Study Companion: MM07 — Deformation of Axially Loaded Members

A steel wire 100\,\mathrm m long hangs from a tower and carries nothing but its own weight. How much longer is it than when it lay on the ground? Lesson MM04 gave the stretch of a bar pulled at its ends, \delta = PL/(AE), but in the hanging wire the force is different at every height, and in a stepped shaft, a tapered strut or a cable the area or the material changes too. This lesson turns the one formula into a method: cut the member where the force or the section changes, add the stretches of the pieces, integrate when the change is continuous, and turn those stretches into the movement of the points a designer cares about.

Sources: Gere and Goodno, Mechanics of Materials, 9th edition (SI), sections 2.1–2.3 (the syllabus for this lesson); Beer, Johnston, DeWolf and Mazurek, 8th edition, sections 2.1A (normal strain) and 2.1G (deformations of members under axial loading); Liu, Mechanics of Materials: Formulations and Solutions with Python, sections 7.1–7.6; Hearn, Mechanics of Materials 1, section 1.6 with Examples 1.1 and 1.4; Venkataraman, Essential Mechanics, section 3.2.3; Muvdi and Elhouar, section 1.5. Translations: the internal axial force is Gere's N, Beer's P_i, Muvdi's F, Hearn's F_n and Liu's \sigma A; this book writes N for the internal force and P for an applied load. A member's change of length is \delta in every book; the displacement of a point is Liu's u(x), and Gere and Muvdi also call it \delta. This book keeps \delta for a change of length and u for a displacement. Liu's distributed axial load b_f is our p(x). All working is in SI. Tension positive.

How to use this companion

Each nugget is one idea. It opens with a claim, then a picture shows why it holds, a number checks it, and an equation states it, labelled with the kind of statement it is. A worked example with units, a real use, the most common wrong answer and a check question follow. A bridge says why the next idea is needed, and the Narration spine summarises the video beat. The synthesis, computed views, traps and self-check close the lesson.

The puzzle

The hanging steel wire is 100\,\mathrm m long, with density 7850\,\mathrm{kg/m^3} and E = 200\,\mathrm{GPa}. The plausible answer is "put its whole weight at the bottom and use \delta = WL/(EA)", which gives 3.85\,\mathrm{mm}. It is twice too large: the wire stretches 1.93\,\mathrm{mm}, whatever its diameter. Why half, and why does the diameter not matter? The answer is in Nugget 05.

Nugget 01: A Bar Is a Spring of Stiffness EA Over L

Claim. A prismatic bar of a linearly elastic material, loaded through the centroids of its end sections, changes length by \delta = PL/(EA). It behaves exactly like a coil spring of stiffness k = EA/L.

The picture. A coil spring and a bar, each pulled by the same force.

Gere starts with the spring because it is the simplest load-deflection relation there is: P = k\delta, or \delta = fP, where k is the stiffness (force per unit elongation) and f = 1/k the flexibility (elongation per unit force, also called compliance). The bar does the same thing; what the picture asks is which property of the bar plays the part of k. Beer's scaling argument (2.1A) answers it: a bar of twice the area needs twice the load for the same stretch, and a bar of twice the length stretches twice as much under the same load. So the stretch per unit length, the strain \varepsilon = \delta/L, is what a material responds to, and the stretch is proportional to L/A.

Check with numbers. Gere's strut: an aluminium strut 2\,\mathrm m long at a compressive stress of 48\,\mathrm{MPa} with E = 72\,\mathrm{GPa} has \varepsilon = 48/72\,000 = 6.67\times10^{-4} and shortens by 6.67\times10^{-4}\times2000 = 1.33\,\mathrm{mm}: one part in 1500. That is why every calculation in this book uses the original length: the change is far too small to alter the geometry.

The idea. The derivation is the three definitions from Lessons MM02 and MM04 chained together: \sigma = P/A needs a centric load on a prismatic bar (uniform stress away from the ends); \varepsilon = \delta/L needs a homogeneous material (the same strain everywhere); \sigma = E\varepsilon needs linear elasticity (stress below the proportional limit, Beer; "linearly elastic", Gere; the same condition). Then P/A = E\delta/L and \delta = PL/(EA). The product EA is the axial rigidity (Gere). The same steps hold for compression, where \delta is a shortening. Venkataraman (3.2.3, Eq. 3.7) gives the result in exactly this way, as the one "very useful" formula of the section, and Liu derives it as the simplest solution of a differential equation (Nugget 08). Stiffness and flexibility of the bar are the ones computer methods of structural analysis assemble (Gere).

The equation (geometric relationship combined with the material model):

Conditions: prismatic, homogeneous, linearly elastic, load through the centroids of the ends.

Worked number. A steel rod 16\,\mathrm{mm} in diameter and 1.5\,\mathrm m long carries 20\,\mathrm{kN}, with E = 200\,\mathrm{GPa}. A = \pi(8\,\mathrm{mm})^2 = 201.06\,\mathrm{mm^2}, so \sigma = 20\,000/201.06 = 99.5\,\mathrm{MPa}, well below yield. Stiffness k = 200\times10^{9}\,\mathrm{Pa}\times201.06\times10^{-6}\,\mathrm{m^2}/1.5\,\mathrm m = 26.81\,\mathrm{MN/m}, flexibility f = 1/k = 0.0373\,\mathrm{mm/kN}, and \delta = 20\,\mathrm{kN}\times0.0373\,\mathrm{mm/kN} = 0.746\,\mathrm{mm}.

Real example (one you can see). A bungee cord is a bar with a very small E; its "spring constant" printed on the packet is the EA/L of this nugget, and a longer cord of the same rubber is softer in exact proportion to its length.

Plausible wrong answer. "\delta = PL/(EA) works for any bar." It needs one N, one A and one E along the whole length; the next nuggets deal with bars where any of them changes.

Check. Two rods of the same steel carry the same load; rod 2 is twice as long and has twice the diameter. Which stretches more? (Rod 1: rod 2 has 2L but 4A, so half the stretch.)

Narration spine. A spring stretches under a weight; a bar replaces it and stretches by the same amount; the three definitions slide in and collapse into \delta = PL/(EA); EA/L is boxed as the bar's k.

Bridge. A steel cable is not a steel bar, even though it is made of steel wires. What changes?

Nugget 02: A Cable Is a Softer Spring Than Its Steel

Claim. A wire rope stretches more than a solid bar of the same nominal diameter for two reasons: it has less steel in its cross section (the effective area), and its helical wires straighten as they tighten (the effective modulus). Both effective values go into \delta = PL/(EA).

The picture. A wire-rope cross section beside the circle of its nominal diameter.

Gere describes the usual wire rope: six strands wound helically around a central strand, each strand itself wound from wires. It carries tension but almost no bending. The gaps between the round wires are why its metallic area is only about 60\,\% of the circle; the helix is why it lengthens partly by the wires closing up, like the fibres of a rope, so the cable's effective modulus is about 140\,\mathrm{GPa} against 210\,\mathrm{GPa} for the wire's steel.

Check with numbers. Gere's Table 2-1, which is for textbook problems only, lists for a 25\,\mathrm{mm} cable an effective area of 304\,\mathrm{mm^2} (the text rounds it to 300) and an ultimate load of 406\,\mathrm{kN}. A 14\,\mathrm m length carrying 38\,\mathrm{kN} with E = 140\,\mathrm{GPa} stretches \delta = 38\,000\times14\,000/(304\times140\,000) = 12.5\,\mathrm{mm}. A solid steel bar of 25\,\mathrm{mm} diameter (491\,\mathrm{mm^2}, E = 200\,\mathrm{GPa}) would stretch only 5.42\,\mathrm{mm}.

The idea. The cable is the clearest case of a general rule: E and A in the formula belong to the member as built, not to the material it is made of. The cable's allowable load is its ultimate load divided by a factor of safety, usually 3 to 10 (Lesson MM06); here 406/38 = 10.7. Gere notes that the wire stress at breaking can reach about 1400\,\mathrm{MPa}; Table 2-1's 406\,\mathrm{kN} on 304\,\mathrm{mm^2} is 1336\,\mathrm{MPa}.

The equation (material model: effective properties of a built-up member):

Table 2-1 (Gere, nominal diameter, effective area, ultimate load): 12\,\mathrm{mm}, 76.7\,\mathrm{mm^2}, 102\,\mathrm{kN}; 20\,\mathrm{mm}, 173\,\mathrm{mm^2}, 231\,\mathrm{kN}; 25\,\mathrm{mm}, 304\,\mathrm{mm^2}, 406\,\mathrm{kN}; 32\,\mathrm{mm}, 481\,\mathrm{mm^2}, 641\,\mathrm{kN}; 38\,\mathrm{mm}, 697\,\mathrm{mm^2}, 930\,\mathrm{kN}.

Worked number. A passenger lift hangs on four 12\,\mathrm{mm} cables, each 40\,\mathrm m long, with A_{\text{eff}} = 76.7\,\mathrm{mm^2} and E_{\text{eff}} = 140\,\mathrm{GPa}. Eight people, 6\,\mathrm{kN} in all, step in. The added stretch is \delta = 6000\times40\,000/(4\times76.7\times140\,000) = 5.59\,\mathrm{mm}.

Real example (one you can see). That 5.6\,\mathrm{mm} is the small dip you feel when a crowd steps into a lift on a high floor. Modern lifts re-level the car to the landing for exactly this reason; in a low-rise lift with short cables, the dip is too small to notice.

Plausible wrong answer. "Use the circle of the nominal diameter and the steel's 210\,\mathrm{GPa}." For the lift that gives 2.53\,\mathrm{mm}, less than half the real stretch.

Check. A 20\,\mathrm{mm} cable is replaced by a solid steel rod of the same diameter. Does the stretch go up or down? (Down: more area and a higher modulus.)

Narration spine. A solid circle fades into seven strands; the area number drops from 491 to 304; a lift car dips by a few millimetres as figures step in.

Bridge. So far the force was the same along the whole member. In a real shaft or column loads enter part-way along, and the section changes.

Nugget 03: Stepped Bars: Cut Each Segment, Then Add the Stretches

Claim. When the internal force, the area or the material changes from one segment to the next, apply \delta = NL/(EA) to each segment with its own internal force N_i, found by a cut, and add the results with their signs: \delta = \sum N_iL_i/(E_iA_i).

The picture. A stepped bar fixed at A, with loads at B, C and D, cut once inside each segment.

Each cut keeps the part to its right (Beer's choice; any free body gives the same N). Cut in CD: only the 30\,\mathrm{kN} at D acts, so N_3 = +30\,\mathrm{kN}. Cut in BC: 30 - 45 = -15\,\mathrm{kN}, compression. Cut in AB: 30 - 45 + 75 = +60\,\mathrm{kN}. The sign of N follows the deformation convention of Lesson MM01: tension positive, whatever direction the arrow points in.

Check with numbers. Steel, E = 200\,\mathrm{GPa}, so E in \mathrm{N/mm^2} is 200\,000:

Total \delta = 0.1552 - 0.0388 + 0.3158 = 0.4322\,\mathrm{mm} (elongation). The middle segment shortens while the whole bar lengthens; only a signed sum gets this right.

The idea. This bar is Beer's Concept Application 2.1 with its US data rounded to SI (Beer's own numbers, 0.9 and 0.3\,\mathrm{in^2}, 12, 12 and 16\,\mathrm{in} and 75, 45 and 30 kips, give 75.9\times10^{-3}\,\mathrm{in} = 1.93\,\mathrm{mm} by the same steps). Gere sets out the same procedure (2.3, Eq. 2-5): number the segments, find each N_i from a free body, write N to keep the internal force apart from the external loads P, compute each \delta_i and add them algebraically. Within a segment N is constant, so the displacement grows linearly along it. Venkataraman's Example 3.4 (steel then brass) and Hearn's Example 1.1 (three steel sections) are the same sum with one force; Muvdi (Eq. 1.29) writes it for any number of concentrated loads.

The equation (geometric relationship: compatibility of segments in series):

Worked number. Each segment's stress is under 250\,\mathrm{MPa}, so the law applies; the bar grows 0.432\,\mathrm{mm}. If instead the loads were all moved to D, the force would be 75 - 45 + 30 = 60\,\mathrm{kN} in every segment and the answer would be 60\,000(300/580 + 300/580 + 400/190)/200\,000 = 0.942\,\mathrm{mm}: where a load enters matters.

Real example (one you can see). A drill string, a stepped machine shaft or a multi-storey column: each storey's column carries the loads of all the floors above it and has a different section, so its shortening is the sum over storeys.

Plausible wrong answer. "Use the net applied load in every segment." That treats the bar as if all the loads acted at the end; it gives 0.942\,\mathrm{mm} here instead of 0.432\,\mathrm{mm}.

Check. In the table, which segment controls the stress, and which the stretch? (CD for both: smallest area, largest length.)

Narration spine. Three cuts slice the bar; at each, the arrows to the right are added and an N appears; three small stretches, one negative, add into a total.

Bridge. The sum gives the movement of the end. What about every point in between, and how does one see the whole distribution at once?

Nugget 04: Displacement Diagrams: Force Is the Slope, Stretch Is the Area

Claim. Plot the internal force N(x) along the bar (the axial force diagram) and the axial displacement u(x) (the axial displacement diagram). The slope of the displacement diagram at any point is N/(EA) there, and the change in displacement between two points is the area under the force diagram between them, divided by EA.

The picture. The bar of Nugget 03, its force diagram and its displacement diagram.

The two rules are the integral form of d\delta = N\,dx/(EA): a short element of length dx stretches by N\,dx/(EA), and the displacement of a point is the sum of the stretches of all the elements between it and the fixed end. So du/dx = N/(EA), which is the strain (Beer's strain at a point, \varepsilon = d\delta/dx, Eq. 2.2; Muvdi Eq. 1.25). Where N is negative the line falls; where EA is small the line is steep.

Check with numbers. Slope in AB: 60\,000/(580\times200\,000) = 5.17\times10^{-4}, and over 300\,\mathrm{mm} that is 0.1552\,\mathrm{mm}, the value at B. Area under the force diagram from B to C: -15\,000\times300 = -4.5\times10^{6}\,\mathrm{N\,mm}; divided by EA = 1.16\times10^{8}\,\mathrm N it is -0.0388\,\mathrm{mm}, so u_C = 0.1552 - 0.0388 = 0.1164\,\mathrm{mm}. Then u_D = 0.1164 + 0.3158 = 0.4322\,\mathrm{mm}, the total of Nugget 03.

The idea. Gere (2.3) gives the two rules and draws the diagrams for a bar A–D with segments L/4, L/4 and L/2, loads 3P, 2P (backwards) and P: force +2P, -P, +P; displacements u_B = PL/(2EA), u_C = PL/(4EA), u_D = 3PL/(4EA), the largest. Where N is constant the displacement is a straight line; where N varies linearly it is a parabola (Nugget 05). Beer (2.1G, Eq. 2.12) adds the matching distinction: the deformation of a member is the relative displacement of its ends, \delta_{B/A} = u_B - u_A, and equals PL/(AE); it equals the displacement of the free end only when the other end is fixed. In this bar the deformation of BC is u_C - u_B = -0.0388\,\mathrm{mm} although both B and C move to the right.

Gere uses two sign conventions at once: tension positive for N (deformation convention) and "positive in the +x direction" for displacements (statics convention), so in his hanging-column example the displacement functions come out negative while the text quotes downward magnitudes. This book does the same and says which one a number uses.

The equation (geometric relationship):

Worked number. Gere's diagram with P = 10\,\mathrm{kN}, L = 2\,\mathrm m, E = 200\,\mathrm{GPa}, A = 500\,\mathrm{mm^2} (EA = 1.0\times10^{8}\,\mathrm N): u_B = 10\,000\times2/(2\times10^{8}) = 0.10\,\mathrm{mm}, u_C = 0.05\,\mathrm{mm}, u_D = 0.15\,\mathrm{mm}; stresses +40, -20 and +20\,\mathrm{MPa}.

Real example (one you can see). A long bolt tightened through two plates and a spacer: the bolt's force diagram is constant between the nut and the head, and its displacement diagram is the straight line that a bolt-extension gauge reads at its end.

Plausible wrong answer. "The point that moves most is where the force is largest." In this bar AB has the largest force, but D moves most, because displacement accumulates.

Check. In a segment the displacement diagram is horizontal. What is N there? (Zero.)

Narration spine. The force diagram builds step by step under the bar; below it, the displacement line is drawn with slopes taken from the steps; areas are shaded and labelled with the jumps.

Bridge. The steps came from point loads. A bar's own weight, or friction along a pile, changes the force continuously; the sum becomes an integral.

Nugget 05: When the Force Varies Along the Bar, Integrate

Claim. If the internal force varies continuously, N = N(x), slice the bar into elements, each short enough to have a constant force, and add their stretches: \delta = \int_0^L N\,dx/(EA). A bar hanging under its own weight stretches exactly half as much as it would with the whole weight hung at its end.

The picture. A hanging pipe, a slice at depth x, and the force diagram.

The slice at depth x carries the weight of everything below it, w(L - x), where w = \gamma A is the weight per unit length. The force diagram is a triangle, so its area is half the rectangle WL that a weight W = wL hung at the end would give. By the area rule of Nugget 04, the stretch is half.

Check with numbers. The puzzle's wire: \gamma = \rho g = 7850\times9.81 = 77.0\,\mathrm{kN/m^3}, L = 100\,\mathrm m, E = 200\,\mathrm{GPa}. Then \delta = \gamma L^2/(2E) = 77\,009\times100^2/(2\times200\times10^{9}) = 1.93\times10^{-3}\,\mathrm m = 1.93\,\mathrm{mm}; the whole weight at the end would give 3.85\,\mathrm{mm}. The area cancels: a thicker wire is heavier in exact proportion to its stiffness. The stress at the top is \gamma L = 7.70\,\mathrm{MPa}.

