- #1

lizr1

- 2

- 0

An axial load of 28.5kN is applied to a bar of length280mm, width 38mm and thickness 21mm. After the load is applied, the bar elongatesby 4.85mm and the width contracts by 2.98mm. At he prescribed load, note thatthe stress in the bar is less than its proportional limit.

1) Calculate the modulus of elasticity

2) Determine Poisson’s Ratio

**1) Calculate the modulus of elasticity**

Cross Sectional Area = 38 *21 = 798mm

^{2}

Stress(σ) = (F)/(A) = 28.5 / 798 = 35.71 MPa

Strain(ε) = (e)/(L) = 4.85 / 280 = 0.01732 mm/mm

Modulus of elasticity: σ = Eε

35.71 = (E)(0.01732) =

__2.06 GPa__

**2) DeterminePoisson’s Ratio**

ε

_{lat}= Δwidth/width = -2.98 / 38 = -0.0784

v = -ε

_{lat}/ ε

_{long}= -(-0.0784 / 0.01732) =

__4.53__