Study Companion: MM02 — Normal Stress and Strain

Two steel rods hang from the same beam, each with a 12\,\mathrm{kN} load on its end. One is 16\,\mathrm{mm} in diameter, the other 10\,\mathrm{mm}. The internal force in both is exactly 12\,\mathrm{kN} (Lesson MM01), yet the thin rod will yield first and the thick one will not. What number says so, and by how much does the thin rod suffer more? That number is stress, and its companion, strain, says how much a rod stretches for its length. This lesson builds both for the simplest member there is, a straight bar pulled or pushed along its axis, and shows when the simple formula can be trusted.

Sources: Gere and Goodno, Mechanics of Materials, 9th edition (SI), section 1.4 (the syllabus for this lesson); Beer, Johnston, DeWolf and Mazurek, 8th edition, section 1.2 (the axial-stress part; Beer's definition of strain is in 2.1 and is used here); Liu, Mechanics of Materials: Formulations and Solutions with Python, sections 3.1–3.3; Hearn, Mechanics of Materials 1, 3rd edition, sections 1.1–1.4; Venkataraman, Essential Mechanics, section 3.1; Muvdi and McNabb, sections 1.3.1–1.3.2 and 1.4.1–1.4.3. The references agree on the content and differ in emphasis and symbols. The internal axial force is N in Gere and Liu, F in Beer and Muvdi and P in Hearn and Venkataraman; this book writes N for the force at a cut and P for an applied load. "Axial stress" (Beer), "direct stress" (Hearn) and "normal stress" (the rest) are one quantity. Stress is quoted in pascals (Pa), kPa, MPa (10^6\,\mathrm{Pa}) and GPa; 1\,\mathrm{MPa} is the same as 1\,\mathrm{N/mm^2} (Gere notes the N/mm² form is not recommended in strict SI, but it is the working unit for stress in millimetres and newtons, and it is used with MPa here); Hearn's \mathrm{MN/m^2} and \mathrm{GN/m^2} are the same numbers as MPa and GPa. The US examples of the references are converted: 1 ksi = 6.895 MPa, 1 in = 25.4 mm.

Sign convention: tension positive, compression negative, for stress and strain alike.

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

Two steel rods carry 12\,\mathrm{kN} each, one 16\,\mathrm{mm} in diameter and one 10\,\mathrm{mm}. The plausible answer is "they carry the same, so they are equally loaded". They are not. The answer is in Nugget 01.

Nugget 01: Stress Is Force Per Area, Because Force Alone Cannot Rank Two Bars

Claim. What a piece of material feels is not the force through it but the force per unit of cross-sectional area. For a straight bar of constant section (a prismatic bar) pulled along its axis through the centroid, that average is the normal stress \sigma = N/A.

The picture. The same 12\,\mathrm{kN} spread over two different areas.

Each cut face carries the same total, 12\,\mathrm{kN}, drawn as the same three arrows. On the large face the arrows are far apart; on the small face they are crowded. Crowding is what loads the material: more force to each small patch.

Check with numbers. Areas: A = \pi d^2/4 gives 201.06\,\mathrm{mm^2} for the 16\,\mathrm{mm} rod and 78.54\,\mathrm{mm^2} for the 10\,\mathrm{mm} rod. Stresses: 12\,000\,\mathrm N / 201.06\,\mathrm{mm^2} = 59.7\,\mathrm{N/mm^2} = 59.7\,\mathrm{MPa} and 12\,000 / 78.54 = 152.8\,\mathrm{MPa}. The ratio is (16/10)^2 = 2.56: halving the diameter quarters the area and quadruples the stress.

The idea. Gere (1.4) introduces the bar and its cut, finds the segment in equilibrium and says the cut face carries a continuous distribution of stress whose resultant is the internal force. Intensity of force is the natural measure because it removes the size of the bar from the question: two bars of the same material fail at the same stress however big they are, which is why strength is quoted in MPa, not kN.

The equation (definition: average normal stress on a section normal to the axis; valid when the stress is uniform, Nugget 03):

Worked number. The boom-and-rod frame of Lesson MM01 puts 50\,\mathrm{kN} of tension in the rod BC. For a rod 20\,\mathrm{mm} in diameter, A = 314.16\,\mathrm{mm^2} and \sigma = 50\,000/314.16 = 159.2\,\mathrm{MPa}. For a steel allowing 165\,\mathrm{MPa} this is safe; the allowable stress is the subject of Lesson MM06.

Real example (one you can see). Press a thumb and then a sharp pin against a wall with the same force. The thumb does nothing; the pin marks the wall. The force is the same, the area is not.

Plausible wrong answer. "Two bars with the same load are equally likely to break." The thin one is closer to failing, by the ratio of the areas.

Check. A 20\,\mathrm{mm} rod replaces the 10\,\mathrm{mm} rod under the same 12\,\mathrm{kN}. By what factor does the stress fall? (By 4, to 38.2\,\mathrm{MPa}.)

Narration spine. Two bars, same arrow, a cut shows a thick disc and a thin disc; the same three arrows fill each; the number under each disc is force over area.

Bridge. We said the cut face carries "a distribution". What exactly does that distribution add up to, and does it have to be uniform?

Nugget 02: The Force on a Cut Is the Sum of Stress Over the Area

Claim. Stress is defined at a point as the limit of force per area, and the internal force is the sum of the stress over the whole face: N = \int \sigma\,dA. The average is N/A, and it equals the local value only if the stress is uniform.

The picture. A cut face broken into small elements.

Both shapes in the graph enclose the same area, which equals N. Statics tells us only that total. It cannot say which shape the real stress takes: the distribution of a force over a section is statically indeterminate, and what fixes it is how the material deforms (Nugget 03).

Check with numbers. Take a 20\,\mathrm{mm} round bar pulled with 1500\,\mathrm N (Venkataraman's Example 3.1). A = \pi(10\,\mathrm{mm})^2 = 314.16\,\mathrm{mm^2}, so the average is 1500/314.16 = 4.77\,\mathrm{MPa}. (The book prints 4.77\times10^{5}\,\mathrm{Pa}; the correct power is 10^{6}.) A peaked distribution would be higher than 4.77 in the middle and lower at the rim, and would still add up to 1500\,\mathrm N.

The idea. Beer (1.2A) and Venkataraman (3.1.1) define stress at a point as the limit of the force \Delta F on a small area \Delta A as \Delta A \to 0, and state that the resultant of the elementary forces must equal P. Muvdi writes the same limit as dF/dA. A second route, Liu's (3.2–3.3), skips "average" altogether: stress is a field, and for the simplest bar it follows from equilibrium of a short element; this route is used in Nugget 06.

The equation (definition of point stress, and physical law: the resultant of the elementary forces equals the internal force):

Worked number. Suppose the stress across a 60\,\mathrm{mm} wide, 10\,\mathrm{mm} thick strip were \sigma(y) = 80\,\mathrm{MPa} in the central 30\,\mathrm{mm} and 40\,\mathrm{MPa} in the two outer 15\,\mathrm{mm} strips. Then N = 80 \times (30\times10) + 40 \times (2 \times 15\times10) = 24\,000 + 12\,000 = 36\,000\,\mathrm N = 36\,\mathrm{kN}, and the average is 36\,000/600\,\mathrm{mm^2} = 60\,\mathrm{MPa}, lower than the central value and higher than the edge value, as an average must be.

Real example (one you can see). A crowd leaning on a barrier: the total push is the sum of every person's push, but the barrier's posts feel the local crowding, not the average.

Plausible wrong answer. "The stress at every point of the cut is N/A." That is true only when the distribution is uniform. N/A is always the average.

Check. If the stress is 50\,\mathrm{MPa} on a face of 400\,\mathrm{mm^2}, what is N if it is uniform? (20\,\mathrm{kN}.)

Narration spine. The face is gridded; each cell gets a small arrow; the arrows are summed into one; two stress shapes with the same area under them are drawn.

Bridge. If statics cannot tell us the shape, when can we safely assume it is flat?

Nugget 03: The Average Is the Truth Only for a Centric Load, Away From the Ends

Claim. \sigma = N/A is the actual stress when the bar is prismatic, the material is homogeneous, the load line passes through the centroid of the section and the section is far from the ends. Each condition has a reason, and the centroid one can be proved.

The picture. A bar with a load through the centroid and a load off to one side.

