Study Companion: S16 — Frames and Machines

The trusses of S13–S15 had one convenient property: every member was a two-force member, so every member force lay along the member and one number described it. Most real structures do not have that luxury. A crane's boom is loaded along its length, a pair of pliers squeezes at the jaws, pivots at the rivet and is gripped at the handles, an excavator's arm is pushed by cylinders partway along it. Members like these carry three or more forces, or forces and couples, and their pin forces point in directions that must be found, not assumed. Structures containing them are frames (built to carry loads) and machines (built to turn an input force into an output force). This lesson takes them apart, one member at a time, with Newton's third law holding every connection together.

Sources: Meriam, Kraige and Bolton, Engineering Mechanics: Statics, article 4/6 ("Frames and Machines"), with Sample Problems 4/6–4/9 and the Chapter Review (4/7, part b). Harper, Solving Statics Problems in MATLAB, "Frames and Machines" (Problems 4/94 and 4/143). Meriam's Sample Problem 4/7 is in US units (lb, inches), converted here: 1 lbf = 4.448\,\mathrm N, 1 in = 25.4\,\mathrm{mm}. Axes: x to the right, y up; moments counter-clockwise 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. Traps and a self-check with full solutions close the lesson.

The puzzle

Meriam's Sample Problem 4/9: an excavator digs, its bucket teeth pulling with 20\,\mathrm{kN} parallel to the ground. Its boom is raised and lowered about the pin O on the machine by two hydraulic cylinders AC, side by side, acting at A; the arm carrying the bucket is swung about the pin B by one cylinder DE.

How hard must each boom cylinder pull, and what oil pressure does that take in a cylinder of 95\,\mathrm{mm} effective diameter? The plausible guess is "about 20\,\mathrm{kN} in total, like the digging force, shared by the two cylinders". The answer is in Nugget 06 and the Synthesis.

Nugget 01: A Multiforce Member Carries Forces That Do Not Follow It

Claim. A member with three or more forces on it, or with forces and a couple, is a multiforce member. The forces at its pins are not along the member, so each pin force has two unknown components. A structure with at least one such member is a frame or a machine, and the truss methods do not apply to it.

The picture. Meriam's Sample Problem 4/6: a frame of three members carrying a 400\,\mathrm{kg} load through a pulley, and the free-body diagram of its member BF.

BF is loaded at three places: the pin B, the strut CE at E, and the pulley's shaft at F. Not one of these forces points along BF. If it were treated as a truss member, its one "member force" would have no meaning.

Check with numbers. The weight is W = 400 \times 9.81 = 3924\,\mathrm N = 3.92\,\mathrm{kN}. The pulley, a body of its own, carries the cable's 3.92\,\mathrm{kN} pull to the left at its top and the 3.92\,\mathrm{kN} weight at its right edge, so its shaft pushes on BF with 3.92\,\mathrm{kN} to the left and 3.92\,\mathrm{kN} down. On BF (B at the origin, E 3\,\mathrm m along, F 5\,\mathrm m):

• \Sigma M_B = 0: 3.92(5) - \frac12E_x(3) = 0, so E_x = 13.08\,\mathrm{kN} and E_y = \frac12E_x = 6.54\,\mathrm{kN} (the strut CE rises 1.5\,\mathrm m in 3\,\mathrm m);
• \Sigma F_y = 0: B_y + 3.92 - 6.54 = 0, B_y = 2.62\,\mathrm{kN} (down on BF);
• \Sigma F_x = 0: B_x + 3.92 - 13.08 = 0, B_x = 9.15\,\mathrm{kN} (to the left on BF).

The pin force at B is \sqrt{9.15^2 + 2.62^2} = 9.52\,\mathrm{kN} at 16^\circ below the horizontal: nothing about the member's direction predicted that.

The idea. Meriam (4/6): a structure is a frame or machine if at least one of its members is a multiforce member, one with three or more forces, or two or more forces and one or more couples. Frames are designed to support loads and are usually fixed in position; machines contain moving parts and are designed to transmit input forces or couples to output forces or couples (some structures are both). Because the forces in multiforce members are in general not along the members, the methods of Articles 4/3–4/5 (joints and sections for two-force members) cannot be used. Instead each member is isolated with its own free-body diagram and the equilibrium equations are applied to it, with the principle of action and reaction observed at every connection. Only statically determinate frames and machines are treated.

The equation (physical law: S11's three equations, applied to each member separately):

Assumptions: rigid members, frictionless pins, weights neglected unless stated.

Worked number. The count for this frame: members AD and BF give 3 + 3 = 6 equations (the pulley's own three were used to find its shaft force); the unknowns are A_x, A_y, D, B_x, B_y and the strut force: six. Solving all six together reproduces every value above: D = 4.32\,\mathrm{kN}, A_x = 4.32\,\mathrm{kN}, A_y = 3.92\,\mathrm{kN}, strut CE = \sqrt{13.08^2 + 6.54^2} = 14.62\,\mathrm{kN} in compression.

