Structural Analysis
Before a truss, frame, or machine can be designed, the force carried by every member must be known. This chapter develops the systematic tools — the method of joints, the method of sections, and the dismembering of multiforce structures — that turn equilibrium into member forces.
Learning objectives
- Determine the forces in truss members using the method of joints and the method of sections.
- Analyze the forces acting on the members of frames and machines built from pin-connected members.
- Classify two-force and multiforce members from their loading and connections.
- Use equilibrium and member geometry to calculate and label tension or compression.
- Verify member-force signs with an independent joint or moment equation.
Simple Trusses
A truss is one of the most economical ways to span a distance: a lattice of slender bars that carries load chiefly through axial tension and compression rather than bending. Roof and bridge structures are the classic examples, and because their loading lies in a single plane they can usually be reduced to a two-dimensional problem.
Truss
A structure built from slender members joined only at their end points. When all members and loads lie in one plane it is a planar truss, and the member forces can be analyzed two-dimensionally. Loads typically reach the truss at its joints — for a roof through purlins, for a bridge through stringers and floor beams.
Simple Truss
A truss formed by starting with a rigid triangle of three pin-connected members and enlarging it by repeatedly adding two members and one new joint. Any truss that can be grown this way from a basic triangle is called simple, and is guaranteed to be rigid.
The Two Design Assumptions
To find the force in each member, two idealizations are made. Together they reduce every member to a two-force member whose force acts along its own axis.
Assumption 1 — Loads act at the joints
External loads and reactions are applied only at the joints. Member self-weight is usually neglected because the axial force it carries dwarfs its own weight; if weight must be included, half is applied as a vertical load at each end of the member.
Assumption 2 — Members are pinned by smooth pins
Joints are treated as frictionless pins. Real connections — bolted or welded gusset plates, or a single through-bolt — act like pins as long as the centerlines of the joining members meet at a common point.
Because each truss member is loaded only at its two pinned ends and carries no weight, it is a two-force member: the forces at its ends are equal in magnitude, opposite in direction, and directed along the member's axis. A force that elongates the member is tensile (T); a force that shortens it is compressive (C). Compression members are often made thicker than tension members to resist buckling.
In practice, always state T or C next to every member-force answer — a magnitude alone is only half the result. Examiners and design codes both treat an unlabeled sign as an incomplete answer.
Before moving on
- Choose a joint, section, or separated member that exposes the desired force.
- Use a force line or moment center to eliminate unknowns deliberately.
- Report each member result with magnitude, units, and tension or compression.
The Method of Joints
If the whole truss is in equilibrium, then so is every individual joint. The method of joints exploits this: draw a free-body diagram of a single pin, and because every force there is concurrent and coplanar, only two equilibrium equations are needed.
Method of Joints
A technique for finding all member forces by isolating each joint as a free body. Since the forces at a pin form a concurrent, coplanar system, equilibrium reduces to two scalar equations per joint, which solve for at most two unknown member forces.
Draw every unknown member force as if it pulls on the pin (tension). After solving, a positive result confirms tension; a negative result means the member is actually in compression. Once a force is known, carry its correct magnitude and sense (T or C) onto the next joint.
Starting Point and Strategy
Begin at a joint with at least one known force and no more than two unknowns. A member pulling on a pin is in tension; a member pushing on a pin is in compression. If the geometry makes the sense obvious "by inspection," use it; otherwise assume tension and let the algebra decide.
Procedure for Analysis
Step-by-step
1. Choose a joint with one known force and at most two unknowns (find support reactions first if needed). 2. Assume each unknown is tensile. 3. Align the x–y axes so forces resolve cleanly, then apply \( \sum F_x = 0 \) and \( \sum F_y = 0 \). 4. Move to an adjacent joint, again keeping to two unknowns, until all members are solved.
When a joint has three or more unknowns, stop and find the support reactions first — students lose the most time trying to force a joint that simply cannot be solved yet. Sketch the whole-truss FBD, get the reactions, then hunt for the joint with two unknowns.
Zero-Force Members
Spotting members that carry no load before doing any algebra dramatically shortens a joint analysis. These zero-force members are not useless — they brace the truss during construction and stand ready if the loading pattern changes.
Zero-Force Member
A member that carries no force under the given loading. They can usually be identified by inspecting individual joints, and removing them from consideration simplifies the remaining equilibrium equations.
If only two non-collinear members meet at a joint and no external load or reaction acts at that joint, then both members are zero-force members.
If three members meet at a joint and two of them are collinear, then the third member is a zero-force member — provided no external force or reaction has a component acting along that third member.
