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.

Truss, section, and frame analysis views

Truss joints expose concurrent member forces, sections isolate selected member actions, and frames require explicit interaction forces between connected rigid bodies.

$Planar Truss$

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.

Simulation

Joint A residual

0.00e+0 kN

66.667 kN C

53.333 kN T

80 kN

3.00 m

determinate

\sum F_x=0,\qquad \sum F_y=0,\qquad \text{and for sections }\sum M_O=0

$Concept Check 1$

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.

Simulation

Normalized section residual

0.00e+0

Cut members

AC · BC · BD

Isolated side

A · B

Cut boundary validity

valid physical boundary

Classification

determinate

80 kN

3.00 m

determinate

\sum F_x=0,\qquad \sum F_y=0,\qquad \text{and for sections }\sum M_O=0

$Concept Check 2$

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.

Simulation

Zero-force members

Classification

determinate

80 kN

determinate

\sum F_x=0,\qquad \sum F_y=0,\qquad \text{and for sections }\sum M_O=0

$Concept Check 3$

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.

Simulation

Classification

determinate

m + r − 2j

Equilibrium rank

10/10

Structural state

stable determinate

80 kN

3.00 m

determinate

\sum F_x=0,\qquad \sum F_y=0,\qquad \text{and for sections }\sum M_O=0

$Concept Check 4$

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.

Simulation

Current max |F|

106.67 kN

Envelope governing member

BD · 106.67 kN at C

Max joint residual

1.42e-14

80 kN

3.00 m

determinate

\sum F_x=0,\qquad \sum F_y=0,\qquad \text{and for sections }\sum M_O=0

$Concept Check 5$

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.

Simulation

Link force

48.074 kN compression

Pin A

(26.67, 40) kN

Force residual

0.00e+0

Moment residual

0.00e+0

Pin B action/reaction residual

0.00e+0

80 kN

6.0 m

3.0 m

2.0 m

3.0 m

frame equilibrium

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

$Concept Check 6$

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.

Simulation

Link force

48.074 kN compression

Pin A

(26.67, 40) kN

Force residual

0.00e+0

Moment residual

0.00e+0

Pin B action/reaction residual

0.00e+0

80 kN

6.0 m

3.0 m

2.0 m

3.0 m

frame equilibrium

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

$Concept Check 7$

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.

Simulation

Input rope force

20.000 kN

Mechanical advantage

Single frame transfer nT

80.000 kN

Equilibrium residual

0.00e+0

80 kN

6.0 m

rope equilibrium

nT=W,\qquad T=\frac{W}{n}

$Concept Check 8$

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.

Simulation

Output force

180.000 N

Mechanical advantage

3.600

Transmitted input moment

90.000 N·m

Output moment

90.000 N·m

Moment residual

0.00e+0

50 N

2.0 m

0.5 m

0.90

moment-balanced

F_{in}r_{in}\eta=F_{out}r_{out}

$Concept Check 9$

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.

Simulation

Axial link force

48.074 kN compression

Link-axis angle

-56.31°

Two-force condition

equal · opposite · collinear

Parent-frame residual

0.00e+0

80 kN

6.0 m

3.0 m

2.0 m

3.0 m

two-force linkage

\mathbf F_B\parallel \overline{BC},\qquad \mathbf F_C=-\mathbf F_B

$Concept Check 10$

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.

Structural Analysis Method Selection

Select joints, sections, or rigid-body disassembly while distinguishing truss equilibrium from frame and machine equilibrium.

Classify the structural system and member idealizations → Pin-jointed truss with joint loads and two-force members?; Pin-jointed truss with joint loads and two-force members? — Yes → Solve truss support reactions from whole-structure equilibrium; Pin-jointed truss with joint loads and two-force members? — No → For a frame or machine, draw the whole-assembly FBD and solve external reactions when possible; Solve truss support reactions from whole-structure equilibrium → Only a few member forces required?; Only a few member forces required? — Yes → Use method of sections with a cut containing a solvable set of unknowns; Only a few member forces required? — No → Use method of joints; start where the unknown set is solvable; Use method of sections with a cut containing a solvable set of unknowns → Every isolated body or joint satisfies equilibrium and member assumptions?; Use method of joints; start where the unknown set is solvable → Every isolated body or joint satisfies equilibrium and member assumptions?; For a frame or machine, draw the whole-assembly FBD and solve external reactions when possible → Disassemble the frame or machine into individual rigid bodies; Disassemble the frame or machine into individual rigid bodies → Show equal-and-opposite internal interactions and identify two-force members; Show equal-and-opposite internal interactions and identify two-force members → Every isolated body or joint satisfies equilibrium and member assumptions?; Every isolated body or joint satisfies equilibrium and member assumptions? — Yes → Member actions verified; Every isolated body or joint satisfies equilibrium and member assumptions? — No → Correct reactions, member idealization, cut, or interaction forces; Correct reactions, member idealization, cut, or interaction forces → Classify the structural system and member idealizations

Classify the structural system and member idealizations: terminator
Pin-jointed truss with joint loads and two-force members?: decision
Solve truss support reactions from whole-structure equilibrium: process
Only a few member forces required?: decision
Use method of sections with a cut containing a solvable set of unknowns: process
Use method of joints; start where the unknown set is solvable: process
For a frame or machine, draw the whole-assembly FBD and solve external reactions when possible: process
Disassemble the frame or machine into individual rigid bodies: process
Show equal-and-opposite internal interactions and identify two-force members: process
Every isolated body or joint satisfies equilibrium and member assumptions?: decision
Correct reactions, member idealization, cut, or interaction forces: process
Member actions verified: terminator

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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