Real example. A wall-mounted folding shelf bracket: the shelf arm carries the load along its length and is held by a hinge and a strut, exactly like BF. Its hinge screws are pulled out by a force at an angle, not along the arm.

Plausible wrong answer. "The pin at B pushes along BF." It pushes at 16^\circ to BF, because BF carries three forces, not two.

Check. Is the pulley in SP 4/6 a two-force member? (No: cable, weight and shaft, three forces.)

Narration spine. The frame with its hanging load. Member BF lifts out of the picture; three arrows appear at B, E and F, each pointing its own way, and a dashed line along BF shows that none follows it.

Bridge. Where to start? For trusses it was the reactions. For frames it depends on whether the frame holds its shape on its own.

Nugget 02: Rigid by Itself? Reactions First. Collapsible? Take It Apart First

Claim. If a frame stays rigid when lifted off its supports, treat it as one body and find all the external reactions first. If it would fold without its supports, the whole-frame equations give only some reactions; the rest come from the members after the frame is taken apart.

The picture. Meriam's Fig. 4/14.

Lifted off its supports, the A-frame (a) keeps its shape: it is one rigid body with three unknown reactions and three equations. Frame (b) contains a four-bar loop BDEF that would deform: its supports are part of what holds it in shape, so it has four reaction components (two at each pin) and the whole frame gives only three equations.

Check with numbers. SP 4/6 is type (a): from the whole frame, moments about A: 5.5(3.92) - 5D = 0, D = 4.32\,\mathrm{kN}; then A_x = 4.32\,\mathrm{kN} (to the left) and A_y = 3.92\,\mathrm{kN}, all before touching a member. SP 4/7 is type (b): a frame with supports at A and C, 0.762\,\mathrm m (30 in) apart, loaded by 133.4\,\mathrm N (30 lb) horizontally at the top B, 1.016\,\mathrm m (40 in) up, and 222.4\,\mathrm N (50 lb) on member ED. Moments about C find only the vertical reactions: A_y = 266.9\,\mathrm N (60 lb) and C_y = 444.8\,\mathrm N (100 lb). The horizontal A_x and C_x wait until the members are solved: A_x = 66.7\,\mathrm N (15 lb) from member AB, and C_x = 333.6\,\mathrm N (75 lb) from member BC.

The idea. Meriam (4/6): if the frame or machine is a rigid unit by itself when removed from its supports (Fig. 4/14a), begin by finding all forces external to it as a single rigid body, then dismember it and consider each part. If it is not rigid by itself (Fig. 4/14b), the external reactions cannot be completed until the structure is dismembered and the parts analysed. The equations of the separate parts are linked through the forces of interaction. In SP 4/7, Meriam notes that BDEF is a movable quadrilateral, not a rigid triangle.

The equation (mathematical identity: counting the whole frame):

Worked number. SP 4/7's two vertical reactions from the whole frame, in SI: \Sigma M_C = 0: 222.4(0.3048) + 133.4(1.016) - 0.762A_y = 0, so A_y = (67.8 + 135.5)/0.762 = 266.9\,\mathrm N; \Sigma F_y = 0: C_y = 222.4(4/5) + 266.9 = 444.8\,\mathrm N.

Real example (one you can see). A stepladder is a type (b) frame: its two legs are hinged at the top and held by a spreader. Lift it with the spreader unlatched and it folds; the floor's friction and the spreader are part of what keeps it a structure.

Plausible wrong answer. "Three equations for the whole frame give all four reactions if you choose the moment point well." No choice of point gives four independent equations for one body; the fourth comes from a member.

Check. A frame of two members pinned together at the top and each pinned to the ground at its foot: rigid off its supports? (No: two bars and a pin swing freely; with the ground it is a triangle, so it is type (b), four reaction components.)

Narration spine. Two frames on their supports. A hand lifts each off: the A-frame rises intact; the second folds as it lifts, its quadrilateral glowing. Back on the supports, the A-frame's three reactions appear at once; the second shows only two vertical ones, the horizontal ones waiting with question marks.

Bridge. Taking a frame apart creates pairs of forces at every pin. They must be drawn consistently.

Nugget 03: Action and Reaction Must Match on Every Diagram

Claim. At a pin joining two members, whatever force the pin exerts on one member it exerts, reversed, on the other. Draw each component in opposite senses on the two diagrams. If a component comes out negative, it is reversed on both.

The picture. Meriam's Figs. 4/15 and 4/16.

In a plane, a pin's two components A_x, A_y act one way on the first member and the opposite way on the second. A ball-and-socket joint in a space frame has three components, each reversed between the two members. In vector notation the force is +\mathbf A on one member and -\mathbf A on the other.

