Compare strain energy in both cases shown in figure. Bars have sa
Compare strain energy in both cases shown in figure. Bars have same Area, length and material.
U1 = strain energy of bar with P1 alone, U2 = strain energy of bar with P2 alone, U = strain energy of bar with P1 and P2 simultaneously
Case 1:
Case 2:
A. <span class="math-tex">\(\;U = \;{U_1} + {U_2}\)</span>
B. <span class="math-tex">\(\;U < \;{U_1} + {U_2}\)</span>
C. <span class="math-tex">\(\;U > \;{U_1} + {U_2}\)</span>
D. <span class="math-tex">\(\;U = \;{U_1} = {U_2}\)</span>
Please scroll down to see the correct answer and solution guide.
Right Answer is: C
SOLUTION
Concept:
Strain energy per unit volume is given as
\({\rm{Strain\;energy\;volume\;}} = \frac{1}{2}\sigma E = \frac{{{\sigma ^2}}}{{2E}}\)
Calculation:
\({\rm{Strain\;energy}} = \frac{{{\sigma ^2}}}{{2E}}AL\)
\({\rm{Strain\;energy\;}} = \frac{{{P^2}AL}}{{2{A^2}E}} = \frac{{{P^2}L}}{{2AE}}\)
\(∴ {U_1} = \frac{{P_1^2L}}{{2AE}}\)\(∴ {U_2} = \frac{{P_2^2L}}{{2AE}}\)
Now,
\(U = \frac{{{{\left( {{P_1} + {P_2}} \right)}^2}L}}{{2AE}}\)
\(∴ U = \frac{{P_1^2L}}{{2AE}} + \frac{{P_2^2L}}{{2AE}} + \frac{{2{P_1}{P_2}L}}{{2AE}}\)
\(∴ \;\;U = {U_1} + {U_2} + \frac{{2{P_1}{P_2}L}}{{2AE}}\;\)
∴ U > U1 + U2