Indeterminate Structures — Question 3

Question 3

A short fourth leg creates diagonal load sharing

A rigid square platform has four vertical spring legs at $(x,y)=(\pm a,\pm a)$, each stiffness $k$. Its central downward load is $W$. The $(+a,+a)$ leg has an initial gap $g$; the other three just touch. Assume all four contact and small displacement.

1. Write a planar displacement $\delta=d+s(x/a)+t(y/a)$ and solve for $d,s,t$.

2. Find the four reactions and the condition for positive force in every leg.

Question 3 - Solution

A short fourth leg creates diagonal load sharing

(a) Derivation and calculation

Write $\delta=d+s(x/a)+t(y/a)$. For the short leg, $$R_{++}=k(d+s+t-g),$$ while the other three have $R_{\pm\pm}=k(d\pm s\pm t)$. Summing forces and moments gives $$4kd-kg=W,\qquad4ks-kg=0,\qquad4kt-kg=0.$$ Therefore $$\boxed{d=(W/k+g)/4},\qquad\boxed{s=t=g/4}.$$

(b) Result and interpretation

Substitution yields the diagonal reaction pairs $$\boxed{R_{++}=R_{--}=(W-kg)/4},$$ $$\boxed{R_{+-}=R_{-+}=(W+kg)/4}.$$ Every leg carries positive force only if $\boxed{W>kg}$. At $W=kg$, both legs of the first diagonal have zero reaction; below it, the assumed four-contact regime is inadmissible.

Physical check

The four reactions sum to $W$, and each opposite pair has equal force, satisfying both central-load moment equations.