California State University, Long Beach
Mechanical and Aerospace Engineering Department
MAE 361: Materials and Properties Laboratory
Torsion Test of Metals and Polymers

Date: October 29, 2013
Submitted To: Dr. Parvin Shariat
Submitted By:
Ryan Kim D. Lim - 008142015
Steven Reid - 007777066
Thomas Leininger- 008403601
Marco Evangelista – 007774076

Objective:
The objective of this experiment is to determine the shear stress, shear strain, and shear modulus of elasticity of solid cylindrical specimens of steel, brass, aluminum, and a polymer. Procedure:
1. Plug the cord for the machine into the outlet hanging from the ceiling.
2. Zero the drum angle using start, forward/reverse, and speed on machine control. When start is triggered, you must specify the direction of movement, either forward or reverse. There are values on the drum which will indicate when you zero it out.
3. Measure diameter of the sample using a caliper and make sure to measure three different areas, preferably the two ends and the middle for an accurate measurement. Record the average of the three measurements.
4. Mount the specimen in the self tightening chucks of the machine:
a. Slide weighing chuck horizontally of the machine and open jaws of the chuck to permit insertion of the specimen.
b. Insert the specimen into the loading chuck and then slide the weighing chuck assembly to the right until it is stopped by the specimen ends. Move the weighing chuck around ½ inch away from one the ends and place specimen midway between the chucks. Then tighten the jaws of the chuck so that each pin is on the corresponding reference lines. Before starting the torsion test, measure the specimen from jaw to jaw.
5. Before starting the test, you must press clear & step so you can replace idle on the bottom right of the screen to running. Start the test using start/stop and forward switch. The speed knob was halfway when our group conducted the experiment so that it was not too fast or too slow.
12-14° is the point where the graph should deviate from linearity or sometimes it can occur before that degree of twist. Once this happens, you must stop the test and then repeat the test in the reverse direction so that it can return back to zero.
6. Once the graph has been returned to zero, print the graph by pressing the print key on the control panel. Before printing the graph you must press the step key so that the bottom right corner of the screen displays done. The graph that is printed will give the values of torque and twist angle for the calculations portion of the experiment.
7. Unload the specimen from the machine by loosening the jaws and sliding the weighing chuck away from the specimen. Take another material specimen and follow the same procedure. List of Apparatus:
1. Tinus Olson LoTorq Torsion Testing Machine
2. Pittsburgh Caliper
3. Ruler
Data & Results:
A. Torsion Test Data & Graphs
Material
Torque T (in-lbs)
Angle θ (degrees)
Shear Stress (psi)
Shear Strain (radians)
Steel
40
0
482.5806452
0

200
0.72
2412.903226
0.000354524

400
1.78
4825.806452
0.000876462

600
2.93
7238.709677
0.001442715

775.186
4.67433
9352.244
0.002301613

600
4.2
7238.709677
0.002068055

400
3.6
4825.806452
0.001772619

200
2.98
2412.903226
0.001467335

0
1.68
0
0.000827222

-100
0.72
-1206.451613
0.000354524

-135
0.36
-1628.709677
0.000177262
Material
Torque T (in-lbs)
Angle θ (degrees)
Shear Stress (psi)
Shear Strain (radians)
Brass
12
0
675
0

60
0.72
3375
0.000286177

120
1.68
6750
0.000667747

180
2.76
10125
0.001097013

240
4.08
13500
0.001621671

280.719
5.14096
15790.44375
0.002043369

180
3.96
10125
0.001573975

120
3.12
6750
0.001240101

60
2.16
3375
0.000858532

0
1.2
0
0.000476962

-63
0
-3543.75
0

Material
Torque T (in-lbs)
Angle θ (degrees)
Shear Stress (psi)
Shear Strain (radians)
Aluminum
0
0
0
0

100
0.96
1209.677419
0.000352314

200
2.04
2419.354839
0.000748668

300
3.36
3629.032258
0.0012331

400
5.28
4838.709677
0.001937728

461.297
6.68172
5580.205645
0.002452151

400
6
4838.709677
0.002201964

200
3.96
2419.354839
0.001453296

100
2.8
1209.677419
0.001027583

0
1.8
0
0.000660589

-95
0.58
-1149.193548
0.000212856
Material
Torque T (in-lbs)
Angle θ (degrees)
Shear Stress (psi)
Shear Strain (radians)
Polymer
0
0
0
0

55
2.4
262.6635514
0.001450432

105
4.8
501.4485981
0.002900863

152
7.2
725.9065421
0.004351295

200
9.6
955.1401869
0.005801727

250.177
12.1034
1194.770533
0.007314648

242
7.2
1155.719626
0.004351295

160
4.8
764.1121495
0.002900863

103
3.12
491.8971963
0.001885561

40
1.4
191.0280374
0.000846085

0
0
0
0

These numbers were based off the following graphs: Aluminum Brass

Polymer Steel

B. Shear Stress vs. Shear Strain Graphs

Sample Calculations:
To calculate shear stress, we use the equation τ = T*c/J where τ is the shear stress, T is the applied torque, c is the radius of the specimen, and J is the polar moment of inertia. J = π*c4/2

For Brass: T = 280.719 lb*in, c = 0.225 in, J = 0.0040257792 in4

τ = 280.719 lb*in * 0.225 in / 0.0040257792 in4

τ = 15794.391psi

To calculate shear strain, we use the equation γ = c*θ /L where c is the radius of the specimen, θ is the angle of twist, and L is the length of the specimen.

For Brass: c = 0.225 in, θ = 0.0897266787 radians, L = 9.875 in.

γ = 0.225 in * 0.0897266787 radians / 9.875 in.

γ = 0.0020444053 radians

To calculate shear modulus of elasticity, we use the equation G = τ/γ where τ is the shear stress and γ is the shear strain.