Why the centroid. If the stress is uniform it can be replaced by the single force N = \sigma A at the centroid of the area, because for equal stress on equal pieces of area the "centre of the push" is the centre of the area. Turn that around: for the stress to be uniform the load line must pass through the centroid.

Check with numbers. Take a short pier whose section is a 300\,\mathrm{mm} \times 200\,\mathrm{mm} rectangle with a 100\,\mathrm{mm}\times100\,\mathrm{mm} corner removed (removed between x = 200 to 300\,\mathrm{mm} and y = 100 to 200\,\mathrm{mm}, origin at the lower left corner). Its area is 60\,000 - 10\,000 = 50\,000\,\mathrm{mm^2}. Its centroid: \bar x = (60\,000 \times 150 - 10\,000 \times 250)/50\,000 = 130\,\mathrm{mm} and \bar y = (60\,000\times100 - 10\,000\times150)/50\,000 = 90\,\mathrm{mm}. A 100\,\mathrm{kN} load through (130, 90)\,\mathrm{mm} gives a uniform \sigma = 100\,000/50\,000 = 2.0\,\mathrm{MPa}. Place the same load through the middle of the rectangle, (150, 100)\,\mathrm{mm}, which is 20\,\mathrm{mm} and 10\,\mathrm{mm} off the centroid, and it brings moments of 100\,\mathrm{kN}\times0.020\,\mathrm m = 2.0\,\mathrm{kN\,m} and 1.0\,\mathrm{kN\,m}: the stress is no longer 2.0\,\mathrm{MPa} everywhere. The proper treatment is in Lessons MM24 and MM27.

The idea. Gere (Eqs. 1-8, 1-9): let the load act at (x_P, y_P) and the stress be uniform. Moments of the stress about the axes must equal the moments of N: N y_P = \int \sigma y\,dA, and with \sigma = N/A constant, y_P = \int y\,dA / A = \bar y, and likewise x_P = \bar x. This is the first proof in this lesson that rests on an assumption (that the stress is uniform): it shows what that assumption demands, not that it holds.

Why far from the ends. Near a concentrated load or a pin the stress peaks and falls, as the picture's lower bar shows. Saint-Venant's principle says that at a distance of about the bar's largest lateral dimension from the load, only the resultant matters and the stress is nearly uniform; Gere states this as "at least as far as the largest lateral dimension" and Muvdi as "at least d". No reference offers a stricter number. Venkataraman and Gere add that homogeneous material is assumed; a stiff fibre in a soft matrix does not share the stress equally.

The equation (approximation with stated conditions):

Worked number. For the pier above, \sigma = 2.0\,\mathrm{MPa} (compression, so -2.0\,\mathrm{MPa}) at any section more than about 300\,\mathrm{mm} from the ends.

Real example (one you can see). A table leg pressed straight down is easy; push down on a corner of a stool and it tilts. The off-centre push adds a bending effect that the leg's stress must now resist besides the squeeze.

Plausible wrong answer. "P/A is the stress near the pin too." Right next to a pin, the stress at the hole can be several times P/A (stress concentration, Lesson MM12).

Check. A 30\,\mathrm{mm} wide bar is pulled by a pin through its end. How far from the end can N/A be used? (About 30\,\mathrm{mm} or more.)

Narration spine. Arrows across a section start tall at one edge; as the load is slid toward the centroid the arrows level out and the number under the picture settles at N/A; a pin appears and the arrows spike near it.

Bridge. We now have a stress. To write down how it acts on a piece of the material, we need a way to record the plane it acts on.

Nugget 04: A Stress Element Records Which Plane the Stress Acts On

Claim. A stress is attached to a plane. A tiny cube of material, the stress element, shows the stresses on its faces; for a bar in tension it has just two loaded faces, the ones normal to the axis.

The picture. A compressed bar and its element.

Check with numbers. A bar 30\,\mathrm{mm}\times20\,\mathrm{mm} carries a compression of 60\,\mathrm{kN}: \sigma = -60\,000/600 = -100\,\mathrm{MPa}. The element has \sigma_x = -100\,\mathrm{MPa} on its two x faces, with arrows pointing into the element, and zero on the others: a uniaxial state.

The idea. Muvdi (1.3.2, 1.3.6) names a plane by its outward normal (X plane, Y plane) and defines a plane as positive when its outward normal points along a positive axis. A normal stress is positive when it points along the outward normal of its plane, so tension is positive on both faces and the arrows leave the cube. Liu writes the same stress \sigma_{xx} with two subscripts: the first is the direction of the face normal and the second the direction of the stress; for a normal stress they are equal, and a bar's stress is \sigma_{xx} or simply \sigma. The element is the tool that, in Part 6, will be rotated to find stresses on inclined planes.

The equation (definition, sign rule):

Worked number. Same bar in tension, 60\,\mathrm{kN}: \sigma_x = +100\,\mathrm{MPa}, arrows pointing out of the cube; the other four faces unchanged at zero.

Real example (one you can see). A cube of foam in a vice is compressed on two faces and free on the rest. It bulges sideways because the free faces cannot hold it in (Lesson MM04).

Plausible wrong answer. "Compression is positive because the load is pushing." The sign is tied to the outward normal: a push inward is negative.

Check. What is the stress on the y faces of a bar pulled along x? (Zero.)

Narration spine. A cube is lifted out of the bar; arrows grow on two opposite faces and the other faces stay empty; the sign rule flips the arrows for compression.

Bridge. Stress is half of the story. The other half is how much the bar stretches.

Nugget 05: Strain Is Stretch per Unit Length

Claim. The stretch of a bar depends on its length as well as the load, so the meaningful measure of how much a material is deformed is the stretch per unit length, the normal strain \varepsilon = \delta/L.

The picture. A bar before and after being stretched, and a half-length bar beside it.

Cut the bar in two equal halves: each half takes half the stretch. Stretch per unit length is the same whichever piece we look at, and so it is a property of the material at that state, not of how long a piece happened to be.

Check with numbers. A bar 2.0\,\mathrm m long stretches by 1.4\,\mathrm{mm}: \varepsilon = 1.4\,\mathrm{mm}/2000\,\mathrm{mm} = 7.0\times10^{-4}. A cable 0.52\,\mathrm m long that stretches by 0.26\,\mathrm{mm} has \varepsilon = 0.26/520 = 5.0\times10^{-4}.

The idea. Gere (Eq. 1-7) and Beer (2.1A) argue as in the picture: for the same load, a bar of double length stretches double, so \delta/L is what stays the same. Strain is dimensionless; it is written as m/m, mm/m, microstrain (\mu\varepsilon = 10^{-6}, written \mu\mathrm{m/m} by gauge makers) or percent. Tension gives positive strain, compression negative (Hearn, Muvdi).

The equation (definition: average normal strain; geometric relationship):

Worked number. \varepsilon = 7.0\times10^{-4} = 700\,\mu\varepsilon = 0.07\,\%. Venkataraman's Example 3.1 has a 100\,\mathrm{mm} rod that shortens 0.1\,\mathrm{mm}: \varepsilon = -1.0\times10^{-3} (the book prints +0.001 and ignores the sign).

Engineering and true values. Venkataraman (3.1.2) distinguishes the engineering strain \delta/L_0 and stress N/A_0 (original dimensions) from the true strain \ln(L/L_0) and true stress N/A (current area, valid while the volume is constant). For a 1.5\,\mathrm{mm} stretch of 100\,\mathrm{mm} they are 0.015 and \ln 1.015 = 0.01489, a 0.7\,\% difference; for L/L_0 = 1.5 they are 0.5 and 0.405. Design uses engineering values, and the difference is felt only at large strain (Lesson MM03).

Real example (one you can see). A strain gauge on a bridge member reads of order 100 to 1000\,\mu\varepsilon in service: a metre of steel is stretched by 0.1 to 1\,\mathrm{mm}.

Plausible wrong answer. "A bar that stretched more is more strained." Only if the lengths are equal. A long bar with a small strain can stretch more than a short bar with a large strain.

Check. What is the strain of a 50\,\mathrm{mm} gauge length that stretches by 0.04\,\mathrm{mm}? (8.0\times10^{-4} = 800\,\mu\varepsilon.)

Narration spine. A bar is marked into equal lengths; as it stretches each mark moves equally; the ratio stretch over length is written.

Bridge. The simple formulas assume a constant force and area along the bar. Real members change both.

Nugget 06: The Stress Changes Along a Bar When the Force or the Area Does

Claim. Compute the stress section by section: at each section take the internal force from a free body and divide by the area there. The largest stress is where the ratio N/A is largest, which need not be where the load is largest.

The picture. A stepped bar of three sections, and a hanging wire.

Check with numbers. Stepped bar. The same 15\,\mathrm{kN} passes through every section, but the areas differ: 16\,\mathrm{mm} round: A = 201.06\,\mathrm{mm^2}, \sigma = 74.60\,\mathrm{MPa}; 24\,\mathrm{mm} square: A = 576\,\mathrm{mm^2}, \sigma = 26.04\,\mathrm{MPa}; 10\,\mathrm{mm} round: A = 78.54\,\mathrm{mm^2}, \sigma = 190.99\,\mathrm{MPa}. The thin section governs.

Hanging wire (8\,\mathrm{mm} steel wire, weight density \gamma = 77\,\mathrm{kN/m^3}; a 300\,\mathrm N load hangs 5\,\mathrm m below the support and a 600\,\mathrm N load at the end, 4\,\mathrm m further down, so the wire is 9\,\mathrm m long): the internal force is largest at the top, equal to the loads plus the weight of the whole wire. A = 50.27\,\mathrm{mm^2}. The loads alone give 900\,\mathrm N / 50.27\,\mathrm{mm^2} = 17.91\,\mathrm{MPa}; the wire's weight adds \gamma L = 77\,000\,\mathrm{N/m^3} \times 9\,\mathrm m = 693\,000\,\mathrm{Pa} = 0.693\,\mathrm{MPa}, so \sigma_{\max} = 18.60\,\mathrm{MPa}, 3.7\,\% more than the loads alone suggest. (Gere's own example, with 380\,\mathrm N and 700\,\mathrm N at 6\,\mathrm m and a further 5\,\mathrm m on a 9.5\,\mathrm{mm} wire, gives 15.24 + 0.847 = 16.08\,\mathrm{MPa}, about 5.5\,\% more than the loads alone.)

The idea. Hearn (Example 1.1), Venkataraman (Example 3.2) and Gere (Example 1-6) all do this one section at a time. Liu's route, a second view of the hanging bar, solves the equilibrium of a thin slice of length dx with a body force b per unit volume (here weight density \gamma). The slice has N on one face and N + dN on the other, and b A\,dx acting on it, so dN/dx + bA = 0, and for a bar of constant area d\sigma/dx + b = 0. With the top at x = 0 fixed, the free end at x = L and no end load, \sigma(L) = 0 fixes the constant: \sigma(x) = b(L - x), a straight line from bL at the top to zero at the free end.

Reconcile. For the wire's own weight, b = \gamma and L = 9\,\mathrm m: Liu gives \sigma(0) = 77\,000 \times 9 = 693\,\mathrm{kPa}, and at x = 4.5\,\mathrm m half of it, 346.5\,\mathrm{kPa}. That is exactly the weight term \gamma L in the section-by-section calculation: the two routes agree. (Liu's step d\sigma/dx = dN/dx / A assumes a prismatic bar; for a varying section the equation is d(A\sigma)/dx + bA = 0.)

The equation (physical law: equilibrium of a slice; geometric relationship between area and stress):

Worked number. Stepped bar: \sigma_{\max} = 191.0\,\mathrm{MPa} in the 10\,\mathrm{mm} section. Wire: \sigma_{\max} = 18.60\,\mathrm{MPa} at the top.

Real example (one you can see). A long hanging cable on a mountain lift is thicker near the top than near the bottom, because the top holds the rest of the cable as well as the cabin.

Plausible wrong answer. "The stress is greatest where the load is applied." For a hanging wire the load is applied at the bottom and the stress peaks at the top.

Check. If the wire were twice as long, by what factor does the self-weight term \gamma L change? (Doubles; and so does the length where it begins to matter, which is why very long cables need tapering.)

Narration spine. A bar is cut at several places; each cut shows its own force and area; the stress is plotted in a staircase; then the wire's weight is added slice by slice and the staircase becomes a line.

Bridge. A load can also change which members carry it as it grows. The last example shows one case.

Nugget 07: A Load That Changes the Path: the Spring in the Tubes

Claim. When a structure's load path changes with the load, the stresses must be found in each regime, and the internal force in a member can be a different function of P in each.

