Analysis of Structures - Theory & Concepts

Analysis of Structures

Learning Objectives

Model planar trusses using pin joints, joint loads, and two-force members.
Calculate support reactions and member forces by the methods of joints and sections.
Detect zero-force members and distinguish determinacy counts from geometric stability.
Disassemble frames and machines into member free-body diagrams with consistent action-reaction forces.
Identify two-force and multi-force members and calculate mechanical advantage.

Planar Truss

A framework of straight, slender members joined by ideal pins, with external loads and support reactions applied only at joints.

Governing Assumptions for Trusses

Members are straight and connected only at their ends by frictionless pins.
Loads and reactions act at joints; member self-weight is neglected or converted to equivalent joint loads.
Every qualifying member is a two-force member, so its internal force acts along its axis.
The visualization is a rigid-body statics model. Any displayed movement is schematic and is not elastic deformation.

Truss Force Sign Convention

A positive solved member force is tension and pulls away from each joint. A negative solved member force is compression and pushes toward each joint. A force within the numerical tolerance is classified as zero.

Planar Joint Equilibrium

Every joint of a stable truss must independently satisfy two scalar equilibrium equations.

\sum F_x=0,\qquad \sum F_y=0

Variables

Symbol	Description	Unit
$F_x$	Horizontal force component at a joint	kN
$F_y$	Vertical force component at a joint	kN

Planar Truss Determinacy Count

A necessary counting check for a planar pin-jointed truss.

m+r=2j

Variables

Symbol	Description	Unit
$m$	Number of truss members	-
$r$	Number of independent support-reaction components	-
$j$	Number of joints	-

Counting Does Not Prove Stability

The equality $m+r=2j$ is necessary for a simple statically determinate truss, but improper geometry or concurrent support reactions can still create a mechanism. A rank or geometric stability check is also required.

Guided Example: Symmetric Five-Joint Bridge Truss

STEP-BY-STEP

Treat the complete truss as one rigid body and solve the pin and roller reactions.
At a support joint with at most two unknown members, assume both unknown forces act in tension.
Apply joint equilibrium and retain the algebraic signs of the answers.
Continue joint by joint until all member forces are known.
Verify every joint by calculating the residual vector; a correct solution has a residual near zero.
Pass a cut through three selected members and verify the same forces using rigid-body equilibrium of one isolated side.

Simulation 1 Instructions: Method of Joints

Select a joint, predict the first incident member as tension, compression, or zero, then reveal the calculated member forces and joint residual. Change the joint load and truss height to examine how geometry affects axial force.

Interactive engineering simulation

Method-of-Joints Truss Solver

Select a joint, inspect its free-body diagram, and verify the two scalar equilibrium equations.

Rigid-body statics

explore task

Choose a joint with no more than two unknown member forces. Predict tension or compression before revealing values.

Joint load80 kN

10–180

Drag for exploration or enter an exact value. Press Enter to apply; Escape restores the current value.

Truss height3.00 m

1.50–6.00

Drag for exploration or enter an exact value. Press Enter to apply; Escape restores the current value.

Joint A residual: 0.00e+0 kN. First incident result: AB = 66.7 C kN.

Max joint residual

1.42e-14

Assumption

Pinned, axial-only

Diagrams are schematic equilibrium models. Member lines do not show elastic deformation, buckling, connection flexibility, or second-order effects.

Model scope and verification

Use the displayed units and idealizations, then verify the governing balance or compatibility equation before interpreting the result.

Concept Check 1

Why should a joint with more than two unknown member forces usually be postponed when using only $\sum F_x=0$ and $\sum F_y=0$ ?

Method of Sections

A truss-analysis method that exposes selected member forces by cutting through the truss and applying three rigid-body equilibrium equations to either isolated side.

Section Equilibrium

The isolated portion of a planar truss must satisfy complete rigid-body equilibrium.

\sum F_x=0,\qquad \sum F_y=0,\qquad \sum M_O=0

Variables

Symbol	Description	Unit
$M_O$	Moment about any convenient point O	$kN\cdot m$

Simulation 2 Instructions: Method of Sections

Choose a section crossing no more than three unknown members. Compare two available cuts and inspect the section-equilibrium residual for the highlighted isolated side.

