Name: Kevin Lundstrom
Class: Monday
Group: 3
Lab Partners: Rusty Flocken
Oliver Beres

Torsion

Introduction

The purpose of this experiment was to ultimately determine the shear modulus, 0.2% offset shear yield point, and ultimate shear strength of two different materials: aluminum and brass. This was accomplished using a torsion test machine connected to a computer that recorded the torsion data. After determining these three material properties, the measured experimental values could be compared to published data to determine the accuracy of the test.

Experimental Procedure

In this experiment the specimens underwent torsion by means of a torsion test machine. This machine consisted of two drill chucks: one was attached to a rotating chain-driven wheel, and the other was held stationary. A potentiometer was used to determine the number of revolutions the rotating wheel had undergone. Strain gauges were attached to the stationary drill chuck. LabVIEW was used to measure and record the data from the potentiometer as well as the strain gauges. Each sample was measured until it failed. The length of each specimen as well as the diameter was recorded in order to perform calculations. Also, the diameter of the potentiometer as well as the drill chuck, as well as the minimum and maximum resistance of the potentiometer was recorded for calculations.

Results/Discussion

The recorded measurements are shown in the table below. These were necessary to perform the calculations.

Rmin Potentiometer | 1.4212 ohms | | Rmax Potentiometer | 1985.2 ohms | | DChuck | 2.045" | | Dpotentiometer | 1" | | | Aluminum | Brass | Length | 4.85" | 5.04" | Diameter | 0.379" | 0.374" |
Table 1 – Recorded Values

Data was provided relating the torque to the twisting moment. The units of torque were converted into Nm from lb-in and the strain values were plotted against these torque values. The strain values were calculated by summing the absolute values of every strain for each provided torque. The resulting plot is shown below.

Fig. 1 – Torque vs. Strain Chuck

A trend line was fit to the data with the intercept set to zero in order to determine their relationship. The shear stress could then be calculated using the moment equation shown below.

Mt=τxyJr (1)

This can be solved for shear stress resulting in the following equation.

τxy= Mt*rJ (2)

Next, the principle stresses were calculated. The following equation is used to calculate the principle strains.

ε1,2=εx+εy2±εx-εy22+εxy2 (3)

However, in order to calculate the principle strains the xy strain needed to be calculated. This was accomplished using the equation below. εxy=εb-εa+εc2 (4)

Finally, in order to plot a stress strain curve, gamma had to be calculated. This was accomplished using the following equation.

γ=tan-1rspecimen*θspecimenlspecimen (5)

After calculating τxy and γ, the stress strain curve was plotted for both specimens. For the aluminum rod, the following stress-strain curve resulted.

Fig 2. – Stress-Strain Curve – Aluminum - Overview

As it is difficult to see the offset yielding values, the data was re-plot in order to focus on the values around the yield point.

Fig 3. – Stress-Strain Curve – Aluminum - Excess Data Removed

From the plot, the modulus of elasticity was determined to be 20.1 GPa, the offset yielding was determined to be roughly 190 MPa, and the ultimate shear strength was determined to be 300 MPa. After the specimen failed it was noticed that there were visible twisting marks on the exterior. On the inside there was circular marks around a center core that did not contain these marks. As aluminum is a ductile material it is hypothesized that the center did not experience nearly as much torsion, and the exterior shell rotated around the center core.

A stress-strain plot was also created for the brass data. This plot can be seen below.

Fig 4. – Stress-Strain Curve – Brass - Overview

As it is difficult to see the offset yielding values, the data was re-plot in order to focus on the values around the yield point.

Fig 5. – Stress-Strain Curve – Brass - Excess Data Removed

From this plot, the modulus of elasticity was determined to be 30.38 GPa, the 0.2% offset yield is roughly 350 MPa, and the ultimate shear strength is 440 MPa. When the broken specimen was examined there was much less visible twisting on the exterior. This could be due to the fact that the rod experienced a significant less number of rotations. When looking at the cross-section, the break was clean. There was not a center core that appeared to experience less torsion as was seen on the aluminum sample. Brass is significantly less ductile which is probably the reason for this.

Conclusion

When the experimental data was compared to published data, the differences were quite large. A comparison of these values can be seen below.

| Aluminum | Brass | | Shear Modulus | 0.2% Offset | Ultimate Shear Strength | Shear Modulus | 0.2% Offset | Ultimate Shear Strength | Experimental | 20.1 GPa | 190 MPa | 300 MPa | 30.38 GPa | 350 MPa | 440 MPa | Actual | 26 GPa | | 207 MPa | 37 GPa | | 260 MPa |
Table 2 – Experimental vs. Published Results

While neither of the shear moduli was incredibly off, both ultimate shear strengths differed greatly. A possible reason for this is that the potentiometer wasn’t able to provide enough contact with the chuck to accurately measure the rotation. This would result in the stress-strain curve not being accurate. Usually imperfections in the specimen created during the manufacturing cause the specimen to deviate from the accepted values, however imperfections would result in a lower shear modulus and ultimate shear strength. While this is seen in the shear modulus, it is not seen in the ultimate sear strength. Therefore the discrepancy must be due to the inability in the potentiometer to accurately measure the rotations.