Check with numbers. In SP 4/7 Meriam draws B_y downward on AB and upward on BC, a guess. Member AB gives B_y = 50(4/5) - 60 = -20 lb: the guess was wrong, and B_y is 20\,\mathrm{lb} = 89.0\,\mathrm N upward on AB and downward on BC. Carried into BC with its sign, -20, it closes BC's two check equations: \Sigma F_y: 100 + (-20) - 100(4/5) = 0 and \Sigma M_C: (30 - 15)(40) + (-20)(30) = 0 (in lb and inches). In SP 4/6 the same care at B checks member AD, all of whose forces are then known: A_x = 4.32 left, D = 4.32 right, B_x = 9.15 right (the reverse of its sense on BF), the strut pushing C with 13.08 left, and the cable pulling with 3.92 right: -4.32 + 4.32 + 9.15 - 13.08 + 3.92 = 0.

(Meriam prints this check as 4.32 - 13.08 + 9.15 + 3.92 + 4.32 = 0, which sums to 8.63: the first term, A_x acting to the left, must be -4.32.)

The idea. Meriam (4/6): forces are best represented by rectangular components, which makes moment arms easy; vector notation helps in three dimensions when moments are taken about axes not parallel to the coordinate axes. It is not always possible to assign every component's true sense at once; an arbitrary assignment is fine, but it is absolutely necessary that a force be represented consistently on the diagrams of the interacting bodies. A negative result means the component acts opposite to the arrow: reverse it on both members (and the sign of its terms), or leave the arrows and let the minus sign carry the meaning. In vector notation, use a plus sign for the action and a minus sign for the reaction. SP 4/7 adds a subtlety: a load applied at a pin shared by two members can be assigned to either member, which changes that member's pin reaction. Meriam puts the 30 lb force on BC; putting it on BA would change B_x.

The equation (physical law: Newton's third law at a connection):

Worked number. A ball joint between two members of a space frame exerts (2, -1, 3)\,\mathrm{kN} on member 1. On member 2 it exerts (-2, 1, -3)\,\mathrm{kN}: all three components reversed.

Real example. A loaded scissor lift's pins are designed from exactly these paired forces: each pin's force is computed once and applied, reversed, to the two arms it joins.

Plausible wrong answer. "B_y came out negative, so I drew it wrong on AB only." It must be reversed on BC too, or BC will not balance.

Check. If A_x is drawn to the right on member 1, which way is it drawn on member 2? (To the left.)

Narration spine. Two members pull apart at a pin; the pin's arrows split into a pair pointing opposite ways. One pair turns red with a minus sign and both arrows flip together. Then a ball joint with three arrows, each paired, and finally +\mathbf A and -\mathbf A.

Bridge. Some members of a frame carry only two forces. Spotting them first saves the most work.

Nugget 04: Find the Two-Force Members Before Anything Else

Claim. A member loaded only at two points, with no couple, is a two-force member, however it is shaped: its force lies along the straight line joining those two points. Recognising it turns two unknown components into one known direction and one unknown magnitude.

The picture. An L-shaped member, and Harper's frame where one is used.

The bent member BD is pinned at both ends and loaded nowhere else. Its two pin forces must be equal, opposite and collinear, so they lie along the chord BD, a 3–4–5 line (300\,\mathrm{mm} across, 400\,\mathrm{mm} down), not along either leg.

Check with numbers. Harper's Problem 4/143: member AC (1.2\,\mathrm m) is pinned to a wall at A, held at its midpoint B by BD, and loaded at C by 3\,\mathrm{kN} at an angle \theta from the vertical. Because BD is a two-force member, AC has only three unknowns: A_x, A_y and the magnitude B along (-3, 4)/5.

• \Sigma M_A = 0: \frac45B(0.6) - 3\cos\theta(1.2) = 0, so B = 7.5\cos\theta\ \mathrm{kN};
• \Sigma F_x = 0: A_x - \frac35B + 3\sin\theta = 0, so A_x = 0.6B - 3\sin\theta;
• \Sigma F_y = 0: A_y + \frac45B - 3\cos\theta = 0, so A_y = 3\cos\theta - 0.8B.

At \theta = 0 (load straight down): B = 7.5\,\mathrm{kN}, A_x = 4.5, A_y = -3\,\mathrm{kN}, A = 5.41\,\mathrm{kN}.

The idea. Meriam (SP 4/6, hint 2): without noticing that CE is a two-force member, the solution would be much longer, because BF's three equations would contain four unknowns, B_x, B_y, E_x, E_y. "The direction of the line joining the two points of force application, and not the shape of the member, determines the direction of the forces acting on a two-force member." In SP 4/7 the member EF is the two-force member, fixing the direction of the forces at E and F. Harper (P 4/143) likewise notes first that BD is a two-force member, so the force \mathbf B points from B to D. In Harper's hand pump (P 4/94, Nugget 05) the link AB plays the same role.

