Chapter 7, Problem 50P

To determine

Find the vertical deflection at joint E of the frame using virtual work method.

Answer to Problem 50P

The vertical deflection at joint E of the frame is $\underset{\_}{\text{23}\text{mm}\downarrow }$.

Explanation of Solution

Given information:

The frame is given in the Figure.

The value of E is 200 GPa and I is $350\times {10}^6{\text{mm}}^4$.

Apply the sign conventions for calculating reactions, forces and moments using the three 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 $\left(\downarrow -\right)$.
• For summation of moment about a point is equal to zero $\left(\sum M_{\text{at}\text{a}\text{point}}=0\right)$, consider the clockwise moment as negative and the counter clockwise moment as positive.

Calculation:

Consider the real system.

Draw a diagram showing all the given real loads acting on it.

Let the bending moment due to real load be $M$.

Sketch the real system of the frame as shown in Figure 1.

Find the reactions at the supports A and B:

Summation of moments about A is equal to 0.

$\begin{array}{l}\sum M_A=0\\ B_y\left(10\right)-15\left(15\right)\left(4.5\right)-65\left(3\right)=0\\ B_y=120.75\text{kN}\uparrow \end{array}$

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

$\begin{array}{l}+\uparrow \sum F_y=0\\ A_y+B_y-15\left(15\right)=0\\ A_y+120.75-15\left(15\right)=0\\ A_y=104.25\text{kN}\uparrow \end{array}$

Consider ACDE, summation of moments about E is equal to 0.

$\begin{array}{l}\sum M_E=0\\ 15\left(8\right)\left(4\right)+65\left(3\right)-A_x\left(6\right)-A_y\left(5\right)=0\\ 480+195-A_x\left(6\right)-104.25\left(5\right)=0\\ -6A_x+153.75=0\\ A_x=25.625\text{kN}\leftarrow \end{array}$

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

$\begin{array}{l}+\to \sum F_x=0\\ -A_x-B_x+65=0\\ -25.625-B_x+65=0\\ B_x=39.375\text{kN}\leftarrow \end{array}$

Sketch the member end forces of the real system as shown in Figure 2.

Consider the virtual system.

Draw a diagram of frame without the given real loads. For vertical deflection apply unit load at the joint in the vertical direction.

Let the bending moment due to virtual load be $M_v$.

Sketch the virtual system of the frame with unit load at joint E as shown in Figure 3.

Find the reactions at the supports A and B:

Summation of moments about A is equal to 0.

$\begin{array}{l}\sum M_A=0\\ B_y\left(10\right)-1\left(5\right)=0\\ B_y=\frac{1}{2}\text{kN}\uparrow \end{array}$

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

$\begin{array}{l}+\uparrow \sum F_y=0\\ -A_y+\frac{1}{2}-1=0\\ A_y=\frac{1}{2}\text{kN}\uparrow \end{array}$

Consider ACDE, summation of moments about E is equal to 0.

$\begin{array}{l}\sum M_E=0\\ A_x\left(6\right)-A_y\left(5\right)=0\\ A_x\left(6\right)-\frac{1}{2}\left(5\right)=0\\ A_x=\frac{5}{12}\text{kN}\to \end{array}$

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

$\begin{array}{l}+\to \sum F_x=0\\ A_x-B_x=0\\ \frac{5}{12}-B_x=0\\ B_x=\frac{5}{12}\text{kN}\leftarrow \end{array}$

Sketch the member end forces of the virtual system as shown in Figure 4.

Find the equations for M and $M_v$ for the 5 segments of the frame as shown in Table 1.

Segment	x-coordinate		M $\left(\text{kN}\cdot \text{m}\right)$	$M_v$ $\left(\text{kN}\cdot \text{m}\right)$
	Origin	Limits (m)
AH	A	$0-3$	$25.625x$	$-\frac{5x}{12}$
HD	A	$3-6$	$25.625x-65\left(x-3\right)$	$-\frac{5x}{12}$
BF	B	$0-6$	$39.375x$	$\frac{5x}{12}$
DE	D	$0-5$	$59.25x-108.75-7.5x^2$	$\frac{x}{2}-\frac{5}{2}$
FE	F	$0-5$	$90.75x-266.25-7.5x^2$	$\frac{x}{2}-\frac{5}{2}$

Find the vertical deflection at E using the virtual work expression:

${\Delta }_E={\displaystyle {\int }_0^L\frac{M_{v}M}{EI}dx}$ (1)

Moment of inertia of span AH is I, moment of inertia of span HD is I, moment of inertia of span BF is I, moment of inertia of span DE is 2I, and moment of inertia of span FE is 2I.

Rearrange Equation (1) for the limits $0-3$, $3-6$, $0-6$, $0-5$, and $0-5$ as follows.

${\Delta }_E=\frac{1}{EI}\left[{\displaystyle {\int }_0^3\left(M_{v}M\right)dx}+{\displaystyle {\int }_3^6\left(M_{v}M\right)dx}+{\displaystyle {\int }_0^6\left(M_{v}M\right)dx}+\frac{1}{2}{\displaystyle {\int }_0^5\left(M_{v}M\right)dx}+\frac{1}{2}{\displaystyle {\int }_0^5\left(M_{v}M\right)dx}\right]$

Substitute $-\frac{5x}{12}$ for $M_v$, $25.625x$ for $M$ for the limits $0-3$, $-\frac{5x}{12}$ for $M_v$, $25.625x-65\left(x-3\right)$ for $M$ for the limits $3-6$, $\frac{5x}{12}$ for $M_v$, $39.375x$ for $M$ for the limits $0-6$, $\frac{x}{2}-\frac{5}{2}$ for $M_v$, $59.25x-108.75-7.5x^2$ for $M$ for the limits $0-5$, $\frac{x}{2}-\frac{5}{2}$ for $M_v$, $90.75x-266.25-7.5x^2$ for $M$ for the limits $0-5$.

