Study Companion: MM01 — Internal Forces and the Method of Sections

A crane hook is rated for a load, a bolt for a shear, a rod for a pull. None of those ratings can be checked until we know how hard the material inside the member is being pushed, pulled, sheared or bent. The loads we are given act on the outside. This lesson is the one tool that reaches inside: cut the member in imagination, draw what is left, and let equilibrium say what the removed part must have been doing to the part we kept. The answer is a short list of internal forces, N, V, M and T, and the rest of Mechanics of Materials is about what those forces do to the material.

Sources: Gere and Goodno, Mechanics of Materials, 9th edition (SI), sections 1.1–1.3 (the syllabus for this lesson); Beer, Johnston, DeWolf and Mazurek, Mechanics of Materials, 8th edition, section 1.1; Venkataraman, Essential Mechanics, opening of chapter 3 and section 3.1; Muvdi and McNabb, Mechanics of Materials, sections 1.1–1.2. Venkataraman 3.1 does not cut members; it is used only in the last nugget, as the bridge from an internal force to stress. All working is in SI; where a reference uses kips and feet the numbers here are our own SI examples (1 kip = 4.448 kN, 1 ft = 0.3048 m).

Sign conventions used throughout (Gere's, drawn on the figures): tension is positive. An internal force is drawn on the cut face in its assumed positive sense and a negative answer means it acts the other way. On the left piece of a cut beam, positive shear V acts downward and positive moment M counter-clockwise on the cut face; on the right piece, both are reversed. For reactions and applied loads the ordinary statics convention is used: forces positive along +x and +y, moments counter-clockwise positive. Section 1.2 of Gere's problem-solving approach is four steps (conceptualise, categorise, analyse, finalise); every worked example below follows them, and final answers keep three significant digits (four when the leading digit is 1).

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, traps and self-check close the lesson. Two computed views follow the synthesis: a Python script and an Octave-compatible MATLAB script that reproduce the numbers, and a SOLIDWORKS Simulation exercise with a hand number to match. They check the derivations; they never replace them.

The puzzle

A 2\,\mathrm{kN} drill head hangs on the end of an 8\,\mathrm m steel rod, and the rod hangs from a gantry. The rod has a mass of 6\,\mathrm{kg} per metre. Where does the rod carry the most force, and how much is it? The plausible answer is "2\,\mathrm{kN}, the load on the end" — it is the only load anyone mentioned. The answer is in Nugget 05.

Nugget 01: A Cut Shows the Force the Material Was Already Carrying

Claim. A member in equilibrium is also in equilibrium piece by piece. Cut it by an imaginary plane normal to its axis and the material that was removed must have been exerting forces on the face that remains. Those forces reduce to a few internal forces (also called stress resultants): an axial force N, a shear force V, a bending moment M and, for a shaft, a torsional moment T.

The picture. A straight bar pulled at its ends, with the cut drawn.

Look at the left piece alone. Its own load P pulls it left; it is at rest, so something pulls it right with an equal force. That something is the right-hand material, acting across the cut. The right piece sees the mirror image. The two cut-face forces are a Newton's-third-law pair, so they are equal and opposite, and each one has the size needed to balance the load on its own piece.

Check with numbers. With P = 12\,\mathrm{kN}: the left piece needs N - 12\,\mathrm{kN} = 0, so N = 12\,\mathrm{kN}. The same arithmetic holds wherever the bar is cut, because nothing else acts on it.

The idea. Gere (1.1) frames the subject: Mechanics of Materials asks what the loads do inside a body, so stresses, strains and displacements, using theory and experiment together. Statics comes first because it supplies the loads and reactions; the deformation analysis later supplies extra equations when statics alone is not enough. At a cut in a general planar member the internal forces are a force along the axis (N), a force across it (V) and a couple (M); in a shaft the twisting couple is T. In three dimensions each cut has up to six components, three forces and three moments.

The equation (physical law: static equilibrium of one piece of a body):

Worked number. A cable is pulled by P = 40\,\mathrm{kN} at each end. Cut it anywhere: N = 40\,\mathrm{kN} in tension. The cable has not "lost" any pull by being cut; the cut face carries it.

Real example (one you can see). Pull a rubber band between your fingers. Every point along it is doing the same job: each short length is held in tension by the neighbours on either side. Cut it with scissors and the ends snap back, which is the internal force suddenly having nothing to push against.

Plausible wrong answer. "Internal forces appear only where something is attached." The cut can go anywhere and the internal force exists everywhere; the cut merely reveals it.

Check. A tow rope pulls a car with 6\,\mathrm{kN}. What is the internal force in the rope at its midpoint? (6\,\mathrm{kN} tension, the same as at either end, if the rope's own weight is ignored.)

Narration spine. A bar is pulled; a plane slices it and the halves part. Arrows appear on the two cut faces, equal and opposite, labelled N.

Bridge. The cut needs a free body to be useful, and a free body needs the outside forces first. Next: get the loads and reactions from statics.

