Title: The Tensile test
Objective: •
To become familiar with tensile testing system
•
To learn how to develop stress-strain curve from load-displacement curve for materials
•
To learn how to extract important material properties from these curve such as engineering stress, strain, yield stress, percentage of elongation…etc
Introduction: In order to compare and select materials for various applications, one must have access to the important properties of that material i.e. material constant. One One of the the most most often often used used tests tests perfo perform rmed ed to deter determin mine e a numb number er of impor importan tantt mech mechani anical cal prope properti rties es of mater material ial is tensil tensile e test. test. This This test test is conducted to specimens that have uniform horizontal cross section in the Universal Testing Machine that can show the load applied for any elongation that is obtained from the gauge length Based on the load and displacement obtained from the experiment, we can get: (a) Engineering Engineering tensile tensile stress, ε = F/Ao (N/mm²) (b) Strain, ε = L1 – Lo / Lo (c) Yield stress, σy = Yield load / initial cross-sectional area
(N / mm²)
(d) Ultimate tensile stress, σm = Maximum load / initial cross-sectional area (N / mm²) mm²) (e) Elongation percentage, %EL = (L1 – Lo / Lo) × 100 (%)
(f) Reducti Reduction on perce percenta ntage ge in area area (%) = (Ao – A1 / Ao) × 100
Where: F = Load (N) Ao = Initial horizontal cross- sectional area, (mm²) A1 = Final horizontal cross- sectional area, (mm²) Lo = Initial gauge length, (mm) L1 = Final gauge length, (mm)
Yield load and maximum load can be obtained from the load-displacement curve produced from each experiment is (1)
Mild Steel Fy = Yield load, (N) Fm = Maximum load, (N) Fp = Load at break, (N)
Load (N) Fm
Fp Fy
Displacement (mm)
(2) For aluminium alloys or specimen that does not have certain yield load, it can be obtained with “offset strain 0.2 % method” and known as ‘pruf stress’. Fm = Maximum load (N) Fp = Load at break (N) F0.2% = pruf stress
Load (N)
Fm
Fp F0.2%
Displacement (mm)
Specimen: •
Mild Steel
•
Aluminiun Alloy
Apparatus: Universal testing Machine Procedure: (a) The diameter diameter and gauge length for all the specimen provided provided is taken (b) The specimen is installed inside the testing machine and load is added
little by little until the specimen break, Value of the load applied and elongation is recorded
(c) The diameter and the gauge length of the specimen at the fractured part is
once again measured and recorded (d) The it fracture for both specimen is observed
Result: Diameter of the specimen Specimen
Reading / mm
(i)
(ii)
(iii)
Average
Aluminium alloy
Initial diameter
9.70
9.80
9.80
9.77
Final diameter Diameter difference
6.80
Initial diameter
9.80
9.90
9.90
9.87
Final diameter
6.50
6.40
6.60
6.50
Mild Steel
Diameter difference
6.80 6.80 9.77 – 6.80 = 2.97
6.80
9.87 - 6.50 = 3.37
Specimen gauge length: Reading/ mm Final length 2 3
Difference
Specimen
Initial length
Aluminium alloy
50.00
61.10
61.50
61.10
61.23
11.23
Mild steel
50.00
71.70
68.00
73.70
71.00
21.00
1
Calculation: Aluminium alloy Initial cross-sectional area, A o = π(d²/4) = π (9.77²/4) = 74.97mm² Final cross-sectional area, A = π(d²/4) = π (6.80²/4) = 36.32mm² Based on the computer calculated data;
Average
Title: Mechanical properties of plastic
Objective: To obtain yield tensile stress (σ y), breaking stress (σ b) and elongation (ε) for plastic. (i)PE (Polyethelene), (ii) ABS (Acrylonitrile Butadiene Styrene)
Introduction: Refer to the figure to obtain:
Figure F
Lo
L
F
(1) Tensile stress σ = F/A σ = stress- tensile (N/m²) F = force (N) A = area (m²) (2) Elongation ε = (L-L o/Lo) × 100% ε = elongation Lo = initial length L = final length (3) Stress- elongation curve i.e. polyethelene ( PE)
σ
σb
Stress MN/m
Elongation %
σy = yield stress σb = breaking stress
Procedure: (a)The dimension of the plastic specimen given is measured (b) Change the shape until the specimen breaks (c) Elongation and the load is obtained to calculate tensile stress
Result and Discussion: (1) PE (polyeth (polyethelyne elyne)) Lo= 50 mm
Reading/m m
Final length(L 1)
Initial width(w o)
Final width (w1)
1 2 3 Average
211.0 210.0 210.0 210.3
12.85 12.85 12.85 12.85
5.25 6.40 5.25 5.63
(2) ABS (Acrylonitrile butadiene styrene)
Initial thickness(t o ) 3.30 3.30 3.30 3.30
Final thickness(t 1) 1.10 1.10 1.10 1.10
Lo = 50 mm
Reading/m m
Final length(L 1)
Initial width(w o)
Final width (w1)
1 2 3 Average
50.0 50.0 50.0 50.0
13.15 13.10 13.15 13.13
13.15 13.10 13.15 13.13
Calculation: PE:
Ao = wo× to = 12.85×3.30 = 42.405mm²
ABS: Ao = wo× to = 13.13×3.27 = 42.935mm²
For PE plastic: Maximum load (based on computer data), F = 807 N Yield stress, σy= F/Ao = 807/ 42.405 = 19.03 N/mm ²
Breaking load (based on computer data), F = 51 N Breaking stress, σb = F/Ao = 51/ 42.405 = 1.12 N/mm ²
Elongation, ε =
L1
−
L0
L0
x 100%
= [(210.3 – 50.0)/50.0] × 100% = 320.6% For ABS plastic:
Initial thickness(t o ) 3.30 3.25 3.25 3.27
Final thickness(t 1) 3.30 3.25 3.25 3.27
Maximum load (based on computer data), F = 1689 N Yield stress, σy= F/Ao = 1689/ 42.935 = 39.34 N/mm ²
Breaking load ( based on computer data), F = 1630 Breaking stress, σb = F/Ao = 1630/ 42.935 = 37.96 N/mm ²
Elongation, ε =
L1
−
L0
x 100%
L0
= [(50.0 – 50.0)/50.0] × 100% = 0.00% Conclusion: PE plastic is more ductile compared to ABS plastic. The difference in property of PE plastic is because it can absorb more energy compare to ABS plastic. PE plastic has larger area under the curve thus it shows that it can absorb more energy. Furthermore, the higher value of elongation percentage,ε for PE plastic compare to ABS plastic suggests that PE plastic is more elastic than ABS plastic.