Chapter 7, Problem 9P

To determine

Calculate the horizontal deflection at joint H of the truss.

Answer to Problem 9P

The horizontal deflection at joint H is $\underset{\_}{23\text{mm}\to }$.

Explanation of Solution

Given information:

The truss is given in the Figure.

The value of E is 200 GPa.

Procedure to find the deflection of truss by virtual work method is shown below.

• For Real system: If the deflection of truss is determined by the external loads, then apply method of joints or method of sections to find the real axial forces (F) in all the members of the truss.
• For virtual system: Remove all given real loads, apply a unit load at the joint where is deflection is required and also in the direction of desired deflection. Use method of joints or method of sections to find the virtual axial forces $\left(F_v\right)$ in all the member of the truss.
• Finally use the desired deflection equation.

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.

Method of joints:

The negative value of force in any member indicates compression (C) and the positive value of force in any member indicates tension (T).

Condition for zero force members:

1. 1. If only two non-collinear members are connected to a joint that has no external loads or reactions applied to it, then the force in both the members is zero.
2. 2. If three members, two of which are collinear are connected to a joint that has no external loads or reactions applied to it, then the force in non-collinear member is zero.

Calculation:

Consider the real system.

Find the member axial force (F) for the real system using method of joints:

Let $A_x$ and $A_y$ be the horizontal and vertical reactions at the hinged support A.

Let $C_y$ be the vertical reaction at the hinged support C.

Sketch the resultant diagram of the truss as shown in Figure 1.

Find the reactions at the supports using equilibrium equations:

Summation of moments about A is equal to 0.

$\begin{array}{l}\sum M_A=0\\ C_y\left(4\right)-150\left(4\right)-75\left(8\right)=0\\ C_y=300\text{kN}\uparrow \end{array}$

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

$\begin{array}{l}+\uparrow \sum F_y=0\\ -A_y+C_y=0\\ -A_y+300=0\\ A_y=300\text{kN}\downarrow \end{array}$

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

$\begin{array}{l}+\to \sum F_x=0\\ -A_x+150+75=0\\ A_x=225\text{kN}\leftarrow \end{array}$

Find the member forces using method of joints:

Apply equilibrium equation to the joint A:

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

$\begin{array}{l}+\to \sum F_x=0\\ -225+F_{AB}=0\\ F_{AB}=225\text{kN}\end{array}$

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

$\begin{array}{l}+\uparrow \sum F_y=0\\ -300+F_{AD}=0\\ F_{AD}=300\text{kN}\end{array}$

Apply equilibrium equation to the joint C:

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

$\begin{array}{l}+\to \sum F_x=0\\ F_{BC}=0\end{array}$

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

$\begin{array}{l}+\uparrow \sum F_y=0\\ 300+F_{CF}=0\\ F_{CF}=-300\text{kN}\end{array}$

Apply equilibrium equation to the joint B:

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

$\begin{array}{l}+\to \sum F_x=0\\ -F_{AB}-F_{BD}\cos 63.435°+F_{BF}\cos 63.435°=0\\ -225-F_{BD}\cos 63.435°+F_{BF}\cos 63.435°=0\\ -F_{BD}\cos 63.435°+F_{BF}\cos 63.435°=225\end{array}$ (1)

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

$\begin{array}{l}+\uparrow \sum F_y=0\\ F_{BD}\sin 63.435°+F_{BF}\sin 63.435°=0\end{array}$ (2)

Solve Equation (1) and Equation (2).

$\begin{array}{l}{F}_{BD}=-251.56\text{kN}\\ F_{BF}=251.56\text{kN}\end{array}$

Apply equilibrium equation to the joint D:

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

$\begin{array}{l}+\to \sum F_x=0\\ 150+F_{DE}+F_{BD}\cos 63.435°=0\\ 150+F_{DE}+\left(-251.56\right)\cos 63.435°=0\\ F_{DE}=-37.5\text{kN}\end{array}$

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

$\begin{array}{l}+\uparrow \sum F_y=0\\ -F_{AD}+F_{DG}-F_{BD}\sin 63.435°=0\\ -300+F_{DG}-\left(-251.56\right)\sin 63.435°=0\\ F_{DG}=75\text{kN}\end{array}$

Apply equilibrium equation to the joint F:

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

$\begin{array}{l}+\to \sum F_x=0\\ -F_{EF}-F_{BF}\cos 63.435°=0\\ -F_{EF}-\left(251.56\right)\cos 63.435°=0\\ F_{EF}=-112.5\text{kN}\end{array}$