The idea. Gere (2.3, Eqs. 2-6 and 2-7) writes the element's stretch as d\delta = N(x)\,dx/(EA(x)) and integrates; the distributed load p(x) (N/m) may be friction, centrifugal force or weight, and equilibrium of the element gives dN/dx = -p. Gere's Example 2-3 is a drill pipe of outer diameter d and wall t hanging in a well: A = \pi t(d - t), N(x) = w(L - x), and

The displacement diagram is a parabola, steepest at the top (slope W/(EA)) and flat at the bottom where N = 0. Beer (Eq. 2.11) gives the same integral "for a rod hanging under its own weight", and Muvdi (Eq. 1.30) for any variation of force, area or modulus, found first. Muvdi's Example 1.8 hangs a 100\times200\,\mathrm{mm} bar 3.0\,\mathrm m long (\rho = 8800\,\mathrm{kg/m^3}, E = 180\,\mathrm{GPa}) with 5\,\mathrm{kN} at its end: p = 0.02\times8800\times9.81 = 1726.6\,\mathrm{N/m}, N = 5000 + 1726.6x (with x up from the end), and \delta = 5000\times3/(0.02\times180\times10^{9}) + 1726.6\times3^2/(2\times0.02\times180\times10^{9}) = (4.167 + 2.158)\times10^{-6} = 6.32\times10^{-6}\,\mathrm m. His Example 1.7 integrates a load spread over one segment the same way.

The equation (geometric relationship, integrated):

Worked number. A steel drill pipe 1500\,\mathrm m long, d = 100\,\mathrm{mm}, t = 8\,\mathrm{mm}, \gamma = 77\,\mathrm{kN/m^3}, E = 210\,\mathrm{GPa}: A = \pi\times8\times92 = 2312\,\mathrm{mm^2}, W = 77\,000\times2312\times10^{-6}\times1500 = 267\,\mathrm{kN}, u(L) = 77\,000\times1500^2/(2\times210\times10^{9}) = 0.4125\,\mathrm m = 412.5\,\mathrm{mm}. At mid-depth u = 412.5\times(1 - 0.25) = 309.4\,\mathrm{mm}. The top stress is \gamma L = 115.5\,\mathrm{MPa}.

Real example (one you can see). A driller lowering a string of pipe sees the depth reading run ahead of the true pipe length by tens of centimetres; deep-well depth logs correct for exactly this self-weight stretch.

Plausible wrong answer. "Put the weight at the end." It doubles the answer: 825\,\mathrm{mm} instead of 412.5\,\mathrm{mm} for the pipe.

Check. A pile is held by friction spread uniformly along its length and loaded by P at the top. Is its shortening PL/(EA) or PL/(2EA)? (PL/(2EA): the force falls linearly to zero.)

Narration spine. A pipe hangs; a slice is picked out and the weight below it is shaded; the force diagram fills a triangle beside it; the triangle is half the rectangle, and the stretch halves.

Bridge. The integral also handles the other thing that can vary continuously: the area.

Nugget 06: Tapered Bars: Integrate the Area, and Know When P Over A Fails

Claim. For a bar whose cross section changes smoothly, use \delta = \int N\,dx/(EA(x)). A flat bar of linear width gives a logarithm, a cone gives 4PL/(\pi E d_1d_2), and the method is accurate only while the taper is gentle.

The picture. A flat bar tapering from width b_2 at the fixed end to b_1 at the free end.

The width at distance x from A is b(x) = b_2 - (x/L)(b_2 - b_1), so A(x) = t\,b(x), and the force is P everywhere. The narrow end stretches more per unit length, because its stress is higher.

Check with numbers. Gere's Example 2-5: P = 275\,\mathrm{kN}, E = 70\,\mathrm{GPa}, t = 19\,\mathrm{mm}, b_2 = 150\,\mathrm{mm}, b_1 = 100\,\mathrm{mm}, L = 1.5\,\mathrm m.

At B, the mid-length width is b_B = 125\,\mathrm{mm} and \delta_B = [PL/(Et(b_2 - b_1))]\ln(2b_2/(b_1 + b_2)) = 6.203\times\ln1.2 = 1.131\,\mathrm{mm}. A prismatic bar of the narrow width would give 3.10\,\mathrm{mm}; one of the mid width would give 2.48\,\mathrm{mm}, 1.4\,\% low.

The idea. Step the source skips: with u = b_2 - (x/L)(b_2 - b_1), du = -(b_2 - b_1)\,dx/L, so \int_0^L dx/(t\,b(x)) = [L/(t(b_2 - b_1))]\ln(b_2/b_1). As b_1 \to b_2 the logarithm tends to (b_2 - b_1)/b_1 and the result tends to PL/(Eb_1t), the prismatic bar (Gere's Note 2, by L'Hospital's rule). Gere then asks what load P_B, pushing back at B, would bring C back to zero: P_B = 611.6\,\mathrm{kN}, leaving stresses of -118.1, 115.8 and 144.7\,\mathrm{MPa} near A, B and C, all below the 480\,\mathrm{MPa} yield of 7075-T6 aluminium.

Hearn's cone. Hearn (Example 1.4) extends a circular tapered bar, diameter d to D over L, to the apex of its cone; similar triangles give the radius, and integrating \sigma\,dx/E gives

the prismatic formula with the geometric mean of the end areas. For a steel bar (E = 200\,\mathrm{GPa}) 1\,\mathrm m long, tapering from 20 to 40\,\mathrm{mm} under 50\,\mathrm{kN}: \delta = 4\times50\,000\times1000/(\pi\times200\,000\times20\times40) = 0.398\,\mathrm{mm}, between the 0.796\,\mathrm{mm} of a 20\,\mathrm{mm} rod and the 0.199\,\mathrm{mm} of a 40\,\mathrm{mm} one.

The limitation. Only Gere warns that \sigma = N/A is itself approximate in a tapered bar: the stress is no longer uniform over a cross section. When the sides meet at 20\,{}^\circ the N/A value is 3\,\% below the exact one, less for smaller angles, so the integral is satisfactory only for a small taper. Beer, Hearn and Muvdi integrate without the warning; this book keeps it.

The equation (approximation: uniform stress over each section):

Worked number. \delta_C = 2.515\,\mathrm{mm}, \delta_B = 1.131\,\mathrm{mm} (flat bar); 0.398\,\mathrm{mm} (cone).

Real example (one you can see). A lamp post or a crane jib is tapered to put material where the force is: its axial shortening is an integral of this kind.

