Chapter 8, Problem 8.1P

Determine the vertical displacement of joint A. Assume the members are pin connected at their end points. Take A = 3 in² and E = 29(10³) ksi for each member. Use the method of virtual work.

To determine

Vertical displacement of Joint A

Answer to Problem 8.1P

   ${\Delta }_{\text{A}}=2.946\times {10}^{-4}\text{ in}$

Explanation of Solution

Given information:

The members are pin connected at end points.

A = 3 in²

E = 29(10³) ksi

Virtual Work Method

Virtual external work done = Sum of the internal work done

   $\begin{array}{l}\text{P}\times \Delta ={\displaystyle {\sum }_{\text{i}}\frac{{\text{n}}_{\text{i}}{\text{N}}_{\text{i}}{\text{L}}_{\text{i}}}{\text{EA}}}\\ \text{where i = force members}\\ \text{n = force in members due to virtual unit load}\\ \text{N = force in members due to original load}\\ \text{L= length of the force members}\end{array}$

Calculation:

The virtual and real member forces are calculated by method of joints.

  

Member forces when original load is applied:

Taking moment about B, we get,

   $\begin{array}{l}-(12\times 3)+(6\times 6)=-{\text{C}}_{\text{x}}\times 8\\ {\text{C}}_{\text{x}}=0\\ {\text{B}}_{\text{x}}+{\text{C}}_{\text{x}}=0\\ {\text{B}}_{\text{x}}=0\\ {\text{B}}_{\text{y}}=-3\text{ kN}\\ {\text{C}}_{\text{y}}=-6\text{ kN}\end{array}$

Joint A:

   $\begin{array}{l}{\displaystyle \sum {\text{F}}_{\text{y}}}=0\\ {\text{N}}_{\text{AD}}\text{sin 53}{\text{.13}}^{\text{o}}=-3\\ {\text{N}}_{\text{AD}}=-3.75\text{ kN}\\ {\displaystyle \sum {\text{F}}_{\text{x}}}=0\\ -{\text{N}}_{\text{AD}}\text{cos 53}{\text{.13}}^{\text{o}}={\text{N}}_{\text{AB}}\\ {\text{N}}_{\text{AB}}=2.25\text{ kN}\\ \end{array}$

Joint B:

   $\begin{array}{l}{\displaystyle \sum {\text{F}}_{\text{y}}}=0\\ {\text{N}}_{\text{BD}}\text{sin 53}{\text{.13}}^{\text{o}}={\text{B}}_{\text{y}}\\ {\text{N}}_{\text{BD}}=3.75\text{ kN}\\ {\displaystyle \sum {\text{F}}_{\text{x}}}=0\\ {\text{N}}_{\text{BD}}\text{cos 53}{\text{.13}}^{\text{o}}=-{\text{N}}_{\text{BA}}-{\text{B}}_{\text{x}}\\ {\text{N}}_{\text{BA}}=-2.25\text{ kN}\\ \end{array}$

Joint D:

   $\begin{array}{l}{\displaystyle \sum {\text{F}}_{\text{x}}}=0\\ {\text{N}}_{\text{BD}}\text{cos 53}{\text{.13}}^{\text{o}}=-{\text{N}}_{\text{DC}}{\text{+N}}_{\text{DA}}\text{cos 53}{\text{.13}}^{\text{o}}\\ {\text{N}}_{\text{DC}}=0\\ \end{array}$

Member forces when virtual load is applied:

Taking moment about B, we get,

   $\begin{array}{l}{\text{C}}_{\text{x}}=0\\ {\text{B}}_{\text{x}}+{\text{C}}_{\text{x}}=0\\ {\text{B}}_{\text{x}}=0\\ {\text{B}}_{\text{y}}=-1\text{ kN}\\ {\text{C}}_{\text{y}}=0\text{ kN}\end{array}$

Joint A:

   $\begin{array}{l}{\displaystyle \sum {\text{F}}_{\text{y}}}=0\\ {\text{n}}_{\text{AD}}\text{sin 53}{\text{.13}}^{\text{o}}=-1\\ {\text{n}}_{\text{AD}}=-1.25\text{ kN}\\ {\displaystyle \sum {\text{F}}_{\text{x}}}=0\\ -{\text{n}}_{\text{AD}}\text{cos 53}{\text{.13}}^{\text{o}}={\text{n}}_{\text{AB}}\\ {\text{n}}_{\text{AB}}=0.75\text{ kN}\\ \end{array}$

Joint B:

   $\begin{array}{l}{\displaystyle \sum {\text{F}}_{\text{y}}}=0\\ {\text{n}}_{\text{BD}}\text{sin 53}{\text{.13}}^{\text{o}}=-{\text{B}}_{\text{y}}\\ {\text{n}}_{\text{BD}}=1.25\text{ kN}\\ {\displaystyle \sum {\text{F}}_{\text{x}}}=0\\ {\text{n}}_{\text{BD}}\text{cos 53}{\text{.13}}^{\text{o}}=-{\text{n}}_{\text{BA}}-{\text{B}}_{\text{x}}\\ {\text{n}}_{\text{BA}}=-0.75\text{ kN}\\ \end{array}$

Joint D:

   $\begin{array}{l}{\displaystyle \sum {\text{F}}_{\text{x}}}=0\\ {\text{n}}_{\text{BD}}\text{cos 53}{\text{.13}}^{\text{o}}=-{\text{n}}_{\text{DC}}{\text{+n}}_{\text{DA}}\text{cos 53}{\text{.13}}^{\text{o}}\\ {\text{n}}_{\text{DC}}=0\\ \end{array}$

   $\begin{array}{lllll}Member\hfill & n\left(kN\right)\hfill & N\left(kN\right)\hfill & L\left(ft\right)\hfill & nNL\hfill \\ AB\hfill & 0.75\hfill & 2.25\hfill & 12\hfill & 20.25\hfill \\ AD\hfill & -1.25\hfill & -3.75\hfill & 10\hfill & 46.875\hfill \\ BD\hfill & 1.25\hfill & 3.75\hfill & 10\hfill & 46.875\hfill \\ DC\hfill & 0\hfill & 0\hfill & 6\hfill & 0\hfill \end{array}$

   $\begin{array}{l}\text{P}\times {\Delta }_{\text{A}}={\displaystyle {\sum }_{\text{i}}\frac{{\text{n}}_{\text{i}}{\text{N}}_{\text{i}}{\text{L}}_{\text{i}}}{\text{EA}}}\\ {\displaystyle {\sum }_{\text{i}}\frac{{\text{n}}_{\text{i}}{\text{N}}_{\text{i}}{\text{L}}_{\text{i}}}{\text{EA}}}=\frac{114\text{ kN}}{29\times {10}^3\times 3}=\frac{\text{25}\text{.628 kpf}}{29\times {10}^3\times 3}=2.946\times {10}^{-4}\text{ in}\\ \text{1 kN }\times {\Delta }_{\text{A}}=2.946\times {10}^{-4}\text{ in}\\ {\Delta }_{\text{A}}=2.946\times {10}^{-4}\text{ in}\end{array}$

Conclusion:

   ${\Delta }_{\text{A}}=2.946\times {10}^{-4}\text{ in}$