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

$\begin{array}{l}+\uparrow \sum F_y=0\\ -F_{CF}+F_{FH}-F_{BF}\sin 63.435°=0\\ -\left(-300\right)+F_{FH}-\left(251.56\right)\sin 63.435°=0\\ F_{FH}=-75\text{kN}\end{array}$

Apply equilibrium equation to the joint E:

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

$\begin{array}{l}+\to \sum F_x=0\\ -F_{DE}+F_{EF}-F_{EG}\cos 63.435°+F_{EH}\cos 63.435°=0\\ -\left(-37.5\right)+\left(-112.5\right)-F_{EG}\cos 63.435°+F_{EH}\cos 63.435°=0\\ -F_{EG}\cos 63.435°+F_{EH}\cos 63.435°=75\end{array}$ (3)

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

$\begin{array}{l}+\uparrow \sum F_y=0\\ F_{EG}\sin 63.435°+F_{EH}\sin 63.435°=0\end{array}$ (4)

Solve Equation (3) and Equation (4).

$\begin{array}{l}{F}_{EG}=-83.85\text{kN}\\ F_{EH}=83.85\text{kN}\end{array}$

Apply equilibrium equation to the joint G:

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

$\begin{array}{l}+\to \sum F_x=0\\ 75+F_{GH}+F_{EG}\cos 63.435°=0\\ 75+F_{GH}+\left(-83.85\right)\cos 63.435°=0\\ F_{GH}=-37.5\text{kN}\end{array}$

Consider the virtual system:

For horizontal deflection apply l kN at joint H in the horizontal direction.

Find the member axial force $\left(F_v\right)$ due to virtual horizontal load using method of joints:

Sketch the resultant diagram of the virtual system $\left(F_v\right)$ as shown in Figure 2.

Find the reactions at the supports using equilibrium equations:

Summation of moments about A is equal to 0.

$\begin{array}{l}\sum M_A=0\\ C_y\left(4\right)-1\left(8\right)=0\\ C_y=2\text{kN}\uparrow \end{array}$

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

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

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

$\begin{array}{l}+\to \sum F_x=0\\ -A_x+1=0\\ A_x=1\text{kN}\leftarrow \end{array}$

Find the member forces using method of joints:

Apply equilibrium equation to the joint A:

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

$\begin{array}{l}+\to \sum F_x=0\\ -1+F_{AB}=0\\ F_{AB}=1\text{kN}\end{array}$

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

$\begin{array}{l}+\uparrow \sum F_y=0\\ -2+F_{AD}=0\\ F_{AD}=2\text{kN}\end{array}$

Apply equilibrium equation to the joint C:

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

$\begin{array}{l}+\to \sum F_x=0\\ F_{BC}=0\end{array}$

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

$\begin{array}{l}+\uparrow \sum F_y=0\\ 2+F_{CF}=0\\ F_{CF}=-2\text{kN}\end{array}$

Apply equilibrium equation to the joint B:

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

$\begin{array}{l}+\to \sum F_x=0\\ -F_{AB}-F_{BD}\cos 63.435°+F_{BF}\cos 63.435°=0\\ -1-F_{BD}\cos 63.435°+F_{BF}\cos 63.435°=0\\ -F_{BD}\cos 63.435°+F_{BF}\cos 63.435°=1\end{array}$ (5)

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

$\begin{array}{l}+\uparrow \sum F_y=0\\ F_{BD}\sin 63.435°+F_{BF}\sin 63.435°=0\end{array}$ (6)

Solve Equation (5) and Equation (6).

$\begin{array}{l}{F}_{BD}=-1.12\text{kN}\\ F_{BF}=1.12\text{kN}\end{array}$

Apply equilibrium equation to the joint D:

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

$\begin{array}{l}+\to \sum F_x=0\\ F_{DE}+F_{BD}\cos 63.435°=0\\ F_{DE}+\left(-1.12\right)\cos 63.435°=0\\ F_{DE}=0.5\text{kN}\end{array}$

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

$\begin{array}{l}+\uparrow \sum F_y=0\\ -F_{AD}+F_{DG}-F_{BD}\sin 63.435°=0\\ -2+F_{DG}-\left(-1.12\right)\sin 63.435°=0\\ F_{DG}=1\text{kN}\end{array}$

Apply equilibrium equation to the joint F:

