Experiment #2 Torsion of Cylindrical Rods Stephen Mirdo Performed on October 4, 2010 Report due October 11, 2010
Table of Contents
Object ………………………………………..………………………….………….…. p. 1
Theory …………………………………………………………………………..…pp. 1 - 2
Procedure ………………………….……………… ………………………….……………………………………..…… ……………………..……..... ..... pp. 3 - 4
Results ….............................................................................................................. pp. 5 - 10
Discussion and Conclusion ……………………...…………………….......…... pp. 11 - 12
Appendix ………………………………………………………..….……..…... pp. 13 - 18
Object The object of this experiment was to determine the relationship between the angle of twist and applied torque for a rod as well as the relationship between the deflection and length of the rod at a specified torque.
Theory Torque is defined as a moment that acts about a member’s longitudinal axis. A member that has had torque applied to it such that it deforms along its longitudinal axis is said to be under torsion. The torsion presents itself as shear strain, γ, which is equal to the angle of twist along the longitudinal axis denoted by the symbol φ. For a diagram of this relationship, see Figure 1. Because this experiment uses cylindrical specimens, the theory discussed will pertain only to members of circular cross-section.
Figure 1: Diagram of torsional deformation of a cylindrical specimen. (Adapted from Mechanics of Materials, Wiley, 2011) As a member of circular cross-section is twisted along its longitudinal axis, the cross-sectional area remains constant without deforming. deforming. Because there is no deformation in the plane of cross-section, it is implied that there is no strain in the member’s latitudinal direction. Therefore, the components of shear stress act only in the radial direction. The shear strain of a member is the product of a shear stress denoted by the symbol τ. τ. Shear stress occurs in a cylindrical member when torque, T, is applied. The shear stress is a function of the radius of the circular cross section as seen in Figure 2 below. Along the longitudinal longitudinal axis, the shear stress is null. At any given point along the radius of the cross-section, the shear stress is a function of the radius ρ, denoted by τρ, where J is the moment of inertia. The further out from from the axis a measurement is taken for ρ, the larger larger the calculated shear stress will be. At the circumference circumference of the member, the shear stress will be at its greatest value, τmax.
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Figure 2: Diagram of shear stress, τ, as a function of the radius, ρ, of a cross sectional area. (Adapted from Mechanics of Materials, Wiley, 2011)
τρ = Tρ / J
(Equation 1)
To calculate the polar moment of inertia, J, for a solid circular shaft, employ the following equation where D is the diameter of the cylindrical member.
J = (π/32)D4
(Equation 2)
If the shear stress induced in the member is below the proportional limit of the material, then Hooke’s Law may be applied so as to calculate the material’s modulus of rigidity. In other words, words, if the stress causes only elastic or non-permanent non-permanent deformation, deformat ion, the materials torsional stiffness stiffness can be determined. To apply Hooke’s Law, the shear stress is related to the shear strain by the following expression:
τ = Gγ
(Equation 3)
where G is the modulus of rigidity and γ is the shear strain. strain. Referencing Figure 1, a geometric interpretation of the shear stain is represented as:
γ = ρ (Δφ / Δ x)
γ = ρ (dφ / d x)
(Equation 4)
Equation 4 implies that the shear strain is proportional to the product of the radius of the member and the change in the angle of twist with respect to the longitudinal axis. Substituting Equations 1 and 4 into Hooke’s Law (Equation 3) yields the separable equation:
Tρ / J = G ρ (dφ / d x)
T/J = G (dφ / d x)
T / GJ = dφ / d x
(Equation 5)
Separating Equation 5 and integrating with respect to the longitudinal axis will yield an expression that describes the angle of tw ist, φ, for a member that is prismatic and experiencing a constant internal torque.
∫ dφ = ∫L T/GJ d x 2
φ = TL / GJ
(Equation 6)
Procedure Equipment: TQ Hi-Tech Torsion Experiment Apparatus (SN: HFC.2) Brass Rod Steel Rod Aluminum Rod Weights in 2 lb increments up to 12 lbs.
