1. The problem statement, all variables and given/known data 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? 2. Relevant 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 3. 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!!