Answer:
The resultant strain in the aluminum specimen is [tex]4.14 \times {10^{ - 3}}[/tex]
Explanation:
Given that,
Dimension of specimen of aluminium, 9.5 mm × 12.9 mm
Area of cross section of aluminium specimen,
[tex]A=9.5\times 12.9=123.84\times 10^{-6}\ mm^2A=122.55\times 10^{-6}\ m^2[/tex]
Tension acting on object, T = 35000 N
The elastic modulus for aluminum is,[tex]E=69\ GPa=69\times 10^9\ Pa[/tex]
The stress acting on material is proportional to the strain. Its formula is given by :
[tex]\epsilon=\dfrac{\sigma}{E}[/tex]
[tex]\sigma[/tex] is the stress
[tex]\epsilon=\dfrac{F}{EA}\\\epsilon=\dfrac{35000}{69\times 10^9\times 122.5\times 10^{-6}}\\\epsilon=4.14\times 10^{-3}[/tex]
Thus, The resultant strain in the aluminum specimen is [tex]4.14 \times {10^{ - 3}}[/tex]
Answer:
Strain = 4.139 x 10^(-3)
Explanation:
We are given;
Dimension; 9.5 mm by 12.9 mm = 0.0095m by 0.0129m
Elastic Modulus; E = 69GPa = 69 x 10^(9)N/m²
Force = 35,000N
Now, Elastic modulus is given by;
E = σ/ε
Where
E is elastic modulus
σ is stress
ε is strain
Now, stress is given by the formula;
σ = F/A
Area = 0.0095m x 0.0129m = 0.00012255 m²
Thus, σ = 35000/0.00012255 = 285597715.218 N/m²
Now, we are looking for strain.
Let's make ε the subject;
E = σ/ε, Thus, ε = σ/E = 285597715.218/69 x 10^(9) = 0.00413909732 = 4.139 x 10^(-3)
