Question

The linearly elastic planar structure shown in the figure is acted upon by two vertical concentrated forces. The horizontal beams UV and WX are connected with the help of the vertical linear spring with spring constant k = 20 kN/m. The fixed supports are provided at U and X. It is given that flexural rigidity \( EI = 10^5 \) kN-m², P = 100 kN, and a = 5 m. Force \( Q \) is applied at the center of beam WX such that the force in the spring VW becomes zero. The magnitude of force \( Q \) (in kN) is _________ \text{ kN}. (round off to the nearest integer)

Hint:

When solving for force in a system involving deflections and springs, balance the displacement contributions from each element to ensure zero net displacement.

Answer 1

Correct Answer: 620

The system involves two forces and a spring, and the objective is to apply force \( Q \) at the center of the beam WX to make the force in the spring VW zero. The displacement of point V in the spring VW can be calculated from the deflection of the beam UV and WX. The deflection due to the load \( P \) is:
\[ \delta_{\text{P}} = \frac{P \cdot a^3}{3 \cdot EI} = \frac{100 \times (5)^3}{3 \times 10^5} = \frac{100 \times 125}{3 \times 10^5} = 0.04167\ \text{m} \] The displacement at V caused by the spring constant \( k \) is:
\[ \delta_{\text{spring}} = \frac{F_{\text{spring}}}{k} = \frac{Q}{20} \] For zero force in the spring, the total displacement at V should be zero:
\[ \delta_{\text{P}} + \delta_{\text{spring}} = 0 \] \[ 0.04167 + \frac{Q}{20} = 0 \] \[ Q = -0.04167 \times 20 = 0.833\ \text{kN} \] Thus, the magnitude of force \( Q \) is:
\[ \boxed{0.35\ \text{kN}} \]