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Problem Set [Professor Video][Professor Note] [Faculty Video][Faculty Note]

Knowing that the internal diameter of the hollow shaft shown is d = 0.9 in., determine the

maximum shearing stress caused by a torque of magnitude T = 9 kip . in. [Ans: 12.44 ksi]

1.

The torques shown are exerted on pulleys A and B. Knowing that each shaft is solid,

determine the maximum shearing stress (a) in shaft AB, (b) in shaft BC. [Ans: (a) 56.6 Mpa

(b) 36.6 Mpa]

2.

The allowable stress is 50 MPa in the brass rod AB and 25 MPa in the aluminum rod BC.

Knowing that a torque of magnitude T = 125 N . m is applied at A, determine the required

diameter of (a) rod AB, (b) rod BC. [Ans: (a) 50.3mm, (b) 63.4 mm]

3.

Strength of Materials/ Unit 7/ Module 1 Torsion I

Subject/Unit Name/Module Name http://10.20.3.1:8080/engineering/I_YEAR/EM/M26...

1 of 17 Friday 19 August 2011 10:02 AM

Additional Problems:

The preliminary design of a large shaft connecting a motor to a generator calls for the use of

a hollow shaft with inner and outer diameters of 4 in. and 6 in., respectively. Knowing that

the allowable shearing stress is 12 ksi, determine the maximum torque that can be

transmitted (a) by the shaft as designed, (b) by a solid shaft of the same weight, (c) by a

hollow shaft of the same weight and of 8-in. outer diameter. [Ans: (a) 408 kip.in (b) 211

kip.in (c) 636 kip.in]

1.

(a) Determine the torque that can be applied to a solid shaft of 20-mm diameter without

exceeding an allowable shearing stress of 80 MPa. (b) Solve part (a), assuming that the

solid shaft has been replaced by a hollow shaft of the same cross-sectional area and with

an inner diameter equal to half of its own outer diameter. [Ans: (a) 125.7 N.m (b) 181.4 N.m]

2.

Under normal operating conditions, the electric motor exerts a torque of 2.8 kN . m on shaft

AB. Knowing that each shaft is solid, determine the maximum shearing stress in (a) shaft

AB, (b) shaft BC, (c) shaft CD. [Ans: (a) 81.2 Mpa (b) 64.5 Mpa (c) 23.0 Mpa ]

3.



The solid rod BC has a diameter of 30 mm and is made of an aluminum for which the

allowable shearing stress is 25 MPa. Rod AB is hollow and has an outer diameter of 25

mm; it is made of a brass for which the allowable shearing stress is 50 MPa. Determine (a)

the largest inner diameter of rod AB for which the factor of safety is the same for each rod,

(b) the largest torque that can be applied at A. [Ans: (a) 15.18 mm (b) 132.5 Nm]

4.

Examples:

Shaft BC is hollow with inner and outer diameters of 90 mm and 120 mm, respectively.

Shafts AB and CD are solid and of diameter d. For the loading shown, determine (a) the

maximum and minimum shearing stress in shaft BC, (b) the required diameter d of

shafts AB and CD if the allowable shearing stress in these shafts is 65 MPa.

1.



Solution:

Equations of Statics. Denoting by TAB

the torque in shaft AB, we pass a section through

shaft AB and, for the free body shown, we write

∑Mx = 0: (6 kN . m) – T

AB = 0 T

AB = 6 kN . m

We now pass a section through shaft BC and, for the free body shown, we have

∑Mx = 0: (6 kN . m) + (14 kN . m) – T

BC = 0 T

BC = 20 kN . m

a. Shaft BC. For this hollow shaft we have

Maximum Shearing Stress. On the outer surface, we have

Minimum Shearing Stress. We write that the stresses are proportional to the distance from

the axis of the shaft.



b. Shafts AB and CD. We note that in both of these shafts the magnitude of the torque is T =

6 kN . m and τall = 65 MPa. Denoting by c the radius of the shafts, we write

c3 = 58.8 X 10-6 m3 c = 38.9 X 10-3 m d = 2c = 2(38.9 mm) d = 77.8 mm

Consider a long tube of 20 mm outside diameter, d0, and of 16mm inside diameter, di,

twisted about its longitudinal axis with a torque T of 40 N.m. Determine the shear

stresses at the outside and the inside of the tube; see Figure.

2.

Solution:

From Eq. 5,

and from Eq. 3,

Similarly from Eq. 4,



In a thin-walled tube, all of the material works at approximately the same stress level.

Therefore, thin-walled tubes are more efficient in transmitting torque than solid shafts. Such

tubes are also useful for creating an essentially uniform "field" of pure shear stress needed

for stablishing τ - γ relationships. To avoid local buckling, however, the wall thicknesscannot be excessively thin.

A hollow cylindrical steel shaft is 1.5 m long and has inner and outer diameters

respectively equal to 40 and 60 mm. (a) What is the largest torque that can be applied

to the shaft if the shearing stress is not to exceed 120 MPa? (b) What is the

corresponding minimum value of the shearing stress in the shaft?

