Asb Ketegangan B.i

TITLE : Tensile tests OBJECTIVE : Test the tensile of materials.From this experiment we will know the strain and the stress of material towards the force. INTRODUCTION: Strain is defined as against the outer force which is elongate at a solid surface. It’s belong as a part of stress which force,F was pulled at both end of the solid. Strain also is an ability of material to stop the outer force which acted on it, so their shape and size were never change. Therefore, the materials with a high strain are not easier to crooked or limber. This experiments were done on a specimen which have a uniform cross-sectional cutting in a testing machine (Universal Testing Machine).

The load and elongation value obtained can be assume as : a. Engineering stress; σ = F/Ao (kN/mm2) b. Strain; ε = L1 – L0 L0

c. Yield strain; σy =

Yield load (kN/mm2) Initial cross-sectional area

d. Ultimate stress; σm =

Maximum load (kN/mm2) Initial cross-sectional area

e. Percentage elongation; = L1 – L0 x 100% L0 f. Percentage reduction in area = A0 - A1 x 100% A0

Where; F

=

Force, N Initial cross-sectional area, (mm2)

A0 = A1

=

Final cross-sectional area, (mm2)

L0

=

Initial gauge length, (mm)

L1

=

Final gauge lenght, (mm)

Yield load and maximum load can be obtain from the curve of load elongation which was produced from the experiments. i.

Mild steel Fy Fm Fp

= Yield load, N = Maximum load, N = Breaking load, N

load Fm Fp Fy displacement

ii.

For alluminium or materials which not have a yield load, it would be found by “offset strain 0.2% technique” and know as pruff stress. load

Fm Fp

SPECIMEN: 1) Mild steel 2) Aluminum alloy 3) PE (Polyethelene) 4) ASB (Acrylonitrile Butadiene Styrene) APPARATUS: Universal Testing Machine Universal Testing Machine Load

Extensometer

Specimen

Go to down

The specimen shape used: A0

d

L0

A1 L1

PROCEDURE: 1) Center line and gauge length for all the specimens were obtained. 2) Specimens were set in testing machine and the load were pulled up little by little until it’s broken. The load and elongation value were recorded. 3) Center line and gauge length at the broken specimen were measured. 4) The broken way for the specimens were observed.

RESULT:

1) Mild steel L0 = 50 mm Overall length = 204 mm Readings

Initial diameter (D0),mm 9.80 9.90 9.90 9.87

1 2 3 Average

Final diameter (D1),mm 6.50 6.40 6.50 6.50

A0 = Л(D/2)²

A1 = Л(D/2)²

= Л(9.87/2)²

= Л(6.50/2)²

= 76.51 mm²

= 33.18 mm²

Engineering stress; σ = F/Ao (kN/mm2)

= 29520/76.51

= 385.83 N/mm² Strain; ε = L1 – L0

= 71.0 – 50.0 = 0.42 mm

L0

50.0

Yield stress; σy = =

Yield load (kN/ mm2) Initial cross-sectional area 30255 = 395.44 N/mm² Л(9.87/2)²

Ultimate stress; σm =

Maximum load

(kN/mm2)

Length (L1),mm 71.70 68.00 73.70 71.00

Initial cross-sectional area =

41020 = 536.14 N/mm² Л(9.87/2)²

Percentage elongation; = L1 – L0 x 100% L0

= 71 – 50 x 100% 50 = 42 %

Percentage reduction in area = A0 - A1 x 100% A0

= Л(9.87/2)² - Л(6.50/2)² Л(9.87/2)² = 56.63 %

Mild steel : LOAD (kN) 0.0 16.0 31.0 31.5 32.0 33.0 34.5 36.0 37.0 38.0 39.0 39.5 40.0 40.5 40.5 41.0 41.0

DISPLACEMENT(mm) STRESS (kN/mm²) 0.0 0.00 2.5 209.15 5.0 405.23 7.5 411.76 10.0 405.23 12.5 431.37 15.0 450.98 17.5 470.59 20.0 483.66 22.5 496.73 25.0 509.80 27.5 516.34 30.0 522.88 32.5 529.41 35.0 529.41 37.5 535.95 40.0 535.95

STRAIN 0.00 0.05 0.10 0.15 0.20 0.25 0.30 0.35 0.40 0.45 0.50 0.55 0.60 0.65 0.70 0.75 0.80

Graph stress versus strain

stress (kN/mm2)

600 500 400 300 200 100 0 1

3

5

7

9

11

13

15

17

strain

2) Aluminium L0 = 50 mm Readings 1 2 3 Average

Initial diameter (D0),mm 9.70 9.80 9.80 9.77

A0 = Л(D/2)²

Final diameter (D1),mm 6.80 6.80 6.80 6.80 A1 = Л(D/2)²

= Л(9.77/2)²

= Л(6.80/2)²

= 74.97 mm²

= 36.32 mm²

Engineering stress; σ = F/Ao (kN/mm2)

