Lab Reort

 

Tensile Test – Lab Report

by Jannatul Ferdous

ENGL 21007 R[24989]

September 15, 2020

 

 

ABSTRACT:

The purpose of this experiment is to understand tension load and how the cross-section area and the original length relate to elongation. The loads P used for each samples and elongation     ΔL = L – L0, is used to find the relation after plotting them on the graph. The relationship between stress σ = P/A and strain ε = ΔL/ L0 is also found after plotting them on the graph. The material that will be used to do the tensile test on is rubber.

 

INTRODUCTION:

Tension is created when a structural member is pulled from both sides. Under the effect of the tension the member’s length will increase and the cross-sectional area, will decrease. The elongation is a function of the load applied, the cross-sectional area, the length, and the type of the material the member is made of.

Hooke’s law states that the elongation is proportional to the force which describes the elastic, linear behavior of a material. The relationship is expressed as

σ = Eε

The normal stress σ = P/A is determined by dividing the magnitude P of the force by the cross-sectional area A of the member perpendicular to direction of the applied force. The strain       ε = ΔL/ L0 is the ratio of change in length of loaded member to the original length L0 of unloaded piece. The proportionality constant E is called the modulus of elasticity of the material.

The elongation, ΔL = L – L0, of a structural member with uniform cross section subjected to axial loading can be determined by

ΔL = PL0 / AE

 

MATERIALS:

  1. Test specimens of rubber
  2. Test weights and weight holder
  3. Rulers
  4. Support stand and clamps

METHODS:

  • Sample 1: Case 1 – One 10-inch long rubber rod, 0.05 in diameter:
  1. Hang the rubber rod from the top of the sand using the clamps.
  2. Hang the weight holder through the hole drilled at the bottom and the top of the rod.
  3. Measure the initial length L0 between the holes at the bottom and the top of the rod.
  4. Apply a 5 lb weight and measure the elongated length ΔL = L – L0

Case 2 – One 10-inch long rubber rod, 0.05 in diameter:

  • Repeat the steps described above using a 7 lb weight.

Case 3 – One 10-inch long rubber rod, 0.05 in diameter:

  • Repeat the steps of Case 1 described above using a 9 lb weight

 

  • Sample 2: Case 1 – One 15-inch long rubber rod, 0.05 in diameter:
  • Repeat the steps described in Sample 1 case 1 using a 5 lb weight.

Case 2 – One 15-inch long rubber rod, 0.05 in diameter:

  • Repeat the steps described in Sample 1 case 1 using a 7 lb weight.

Case 3 – One 10-inch long rubber rod, 0.05 in diameter:

  • Repeat the steps described in Sample 1 case 1 using a 9 lb weight.

 

  • Sample 3: Case 1 – One 20-inch long rubber rod, 0.05 in diameter:
  • Repeat the steps described in Sample 1 case 1 using a 5 lb weight.

Case 2 – One 20-inch long rubber rod, 0.05 in diameter:

  • Repeat the steps described in Sample 1 case 1 using a 7 lb weight.

Case 3 – One 20-inch long rubber rod, 0.05 in diameter:

  • Repeat the steps described in Sample 1 case 1 using a 9 lb weight.

 

RESULT:

Determine the area A = (pi(D)^2)/4 in^2 of the test specimens in each case. Record the data in a table with the first column load P = mg/gc lbf, the second column is the area A = (pi(D)^2)/4 in^2 , third column stress σ = F/A psi, the fourth column is the original length, the fifth column is the length after the load is applied, the sixth column is the measured elongation ΔL in, the seventh column is the strain ε = ΔL/ L0, and the eighth column is the calculated E.

Plot the three cases described above on the same graph showing the relationship between the applied loads (on the vertical axis) and the measured elongations (on the horizontal axis).

Also, plot the relationship between the applied stress a (on the vertical axis) and the strain (on the horizontal axis) on another graph sheet for the three cases.

 

Sample

 P (lb)

A (in2)

 σ (psi) L0 (in) L (in) ΔL (in) ε E (psi)
1 5 0.0019634954 2546.4791

10

10.005

0.005

5.0 x 10^-4

 5092958.2
7 0.0019634954 3565.070741

10

10.006

0.006

6.0 x 10^-4

 5941784.568
9 0.0019634954 4583.662381

10

10.007

0.007

7.0 x 10^-4

 6548089.116
2 5 0.0019634954 2546.4791 15

15.03

0.03

0.002

 1273239.55
7 0.0019634954 3565.070741 15

15.04

0.04

0.00267

 1336910.511
9 0.0019634954 4583.662381 15

15.05

0.05

0.0033

 1375098.728
3 5 0.0019634954 2546.4791 20

20.05

0.05

0.0025

 1018591.64
7 0.0019634954 3565.070741 20

20.06

0.06

0.003

 1188356.914
9 0.0019634954 4583.662381 20 20.07 0.07 0.0035

1309617.823

 

DISCUSSION:

The relationship between the applied load and elongation is directly proportional to each other. The graphs for stress vs strain and load vs elongation are like each other in a way that stress and strain are also directly proportional to each other. In this case if we had steel instead of rubber the results of elongation would have been very different because steel is much stronger. The slope does not converge to same value for the three cases. When the member passes its yield strength and enters its ultimate strength that is when material is at its weakest state, which will eventually cause it to break.

 

CONCLUSION:

After doing the experiment we can come to an understanding of tension loading, which is the ability of a material to withstand a pulling force, in this case the pulling force would be the stress. It can also be concluded that the relation between the cross-section area to elongation is directly proportional and the relation between the original length to elongation is inversely proportional. Tensile testing on materials is very important because, by measuring the force required to elongate a material, in our case rubber, to breaking point, material properties can be determined that will allow quality managers to predict how materials and products will behave in their intended applications.

 

ACKNOWLEDGEMENT:

Special thanks to Professor Brown for her assist in doing the lab report.

 

REFERENCES: