Chapter 15, Problem 1P

To determine

Calculate the horizontal and vertical components of deflection at joint 1 by solving the required equations of equilibrium.

Answer to Problem 1P

The horizontal deflection at joint 1 is $\underset{\_}{{\Delta }_x=-\frac{96L}{AE}}$.

The vertical deflection at joint 1 is $\underset{\_}{{\Delta }_y=-\frac{172L}{AE}}$.

Explanation of Solution

Apply the sign conventions for the equations of equilibrium as shown below.

• For summation of forces along x-direction is equal to zero $\left(\sum F_x=0\right)$, consider the forces acting towards right side as positive $\left(\to +\right)$ and the forces acting towards left side as negative $\left(\leftarrow -\right)$.
• For summation of forces along y-direction is equal to zero $\left(\sum F_y=0\right)$, consider the upward force as positive $\left(\uparrow +\right)$ and the downward force as negative $\left(\downarrow -\right)$.

Calculation:

Refer to the given Figure.

Resolve the inclined load of 80 kN in joint 1 into horizontal and vertical component.

$\begin{array}{l}\text{Horizontal}=80\left(\frac{3}{5}\right)\\ =48\text{ kN}\\ \text{Vertical}=80\left(\frac{4}{5}\right)\\ =64\text{ kN}\end{array}$

Sketch the Free Body Diagram of the structure into a number of basic cases as shown in Figure 1.

Refer to Figure 1.

Calculate the values of $\cos {\varphi }_x$ and $\sin {\varphi }_x$ for the bars as shown below.

Bar	$\cos {\varphi }_x$	$\sin {\varphi }_x$
1	1	0
2	$-\frac{3}{5}$	$\frac{4}{5}$

Use Equilibrium equations.

Summation of forces along x-direction is equal to 0.

$\begin{array}{l}+\to \sum F_x=0\\ K_{11}{\Delta }_x+K_{12}{\Delta }_y=-48\end{array}$        (1)

Summation of forces along y-direction is equal to 0.

$\begin{array}{l}+\uparrow \sum F_y=0\\ K_{21}{\Delta }_x+K_{22}{\Delta }_y=-64\end{array}$        (2)

Calculate the stiffness coefficients due to the unit horizontal and vertical displacements as shown below.

$\begin{array}{c}{K}_{11}={\displaystyle \sum \frac{AE}{L}{\cos }^2{\varphi }_x}\\ =\frac{AE}{L}\left(1^2+{\left(-\frac{3}{5}\right)}^2\right)\\ =\frac{34AE}{25L}\end{array}$

$\begin{array}{c}{K}_{21}={\displaystyle \sum \frac{AE}{L}\cos {\varphi }_x\sin {\varphi }_x}\\ =\frac{AE}{L}\left(1\times 0+\left(-\frac{3}{5}\right)\left(\frac{4}{5}\right)\right)\\ =-\frac{12AE}{25L}\end{array}$

$\begin{array}{c}{K}_{12}={\displaystyle \sum \frac{AE}{L}\sin {\varphi }_x\cos {\varphi }_x}\\ =\frac{AE}{L}\left(1\times 0+\left(-\frac{3}{5}\right)\left(\frac{4}{5}\right)\right)\\ =-\frac{12AE}{25L}\end{array}$

$\begin{array}{c}{K}_{22}={\displaystyle \sum \frac{AE}{L}{\sin }^2{\varphi }_x}\\ =\frac{AE}{L}\left(0+{\left(\frac{4}{5}\right)}^2\right)\\ =\frac{16AE}{25L}\end{array}$

Solve Equations (1) and (2) to get the values of ${\Delta }_x$ and ${\Delta }_y$ as shown below.

$\begin{array}{l}\frac{34AE}{25L}{\Delta }_x+\left(-\frac{12AE}{25L}\right){\Delta }_y=-48\\ 34{\Delta }_x-12{\Delta }_y=-1,200\frac{L}{AE}\\ \left(-\frac{12AE}{25L}\right){\Delta }_x+\frac{16AE}{25L}{\Delta }_y=-64\\ -12{\Delta }_x+16{\Delta }_y=-1600\frac{L}{AE}\end{array}$

$\begin{array}{l}{\Delta }_x=-\frac{96L}{AE}\\ {\Delta }_y=-\frac{172L}{AE}\end{array}$

Therefore, the horizontal deflection at joint 1 is $\underset{\_}{{\Delta }_x=-\frac{96L}{AE}}$ and the vertical deflection at joint 1 is $\underset{\_}{{\Delta }_y=-\frac{172L}{AE}}$.