DERIVATION OF TORSION THEORY

In this section we will discuss the effect of applying a torsional loading to a long straight member such as shaft or tube. We will consider the member to have circular cross section and show the derivation of the torsion formula.

ASSUMPTIONS

The material is homogenous and isotropic and obeys Hooke’s law Sections that are originally plane to the torsional axis remains plane after

deformation Initially straight radii remain straight after deformation Angle of twist per unit length along the bar is constant

Fig 1-1a&b show physically what happens when a torque is applied to a circular bar to illustrate the above assumptions made. Twisting causes the circles to remain circles and each longitudinal grid lines deforms into a helix that intersects the circles at equal angles. Also cross sections at the ends remain flats i.e they do not wrap or bulge in or out.

Figure 1-1

RELATION BETWEEN TORQUE AND SHEAR STRESS

If the shaft is fixed at one end and a torque is applied at the other end as shown in Fig 1-2, radial line located on the cross-section at a distance x will rotate through an angle (x). This angle is defined as the angle of twist depends on the position of x and varies along the shaft as shown.

Consider a small element located at a distance from the axis of the shaft (Figure 1-3). Due to the deformation of the element , there will be shear strain .

BD = = x

Thus, γ=ρ

ΔφΔx =

ρdφdx (1-1)

If we let = d and x = dx

At any points on the cross section and x are the same, then

dφdx is a constant.

Eqn 1-1 states that the magnitude of the shear strain for any elements varies only with the radial distance from the axis of the shaft (Fig 1-4). Hence, the shear stress, vary from zero at the shaft centre to maximum value max at its outer surface as shown in Figure 1-5

On any bar cross section, the resultant stress distribution must be equal to the applied torque T (see Fig 1-6) That is

T=∫ ρ (τ dA )=∫ ρ( ρc

τmax )dA

(1-2)

The integration proceeds over the entire area of the cross section. At any given section, the maximum shearing stress max

and the distance c from the centre are constant. Hence the above expression can be written as

T =

τmax

c∫ ρ2 dA

(1-3)

Fig 1-2

Fig 1-3

in which ∫ ρ2 dA = J is the polar moment of inertia of the cross-section.

For a solid circular section of radius c ,

J = ∫0

C

ρ22 d=¿c4/2.

Thus,

Equation 1-4 is the torsion formula for circular bars. The shear stress at a distance from the centre is :

Fig 1-4

max=

TcJ (1-4)

=

TρJ (1-5)

Figure 1-5

This transverse shear stress obtained by equation (1-4) or (1-5) is accompanied by a longitudinal shear stress of equal value as shown in the figure (Fig 1-5b).

In many practical applications, shafts are used to transmit power. From dynamics, power transmitted P by a constant torque T rotating at a constant angular speed is given by

(1-6)

Where P in watt, T in N-m and is measured in radian per sec.

If the shaft is rotating at rev per minutes (rpm), then = 2 /60.

Exercises

1. Determine the torque carrying capacity of a shaft, 50 mm diameter if its shear strength is limited to 56 MPa.

2. A solid shaft of radius R is subjected to torque T, determine the fraction of the torque T that is resisted by the material contained within the outer region of the shaft which has an inner radius of ½ R and outer radius R

3. A tubular shaft (hollow shaft) having inner diameter 30 mm and outer diameter 42 mm is to transmit 90 kW of power. Determine the frequency of rotation (no. of revolution per minute, rpm) so that the shear stress will not exceed 50 MPa.

See also Hibbeler chapter 5

P = T.

RELATION BETWEEN TORQUE AND ANGLE OF TWIST

According to Hooke’s law, max = max/G; introducing the torsion formula eqn 1-4, we obtain

max=

TcGJ (1-7)

Where G is the modulus of rigidity or shear modulus. For small deformation

max=

cφL (1-8)

where L is the length of the bar

From equation 1-6 and 1-7 lead to the angle of twist, the angle through which one cross section of a circular bar rotates with respect to another:

(1-9)

Angle is measured in radians. The product GJ is termed as torsional rigidity of the member. Note that eqn 1-4 through 1-9 are valid for both solid and hollow circular bars.

Combining Equations 1-5 and 1-9, the torsion formula derived as:

(1-10)

Note:

The polar moment of inertia, J for hollow circular section with inner radius = Ci and outer radius = Co

J=∫C i

C o

ρ22 πρdρ=2 πρ4

4 |Ci

Co

=π(co

4−c i4 )

2

TJ= τ

ρ=G

∅L

=

T LG J

EXAMPLE.

