Answer: ε₁+ε₂+ε₃ = 0
Explanation: Considering the initial and final volume to be constant which gives rise to the relation:-
l₀l₀l₀=l₁l₂l₃
taking natural log on both sides
Considering the logarithmic Laws of division and multiplication :
ln(AB) = ln(A)+ln(B)
ln(A/B) = ln(A)-ln(B)
Use the image attached to see the definition of true strain defined as
ln(l1/1o)= ε₁
which then proves that ε₁+ε₂+ε₃ = 0
hooooooooooooooooooooooooooooooooooooooooooooooooooooooe
Answer:
The compressive stress of aplying a force of 708 kN in a 81 mm diamter cylindrical component is 0.137 kN/mm^2 or 137465051 Pa (= 137.5 MPa)
Explanation:
The compressive stress in a cylindrical component can be calculated aby dividing the compressive force F to the cross sectional area A:
fc= F/A
If the stress is wanted in Pascals (Pa), F and A must be in Newtons and square meters respectively.
For acylindrical component the cross sectional area A is:
A=πR^
If the diameter of the component is 81 mm, the radius is the half:
R=81mm /2 = 40.5 mm
Then A result:
A= 3.14 * (40.5 mm)^2 = 5150.4 mm^2
In square meters:
A= 3.14 * (0.0405 m)^2 = 0.005150 m^2
Replacing 708 kN to the force:
fc= 708 kN / 5150.4 mm^2 = 0.137 kN/mm^2
Using the force in Newtons:
F= 70800 N
Finally the compressive stress in Pa is:
fc= 708000 / 0.005150 m^2 = 137465051 Pa = 137 MPa