The picture. A spring inside two nested tubes.

Until the cap has moved down s = 3.2\,\mathrm{mm}, the spring carries all of P and tube A carries nothing. After that the spring force stays at ks (the spring is as compressed as it can be, held against the cap), and every extra newton goes into tube A.

Check with numbers. Gere's Example 1-5: spring stiffness k = 4200\,\mathrm{kN/m}, gap s = 3.2\,\mathrm{mm}; outer diameter 75\,\mathrm{mm}; tube A inner diameter 62\,\mathrm{mm}, tube B inner diameter 57\,\mathrm{mm}. Force to close the gap: P = ks = 4200\,\mathrm{kN/m}\times3.2\times10^{-3}\,\mathrm m = 13.44\,\mathrm{kN}. Areas: A_A = \frac{\pi}{4}(75^2 - 62^2) = 1399\,\mathrm{mm^2}, A_B = \frac{\pi}{4}(75^2 - 57^2) = 1866\,\mathrm{mm^2}. (a) P = 11\,\mathrm{kN} (< 13.44): the cap has moved 11/4200 = 2.62\,\mathrm{mm} < s, N_A = 0, \sigma_A = 0, \sigma_B = 11\,000/1866 = 5.89\,\mathrm{MPa}. (b) P = 22\,\mathrm{kN}: N_A = 22 - 13.44 = 8.56\,\mathrm{kN}, \sigma_A = 8560/1399 = 6.12\,\mathrm{MPa}, \sigma_B = 22\,000/1866 = 11.79\,\mathrm{MPa}. (c) For \sigma_A = 5.5\,\mathrm{MPa}: N_A = 5.5 \times 1399 = 7.69\,\mathrm{kN}, P = 7.69 + 13.44 = 21.13\,\mathrm{kN} and \sigma_B = 21\,134/1866 = 11.33\,\mathrm{MPa}.

The idea. Gere's four steps (conceptualise, categorise, analyse, finalise) frame it: a free body for each regime, and a check on the assumption. The "finalise" point is that if tube A were elastic it would be another spring in parallel and N_A \neq P - ks; the true split then depends on stiffnesses (Lesson MM08).

The equation (physical law: equilibrium in each regime, rigid tubes):

Worked number. \sigma_B is 5.89, 11.79\,\mathrm{MPa} for P = 11, 22\,\mathrm{kN}; tube A is unstressed until P = 13.44\,\mathrm{kN}.

Real example (one you can see). A car's bump stop: the spring takes the first part of the bump, then the stop takes the rest. The spring rate changes at the moment of contact.

Plausible wrong answer. "Tube A always carries a share." Not until the gap is closed.

Check. At exactly P = 13.44\,\mathrm{kN}, what is \sigma_A? (Zero: the tube is about to touch.)

Narration spine. The load slowly rises; the cap descends; at the gap's closing the second tube lights up and a kink appears in the plot of N_A against P.

Synthesis

What no single nugget shows. Stress and strain are the two ways of removing size from a question: stress removes the size of the cross-section, strain removes the length. Between them sits the material, which Lessons MM03 and MM04 will connect through the stress-strain curve and Hooke's law. The formulas are averages that become exact under stated conditions: centric load, section far from the ends, constant section, homogeneous material.

• Central idea. Stress is force per area and strain is stretch per length; both remove the size of the member from the question of how hard the material is working.
• Key relationship. \sigma = N/A and \varepsilon = \delta/L, with N = \int \sigma\,dA the statement that stress adds up to the internal force.
• Main assumption. Uniform stress on the section: prismatic, homogeneous, load through the centroid, away from the ends.
• What to explore next. How stress and strain are related for a real material (Lessons MM03 and MM04).

Computed views

Python (numpy). MM02_stress_strain.py computes the stresses and strains above, the hanging wire with and without its weight, Liu's body-force bar and the notched-section centroid.

rod 10 mm, 2470.9 N: 31.46 MPa
rod 20 mm, 1500 N: 4.77 MPa
boom rod 20 mm, 50 kN: 159.2 MPa
stepped bar 16 mm round, 15 kN: 74.60 MPa
stepped bar 10 mm round, 15 kN: 190.99 MPa
stepped bar 24 mm square, 15 kN: 26.04 MPa
strain 1.4 mm in 2.0 m: 700 microstrain
strain 0.26 mm in 0.52 m: 500 microstrain
hanging wire, own weight included: 18.60 MPa
hanging wire (Gere data): 16.08 MPa
body-force bar: sigma(0) = 693.0 kPa, mid = 346.5 kPa
notched section: A = 50000 mm2, centroid (130, 90) mm
true strain for 1.5 mm in 100 mm: 0.014889

The first line is the rod of Lesson MM01 (10\,\mathrm{mm}, 2470.9\,\mathrm N at the top): the same number reappears here as a stress.

