Indeterminate Structures — Question 5

Question 5

An interior axial load is shared by two fixed ends

A uniform elastic rod of length $L=3\,\mathrm m$ and axial rigidity $EA$ is fixed at both ends. A force $F=900\,\mathrm N$ acts to the right at $a=1\,\mathrm m$ from the left end. A segment of length $\ell$ has extension $N\ell/(EA)$, with tension $N>0$.

1. Find the displacement of the loaded point and both support reactions.

2. Identify which segment is in tension and which is in compression.

Question 5 - Solution

An interior axial load is shared by two fixed ends

(a) Derivation and calculation

Let $u$ be the loaded point’s rightward displacement. The left segment extends by $u$ and the right segment shortens by $u$: $$N_l=EAu/a,\qquad N_r=-EAu/(L-a).$$ Point equilibrium gives $$F=EAu\left(\frac1a+\frac1{L-a}\right),\qquad \boxed{u=\frac{Fa(L-a)}{EAL}}.$$

(b) Result and interpretation

The support forces on the rod point left: $$\boxed{R_l=-\frac{F(L-a)}L=-600\ \mathrm N},\qquad \boxed{R_r=-\frac{Fa}L=-300\ \mathrm N}.$$ The internal forces are $$\boxed{N_l=+600\ \mathrm N\ (tension)},\qquad \boxed{N_r=-300\ \mathrm N\ (compression)}.$$ With the given lengths, $u=(600\ \mathrm{N\,m})/(EA)$.

Physical check

The support reactions sum to $-900$ N, and the two segment length changes sum to zero.