Torsion Testing Experiment

IntroductionTorsion is the twisting action in a member caused by two opposing moments along the longitudinal axis of a member (see diagram below).

Practical Applications- Torsion Bars in Automobiles This twisting can be put to good use in producing torsional springs. “Torsion Bars” on cars are good example of use of this idea and form a compact and simple spring system. A long Spring steel rod or flat bar attached in such a way that one end is anchored while the other is free to twist. One end is fastened to the frame at one end and to a suspension part at the other (see Figure 2). If an arm is attached at right angles, to the free end, any movement of the arm will cause the rod or bar to twist the bar's resistance to twisting provides a spring action. The torsion bar replaces both Coil spring and Leaf springs in some suspension systems.

In other cases torsion deflection can be undesirable. For instance too much twist in a car steering column would make the car uncontrollable.

Give any other application of studying torsion?

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The Torsion TestThe Torsion Test allows the student to investigate the relationships between the moment (torque) applied to a member, the material, member length and torsional deflection. Theoretically it can be proved that:

TJ=GθL

Where:

G = Shear Modulus (pa)T = Applied Torque or moment (Nm)J = Polar Moment of Inertia (m)θ = Angle of twist (rads)L = Effective length of member (m)

Lab ObjectivesThis Lab consists of two experiments. The objectives of each experiment is discussed below

1. To twist two different materials to find shear modulus “G”, and the relationship between the applied moment and the angle of twist

2. To twist a material to see how the length affects the angle of twist.

AssumptionsIn our experiments we take the following assumptions:

1. Material is homogenous2. Circular section remains circular and do not deform.3. A plane section of a material perpendicular to it's longitudinal axis remain plane and do not

deform after the torque is applied.4. Shaft is loaded by a couple or torque in a plane perpendicular to the longitudinal axis of the

plane.5. the torsion of the drill chucks or the cone shaped mandrels is negligible compared to the torsion

of the test bars.6. There is negligible friction between the supporting rod and the chuck.7. Shear stress is proportional to shear strain, it means Hook's Law is applicable.8. In circular shafts subjected to torque shearing strain varies linearly.

τR

=TJ=GθL

Where τ, Shearing stress in MPa,r = Radius of shaft in mm

T = Twisting MomentJ = Polar moment of inertia.G = Modulus of rigidity.θ = Angle of TwistL = Length of the specimen

Safety Warning:Ensure that the chucks are as tight as possible to avoid slipping.

The EquipmentThe device allows the fundamental relationships of elastic deformation associated with torsion to be measured and investigated. It is called torsion tester. In this experiment a load is applied to different test bars and the resulting deformation measured..

The equipment consists of a loading arm with a protractor the is allowed to pivot freely and torsionally fixed end, which can be moved up and down to give various effective lengths. Drill type chucks at each end grip the specimen.

Accessories Provided1. Measuring Tape2. 2 Specimen Rods (Steel, Brass)3. Micrometer Screw Guage4. Weight Hanger5. Total 50 weights of 10g each6. Chuck key

Reference MaterialSome Additional Information:

Shear Moduli:

Steel = 79.6 GPaBrass = 38.0 GPa

Nominal specimen diameter = 3 mm

J= π D4

32 for a circular cross section

Where:D = Diameter of a section

Demo for studentsThe stresses caused by the torsional forces in a material are complicated. Firstly, there are shear stresses along the axis member and secondly, there are compressive and tensile stresses at 45° to the axis. We’ll not calculate it but simply demonstrate to proof that they exist.

Procedure for Shear Stress1. You are given two pieces of pipe one is solid where as the second have a slit along its length2. Grip and twist both pieces and compare.

What do you observe?the slatted piece will twist easily and show the shear forces created

Procedure for Compressive and Tensile Stresses1. You are given a piece of ordinary circular section board chalk.2. Twist it gently until it breaks.

What angle does it breaks?The chalks breaks roughly 45 degrees to the axis (forming a helix) indicating the maximum stress plane (in tension for chalk)

LAB 1a

Objective:1. To prove the general torsion formulae2. To measure the shear modulus of material3. To show how torque and the material type affect the torsional deflection.

Basic Method:1. Measure the diameter if the steel specimen using micrometer screw gauge.2. Set the specimen length to 450 mm.3. Zero the protractor using the pointer arm.4. Tighten the chuck using the chuck keys.5. Add masses in 50g increments to the load arm noting the angle of twist (to the nearest 0.25°) on

the protractor until either the load arm hits the end stop or a maximum load of 500 g is used.