For Brass: τ = 15794.391 psi, γ = 0.0020444053 radians

G = 15794.391 psi / 0.0020444053 radians

G = 7742871.677 psi

Shear Stress for Aluminum: T = 461.297 lb*in, c = 0.375 in, J = 0.031619 in4

τ = 461.297 lb*in * 0.375 in / 0.031619 in4

τ = 5568.868 psi

Shear Strain for Aluminum: c = 0.375 in, θ = 0.116618 radians, L = 17.825 in.

γ = 0.375 in * 0.116618 radians / 17.825 in.

γ = 0.00246 radians

Shear Modulus of Elasticity for Aluminum: τ = 5568.868 psi, γ = 0.00246 radians

G = 5568.868 psi / 0.00246 radians

G = 2269862.655 psi

Shear Stress for Polymer: T = 250.177 lb*in, c = 0.511 in, J = 0.107 in4

τ = 250.177 lb*in * 0.511 in / 0.107 in4

τ = 1193.616 psi

Shear Strain for Polymer: c = 0.511 in, θ = 0.210 radians, L = 14.750 in.

γ = 0.511 in * 0.210 radians / 14.750 in.

γ = 0.007 radians

Shear Modulus of Elasticity for Polymer: τ = 1193.616 psi, γ = 0.007 radians

G = 1193.616 psi / 0.007 radians

G = 163839.406 psi

Shear Stress for Steel: T = 775.186 lb*in, c = 0.374 in, J = 0.031 in4

τ = 775.186 lb*in * 0.374 in / 0.031 in4

τ = 9433.465 psi

Shear Strain for Steel: c = 0.374 in, θ = 0.082 radians, L = 13.250 in.

γ = 0.374 in * 0.82 radians / 13.250 in.

γ = 0.002 radians

Shear Modulus of Elasticity for Steel: τ = 9943.465 psi, γ = 0.002 radians

G = 9943.465 psi / 0.002 radians

G = 4096554.868 psi

Discussion:
In this experiment, students measured shear stress of rods made of different kinds of materials within a range of twist angles. With these values, the shear modulus of elasticity (G) of each material was calculated. Based on our calculations, the experimentally obtained shear modulus of elasticity of brass, aluminum, polymer, and steel are 7,742,872 psi, 2,269,863 psi, 163,839 psi, and 4,096,555 psi, respectively. According to The Engineering ToolBox, the published values for shear modulus of elasticity of these materials are 5.8x106 psi, 3.8x106 psi, 0.33x106 psi, and 11x106 psi, respectively.
A comparison of the experimentally obtained shear modulus values and the published values highlights the disparity between the two sets of values. To make it quantifiable, the percent error of the values can be calculated. This is done by finding the difference between the experimental and published values and then the quotient of the difference and published value. The percent error of the experimentally obtained values for brass, aluminum, polymer, and steel are 33.5%, 40.3%, 50.4%, and 62.8%, respectively.
The percentage errors for the experimentally obtained values in this lab experiment are quite large and are most likely caused by many sources of error. One possible source of error is human error during calculations. Though least likely, it is also possible that faulty machinery could have caused error. Other more likely sources of errors include human error in operating the LoTorq machine, not tightening the shaft clamps hard enough, misaligning the rod specimens on the shaft holder, going over the twist angle limits, and using fatigued specimens that have been used by previous groups. Answers to Instructor’s Questions:
1. Specify three assumptions that have been made in deriving the equations used in this experiment.
1. Throughout the entire testing, there should be no plastic deformation.
2. Throughout the entire testing, both the length and the radius of each specimen remain the same, as well as the cross section of the specimen, which will remain circular.
3. Throughout the entire testing, the specimen should not undergo any compression or tension forces. Bending moments should also not occur as well.
2. Name three limitations of torsion formulae.
1. There must be pure torsion on the specimen, meaning that there must be no tensile of bending stress on the specimen.
2. The specimen must have a circular cross-section.
3. The specimen must have its deformation remain within the elastic region.
3. Can the same equations be used with a square shaft? Explain.
No, a square shaft will not yield the polar moment of inertia that was calculated for a shaft with a circular cross section. As such, these equations would not work. The formula that was used in this experiment were for shafts with circular cross sections and could have been solid or hollow. If it was hollow, then only the polar moment of inertia would have to be recalculated again since the diameter of the inner radius affects the results. Regardless, a square shaft cannot use these equations.
4. Can the same formulae be used for a hollow shaft? Explain.
The same formulae can be used for a hollow shaft because the equations used are for shafts with circular cross sections. Using a hollow shaft with the same radius and same length as a solid shaft would have minimal effect on the results since only the polar moment of inertia would need to be recalculated. In reality, the areas of stress concentration of a shaft, solid or hollow, subjected to torsion are towards the circumference of the part
5. How does shearing strain vary with the distance from the axis of the shaft?
The shearing strain is directly proportional to the distance from the axis of the shaft. As the distance from the axis of the shaft increases, the shearing strain increases.

6. Large shafts for ocean liners are often made hollow. Explain the reason for this.
Large shafts for ocean liners are often made hollow to reduce the weight of the part. This reduces the amount of power needed by the engine to rotate the shaft, as well as the cost of production of the part. A hollow shaft does not compromise its structural integrity since the areas of stress concentration of a rod are located towards the circumference of the part regardless if it is hollow or solid. Recommendations and Conclusions:
Some recommendations for this experiment would be a more detailed explanation of the graphs produced when performing the torsion test. The screen was small and quite difficult to analyze when the graph began to deviate. As such, our group was required to perform more torsion tests than expected. Another issue was that sometimes, the shafts were not fully tightened when we thought they were and thus affected our initial results. Other than that, the experiment went smoothly.