Nugget 02: Statics First: the Free Body, the Supports and the Count

Claim. Before a member can be cut, the forces acting on its outside must be known. They come from a free-body diagram and the equilibrium equations, and a body is solvable by statics alone only if the number of unknown reactions equals the number of independent equations.

The picture. A beam on a pin and a roller with one load, drawn as a free body.

Draw the beam alone, remove the supports, and put in what they did: the pin can push in any direction in the plane, so it supplies A_x and A_y and no moment; the roller supplies one force perpendicular to the surface it rolls on. Loads come as concentrated forces (N) and couples (N·m), line loads (N/m, constant or varying), surface pressures (Pa) and body forces such as weight (N/m³). A distributed load, such as a weight along a beam, may be replaced by its resultant at its centroid when finding reactions. (When the cut passes through the loaded span the part of the load on the kept piece stays in the free body; see Nugget 07.)

Check with numbers. Beam length 4\,\mathrm m, load 12\,\mathrm{kN} at 1.5\,\mathrm m from A. Moments about A: B_y(4\,\mathrm m) - 12\,\mathrm{kN}(1.5\,\mathrm m) = 0, so B_y = 4.5\,\mathrm{kN}. Vertical forces: A_y = 12 - 4.5 = 7.5\,\mathrm{kN}. There is no horizontal load, so A_x = 0.

The idea. Gere 1.3 lists the equations and the support catalogue. For a planar body, three scalar equations; in space, six.

The equation (physical law: equilibrium of a rigid body):

Moments may be taken about any point; choosing the point where two unknowns cross removes both from the equation. A body is statically determinate when the unknowns equal the independent equations and nothing can move as a rigid body; extra supports make it indeterminate and statics alone cannot finish it (the deformation analysis supplies the missing equations, from Part 2 onwards).

The supports used most often, in the plane:

In space a pin becomes a ball-and-socket (3 forces), a fixed end carries 3 forces and 3 moments, and a journal or thrust bearing resists all but rotation about its own axis. A truss with m members, r reaction components and j joints is determinate when m + r = 2j in the plane.

Worked number. Our beam has 3 unknowns (A_x, A_y, B_y) and 3 equations, so it is determinate. Replace the roller with a second pin: 4 unknowns, 3 equations, one degree of indeterminacy. The triangular truss of Nugget 04 has m = 5, r = 3, j = 4: 5 + 3 = 8 = 2 \times 4, determinate.

Real example (one you can see). A kitchen shelf on two wall brackets is a beam on two supports; a shelf screwed to three brackets is indeterminate, which is why it can rock the shelf's weight unevenly onto the middle one if the wall is not flat.

Plausible wrong answer. "The pin at A can carry a moment because it is bolted." An ideal pin turns freely, so it supplies no moment. A bolt through a tight hole in a thick plate is nearer to a fixed end, and the idealisation must match what the joint can really do.

Check. How many reaction components has a cantilever's fixed end in the plane? (Three: two forces and a moment.)

Narration spine. The beam is isolated from the wall and floor; the supports are replaced by arrows; moments about A solve B_y first, then A_y.

Bridge. Some members need no free-body arithmetic at all. If a member is pinned at its ends and loaded only there, the force through it is already known to lie along its axis.

Nugget 03: A Two-Force Member Carries Its Pin Force Straight Through

Claim. A member that is loaded only at its two ends by pins has just one force direction available: the line joining the pins. The cut force anywhere along it is therefore equal to that pin force, and it is tension or compression depending on whether the member is being pulled apart or squeezed.

The picture. A boom AB and a rod BC share a pin at B; a 30\,\mathrm{kN} load hangs from B.

The pin B is in equilibrium under three forces: the load, the boom force along the horizontal line AB and the rod force along the sloping line CB. Three forces in equilibrium close into a triangle, here a similar copy of the 3–4–5 triangle of the geometry.

Check with numbers. The rod runs 0.8\,\mathrm m across and 0.6\,\mathrm m up, length 1.0\,\mathrm m. At pin B: vertical, 0.6\,F_{BC} = 30\,\mathrm{kN}, so F_{BC} = 50\,\mathrm{kN}; horizontal, F_{AB} = 0.8\,F_{BC} = 40\,\mathrm{kN}. The rod pulls the pin up and toward C, so it is in tension; the boom pushes the pin away from A, so it is in compression. Cut the rod anywhere and its piece in the cut carries 50\,\mathrm{kN} tension; cut the boom and it carries 40\,\mathrm{kN} compression.

The idea. Beer (1.1) solves the same problem the long way too: take the whole frame, three equations leave A_y + C_y = 30\,\mathrm{kN}, which cannot separate the two. Dismember the frame: the boom alone, with moments about B, forces A_y = 0, so C_y = 30\,\mathrm{kN} and the pin force at C is \sqrt{40^2 + 30^2} = 50\,\mathrm{kN} along the rod. Same answer, one step longer. The force triangle is the shortcut when each member is a two-force member.