$\begin{array}{c}{\Delta }_E=\frac{1}{EI}\left[\begin{array}{l}{\displaystyle {\int }_0^3\left(-\frac{5x}{12}\right)\left(25.625x\right)dx}+{\displaystyle {\int }_3^6\left(-\frac{5x}{12}\right)\left(25.625x-65\left(x-3\right)\right)dx}\\ +{\displaystyle {\int }_0^6\left(\frac{5x}{12}\right)\left(39.375x\right)dx}+\frac{1}{2}{\displaystyle {\int }_0^5\left(\frac{x}{2}-\frac{5}{2}\right)\left(59.25x-108.75-7.5x^2\right)dx}\\ +\frac{1}{2}{\displaystyle {\int }_0^5\left(\frac{x}{2}-\frac{5}{2}\right)\left(90.75x-266.25-7.5x^2\right)dx}\end{array}\right]\\ =\frac{1}{EI}\left[\begin{array}{l}{\displaystyle {\int }_0^3\left(-10.677x^2\right)dx}+{\displaystyle {\int }_3^6\left(-\frac{5x}{12}\right)\left(-39.375x+195\right)dx}+{\displaystyle {\int }_0^6\left(16.406x^2\right)dx}\\ +\frac{1}{2}{\displaystyle {\int }_0^5\left(29.625x^2-54.375x-3.75x^2-148.125x+271.875+18.75x^2\right)dx}\\ +\frac{1}{2}{\displaystyle {\int }_0^5\left(45.375x^2-133.125x-3.75x^2-226.875x+665.625+18.75x^2\right)dx}\end{array}\right]\\ =\frac{1}{EI}\left[\begin{array}{l}{\displaystyle {\int }_0^3\left(-10.677x^2\right)dx}+{\displaystyle {\int }_3^6\left(16.406x^2-81.25x\right)dx}+{\displaystyle {\int }_0^6\left(16.406x^2\right)dx}\\ +\frac{1}{2}{\displaystyle {\int }_0^5\left(44.625x^2-202.5x+271.875\right)dx}\\ +\frac{1}{2}{\displaystyle {\int }_0^5\left(60.375x^2-360x+665.625\right)dx}\end{array}\right]\\ =\frac{1}{EI}\left[\begin{array}{l}{\left(\frac{-10.677x^3}{3}\right)}_0^3+{\left(\frac{16.406x^3}{3}-\frac{81.25x^2}{2}\right)}_3^6+{\left(\frac{16.406x^3}{3}\right)}_0^6\\ +\frac{1}{2}{\left(\frac{44.625x^3}{3}-\frac{202.5x^2}{2}+271.875x\right)}_0^5\\ +\frac{1}{2}{\left(\frac{60.375x^3}{3}-\frac{360x^2}{2}+665.625x\right)}_0^5\end{array}\right]\end{array}$

$\begin{array}{c}{\Delta }_E=\frac{1}{EI}\left[\begin{array}{l}\frac{-10.677{\left(3\right)}^3}{3}+\frac{16.406{\left(6\right)}^3}{3}-\frac{81.25{\left(6\right)}^2}{2}-\frac{16.406{\left(3\right)}^3}{3}+\frac{81.25{\left(3\right)}^2}{2}+\frac{16.406{\left(6\right)}^3}{3}\\ +\frac{1}{2}\left(\frac{44.625{\left(5\right)}^3}{3}-\frac{202.5{\left(5\right)}^2}{2}+271.875\left(5\right)\right)\\ +\frac{1}{2}{\left(\frac{60.375{\left(5\right)}^3}{3}-\frac{360{\left(5\right)}^2}{2}+665.625\left(5\right)\right)}_0^5\end{array}\right]\\ =\frac{1,607.8125{\text{kN}}^2\cdot {\text{m}}^{\text{3}}}{EI}\end{array}$

Substitute 200 GPa for E and $350\times {10}^6{\text{mm}}^4$ for I.

$\begin{array}{c}{\Delta }_E=\frac{1,607.8125{\text{kN}}^2\cdot {\text{m}}^{\text{3}}}{\left(200\text{GPa}\right)\left(350\times {10}^6{\text{mm}}^4\right)}\\ =\frac{1,607.8125{\text{kN}}^2\cdot {\text{m}}^{\text{3}}\times \frac{{10}^9{\text{mm}}^{\text{3}}}{1{\text{m}}^3}}{\left(200\text{GPa}\times \frac{1{\text{kN/mm}}^2}{1\text{GPa}}\right)\left(350\times {10}^6{\text{mm}}^4\right)}\\ =22.97\text{mm}\\ \approx \text{23}\text{mm}\downarrow \end{array}$

Therefore, the vertical deflection at E is $\underset{\_}{\text{23}\text{mm}\downarrow }$.