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Tensile Test of A-36 Low Carbon Steel By Richard Balbuena ME 312 California State University Los Angeles 10-02-15
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Page 1: ME 312 lab report 1

Tensile Test of A-36 Low Carbon Steel

By

Richard Balbuena

ME 312

California State University Los Angeles

10-02-15

Page 2: ME 312 lab report 1

ObjectiveThe purpose of this experiment was to conduct a tensile test on standard

structural steel in order to demonstrate Hooke’s Law.

Theory

Hooke’s Law is given by the equation F=-kx, where x is the extension of the

spring, being the extended length minus the unstretched length. Through relation, f,

the force used to compress or extend the spring a distance is directionally

proportional to its extension. K is the spring constant that may tell us how stiff a

spring is. The negative sign is there to act against the displacement of the spring in

order to restore the spring back to an equilibrium position.

Young’s Modulus is given by the equation

.

The point of this equation is to measure the how stiff a material is and can tell how

much a material can withstand a change in length when it is under compression or

tension.

Procedure

Before we began conducting the experiment, we first measured the diameter

of the steel sample using a dial caliper. After that was done we placed the sample

under a center punch fixture and aligned it in order to puncture the sample. After

this we placed the sample in the Tinius-Olsen UTM machine to begin the

experiment. Initially we placed a small load of around 100 lbs-200 lbs. but soon

Page 3: ME 312 lab report 1

after we increased the load to a bigger one. Gradually we began to increase the load

rate to up to 2000 lbs. when finally the load on the steel sample snapped it in half.

After removing the sample we measured the final diameter and completed our

experiment.

Diagrams

Tinius-Olsen UTM

A-36 Sample after the fracture

Page 4: ME 312 lab report 1

Results and Data

Proportional limit 47,000 psi

Yield Point 46,000 psi

Yield Strength 52,600 psi

Ultimate Strength 75,400 psi

Rupture Strength 15,110 psi

Modulus of Elasticity 31, 300psi

Modulus of Resilience 9,160 psi

Percent elongation in gauge length 0.072%

Percent reduction in area at fracture 64%

0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.80

1020304050607080

9.1615.1

31.3

46 4752.6

75.4

Stress Vs Strain

Stress

Stra

in (

psi

in t

ho

usa

nd

s)

Discussion

Page 5: ME 312 lab report 1

Based on the numbers attained from this experiment, it was done decently

but not the best. When you compare the modulus of elasticity and resilience to the

actual publicized results listed on Matweb, they are somewhat similar. The ultimate

strength falls within the listed range however the yield strength is a bit off. Some

discrepancies of the test may be because the loading on the Tinius-Olsen machine

may have been done wrong or we could have attained a wrong reading on the

caliper while taking the diameter, resulting in wrong results in the percent

elongation and reduction. Other than that, the experiment went fairly okay. We may

use this steel sample with respect to its properties to use in structures because of its

calculated modulus of elasticity and yield point, which could compete with other

materials out there in the market. Most companies look for materials that will have

these things in order to get the best materials for their money that will insure a good

structure or item.

Conclusion

From this experiment, we have learned to calculate many things from a sample steel

bar. From using the Tinius-Olsen machine, we were able to determine the samples

fracture point as well as other things such as the modulus of elasticity and ultimate

strength.