The equation (physical law: equilibrium of a particle, the pin):

Worked number. The internal force in BC is 50\,\mathrm{kN} (T); in AB it is 40\,\mathrm{kN} (C). Stress, which depends on the area of each member, comes in Lesson MM02.

Real example (one you can see). A camping-tent guy rope and the pole it holds up: the rope is a two-force member (pulled at each end), and you can feel in the rope exactly the force the pole end puts on it.

Plausible wrong answer. "The boom carries 30\,\mathrm{kN} because that is the load." The load is vertical; the boom is horizontal. A member can only carry force along its own line, and the rod supplies the vertical push, which is why the rod carries more than the load.

Check. The same boom 0.8\,\mathrm m long with a rod rising 0.3\,\mathrm m to C (length 0.854\,\mathrm m) carries 30\,\mathrm{kN}. Is the rod force larger or smaller than before? (Larger: the flatter the rod, the more it must pull to hold the same vertical load.)

Narration spine. The frame is pulled apart at the pin; arrows form a closed triangle, 30 up from the load, 40 sideways, 50 along the rod; each is copied onto its member as tension or compression.

Bridge. When many two-force members are joined, the same idea is used joint by joint, or the cut is made through three members at once. That is a truss.

Nugget 04: A Truss Member's Force: Cut Through the Member You Want

Claim. In a truss every member is a two-force member, so every member force can be found by isolating pins (method of joints) or by cutting through at most three members and applying equilibrium to one side (method of sections). The cut is the better choice when only a few members are wanted.

The picture. A triangular truss A–C–B with the midpoint D on the base and a vertical member CD, loaded at C.

The dashed cut crosses AC, CD and DB, leaving B and C on one side. Three cut members give three unknown forces; the kept piece has three equations. Moments are taken about the point where two unknown lines meet, so each equation holds one unknown.

Check with numbers. Joints are 4\,\mathrm m apart on the base and C is 3\,\mathrm m above D, so AC = BC = 5\,\mathrm m (3–4–5). The load is 20\,\mathrm{kN}, so each support gives 10\,\mathrm{kN} by symmetry. Method of joints. At A: the vertical component of AC must cancel 10\,\mathrm{kN}, so F_{AC}(3/5) = 10 and F_{AC} = 16.667\,\mathrm{kN} in compression (it pushes up on the pin). The horizontal part, 16.667 \times 4/5 = 13.333\,\mathrm{kN}, is balanced by AD in tension. At D, AD and DB are collinear and equal, leaving nothing for CD to do: F_{CD} = 0, a zero-force member. At C: 2 \times 16.667 \times 3/5 = 20\,\mathrm{kN} balances the load, which closes the check.

Method of sections. On the kept piece, with each cut member assumed in tension and drawn pulling away from the piece: moments about C give 10\,\mathrm{kN}(4\,\mathrm m) - 3\,\mathrm m\,F_{DB} = 0, so F_{DB} = 13.333\,\mathrm{kN} (T), with CD and AC both passing through C. Moments about D (through which CD and DB pass) give F_{AC} = -16.667\,\mathrm{kN}, a minus sign meaning compression. Vertical forces on the piece then give F_{CD} = 0. The two methods agree.

The idea. Gere 1.3 (Example 1-1) uses both: joints to find every member, sections to isolate selected ones. Choose the cut that crosses the member you want and as few others as possible.

The equation (physical law: equilibrium of the joint, or of the piece cut from the truss):

Worked number. F_{AC} = -16.7\,\mathrm{kN} (C), F_{AD} = F_{DB} = 13.3\,\mathrm{kN} (T), F_{CD} = 0.

Real example (one you can see). Look up at a roof truss in a car park or a railway shed: diagonal members are in compression, the bottom chord is a tie in tension, and short verticals under a symmetric load may carry nothing, as CD does here.

Plausible wrong answer. "A zero-force member can be removed." It carries nothing under this load, but it holds joint D in place and carries load if the loading changes; also it can shorten the unsupported length of a compression chord.

Check. The truss has load 30\,\mathrm{kN} at C instead. What is F_{AC}? (15\,\mathrm{kN} \times 5/3 = 25\,\mathrm{kN}, compression.)

Narration spine. A dashed line crosses three members; the left half fades; three arrows are drawn on the cut ends; a moment equation about one joint drops two unknowns and solves the third.

Bridge. The bar and the truss member carry a single axial force. When the load is spread along the bar, that force varies from section to section and a single cut is not enough: the cut position becomes a variable.

Nugget 05: Axial Force Along a Bar Under Point and Distributed Loads

Claim. Under several axial loads the internal force is constant between loads and jumps at each point load by that load. Under a distributed axial load p(x) it changes continuously, so the internal force is found by integrating the load.

The picture. A bar with four axial loads, the cuts, and the force diagram beneath.