Part I: 1) Place the torsion experiment apparatus on a hard, flat surface with the pulley and load hanger over the edge of the table as seen in Figure 3. 2) Load a cylindrical specimen into the device by tightening the rod into the chuck and clamp at either end of the apparatus. 3) Wind the cord of the load hanger in a clockwise fashion so as to ensure the applied load on the hanger will induce a torque. 4) Position the two angle scales of the torsion experiment apparatus at an arbitrary distance from one another another on the rod. It is easier to use a round number, such as 8 or 10 inches. inches. Record this value as it will be used as the length, L, needed to calculate the modulus of rigidity of the material. 5) Calibrate the pointers of the angle meters by placing the needles at zero. 6) Add two pounds of weight to the load hangar suspended from the pulley. Ensure that the cord is is still wrapped clockwise around the pulley. Record the indicated angles on both of the angle scales, as this is the angle of twist. 7) Increase the load of the hangar by two pound and again record the indicated angles on both angle indicators. 8) Continue to add two pound increments to the load hanger and record the indicated angles until a total weight of twelve pounds has been achieved. 9) Repeat steps 5 through 8 for two to four four trials. Take the average of the angles for use in calculation. 10) After completing the desired number of trials for the specimen, change out the cylindrical rod with with another material specimen. Repeat steps 2 through 9 for each material.
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Part II: 1) Repeat Steps 1 through 3 for the aluminum specimen. 2) Position the load hanger ensuring the cord is wrapped in clockwise fashion around the pulley. Zero the angle indicator indicators. s. 3) Place 10 pounds of weight on the hangar. 4) Record the position of the angle indicators and the indicated angles on each. 5) Reposition the angle scales in increments of two inches up to twelve inches apart while maintaining a load of ten pounds. 6) Repeat Steps 2 through 5 for a desired number of trials.
Figure 3: Diagram of torsion experiment apparatus (Adapted from Materials Laboratory Manual, University of Memphis, Department of M.E.)
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Results Part I: Brass Specimen Table 1: Initial measurements of Brass rod and pulley 10 in. Length of Twist -4 4 1.51 x 10 in Polar Moment of Inertia 3 in. Diameter of Pulley 0.198 in. Diameter of Test Rod Table 2: Measurements of weight induced torsion and calculated modulus of rigidity for cylindrical specimen of Brass. Note: Twist in A and B are the averages of two trials. Weight Twist at A Twist at B Applied Torque ΔAB (lbs) (in degrees) (in degrees) (in radians) (lbf * in) 2 5 3.5 0.02618 3 4 10 5.5 0.07854 6 6 14 7.5 0.11345 9 8 18.25 9.5 0.15272 12 10 22.5 11.25 0.19635 15 12 27.5 14 0.23562 18
Brass Cylinder 18 y = 77.232x 16 ) n 14 i * f b l ( 12 e u q r o 10 T d e i l 8 p p A
6 4 2 0 .01000
0.060 00
0.1 100 0
0 .1 6000
0.2100 0
Angle Ang le of Tw ist (in radians)
Figure 4: Graph of angle of twist vs. applied torque of a cylindrical brass specimen containing a trend line from which the experimental modulus of rigidity will be calculated.
