At a point in a loaded material subjected to plane strains: ϵ x =
A. Maximum shearing strain is 406.06 × 10<sup>-6</sup>
B. Maximum principal strain is 543. 03 × 10<sup>-6</sup>
C. Minimum principal strain is 203.03 × 10<sup>-6</sup>
D. Maximum shearing strain is 203.03 × 10<sup>-6</sup>
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Right Answer is:
SOLUTION
Concept
Principle strains are given by
\({\epsilon_{1\;}} = \frac{{{\epsilon_x} + {\epsilon_y}}}{2} + \sqrt {{{\left( {\frac{{{\epsilon_x} - {\epsilon_x}}}{2}} \right)}^2} + {{\left( {\frac{{{\phi _{xy}}}}{2}} \right)}^2}\;} \;\;\)
\({\epsilon_{2\;}} = \frac{{{\epsilon_x} + {\epsilon_y}}}{2} - \sqrt {{{\left( {\frac{{{\epsilon_x} - {\epsilon_x}}}{2}} \right)}^2} + {{\left( {\frac{{{\phi _{xy}}}}{2}} \right)}^2}\;} \)
Maximum shearing strain is given by
\(\left| {\frac{{{\phi _{max}}}}{2}} \right| = \left| {\frac{{({\epsilon_1} - {\epsilon_2})}}{2}} \right|\)
Calculation
Given,
ϵx = 500 × 10-6, ϵy = 180 × 10-6 and ϕxy = 250 × 10-6
Principle strains are given by
\({\epsilon_{1\;}} = \frac{{{\epsilon_x} + {\epsilon_y}}}{2} + \sqrt {{{\left( {\frac{{{\epsilon_x} - {\epsilon_x}}}{2}} \right)}^2} + {{\left( {\frac{{{\phi _{xy}}}}{2}} \right)}^2}\;} \;\;\)
\({\epsilon_{1\;}} = \frac{{\left( {500 + 180} \right) \times {{10}^{ - 6}}}}{2} + \sqrt {{{\left( {\frac{{500 - 180}}{2}} \right)}^2} \times {{10}^{ - 12}} + {{\left( {\frac{{250}}{2}} \right)}^2} \times {{10}^{ - 12}}\;} \;\;\)
= 340 × 10-6 + 203.03 × 10-6
= 543. 03 × 10-6
\({\epsilon_{2\;}} = \frac{{{\epsilon_x} + {\epsilon_y}}}{2} - \sqrt {{{\left( {\frac{{{\epsilon_x} - {\epsilon_x}}}{2}} \right)}^2} + {{\left( {\frac{{{\phi _{xy}}}}{2}} \right)}^2}\;}\)
\({\epsilon_{2\;}} = \frac{{\left( {500 + 180} \right) \times {{10}^{ - 6}}}}{2} - \sqrt {{{\left( {\frac{{500 - 180}}{2}} \right)}^2} \times {{10}^{ - 12}} + {{\left( {\frac{{250}}{2}} \right)}^2} \times {{10}^{ - 12}}\;} \)
= 340 × 10-6 - 203.03 × 10-6
= 136.97 × 10-6
Maximum shearing strain is given by
\(\left| {\frac{{{\phi _{max}}}}{2}} \right| = \left| {\frac{{({\epsilon_1} - {\epsilon_2})}}{2}} \right|\)
\({\phi _{max}} = {\epsilon_1} - {\epsilon_2}\)
= 543. 03 × 10-6 - 136.97 × 10-6
= 406.06 × 10-6