The equation (geometric relationship: S11's two-force member):

Worked number. SP 4/6's strut CE runs from C to E, 3\,\mathrm m across and 1.5\,\mathrm m up, so E_y = \frac{1.5}{3}E_x = \frac12E_x before any equation is written. With E_x = 13.08\,\mathrm{kN}: E_y = 6.54\,\mathrm{kN} and CE = 14.62\,\mathrm{kN}.

Real example (one you can see). The gas strut of a car's tailgate is a two-force member pinned at both ends: whatever its housing looks like, it pushes along the line between its two ball sockets.

Plausible wrong answer. "A bent member pushes along its legs." BD's force is along the chord at 53.1^\circ to the horizontal, not along the 300\,\mathrm{mm} or 400\,\mathrm{mm} leg.

Check. A curved bracket is pinned at both ends and also carries a lamp in its middle. Is it a two-force member? (No: three forces.)

Narration spine. An L-shaped bar between two pins. Its legs fade, a dashed chord appears, and the pin forces slide onto the chord. Then Harper's frame: the bent member's force arrow drops onto member AC already aligned, and AC's unknowns count down from four to three.

Bridge. In a machine the point is not the reactions but the output: how much force comes out for the force put in.

Nugget 05: A Machine Turns an Input Force Into an Output Force

Claim. In a machine, the equilibrium of its members relates an output force to an input force through the machine's geometry. Moments about the right pin give the ratio; symmetry and two-force members keep the equations few.

The picture. Harper's Problem 4/94: a hand pump. The handle is pinned at C to the piston rod and at A to the link AB, which is pinned to the base. A hand pushes the grip, 200\,\mathrm{mm} beyond C, with P = 120\,\mathrm N perpendicular to the handle; A is 80\,\mathrm{mm} on the other side of C; the link is 200\,\mathrm{mm} long and its foot B is 40\,\mathrm{mm} from the piston's axis.

The link AB is a two-force member: its force F_A acts along it, at an angle \beta to the vertical. Moments about C leave F_A alone; the vertical force sum then gives the piston force C_y, and the pressure is C_y divided by the piston's area.

Check with numbers (handle at \theta below the horizontal; Harper's formulation). The link's angle follows from the geometry: \sin\beta = (80\cos\theta - 40)/200.

• \Sigma M_C = 0: F_A\cos(\theta + \beta)(80) - P(200) = 0, so F_A = 5P/[2\cos(\theta + \beta)];
• \Sigma F_y = 0: C_y - P\cos\theta - F_A\cos\beta = 0, so C_y = P\cos\theta + F_A\cos\beta;
• p = C_y/A_p, with A_p = \pi(0.046)^2/4 = 1.662\times10^{-3}\,\mathrm{m^2}.

At \theta = 0: \beta = 11.54^\circ, F_A = 306.2\,\mathrm N, C_y = 120 + 300 = 420.0\,\mathrm N, p = 252.7\,\mathrm{kPa}. At 15^\circ: 266.6\,\mathrm{kPa}; at 30^\circ: C_y = 482.7\,\mathrm N, p = 290.5\,\mathrm{kPa}. The 120\,\mathrm N push becomes 420\,\mathrm N to 483\,\mathrm N on the piston, a gain of 3.5 to 4.

The idea. Meriam's machine example, Sample Problem 4/8, is an overload release: a soft shear pin S joins two hinged halves until the pull T exceeds a set value, when the pin breaks and rollers release the eye bolt. By symmetry, the forces at S and at the hinge A have no x components (an action on one half along +x would need its reaction on the other half along -x, breaking the mirror image). The links BD and CD are two-force members with equal forces B = C; equilibrium of the connection D gives 2B\cos\theta = T, with \tan\theta = 5/12. Moments about A on the upper half, with S = 800\,\mathrm N (lengths in mm):

Then \Sigma F_y: 800 - \frac{1477}{2(12/13)}\cdot\frac{5}{13} - A = 0, A = 492\,\mathrm N. Meriam's hint warns not to forget the moment of B's y component. Harper's pump keeps the handle angle as a variable and lets MATLAB compute \beta, F_A, C_y and p over 0 \le \theta \le 30^\circ.

The equation (physical law: equilibrium of the machine's members, giving the output in terms of the input):

Worked number. A pair of pliers: the grip 150\,\mathrm N acts 100\,\mathrm{mm} from the rivet, the jaws 25\,\mathrm{mm} on the other side. Moments about the rivet for one handle: 150(100) = J(25), so the jaw force is J = 600\,\mathrm N, four times the grip. The rivet carries 150 + 600 = 750\,\mathrm N, more than either.

