Now Apply the equilibrium equations for the equilibrium of joint C;

Figure 7-9(c)

As the angle between CA and CB is 60 degree, we get the same values of the forces as in the previous joint at B. Therefore we get;

CB = 2.31 kN (Tension) CA= 4.62 compressive.

In the method of virtual work we remove all the acting loads and apply 1 kN virtual load at the point of required displacement in the required direction. In this question we have to calculate the horizontal displacement at joint C. Therefore we apply 1 kN load in the horizontal direction at C and then calculate the support reactions and the member forces (n) due to virtual load.

Figure 7-9(d)

Apply equilibrium equations to calculate support reactions. We find that there will be no vertical reaction. We will get only horizontal reaction which will be at support B. Hence Reaction at B is 1 kN horizontal in the opposite direction of applied force of 1 kN.

Now we consider the equilibrium of joint C to calculate the forces in the truss members CB and CA.

Figure 7-9(e)

Applying Σ F_x = 0;

We can easily calculate the force in CB equal to 1 kN tensile

By applying Σ F_y = 0; we get CA = 0

We do the same at joint B to get the forces in the members connected at B.

Figure 7-9(f)

We get the force in BA = 0 and BC = CB = 1 kN tensile

Now we prepare a table and enter the values of actual forces (N) and virtual forces (n) in that table as shown below.

Table 7-9(a)

Now we apply the principle of virtual work

1 kN . Δ_C = Σ (NnL/AE)

Area of each member = 1 cm² = 1 x 10^-4 m²

Therefore , 1 kN. Δ_C =(4.62 kN. kN.m) / (10^-4 m² x 200 x 10⁶ kN/m²)

We get on simplification

Δ_C = 4.62 m / (10^-4 m² x 200 x 10⁶) = 2.31 x 10^-4 = 2.31 x 10^-1mm = 0.23 mm

The positive sign of displacement means that the displacement is in the direction of 1 kN virtual force in this case in the right hand direction.

Example 3-1 Calculation of Truss member forces

Example 5-3 SFD and BMD for Overhanging beam