Elongation in bar with unequal applied forces

Question

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)?

Answer 1

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

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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.

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

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References:

• Stress and Deformation Analysis of Linear Elastic Bars in Tension
• Introduction to Elasticity

Answer 2

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

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