Explanation :
Analysis of truss under external loading
Analysis of truss under unit load in desired direction
Member |
Forcedue to external loading |
Force due to unit load in the desired direction (F') |
Length(l) |
Axial rigidity |
$ \sum\frac{\mathrm{FF}'\mathcal l}{\mathrm{AE}} $ |
BC |
-3P |
-1 |
L |
AE |
$ \frac{3\mathrm{PL}}{\mathrm{AE}} $ |
AC |
$ \mathrm P\sqrt2 $ |
$ \sqrt2 $ |
$ \mathrm L\sqrt2 $ |
2AE |
$ \frac{2\sqrt2\mathrm{PL}}{2\mathrm{AE}} $ |
AB |
-P |
-1 |
L |
AE |
$ \frac{\mathrm{PL}}{\mathrm{AE}} $ |
$ \therefore $ Horizontal displacement joint
$ \begin{array}{l}\;\mathrm C=\frac{5.414\;\mathrm{PL}}{\mathrm{AE}}\\(\mathrm P=1\mathrm{kN},\;\mathrm L=1\mathrm m,\;\mathrm A=10\;\mathrm{mm}^2,\;\mathrm E=2\times1011\;\mathrm N/\mathrm m^2)\\\therefore\triangle=\frac{5.414\times10^3\times10^3}{(10\times10)\times2\times10^5}=2.7\;\mathrm{mm}\end{array} $