5.TORSION

INTRODUCTION

In the preceding chapters we studied how to calculate the

stresses and strains in structural members to axial loads, that is,

to forces directed along the axis of the member. In this chapter

structural members and machine parts that are in torsion will be

considered. More specifically, we will analyze the stresses and

strains in members of circular cross section subjected to twisting

couples or torques. These couples have a common magnitude T,

and opposite sences. They are vector quantities and can be

represented by curved arrows as in figure (a), or by couple

vectors as in figure(b).

Members in torsion are encountered in many engineering

applications. The most common application is provided by

transmission shafts, which are used to transmit power from one

point to another. These shafts can be either solid, or hollow.

1 Beer&Johnston

An important property of circular shafts should be noted: when a

circular shaft is subjected to torsion, every cross section remains

plane and undistorted. In other words, while the various cross

sections along the shaft rotate through different amounts, each

cross section rotates as a solid rigid slab.(when a bar of square

cross section is subjected to torsion, its various cross sections

warp and do not remain plane.)

Consider a circular shaft that is attached to a fixed support at

one end. If a torque T is applied to the other end, the shaft will

twist. Here it is seen that a radial line located on the cross

section at a distance x from the fixed end of the shaft will rotate

through an angle ∅(𝑥). The angle ∅ 𝑥 ,so defined, is called the

angle of twist. It depends on the position x and will vary along

the shaft as shown.

2

5.1 TORSIONAL DEFORMATION OF A CIRCULAR SHAFT

Beer&Johnston

We will now determine the distribution of shearing strains in a circular

shaft of length L and radius c which has been twisted through an angle

∅(𝑥). Detaching from the shaft a cylinder of radius ρ, we consider the

small square element formed by two adjacent circles and two adjacent

straight lines traced on the surface of the cylinder before any load is

applied. As the shaft is subjected to a torsional load, element deforms

into a rhombus (figure c).

We now recall from previous sections that the shearing strain γ in a

given element is measured by the change in the angles formed by the

sides of that element. Since the circles defining two of the sides of the

element considered here remain unchanged, the shearing strain γ must

be equal to the angle between lines AB and A´B. We can express the arc

length AA´ as AA´= Lγ. But on the other hand, we have AA´= ρ. It

follows that Lγ = ρ, or

𝛾 =𝜌∅

𝐿 and 𝛾𝑚𝑎𝑥 =

𝑐∅

𝐿

Where γ and are both expressed in radians. The equation obtained

shows, that the shearing strain γ at a given point of a shaft in torsion is

proportional to the angle . It is also shows that γ is proportional to the

distance ρ from the axis of the shaft to the point under consideration.

Thus the shearing strain in a circular shaft varies linearly with the

distance from the axis of the shaft

3

Beer&Johnston

5.2 THE TORSION FORMULA

Recalling the Hooke’s law for shearing stress and strain we write,

where G is the modulus of rigidity or shear modulus of the material. Using previous equation and hooke’s law we can write,

𝜏 =𝜌

𝑐𝜏𝑚𝑎𝑥 (1)

The equation expresses the shear-stress distribution as a function of the radial position ρ of the element; in other words, it defines the stress distribution in terms of the geometry of the shaft. We recall that the sum of the elementary forces exerted on any cross section of the shaft must be equal to the magnitude T on the torque exerted on the shaft:

𝜌 𝜏𝑑𝐴 = 𝑇 (2)

Substituting for 𝜏 from (1) into (2) we write

𝑇 = 𝜌 𝜏𝑑𝐴 =𝜏𝑚𝑎𝑥𝑐 𝜌2 𝑑𝐴

4 Beer&Johnston

THE TORSION FORMULA

𝑇 = 𝜌 𝜏𝑑𝐴 =𝜏𝑚𝑎𝑥𝑐 𝜌2 𝑑𝐴

The integral in this equation depends only on the geometry of the shaft. It represents the

polar moment of inertia of the shaft’s cross-sectional area computed about the shaft’s

longitudinal axis. We will symbolize its value as J, and therefore the above equation can be

written in a more compact form, namely

𝜏𝑚𝑎𝑥 =𝑇𝑐

𝐽

where

𝜏𝑚𝑎𝑥 = the maximum shear stress in the shaft, which occurs at the outer surface

T = the resultant internal torque acting at the cross section. Its value is determined from

the method of sections and the equation of moment equlibrium applied about the shaft’s

longitudinal axis

J = the polar moment of inertia of the cross-sectional area

c = the outer radius of the shaft

The shear stress at the intermediate distance ρ can be determined from a similar equation

𝜏 =𝑇𝜌

𝐽

Either of the two above equations is often referred to as the torsion formula. 5

6

Polar moment of inertia for solid cross section: Polar moment of inertia for tubular cross section:

𝐽 =𝜋

2𝑐4 𝐽 =

𝜋

2𝑐24 − 𝑐1

4

Typical elements, when isolated from the shaft in figure, must be subjected to shear stresses directed as

shown. Hence, not only does the internal torque T develop a linear distribution of shear stress along

each radial line in the plane of the cross-sectional area, but also an associated shear-stress distribution is

developed along an axial plane.

