   # Use the virtual work method to determine the horizontal deflection at joint H of the truss shown in Fig. P7. 9.

#### Solutions

Chapter
Section
Chapter 7, Problem 9P
Textbook Problem
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## Use the virtual work method to determine the horizontal deflection at joint H of the truss shown in Fig. P7. 9. To determine

Calculate the horizontal deflection at joint H of the truss.

### 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 (Fv) 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 (Fx=0), consider the forces acting towards right side as positive (+) and the forces acting towards left side as negative ().
• For summation of forces along y-direction is equal to zero (Fy=0), consider the upward force as positive (+) and the downward force as negative ().
• For summation of moment about a point is equal to zero (Matapoint=0), 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 Ax and Ay be the horizontal and vertical reactions at the hinged support A.

Let Cy 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.

MA=0Cy(4)150(4)75(8)=0Cy=300kN

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

+Fy=0Ay+Cy=0Ay+300=0Ay=300kN

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

+Fx=0Ax+150+75=0Ax=225kN

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.

+Fx=0225+FAB=0FAB=225kN

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

Apply equilibrium equation to the joint C:

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

+Fx=0FBC=0

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

+Fy=0300+FCF=0FCF=300kN

Apply equilibrium equation to the joint B:

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

+Fx=0FABFBDcos63.435°+FBFcos63.435°=0225FBDcos63.435°+FBFcos63.435°=0FBDcos63.435°+FBFcos63.435°=225 (1)

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

+Fy=0FBDsin63.435°+FBFsin63.435°=0 (2)

Solve Equation (1) and Equation (2).

FBD=251.56kNFBF=251.56kN

Apply equilibrium equation to the joint D:

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

+Fx=0150+FDE+FBDcos63.435°=0150+FDE+(251.56)cos63.435°=0FDE=37.5kN

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

Apply equilibrium equation to the joint F:

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

+Fx=0FEFFBFcos63.435°=0FEF(251.56)cos63.435°=0FEF=112.5kN

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

+Fy=0FCF+FFHFBFsin63.435°=0(300)+FFH(251.56)sin63.435°=0FFH=75kN

Apply equilibrium equation to the joint E:

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

+Fx=0FDE+FEFFEGcos63.435°+FEHcos63.435°=0(37.5)+(112.5)FEGcos63.435°+FEHcos63.435°=0FEGcos63.435°+FEHcos63.435°=75 (3)

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

+Fy=0FEGsin63.435°+FEHsin63.435°=0 (4)

Solve Equation (3) and Equation (4).

FEG=83.85kNFEH=83.85kN

Apply equilibrium equation to the joint G:

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

+Fx=075+FGH+FEGcos63.435°=075+FGH+(83.85)cos63.435°=0FGH=37.5kN

Consider the virtual system:

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

Find the member axial force (Fv) due to virtual horizontal load using method of joints:

Sketch the resultant diagram of the virtual system (Fv) as shown in Figure 2.

Find the reactions at the supports using equilibrium equations:

Summation of moments about A is equal to 0.

MA=0Cy(4)1(8)=0Cy=2kN

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

+Fy=0Ay+Cy=0Ay+2=0Ay=2kN

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

+Fx=0Ax+1=0Ax=1kN

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

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