Indeterminate Structures — Question 9

Question 9

An energy minimum recovers displacement compatibility

Two linear springs of stiffnesses $k_1,k_2>0$ support the same load point with total load $P$. For this system the physical force sharing minimizes the complementary energy $C=R_1^2/(2k_1)+R_2^2/(2k_2)$ subject to $R_1+R_2=P$.

1. Derive the reactions by constrained minimization.

2. Show that the minimum is unique and interpret the stationary condition physically.

Question 9 - Solution

An energy minimum recovers displacement compatibility

(a) Derivation and calculation

Eliminate $R_2=P-R_1$ to obtain $$C(R_1)=\frac{R_1^2}{2k_1}+\frac{(P-R_1)^2}{2k_2}.$$ Stationarity requires $$\frac{dC}{dR_1}=\frac{R_1}{k_1}-\frac{P-R_1}{k_2}=0.$$ Solving, $$\boxed{R_1=\frac{Pk_1}{k_1+k_2}},\qquad\boxed{R_2=\frac{Pk_2}{k_1+k_2}}.$$

(b) Result and interpretation

The second derivative is everywhere positive: $$\boxed{\frac{d^2C}{dR_1^2}=\frac1{k_1}+\frac1{k_2}>0}.$$ Therefore the stationary point is the unique minimum. Physically, $$\frac{R_1}{k_1}=\frac{R_2}{k_2}=\frac{P}{k_1+k_2}=\delta,$$ which is the common load-point displacement required by compatibility.

Physical check

The stiffer spring carries the larger force, but both undergo the same displacement.