# Elongation in bar with unequal applied forces

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)? 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. $$\delta = \delta_\text{static} + \delta_\text{dynamic} \qquad = \frac {F_\text{less}L}{EA} + \frac{(F_\text{more} - F_\text{less})L}{2AE}$$

During changes in acceleration (when $\frac{\mathrm{d\;a(t)}}{\mathrm{dt}} \neq 0$), forces and accelerations within the damped solid body are transient, where $a(x,t)$, until they reach steady-state, where $\frac{\partial{\;a(x,t)}}{\partial{x}} = 0$. Transient deformations in solid bodies are illustrated by a mass/spring system, where each mass element can be thought to represent differential mass element.

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.

Dynamic Deformation:

A Free Body Diagram is taken at an arbitrary cross-section of the bar, where the mass of the split body is $m = (\frac{M}{L})x$. Summation of forces acting on $m$ is solved for $T(x)$. $$\sum F \;\text{on split body:} \qquad F_{o} - T = ma$$ $$F_{o} - T = \overbrace{\left(\frac{M}{L}x\right)}^\text{m} \overbrace{\left(\frac{F_{o}}{M}\right)}^\text{a} = \frac{F_{o}}{L}x \qquad \Rightarrow \qquad T = F_{o} - \frac{F_{o}x}{L}$$ $$\therefore T = F_{o}\left(1-\frac{x}{L}\right)$$ The static axial deformation ($\delta = \frac{FL}{AE}$) written in differential form:
$$\mathrm{d \delta} = \frac{T \mathrm{dx}}{AE} = \frac{[F_{o}(1-\frac{x}{L})]\mathrm{dx}}{AE}$$ Integrate differential deformation over the length of the bar to determine total deformation: $$\delta = \int_0^L \mathrm{d \delta} \; = \frac{F_{o}}{AE} \int_0^L 1-\frac{x}{L} \mathrm{dx} \implies \; \delta = \frac{F_{o}L}{2AE}$$

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.

$$\therefore \delta = \; \overbrace{\frac{F_\text{more}L}{2AE} + \frac{F_\text{less}L}{2AE}}^\text{Integration} \;=\; \overbrace{\frac{F_\text{less}L}{AE} + \frac{(F_\text{more}-F_\text{less})L}{2AE}}^\text{Superposition}$$

References:

$$T= F_\text{more} - \frac{(F_\text{more}-F_\text{less})x}{L}$$

where $$T$$ is tension in wire, $$x$$ is distance from $$F_\text{more}$$

Consider the same part,

Now,

Stress $$\frac{T}{A} = Y\frac{dn}{dx}\;,$$

where $$n$$ is elongation in wire.

$$\frac{T}{A}= F_\text{more} - \frac{(F_\text{more}-F_\text{less})}{LA}\;x= Y\frac{dn}{dx}\\\implies \frac{(F_\text{more} - F_\text{less})}{L}\; xdx= YdnA$$

On integrating both sides

$$F_\text{more}\;x - \frac{(F_\text{more}-F_\text{less})}{L}\;\frac{x^2}{2} = YNA$$

Putting the limits

$$F_\text{more}\;L - (F_\text{more}-F_\text{less})\;\frac{L}{2}= YNA\\ \implies \frac{F_\text{more}- \dfrac{(F_\text{more}- F_\text{less})}{2}}{A} = \frac{YN}{L}= \text{stress in wire}\;.$$

Hence elongation $$N =\frac{L\left(F_\text{more} - \dfrac{(F_\text{more}-F_\text{less})}{2}\right)}{AY}\;.$$

• This is the weirdest combination of text and math formatting combined. Instead of T=$F_{more} - (F_{more}-F_{less})x/L$ use $$T = F_{more} - (F_{less}-F_{more}) \frac{x}{L}$$ for example which renders as $$T = F_{more} - (F_{less}-F_{more}) \frac{x}{L}$$ Dec 11, 2015 at 19:35