Homework 2 Solutions

Homework 2 Solutions - Problem 2.8 E = 210 GPa t = 0.002 l...

Info iconThis preview shows pages 1–4. Sign up to view the full content.

View Full Document Right Arrow Icon
Problem 2.8 E = 210 GPa ε t = 0.002 ε l = 0.005 " = # t l = 0.002 0.005 = 0.4 Generalized Hooke’s Law with considerations to Poisson’s ratio t = 1 E t $ %# l [ ] l = 1 E l $ t [ ] Solving the equation for σ l l = E l + $# t ( ) 1 % $ 2 = 210 0.005 + 0.4 0.002 ( ) ( ) 1 % 0.4 ( ) 2 = 1.45 GPa t = E t + $" l = 210 0.002 ( ) + 0.4 1.45 ( ) = 1 GPa Without consideration to Poisson's ratio l = E 1 = 210 0.005 = 1.05 GPa t = E t = 210 0.002 = 0.42 GPa Error in stress calculation if Poisson's ratio is not taken into consideration: l = 1.45 % 1.05 = 400 MPa t = 1 % 0.42 = 580 MPa or l = 1.05 % 1.45 1.45 = % 27.6% l = 0.42 % 1.0 1.0 = % 58%
Background image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Problem 2.9
Background image of page 2
Problem 2.11 However if the material is assumed to be cubic instead of isotropic
Background image of page 3

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Background image of page 4
This is the end of the preview. Sign up to access the rest of the document.

Page1 / 4

Homework 2 Solutions - Problem 2.8 E = 210 GPa t = 0.002 l...

This preview shows document pages 1 - 4. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online