Interactive engineering simulation

Method-of-Sections Cut Explorer

Choose a section cut and verify that the isolated truss half satisfies force and moment equilibrium.

Rigid-body statics

explore task

Select a cut crossing no more than three unknown members, then choose the simpler side for equilibrium.

Joint load80 kN

10–180

Drag for exploration or enter an exact value. Press Enter to apply; Escape restores the current value.

Truss height3.00 m

1.50–6.00

Drag for exploration or enter an exact value. Press Enter to apply; Escape restores the current value.

\sum F_x=0,\quad \sum F_y=0,\quad \sum M_O=0

Section equilibrium residual

0.00e+0 kN-equivalent

Cut members: AC, BC, BD. The highlighted forces act on the isolated side; the opposite half receives equal-and-opposite forces.

Max joint residual

1.42e-14

Assumption

Pinned, axial-only

Diagrams are schematic equilibrium models. Member lines do not show elastic deformation, buckling, connection flexibility, or second-order effects.

Model scope and verification

Use the displayed units and idealizations, then verify the governing balance or compatibility equation before interpreting the result.

Concept Check 2

How can taking moments about the intersection of two cut-member lines isolate the force in the third cut member?

Zero-Force Member

A truss member whose axial force is zero for the current loading arrangement, although it may still be essential for stability or another load case.

Zero-Force-Member Inspection Rules

At an unloaded, unsupported joint with only two non-collinear members, both members are zero-force members.
At an unloaded, unsupported joint with three members, if two are collinear, the non-collinear member is a zero-force member.
Reapply the rules after each identified zero-force member because removing it from the force analysis may expose another rule.
Do not physically remove the member from the real structure without checking stability, buckling restraint, and other load cases.

Simulation 3 Instructions: Zero-Force Members

Inspect the unloaded middle joint before revealing the rule-based detector. The highlighted member is classified by geometry, not by an arbitrary force threshold alone.

Interactive engineering simulation

Zero-Force-Member Detector

Apply the two inspection rules before revealing the solver classification.

Rigid-body statics

explore task

Inspect unloaded joints first. Look for two non-collinear members or three members with two collinear.

Joint load80 kN

10–180

Drag for exploration or enter an exact value. Press Enter to apply; Escape restores the current value.

Detected zero-force members

At unloaded joint B, BD is non-collinear with the other two members and is zero-force.

Max joint residual

0.00e+0

Assumption

Pinned, axial-only

Diagrams are schematic equilibrium models. Member lines do not show elastic deformation, buckling, connection flexibility, or second-order effects.

Model scope and verification

Use the displayed units and idealizations, then verify the governing balance or compatibility equation before interpreting the result.

Concept Check 3

Why can a zero-force member under gravity loading become important when wind or a moving load changes the joint-loading pattern?

Static Determinacy

A condition in which all reactions and member forces can be calculated from independent equilibrium equations alone.

Stability, Determinacy, and Invalid Geometry

A positive value of $m+r-2j$ indicates extra unknowns and static indeterminacy. A negative value indicates too few force unknowns for the joint equations. Even when the count is zero, a rank-deficient equilibrium matrix identifies a geometric mechanism. Duplicate members, zero-length members, missing joints, and invalid support directions are invalid model inputs rather than valid structural classifications.

Simulation 4 Instructions: Stability and Determinacy

Switch among a stable determinate truss, an unbraced square mechanism, and a redundant truss. Compare the determinacy index with the equilibrium-matrix rank.

Interactive engineering simulation

Truss Stability and Determinacy

Compare force-counting with equilibrium-matrix rank using valid, unique-member geometries.

Count + rank

m + r − 2j

Matrix rank

10 of 10 equilibrium rows

Interpreted condition

stable determinate

Why counting is insufficient: a zero count does not guarantee stable geometry, and a positive count does not prevent a rank-deficient mechanism.

Predict the condition

Model scope and verification

Use the displayed units and idealizations, then verify the governing balance or compatibility equation before interpreting the result.

Concept Check 4

Why does adding one diagonal to an unbraced rectangular panel change its geometric stability even before any loads are applied?

