Form the matrices required to solve completely the plane truss shown in Fig. P.6.3 and determine the force in member 24. All members have equal axial rigidity.

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Referring to Fig. P.6.3 and Fig. 6.3

[latex]\begin{array}{lccccc} ext { Member } & 12 & 23 & 34 & 45 & 24 \ ext { Length } & l & l & l & l & l \\lambda(\cos heta) & -1 / 2 & 1 / 2 & -1 / 2 & 1 / 2 & 1 \\mu(\sin heta) & \sqrt{3} / 2 & \sqrt{3} / 2 & \sqrt{3} / 2& \sqrt{3} / 2 & 0\end{array}[/latex]

From Eq. (6.30) the member stiffness matrices are

[latex]\left[K_{i j} ight]=\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} ight][/latex] (6.30)

[latex]\begin{aligned}&{\left[K_{12} ight]=\frac{A E}{l}\left[\begin{array}{cccc}1 / 4 & -\sqrt{3} / 4 & -1 / 4 & \sqrt{3} / 4 \-\sqrt{3} / 4 & 3 / 4 & \sqrt{3} / 4 & -3 / 4 \-1 / 4 & \sqrt{3} / 4 & 1 / 4 & -\sqrt{3} / 4 \\sqrt{3} / 4 & -3 / 4 & -\sqrt{3} / 4 & 3 / 4\end{array} ight]} \&{\left[K_{23} ight]=\frac{A E}{l}\left[\begin{array}{cccc}1 / 4 & \sqrt{3} / 4 & -1 / 4 & -\sqrt{3} / 4 \\sqrt{3} / 4 & 3 / 4 & -\sqrt{3} / 4 & -3 / 4 \-1 / 4 & -\sqrt{3} / 4 & 1 / 4 & \sqrt{3} / 4 \-\sqrt{3} / 4 & -3 / 4 & \sqrt{3} / 4 & 3 / 4\end{array} ight]} \&{\left[K_{34} ight]=\frac{A E}{l}\left[\begin{array}{cccc}1 / 4 & -\sqrt{3} / 4 & -1 / 4 & \sqrt{3} / 4 \-\sqrt{3} / 4 & 3 / 4 & \sqrt{3} / 4 & -3 / 4 \-1 / 4 & \sqrt{3} / 4 & 1 / 4 & -\sqrt{3} / 4 \\sqrt{3} / 4 & -3 / 4 & -\sqrt{3} / 4 & 3 / 4\end{array} ight]} \&\left\lfloor K_{45} ight floor=\frac{A E}{l}\left[\begin{array}{cccc}1 / 4 & \sqrt{3} / 4 & -1 / 4 & -\sqrt{3} / 4 \\sqrt{3} / 4 & 3 / 4 & -\sqrt{3} / 4 & -3 / 4 \-1 / 4 & -\sqrt{3} / 4 & 1 / 4 & \sqrt{3} / 4 \-\sqrt{3} / 4 & -3 / 4 & \sqrt{3} / 4 & 3 / 4\end{array} ight]\end{aligned}[/latex]

[latex]\left[K_{24} ight]=\frac{A E}{l}\left[\begin{array}{rrrr}1 & 0 & -1 & 0 \0 & 0 & 0 & 0 \-1 & 0 & 1 & 0 \0 & 0 & 0 & 0\end{array} ight][/latex]

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

[latex]\{F\}=\left[K_{i j} ight]\{\delta\}[/latex] (6.29)

[latex]\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} ight\}=\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} ight]\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} ight\}[/latex] (i)

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

[latex]\begin{aligned}F_{x, 2} &=0=\frac{A E}{4 l}\left(6 u_2-4 u_4 ight) (ii) \F_{y, 2} &=0=\frac{A E}{4 l}\left(6 v_2 ight) (iii)\F_{x, 4} &=0=\frac{A E}{4 l}\left(-4 u_2+6 u_4 ight) (iv)\F_{y, 4} &=-P=\frac{A E}{4 l}\left(6 v_4 ight) (v)\end{aligned}[/latex]

From Eq. (v)

[latex]v_4=-\frac{2 P l}{3 A E}[/latex]

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

Hence, from Eq. (6.32)

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

[latex]S_{24}=\frac{A E}{l}\left[\begin{array}{ll}1 & 0\end{array} ight]\left\{\begin{array}{c}0-0 \\frac{-2 P l}{3 A E}-0\end{array} ight\}[/latex]

which gives
[latex]S_{24} = 0[/latex]

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

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