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NAME: Alison Siegmann ____________________________ GROUP No.__3___________
COURSE No._MECH2500-05____ COURSE TITLE. Mechanics of Materials ___________
DATE: 10/24/2015_______ LABORATORY MEETING DAY: 10/14/2015__________
LABORATORY MEETING TIME: 10:00 AM____________
WENTWORTH INSTITUTE of TECHNOLOGY
EXPERIMENT No. _____5_______
EXPERIMENT TITLE: _Instron Torsion Test on Solid Shafts________________
INSTRUCTOR: __Professor Xiaobin Le ______________________________________
GROUP MEMBERS: GRADING:
LEADER Alison Siegmann ABSTRACT: _______________
Matt Leader INTRODUCTION: _______________
Michael O’Keefe_ PROCEDURE: ___________________
Aaron Paternoster RESULTS: _______________________
________________ DATA ___________________
________________ SAMPLE CALC’S ___________________
DIAGRAMS ___________________
WRITING CENTER FACILITY
CONSULTATION: _____________ GRAPHS ___________________
CONCLUSIONS: _________________
DATE REPORT
SUBMITTED: 10/24/2015_______ ENGLISH MECHANICS: __________
GRADE: ____________________ GROUP LEADER: _________________
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Abstract
The purpose of this experiment was to calculate and verify the proportional limit, shear
yield strength, shear ultimate strength, and shear modulus of elasticity of shafts made of steel and
aluminum under torsional load. This was accomplished by applying increasing torsional loads on
the shafts using the Instron machine and plotting a shear stress vs. shear strain diagram.
Introduction
A shaft is defined as a bar whose central axis is a straight line. When a force is applied
tangent to the face of the shaft and perpendicular to the central axis, torque is generated. A shaft
under torsional load exhibits both shear stress and shear strain. The shear stress is dependent on
both the magnitude of the force applied as well as the radius of the shaft [Hibbler]:
𝜏𝑚𝑎𝑥 =𝑇∗𝑐
𝐽, where T is the torque, c is the outer radius, and J is the polar moment of inertia or
𝐽 =𝜋
2𝑐4
Similarly, the shear strain dependent on both the force, radius, and the shear modulus of
the shaft [Hibbler]:
𝛾 =𝜙 ∗ 𝑐
𝐿, 𝜙 = 𝑡𝑤𝑖𝑠𝑡 𝑎𝑛𝑔𝑙𝑒, 𝐿 = 𝑙𝑒𝑛𝑔𝑡ℎ 𝑜𝑓 𝑠ℎ𝑎𝑓𝑡
With these formulae and by running torsional experiments, it is possible to calculate
various mechanical properties of a shaft material, such as the proportional limit, shear yield
strength, shear ultimate strength, and shear modulus of elasticity by plotting the shear stress
against the shear strain.
Experimental Procedure
The beginning of the lab starts with turning the power on both the power unit and the instron
torsion machine on their backsides. Then “Bluehill3” should be opened on the computer attached. The
test being run is called “MECH302-torsion_circular” which can be selected in the program. Firstly the
initial diameter of the specimen being tested should be measured. The first specimen should then be
placed in the right clamp of machine. Both ends of the specimen need to be extended out the clamps by at
least a half an inch. Once the end is put inside the three jaw clamp fully, that end should be tightened until
secure. The clamp should then be moved towards the left clamp until the specimen is in the jaw, Where it
should be tightened securely. Once both ends are clamped satisfactorily, the initial gage length should be
measured. Then the safety guard should be pushed downwards to cover the specimen. The test should
then be run. Once finished, the specimen pieces should be removed and the final diameter and final length
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of the specimen should be measured. The yield and maximum torque should be recorded as well. The
program should compact the data into a file that can be extracted.
Data and Results
The first graph is from the Instron machine and shows the results of the Aluminum
specimen. The second graph is the shear stress vs. strain diagram for the aluminum specimen
using the data from the torsion test.