Moving Loads on Bridge Trusses

A vehicle idealized as a joint load changes support reactions and member forces as it moves across the deck. Under rigid-body statics assumptions, each load position is a separate equilibrium case; the displayed envelope records the largest absolute force from the sampled positions.

Simulation 5 Instructions: Bridge Moving Load

Move the load among deck joints and compare the current member forces with the load-position envelope. Predict where the greatest absolute member force occurs before reading the bars.

Interactive engineering simulation

Bridge-Truss Moving-Load Envelope

Move a joint load across the deck and calculate every envelope case using the same displayed truss height.

Height-consistent envelope

Member	Envelope \|F\|	Governing load position	State
BD	106.667 kN	C	compression
AB	66.667 kN	C	compression
BC	66.667 kN	C	tension
CD	66.667 kN	C	tension
DE	66.667 kN	C	compression
AC	53.333 kN	C	tension
CE	53.333 kN	C	tension

Joint load80 kN

10–200

Drag for exploration or enter an exact value. Press Enter to apply; Escape restores the current value.

Truss height3.00 m

1.50–6.00

Drag for exploration or enter an exact value. Press Enter to apply; Escape restores the current value.

Classification

determinate

Maximum joint residual

1.42e-14

Governing envelope

BD: 106.667 kN at C

Model scope and verification

Use the displayed units and idealizations, then verify the governing balance or compatibility equation before interpreting the result.

Concept Check 5

Why can a member change from tension to compression as the same vehicle load moves from one side of a bridge truss to the other?

Frame

A stationary assembly containing at least one multi-force member and intended to support external loads.

Machine

An assembly of connected rigid bodies intended to transmit or transform forces and motion.

Two-Force Member

A member subjected to forces only at two points, with no applied couple; equilibrium requires the two forces to be equal, opposite, and collinear with the member axis.

Multi-Force Member

A rigid member acted on by three or more forces, or by forces plus an applied couple, requiring complete rigid-body equilibrium.

Action-Reaction at Internal Pins

When a frame is disassembled, the force exerted by member 1 on a shared pin is equal in magnitude and opposite in direction to the force exerted by the pin on member 1 or by member 2 on the same pin. The pair must not be counted twice on the whole-system free-body diagram.

Internal Pin Action-Reaction Pair

Forces at the same ideal pin appear as equal-and-opposite vectors on the separated member diagrams.

\mathbf{F}_{B,1}=-\mathbf{F}_{B,2}

Variables

Symbol	Description	Unit
$\mathbf{F}_{B,1}$	Force at pin B acting on member 1	kN
$\mathbf{F}_{B,2}$	Force at pin B acting on member 2	kN

Frame and Machine Disassembly Procedure

STEP-BY-STEP

Draw the whole-system free-body diagram and solve available external reactions.
Separate every rigid member and any multi-member pin that requires its own free-body diagram.
Mark each shared pin-force pair with equal magnitude and opposite direction.
Identify true two-force members before assigning unnecessary force components.
Apply $\sum F_x=0$ , $\sum F_y=0$ , and $\sum M=0$ to each multi-force member.
Check that the assembled and disassembled solutions have near-zero force and moment residuals.

Simulation 6 Instructions: Frame Disassembly

Toggle between the assembled system and individual member free-body diagrams. Verify the equal-and-opposite pin-force pair and identify the inclined two-force link.

Interactive engineering simulation

Pin-Connected Frame Disassembly

Separate a frame into member free-body diagrams and preserve equal-and-opposite pin forces.

Rigid-body statics

explore task

Click Disassemble and verify that every shared pin-force pair is equal and opposite.

Supported load24 kN

5–80

Drag for exploration or enter an exact value. Press Enter to apply; Escape restores the current value.

Two-force link

15.62 kN compression

Pin-pair check

0.00000 kN

Equilibrium residual

0.000000

\mathbf{F}_{B,1}=-\mathbf{F}_{B,2}

A member pinned at only two ends with no intermediate load is treated as a two-force member. Otherwise, retain all pin components and moments required by equilibrium.

Diagrams are schematic equilibrium models. Member lines do not show elastic deformation, buckling, connection flexibility, or second-order effects.

