How is elongation in a uniform rod with unequal forces acting on opposite sides calculated? If applied forces are equal and opposite, the elongation is defined by the formula ($\delta = \frac{FL}{AE}$). How does the solution change for the case when forces are unequal (as shown)?
Newtonian Mechanics – Elongation in a Bar with Unequal Applied Forces
elasticityforcesnewtonian-mechanics
Best Answer
As you correctly note, the solution is different when the applied forces are not equal. The bar is not in static equilibrium: Both static and dynamic forces deform the bar in motion. These concepts are illustrated by superposition. $$\delta = \delta_\text{static} + \delta_\text{dynamic} \qquad = \frac {F_\text{less}L}{EA} + \frac{(F_\text{more} - F_\text{less})L}{2AE}$$
Newton's Second Law requires that the bar (of mass $M$) accelerate in the direction of $F_{net}$. $$\sum F \;\text{on bar:} \qquad F_\text{more}-F_\text{less} = Ma \qquad \Rightarrow \;\therefore a = \frac{F_\text{more} - F_\text{less}}{M}$$
The derivation of deformation is shown when a single force acts on the bar.
The derivation can be generalized to include both forces, where integration of $T(x) = F_\text{more} - \dfrac{F_\text{more}-F_\text{less}}{L}x$ results in the same solution given by superposition.
$$\therefore \delta = \; \overbrace{\frac{F_\text{more}L}{2AE} + \frac{F_\text{less}L}{2AE}}^\text{Integration} \;=\; \overbrace{\frac{F_\text{less}L}{AE} + \frac{(F_\text{more}-F_\text{less})L}{2AE}}^\text{Superposition}$$
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