Laboratory 4 Torsion testing - Suranaree University of ...eng.sut.ac.th/metal/images/stories/pdf/Lab_4Torsion_Eng.pdf · Laboratory 4: Torsion testing ... • Students are able to

Laboratory 4: Torsion testing

Mechanical metallurgy laboratory 431303 1

T. Udomphol

LLaabboorraattoorryy 44

Torsion Testing

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Objectives

• Students are required to understand the principles of torsion testing, practice their

testing skills and interpreting the experimental results of the provided materials

when failed under torsion.

• To determine the maximum shearing stress, shear stress at proportional limit, shear

modulus or modulus of rigidity and relationships between torque and degree of

rotation of the tested materials.

• Students are able to differentiate the ability of materials such as cast iron and brass

to withstand torque prior to torsion failure. Analysis and interpretation of the test

parameters obtained should be carried out in relation to the failure nature of each

material.

• Students are capable of selecting materials for engineering applications associated

with torsion.



T. Udomphol

1. Literature review

In many areas of engineering applications, materials are sometimes subjected to torsion in

services, for example, drive shafts, axles and twisted drills. Moreover, structural applications such as

bridges, springs, car bodies, airplane fuselages and boat hulls are randomly subjected to torsion. The

materials used in this case should require not only adequate strength but also be able to withstand

torque in operation. Even though torsion test is not as universal as tension test and do not have any

standardized testing procedure, the significance lies on particular engineering applications and for the

study of plastic flow in materials. Torsion test is applicable for testing brittle materials such as tool

steels and the test has also been used to determine the forgeability of the materials by means of torsion

testing at elevated temperatures.

Figure 1: Torsion in cylindrical bar [1].

1.1 Torsion test

Generally, torsion occurs when the twisting moment or torque is applied to a member

according to figure 1. The torque is the product of tangential force multiplied by the radial distance

from the twisting axis and the tangent, measured in a unit of N.m. In torsion testing, the relationship

between torque and degree of rotation is graphically presented and parameters such as ultimate

torsional shearing strength (modulus of rupture), shear strength at proportional limit and shear

modulus (modulus of rigidity) are generally investigated. Moreover, fracture surfaces of specimens

tested under torsion can be used to determine the characteristics of the materials whether it would fail

in a brittle or a ductile manner.



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Figure 2: Torsion testing machine [2].

In order to study the response of materials under a torsional force, the torsion test is

performed by mounting the specimen onto a torsion testing machine as shown in figures 2-4, and then

applying the twisting moment till failure. The torque and degree of rotation are measured and plotted

as shown in figure 5. It can be seen that higher torsional force is required at the higher degrees of

rotation. Normally, the test specimens used are of a cylindrical rod type since the stress distribution

across the section of the rod is the simplest geometry, which is easy for the calculation of the stresses.

Both ends of the cylindrical specimen are tightened to hexagonal sockets in which one is fitted to a

torque shaft and another is fitted to an input shaft. The twisting moment is applied by turning the

input handwheel as illustrated in figure 3 to produce torque until the specimen fails.



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Figure 3: Schematic diagram showing details of torsion testing machine [2]

Figure 4: Torsion specimen [2]

When the twisting moment is applied, the torque is reacted by a torque shaft, which moves in

relation to the deflection arm. The movement of the deflection arm is measured by a linear

potentiometer, which is connected to a calibrated TQ digital torque meter to give a readout of the

torque in a unit of Nm or lb.in. The more we turn the input handwheel clockwise to increase the

degree of rotation, the more torque is produced. At the initial stage, the graphical relationship of the

torque and degree of rotation measured is linear as demonstrated in figure 5. The specimen is



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elastically deformed and the recovery of the specimen to its original shape is possible if the specimen

is unloaded. However, if a high degree of rotation is applied passing a proportional limit, the

specimen starts to deform plastically and will not return to its original shape when the input

handwheel is turned anti-clockwise.

The degree of rotation can be read out from two protractors, elastic and plastic range

protractors. First, an elastic range protractor scale is fitted on the input handwheel connected to the

input shaft of the gear box. This protractor scale provides an accurate reading of 0.1o and thus one

revolution represents 6o. A resettable revolution counter is fitted to the gearbox to record the overall

input revolutions. When the twisting moment is large, the second protractor scale is required for the

readouts in the plastic range of deformation. The second protractor is fitted onto the output shaft and

provides a reading of 1o with one revolution representing 360

o.

Figure 5: Relationship between torque and angle of twist [2].