MATLAB (Octave-compatible). MM02_stress_strain.m tabulates the stress in a 12\,\mathrm{kN} bar for five diameters, then the strains and the stepped bar.

diameter (mm)   area (mm^2)   stress (MPa)
       8          50.27         238.73
      10          78.54         152.79
      12         113.10         106.10
      16         201.06          59.68
      20         314.16          38.20
strain (microstrain): 700 500 15000
last one in percent: 1.5
true strain of the last: 0.014889
stepped bar stresses (MPa): 74.60 26.04 190.99
hanging wire max stress (MPa): 18.60

SOLIDWORKS Simulation exercise (our own; uniform tension of a bar). A bar in uniform tension should show the stress P/A away from the ends.

• Geometry. Solid square bar 20\,\mathrm{mm}\times20\,\mathrm{mm}, 300\,\mathrm{mm} long, steel.
• Restraint. Fixed geometry on one end face.
• Load. 30\,\mathrm{kN} tensile force normal to the other end face.
• Mesh. Standard solid mesh, element size 4\,\mathrm{mm}; refine once and watch the value settle.
• Hand number the result must match. \sigma_x = 30\,000/400 = 75\,\mathrm{MPa} at mid-length.
• What to look for. Plot the x-normal stress (SX) and probe the middle cross-section: it should read 75\,\mathrm{MPa} within about 1\,\%, and nearly uniformly. Probe near the fixed face and you will see the stress depart from 75\,\mathrm{MPa} (the restraint prevents the sideways contraction that the next lesson introduces), and it settles within about one bar width (20\,\mathrm{mm}) from the end — Saint-Venant's principle made visible.

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

Traps

• Treating N/A as the stress at every point. It is the average, equal to the local value only when the stress is uniform. (Nugget 02.)
• Applying N/A right at a pin or a load. The stress peaks there; use sections at least one width from the ends. (Nugget 03.)
• Taking the load line off the centroid for granted. An offset adds bending to the squeeze. (Nugget 03.)
• Mixing areas and units. N in newtons over A in mm² gives MPa; over m² gives Pa. A factor of 10^6 hides there. (Nugget 01.)
• Dividing by the wrong area. For a tube use \frac{\pi}{4}(d_o^2 - d_i^2), not the outer circle. (Nugget 07.)
• Forgetting the member's own weight. Over long members the self-weight term \gamma L is part of the maximum stress. (Nugget 06.)
• Comparing stretch instead of strain. Compare \delta/L, not \delta. (Nugget 05.)
• Reading the largest load as the largest stress. Stress is N/A; a smaller force on a much smaller area can govern. (Nugget 06.)
• Wrong sign on compression. Compression gives negative stress and negative strain; a sign error hides a failure by buckling or crushing. (Nugget 04.)