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To calculate the modulus of rigidity of the brass rod, the slope of the line in the plot of angle of twist and applied torque must must be obtained. A linear trend line passing through the origin is fitted to the plotted data and, using Excel, an equation of the line is generated. The term in the equation before the variable variable x is the slope, m, of the line. line. The slope is equivalent to the product of the modulus of rigidity and the polar moment of inertia divided by the length between the angle indicators of the testing apparatus. m = GJ / L
(Equation 7)
Rearranging this expression algebraically to solve for the modulus of rigidity yields the following expression: G = mL / J
(Equation 8)
Using Equation 8 to solve for the modulus of rigidity is as follows: -4
4
6
GBrass = (77.232 lbf in * 10 in.) / 1.51 x 10 in = 5.12 x 10 lbf/in
6
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Steel Specimen Table 3: Initial measurements of Steel rod and pulley 10 Length of Twist (in) 4 3.96E-04 Polar Moment of Inertia (in ) 3 Diameter of Pulley (in) 0.252 Diameter of Test Rod (in) Table 4: Measurements of weight induced torsion and calculated modulus of rigidity for cylindrical specimen of Steel. Note: Twist in A and B are the averages of two trials. Weight Twist at A Twist at B Applied Torque ΔAB (lbs) (in degrees) (in degrees) (in radians) (lbf * in) 2 1 0.25 0.01309 3 4 1.75 0.75 0.01745 6 6 2 0.75 0.02182 9 8 3 1.25 0.03054 12 10 3.5 1.5 0.03491 15 12 4 1.75 0.03927 18
Steel Cylinder 20 18 y = 417.04x
16
) n i * 14 f b l (
12
e u q r 10 o T d 8 e i l p 6 p A
4 2 0
0.01000
0.01500
0.02000
0.02500
0.03000
0.03500
0.04000
0.04500
Angle of Twist (in radians)
Figure 5: Graph of angle of twist vs. applied torque of a cylindrical Steel specimen containing a trend line from which the experimental modulus of rigidity will be calculated. Again, using the slope of the trend line passing through the origin of the plot of angle of twist and applied torque, the modulus of rigidity is calculated for the steel specimen. Using Equation 8 to solve for the shear shear modulus is as follows: GSteel = (417.04 lbf in * 10 in) / 3.96 x 10 -4 in4 = 10.5 x 10 6 lbf/in2
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Aluminum Specimen Table 5: Initial measurements of Aluminum rod and pulley 10 Length of Twist (in) 4 1.51E-04 Polar Moment of Inertia (in ) 3 Diameter of Pulley (in) 0.198 Diameter of Test Rod (in) Table 6: Measurements of weight induced torsion and calculated modulus of rigidity for cylindrical specimen of Aluminum. Note: Twist in A and B are the averages of two trials. ΔAB Weight Twist at A Twist at B Applied Torque (lbs) (in degrees) (in degrees) (in radians) (lbf * in) 2 3.25 1.5 0.03054 3 4 6.75 3.25 0.06109 6 6 10.5 5 0.09599 9 8 14.25 6.75 0.13090 12 10 18 8.75 0.16144 15 12 21 10.25 0.18762 18
Aluminum Cylinder 20 18 y = 94.189 x
) 16 n i * 14 f b l ( 12 e u q 10 r o T 8 d e i l p 6 p A 4
2 0 0.02000
0.04000
0.06000
0.08000
0.10000
0.12000
0.14000
0.16000
0.18000
0.20000
Angle Angle of Twist (in radians)
Figure 6: Graph of angle of twist vs. applied torque of a cylindrical Aluminum specimen containing a trend line from which the experimental modulus of rigidity will be calculated. Using the slope of the trend line passing through the origin and Equation 8 to calculate the modulus of rigidity of the aluminum specimen is as follows: Galuminum = (94.189 lbf in * 10 in) / 1.51 x 10 -4 in4 = 6.24 x 10 6 lbf/in2
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Part II: Table 7: Measurements of 15 lbf*in torque over a variable distance on aluminum specimen. Note: Twist in A and B are the averages of two trials. Distance Between Indicators A and B
Angle A
Angle B
ΔA-B ΔA-B (In radians)
2 in 4 in 6 in 8 in 10 in 12 in
17 17 17 17 17 17
15 13 11 8.5 7 5
0.034906585 0.06981317 0.069813 17 0.104719755 0.148352986 0.174532925 0.20943951
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12 y = 56.642x ) s e h c n i n i ( h t g n e L
10
8
6
4
2
0 0
0.05
0.1
0.15
0.2
0.25
Angle of Twist (in radians)
Figure 7: Plot of angle of twist vs. length over which a ten pound force is applied to a cylindrical specimen of aluminum. It is noted that the trend line passing through the origin fitted to the data set seen in Table 7 yields a linear function. function. An increase in the length under which torque is applied will yield an increase increase in the angle angle of twist of the test specimen. specimen. This occurrence indicates that the specimen is under pure torsion. To determine if the results of Part I of the experiment were statistically significant, a Student’s t-Test was performed. There will be be three tests performed: performed: one with brass and steel, one with brass and aluminum and another with steel and aluminum.