A solid shaft in a rolling mill transmits 20 kW at 120 rpm. Determine the diameter of the shaft if the shearing stress is not to exceed 40 MPa and the angle of twist is limited to 6o in a length of 3 m. Use G = 83 GPa.

Solution

This problem illustrates a design that must possess sufficient strength as well as rigidity.

First determine the torque from eqn (1-9)

T =

60 P2π =

(60 )( 20 x103 )(2 π ) (120 ) = 1591.6 N-m

To satisfy the condition of strength we apply the torsion formula Eq (1-4)

max=

TcJ

40x 106 =

1591 .5 c

πc4

2 =

From which c3 = 2.53x 10-5 , and c = 0.0294 m = 29.4 mm

We next apply the angle of twist relation, Eq (1-8)

=

T LG J , 6/180 =

1591 .5 (3)

(83 x109 )πc4

2

From which, c = 24.3 mm

The larger size, c = 29.4 mm, will satisfy both strength and rigidity.

Gauge length

Test specimen. The heads or ends could be square or hexagonal

FUNDAMENTAL ASPECT OF TORSION TEST

Torsion tests are carried out to determine the mechanical properties of engineering materials such as modulus of rigidity G, yield shear strength, ultimate shear strength, modulus of rupture in shear and ductility.

Torsion test can also be carried out on full-size parts such as shafts, axles, drills, etc. In torsion test large strains can be applied to the sample before plasticity, unlike tension and compression tests. Because of this, torsion test is most preferred to obtain flow test data and the workability of the material.

Torsion tests are most carried out on prismatic bars of circular cross-section by applying the torque about the longitudinal axis. The shear modulus vs the shear strain can be determined from the torque and the angle of twist of the test specimen over a predetermined gauge length.

TEST STANDARD

The standard testing procedures for the determination of surface shear are given in ASTM E 143. “Standard Test Method for Shear Modulus at Room temperature.”

The expected load or torque T is calculated based on the torsion theory using established ultimate shear strength value ( see Table 1-1). In preparing the specimens, the machining operations should be chosen carefully to prevent changes in the surface structure of the material. Cast materials are recommended to be tested as received without surface machining. The torsion test is best carried out with on uniform cylindrical bars with length of 8 to 10 diameters with gauge length at least two diameters long and a distance equal to two diameters from each end. Usually the tests are carried out to failure or to a predetermined load.

The test machine and fixture must ensure the specimen is mounted and aligned free of any axial and bending loads at the start of the test and the rate of load application during

the test must be constant as in tensile testing. Square, triangular or hexagonal ends are most common when test are to be run under displacement control to prevent changes in gauge section that may result from axial extension. Also hexagonal ends has added advantage of allowing rapid specimen removal for purposes of quenching.

In torsion test, the torque and the deformation (angle of twist) are recorded using suitable instrumentations. There are many types of devices or sensors to measure the deformation. Ideal testing allows measurement to be taken at the gauge section of the specimen.

A typical shear stress-strain plot is as shown below. The shear stress is determined via eqn 1.4 while the shear strain,, obtained from the angle of twist via Eqn 1-7.

DATA REPORTING

Materials Conditions.Composition of the test piece, thermomechanical history, dimensions, and surface conditions should be reported.

Test ConditionsSuch as test speed, duration of testing (whether to failure or not) instrumentation and environmental conditions should be noted. The nature of fracture (brittle, spiral convolution, buckling, tearing, etc) and parameters defining failure (eg load drop, large deformation, etc) should be recorded.

Data InterpretationThe objective of the experiment is to obtain reliable data and results. However, is not unusual to have scatter in data for almost identical specimen. Table below lists typical ultimate shear stress values in torsion for some materials. Modulus data are subjected to interpretation since most materials do not show a completely straight plot. The yield point must also be arbitrary defined. Most frequently obtained from the shear stress-strain plot by the intersect of the secant line drawn from the initial slope.

Table 1-1. Typical values of ultimate shear stress in torsion (Source: Metal handbook,9th edi, vol 8)

Material ConditionShear stressMPa Ksi

Wrought iron . . . 262 38Gray iron G4000 345 50Ductile iron D7003 414 60SAE 1020 as hot

rolled358.5 52

SAE 1095 normalised 724 105SAE 4140 normalised 772 112AISI 302 stainless steel annealed 448 65Aluminium 2024-T4 283 41Aluminium 6061-T6 214 31Aluminium 360-F 221 3285% red brass C23000 207 - 448 30 - 6565% yellow brass C27000 227 - 276 33 - 40