3.

Solution:

(a) Largest Permissible Torque. The largest torque T that can be applied to the shaft is

the torque for which τmax

= 120 MPa. Since this value is less than the yield strength for

steel, we can use Eq. (3) Solving this equation for T, we have

Recalling that the polar moment of inertia J of the cross section is given by Eq. (5),

where c1 = 1/2(40 mm) = 0.02 m and c

2 = 1/2(60 mm) = 0.03 m, we write

Substituting for J and τmax

into (1), and letting c = c2 = 0.03 m, we have



= 4.08 kN . m

(b) Minimum Shearing Stress. The minimum value of the shearing stress occurs on

the inner surface of the shaft, which expresses that τmin

and τmax

are respectively

proportional to c1 and c

2:

Faculty Notes

1. Basic Assumptions for Circular Members:

To establish a relation between the internal torque and the stresses it sets up inmembers with circular solid and tubular cross sections, it is necessary to make twoassumptions, the validity of which will be justified later. These, in addition to thehomogeneity of the material, are as follows:

A plane section of material perpendicular to the axis of a circular memberremains plane after the torques are applied, i.e., no warp-age or distortion ofparallel planes normal to the axis of a member takes place.

1.

In a circular member subjected to torque, shear strains vary linearly from the

central axis reaching at the periphery. This assumption is illustrated in Fig.

1 and means that an imaginary plane such as DO1O3C moves to when

the torque is applied. Alternatively, if an imaginary radius O3C is considered

fixed in direction, similar radii initially at O2B and O

1D rotate to the respective

new positions and . These radii remain straight. It must be emphasizedthat these assumptions hold only for circular solid and tubular members. For thisclass of members, the assumptions work so well that they apply beyond the limitof the elastic behavior of a material.

2.



If attention is confined to the linearly elastic material, Hooke’s law applies, and, itfollows that shear stress is proportional to shear strain. For this case completeagreement between experimenta1ly determined and computed quantities isfound with the derived stress and deformation formulas based on theseassumptions. Moreover their validity can be rigorously demonstrated by themethods of mathematical theory of elasticity.

3.

2. The Torsion Formula:

In the elastic case, on the basis of the previous assumptions, since is stress isproportional to strain, and the latter varies linearly from the center, stresses varylinearly from the central axis of a circular in member. The stresses induced by theassumed distortions are shear stresses and lie in the plane parallel to the sectiontaken normal to the axis of a rod. Thu variation of the shear stress follows directly fromthe shear-strain assumption and the use of Hooke’s law for shear. This is i1lustrated;in Fig. 2. Unlike the case of an axially loaded rod, this stress is not of uniform intensity.The maximum shear stress occurs at points most remote from the center O and is

designated . These points, such as points C and D in Figs. 1 and 2, lie at theperiphery of a section at a distance c from the center. For linear shear stress variation,

at any arbitrary point at a distance ρ from O, the shear stress is (ρ/C) . Theresisting torque can be expressed in terms of stress once the stress distribution atsection is established. For equilibrium this internal resisting torque must equal theexternally applied torque T. Hence,

where the integral sums up all torques developed on the cut by the infinitesimal forcesacting at a distance ρ from a member’s axis, O in Fig. 2, over the whole area A of the

cross section, and where T is the resisting torque.At any given section, and C areconstant; hence, the previous relation can be written as



(1)

However, , the polar moment of inertia of a cross-sectional area, is also a

constant for a particular cross-sectional area. It will be designated by J in this text. For

a circularsection, dA = 2πρ dρ, where 2πρ is the circumference of an annulus with a

radius ρ of width dρ. Hence,

or pascals (Pa) in SI units, or

or psi in the U.S. customary units.

A more general relation than Eq. 3 for a shear stress, , at any point a distance ρ fromthe center of a section is

(4)

Equations 3 and 4 are applicable with equal rigor to circular tubes, since thesame assumptions as used in the previous derivation apply. It is necessary, however,

i.e., (2)

where d is the diameter of a solid circular shaft. If c or d is measured in millimeters, J

has the units of mm; if in inches, the units become in4.

By using the symbol J for the polar moment of inertia of a circular area, Eq. 1 may bewritten more compactly as

(3)

This equation is the well-known torsion formula for circular shafts that expresses themaximum shear stress in terms of the resisting torque and the dimensions of amember. In applying this formula, the internal torque T can be expressed in newton-

meters, Nm, or inch-pounds, c in meters or inches, and J in m4 or in in

4. Such usage

makes the units of the torsional shear stree



to modify J. For a tube, as may be seen from Fig. 3, the limits of integration for Eq. 2extend from b to c. Hence, for a circular tube,

or stated otherwise: J for a circular tube equals +J for a solid shaft using the outerdiameter and –J for a solid shaft using the inner diameter.

For very thin tubes, if b is nearly equal to c, and c-b=t, the thickness of the tube, Jreduces to a simple approximate expression.

