Answer:
[tex]X_t=2.17391304*10^{-4}[/tex]
[tex]X_r=2.89855072*10^{-4}[/tex]
[tex]e_t=0.0026[/tex]
[tex]e_r=0.0035[/tex]
Explanation:
From the question we are told that:
Dimension [tex]12*12[/tex]
Thickness [tex]l_t=5mm=5*10^-3[/tex]
Normal tensile force on top side [tex]F_t= 15kN[/tex]
Normal tensile force on right side [tex]F_r= 20kN[/tex]
Elastic modulus, [tex]E=115Gpap=>115*10^9[/tex]
Generally the equation for Normal Strain X is mathematically given by
[tex]X=\frac{Force}{Area*E}[/tex]
Therefore
For Top
[tex]X_t=\frac{Force_t}{Area*E}[/tex]
Where
[tex]Area=L*B*T[/tex]
[tex]Area=12*10^{-2}*5*10^{-3}[/tex]
[tex]Area=6*10^{-4}[/tex]
[tex]X_t=\frac{15*10^3}{6*10^{-4}*115*10^9}[/tex]
[tex]X_t=2.17391304*10^{-4}[/tex]
For Right side[tex]X_r=\frac{Force_r}{Area*E}[/tex]
Where
Area=L*B*T
[tex]Area=12*10^{-2}*5*10^{-3}[/tex]
[tex]Area=6*10^{-4}[/tex]
[tex]X_r=2.89855072*10^{-4}[/tex]
[tex]X_r=2.89855072*10^{-4}[/tex]
Generally the equation for elongation is mathematically given by
[tex]e=strain *12[/tex]
For top
[tex]e_t=2.17391304*10^{-4}*12[/tex]
[tex]e_t=0.0026[/tex]
For Right
[tex]e_r=2.89855072*10^{-4} *12[/tex]
[tex]e_r=0.0035[/tex]