Model scope and verification

Use the displayed units and idealizations, then verify the governing balance or compatibility equation before interpreting the result.

Concept Check 6

Why must the internal pin forces disappear when the complete assembled frame is treated as one rigid body?

Simulation 7 Instructions: Compound Frame

Change the inclined-link geometry and observe how the axial link force and support-pin components satisfy the loaded member equilibrium.

Interactive engineering simulation

Compound Structural Frame

Analyze a loaded member restrained by a two-force link and a pin support.

Rigid-body statics

explore task

Identify the two-force link before writing the three equilibrium equations for the loaded member.

Supported load24 kN

5–80

Drag for exploration or enter an exact value. Press Enter to apply; Escape restores the current value.

Link horizontal run2.50 m

0.50–5.00

Drag for exploration or enter an exact value. Press Enter to apply; Escape restores the current value.

Link vertical drop3.00 m

0.50–5.00

Drag for exploration or enter an exact value. Press Enter to apply; Escape restores the current value.

Two-force link

15.62 kN compression

Pin-pair check

0.00000 kN

Equilibrium residual

0.000000

\mathbf{F}_{B,1}=-\mathbf{F}_{B,2}

A member pinned at only two ends with no intermediate load is treated as a two-force member. Otherwise, retain all pin components and moments required by equilibrium.

Diagrams are schematic equilibrium models. Member lines do not show elastic deformation, buckling, connection flexibility, or second-order effects.

Model scope and verification

Use the displayed units and idealizations, then verify the governing balance or compatibility equation before interpreting the result.

Concept Check 7

What happens to the required link force as the two-force link becomes nearly horizontal and loses vertical force effectiveness?

Pulley Forces in a Supporting Frame

For an ideal massless rope and frictionless pulley, the rope tension is the same in every continuous segment. The moving block is supported by the vector sum of its rope segments, while the frame receives the resultant force transmitted through the pulley axle.

Ideal Pulley Mechanical Advantage

For parallel supporting rope segments, the ideal input force equals the load divided by the number of supporting segments.

F_{\mathrm{in}}=\frac{W}{n},\qquad \mathrm{MA}=\frac{W}{F_{\mathrm{in}}}=n

Variables

Symbol	Description	Unit
$W$	Supported load	kN
$n$	Number of rope segments supporting the moving block	-
$F_{\mathrm{in}}$	Ideal rope input force	kN

Simulation 8 Instructions: Pulley-Supporting Frame

Change the number of supporting rope segments and track rope tension, frame force transfer, mechanical advantage, and equilibrium residual.

Interactive engineering simulation

Pulley-Supporting Frame

Track rope tension, support force, and pulley mechanical advantage without double-counting forces.

Rigid-body statics

explore task

Count rope segments supporting the moving block, then transfer the resulting pulley force to the frame.

Supported load24 kN

5–80

Drag for exploration or enter an exact value. Press Enter to apply; Escape restores the current value.

Supporting rope segments4

1–6

Drag for exploration or enter an exact value. Press Enter to apply; Escape restores the current value.

Mechanical advantage

4.00

Input force

6.00 kN

Equilibrium residual

0.000000

\mathbf{F}_{B,1}=-\mathbf{F}_{B,2}

A member pinned at only two ends with no intermediate load is treated as a two-force member. Otherwise, retain all pin components and moments required by equilibrium.

Diagrams are schematic equilibrium models. Member lines do not show elastic deformation, buckling, connection flexibility, or second-order effects.

Model scope and verification

Use the displayed units and idealizations, then verify the governing balance or compatibility equation before interpreting the result.

Concept Check 8

Why should the number of rope segments around a fixed pulley not automatically be counted as the mechanical advantage of the moving load?

Lever Moment Balance

For ideal static equilibrium, input and output moments balance about the pivot.

F_{\mathrm{in}}d_{\mathrm{in}}=F_{\mathrm{out}}d_{\mathrm{out}}

Variables

Symbol	Description	Unit
$d_{\mathrm{in}}$	Input force moment arm	m
$d_{\mathrm{out}}$	Output force moment arm	m

Simulation 9 Instructions: Lever or Bolt Cutter

Change the input and output moment arms and compare output force, idealized efficiency, mechanical advantage, and moment residual.

