Mechanical Laboratory 2 Lab #9: Torsion Test

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Mechanical Laboratory 2

Lab #9: Torsion Test

Ante Kulas

MMAE-419-L01

Professor: Dr. Murat Vural

TA: George Kim

11/02/2018

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ABSTRACT:

In process of determining if structure or system is sustainable for service torsion testing plays major role. The purpose of torsion testing is to introduce students to mechanical properties of for metals while in shear. Load is applied to the steel, aluminum and brass specimens until they fractured. Students were able to determine offset torsional yield strength, proportional limit, modulus of rupture,

modulus of rigidity and ductility for materials from torque and angle that was recorded. Understanding

of mechanical properties of material is crucial for engineering because choosing right material for design

will decide if design will give optimal results and minimize failure of systems and machines.

INTRODUCTION:

Main goal of this lab is better understanding of mechanical properties of metals by determining offset

torsional yield strength, proportional limit, modulus of rupture, modulus of rigidity and ductility for

materials from torque and angle that was recorded. Four materials were tested, structural steel

1045HR,1018HR, aluminum 6061-T6 and naval brass. After testing was done data for time, torque and

angle was recorded and from there mechanical properties were calculated. First step of calculations

included translating data for angle that was recorded in degrees to radians and it was called angle of

twist and notation for it was φ’. Same thing was done for done with torque since data was translated

from ft-lb to in-kip and notation for it is T. To calculate central difference slope for each point equation

(1) was used, where i represent position of each point and N represent how far is upper and lower

position of data.

𝒅𝑻

𝒅𝛟′(

𝑪𝑫

) = 𝑻(𝒊+𝑵)−𝑻(𝒊−𝑵)

𝛟′(𝒊+𝑵)−𝛟′(𝒊−𝑵) (1)

After central difference is found that have notation 𝒅𝑻 , average central difference slope was found for

each data point by using equation (2), where n is number of central difference slope data points to take

average of N and n should be carefully picked because too small or too large will not help in smoothing

curve and may have contra effect. For most parts it was found that optimal N was around 20, and

optimal n was around 15.

𝒅𝑻

𝒅𝛟′(

𝑪𝑫𝒂𝒗𝒆

) = ∑

𝒅𝑻

𝒅𝛟′(𝑪𝑫

)(𝒋)𝒊+𝒏𝒋=𝒊−𝟏

𝟐𝒏+𝟏 (2)

Than re-constructed torque was found using (3) and plotted in Figures 10, 14, 18 and 22 v. angle of

twist per unit length, where 𝒅𝑻𝒂 is average central difference slope.

𝑻(𝒊) = 𝑻(𝒊−𝟏) + 𝒅𝑻𝒂(𝒊−𝟏) ∙ ∆𝝓′(𝒊) (3)

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Both slopes, 𝒅𝑻𝒂 and 𝒅𝑻 were plotted with respect to angle of twist per unit length in Figures 11, 15,

19 and 23 and later were compared. Shear strain was calculated using equation (4), where r is radius of

specimen.

𝜸 = 𝒓 ∙ 𝝓′ (4)

Shear stress for elastic part is calculated using equation (5) and for plastic part is calculated using (6),

and both shear stresses were plotted vs. shear strain in Figures 12, 13, 16, 17, 20, 21, 24 and 25.

D stands for outside diameter of specimen.

𝝉𝒎𝒂𝒙 = 𝟏𝟔 ∙ 𝑴𝑻

𝝅 ∙ 𝑫𝟑 (5)

𝝉 = 𝟏

𝟐∙𝝅∙𝒓𝟑 ∙ [𝝓 ∙ 𝒅𝑻𝒂 + 𝟑𝑴𝑻] (6)

PROCEDURE:

Torsion test machine (Figure 1) had one head that is stationary and other that is applying twisting

moment (torque) to specimen. Dimensions of specimens are shown in (Figure 2-3). On each specimen

two marks were placed with distance between them being two inches and it was gage length. Specimen

was placed in torsion machine and torque sensor is placed on specimen. Machine was then calibrated,

and twisting moment was applied. Experiment was repeated for four materials (structural steel

1045HR,1018HR, aluminum 6061-T6 and naval brass) and data for time, torque and angle was recorded

and from there mechanical properties were calculated. Specimens fractured surface can be sin in

(Figures 6-9), and on Figure 4 (1018HR steel and naval brass).