Real example. Bolt cutters, nutcrackers and car jacks are machines analysed exactly this way; the rivet or pivot always carries the sum of the input and output forces, which is why pivots are the first parts to wear.

Plausible wrong answer. "The pump's piston gets the hand's 120\,\mathrm N times the lever ratio 200/80 = 2.5, that is 300\,\mathrm N." That is only the link's share (F_A\cos\beta = 300\,\mathrm N at \theta = 0); the handle also pushes down at C with the hand's own P\cos\theta, giving 420\,\mathrm N.

Check. In SP 4/8, why can the forces at S and A have no x components? (Symmetry: the two halves are mirror images about the x-axis, so an x action on one would need a mirror x reaction on the other, in the same direction, which Newton's third law forbids.)

Narration spine. The hand pump: the handle tilts through 30^\circ while the grip force stays at 120\,\mathrm N. The link tilts too, its force arrow growing; the piston arrow grows from 420 to 483\,\mathrm N, and a pressure gauge climbs from 253 to 290\,\mathrm{kPa} while the graph traces itself.

Bridge. For each of these the free body was obvious. In a large machine, the choice of free body is the whole problem.

Nugget 06: Cut Where the Force You Want Becomes External

Claim. To find the force in one connecting part, such as a hydraulic cylinder, choose a free body that the part's force acts on from outside, and take moments about the pin that removes the other unknowns. Then force equals pressure times area turns the force into the pressure the hydraulics must supply.

The picture. The puzzle's arm assembly for the boom cylinders, and the arm alone for the bucket cylinder.

For the boom cylinders the free body is the whole arm assembly (boom, arm, bucket and the cylinder DE inside it), pinned at O: the cylinders AC act on it from outside, and moments about O remove O_x and O_y. For the cylinder DE the free body is the arm EBIF with the bucket, cut at B and through DE: moments about B remove B_x and B_y.

Check with numbers (Meriam's solution).

• (a) \alpha = \tan^{-1}[(0.4 + 0.28 + 0.55)/1.4] = 41.3^\circ. \Sigma M_O = 0: -20\,000(3.95) - 2F_{AC}\cos41.3^\circ(0.68) + 2F_{AC}\sin41.3^\circ(2) = 0, so F_{AC} = 48\,800\,\mathrm N = 48.8\,\mathrm{kN} in each cylinder. Pressure: p_{AC} = 48\,800/(\pi(0.095)^2/4) = 6.89\times10^6\,\mathrm{Pa} = 6.89\,\mathrm{MPa}.
• (b) \beta = \tan^{-1}[(0.1 + 0.4)/(0.4 + 2.1)] = 11.31^\circ. \Sigma M_B = 0: -20\,000(3.5) + F_{DE}\cos11.31^\circ(0.73) + F_{DE}\sin11.31^\circ(0.4) = 0, so F_{DE} = 88\,100\,\mathrm N = 88.1\,\mathrm{kN}, and p_{DE} = 88\,100/(\pi(0.105)^2/4) = 10.18\,\mathrm{MPa}.

The idea. Meriam (SP 4/9): begin with a free-body diagram of the entire arm assembly, including only the dimensions that this part needs; the details of the other cylinders do not matter yet. For the cylinder DE, "cut" the assembly where the wanted cylinder's force becomes external to the free body: isolate the arm with the bucket. The general advice of 4/6: most simultaneous equations can be avoided by a careful choice of the member or group of members for the free body, and of moment axes that eliminate unwanted terms. Recall that force equals pressure times area.

The equation (physical law for each free body, with the definition of pressure):

Worked number. The answer to the puzzle: each boom cylinder pulls with 48.8\,\mathrm{kN}, 97.6\,\mathrm{kN} for the pair, nearly five times the 20\,\mathrm{kN} digging force, at an oil pressure of 6.89\,\mathrm{MPa}. The cylinder sits close to the pin O, with a short moment arm, while the digging force acts 3.95\,\mathrm m below it. The bucket cylinder needs 88.1\,\mathrm{kN} and 10.18\,\mathrm{MPa}.

Real example (one you can see). Watch any excavator dig: the boom and arm cylinders are thick and short, mounted close to the pivots. That geometry trades force for speed of swing, which is why the oil pressures run to tens of megapascals.

Plausible wrong answer. "Each cylinder pulls about 10\,\mathrm{kN}, half the digging force." Moments, not forces, must balance about O, and the cylinder's arm is short: 48.8\,\mathrm{kN} each.

Check. Why does the cylinder DE not appear in part (a)'s equation? (It is inside the free body, joining two of its parts: its forces cancel in pairs.)