Example 5.1(beer&johnston)

7

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?

Example 5.2(hibbeler)

8

The pipe shown in the figure has an inner diameter of 80 mm and an outer diameter of

100 mm. If its end is tightened against the support at A using a torque wrench at B,

determine the shear stress developed in the material at the inner and outer walls along

the central portion of the pipe when the 80-N forces are applied to the wrench.

9

The torsion formulas were derived for a shaft of uniform

circular cross section subjected to torques at its ends. However,

they can also be used for a shaft of variable cross section or for a

shaft subjected to torques at locations other than its ends.

The distribution of shearing stresses in a given cross section S of

the shaft is obtained from torsion formula, Tc/J, where J denotes

the polar moment of inertia of that section, and where T

represents the internal torque in that section. The value T is

obtained by drawing the free-body diagram of the portion of

shaft located on one side of the section and writing that the sum

of torques applied to that portion, including the internal torque T,

is zero.

If the absolute maximum torsional stress is to be determined,

then it becomes important to find the location where the ratio

Tc/J is maximum. Beer&Johnston

10

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.

Example 5.3 (Beer&Johnston)

11

The solid 30-mm-diameter shaft is used to transmit the torques applied to

the gears. Determine the absolute maximum shear stress on the shaft.

Example 5.4 (Beer&Johnston)

12

The motor delivers a torque of 50 N.m to the shaft AB. This torque is transmitted to

shaft CD using the gears at E and F. Determine the equilibrium torque T’ on shaft CD

and the maximum shear stress in each shaft. The bearings B, C, and D allow free

rotation of the shafts.

Example 5.5 (Hibbeler)

5.3 POWER TRANSMISSION

Shafts and tubes having cross sections are often used to transmit power developed by a machine. When used for this

purpose they are subjected to torques that depend on the power generated by the machine and the angular speed of

the shaft. Power is defined as the work performed per unit of time. The work transmitted by a rotating shaft equals

the torque applied times the angle of rotation. Therefore, if during an instant of time dt an applied torque T will

cause the shaft to rotate dθ, then the instantaneous power is,

𝑃 =𝑇𝑑𝜃

𝑑𝑡

Since the shaft’s angular velocity 𝜔 = 𝑑𝜃/𝑑𝑡, we can also

express the power as,

𝑃 = 𝑇𝜔

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POWER TRANSMISSION

In the SI system, power is expressed in watts when torque is measured in newton-meters (N.m) and ω is in radians per

second (rad/s). In the FPS system, the basic units of power are foot-pounds per second; however, horsepower (hp) is

often used in engineering practice, where 1 ℎ𝑝 = 550 ft.lb/s

For machinery, the frequency of a shaft’s rotation, f, is often reported. This is a measure of the number of revolutions or

cycles the shaft makes per second and is expressed in hertz (1 Hertz = 1 cycle/s). Since 1 cycle = 2𝜋 rad, then 𝜔 = 2𝜋𝑓

and the above equation for power becomes

𝑃 = 2𝜋𝑓𝑇

Shaft design

When the power transmitted by a shaft and its frequency of rotation are known, the torque developed in the shaft

can be determined from the equation above, that is 𝑇 = 𝑃/2𝜋𝑓. Knowing T and the allowable shear stress for

the material, 𝜏𝑎𝑙𝑙𝑜𝑤 , we can determine the size of the shaft’s cross section using the torsion formula, provided the

material behavior is linear-elastic. Specifically, the design or geometric parameter J/c becomes

𝐽

𝑐=

𝑇

𝜏𝑎𝑙𝑙𝑜𝑤

For a solid shaft, 𝐽 = 𝜋/2 𝑐4, and thus, upon substitution, a unique value for the shaft’s radius c can be

determined. If the shaft is tubular, so that design permits a wide range of possibilities for the solution since an

arbitrary choice can be made for either, 𝑐0 or 𝑐𝑖 and the other radius is determined from the torsion formula.

14

15

The drive shaft AB of an automobile is to be designed as a thin-walled tube. The

engine delivers 150 hp when the shaft is turning at 1500 rev/min. Determine the

minimum thickness of the shaft’s wall if the shaft’s outer diameter is 2.5 in. The

material has an allowable shear stress of 𝜏𝑎𝑙𝑙𝑜𝑤 = 7 ksi

Example 5.6 (Hibbeler)

16

The motor A develops a power of 300 W and turns its connected pulley at

90 rev/min. Determine the required diameters of the steel shafts on the

pulleys at A and B if the allowable shear stress is 𝜏𝑎𝑙𝑙𝑜𝑤 = 85 MPa.

Example 5.7 (Hibbeler)