Considering a cylindrical bar with one end being twisted as shown in figure 6, the twisting

moment MT is resisted by the shear stress τ existing across the specimen section. This shear stress is

zero at the center of the bar, increases linearly with its radius and finally reaches its maximum value at

the peripheral of the bar. If the cylindrical bar with a length of L, the twisting moment can be related

to the shear stress as follow



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rL

G

J

MT

τθ== 9(1)

The shear strain, γ, can be calculated from equation 2

L

rθθγ == tan 9(2)

where J is the polar moment of inertia, mm2 or in

2

G is the shear modulus, N/mm2 or lbf/in

2

θ is degree of rotation, radian

r is the radius of the cylindrical bar, mm or in

L is the length of the cylindrical bar, mm or in

τ is the shear stress, N/mm2 or lbf/in

2

Figure 6: Torsion of a solid bar [3].

According to the graphical relationship of torque and degree of rotation, we can notice that

the torsion specimen deformed elastically and then plastically similar to the case of the tension tested

specimen. The initial stage of elastic behavior shows a linear relationship of torque and degree of

rotation with its slope representing the shear modulus of the modulus of rigidity, G. The stress at the

proportional limit is frequently determined at 0.04 rad.m-1 of the gauge length. Beyond the

proportional limit, specimen deformed in a plastic manner and the relationship between the torque and

the degree of rotation is no longer linear. However the determination of the proportional limit carried

out using a torsional specimen of a thin-wall tube type will provides a more accurate value in

comparison to that obtained from a cylindrical rod type specimen. Since the stresses vary across the

section of the specimen from the center toward the peripheral of the specimen as mentioned



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previously, the reduced effect of stress distribution in the thin-walled specimen is therefore beneficial

for the calculation of the stress. Within the elastic range of deformation, the shear stress can be

calculated according to equation 1

J

rMT=τ 9(3)

For a solid cylindrical specimen, the polar moment J = πD4/32, we can therefore determine

the shear stress as shown in equation 4

34

16

32/

2/

D

M

D

DMTT

ππτ == 9(4)

For a tube specimen, the maximum shear stress at the peripheral of the tube can be calculated

from equation 5

)(

164

2

4

1

1

DD

DMT

−=

πτ 9(5)

where D1 is the outer diameter of the tube

D2 is the inner diameter of the tube

Therefore, if the torque and the degree of rotation are known according to the experimental

result, the shear stress and the shear strain can be determined from equations 2 and 4. The obtained

information is then used for the construction of the graphical relationship between the modular shear

stress (16MT/aD

3) and the shear strain (θr/L) as illustrated in figure 7. The curve is somewhat similar

to those typical stress-strain curves tested under tension, giving elastic and plastic ranges with respect

to the torsional stress applied.

Nevertheless, the calculated shear stress according to equation 4 is only suitable for the

evaluation of the stresses in the elastic range. The plastic stress obtained from the shear stress-shear

strain curve is therefore larger than the real stress. Furthermore, in the case that there is a large

amount of plastic deformation involved, the length of the specimen is considerably changed, which

can result in the superposition of the longitudinal stresses on the torsional shear stresses. Even



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though, the former is considered to be small and can be neglected, they might also affect the torsional

failure strain of the specimen. Nadai [3] has proposed the method for the calculation of the stress in

the plastic range from a known torque-degree of rotation curve as expressed in equation 6

3

max

2

3

a

M

u

πτ = 9(6)

where Mmax

is the maximum twisting moment

a is the distance from the center to the peripheral of the specimen according

to figure 4.

Figure 7: Relationship between modular shear stress and shear strain [2].

1.2 Types of torsion failures

Torsion failures are different from tension failures and normally provide little deformation or

elongation. The characteristic of the fracture surface is related to the state of stress at the point on the

bar surface, which can be described as shown in figure 8. It can be seen that the maximum shear

stresses exist along two planes, which are perpendicular to each other. One is perpendicular to the

longitudinal axis (yy) and another is aligned parallel to the longitudinal axis (xx). The principle



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stresses σ1 and σ

3 are inclined at 45

o to the longitudinal axis and have their magnitudes equal to those

of the shear stresses. The principle stress σ1 is tensile while the principle stress σ

3 is compressive.

The intermediate stress σ2 is zero under torsion.