Cut between A and B and keep the left piece. Only the 30\,\mathrm{kN} load at A acts on it, pushing into the cut, so the cut face must push back: the bar is in compression there. Move the cut to the right past B: the 50\,\mathrm{kN} load now also acts on the left piece, and the picture flips to tension. Each point load moves the diagram by its own size.

Check with numbers. Loads (positive to the right): +30, -50, +40, -20\,\mathrm{kN}, which sum to 0, so the bar is in equilibrium. Left piece, tension positive, the cut face force points away from the cut (+x on the left piece), so N + \sum(\text{loads on the piece}) = 0: AB: N = -30. BC: N + 30 - 50 = 0, N = +20. CD: N + 30 - 50 + 40 = 0, N = -20 (all in \mathrm{kN}). From the right piece at CD: the only load is 20\,\mathrm{kN} to the left, and the cut face force points to -x, so -N - 20 = 0, N = -20\,\mathrm{kN}. The same value from both sides.

The idea. Muvdi (1.2) handles the distributed case with an element of length dx cut from the bar: the force on its left face is N, on its right face N + dN, and the load on it is p\,dx. Equilibrium of the element gives the rate of change; integration gives the total.

The equation (physical law: equilibrium of a bar element; the sign is for a load p that acts toward +x on a bar whose x increases to the right and whose cut force is tension positive):

A uniform load p over length \ell is the resultant p\ell, applied at the centre, and the internal force varies linearly across it. A load that points the other way flips the sign, so derive the sign for the direction of the load in each problem rather than copy it.

Worked number. The puzzle's rod: 8\,\mathrm m long, 6\,\mathrm{kg/m}, so the weight per metre is w = 6 \times 9.81 = 58.86\,\mathrm{N/m} and the whole rod weighs 8 \times 58.86 = 470.9\,\mathrm N. Take y downward from the support and keep the piece below the cut: that piece carries the tip load and the weight of the length below, so

At the top, y = 0: N = 2000 + 470.9 = 2470.9\,\mathrm N. At y = 3\,\mathrm m: 2000 + 58.86 \times 5 = 2294.3\,\mathrm N. At the bottom, y = 8\,\mathrm m: 2000\,\mathrm N, the tip load. So the rod carries the most force at the top, 2.47\,\mathrm{kN}, not 2\,\mathrm{kN}: the puzzle's rod must also hold up its own weight. This variation is why a very long hanging cable is thicker at the top.

Step the source skips: the integral. The weight below the cut is \int_y^8 w\,dy' = 58.86\,(8 - y), linear in y, which is the straight line of the diagram.

Real example (one you can see). Climb a rope hand over hand: your hands feel less pull at the bottom of the rope than at the top, because the top has to hold the rope's own weight as well.

Plausible wrong answer. "The internal force is the same everywhere along the bar." It is constant only between loads. At every point load it jumps, and under a distributed load it varies continuously.

Check. A 2\,\mathrm m bar is held at the wall end (x = 2\,\mathrm m) and loaded toward the wall with p = 300x^2\,\mathrm{N/m} (x from the free end). What is the internal force at the wall? (-\int_0^2 300x^2\,dx = -800\,\mathrm N, compression.)

Narration spine. Cuts slide left to right across the bar; at each point load the plotted value jumps. Then a stack of tiny loads fills a hanging rod and the diagram becomes a slope.

Bridge. We have been drawing "away from the cut" arrows and reading signs from the answers. That habit needs one rule behind it, so that left piece and right piece always agree.

Nugget 06: One Sign Convention for the Cut, Kept Apart From the One for Reactions

Claim. Reactions and applied loads use the ordinary statics signs (along +x, +y, counter-clockwise). Internal forces use a separate, physical sign: tension positive, and each internal force drawn on the cut in its positive sense. A negative answer then means the force acts the opposite way, and the left and right pieces give the same number for the same cut.

The picture. Positive internal forces on the two faces of one cut.

Check with numbers. Nugget 05's bar, cut between B and C, left piece: assume N positive and point it +x: N + 30 - 50 = 0 \Rightarrow N = +20\,\mathrm{kN}, so it really is tension. For the AB cut the same recipe gave -30\,\mathrm{kN}, so there the force points the other way: compression. Writing a compression answer as "30\,\mathrm{kN} (C)" is Muvdi's habit and is equivalent.

The idea. Two sign systems are in play and they should not be mixed. Gere calls them the statics convention (for reactions) and the deformation convention (for internal forces); Muvdi calls the internal one the physical convention and the sign each force has in an equilibrium equation the equations convention. Same idea, two vocabularies. Beer's section 1.1 draws every force with an explicit sense and reads the result off, with no named convention. This lesson uses Gere's names. Muvdi's advice is the working rule: always assume the unknown internal force positive, solve, and read the sign afterwards.