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Table 8: t-Test for Brass and Steel t-Test Mean σ
df = t stat = t crit =
Brass 5.11E+06 57640.72812
Steel 1.09E+07 3069965.933
2 2.646 4.303
The calculated values for the modulus of rigidity of brass and steel for accepted as tstat < tcrit. Table 9: t-Test for Brass and Aluminum t-Test Mean σ
df = t stat = t crit =
Brass 5.11E+06 57640.72812
Aluminum 6.26E+06 415200.692
2 3.883 4.303
The calculated values for the modulus of rigidity of brass and aluminum for accepted as tstat < tcrit. Table 10: t-Test for Steel and Aluminum t-Test Mean σ
df = t stat = t crit =
Steel 1.09E+07 3069965.933
Aluminum 6.26E+06 415200.692
2 2.097 4.303
The calculated values for the modulus of rigidity of steel and aluminum for accepted as tstat < tcrit.
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Discussion & Conclusion A material’s modulus of rigidity will influence how much twist will occur in a member when a given torque is applied. A higher modulus modulus will yield less of an angle of twist for a specified torque. Conversely, a lower modulus of rigidity for a material sustaining the same specified torque will deform to a greater extent along its longitudinal axis. It was noted post-experiment that the calculated modulus of rigidity, G, for the aluminum specimen was not not in agreement with the theoretical value. A percent error calculation was then performed on all three test specimens to ensure Equation 8 performed as expected. Table 8: Percent error in calculation of modulus of rigidity for test materials Material Brass Steel Aluminum
Calculated Modulus of Rigidity (lbf/in 2)
Theoretical Modulus of Rigidity (lbf/in 2)
% Error
5.12E+06 1.05E+07 6.24E+06
5.50E+06 1.15E+07 3.80E+06
6.91% 8.70% 64.21%
As seen in Table 8, the calculated modulus of rigidity for brass and steel are within acceptable values. However, the aluminum’s aluminum’s experimental experimenta l value for modulus modulus of rigidity rigidit y is well outside of any acceptable error range. There are a multitude of reasons for the discrepancy between the theoretical and calculated values of the modulus modulus of rigidity for the aluminum specimen. A possible source of this error could be that the material material tested was, in fact, not aluminum. A for the 6 2 modulus of rigidity value of 6.26 x 10 lbf/in indicated that perhaps the test material was a titanium alloy or zinc. Without further material analysis, it is impossible to state state with certainty what material was used in this experiment. Another possible cause for the calculated value for the modulus of rigidity, G, being 64.2% different from the accepted value for aluminum was perhaps the test rod specimen wasn’t of a consistent diameter. A varying diameter throughout the length of the test specimen would interfere with an accurate calculation by varying the polar moment of inertia, J. Instances of a varying J value would yield unique G values. Other sources of error are present present in this experiment experiment.. The indicated indicated angles are only read to the nearest half degree. This limitation will prevent an accurate measurement for the angle of twist of a specimen under torsion. Another source of error in this experiment is that the samples tested have no record of use. The specimens may have been put through other tests such as heat treatment or elongation that may have altered their material properties. Improvements for this experiment are in order. From an educational point of view, the experiment would be more interesting if the calculated modulus was used to 11
identify the test material. Another improvement would be to have clearly identified test specimens. As seen with the discrepancy in theoretical and experimenta experimentall G values for the aluminum, it is difficult to perform this experiment successfully without the proper materials present. One final improvement that should be made to this experiment would be to use a strain strain gage and Hooke’s Law to calculate the modulus of rigidity. rigidity. The use of precise, digital instruments increases the attractiveness of an experiment.
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Appendix Data Usage Sample calculation of the polar moment of inertia for the brass test rod: 2
-4
4
J = (π/32) (π/32) * (0.198 in) = 1.51 x 10 in Sample calculation of the applied torque on brass test rod at weight of 10 lbs:
T = (3.0 in / 2 ) * 10 lbf = 15 lbf*in Sample calculation of the modulus of rigidity of the steel test specimen: -4
4
6
2
GSteel = (417.04 lbf in * 10 in) / 3.96 x 10 in = 10.5 x 10 lbf/in
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Bibliography nd
Mechanics of Materials, 2 Edition Timothy A. Philpot (2011) Fundamentals of Material Science and Engineering: An Integrated Approach W.D. Callister, Jr and D.G. Rethwish (2008) Materials Laboratory Manual, Fall 2010 University of Memphis, Department of Mechanical Engineering Material Properties Table - An A-Z of engineering metal properties http://www.engineerstudent.co.uk/material_properties_table.html
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