As mentioned previously, the characteristics of torsion fractures are influenced by torsional

and tensile forces. These result in two types of torsion failures; 1) ductile failure due to the shear

stresses and 2) brittle failure due to the tensile stresses. The former produces the fracture surface

along the plane of the maximum shear stress and more frequently normal to the longitudinal axis as

shown in figures a) and b). The latter exhibits the fracture planes normal to the directions of the

tensile stresses, which are 45o to the longitudinal axis. Figure 9 shows brittle and ductile fracture

surfaces observed from a driveshaft and a brass rod respectively failed under torsion. This means the

driveshaft failed under a brittle manner influenced by the tensile stresses while the ductile failure of

the brass rod was affected by the shear stresses. However, if the specimen fails into little pieces

where the fracture plane cannot be determined, the fracture in this case is considered to start along the

maximum shear stress parallel to the longitudinal axis of the specimen. In addition, it has also been

shown in a study, which has indicated that the fracture characteristic is associated with the hardness of

the tool steels. If the tool steel possesses the hardness values of higher than 720 VHN, failure is

expected to be influenced by tensile stress. Conversely, if the hardness of the tool steel is lower than

720 VHN, the shear stress is responsible for such a failure.

Figure 8: Types of failure in torsion [3].



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Figure 9: Fracture surfaces of a driveshaft and a brass-rod failed in brittle and ductile

manners under torsion.

1.3 Calibration of the torsion test machine

Prior to torsion testing, calibration of the testing machine should be carried out as follows

1.3.1 Put the calibration arm onto the square end of the torque shaft and level it by adjusting

the handwheel as demonstrated in figure 10. Zero the dial gauge.

1.3.2 Set the digital meter for torque measurement to SI unit and adjust the readout to zero by

turning the knob at the rear of the device.

1.3.3 Add a 5 kg load to the calibration arm and adjust the dial gauge to zero by rotating the

handwheel. The reading on the digital meter should be 24.5± 0.5 Nm. If the error is higher than 0.5

Nm, the CAL screw at the rear of the instrument should be adjusted to set the reading back to 24.5

Nm. (Note: the calibration arm = 500 mm, load = 5 kg, thus the torque = 5 x 9.81 x 0.5 = 24.5 Nm).

1.3.4 Remove the load and check that the reading now returns to zero.

1.3.5 The full range calibration can be carried out by putting the available weights of 500 g, 1

kg and 2 kg incrementally while the reading of torque of each increment should be recorded. The

obtained values should be within the tolerance of ± 0.5 Nm of the calculated values.

1.3.6 Plot the graph of the torque reading against the applied torque (0.5 x load x 9.81),

giving a linear relationship. Calculate the slope which should be approximately unity.



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Figure 10: Calibration of the torsion testing machine [2].



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2. Materials and equipment

2.1 Test specimens

2.2 Micrometer or vernia caliper

2.3 Permanent pen

2.4 Torsion testing machine

3. Experimental Procedure

3.1 Measure initial diameter, initial length and initial gauge length of the specimen. Record these

parameters on the table provided.

3.2 Draw a line using a permanent pen along the length of the test specimen. This line will help

to notice the degree of rotation during applying the twisting moment.

3.3 Calibrate the testing equipment following section 1.3.

3.4 Grip the test specimen on to the torsion testing machine using hexagonal sockets and make

sure the specimens are firmly mounted. Fit both ends of the specimen to input and torque

shafts and set reading on the torque meter to zero.

3.5 Start twisting the specimen at strain increment of 0.5o until failure occurs. Record the

received data rotation in the table provided for the construction of torque and degree

relationship

3.6 Construct the relationship between shear stress and shear strain. Determine maximum shear

stress, shear stress at proportional limit and modulus of rigidity.

3.7 Sketch fracture surfaces of failed specimens and described their natures in the table provided.

3.8 Discuss and conclude the obtained experimental results.



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4. Results

Details Cast iron Brass

Initial diameter (mm)

Final diameter (mm)

Initial gauge length (mm)

Final gauge length (mm)

Initial overall length of specimen (mm)

Final overall length of specimen (mm)

Strain increment (degree)

Maximum torque (Nm)

Maximum shear stress (MPa)

Shear stress at proportional limit (MPa)

Modulus of rigidity (GPa)

Fracture surface details

Table 1 Experimental results for torsion test of cast iron and brass.



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5. Discussion

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6. Conclusions

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7. Questions

7.1 Did cast iron and brass fail in different fracture manners? Explain.

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7.2 What are the differences between torsion and tension tests?

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7.3 Which engineering applications do you think that torsion test is vital? Give three examples.

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8. References

8.1 www.doitpoms.ac.uk

8.2 SM1/2 Torsion testing machine, TecQuipment Ltd 2000.

8.3 Dieter, G.E., Mechanical metallurgy, 1988, SI metric edition, McGraw-Hill, ISBN 0-07-

100406-8.

Documents

Laboratory 4 Torsion testing - Suranaree University of ...eng.sut.ac.th/metal/images/stories/pdf/Lab_4Torsion_Eng.pdf · Laboratory 4: Torsion testing ... • Students are able to