Plausible wrong answer. "Use the average area." It is close for a gentle taper (1.4\,\% low here) but wrong in principle, and for a cone tapering 1{:}2 it gives 0.354\,\mathrm{mm} instead of 0.398\,\mathrm{mm}, 11\,\% low.

Check. Why does the narrow half of the flat bar stretch more than the wide half? (\delta_B = 1.131\,\mathrm{mm} in the wide half and 1.384\,\mathrm{mm} in the narrow half: higher stress, same modulus.)

Narration spine. The bar narrows; slices are drawn with their areas; each slice's stretch appears as a bar growing toward the narrow end; the slices sum into a logarithm.

Bridge. All of these were the stretches of members. In a structure, members hold up rigid parts, and what matters is how those parts move.

Nugget 07: Rigid Bars on Deformable Supports: Deformation Becomes Movement

Claim. When a rigid bar is held by deformable members, find each member's force by statics, each member's change of length by NL/(EA), and then the movement of any point of the rigid bar from the geometry of small rotations (similar triangles).

The picture. Beer's Sample Problem 2.1: a rigid bar BDE held by two links.

The bar does not bend; it only translates and rotates. So once two of its points are known to have moved, every other point is found by a straight line through them.

Check with numbers. Moments about B: F_{CD}\times0.2 = 30\times0.6, F_{CD} = +90\,\mathrm{kN} (tension); about D: F_{AB} = -60\,\mathrm{kN} (compression). Link AB (aluminium, E = 70\,\mathrm{GPa}, 500\,\mathrm{mm^2}, 0.3\,\mathrm m): \delta_{AB} = -60\,000\times300/(500\times70\,000) = -0.514\,\mathrm{mm}, so B rises 0.514\,\mathrm{mm}. Link CD (steel, 200\,\mathrm{GPa}, 600\,\mathrm{mm^2}, 0.4\,\mathrm m): \delta_{CD} = 90\,000\times400/(600\times200\,000) = 0.300\,\mathrm{mm}, so D falls 0.300\,\mathrm{mm}. The bar pivots about H between B and D, a distance x from D with 0.514/0.300 = (200 - x)/x, x = 73.7\,\mathrm{mm}. E is 400 + 73.7 = 473.7\,\mathrm{mm} from H, so it falls 0.300\times473.7/73.7 = 1.93\,\mathrm{mm}.

The idea. The rotations are tiny (here 0.814\,\mathrm{mm} over 200\,\mathrm{mm}, about 0.004\,\mathrm{rad}), so the points move perpendicular to the bar and the angles can be taken from the displacements directly. Gere (Example 2-1) does the same with springs instead of links: a bar ABC on a spring at B and a spring CD, linked to a second bar with a rotational spring; for P = 12\,\mathrm{kN} and k = 2\,\mathrm{MN/m} the results \delta_D = 4P/(3k) = 8.0\,\mathrm{mm} and \delta_C = 28P/(15k) = 11.2\,\mathrm{mm} come from two moment equations and the geometry \delta_B = \delta_C/2. Gere's Example 2-2 runs the idea backwards, as design: a rigid beam on a steel column (A = 8500\,\mathrm{mm^2}, L = 725\,\mathrm{mm}, E = 205\,\mathrm{GPa}) whose end C may move at most 0.75\,\mathrm{mm}. If the beam translates, P = (EA/L)\delta_C = 1803\,\mathrm{kN}; if it rotates about a pin with B midway, the column carries 2P/3 and moves half as far, so P = \frac{3}{4}(EA/L)\delta_C = 1352\,\mathrm{kN}. Muvdi's Example 1.6 puts a rod and a spring in series under a rigid beam: the 50\,\mathrm{kN} rod force compresses the spring (k = 5000\,\mathrm{kN/m}) by 10\,\mathrm{mm} and stretches the 20\,\mathrm{mm} aluminium rod (2.0\,\mathrm m, 70\,\mathrm{GPa}) by 50\,000\times2000/(314.16\times70\,000) = 4.55\,\mathrm{mm}, so B moves 14.55\,\mathrm{mm} (Muvdi prints the rod's stretch rounded to 4\,\mathrm{mm} and B as 14\,\mathrm{mm}).

The equation (geometric relationship: rigid-body kinematics for small rotations):

Worked number. u_E = 0.300 + (0.300 + 0.514)\times400/200 = 1.93\,\mathrm{mm} downward, the same as the similar-triangle result.

Real example (one you can see). A car's rear axle on two springs, or a diving board on its fulcrum: a small compression at the support becomes a much larger movement at the far end.

Plausible wrong answer. "E moves by the sum of the two link deformations." It ignores the lever: the true 1.93\,\mathrm{mm} is more than twice the 0.814\,\mathrm{mm} sum.

Check. If link AB were made rigid, how far would E fall? (0.300\times600/200 = 0.900\,\mathrm{mm}.)

Narration spine. The bar and its links; the links change length by their numbers; the bar tilts through the new positions of B and D; a line extended to E shows its drop.

Bridge. Every nugget so far was the same three steps: equilibrium, Hooke's law, geometry. Liu writes the three as one differential equation.

Nugget 08: A Second View: the Bar as a Differential Equation

Claim. Equilibrium of a slice, Hooke's law and \varepsilon = du/dx combine into one second-order equation, EA\,u'' = -p(x). Two end conditions fix its two constants, and every result of this lesson is a solution of it.

The picture. A slice of a bar with a distributed load, and the chain of three relations.

Liu (7.1) writes it as \partial^2u/\partial x^2 = -b_f/(EA) with b_f the body force per unit length. The equation is second order, so it has two constants of integration, and a bar has two ends: one condition at each. A fixed end prescribes the displacement, u(0) = 0; a loaded end prescribes the force, EA\,u'(L) = F; a displacement-driven end prescribes u(L) = u_R, as a testing machine does (Liu 7.5).