Self-check

1. A hollow aluminium tube has an outer diameter of 60\,\mathrm{mm} and a wall of 5\,\mathrm{mm}. It carries a compression of 25\,\mathrm{kN}. Find the stress.
2. The stepped bar of Nugget 06 is changed to carry 20\,\mathrm{kN} instead of 15\,\mathrm{kN}, and its allowable stress is 200\,\mathrm{MPa}. Does the 10\,\mathrm{mm} section still pass, and which section governs?
3. A cable 0.52\,\mathrm m long stretches 0.26\,\mathrm{mm} under load. Find the strain in microstrain and percent.
4. A steel wire carries 300\,\mathrm N at 5\,\mathrm m and 600\,\mathrm N at 9\,\mathrm m below the support; diameter 8\,\mathrm{mm}, \gamma = 77\,\mathrm{kN/m^3}. Find \sigma_{\max} and the share due to the wire's weight.
5. For the notched section of Nugget 03, with a load of 100\,\mathrm{kN}, find the stress and the coordinates of the load line that keeps it uniform.
6. A 30\,\mathrm{mm}\times20\,\mathrm{mm} bar is compressed by 60\,\mathrm{kN}. Draw the stress element, with values.
7. A hanging bar under its own weight has \gamma = 77\,\mathrm{kN/m^3} and length 9\,\mathrm m. Using Liu's equation, find \sigma at the top and at the middle.
8. Convert: 18\,\mathrm{kN} on 120\,\mathrm{mm^2} to MPa; and 36\,\mathrm{ksi} to MPa.
9. For a spring (k = 3000\,\mathrm{kN/m}, gap s = 4\,\mathrm{mm}) inside rigid tubes A (1200\,\mathrm{mm^2}) and B (1500\,\mathrm{mm^2}) as in Nugget 07, find the closing force and the stresses at P = 9\,\mathrm{kN} and P = 24\,\mathrm{kN}.
10. A 100\,\mathrm{mm} gauge length stretches 1.5\,\mathrm{mm}. Find the engineering and true strain.
11. A rod BC carries 50\,\mathrm{kN} and the allowable stress is 165\,\mathrm{MPa}. What is the smallest diameter, and does a 20\,\mathrm{mm} rod pass?

Fully worked solutions

1. Inner diameter 60 - 2\times5 = 50\,\mathrm{mm}. A = \frac{\pi}{4}(60^2 - 50^2) = \frac{\pi}{4}(1100) = 863.94\,\mathrm{mm^2}. \sigma = -25\,000/863.94 = -28.9\,\mathrm{MPa} (compression).

2. \sigma = 20\,000/78.54 = 254.6\,\mathrm{MPa} > 200\,\mathrm{MPa}: the 10\,\mathrm{mm} section fails and governs (the others: 99.5 and 34.7\,\mathrm{MPa}).

3. \varepsilon = 0.26/520 = 5.0\times10^{-4} = 500\,\mu\varepsilon = 0.05\,\%.

4. A = \frac{\pi}{4}8^2 = 50.27\,\mathrm{mm^2}. Loads: 900/50.27 = 17.91\,\mathrm{MPa}. Weight: 77\,000 \times 9 = 693\,000\,\mathrm{Pa} = 0.693\,\mathrm{MPa}. \sigma_{\max} = 18.60\,\mathrm{MPa}; the weight is 0.693/18.60 = 3.7\,\% of it.

5. \sigma = 100\,000/50\,000 = 2.0\,\mathrm{MPa} (compressive if it is a pier). Load line through the centroid, (\bar x, \bar y) = (130, 90)\,\mathrm{mm} from the lower left corner.

6. \sigma_x = -60\,000/600 = -100\,\mathrm{MPa} on the two faces normal to the axis, arrows pointing into the element; the other faces stress-free.

7. \sigma(x) = b(L - x): \sigma(0) = 77\,000\times9 = 693\,\mathrm{kPa}; at x = 4.5\,\mathrm m, 77\,000 \times 4.5 = 346.5\,\mathrm{kPa}.

8. 18\,000/120 = 150\,\mathrm{N/mm^2} = 150\,\mathrm{MPa}. 36 \times 6.895 = 248.2\,\mathrm{MPa}.

9. Closing force ks = 3000\,\mathrm{kN/m}\times0.004\,\mathrm m = 12\,\mathrm{kN}. At 9\,\mathrm{kN} (<12): N_A = 0, \sigma_B = 9000/1500 = 6.0\,\mathrm{MPa}. At 24\,\mathrm{kN}: N_A = 24 - 12 = 12\,\mathrm{kN}, \sigma_A = 12\,000/1200 = 10.0\,\mathrm{MPa}, \sigma_B = 24\,000/1500 = 16.0\,\mathrm{MPa}.

10. Engineering strain 1.5/100 = 0.0150; true strain \ln(1.015) = 0.01489.

11. A \ge 50\,000/165 = 303.0\,\mathrm{mm^2}, d \ge \sqrt{4\times303.03/\pi} = 19.6\,\mathrm{mm}. A 20\,\mathrm{mm} rod: \sigma = 50\,000/314.16 = 159.2\,\mathrm{MPa} < 165\,\mathrm{MPa}, so it passes.