Question p.6.3: Form the matrices required to solve completely the plane tru...

Referring to Fig. P.6.3 and Fig. 6.3

From Eq. (6.30) the member stiffness matrices are

\left[K_{i j}\right]=\frac{A E}{L}\left[\begin{array}{cccc}\lambda^2 & \lambda \mu & -\lambda^2 & -\lambda \mu \\\lambda \mu & \mu^2 & -\lambda \mu & -\mu^2 \\-\lambda^2 & -\lambda \mu & \lambda^2 & \lambda \mu \\-\lambda \mu & -\mu^2 & \lambda \mu & \mu^2\end{array}\right]  (6.30)

The stiffness matrix for the complete truss is now assembled using the method described in Example 6.1. Equation (6.29) then becomes

\{F\}=\left[K_{i j}\right]\{\delta\}  (6.29)

\left\{\begin{array}{l}F_{x, 1} \\F_{y, 1} \\F_{x, 2} \\F_{y, 2} \\F_{x, 3} \\F_{y, 3} \\F_{x, 4} \\F_{y, 4} \\F_{x, 5} \\F_{y, 5}\end{array}\right\}=\frac{A E}{4 l}\left[\begin{array}{cccccccccc}1 &-\sqrt{3} & -1 & \sqrt{3} & 0 & 0 & 0 & 0 & 0 & 0 \\-\sqrt{3} & 3 & \sqrt{3} & -3 & 0 & 0 & 0 & 0 & 0 & 0 \\-1 & \sqrt{3} & 6 & 0 & -1 &\sqrt{3} & -4 & 0 & 0 & 0 \\\sqrt{3} & -3 & 0 & 6 & -\sqrt{3} & -3 & 0 & 0 & 0 & 0 \\0 & 0 & -1 & -\sqrt{3} & 2 & 0 & -1 & \sqrt{3} & 0 & 0 \\0 & 0 & -\sqrt{3} &- 3 & 0 & 6 & \sqrt{3} &- 3 & 0 & 0 \\0 & 0 & -4 & 0 & -1 & \sqrt{3} & 6 & 0 & -1 & -\sqrt{3} \\0 & 0 & 0 & 0 & \sqrt{3} & -3 & 0 & 6 & -\sqrt{3} & -3 \\0 & 0 & 0 & 0 & 0 & 0 & -1 & -\sqrt{3} & 1 & \sqrt{3} \\0 & 0 & 0 & 0 & 0 & 0 & -\sqrt{3} & -3 & \sqrt{3} & 3\end{array}\right]\left\{\begin{array}{c}u_1=0 \\v_1=0 \\u_2 \\v_2 \\u_3=0 \\v_3=0 \\u_4 \\v_4 \\u_5=0 \\v_5=0\end{array}\right\}  (i)

In Eq. (i) F_{x, 2}=F_{y, 2}=0, F_{x, 4}=0, F_{y, 4}=-P .. Thus from Eq. (i)

From Eq. (v)

From Eq. (iii)
v_2=0
and from Eqs (ii) and (iv)
u_2=u_4=0

Hence, from Eq. (6.32)

S_{i j}=\frac{A E}{I}\left[\begin{array}{ll}\lambda & \mu \\& i j\end{array}\right]\left\{\begin{array}{l}u_j-u_i \\v_j-v_i\end{array}\right\}  (6.32)

which gives
S_{24} = 0