Narration spine. The excavator digs. A dashed outline wraps the whole arm and the pin O lights; the two boom cylinders' arrows grow until they balance the 20\,\mathrm{kN} teeth force. The outline shrinks to the arm alone, the pin B lights, and cylinder DE's arrow grows. Two pressure gauges read 6.89 and 10.18\,\mathrm{MPa}.

Bridge. A machine's loads change direction as it works. Its pins must be sized for the worst.

Nugget 07: Pins Must Carry the Worst Direction of the Load

Claim. When a load can act in any direction over a range, each pin force is a function of that direction. Its maximum need not occur at an obvious angle; it is found from the derivative, or read from a plot, and each pin is sized for its own worst case.

The picture. Harper's frame of Nugget 04, the 3\,\mathrm{kN} load swung from \theta = -90^\circ to +90^\circ.

B follows 7.5\cos\theta: largest, 7.5\,\mathrm{kN}, with the load straight down. A peaks elsewhere: 6\,\mathrm{kN} at \theta = -26.6^\circ, with the load swung back toward the wall.

Check with numbers. Substituting B = 7.5\cos\theta:

With double angles, A^2/9 = 2.125 - 1.5\sin2\theta + 1.125\cos2\theta, so dA/d\theta = 0 when -3\cos2\theta - 2.25\sin2\theta = 0, that is \tan2\theta = -4/3. Its roots are Harper's four: -\tan^{-1}(1/2), \tan^{-1}2, and these shifted by \pi. At \theta = -\tan^{-1}(1/2) = -26.6^\circ: \sin\theta = -1/\sqrt5, \cos\theta = 2/\sqrt5, so A = 3\sqrt{1 + 3(2/5) + \frac94(4/5)} = 3\sqrt{4} = 6\,\mathrm{kN}, the maximum. At \theta = \tan^{-1}2 = 63.4^\circ: A = 3\sqrt{1 - 3(2/5) + \frac94(1/5)} = 3\sqrt{0.25} = 1.5\,\mathrm{kN}, the minimum.

The idea. Harper (4.5) plots A and B over the range and notes the difference: B's maximum is obvious from \cos\theta (7.5\,\mathrm{kN} at 0^\circ), while the formula for A shows no obvious maximum. It differentiates A symbolically, solves dA/d\theta = 0, and uses the graph to pick the root in range that is the maximum: A_{\max} = 6\,\mathrm{kN} at \theta = -26.6^\circ.

The equation (mathematical identity: the extremum condition on the pin force):

Worked number. Sizing: pin B for 7.5\,\mathrm{kN}, pin A for 6\,\mathrm{kN}. Had pin A been sized for the "obvious" vertical load it would see only 5.41\,\mathrm{kN} and be 10\,\% undersized.

Real example. A crane's slewing jib or a robot arm swings its load through a range of angles. Each pivot is checked over the whole sweep, as here, because different pins peak at different angles.

Plausible wrong answer. "Every pin is worst with the load straight down." Pin A is worst at -26.6^\circ (6\,\mathrm{kN} against 5.41\,\mathrm{kN} at 0^\circ).

Check. At which angle does pin B carry nothing, and why? (\theta = \pm90^\circ: the load is horizontal, along AC, and has no moment about A, so B is not needed.)

Narration spine. The 3\,\mathrm{kN} arrow at C swings like a pendulum from horizontal to horizontal. Two gauges at A and B swing with it; beside them the two curves draw themselves, and markers drop on 7.5\,\mathrm{kN} at 0^\circ and on 6\,\mathrm{kN} at -26.6^\circ.

Bridge. Every structure in this part has been taken apart into members, each in equilibrium.

Synthesis

The excavator's boom cylinders each pull 48.8\,\mathrm{kN}, not about 10\,\mathrm{kN}: the digging force acts almost 4\,\mathrm m below the boom's pin, the cylinders less than 1\,\mathrm m from it, and moments must balance. That took one free body, the whole arm assembly, and one moment equation about O. The bucket cylinder took a second free body, the arm alone, cut where its force became external.

Part 4 has now taken apart every kind of statically determinate structure. Trusses (S13–S15) are made of two-force members, so each joint is a particle and each member one number. Frames and machines contain multiforce members, so each member is a rigid body of its own, and each pin force has two components (three in space). The method is always the same: isolate, draw every force with action and reaction matched, and choose the free body and moment centre that leave the fewest unknowns. A frame or machine is solved by taking it apart: spot the two-force members, keep every action–reaction pair consistent, and cut where the force you want becomes external.

• Central idea: multiforce members carry pin forces in unknown directions; each member gets its own free-body diagram and three equations.
• Key relationship: \mathbf F_{2,1} = -\mathbf F_{1,2} at every connection; force along the line of the pins for a two-force member; p = F/A for a hydraulic cylinder.
• Most important assumption: a statically determinate structure, frictionless pins, member weights neglected.
• Open question: every load so far was concentrated at a point. What about a load spread along a beam, or the weight of a body spread through its volume? (Part 5, distributed forces.)

