Determine the vertical movement of joint D if the length of member BF is increased by 1.5 in.

 Determine the vertical movement of joint D if the length of member BF is increased by 1.5 in.

Consider a vertical load \( P \) at joint D:

Moment about H, \( \Sigma M_H = 0 \)

\( E \times 120 - P \times 40 = 0 \)

∴ \( E = \frac{P}{3} \downarrow \)

\( \tan \Theta = \frac{30}{40} \)

∴ \( \Theta = 36.86^\circ \)

\( \Sigma F_y = 0 \) (↑+)

\( F_{BF} \sin 36.86^\circ - E = 0 \)

\( F_{BF} \sin 36.86^\circ - \frac{P}{3} = 0 \)

∴ \( F_{BF} = \frac{5P}{9} \)

For virtual work, we remove BF and replace it with forces \( F_{BF} \) and \( -F_{BF} \) at F and B respectively. Denoting the virtual displacements of Points B and F as \( \delta r_B \) and \( \delta r_F \) respectively, and noting that \( P \) and \( (\delta D) \) have the same direction.

From the principle of virtual work:

\( \delta U = 0 \)

Or, \( P \delta D + F_{BF} \delta r_F + (-F_{BF}) \delta r_B = 0 \)

Or, \( P \delta D + F_{BF} \delta r_F \cos \Theta_F - F_{BF} \delta r_B \cos \Theta_B = 0 \)

Or, \( P \delta D + F_{BF} (\delta r_F \cos \Theta_F - \delta r_B \cos \Theta_B) = 0 \)

Where \( (\delta r_F \cos \Theta_F - \delta r_B \cos \Theta_B) = \delta BF \), which is the change in BF.

Or, \( P \delta D - F_{BF} \delta BF = 0 \)

Or, \( P \delta D - \left( \frac{5P}{9} \right) \times 1.5 = 0 \)

∴ \( \Delta D = \frac{5}{6} = 0.833 \text{ in.} \downarrow \)