Scan for zero-force members before writing a single equation. Marking them on the sketch first often collapses a daunting truss into just a handful of members that actually carry load.
The Method of Sections
When only a few member forces are wanted, cutting the truss is far faster than walking joint by joint. An imaginary cut exposes internal member forces as external forces on each segment, and because a segment is a rigid body, all three planar equilibrium equations are available.
Method of Sections
A technique based on the fact that any segment of a truss in equilibrium is itself in equilibrium. An imaginary section "cuts" through the members of interest; the free-body diagram of either resulting segment is then solved with the three equations of rigid-body equilibrium. Aim to cut no more than three members with unknown forces.
Sum moments about the point where the lines of action of two of the three unknowns intersect — the third force then follows directly from a single moment equation. If two unknowns are parallel, sum forces perpendicular to them to isolate the third. These choices avoid solving simultaneous equations.
The method of sections shines on exam questions that ask for "the force in member X" — don't reflexively start at a support and grind through joints. One well-placed cut and one moment equation is often the whole solution.
Space Trusses
A space truss carries load in three dimensions. Its building block is the tetrahedron, and the same two solution strategies — joints and sections — apply, now with three force equations per joint instead of two.
Space Truss
A stable three-dimensional structure of members joined at their ends. The simplest unit is a tetrahedron of six members; a simple space truss is grown by adding three members and one joint at a time to form connected tetrahedrons.
Design Assumptions
Members are treated as two-force members when loads are applied only at the joints and the joints behave as ball-and-socket connections — valid when member centerlines meet at a point and self-weight is negligible (or split half to each end).
When the 3-D geometry at a joint is hard to picture, write each member force as a Cartesian vector \( \mathbf{F} = F\,\mathbf{u} \), using the unit vector along the member. Begin at a joint with at most three unknowns to avoid large simultaneous systems.
In 3-D problems, set up the unit vectors carefully — most errors trace back to a wrong sign in a direction cosine, not the equilibrium algebra. Double-check each \( \mathbf{u} \) against the coordinates before summing.
Frames and Machines
Trusses are all two-force members. Frames and machines are not: they contain multiforce members carrying three or more forces, so member forces can no longer be assumed to act along the member axis. The structure must be taken apart and each piece analyzed.
Multiforce Member
A member subjected to three or more forces or couples. Unlike a two-force member, its internal force direction is not known in advance, so its free-body diagram must show full force components.
Frame vs. Machine
Both are pin-connected assemblies of multiforce members. A frame is stationary and built to support loads; a machine contains moving parts and is built to transmit or alter forces (think pliers or a press).
Forces shared by two contacting members are equal in magnitude and opposite in direction on each member (Newton's third law). When members are grouped as one system these forces are internal and not drawn; when members are separated they become external and must appear, equal and opposite, on both free-body diagrams.
Before writing equations, find every two-force member: regardless of its shape, it carries equal, opposite, collinear forces along the line joining its two pins. Recognizing them collapses many unknowns and avoids unnecessary equations.
Procedure for Analysis
Free-body diagrams
Draw the whole structure, a portion, or each member — whichever leads most directly to the answer. Identify two-force members. Internal forces within a chosen group are not shown; contact forces between separated members appear equal and opposite on each.
Equations of equilibrium
Count unknowns against available equations (three per member in 2-D). Sum moments about a point where the lines of action of as many unknowns as possible intersect. A negative result means the assumed sense was reversed.
The single most common frame/machine mistake is forgetting Newton's third law at a shared pin — drawing the contact force the same direction on both members. Pick a sense on one member, then force the mirror image on the other, every time.
Physical meaning
A truss calculation reports the axial action that one joint transmits to another: tension pulls a bar longer, while compression pushes it shorter. A section calculation has the same physical basis, but replaces a chosen set of bars by the forces crossing an imaginary boundary.
Problem-solving strategy
- Classify each member before drawing equations: two-force, three-force, or multiforce.
- Use a whole-structure free-body diagram for reactions, then isolate the smallest joint, cut portion, or member that exposes the requested unknown.
- For a truss joint, begin with at most two unknown member forces; for a section, cut no more than three unknown forces.
- Assume tensile cut forces, retain algebraic signs, and use a moment center through competing unknown lines of action.
- Audit every result against the joint or segment not used to solve it and label its physical sense.
Recognition strategy
The loaded apex is a concurrent-force joint with two equal unknown member forces. Symmetry makes horizontal components cancel, so joint equilibrium is more direct than solving the full truss.
Given information
- Two 45° members are geometrically and support-wise symmetric; the joint load is 12.0 kN downward.
- Members are pin-connected and loaded only at joints; the requested force is axial.
