Indeterminate Structures — Question 1

Question 1

Compatibility determines the third support reaction

A weightless rigid beam rests on three identical vertical springs of stiffness $k$ at $x=-a,0,a$. Their unloaded tops are level. A downward force $W$ acts at the center. Use small displacements and $R=k\delta$.

1. Show why force and moment balance leave one reaction undetermined.

2. Use rigid-beam compatibility to determine all reactions and the displacement.

Question 1 - Solution

Compatibility determines the third support reaction

(a) Derivation and calculation

Force and moment balance give $$R_-+R_0+R_+=W,\qquad a(R_+-R_-)=0.$$ Thus the statically admissible family is $$R_-=R_+=r,\qquad R_0=W-2r.$$ Equilibrium alone leaves $r$ undetermined.

(b) Result and interpretation

A rigid beam has an affine displacement, so $$\delta_0=\frac{\delta_-+\delta_+}{2},\qquad R_i=k\delta_i.$$ Since $R_-=R_+$, all three deflections and reactions are equal: $$\boxed{R_-=R_0=R_+=W/3},\qquad\boxed{\delta=W/(3k)}.$$

Physical check

Equal load sharing follows from equilibrium plus equal stiffness and level initial fit, not from equilibrium alone.