The equation (definition: the positive senses):

Worked number. Muvdi-style check, in SI: loads 25\,\mathrm{kN}\rightarrow at A, 40\,\mathrm{kN}\leftarrow at B, 20\,\mathrm{kN}\rightarrow at C, 5\,\mathrm{kN}\leftarrow at D. Left pieces give -25, +15, -5\,\mathrm{kN} for AB, BC, CD; the right piece at CD gives -N - 5 = 0, N = -5.

Real example (one you can see). Squeeze a sponge between your palms and then stretch a long strip of it. In the first the piece "pushes into" the cut, in the second it "pulls away"; the convention just records which.

Plausible wrong answer. "A minus sign means I made an arithmetic mistake." It means the assumed arrow was drawn against the real direction. The magnitude is right.

Check. If the left piece gives V = -5\,\mathrm{kN}, what does the right piece give? (V = -5\,\mathrm{kN} also, because V is defined positive on each face in the sense that makes the two faces agree.)

Narration spine. A cut opens: two arrows on the bar pull away from each other (tension); on a beam the arrows on the two faces form a down-and-up pair, and curved arrows form a couple.

Bridge. The bar had only N. A beam, shaft or frame can have all four internal forces at once.

Nugget 07: Cutting Beams, Shafts and Frames: N, V, M and T

Claim. The same recipe, cut then equilibrium of one piece, gives every internal force. The only new thing is to keep the load that lies on the kept piece in the free body, and to take moments about the cut.

The picture. A simply supported beam with a uniform load and a point load, cut at x = 4\,\mathrm m.

Check with numbers. Reactions. The total load is 2 \times 6 + 9 = 21\,\mathrm{kN}. Moments about A: B_y(6) = 9(2) + 12(3) = 54, so B_y = 9\,\mathrm{kN} and A_y = 21 - 9 = 12\,\mathrm{kN}. Cut at x = 4\,\mathrm m, left piece. The load on the piece is 9\,\mathrm{kN} (point) and 2 \times 4 = 8\,\mathrm{kN} (uniform, resultant at 2\,\mathrm m). Vertical forces, with V positive downward: 12 - 9 - 8 - V = 0, so V = -5\,\mathrm{kN}. Moments about the cut, M counter-clockwise positive: -12(4) + 9(2) + 8(2) + M = 0, so M = 48 - 18 - 16 = 14\,\mathrm{kN\,m}. No horizontal load, so N = 0. Right piece (length 2\,\mathrm m, upward 9\,\mathrm{kN} at B, 2 \times 2 = 4\,\mathrm{kN} down at 1\,\mathrm m): V + 9 - 4 = 0 gives V = -5\,\mathrm{kN} and -M + 9(2) - 4(1) = 0 gives M = 14\,\mathrm{kN\,m}. Both pieces agree.

The idea. Gere (Examples 1-2 to 1-4) cuts a beam, a stepped shaft and a plane frame. Three points matter. For reactions a distributed load may be replaced by its resultant, but for the cut the part of the load on the kept piece stays on it. For a shaft the loads are torques, and T is positive when its vector points away from the cut face; the internal torque is constant in a segment with no applied torque. For a frame the cut member has its own axes, along and across it, so N and V are measured along the member's axis, not along x and y.

The equation (physical law: equilibrium of one piece, with the sign rule of Nugget 06):

each taken over the loads and reactions on the kept piece together with the internal force or couple drawn on its cut face in the positive sense of Nugget 06. The sign of the answer then says whether that assumed sense was right.

Worked number, shaft. Shaft fixed at A; torques 1200\,\mathrm{N\,m} at B (vector toward -x), 700\,\mathrm{N\,m} at C and 300\,\mathrm{N\,m} at D (toward +x). Equilibrium of the whole shaft: M_{Ax} - 1200 + 700 + 300 = 0, so M_{Ax} = 200\,\mathrm{N\,m}. On the left piece the cut-face torque is positive along +x, so AB: 200 + T = 0, T = -200; BC: 200 - 1200 + T = 0, T = 1000; CD: 200 - 1200 + 700 + T = 0, T = 300 (all in \mathrm{N\,m}). From the right piece of BC: the loads 700 + 300 give T = 1000 as well.

Worked number, bracket. An L-shaped bracket has a column fixed to a wall base and a horizontal arm whose end load of 500\,\mathrm N acts downward, 0.2\,\mathrm m from the column axis. Cut the column anywhere. The piece above the cut carries the load. Along the column axis the load pushes down: N = -500\,\mathrm N (compression). Across it the load is parallel to the column, so V = 0. About the cut, the load has a moment arm of 0.2\,\mathrm m: M = 500 \times 0.2 = 100\,\mathrm{N\,m}, the same at every height. The bracket's column is squeezed and bent at the same time; the vertical load on an offset arm is not just a push.

Real example (one you can see). Hang a heavy bag on the end of a coat hook screwed to a wall: the hook's stem is bent as well as pulled, because the bag sits out from the wall. That is the bracket above.