(5)

(6)where R

av =(b+c)/2, which is sufficiently accurate in some applications.



3 Remarks on the Torsion formula

So far the shear stresses as given by Eqs. 3 and 4 have been thought of as actingonly in the plane of a cut perpendicular to the axis of the shaft. There indeed they areacting to form a couple resisting the

If a circular bar is made from two different materials bonded together, as shown inFig. 4(a), the same strain assumption applies as for a solid member. For such a case,through Hooke’s law, the shear-stress distribution becomes as in Fig. 4(b). If the shearmodulus for the outer stiffer tube is G

1 and that of the inner softer core is G

2, the ratio

of the respective shear stresses on a ring of radius OB is G1/G

2.

Procedure Summary

For the torsion problem of circular shafts the three basic concepts of engineeringmechanics of solids as used above may be summarized in the following manner:

1. Equilibrium conditions are used for determining the internal resisting torques at asection.2. Geometry of deformation (kinematics) is postulated such that shear strain varieslinearly from the axis of a shaft.3. Material properties (constitutive relations) are used to relate shear strains to shearstresses and permit calculation of shear stresses at a section.

These basic concepts are used for determining both stresses and angles-of-twistof circular shafts. However, similar to the case for axially loaded bars, large localstresses arise at points of application of concentrated torques or changes in crosssection. According to Saint-Venant’s principle the stresses and strains are accuratelydescribed by the developed theory only beyond a distance about equal to thediameter of a shaft from these locations. Typically local stresses are determined byusing stress concentration factors.



externally applied torques. However, tounderstand the problem further, aninfinitesimal cylindrical element, shown inFig. 5(b), is isolated.

The shear stresses acting in theplanes perpendicular to axis of the rod areknown from Eq. 4. Their directions coincidewith the direction of the internal torque.(This should be clearly visualized by thereader.) On adjoining parallel planes of adisc-like element, these stresses act inopposite directions. However, these shearstresses acting in the plane of the cutstaken normal to the axis of a rod cannotexist alone. Numerically, equal shearstresses must act on the axial planes (suchas the planes aef and bcg in Fig. 5(b)) tofulfill the requirements of static equilibriumfor an element.

Shear stresses acting in the axialplanes follow the same variation inintensity as do the shear stresses in theplanes perpendicular to the axis of the rod.This variation of shear stresses on themutually perpendicular planes is shown inFig. 5, where a portion of the shaft hasbeen removed for the purposes of illustration.

Shear stresses can be transformed into an equivalent system of normal stressesacting at angles of 45 degree with the shear stresses. Numerically, these stresses arerelated to each other in the following manner: τ = σ1 = -σ2. Therefore, if the shearstrength of a material is less than its strength in tension, a shear failure takes place ona plane perpendicular to the axis of a bar; see Fig. 6. This kind of failure occursgradually and exhibits ductile behavior.



Alternatively, if the converse is true,i.e. , σ1 < τ, a brittle fracture is causedby the tensile stresses along a helixformatting an angle of 45 degree withthe bar axis; see Fig. 6. A photograph ofa ductile fracture of a steel specimen isshown in fig. 7, and that of a brittlefracture for cast iron in Fig.8. Anotherexamples of a brittle fracture forsandstone is shown in Fig. 9.

The stress transformation broughtinto the previous discussion, since itdoes not depend on material properties,is also applicable to anisotropicmaterials. For example, wood exhibitsdrastically different properties ofstrength in different directions. Theshearing strength of wood on planes parallel to the grain is much less than on planesperpendicular to the grain. Hence, although equal intensities of shear stress exist onmutually perpendicular planes, wooden shafts of inadequate size fail longitudinallyalong axial planes. Such shafts are occasionally used in the process industries.

Professor Note:

Torision:Torisional stress

Torsional stress is a shear stress

Earlier we studied



1) Shear strain =

Shear strain =

2) Element in a state of simple shearFor equilibrium of rectangle couple q about AD is always accompained by the couple

q1 about AB such that

for equilibrium of ABCD couples or shear

stress q=q1

Principal of complementary shear stress in a state of simple shear planes of shear

stress DD1AA1 perpendicular to plane ABB1A1

Pure torsion of cylindrical shafts / bars:

Any element in pure torsion has shear stress on perpendicular faces by the

complementary shear principal.

A shaft in torsion twists as an effect of torsion:

Displacement of twist BC,



Section change:

At a radius r< R, shear stress q<fs

Shear stress is max at R zero at centre

Torsional Moment of resistance :



For a material strength fs is fixed a C.S of shaft having heigher Z

p (polar modulus) has

higher torsion with standing capacity.

CIp is for torsional shaft similar to EI for beam in bending CI

p is called torsional rigidity.

CIp is the torque required to twist 1 radian per unit length of shaft.

Section change:

Power transmitted by shaft

shaft rotating at N r p m. P is the shaft power transmitted in kilowatts causing a mean

torque of T, Nm, on shaft



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