Computed views

The same calculations, written twice so you can work in whichever language you prefer. The MATLAB follows Harper's programs; the Python is a translation, and both print the same numbers.

Python (numpy, scipy). S16_frames_and_machines.py solves member BF of Sample Problem 4/6 as a 3\times3 system, with the strut's direction built in. It sweeps Harper's frame over the load angle, finding the maximum of A on the grid and by a bounded search, and tabulates Harper's hand pump. It also checks the overload release and the pliers.

Part 1: Sample Problem 4/6, member BF (kN, m; B at the origin)
  E = (13.08, 6.54), strut CE 14.62 kN
  B = (-9.16, -2.62), |B| = 9.52 kN at 16 deg
Part 2: Harper's frame P 4/143, load angle -90 to 90 deg
  B max 7.50 kN at -0.0 deg
  A max 6.00 kN at -26.6 deg, min 1.50 kN at 63.4 deg
  A at 0 deg 5.41 kN
  bounded search: A max 6.000 kN at -26.57 deg
Part 3: Harper's hand pump P 4/94, P = 120 N
  theta  0: beta 11.54 deg, F_A 306.2 N, C_y 420.0 N, p 252.7 kPa
  theta 15: beta 10.74 deg, F_A 333.0 N, C_y 443.1 N, p 266.6 kPa
  theta 30: beta  8.42 deg, F_A 382.9 N, C_y 482.7 N, p 290.5 kPa
Part 4: overload release and pliers
  SP 4/8: T = 1477 N, A = 492 N
  pliers: jaw 600 N, rivet 750 N

The minus signs on B mean it acts left and down on BF, as Nugget 01 found. Unrounded, B_x = 13.08 - 3.924 = 9.156\,\mathrm{kN}; Meriam and Nugget 01 round it to 9.15. The bounded search confirms that A's largest value is -\tan^{-1}(1/2) = -26.57^\circ, as Nugget 07 found.

MATLAB (Octave-compatible). S16_frames_and_machines.m does the same with the backslash operator, max and min.

SP 4/6: E_x = 13.08, strut 14.62 kN, B = (-9.16, -2.62), |B| = 9.52 kN
B max 7.50 at 0.0 deg; A max 6.00 at -26.6, min 1.50 at 63.4 deg
theta  0: beta 11.54 deg, F_A 306.2 N, C_y 420.0 N, p 252.7 kPa
theta 15: beta 10.74 deg, F_A 333.0 N, C_y 443.1 N, p 266.6 kPa
theta 30: beta  8.42 deg, F_A 382.9 N, C_y 482.7 N, p 290.5 kPa

Traps

• Treating a multiforce member as a truss member. Its pin forces are not along it; give each pin two components. (From Nugget 01.)
• Solving the whole collapsible frame for all reactions. Only three of its four come from the whole; the rest need a member. (From Nugget 02.)
• Drawing a pin force the same way on both members. It is reversed on the second. (From Nugget 03.)
• Reversing a negative result on one member only. Reverse it on both, or keep the sign on both. (From Nugget 03.)
• Forgetting where a load at a shared pin was assigned. It belongs to one member's diagram, and that member's pin reaction includes it. (From Nugget 03.)
• Following a bent member's legs. A two-force member pushes along the line of its pins. (From Nugget 04.)
• Forgetting the moment of a force's second component. In SP 4/8 both components of B have arms about A. (From Nugget 05.)
• Including an internal cylinder in the free body's equations. It is internal; cut through it to make it external. (From Nugget 06.)
• Sizing a pin for the vertical load only. Sweep the load's direction. (From Nugget 07.)

Self-check

1. Meriam's frame (SP 4/6) carries 600\,\mathrm{kg} instead of 400\,\mathrm{kg}. Find D, A_x, A_y, E_x, E_y, B_x and B_y.
2. For each structure, say whether it is rigid off its supports and name its two-force members: (a) SP 4/6; (b) SP 4/7; (c) Harper's frame of Nugget 04; (d) the hand pump.
3. In SP 4/7 in SI, member ED carries 222.4\,\mathrm N at one end, the two-force member EF at the other end E, and the pin D midway, 0.3048\,\mathrm m from each end, all three forces parallel. Find the forces at E and D. Then find B_x on member AB from moments about A, with EF's force applied at F, 0.508\,\mathrm m above A, B at 1.016\,\mathrm m, and EF's horizontal component \frac35 of its force.
4. The overload release (SP 4/8) gets a stronger shear pin that breaks at 1000\,\mathrm N. Find the maximum pull T and the hinge force A.
5. A ball joint exerts (2, -1, 3)\,\mathrm{kN} on member 1 of a space frame. What does it exert on member 2?
6. The excavator (SP 4/9) digs with 30\,\mathrm{kN} instead. Find the force and pressure in the boom cylinders and in the bucket cylinder.
7. The hand pump at \theta = 15^\circ: find \beta, the link force F_A, the piston force C_y and the pressure.
8. Harper's frame (Nugget 04) with the load at \theta = 90^\circ (horizontal, away from the wall): find B and A.
9. Harper's symbolic solution gave four roots for dA/d\theta = 0. What is A at \theta = \tan^{-1}2, and what does that angle mean?
10. Pliers grip with 150\,\mathrm N at 100\,\mathrm{mm} from the rivet; the jaws are 25\,\mathrm{mm} from it. Find the jaw force and the rivet force.
11. In SP 4/6, check member AD in the y direction, and say why it needs no new unknowns.