Plausible wrong answer. "Replace the whole uniform load by its resultant and then cut." For the reactions that is right. For a cut at 4\,\mathrm m the resultant of the whole load, 12\,\mathrm{kN} at 3\,\mathrm m, is not on the kept piece; only the 8\,\mathrm{kN} at 2\,\mathrm m is.

Check. A cantilever carries a load at its tip. Where along it are V and M largest in magnitude? (V is the same everywhere for a tip load; M is largest at the wall, where the load has the longest arm.)

Narration spine. The beam is cut; the right half disappears; the reaction and loads are drawn on the left half; arrows V and M appear on the cut face and two equations solve them; the right half is restored and gives the same.

Bridge. All of this tells how hard the member is loaded in total. The next question is how hard each piece of the material is loaded, and that needs the cut face's area.

Nugget 08: The Internal Force Is the Total of a Stress Spread Over the Cut

Claim. The internal force N is not what the material feels. It is the sum of tiny forces acting on tiny areas of the cut face; their intensity, force per area, is the stress, and N is its total, \int \sigma\,dA = N. When the stress is uniform it is \sigma = N/A.

The picture. The cut face of a rod, with the force drawn as a uniform spread.

Check with numbers. The puzzle's rod at the top carries N = 2470.9\,\mathrm N. If its diameter is 10\,\mathrm{mm}, the area is A = \pi (5\,\mathrm{mm})^2 = 78.54\,\mathrm{mm^2}, so \sigma = 2470.9\,\mathrm N / 78.54\,\mathrm{mm^2} = 31.46\,\mathrm{N/mm^2} = 31.5\,\mathrm{MPa}. Another rod, 20\,\mathrm{mm} in diameter and pulled by 1500\,\mathrm N (Venkataraman's Example 3.1): A = 314.16\,\mathrm{mm^2}, \sigma = 4.77\,\mathrm{MPa}. (The book prints 4.77 \times 10^{5} Pa; the correct power is 10^{6}.)

The idea. Venkataraman 3.1 puts the sum directly: the load P is the total of the stress over the cross-section. Near the point where a load is applied the spread cannot be found by simple means, so an end plate or collar is used to spread it. One bar-width from the end the spread is nearly uniform (Saint-Venant's principle, taken up fully in Lesson MM12), and design commonly takes it as uniform. Venkataraman also contrasts engineering stress and strain, which use the original area and length, with true values: true stress uses the current area, and true strain is \ln(L/L_0); design uses the engineering values. For L/L_0 = 1.5, engineering strain is 0.5 and true strain is \ln 1.5 = 0.405.

The equation (definition of stress, and the geometric relationship for uniform spread):

Worked number. Size a rod for N = 12\,\mathrm{kN} at a stress limit of 80\,\mathrm{MPa}: A \ge 12\,000\,\mathrm N / 80\,\mathrm{N/mm^2} = 150\,\mathrm{mm^2}, so d \ge \sqrt{4 \times 150/\pi} = 13.8\,\mathrm{mm}.

Real example (one you can see). Two ropes can hold the same load, but the thin one frays first. The internal force is the same in both; the area it is spread over is not.

Plausible wrong answer. "Two rods with the same internal force are equally likely to fail." Not so: \sigma depends on A. A 12\,\mathrm{kN} pull gives 59.7\,\mathrm{MPa} on a 16\,\mathrm{mm} rod and 119\,\mathrm{MPa} on an 11.3\,\mathrm{mm} rod, so the thin one is closer to failing.

Check. The rod's diameter is doubled; the internal force is unchanged. By what factor does the stress fall? (By 4, since area scales with diameter squared.)

Narration spine. The cut face is shown end-on; a patch is highlighted, then every patch; equal arrows fill the face and merge into the single force N, then N/A is written.

Synthesis

What no single nugget shows. The method of sections is the same move every time: isolate a piece, put on it everything that acts on its outside, and let equilibrium name the force or moment that the removed part supplied. Reactions come first (Nugget 02), two-force members and trusses use it with the sign rule (Nuggets 03, 04, 06), distributed loads turn it into an integral (Nugget 05), beams, shafts and frames add the shear, moment and torque (Nugget 07), and stress is the distribution of the result across the cut (Nugget 08). Every later lesson takes the internal forces from here and asks what they do to the material.

• Central idea. Cut a member and equilibrium of one piece reveals the internal force there.
• Key relationship. N, V, M, T are found by \sum F = 0 and \sum M = 0 on the kept piece, and they are the totals of a stress spread over the cut: \int \sigma\,dA = N.
• Main assumption. The member is in equilibrium and the loads are known; for a truss, members are pinned two-force members; a distributed load stays on the kept piece. Statics alone is enough only when the structure is determinate.
• What to explore next. The internal force is not enough to say whether a member survives: that needs the stress over the area (Lesson MM02) and the material's strength (Lesson MM03).

