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LDC1972
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Homework Statement
Mild steel bar 40 mm diameter and 100 mm long, subjected to a tensile force along its axis (through central length)
Youngs Modulus = 200GN m^-2
Poisson's ratio = 0.3
Calculate the force required to reduce the diameter to 39.99mm
(there is also a picture saying Use the x- y co-ordinate system above). I shall upload that momentarily if needed?
Homework Equations
Poissons ratio = Transverse strain = - εt / εl
Diameter = 0.04m
Lo = 0.1m
E = 200GN m^-2 = 200x10^9 Pascals (Pa)
V = 0.3
Transverse strain εt = (Reduction in Diameter - Original Diameter) / Original Diameter
Axial strain = Transverse strain / V
Δd = εtd
ε = εt / -V
ε = σ / E
σ = F/A
The Attempt at a Solution
So here's what I did and hoping it is right?
Transverse strain εt = (Reduction in Diameter - Original Diameter) / Original Diameter
(0.3999 - 0.4000) / 0.4000 = -0.00025 Transverse strain (εt)
Axial strain = Transverse strain / V
-0.00025 / 0.3 = .00083333333 Axial strain
I confirmed these by dividing Transverse strain by Axial strain = -0.3
Since decrease in diameter = transverse strain x Original Diameter then:
Δd = εtd = -0.00025 x .40 = 0.1mm which is correct.
As εt = -Vε Then:
-0.00025 = -0.3 x ε
∴
ε = εt / -V
ε = 0.00083333333
ε = σ / E
σ is unknown ∴
σ = ε x E
σ = 16666666 Pa (stress value)
σ = F / A
∴
Force = σ x A (Area)
= 16666666 Pa x 0.00125663706
= 20943.95019 Nm
I think here I can bring the figure down a thousand to 20.944 KNm?
Thank guys for ANY help! This took me ages!