# Consider a linear elastic rectangular thin sheet of metal, subjected to uniform uniaxial tensile stress of 100 MPa along the length direction. Assume

+1 vote
35 views
in General
closed
Consider a linear elastic rectangular thin sheet of metal, subjected to uniform uniaxial tensile stress of 100 MPa along the length direction. Assume plane stress conditions in the plane normal to the thickness. The Young’s modulus E = 200 MPa and Poisson’s ratio ν = 0.3 are given. The principal strains in the plane of the sheet are
1. (0.35, −0.15)
2. (0.5, 0.0)
3. (0.5, −0.15)
4. (0.5, −0.5)

by (19.3k points)
selected

Correct Answer - Option 3 : (0.5, −0.15)

Concept:

${\epsilon_x} = \frac{{{\sigma _x}}}{E} - \mu \left( {\frac{{{\sigma _y}}}{E} + \frac{{{\sigma _z}}}{E}} \right)$

Calculation:

σ1 = 100 MPa, μ = 0.3 and ϵ = 200 MPa

∵ It is uniaxial tensile stress so, σ2 = σ3 = 0

${\epsilon_1} = \frac{{{\sigma _1}}}{E} - \mu \left( {\frac{{{\sigma _2}}}{E} + \frac{{{\sigma _3}}}{E}} \right)$

$\Rightarrow {\epsilon_1} = \frac{{{\sigma _1}}}{E} = \frac{{100}}{{200}} = 0.5$

$\Rightarrow {\epsilon_2} = \frac{{{\sigma _2}}}{E} - \mu \left( {\frac{{{\sigma _1}}}{E} + \frac{{{\sigma _3}}}{E}} \right) = - \mu \frac{{{\sigma _1}}}{E}$

$\Rightarrow {\epsilon_2} = - 0.3\left( {\frac{{100}}{{200}}} \right) = - 0.15\therefore \left( {{\epsilon_1},\epsilon{_2}} \right) = \left( {0.5, - 0.15} \right)$

+1 vote