= 16030 / Л(9.77/2)² = 213.82 N/mm²

Strain; ε = L1 – L0 L0

= 61.2 – 50.0 = 0.224 mm 50.0

Length (L1),mm 61.10 61.50 61.10 61.23

Yield stress; σy = =

Yield load (N/mm2) Initial cross-sectional area 18094 = 241.35 N/mm² Л(9.77/2)²

Ultimate stress; σm = =

Maximum load (kN/mm2) Initial cross-sectional area 21580 = 287.85 N/mm² Л(9.77/2)²

Percentage elongation; = L1 – L0 x 100% L0 = 61.23 – 50.00 x 100% = 22.4% 50.00 Percentage reduction in area = A0 - A1 x 100% = Л(9.77/2)² - Л(6.80/2)² x 100 % A0 Л(9.77/2)² = 51.55 %

Aluminium : STRESS (kN/mm²) 0.00 146.67 240.00 246.67 253.33 266.67 273.33 280.00 286.67 286.67 286.67 293.33 293.33 293.33 266.67 220.00 220.00

STRAIN 0.00 0.05 0.10 0.15 0.20 0.25 0.30 0.35 0.40 0.45 0.50 0.55 0.60 0.65 0.70 0.75 0.80

stress (kN/mm2)

Graph stress versus strain 350 300 250 200 150 100 50 0 1

3

5

7

9 strain

11

13

15

17

3) PE (Polyethelene) L0 = 50 mm Readings (mm) 1 2 3 AVERAGE

Width W0 (mm) 12.85 12.85 12.85 12.85

A0 = W0 x t0

Width W1 (mm) 5.25 6.40 5.25 5.63

Thicky t0 (mm) 3.30 3.30 3.30 3.30

Thicky t1 (mm) 1.10 1.10 1.10 1.10

A1 = W1 x t1

= 12.85 x 3.30

= 5.63 x 1.10

= 42.41 mm²

= 6.19 mm²

Yield stress; σy =

Yield load (kN/mm2) Initial cross-sectional area

= 51 / 42.41 mm² = 1.20 N/mm² Ultimate stress; σm =

Maximum load (kN/mm2) Initial cross-sectional area

= 807 / 42.41 mm² = 19.03 N/mm²

Percentage elongation; = L1 – L0 x 100% L0

= [(210.3 – 50.0) / 50.0] x 100 % = 320.6 %

Length L1(mm) 211.0 210.0 210.0 210.3

Percentage reduction in area = A0 - A1 x 100% A0

= 42.41 mm² - 6.19 mm² x 100 % 42.41 mm² = 85.4 %

4) ABS (Acrylonitrile Butadiene Styrene) L0 = 50 mm Readings (mm) 1 2 3 AVERAGE

Width W0 (mm) 13.15 13.10 13.15 13.13

A0 = W0 x t0

Width W1 (mm) 13.15 13.10 13.15 13.13

Thicky t0 (mm) 3.30 3.25 3.25 3.27 A1 = W1 x t1

= 13.13 x 3.27

= 13.13 x 3.27

= 42.94 mm²

= 42.94 mm²

Yield strain; σy =

Yield load (kN/mm2) Initial cross-sectional area

= 1630 / 42.94 = 37.96 N/mm²

Ultimate stress; σm =

Maximum load (kN/mm2) Initial cross-sectional area

= 1689 / 42.94 = 39.33 N/mm²

Thicky t1 (mm) 3.30 3.25 3.25 3.27

Length L1(mm) 50.0 50.0 50.0 50.0

Percentage elongation; = L1 – L0 x 100% L0

= [(50.0 – 50.0)/50.0] x 100% = 0.00 %

Percentage reduction in area = A0 - A1 x 100% A0

= 42.94 mm² - 42.94 mm² x 100% 42.94 mm² = 0.00 %

DISCUSSION: 1) This experiments are to obtain the characteristics of materials. 2) Wild steel have a high stress compare to the aluminium. It’s because wild steel take more time to change their shape become plastic material than aluminium. This condition were explained through the micro structure of the wild steel. 3) Chink atoms of carbon in steel acted as blocking to the breaking, then reduced the derail on the atom plane. This was increased the hardness of material. 4) Area of the graph is a sum of energy absorb by material when elongated. From the observation area of the graph for aluminium less than mild steel. 5) Mild steel can absorb more energy than aluminium. Therefore mild steel are brittle than aluminium. 6) Errors in these experiments were shown by the different stress value for the materials.

CONCLUSION: From the experiments we know the mechenics properties for the materials such as mild steel, aluminium, polyethelene PE and acrylonitrile butadiene styrene ABS. This properties were show clearly in the result, calculation and discussion.

REFERENCES:  B.S. 18: Part 2: 1971  Rollason: Metallurgy for Engineers, (m.s. 5-10)