-20000
0
20000
40000
60000
80000
100000
120000
140000
160000
180000
-0.2 0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8
Shea
r St
ress
(p
si)
Strain (rad)
Aluminum Specimen Stress vs. Strain
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Aluminum Specimen
Maximum Torque 931.7922 lbf-in
Initial Length 4.4375 in
Final length 4.6875 in
Shear Modulus 68048.7 psi
Shear stress at Yield (Zero slope) 33750.41 Psi
Ultimate Shear Strength 155533.1 Psi
Proportional Limit 99610.427 Psi
This next graph is the Instron machine data for the Steel specimen.
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Steel Specimen
Maximum Torque 1953.124 lbf-in
Initial Length 8.1875 in
Final length 8.25 in
Shear Modulus 5952741 psi
Shear stress at Yield (Zero slope) 74987.58 Psi
Ultimate Shear Strength 163006 Psi
Proportional Limit 7938930.36 Psi
From these graphs we can find some very useful information about the given materials.
Some of this information includes the proportional limit, shear modulus of elasticity, shear yield
strength, and shear ultimate strength. The yield strength for the Aluminum specimen is around
53,423.9 lb/in^2 when using 640 lb-in for the yield torque. The steel had a yielding torque of
around 1,250 lb-in which caused a yield stress of around 103,402.5 lb/in^2. To calculate the
shear stress we are using the equation as follows, with the calculation for yield strength for the
Aluminum being used as an example:
𝑠ℎ𝑒𝑎𝑟 𝑠𝑡𝑟𝑒𝑠𝑠 =𝑇 ∗ 𝐶
𝐽
𝑠ℎ𝑒𝑎𝑟 𝑠𝑡𝑟𝑒𝑠𝑠 =𝑇 ∗ 𝐶
𝑝𝑖 ∗ 𝐶4
2
0
20000
40000
60000
80000
100000
120000
140000
160000
180000
0 0.02 0.04 0.06 0.08 0.1 0.12
Shea
r St
ress
(p
si)
Strain (rad)
Steel Specimen Stress vs. Strain
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𝑠ℎ𝑒𝑎𝑟 𝑠𝑡𝑟𝑒𝑠𝑠 =2 ∗ 𝑇
𝑝𝑖 ∗ 𝐶3
𝑠ℎ𝑒𝑎𝑟 𝑦𝑖𝑒𝑙𝑑 𝑠𝑡𝑟𝑒𝑛𝑔𝑡ℎ =2 ∗ 640
(𝑝𝑖 ∗ 0.2483)
𝑠ℎ𝑒𝑎𝑟 𝑦𝑖𝑒𝑙𝑑 𝑠𝑡𝑟𝑒𝑛𝑔𝑡ℎ = 53,423.9𝑙𝑏
𝑖𝑛2
Where T is Torque, C is the outer radius of the specimen, And J is the polar moment of
inertia for the bar. In the data collected by the Instron machine during the test we can find the
values for the modulus of elasticity which are as follows; Steel is approximated at 5952740.6 psi,
and Aluminum at 68048.7 psi. Next we will calculate the average shear strain in both specimens.