Computed views

Python (numpy). The script solves the cut equilibria of the bar, the beam, the hanging rod, the L-bracket, the boom and rod, and the triangular truss, and prints the numbers argued above. It is MM01_sections.py.

bar segment forces AB, BC, CD (kN): [-30.  20. -20.]
beam A_y, B_y (kN): 12.0 9.0
beam cut at 4 m: V (kN), M (kN m): -5.0 14.0
rod force at top (N): 2470.9
rod force at bottom (N): 2000.0
bracket column N, V (N), M (N m): -500 0.0 100.0
boom, rod forces (kN): 40.0 50.0
truss R, AC, AD (kN): 10.0 -16.667 13.333

MATLAB (Octave-compatible). MM01_sections.m does the same with cumsum and the backslash operator for the pin equations; it prints the same eight lines.

bar segment forces AB, BC, CD (kN): -30 20 -20
beam A_y, B_y (kN): 12 9
beam cut at 4 m: V (kN), M (kN m): -5 14
rod force at top (N): 2470.9
rod force at bottom (N): 2000.0
bracket column N, V (N), M (N m): -500 0 100
boom, rod forces (kN): 40.0 50.0
truss R, AC, AD (kN): 10.0 -16.667 13.333

SOLIDWORKS Simulation exercise (our own; Free Body Force at a cut section). Check the bracket of Nugget 07 against the hand resultants.

• Geometry. L-shaped bracket from a solid 20\,\mathrm{mm} \times 20\,\mathrm{mm} square bar: vertical leg 300\,\mathrm{mm} high, horizontal arm with its load line 200\,\mathrm{mm} from the leg's axis. Use a Split feature to cut the leg at mid-height into two bodies and keep the bodies bonded.
• Restraint. Fixed geometry on the base face of the leg.
• Load. 500\,\mathrm N downward on the end face of the arm.
• Mesh. Standard solid mesh, element size 5\,\mathrm{mm} (refine once to see the value settle).
• Hand number the result must match. N = -500\,\mathrm N (compression), V = 0, M = 100\,\mathrm{N\,m}.
• What to look for. In the results, use Free Body Force on the cut face of one body: the force along the leg should read 500\,\mathrm N, the lateral force near zero, and the moment 100\,\mathrm{N\,m}. A lateral force that is not near zero usually means a load or restraint was placed on the wrong face. The stress is highest where the leg joins the arm; the cut-plane resultants do not change, which is the point.

(This exercise was written for the lesson and has not been run in SOLIDWORKS here; the hand numbers are the target.)

Traps

• Reading the load as the internal force. The load at the end is not the force everywhere along the member; weight and other loads add to it. (Rod of the puzzle: 2.47\,\mathrm{kN} at the top, not 2\,\mathrm{kN}. Nugget 05.)
• Giving a member a force it cannot carry. A pinned two-force member carries force along its own line only; the boom is not loaded by the vertical weight. (Nugget 03.)
• Mixing the two sign systems. Reaction signs follow statics; internal forces follow the tension-positive convention. A minus sign in an internal-force answer is information, not an error. (Nugget 06.)
• Replacing the load by its resultant before cutting through it. Reactions can use the whole resultant; a cut through the loaded span keeps only the load on the kept piece. (Nugget 07.)
• Counting a statically indeterminate structure as solvable. Four reaction unknowns on a plane beam cannot be found from three equations. (Nugget 02.)
• Forgetting the moment of an offset load. A load that does not pass through the column axis bends it as well as compressing it. (Nugget 07.)
• Removing a "zero-force" member without a thought. It carries nothing under this load, but its job is to hold the joint under others. (Nugget 04.)
• Comparing members by internal force alone. The same force on a smaller area is a higher stress. (Nugget 08.)

Self-check

1. A bar has loads 25\,\mathrm{kN}\rightarrow at A, 40\,\mathrm{kN}\leftarrow at B, 20\,\mathrm{kN}\rightarrow at C and 5\,\mathrm{kN}\leftarrow at D. Find the internal force in each segment.
2. A bar of length 2\,\mathrm m is fixed at x = 2\,\mathrm m and loaded toward the wall by p = 300x^2\,\mathrm{N/m} (x from the free end). Find the internal force at x = 1\,\mathrm m and at the wall.
3. A boom AB is 1.5\,\mathrm m long and horizontal; a rod BC rises to a point 2.0\,\mathrm m above A. A load of 20\,\mathrm{kN} hangs at B. Find the forces in the boom and the rod and say which is in tension.
4. A cantilever 3\,\mathrm m long, fixed at its left end, carries 4\,\mathrm{kN/m} over its whole length and 6\,\mathrm{kN} at the tip. Find N, V and M in a cut 1\,\mathrm m from the wall.
5. In the triangular truss of Nugget 04 the load at C is 30\,\mathrm{kN}. Find F_{AC}, F_{AD} and F_{CD}.
6. A shaft fixed at A has torques 900\,\mathrm{N\,m} at B (toward -x), 500\,\mathrm{N\,m} at C and 100\,\mathrm{N\,m} at D (toward +x). Find the torque in each segment.
7. A 16\,\mathrm{mm}-diameter rod carries 12\,\mathrm{kN}. Find the stress. What diameter keeps the stress at 80\,\mathrm{MPa}?
8. An L-bracket's arm carries 800\,\mathrm N down at 0.35\,\mathrm m from the column axis. Find N, V and M in the column.
9. A plane truss has 7 members, 3 reaction components and 5 joints; another has 8 members, 3 reactions and 5 joints. Which is determinate?
10. A 20\,\mathrm m rope of mass 0.5\,\mathrm{kg/m} hangs vertically and carries 6\,\mathrm{kN} at its lower end. Find the force at the midpoint and at the top.
11. A reference gives a force as 25\,\mathrm{kips} (compression). Convert it to SI. Then explain the sign.