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

$\begin{array}{l}+\to \sum F_x=0\\ -F_{EF}-F_{BF}\cos 63.435°=0\\ -F_{EF}-\left(1.12\right)\cos 63.435°=0\\ F_{EF}=-0.5\text{kN}\end{array}$

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

$\begin{array}{l}+\uparrow \sum F_y=0\\ -F_{CF}+F_{FH}-F_{BF}\sin 63.435°=0\\ -\left(-2\right)+F_{FH}-\left(1.12\right)\sin 63.435°=0\\ F_{FH}=-1\text{kN}\end{array}$

Apply equilibrium equation to the joint E:

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

$\begin{array}{l}+\to \sum F_x=0\\ -F_{DE}+F_{EF}-F_{EG}\cos 63.435°+F_{EH}\cos 63.435°=0\\ -\left(0.5\right)+\left(-0.5\right)-F_{EG}\cos 63.435°+F_{EH}\cos 63.435°=0\\ -F_{EG}\cos 63.435°+F_{EH}\cos 63.435°=1\end{array}$ (7)

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

$\begin{array}{l}+\uparrow \sum F_y=0\\ F_{EG}\sin 63.435°+F_{EH}\sin 63.435°=0\end{array}$ (8)

Solve Equation (3) and Equation (4).

$\begin{array}{l}{F}_{EG}=-1.12\text{kN}\\ F_{EH}=1.12\text{kN}\end{array}$

Apply equilibrium equation to the joint G:

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

$\begin{array}{l}+\to \sum F_x=0\\ F_{GH}+F_{EG}\cos 63.435°=0\\ F_{GH}+\left(-1.12\right)\cos 63.435°=0\\ F_{GH}=0.5\text{kN}\end{array}$

The expression to find the deflection $1\left(\Delta \right)$ is shown below:

$1\left(\Delta \right)=\sum F_v\left(\frac{FL}{AE}\right)$

Here, L is the length of the member, A is the area of the member, and E is the young’s modulus of the member.

For E constant, the expression becomes,

$\left(1\right)\Delta =\frac{1}{E}\sum \frac{F_{v}FL}{A}$ (9)

Find the length of member BD, BF, EG, and EH using given Figure.

$\begin{array}{c}L=\frac{4}{\sin 63.435°}\\ 4.47\text{m}\end{array}$

Find the product of $\frac{F_{v}FL}{A}$ for each member as shown in Table 1.

Member	L $\left(\text{m}\right)$	$\begin{array}{l}A\\ \left({\text{m}}^{\text{2}}\right)\end{array}$	$\begin{array}{l}F\\ \left(\text{kN}\right)\end{array}$	$\begin{array}{l}{F}_v\\ \left(\text{kN}\right)\end{array}$	$\begin{array}{l}\frac{F_{v}FL}{A}\\ \left({\text{kN}}^2\text{/m}\right)\end{array}$
AD	4	0.0025	$300$	2	960,000
CF	4	0.0025	$-300$	$-2$	960,000
DG	4	0.0025	75	1	120,000
FH	4	0.0025	$-75$	$-1$	120,000
GH	4	0.0025	$-37.5$	0.5	$-50,000$
AB	2	0.0025	225	1	300,000
DE	2	0.0025	$-37.5$	0.5	$-25,000$
EF	2	0.0025	$-112.5$	$-0.5$	75,000
BD	4.47	0.0025	$-251.56$	$-1.12$	$839,607$
BF	4.47	0.0025	$251.56$	1.12	$839,607$
EG	4.47	0.0025	$-83.85$	$-1.12$	$279,858$
EH	4.47	0.0025	$83.85$	1.12	$279,858$
				$\sum$	$4,548,930$

Find the horizontal deflection at joint H $\left({\Delta }_{HH}\right)$:

Substitute $4,548,930{\text{kN}}^2\text{/m}$ for $\sum \frac{F_{v}FL}{A}$ and $200\text{GPa}$ for E in Equation (9).

$\begin{array}{c}\left(1\text{kN}\right){\Delta }_{HH}=\frac{4,548,930{\text{kN}}^2\text{/m}}{200\text{GPa}}\\ =\frac{4,548,930{\text{kN}}^2\text{/m}\times \left(\frac{1\text{m}}{{10}^3\text{mm}}\right)}{200\text{GPa}\times \frac{1{\text{kN/mm}}^{\text{2}}}{1\text{GPa}}}\\ {\Delta }_{HH}=23\text{mm}\to \end{array}$

Therefore, the horizontal deflection at joint H is $\underset{\_}{23\text{mm}\to }$.