1. Loads, supports and elastic energy

How far will the end move?

A straight bar is fixed at its left end and pulled to the right by a force $F$. If the material is twice as stiff, does the support reaction double? Does the extension halve? We can answer both questions before constructing a numerical model.

Conceptual free-body sketch, not a computed deformation. Both forces act on the bar; their signed sum is zero.

Assume a uniform cross-section $A$, length $L$, an axial load through its center, and slow loading so acceleration is negligible. We describe the interior away from the detailed grip by uniform axial stress. Strains are small, the material is homogeneous, and its axial response is linear elastic.

Three different ingredients

Equilibrium. Taking rightward forces as positive, the complete bar satisfies $R+F=0$. Cut it at a position $x$. The tensile internal force $N(x)$ must balance the end load, so $N=F$ throughout the bar. The axial stress is $\sigma=N/A=F/A$, measured in pascals: $1\,\mathrm{Pa}=1\,\mathrm{N/m^2}$.

Kinematics. Let $u(x)$ be displacement in metres, with $u(0)=0$. Two nearby points initially separated by $\mathrm{d}x$ acquire a separation change $u'(x)\mathrm{d}x$. The axial strain is therefore $\varepsilon=u'(x)$, a dimensionless ratio.

Material law. For the chosen linear elastic material, $\sigma=E\varepsilon$, with Young’s modulus $E$ in pascals. Equilibrium alone cannot supply $E$ or decide whether this law fits a specimen. These scalar definitions and Hooke’s law are developed in Roylance’s Introduction to Elasticity, pp. 4–6.

Combining the three ingredients gives

$$EAu'(x)=F,\qquad u(x)=\frac{F}{EA}x,\qquad \delta=u(L)=\frac{FL}{EA}.$$

The equivalent spring stiffness is $k=EA/L$, in N/m. It depends on geometry as well as material. Doubling $E$ halves $\delta$ under the same applied $F$, but leaves $R=-F$ unchanged. By contrast, if the end displacement $\delta$ is prescribed, the required force $F=EA\delta/L$ doubles.

That distinction is the mechanical form of prescribed values versus natural boundary data. Changing which quantity is controlled changes the experiment.

An independent numerical prediction

Choose illustrative values, not a material specification:

Quantity	Value
$L$	$2\,\mathrm{m}$
$A$	$100\,\mathrm{mm^2}=10^{-4}\,\mathrm{m^2}$
$E$	$200\,\mathrm{GPa}=2\times10^{11}\,\mathrm{Pa}$
$F$	$1000\,\mathrm{N}$

Then $EA=2\times10^7\,\mathrm{N}$, $\sigma=10\,\mathrm{MPa}$, $\varepsilon=5\times10^{-5}$ and

$$\delta=10^{-4}\,\mathrm{m}=0.1\,\mathrm{mm},\qquad R=-1000\,\mathrm{N}.$$

The extension is tiny compared with the bar length, consistent with the small-strain approximation. This calculation says nothing about a material’s yield stress: stiffness and strength answer different questions.

Energy gives the same answer

For a spring loaded gradually from zero, the force rises from zero to $k\delta$. The work is the area under that force–displacement line:

$$U=\int_0^\delta k s\,\mathrm{d}s=\frac12k\delta^2.$$

At equilibrium this is $F\delta/2=0.05\,\mathrm{J}$ for our bar. The final force multiplied by the final displacement, $F\delta$, is twice the stored energy; it is not the work done during gradual loading.

For a fixed external force, form the total potential energy

$$\Pi(\delta)=\frac12k\delta^2-F\delta.$$

A small admissible change $\eta$ gives $\Pi'(\delta)\eta=(k\delta-F)\eta$. Requiring this to vanish for every $\eta$ recovers $k\delta=F$. Because $k>0$, the second derivative is positive: equilibrium minimizes this potential. This is the discrete starting point for the virtual-work formulation in chapter 3; see also Bathe’s linear-analysis lecture 1.

Replace an end load by a distributed load

Let $p$ be a constant axial force per unit length, in N/m, and make the right end free. A small segment gives $N'(x)+p=0$. With $N(L)=0$,

$$N(x)=p(L-x),\qquad u(x)=\frac{p}{EA}\left(Lx-\frac{x^2}{2}\right),\qquad R=-pL.$$

Stress now falls toward the free end, while displacement grows with a decreasing slope. This quadratic shape is exactly the idea behind the two-dimensional square we will run later. A uniform body load is not the same as a load applied only at the tip.

What goes wrong if a support is missing?

If neither end fixes displacement, adding a constant $c$ to $u(x)$ changes no strain and costs no elastic energy. A balanced load can determine extension without determining absolute position. An unbalanced load cannot have a static equilibrium at all. A singular stiffness matrix can be a faithful report of a missing physical constraint.

Likewise, fixing both ends while prescribing a nonzero extension would impose incompatible data. Adding restraints until a solve succeeds is not a substitute for deciding how the body is actually held.

Exercises

1. Double the cross-sectional area under the same end force. Predict stress, strain, extension and reaction before substituting numbers.
2. Prescribe $\delta=0.2\,\mathrm{mm}$ for the numerical bar instead of $F$. Find the new force and energy. Check: $2000\,\mathrm{N}$ and $0.2\,\mathrm{J}$.
3. For the uniform distributed load, integrate $N(x)^2/(2EA)$ to find the stored energy. Check: $p^2L^3/(6EA)$.
4. Explain why replacing that distributed load by an end force $pL$ preserves the reaction but doubles the end displacement.
5. A plot shows an extension of one tenth of the bar length. Which physical assumption would you revisit before interpreting the stress?

Reading

• David Roylance, Introduction to Elasticity, MIT, January 21, 2000, especially pp. 4–7: course module.
• Klaus-Jürgen Bathe, Finite Element Procedures for Solids and Structures, MIT OpenCourseWare, Spring 2010, linear analysis, lecture 1: Some Basic Concepts of Engineering Analysis.

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