Fully worked solutions

1. Left pieces, tension positive: AB: N + 25 = 0, N = -25\,\mathrm{kN}. BC: N + 25 - 40 = 0, N = +15\,\mathrm{kN}. CD: N + 25 - 40 + 20 = 0, N = -5\,\mathrm{kN}. Check with the right piece at CD: -N - 5 = 0, N = -5\,\mathrm{kN}. ✓. Segments: -25, +15, -5\,\mathrm{kN}.

2. Free end at x = 0; the piece on the free-end side of a cut at x carries \int_0^x 300x'^2\,dx' = 100x^3\,\mathrm N toward the wall, pushing into the cut, so N(x) = -100x^3. At x = 1\,\mathrm m: -100\,\mathrm N. At the wall, x = 2\,\mathrm m: -100 \times 8 = -800\,\mathrm N (compression, matching -\int_0^2 300x^2\,dx).

3. The rod runs 1.5\,\mathrm m across and 2.0\,\mathrm m up, so its length is \sqrt{1.5^2 + 2.0^2} = 2.5\,\mathrm m with \sin = 0.8, \cos = 0.6. At B vertical: 0.8\,F_{BC} = 20\,\mathrm{kN}, F_{BC} = 25\,\mathrm{kN}. Horizontal: F_{AB} = 0.6 \times 25 = 15\,\mathrm{kN}. The rod pulls B up and toward C: tension. The boom pushes B away from A: compression.

4. Take the right piece (length 2\,\mathrm m). Loads: 4 \times 2 = 8\,\mathrm{kN} at 1\,\mathrm m and 6\,\mathrm{kN} at 2\,\mathrm m from the cut. With V positive upward on the right face: V - 8 - 6 = 0, so V = 14\,\mathrm{kN}. About the cut with moments counter-clockwise positive: loads give -8(1) - 6(2) = -20; a clockwise-positive M on the right face is -M counter-clockwise, so -M - 20 = 0 and M = -20\,\mathrm{kN\,m} (hogging). N = 0.

5. Reactions 15\,\mathrm{kN}. At A: F_{AC}(3/5) = 15, F_{AC} = 25\,\mathrm{kN} compression. F_{AD} = 25 \times 4/5 = 20\,\mathrm{kN} tension. F_{CD} = 0.

6. M_{Ax} - 900 + 500 + 100 = 0, so M_{Ax} = 300\,\mathrm{N\,m}. Left pieces: AB: 300 + T = 0, T = -300. BC: 300 - 900 + T = 0, T = 600. CD: 300 - 900 + 500 + T = 0, T = 100 (all in \mathrm{N\,m}). Check with the right pieces: BC: 500 + 100 = 600 ✓. CD: 100 ✓.

7. A = \pi(8\,\mathrm{mm})^2 = 201.06\,\mathrm{mm^2}; \sigma = 12\,000/201.06 = 59.7\,\mathrm{MPa}. For 80\,\mathrm{MPa}: A = 150\,\mathrm{mm^2}, d = \sqrt{4 \times 150/\pi} = 13.8\,\mathrm{mm}.

8. N = -800\,\mathrm N (compression), V = 0, M = 800 \times 0.35 = 280\,\mathrm{N\,m}.

9. Seven members: 7 + 3 = 10 = 2 \times 5, determinate. Eight members: 8 + 3 = 11 > 10, one degree of indeterminacy.

10. The half below the midpoint weighs 0.5 \times 10 \times 9.81 = 49.05\,\mathrm N, so the force there is 6000 + 49.05 = 6049\,\mathrm N (about 6.05\,\mathrm{kN}). The whole rope weighs 0.5 \times 20 \times 9.81 = 98.1\,\mathrm N, so at the top the force is 6098\,\mathrm N. The weight is small beside 6\,\mathrm{kN} but the trend is Nugget 05's.

11. 25 \times 4.448 = 111.2\,\mathrm{kN}. The "compression" says the internal force pushes into the cut, i.e. N = -111.2\,\mathrm{kN} in the tension-positive convention.