter 5: Torsion
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Chapter 5: TorsionExample 5.18The stepped shaft is supported by
bearings at A and B. Determine the maximum stress in the shaft due
to the applied torques. The fillet at the junction of each shaft
has a radius of r = 6 mm.Solution:By inspection, moment equilibrium
about the axis of the shaft is satisfiedThe stress-concentration
factor can be determined by the graph using the geometry,Thus, K =
1.3 and maximum shear stress is
Chapter 5: Torsion
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Chapter 5: TorsionInelastic TorsionConsidering the shear stress
acting on an element of area dA located a distance p from the
center of the shaft,
Shearstrain distribution over a radial line on a shaft is always
linear.Perfectly plastic assumes the shaft will continue to twist
with no increase in torque.It is called plastic torque.
Chapter 5: Torsion
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Chapter 5: TorsionExample 5.20A solid circular shaft has a
radius of 20 mm and length of 1.5 m. The material has an
elasticplastic diagram as shown. Determine the torque needed to
twist the shaft = 0.6 rad.Solution:The maximum shear strain occurs
at the surface of the shaft,The radius of the elastic core can be
obtained byBased on the shearstrain distribution, we have
Chapter 5: Torsion
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