Question

A truss EFGH is shown in the figure, in which all the members have the same axial rigidity \( R \). In the figure, \( P \) is the magnitude of external horizontal forces acting at joints F and G. 

Hint:

To calculate displacement in trusses with the same axial rigidity, use \( \Delta = \frac{P L}{R A} \), where \( A \) is the cross-sectional area of the members.

Answer 1

Correct Answer: 0.9

For a truss with the same axial rigidity, the horizontal displacement at joint G can be found using the axial displacement formula: \[ \Delta = \frac{P L}{R A}, \] where \( L = 3 \, \text{m} \), \( R = 500 \times 10^3 \, \text{kN} \), and \( P = 150 \, \text{kN} \). Substituting the values, we get: \[ \Delta = \frac{150 \times 3}{500 \times 10^3} = 0.0009 \, \text{m} = 0.9 \, \text{mm}. \] Thus, the magnitude of the horizontal displacement of joint G is \( \boxed{0.9} \, \text{mm} \).