To do this we will use the following equation and the calculations using the aluminum
specimen’s data:
𝑆ℎ𝑒𝑎𝑟 𝑆𝑡𝑟𝑎𝑖𝑛 =𝜙 ∗ 𝐶
𝐿
𝑆ℎ𝑒𝑎𝑟 𝑆𝑡𝑟𝑎𝑖𝑛 =(
1850° ∗ 𝜋180 ) ∗ 0.248
4.4375
𝑆ℎ𝑒𝑎𝑟 𝑆𝑡𝑟𝑎𝑖𝑛 = 1.8 𝑟𝑎𝑑𝑖𝑎𝑛𝑠
Where φ is angle of the twist in radians, C is the final radius of the specimen, and L is the
initial length of the specimen. After running the calculations for the steel we found that the steel
had an average shear strain of around 0.064 radians. This is much lower than the shear strain for
aluminum. This makes sense seeing as in our experiment we observed the aluminum twisting
nearly 5 complete times while the steel didn’t make one complete revolution. Another variable of
interest is the maximum shear stress on the specimens. Maximum stress for brittle materials is
important since they do not have much of a necking region, but simply fracture. The maximum
stress also known as Ultimate strength, was calculated using the same formula for shear stress as
before, but now the maximum torque recorded by the machine is used as T. We will demonstrate
the calculation using the data for the steel specimen:
𝑈𝑙𝑡𝑖𝑚𝑎𝑡𝑒 𝑆𝑡𝑟𝑒𝑛𝑔𝑡ℎ =2 ∗ 𝑇
𝜋 ∗ 𝐶3
𝑈𝑙𝑡𝑖𝑚𝑎𝑡𝑒 𝑆𝑡𝑟𝑒𝑛𝑔𝑡ℎ =2 ∗ 1953
𝜋 ∗ .24875
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𝑈𝑙𝑡𝑖𝑚𝑎𝑡𝑒 𝑆𝑡𝑟𝑒𝑛𝑔𝑡ℎ = 161564.4𝑙𝑏
𝑖𝑛2
From these results we see that aluminum is much weaker than steel when it comes to
torsion strength, however it is much more ductile. In this experiment there could be many
sources of error. One source of error could stem from taking measurements of the specimen.
When measuring the length of the specimen inside the jaws of the machine a tape measure was
used which greatly limits precision. Another source of error could stem from the specimen
slipping within the jaws of the machine. If not sufficiently tightened or if placed in misaligned,
the specimen will most likely slip before fracturing and skew the results. This in fact happened
when we tried to run the calculation for the steel specimen the first time, causing us to have to
redo the experiment.
Figure 1 displays the initial setup for the aluminum specimen in the Instron machine.
Figure 1 Aluminum Specimen Set-up
Figure 2 and 3 illustrate the aluminum specimen once the torsion test was completed. As
shown in figure 2, the aluminum specimen completed 5 full rotations before fracture. This can be
seen in the spiraling lines on the outer surface. Figure 3 shows the cross-section of the specimen.
The specimen did not completely fracture when the test was completed, so the center of the
specimen had to be broken in order to separate it into two pieces. The deformation of the cross-
section can be seen in the spiraling grain of the aluminum.
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Figure 2 Final Aluminum Specimen
Figure 3 Cross-section of final Aluminum Specimen
Figure 4 displays the initial setup for the steel specimen.
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Figure 4 Steel Specimen set-up
Figures 5 and 6 show the final aluminum specimen after the torsion test was completed.
Originally a horizontal line was drawn on the specimen. After the test, the horizontal line was
deformed, now spiraling over the length of the specimen. This illustrates the amount that the
specimen was deformed during the test. Figure 6 illustrates the final cross-section of the steel
specimen after fracture. The spiraling of the grain shows the deformation of the specimen.
Figure 5 Final Aluminum Specimen
Figure 6 Final cross-section of steel specimen
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Conclusion
The torsion test was performed on both the aluminum specimen and the steel specimen.
One end of the specimen was fixed in the grips while the other end of the specimen was attached
to the motor of the Instron machine. Torque was applied to the specimen by the Instron machine,
up until the yield point there was no noticeable movement in the specimen. Once the specimen
reached its yield point, the machine began to rotate one end of the specimen. This movement
could be seen by the movement of the Instron machine was well as by the deformation lines
formed in the outer surface of the specimen. The aluminum was extremely ductile, it reached its
yield point quickly, but completed 5 full revolutions before fracture. The steel specimen
withstood a higher torque but was much less ductile, it did not complete one revolution before
fracture. The stress vs. strain diagrams were created for both the aluminum and steel specimens.
From these diagrams the proportional limit, shear yield strength, shear ultimate strength, and
shear modulus of elasticity was found. The error between these values and the accepted values
for steel and aluminum was large. This is due to the error in the experiments including the steel
slipping out of the grip half way through the experiment causing the test to be run again from the
beginning.