Interactive engineering simulation

Lever and Bolt-Cutter Machine

Change lever arms and quantify ideal mechanical advantage and equilibrium residual.

Rigid-body statics

explore task

Balance input and output moments about the pivot before interpreting mechanical advantage.

Input force24 kN

5–80

Drag for exploration or enter an exact value. Press Enter to apply; Escape restores the current value.

Input arm0.55 m

0.20–1.20

Drag for exploration or enter an exact value. Press Enter to apply; Escape restores the current value.

Output arm0.12 m

0.05–0.40

Drag for exploration or enter an exact value. Press Enter to apply; Escape restores the current value.

Mechanical advantage

4.22

Input force

24.00 kN

Equilibrium residual

0.000000

\mathbf{F}_{B,1}=-\mathbf{F}_{B,2}

A member pinned at only two ends with no intermediate load is treated as a two-force member. Otherwise, retain all pin components and moments required by equilibrium.

Diagrams are schematic equilibrium models. Member lines do not show elastic deformation, buckling, connection flexibility, or second-order effects.

Model scope and verification

Use the displayed units and idealizations, then verify the governing balance or compatibility equation before interpreting the result.

Concept Check 9

Why does a smaller output moment arm increase output force while reducing the output travel for a given input rotation?

Excavator and Crane Linkages

Hydraulic cylinders and ideal connecting links are commonly treated as two-force members when forces act only through their end pins. Their force direction follows the pin-to-pin axis, while the boom or bucket is a multi-force member requiring moment equilibrium about a convenient pin.

Simulation 10 Instructions: Excavator or Crane Linkage

Change the cylinder-link run and drop. Observe how an unfavorable shallow angle increases axial cylinder force and changes the support-pin reaction pair.

Interactive engineering simulation

Excavator and Crane Linkage

Explore how linkage geometry changes cylinder force and pin-force direction in a crane-like mechanism.

Rigid-body statics

explore task

The cylinder is a two-force member; its force must lie along the line between its pins.

Supported load24 kN

5–80

Drag for exploration or enter an exact value. Press Enter to apply; Escape restores the current value.

Link horizontal run2.50 m

0.50–5.00

Drag for exploration or enter an exact value. Press Enter to apply; Escape restores the current value.

Link vertical drop3.00 m

0.50–5.00

Drag for exploration or enter an exact value. Press Enter to apply; Escape restores the current value.

Two-force link

15.62 kN compression

Pin-pair check

0.00000 kN

Equilibrium residual

0.000000

\mathbf{F}_{B,1}=-\mathbf{F}_{B,2}

A member pinned at only two ends with no intermediate load is treated as a two-force member. Otherwise, retain all pin components and moments required by equilibrium.

Diagrams are schematic equilibrium models. Member lines do not show elastic deformation, buckling, connection flexibility, or second-order effects.

Model scope and verification

Use the displayed units and idealizations, then verify the governing balance or compatibility equation before interpreting the result.

Concept Check 10

Why can a hydraulic cylinder experience a very large axial force when its line of action passes close to the boom pivot?

Common Structural-Analysis Mistakes

Applying a truss load between joints while still treating every member as a two-force member.
Declaring a truss stable from $m+r=2j$ without checking geometry and support directions.
Changing tension and compression sign conventions midway through a joint solution.
Cutting more than three unknown truss members in a planar section without additional information.
Assigning independent pin-force directions to opposite sides of the same internal pin.
Treating a loaded or coupled frame member as a two-force member.
Counting pulley rope forces twice on the assembled-system free-body diagram.

Key Takeaways

Truss forces are axial only when pin-joint and joint-loading assumptions are satisfied.
Joint and section solutions must satisfy equilibrium within a stated numerical tolerance.
Determinacy counting is necessary but geometric rank and invalid-input checks are also required.
Frame disassembly is an analytical step that exposes internal action-reaction pairs.
Two-force members have collinear end forces; multi-force members require full rigid-body equilibrium.
Machines transform force through geometry while preserving moment equilibrium in the ideal statics model.

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