Check with numbers. No body force, end force F: u'' = 0, so u = c_0x + c_1; u(0) = 0 gives c_1 = 0, and EA\,c_0 = F gives u = Fx/(EA) and u(L) = FL/(EA): Nugget 01. Self-weight, p = \gamma A, free end: u = (\gamma/E)(Lx - x^2/2) and u(L) = \gamma L^2/(2E): Nugget 05, with 412.5\,\mathrm{mm} for the drill pipe. Liu's unit-value cases (E = A = L = 1, fixed left end, free right end): constant load q = 1 gives u = x - x^2/2, u(1) = 0.5; a linear load 0.1 + x gives u = -x^3/6 - x^2/20 + 3x/5, u(1) = 23/60 = 0.3833; a sine load \sin(\pi x) gives u = x/\pi + \sin(\pi x)/\pi^2, u(1) = 1/\pi = 0.3183.

The idea. The equation needs a smooth load and, as Liu writes it, a constant EA. Where a point load or a step in section breaks the smoothness, Liu (7.6) splits the bar into sub-bars and joins them by continuity: the displacement at the end of bar 1 is the starting displacement of bar 2. For two point forces, f_R at the end and f_I at the joint, u_1 = (f_R + f_I)x/(EA) and u_2 = (f_R + f_I)l_1/(EA) + f_Rx/(EA), which is Nugget 03's sum written as functions. With f_R = 10\,\mathrm{kN}, f_I = 4\,\mathrm{kN}, l_1 = l_2 = 1\,\mathrm m and EA = 1.0\times10^{8}\,\mathrm N: u(l_1) = 0.14\,\mathrm{mm} and u(2\,\mathrm m) = 0.24\,\mathrm{mm}. A tapered bar would need the general form (EA\,u')' = -p, which Liu does not write; Gere's integral handles A(x) directly. Liu calls AE/L the axial stiffness, L/AE the axial flexibility and AE "the axial stiffness per length"; dimensionally AE is a stiffness times a length, which is why Gere's name, axial rigidity, is used here.

The equation (physical law combined with the material model and a geometric relationship):

Worked number. A displacement-driven test: a steel bar 1.6\,\mathrm m long is stretched 0.8\,\mathrm{mm} by a machine. With p = 0, u = u_Rx/L and \sigma = Eu_R/L = 200\,000\times0.8/1600 = 100\,\mathrm{MPa}, whatever the area: the machine sets the strain, the bar supplies the force.

Real example (one you can see). Every finite-element program for bars starts here: it divides the bar into elements, replaces this equation by a stiffness EA/L per element and solves for the nodal displacements, which is the stepped-bar sum done by machine.

Plausible wrong answer. "One boundary condition is enough because the bar is fixed at one end." The second-order equation needs two; without the end-force condition the constant c_0, and so the whole stretch, is undetermined.

Check. For a free end, what is u' there? (Zero: no force, so no strain.)

Narration spine. A slice shows three relations; they fuse into one equation; two end conditions click into place; the earlier answers reappear one by one as its solutions.

Synthesis

What no single nugget shows. Every deformation problem in this lesson, from a spring to a tapered strut to a lever on two links, is built from the same three ingredients: equilibrium gives the internal force, the material law turns force into strain, and geometry adds the strains up and turns stretches into movements. The sum, the integral, the diagram and the differential equation are four bookkeeping systems for the same addition.

• Central idea. A bar is a spring of stiffness EA/L; a member made of pieces, or with a varying force or section, is those springs in series, added with their signs.
• Key relationship. \delta = \int_0^L N(x)\,dx/(EA(x)), which is \sum N_iL_i/(E_iA_i) for segments and PL/(EA) for one prismatic piece; du/dx = N/(EA).
• Main assumption. Linear elasticity, small displacements and uniform stress over each section; the last is only approximate in a tapered bar.
• What to explore next. When the supports are more than statics can resolve, the same stretches supply the missing equations (Lesson MM08).

Computed views

Python (numpy, scipy, sympy). MM07_axial_deformation.py sums the stepped bar, integrates the hanging pipe and the two tapered bars by quadrature against their closed forms, and solves Liu's boundary-value problem with sympy.dsolve for four loadings.

Part 1: stepped steel bar, E = 200 GPa
  N (kN): [ 60. -15.  30.]
  segment delta (mm): [ 0.1552 -0.0388  0.3158]
  u at B, C, D (mm): [0.1552 0.1164 0.4322]
Part 2: varying force or area
  drill pipe 1500 m: 412.5 mm, gamma L^2 / 2E = 412.5 mm
  flat taper: delta_B 1.1309 mm, delta_C 2.5151 mm, closed form 2.5151 mm
  cone 20 to 40 mm: 0.3979 mm, 4WL/(pi D d E) = 0.3979 mm
Part 3: Liu's boundary-value problem (unit values E = A = L = 1)
  constant q = 1: u = x*(2 - x)/2, u(1) = 0.5000
  linear 0.1 + x: u = x*(-10*x**2 - 3*x + 36)/60, u(1) = 0.3833
  sine sin(pi x): u = (pi*x + sin(pi*x))/pi**2, u(1) = 0.3183
  self-weight as a body force: u(L) = L**2*gamma/(2*E), 412.5 mm

MATLAB (Octave-compatible). MM07_stepped_bar.m tabulates the stepped bar and integrates the flat tapered bar with integral.

segment      N (kN)    L (m)   A (mm^2)  sigma (MPa)   delta (mm)
AB             60.0     0.30        580       103.45       0.1552
BC            -15.0     0.30        580       -25.86      -0.0388
CD             30.0     0.40        190       157.89       0.3158
u_B = 0.1552 mm, u_C = 0.1164 mm, u_D = 0.4322 mm
tapered bar: delta_B 1.1309 mm, delta_C 2.5151 mm
  prismatic 100 mm bar would give 3.1015 mm

SOLIDWORKS Simulation exercise (our own; stepped bar, end displacement against the sum).