Fully worked solutions

1. Every force scales with the load, by 600/400 = 1.5. W = 600 \times 9.81 = 5.886\,\mathrm{kN}; D = 5.5(5.886)/5 = 6.47\,\mathrm{kN} (wall pushes right); A_x = 6.47\,\mathrm{kN} (left); A_y = 5.89\,\mathrm{kN}. On BF: \Sigma M_B: 5.886(5) = \frac12E_x(3), E_x = 19.62\,\mathrm{kN}, E_y = 9.81\,\mathrm{kN}; B_y = 9.81 - 5.886 = 3.92\,\mathrm{kN} down; B_x = 19.62 - 5.886 = 13.73\,\mathrm{kN} left.

2. (a) Rigid off its supports (type a); CE is a two-force member (the pulley is not). (b) Not rigid: BDEF is a movable quadrilateral (type b); EF is the two-force member. (c) Member AC with BD pinned to the wall: BD is the two-force member; off its supports, AC and BD would swing about B, so type (b) as a whole, though here BD's known direction leaves AC with three unknowns. (d) The pump's link AB is a two-force member; the handle is a multiforce member; it is a machine.

3. Member ED: all three forces parallel, D midway. \Sigma M_D: 222.4(0.3048) = E(0.3048), E = 222.4\,\mathrm N (50 lb); \Sigma F: D = 222.4 + 222.4 = 444.8\,\mathrm N (100 lb). Member AB: EF's force, 222.4\,\mathrm N, acts at F with horizontal component \frac35(222.4) = 133.4\,\mathrm N at 0.508\,\mathrm m; \Sigma M_A: 133.4(0.508) - B_x(1.016) = 0, B_x = 66.7\,\mathrm N (15 lb).

4. The equations are linear in S and T, so T = 1477 \times 1000/800 = 1846\,\mathrm N. Then B = T/(2 \times 12/13) = 1000\,\mathrm N and A = 1000 - 1000(5/13) = 615\,\mathrm N.

5. (-2, 1, -3)\,\mathrm{kN}: each component reversed.

6. All forces scale by 1.5: F_{AC} = 73.2\,\mathrm{kN} each, p_{AC} = 10.33\,\mathrm{MPa}; F_{DE} = 132.2\,\mathrm{kN}, p_{DE} = 15.27\,\mathrm{MPa}.

7. \sin\beta = (80\cos15^\circ - 40)/200 = (77.27 - 40)/200 = 0.1864, \beta = 10.74^\circ. F_A = 5(120)/[2\cos25.74^\circ] = 600/1.802 = 333.0\,\mathrm N. C_y = 120\cos15^\circ + 333.0\cos10.74^\circ = 115.9 + 327.2 = 443.1\,\mathrm N. p = 443.1/(1.662\times10^{-3}) = 266.6\,\mathrm{kPa}.

8. \cos90^\circ = 0, \sin90^\circ = 1: B = 0; A_x = -3\,\mathrm{kN}, A_y = 0; A = 3\,\mathrm{kN}. The horizontal load along AC goes straight into the pin A.

9. \sin\theta = 2/\sqrt5, \cos\theta = 1/\sqrt5: A = 3\sqrt{1 - 3(2/5) + \frac94(1/5)} = 3\sqrt{0.25} = 1.5\,\mathrm{kN}, the minimum of A over the range, at 63.4^\circ. The other two roots, shifted by \pi, lie outside -90^\circ to 90^\circ.

10. One handle: \Sigma M_{\text{rivet}}: 150(100) = J(25), J = 600\,\mathrm N. The grip and jaw forces are parallel and on opposite sides of the rivet, so the rivet carries 150 + 600 = 750\,\mathrm N.

11. On AD: A_y = 3.92 up; B_y reversed from BF, 2.62 up; the strut pushes C away from E, so its y component is 6.54 down; D and the cable are horizontal. \Sigma F_y: 3.92 + 2.62 - 6.54 = 0. Every force on AD was already known from the whole frame and from BF, so AD's three equations are pure checks.