Figure 1. Experiment setup

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Figure 2. Specimen dimensions Figure 3. Specimen after fracture

Figure 4. Fractured specimens Figure 5. Torque sensor

Figure 6. 1045HR steel Figure 7. 1045HR steel Figure 8. 6061 AL Figure 9. Naval brass

RESULTS:

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Figure 10. a) Torque vs. twist angle (1045 HR Steel) b) detailed

Figure 11. a) Slope vs. twist angle (1045 HR Steel) b) detailed

Table 1. Calculated and collected values for 1045HR Steel

Offset torsional yield strength [ksi] 28.19

Proportional limit [ksi] 61.56

Modulus of rupture [ksi] 82.12

Modulus of rigidity [ksi] 25622

Ductility (γ) at fracture) [%] 26.27

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Figure 12. Shear Stress vs. Shear Strain (1045 HR Steel) plot

Figure 13. Shear Stress vs. Shear Strain (1045 HR Steel) detailed plot

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Figure 14. a) Torque vs. twist angle (1018 HR Steel) b) detailed

Figure 15. a) Slope vs. twist angle (1018’]\ HR Steel) b) detailed

Table 2. Calculated and collected values for 1018HR Steel






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Figure 16. Shear Stress vs. Shear Strain (1018 HR Steel) plot

Figure 17. Shear Stress vs. Shear Strain (1018 HR Steel) detailed plot

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Figure 18. a) Torque vs. twist angle (6061 Aluminum) b) detailed

Figure 19. a) Slope vs. twist angle (6061 Aluminum) b) detailed

Table 3. Calculated and collected values for 6061Aluminum






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Figure 20. Shear Stress vs. Shear Strain (6061 Aluminum) plot

Figure 21. Shear Stress vs. Shear Strain (6061 Aluminum) detailed plot

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Figure 22. a) Torque vs. twist angle (Naval Brass) b) detailed

Figure 23. a) Slope vs. twist angle (Naval Brass) b) detailed

Table 4. Calculated and collected values for Naval Brass






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Figure 24. Shear Stress vs. Shear Strain (Naval Brass) plot

Figure 25. Shear Stress vs. Shear Strain (Naval Brass) detailed plot

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DISCUSSION:

When torsion from rough data is compared to re-constructed it is clear that reconstructed is smoothed

out version of rough data and it is more usable for further usage. Same goes with plots of central

difference slope and average central difference slope vs, angle of twist per unit length it is clear that

averaged slope is more usable since smoothed out. How much it is to be smoothed depends on values N

and n. Finding optimal values of those values is crucial to get plot smooth enough without ruining data.

Elastic and plastic shear stress vs. shear strain was plotted on the same plot even it is clear that elastic

part is correct only during elastic deformation, and plastic after that. Linear curve fit was done in linear

part of elastic shear stress plot so 2% line can be fitted and from there offset torsional yield strength can

be found so as modulus of rigidity (from slope of fitted line). Transition between plastic and elastic part

of plot is where proportional limit can be found. Modulus of rupture can be found just by finding

maximum shear stress in plastic plot. Ductility (γ at fracture) is shear stress in the moment when fracture

starts to happen.

CONCLUSION:

This lab gives students some inside on how different materials (structural steel 1045HR,1018HR,

aluminum 6061-T6 and naval brass) react under torsional load, and how it is possible to find material

properties that include offset torsional yield strength, proportional limit, modulus of rupture, modulus of

rigidity, ductility (shear strain at fracture) from data collected.

REFERENCES:

Mechanical Behavior of Materials 4th Ed. by Norman E. Dowling (Pearson, 2013 ISBN 0-13-139506-8)

Dr. Murat Vural, MMAE 419 Torsion Testing Handout, IIT, Chicago

Documents

Mechanical Laboratory 2 Lab #9: Torsion Test