Diameter of specimen rod (d) = 3 mm

Length of Rod (L) = 450 mm

Polar Moment of Inertia (J) = 7.95 x 10-12 m4

Length of Lever arm (r) = 0.1m

Prove that the general torsion formula is correct

S. No M (g) T (N) Deg (°) Θ (rad) G= TL/J θ (GPa)

1 50 0.049 2.5 0.044 63.61

2 100 0.098 5 0.087 63.61

3 150 0.147 7 0.122 68.16

4 200 0.196 9.25 0.161 68.77

5 250 0.245 12.5 0.218 63.61

6 300 0.294 13.5 0.236 70.68

7 350 0.343 16 0.279 69.58

8 400 0.392 17.5 0.305 72.70

9 450 0.441 20 0.349 71.57

10 500 0.491 22 0.384 72.29

Calculate Gavg

Gavg=G1+G2+G3…+G9

9

Gavg = 68.46 GPa

Calculate Percentage Deviation of the G obtained from the G given for this material in your textbook.The reference given above is 79.60 Gpa where as our finding indicate it to be 68.46 GPa

%Deviation= ActualValue – ExperimentalValueActualValue

∗100

%Deviation=79.60– 68.4679.60

∗100=13.99%

Suggest a valid reason if its greater than 5%We have used nominal diameter rather using actual diameter due to unavailability of Micrometer Screw Gauge

Lab 1b

Repeat the experiment for brass and note down the angular deflection against different loads

S. No M (g) T (N)Angular

Displacement Degrees (°)

Θ (rad)

1 50 0.04905 4 0.06981317

2 100 0.0981 8 0.13962634

3 150 0.14715 12.5 0.218166156

4 200 0.1962 16.5 0.287979327

5 250 0.24525 21 0.366519143

6 300 0.2943 25 0.436332313

7 350 0.34335 29 0.506145483

8 400 0.3924 32.75 0.57159533

9 450 0.44145 36 0.628318531

Plot a graph to determine G.

0 0.05 0.1 0.15 0.2 0.25 0.3 0.35 0.4 0.45 0.50

0.1

0.2

0.3

0.4

0.5

0.6

0.7

f(x) = 1.45764036103752 x

Brass Loading and Displacement Curve

Series2

Linear (Series2)

Torque (Nm)

Angu

lar D

ispla

cem

ent (

rad)

θ = L/(GJ)*T

L/(GJ) = 1.4576

G = L/1.5656*J

G = 38.8 GPa

Discuss the value of G obtained in comparison with the value of G given in books.Modulus of Rigidity obtained is very much close to the actual value given in books the percentage error is calculated below

%Deviation= ActualValue – ExperimentalValueActualValue

∗100%

%Deviation=38.0– 38.838.0

∗100%=2.11%

Since the error is within 5% range therefore it is acceptable.

After this experiment, explain how you will determine the selection of rod if you are constrained by the length, maximum possible deflection and maximum loading torque.Using Modulus of Rigidity

Lab 2

ObjectiveTo show how the length of a member affects the torsional deflection.

Procedure1. Measure the diameter of the brass rod specimen using a micrometer.2. Set the brass specimen length to 450mm.3. Zero the protractor using the pointer arm.4. Tighten the chuck using the chuck key.5. Add 400 g to the load arm and 6. Note down the angle of twist (to nearest 0.25°) on the protractor against the Length mentioned

in the table below7. Decrease the length by 50 mm and note down the observations in the table below until 250mm

is reached

Length of Lever Arm (R) = 100 mm

Diameter of the Rod (d) = 3 mm

Polar Moment of Inertia (J) = 7.95 x 10-12 m4

Mass (m)= 400 g

Calculate the Torque AppliedApplied Torque (T) = m x g x R/1000

= 0.3924 Nm

S. No Length (m) Angle (°) Angle (rad)

1 0.45 33 0.5760

2 0.4 29.5 0.5149

3 0.35 26 0.4538

4 0.3 23 0.4014

5 0.25 20 0.3491

Plot a graph of Angle versus Length.

0.2 0.25 0.3 0.35 0.4 0.45 0.50.3000

0.3500

0.4000

0.4500

0.5000

0.5500

0.6000

f(x) = 1.13446401379631 x + 0.0619591884457989

Angular Displacement against Length

Series2Linear (Series2)

Length (m)

Angu

lar D

ispla

cem

ent (

rad)

What do you observe?

As the length increases the angle of deflection increases proportionally

Determine G using the Graph

Θ = L*(T/(GJ))

m = T/GJ = 1.1345

G = 1/m * T/J

G = 43.50 GPa