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: _________________

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

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

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.

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

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

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.

Figure 2 Final Aluminum Specimen

Figure 3 Cross-section of final Aluminum Specimen

Figure 4 displays the initial setup for the steel specimen.

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

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.