• Geometry. A steel bar 1.0\,\mathrm m long: 0.6\,\mathrm m of 29\,\mathrm{mm}\times20\,\mathrm{mm} section (580\,\mathrm{mm^2}) followed by 0.4\,\mathrm m of 19\,\mathrm{mm}\times10\,\mathrm{mm} (190\,\mathrm{mm^2}), the two on a common axis.
• Material. E = 200\,\mathrm{GPa}, \nu = 0.3 (linear elastic isotropic).
• Restraints. Fixed geometry on the large end face.
• Load. 30\,\mathrm{kN} normal force, tensile, on the small end face.
• Mesh. Standard solid mesh, 3\,\mathrm{mm} elements, refined at the shoulder.
• Hand numbers the result must match. \delta = 30\,000(600/580 + 400/190)/200\,000 = 0.1552 + 0.3158 = 0.4710\,\mathrm{mm} at the free end and 0.1552\,\mathrm{mm} at the shoulder; stresses 51.7\,\mathrm{MPa} and 157.9\,\mathrm{MPa} away from the shoulder.
• What to look for. The axial displacement plot should be two straight ramps with a kink at the shoulder, the steeper one in the thin part (Nugget 04). The stress near the shoulder is higher than 157.9\,\mathrm{MPa}: a stress concentration, the subject of Lesson MM12, which does not change the end displacement noticeably.

(This exercise was written for the lesson and has not been run in SOLIDWORKS here.)

Traps

• Using \delta = PL/(EA) on a bar whose force, area or material changes. Cut it into segments or integrate. Counterexample: the stepped bar gives 0.942\,\mathrm{mm} with the end load everywhere, 0.432\,\mathrm{mm} correctly. (Nugget 03.)
• Dropping the sign of a compressed segment. The stretches add algebraically; BC's -0.0388\,\mathrm{mm} must be subtracted. (Nugget 03.)
• Confusing the external load with the internal force. P is applied at a point; N is what a cut carries and changes at every load. (Nugget 03.)
• Confusing deformation with displacement. A segment can shorten while both its ends move forward; \delta_{B/A} = u_B - u_A. (Nugget 04.)
• Hanging the self-weight at the end. It doubles the stretch; the force diagram is a triangle. (Nugget 05.)
• Using the nominal area and the steel's modulus for a cable. Use the effective area and modulus: the lift's 5.6\,\mathrm{mm} becomes 2.5\,\mathrm{mm} otherwise. (Nugget 02.)
• Using the average area for a tapered bar. Integrate; for a 1{:}2 cone the average area is 11\,\% low. And remember the integral itself is only good for a small taper. (Nugget 06.)
• Adding link deformations to get a rigid bar's movement. Use the geometry of the rotation. (Nugget 07.)
• Mixing units. N, mm and MPa together, or N, m and Pa; never \mathrm{kN} with \mathrm{GPa} and mm unconverted. (Nugget 01.)
• Solving the bar equation with one boundary condition. It is second order and needs two. (Nugget 08.)

Self-check

1. A steel rod (E = 200\,\mathrm{GPa}) 4\,\mathrm{mm} in diameter and 25\,\mathrm m long may stretch at most 10\,\mathrm{mm} and its stress may not exceed 150\,\mathrm{MPa}. Find the largest load and its stiffness.
2. A 20\,\mathrm{mm} wire rope (A_{\text{eff}} = 173\,\mathrm{mm^2}, E_{\text{eff}} = 140\,\mathrm{GPa}, ultimate load 231\,\mathrm{kN}) 25\,\mathrm m long carries 30\,\mathrm{kN}. Find its stretch and factor of safety, and the stretch of a solid steel rod of the same diameter (E = 200\,\mathrm{GPa}).
3. A bar fixed at A has segments AB (steel, E = 200\,\mathrm{GPa}, 0.50\,\mathrm m, 400\,\mathrm{mm^2}), BC (brass, E = 100\,\mathrm{GPa}, 0.40\,\mathrm m, 400\,\mathrm{mm^2}) and CD (steel, 0.60\,\mathrm m, 250\,\mathrm{mm^2}). Axial loads: 40\,\mathrm{kN} away from A at B, 30\,\mathrm{kN} toward A at C, 20\,\mathrm{kN} away from A at D. Find the internal forces, the stresses and the elongation.
4. A bar fixed at A, with EA = 2.0\times10^{8}\,\mathrm N throughout, has three 0.4\,\mathrm m segments AB, BC, CD. Loads: 30\,\mathrm{kN} away from A at B, 50\,\mathrm{kN} toward A at C, 40\,\mathrm{kN} away from A at D. Draw the force and displacement diagrams; where is the largest displacement, and how large is it?
5. A steel rod 20\,\mathrm{mm} in diameter hangs 30\,\mathrm m down a shaft carrying 10\,\mathrm{kN} at its end (\rho = 7850\,\mathrm{kg/m^3}, E = 200\,\mathrm{GPa}). Find the stretch with and without its own weight, and the largest stress.
6. A concrete pile (E = 25\,\mathrm{GPa}, A = 0.04\,\mathrm{m^2}, L = 12\,\mathrm m) carries 400\,\mathrm{kN} at its top, resisted entirely by skin friction. Find its shortening when the friction is (a) uniform along the length and (b) growing linearly from zero at the top to a maximum at the tip.
7. A steel cone hangs from its base, apex down, 3\,\mathrm m long (\gamma = 77\,\mathrm{kN/m^3}, E = 200\,\mathrm{GPa}). Find its stretch under its own weight.
8. A flat steel bar (E = 200\,\mathrm{GPa}, t = 20\,\mathrm{mm}, L = 1.5\,\mathrm m) tapers from 150 to 100\,\mathrm{mm} wide and carries 125\,\mathrm{kN}. Find its elongation.
9. An aluminium frustum (E = 70\,\mathrm{GPa}) of end radii 10 and 20\,\mathrm{mm}, 0.5\,\mathrm m long, carries 20\,\mathrm{kN}. Find its change of length.
10. In Beer's link-and-bar arrangement of Nugget 07, the load at E is reduced to 24\,\mathrm{kN}. Find the link forces and the movements of B, D and E.
11. A bar fixed at x = 0 (EA = 1.0\times10^{8}\,\mathrm N, L = 2\,\mathrm m) carries a uniform axial body force p = 5\,\mathrm{kN/m} toward the free end and 10\,\mathrm{kN} at the free end. Solve EA\,u'' = -p and find u(L) and the force at x = 0.
12. A testing machine stretches a steel bar (E = 200\,\mathrm{GPa}, A = 200\,\mathrm{mm^2}) 1.6\,\mathrm m long by 0.8\,\mathrm{mm}. Find the stress and the force. Which boundary condition did the machine set?

