MEC2402 Stress Analysis: Aluminium Can Strain Gauge Analysis Report

This report details a stress analysis performed on an aluminum can using strain gauges, referencing a video demonstration. It identifies material properties (E, G, ν) for aluminum, specifies strains (εx, εy, γxy) from gauge measurements, and calculates corresponding stresses (σx, σy, τxy). The report further determines principal stresses (σ1, σ2) and their orientation (θp), discusses the significance of θp being zero, and estimates internal pressure treating the can as a thin-walled pressure vessel, comparing the results with expected values. The assignment solution, contributed by a student, is available on Desklib, a platform offering study tools and resources for students.

STRESS ANALYSIS REPORT
1 | P a g e
Solution 1
(a) Identify reasonable and consistent values for 𝐸, 𝐺, and 𝜈 for aluminium can material
from a material properties table in a textbook or similar reference resource, in combination
with the use of the known 𝐸-𝐺-𝜈 relationship. [10 marks]
Solution :-Properties of aluminium
Property Value Units
Young modulus (E) 70 GPa
Shear modulus (G) 26 GPa
Poisson ratio (𝜗) 0.35 -
Relation 𝑬 = 𝟐𝑮(𝟏 + 𝝑)
(b) Taking the x-axis direction as aligned with the gauge on the 0° axis (the longitudinal directi
the can), and using the measured strains in the 3 axis directions, specify the strains𝜖𝑥, 𝜖𝑦, and𝛾𝑥𝑦.
[10 marks]
Solution :-Values of strain using gauge is
S.No Angle with longitudinal
direction (Degree)
Strain (𝜇 ∈)
1 0 332
2 45 900
3 90 1452
(c) Using the material properties identified in part (a), determine the stresses𝜎𝑥, 𝜎𝑦, and𝜏𝑥𝑦. [15
marks]
From above table
𝜖𝑎 = 332𝜇 ∈
𝜖𝑏 = 900𝜇 ∈
𝜖𝑐 = 1452𝜇 ∈

STRESS ANALYSIS REPORT
2 | P a g e
𝜎𝑥 = 𝐸(1 − 𝜗)𝜀1 + 𝜗𝜀2
(1 + 𝜗)(1 − 2𝜗)
𝜎𝑦 = 𝐸(1 − 𝜗)𝜀2 + 𝜗𝜀1
(1 + 𝜗)(1 − 2𝜗)
∈1= 1
3(𝜀𝑎 + 𝜀𝑏 + 𝜀𝑐) + √2
3 √(𝜀𝑎 − 𝜀𝑏)2 + (𝜀𝑏 − 𝜀𝑐)2 + (𝜀𝑎 − 𝜀𝑐)2
∈1= 1
3(332 + 900 + 1452) +
√2
3 √(332 − 900)2 + (900 − 1452)2 + (1452 − 332)2
∈1= 894.67 + 646.65
∈1= 1541.32𝜇 ∈
∈2= 1
3(𝜀𝑎 + 𝜀𝑏 + 𝜀𝑐) − √2
3 √(𝜀𝑎 − 𝜀𝑏)2 + (𝜀𝑏 − 𝜀𝑐)2 + (𝜀𝑎 − 𝜀𝑐)2
∈2= 894.67 − 646.65
∈2= 248.01𝜇 ∈
𝜎𝑥 = 𝐸(1 − 𝜗)𝜀1 + 𝜗𝜀2
(1 + 𝜗)(1 − 2𝜗)
𝜎𝑥 = 70((1 − 0.35) 1541.32 + 0.35(248.01))
(1 + 0.35)(1 − 2 ∗ 0.35)
𝜎𝑥 = 76.205 𝑀𝑃𝑎
𝜎𝑦 = 𝐸(1 − 𝜗)𝜀2 + 𝜗𝜀1
(1 + 𝜗)(1 − 2𝜗)
𝜎𝑦 = 70((1 − 0.35) 248.01 + 0.35(1541.32))
(1 + 0.35)(1 − 2 ∗ 0.35)
𝜎𝑦 = 121.102 𝑀𝑃𝑎
𝜏𝑥𝑦 = 0
(d) Identify the values of the principal stresses𝜎1 and𝜎2 and the orientation of the principal
stresses𝜃𝑝relative to the x-y axis directions. [15 marks]
𝜎1 = 𝐸(𝜀1 + 𝜗𝜀2)
(1 − 𝜗2)

STRESS ANALYSIS REPORT
3 | P a g e
𝜎1 = 70(1541.32 + 0.35 ∗ 248.01)
(1 − 0.352)
𝜎1 = 129.87 𝑀𝑃𝑎
𝜎2 = 𝐸𝜀2 + 𝜗𝜎1
𝜎2 = 70 ∗ 248.01 + 0.35 ∗ 129.8
𝜎2 = 17.406𝑀𝑃𝑎
𝑡𝑎𝑛2𝜃𝑝
= (𝜖𝑎 − 2𝜖𝑏 + 𝜖𝑐)
𝜖𝑎 − 𝜖𝑐
𝑡𝑎𝑛2𝜃𝑝 = (332 − 2 ∗ 900 + 1452)
−1452 + 332
𝑡𝑎𝑛2𝜃𝑝 = (332 − 2 ∗ 900 + 1452)
−1452 + 332
𝑡𝑎𝑛2𝜃𝑝 = −16
−1120
𝜃𝑝 = 0
(e) Should the value of𝜃𝑝be zero in this case? Discuss this matter. [10 marks]
𝜃𝑝 = 0
𝜃𝑝 value is zero because at maximum principal stress , angle between maximum stress is zero
(f) Treating the can as a thin-walled pressure vessel with a diameter of 65 mm and a wall thick
of 0.1 mm, calculate values for the internal pressure in the unopened can based on results in (
marks]
𝜏𝑚𝑎𝑥 = 𝜎1 − 𝜎2
2 = 𝑝𝑑
8𝑡
𝜏𝜏𝜏𝜏𝜏𝑚𝑎𝑥=
129.87−17.406
2
𝜏𝑚𝑎𝑥 = 56.23 𝑀𝑃𝑎
𝜏𝑚𝑎𝑥 = 𝑝𝑑
8𝑡

STRESS ANALYSIS REPORT
4 | P a g e
56.23 =𝑝 ∗ 65
8 ∗ 0.1
𝒑 = 𝟔𝟗𝟐 𝑲𝑷𝒂
(g) The gauge pressure for the contents of a sealed coke can at 25 ºC is expected to be around
kPa. Discuss the results obtained in part (f). [10 marks]
The value obtained from above formula (f) =692 KPa , which is greater than the expected gauge
pressure 330 KPa , the main difference is due to following reason :-
- The p=692 KPa is absolute pressure value but p =330 KPa is gauge pressure value .So in
gauge pressure , the atmospheric value need to added .
- There always be human error in experimental result .