Axially loaded members

From three basic laws to one formula. A prismatic bar of length $L$, cross-sectional area $A$ and elastic modulus $E$, carrying a constant internal axial force $P$, obeys three relations you already know: the definition of normal stress $\sigma = P/A$, the definition of normal strain $\varepsilon = \delta/L$, and Hooke's law $\sigma = E\varepsilon$. Combine them — $\delta/L = \sigma/E = P/(AE)$ — and rearrange to get the axial-deformation formula that anchors this whole unit:

where: $\delta$ is the axial elongation (positive) or shortening (negative), $P$ is the internal axial force (positive in tension), $L$ is the member length, $A$ is the cross-sectional area, and $E$ is Young's modulus. The product $AE$ is the member's axial stiffness — the larger it is, the less a given $P$ and $L$ deform the bar, exactly the way a stiffer spring stretches less under the same load.

Superposition for varying $P$, $A$, or $E$. Real members are rarely one uniform segment end to end. A stepped shaft changes cross-section; a bar loaded at an intermediate point carries different internal force on either side of that point; a composite bar may splice two materials. In every such case, split the member into segments over which $P$, $A$, and $E$ are each individually constant, find the internal force in each segment by sectioning and equilibrium (not simply the externally applied load), compute $\delta_i = P_iL_i/(A_iE_i)$ for each, and sum algebraically:

Elongations and shortenings carry opposite signs in this sum, so a segment in compression subtracts from one in tension.

Statically indeterminate axial members. When a bar is fixed at both ends (or otherwise over-restrained) and loaded partway along, there are more unknown reactions than independent equilibrium equations, and equilibrium alone cannot solve for them. The missing equation comes from a compatibility condition — a geometric statement about how the deformations of the parts must fit together. For a bar rigidly fixed at both ends, the total elongation from one fixed end to the other must equal zero, because the fixed points cannot move. The solution method has three parts: write the equilibrium equation(s); write the compatibility equation by expressing each segment's deformation as $PL/AE$ in terms of the unknown reactions, and set the sum equal to the known geometric constraint; then solve the two equations simultaneously.

Thermal strain and stress. A free, unrestrained member subjected to a uniform temperature change $\Delta T$ develops a thermal strain

where: $\alpha$ is the material's coefficient of thermal expansion (per degree) and $\Delta T$ is the temperature change. This produces a free thermal deformation $\delta_T = \alpha \Delta T L$ with zero stress, because nothing is resisting the expansion. If instead the member is restrained — for example fixed rigidly between two walls — it cannot execute this free deformation, and the restraint forces a thermal stress to develop. Such problems are solved exactly like mechanical indeterminate problems: write equilibrium, then a compatibility equation that now contains both a mechanical term $PL/AE$ and a thermal term $\alpha\Delta T L$, set to match the physical constraint (typically total deformation equals zero for full restraint).

Rather than a single formula derivation, the reasoning that unlocks every indeterminate or multi-segment axial problem is a fixed three-part logic, and it is worth stating in full because the same structure recurs across statics-of-deformable-bodies problems well beyond this course: the number of independent equations you can write must equal the number of unknowns you are solving for, and when equilibrium alone falls short by exactly the "degree of indeterminacy," compatibility supplies precisely that many extra equations, no more and no less.

Step 1. Count unknowns and equilibrium equations. For a bar with reactions at both fixed ends and one intermediate load, there are two unknown reactions ($R_A$, $R_C$) but only one independent equilibrium equation along the bar's axis (sum of axial forces equals zero). The shortfall is one equation — the bar is indeterminate to the first degree.

Step 2. Section the bar into segments carrying known internal force in terms of the unknowns. Cutting on either side of the load point shows one segment carries $R_A$ and the other carries $-R_C$ (with sign convention for tension positive), each segment's internal force found from a free-body diagram of the portion cut away, not from guessing.

Step 3. Translate the physical restraint into a deformation statement. Because both ends are rigidly fixed, the total distance between them cannot change: the sum of the two segments' elongations, computed from $PL/AE$ using the internal forces from Step 2, must equal zero. This is the compatibility equation — it says nothing about forces directly, only about geometry, which is exactly why it supplies genuinely new information beyond equilibrium.

Step 4. Solve the equilibrium and compatibility equations together as a simultaneous system in the unknown reactions. Once the reactions are known, every segment's internal force, stress, and elongation follow directly from the earlier formulas — the indeterminate problem collapses back into an ordinary determinate calculation once the extra unknown is pinned down.

A stepped steel bar ($E = 200$ GPa) has two segments in series: AB is $0.3$ m long with area $500$ mm², and BC is $0.5$ m long with area $300$ mm². A tensile axial force of $60$ kN acts at the free end C, with the bar fixed at A. Find the total elongation, then extend the same bar to the indeterminate case.

Step 1. Find the internal force in each segment. Since the only load is the $60$ kN tension at C and the bar is otherwise unloaded between A and C, both segments carry the full internal force $P = 60$ kN in tension — sectioning anywhere along either segment and summing forces on the cut piece confirms this.

Step 2. Compute $\delta_{AB}$. With $A_{AB} = 500\text{ mm}^2 = 5\times10^{-4}\text{ m}^2$ and $L_{AB}=0.3$ m,

Step 3. Compute $\delta_{BC}$. With $A_{BC} = 300\text{ mm}^2 = 3\times10^{-4}\text{ m}^2$ and $L_{BC}=0.5$ m,

Step 4. Sum by superposition. $\delta_{total} = \delta_{AB}+\delta_{BC} = 0.180+0.500 = \mathbf{0.680\ mm}$.

Competitive stretch

Now suppose instead the same $0.8$ m total bar (uniform $A = 400$ mm², so the stepped geometry is replaced by a single material and section for this variant) is fixed rigidly at both ends A and C, with a tensile force $P = 90$ kN applied at an intermediate point B located $0.5$ m from A. Let $R_A$, $R_C$ be the tensile reactions at the two fixed ends. Equilibrium at B gives $R_A + R_C = 90$ kN — one equation, two unknowns, indeterminate to the first degree. Compatibility (total elongation A to C is zero) gives, after $A,E$ cancel since they are uniform, $R_A(0.5) = R_C(0.3)$, so $R_A = 0.6R_C$. Substituting: $1.6R_C = 90 \Rightarrow R_C = 56.25$ kN and $R_A = 33.75$ kN. Checking, $R_A(0.5) = 16.875 = R_C(0.3)$ ✓, confirming compatibility is satisfied. The reactions are $R_A = 33.75$ kN and $R_C = 56.25$ kN.

• Using the external applied load directly in $\delta=PL/AE$ for each segment instead of the correct internal force in that segment found by sectioning — these differ whenever loads are applied at intermediate points.
• In indeterminate problems, trying to solve two unknown reactions from one equilibrium equation alone, forgetting that a compatibility equation is required to close the system.
• Sign errors in the compatibility equation — elongation and shortening must be assigned consistent signs so the geometric constraint (e.g. total elongation equals zero) is written correctly.
• For thermal problems, forgetting that a free member under $\Delta T$ develops strain but no stress; thermal stress only arises when deformation is restrained, and the compatibility equation must then combine both mechanical and thermal deformation terms.

Ask the tutor: If you get stuck writing the compatibility equation for a statically indeterminate bar, or want to check your sign convention on a thermal-stress problem, ask the AI tutor to walk through the free-body diagrams step by step.