Fully worked solutions

1. A = \pi(2\,\mathrm{mm})^2 = 12.566\,\mathrm{mm^2}. From the stretch limit, P = \delta EA/L = 10\times200\,000\times12.566/25\,000 = 1005\,\mathrm N. From the stress limit, P = 150\times12.566 = 1885\,\mathrm N. The stretch governs: P = 1.005\,\mathrm{kN} (\sigma = 80.0\,\mathrm{MPa}). Stiffness k = EA/L = 200\,000\times12.566/25\,000 = 100.5\,\mathrm{N/mm}.

2. \delta = 30\,000\times25\,000/(173\times140\,000) = 30.97\,\mathrm{mm}; n = 231/30 = 7.7. The solid rod: A = \pi\times10^2 = 314.16\,\mathrm{mm^2}, \delta = 30\,000\times25\,000/(314.16\times200\,000) = 11.94\,\mathrm{mm}. The cable stretches 2.6 times as much.

3. Cut in CD: N_3 = +20\,\mathrm{kN}. In BC: 20 - 30 = -10\,\mathrm{kN}. In AB: -10 + 40 = +30\,\mathrm{kN}. Stresses: 30\,000/400 = 75\,\mathrm{MPa}, -10\,000/400 = -25\,\mathrm{MPa}, 20\,000/250 = 80\,\mathrm{MPa}. Stretches: 30\,000\times500/(200\,000\times400) = 0.1875\,\mathrm{mm}, -10\,000\times400/(100\,000\times400) = -0.100\,\mathrm{mm}, 20\,000\times600/(200\,000\times250) = 0.240\,\mathrm{mm}. Total 0.1875 - 0.100 + 0.240 = 0.3275\,\mathrm{mm}.

4. The net load away from A is 30 - 50 + 40 = 20\,\mathrm{kN}, so the wall pulls back with 20\,\mathrm{kN}. N_{AB} = +20\,\mathrm{kN}, N_{BC} = 20 - 30 = -10\,\mathrm{kN}, N_{CD} = -10 + 50 = +40\,\mathrm{kN} (equal to the load at D, as it must be). Stretches N\times400/(2.0\times10^{8}\,\mathrm N): 0.04, -0.02, 0.08\,\mathrm{mm}. Displacements u_B = 0.04, u_C = 0.02, u_D = 0.10\,\mathrm{mm}. The force diagram is a step +20, -10, +40\,\mathrm{kN}; the displacement diagram rises, dips, then rises steeply; the largest displacement is 0.10\,\mathrm{mm} at D.

5. A = \pi\times10^2 = 314.16\,\mathrm{mm^2}. End load: 10\,000\times30\,000/(314.16\times200\,000) = 4.775\,\mathrm{mm}. Own weight: \gamma = 7850\times9.81 = 77\,009\,\mathrm{N/m^3}, \gamma L^2/(2E) = 77\,009\times900/(4\times10^{11}) = 1.733\times10^{-4}\,\mathrm m = 0.173\,\mathrm{mm}. Total 4.948\,\mathrm{mm}. Largest stress at the top: 10\,000/314.16 + 77\,009\times30\times10^{-6} = 31.83 + 2.31 = 34.1\,\mathrm{MPa}.

6. (a) Uniform friction: N falls linearly from P to zero, so \delta = PL/(2EA) = 400\,000\times12/(2\times25\times10^{9}\times0.04) = 2.40\times10^{-3}\,\mathrm m = 2.40\,\mathrm{mm}. (b) With y measured up from the tip and friction f_0(1 - y/L), P = f_0L/2 and N(y) = P(2y/L - y^2/L^2); \int_0^L N\,dy = P(L - L/3) = 2PL/3, so \delta = 2PL/(3EA) = 3.20\,\mathrm{mm}. More of the load is carried high up, so more of the pile is highly stressed.

7. At distance s above the apex the area is proportional to s^2 and the cone below weighs \gamma A(s)s/3, so \sigma = \gamma s/3 and \delta = \int_0^L\gamma s\,ds/(3E) = \gamma L^2/(6E) = 77\,000\times9/(6\times200\times10^{9}) = 5.78\times10^{-7}\,\mathrm m: about 0.0006\,\mathrm{mm}. (With the cone's weight W and base diameter d this is 2WL/(\pi d^2E).)

8. \delta = PL\ln(b_2/b_1)/(Et(b_2 - b_1)) = 125\,000\times1500\times0.4055/(200\,000\times20\times50) = 0.380\,\mathrm{mm}.

9. \delta = PL/(\pi E\,r_1r_2) = 20\,000\times500/(\pi\times70\,000\times10\times20) = 0.227\,\mathrm{mm} (the cone formula 4PL/(\pi E d_1d_2) in radii).

10. Moments about B: F_{CD}\times0.2 = 24\times0.6, F_{CD} = 72\,\mathrm{kN}; vertical equilibrium: F_{AB} = -(72 - 24) = -48\,\mathrm{kN}. \delta_{AB} = -48\,000\times300/(500\times70\,000) = -0.411\,\mathrm{mm} (B rises 0.411\,\mathrm{mm}); \delta_{CD} = 72\,000\times400/(600\times200\,000) = 0.240\,\mathrm{mm} (D falls). u_E = 0.240 + (0.240 + 0.411)\times400/200 = 1.543\,\mathrm{mm} down, 24/30 of the 1.93\,\mathrm{mm}: everything is linear.

11. EA\,u' = -px + c_0; EA\,u'(L) = F gives c_0 = F + pL = 10 + 10 = 20\,\mathrm{kN}, which is the force at x = 0. EA\,u = -px^2/2 + c_0x (with u(0) = 0), so u(L) = (-pL^2/2 + c_0L)/(EA) = (FL + pL^2/2)/(EA) = (10\,000\times2 + 5000\times4/2)/10^{8} = 3.0\times10^{-4}\,\mathrm m = 0.30\,\mathrm{mm}.

12. \sigma = E u_R/L = 200\,000\times0.8/1600 = 100\,\mathrm{MPa}; N = 100\times200 = 20\,\mathrm{kN}. The machine set a displacement at the moving end, u(L) = 0.8\,\mathrm{mm}, not a force: a displacement-driven problem (Liu 7.5).