Model and interpretation
Isolate the apex and initially draw both member forces away from it as tension. Use +y upward; a negative solved force means the bars actually push upward on the joint in compression.
Verification
Units: The sine is dimensionless, so 12.0 kN divided by \(2\sin45^\circ\) remains kN, the required unit for an axial member force.
Signs and direction: A negative tension assumption correctly means compression; compression forces at the apex have upward components, not the downward components of the trial arrows.
Plausibility: Each member must exceed 6.00 kN because only 70.7% of its force is vertical; 8.49 kN satisfies that geometric expectation exactly.
Engineering interpretation
Shallower roof members would require larger compression for the same vertical load and may be governed by buckling rather than strength. Connection eccentricity, member weight, wind reversal, and code load combinations need separate design review.
Recognition strategy
Only one chord force is wanted. A section is preferable to propagating joint solutions, and moments about B eliminate the vertical and diagonal cut forces because both lines pass through B.
Given information
- Left cut portion is in equilibrium with \(A_y=24\ \text{kN}\) upward, 3.0 m left of B.
- Chord DE is horizontal and lies 2.0 m below B, giving a 2.0 m perpendicular lever arm; other cut-member lines intersect B.
Model and interpretation
Draw all cut forces as tension on the left portion. Take counterclockwise moments positive about B; the support reaction and a tensile lower-chord DE force create opposite senses.
Verification
Units: Both moment terms are kN·m; dividing the 72.0 kN·m reaction moment by the 2.0 m chord lever arm yields kN.
Signs and direction: The positive result confirms the assumed arrow away from the cut, so DE pulls the isolated left portion and is in tension.
Plausibility: The chord force exceeds the vertical reaction because its 2.0 m lever arm is shorter than the reaction's 3.0 m lever arm, as moment balance requires.
Engineering interpretation
A chord often carries a substantial axial force while visibly appearing lightly loaded. This result is an ideal truss force; gusset-plate capacity, fatigue from traffic cycles, and lateral bracing must be evaluated separately.
Recognition strategy
The arm is a multiforce member, but the brace has pins only at its ends and no other load, so its force has a known 3-4-5 line of action. Moment equilibrium of the arm exposes it directly.
Given information
- \(AC=2.4\ \text{m}\), actuator load \(6.0\ \text{kN}\) at \(x=1.8\ \text{m}\), and brace direction components \((-0.8,\,+0.6)\).
- Pin A supplies \(A_x,A_y\); the brace pulls joint C toward the elevated wall if it is in tension.
Model and interpretation
Isolate the arm with +x right and +y up. Write brace force \(T(-0.8\mathbf i+0.6\mathbf j)\); no brace moment exists because it is a two-force member.
Verification
Units: Each moment term is kN times m and each force equation remains in kN; the dimensionless 3-4-5 direction components introduce no unit conversion.
Signs and direction: The positive brace tension creates leftward and upward forces at C, requiring rightward and upward reactions at A exactly as the solved signs show.
Plausibility: The brace vertical component is 4.50 kN, smaller than the 6.0 kN load because A carries 1.50 kN; its 10.8 kN·m counterclockwise moment balances the load moment.
Engineering interpretation
The brace reduces bending demand in the arm by converting much of the load into axial tension, but it also introduces a 6 kN horizontal anchor demand. A real installation needs anchor, joint, deflection, vibration, and allowable-load checks.
Chapter Review
Simple Truss
Triangular elements joined by pins; every member is a two-force member, so it is in tension, in compression, or carries zero force.
Method of Joints
Each joint is in equilibrium under a concurrent force system. Start where there are at most two unknowns, assume tension, and apply \( \sum F_x = \sum F_y = 0 \). Align an axis with one unknown to solve the other directly; identify zero-force members first.
Method of Sections
Each segment is in equilibrium. Cut through the target members, draw the segment with the fewer forces, and use \( \sum F_x = \sum F_y = \sum M_O = 0 \). Choose moment points and perpendicular directions for direct, single-unknown solutions.
Space Truss
A 3-D truss of tetrahedral elements with ball-and-socket joints, analyzed by the same methods using three force equations per joint.
Frames and Machines
Structures with one or more multiforce members. Dismember and draw a free-body diagram of each part, applying action–reaction at every shared connection. Three equilibrium equations per member in 2-D; always recognize two-force members to simplify.
Master decision rule for the whole chapter: all member forces → joints; a few specific members → sections; any multiforce member present → dismember as a frame/machine. Naming the structure type correctly is half the battle.
Exercises, fundamental problems, and review problems are intentionally omitted from this reference page.