Plane Frame

Structural Analysis · FE Reference Handbook section

Structural Analysis

1 formulas

10 exam-style examples

~47 min

All Structural Analysis lectures

Learning objectives

What you must be able to do before leaving this section.

This chapter section covers Plane Frame within Structural Analysis. Read it the way you would read a textbook chapter: the theory first so the relations mean something, then every equation with its use and its trap, then 10 fully worked examples with the arithmetic shown line by line, and finally a self-check you should be able to answer without notes.

Explain, in your own words, what plane frame describes physically and when it applies.
State every one of the 1 relation the handbook lists here and name each symbol with its unit.
Select the correct relation from the wording of an exam stem within 20 seconds.
Carry a complete solution from givens to a "most nearly" answer with the correct unit.
Recognise the distractors generated by the unit trap: keep kip-ft consistently; EI in kip-in² needs a 1,728 factor.

Lecture

Why this section exists. Plane Frame is the part of Structural Analysis that lets you connect a determinate or one-degree indeterminate structure to a number you can defend. Before any equation is useful you must be able to picture the physical situation it describes; the schematic below is that picture.

How the theory is built. The handbook prints results, not derivations. Each relation in this section comes from one governing principle applied to the idealised system: state the principle, impose the stated assumptions, and the printed equation follows. Knowing which assumption each relation rests on is what lets you reject a wrong answer choice in seconds.

How it is examined. Items from this page are written as reactions, an influence-line ordinate, or one deflection. Roughly two thirds are direct substitution, one third require one intermediate quantity from a neighbouring relation, and a small number are conceptual — testing whether you know the assumption, not the arithmetic.

The habit that earns the points. Unit discipline. keep kip-ft consistently; EI in kip-in² needs a 1,728 factor. Every relation below is dimensionally consistent only when that rule is honoured, and the distractor set is deliberately built from candidates who ignored it. Write the unit next to every number you substitute, every time.

How to study this page. Read the theory, then cover the formula cards and try to reproduce each relation from its description. Then work the examples with the solution hidden, revealing one line at a time. Finish with the self-check questions; if you cannot answer one, return to the matching formula card.

Interior of a steel and glass pedestrian bridge showing the structural framing. — Photo 1. Where this shows up in practice: plane frame.
Wikimedia Commons, CC BY-SA 4.0

Structural Analysis — Plane Frame: reference schematic for orienting the symbols used in this section.

Theory, developed

Read this before the equations — it is what makes them memorable.

The physical situation

Every item from this section describes a determinate or one-degree indeterminate structure. Sketch it before you compute — a labelled sketch with the givens on it converts a wordy stem into a solvable problem and exposes the quantity the examiner left out on purpose.

The governing principle

The 1 relation on this page are consequences of one principle applied to that idealised system. Identify which quantity is conserved, balanced, or defined, and the correct equation follows without memorisation.

Assumptions and limits of validity

Each printed relation carries silent assumptions — linearity, steady state, uniformity, small deformation, or standard conditions, depending on the subject. Conceptual exam items are written by violating exactly one of these, so read the sentence above the equation as carefully as the equation itself.

Solution procedure you should automate

1) Read the last sentence of the stem to identify the requested quantity. 2) Locate the relation on this page whose left-hand side is that quantity. 3) Tabulate the givens with units and mark the missing symbol. 4) If a symbol is missing, find the one relation that produces it. 5) Rearrange symbolically, substitute once, evaluate, and round only at the end.

Notation used in this section

3m + r	Quantity produced by "3m + r = 3j + c Stable and statically determinate" — read its definition and unit from the handbook line directly above the equation.

Handbook notes for this section

Definitions and conditions exactly as the handbook states them.

Static Analysis Classification
3m + r < 3j + c Unstable
3m + r > 3j + c Stable and statically indeterminate
Stability also requires an appropriate arrangement of members and reaction components.

Core formulas for this FE topic

Definitions, applicability, units, assumptions and worked examples for each relation.

Worked exam-style examples

The four ways this section is written on the real exam — thoughts first, then equations, then substitution.

Example 1

Fixed-end moments for a loaded member — Plane Frame

A member of a rigid frame spans 30 ft and carries 2.00 kip/ft. What are the fixed-end moments, and what is the midspan moment once they are applied?

Given

w = 2.00 kip/ft
L = 30 ft
Both ends fixed

Find

FEM and midspan moment

Start with the thinking

Fixed-end moment for a uniform load is wL²/12.
The midspan value is wL²/24 with fixed ends.

Step-by-step solution

Fixed-end moment
$F E M = w L^{2} /12$
Substituting — FEM = 2.00(30)²/12 = 150.0 kip·ft at each end
Midspan
$M_{m} i d = w L^{2} /24$
Substituting — M_mid = 2.00(30)²/24 = 75.0 kip·ft
Check — simple-span value wL²/8 = 225.0 kip·ft is the sum of the end and mid effects ✓

Answer: FEM = 150.0 kip·ft; midspan = 75.0 kip·ft

Why the other options are there

225.0 kip·ft (simple-span value)
900.0 kip·ft (cantilever value)

Reference: FE Reference Handbook — Structural Analysis → Plane Frame

Example 2

Fixed-end moments for a loaded member — Plane Frame (2)

A member of a rigid frame spans 20 ft and carries 1.25 kip/ft. What are the fixed-end moments, and what is the midspan moment once they are applied?

Given

w = 1.25 kip/ft
L = 20 ft
Both ends fixed

Find

FEM and midspan moment

Start with the thinking

Fixed-end moment for a uniform load is wL²/12.
The midspan value is wL²/24 with fixed ends.

Step-by-step solution

Fixed-end moment
$F E M = w L^{2} /12$
Substituting — FEM = 1.25(20)²/12 = 41.7 kip·ft at each end
Midspan
$M_{m} i d = w L^{2} /24$
Substituting — M_mid = 1.25(20)²/24 = 20.8 kip·ft
Check — simple-span value wL²/8 = 62.5 kip·ft is the sum of the end and mid effects ✓

Answer: FEM = 41.7 kip·ft; midspan = 20.8 kip·ft

Why the other options are there

62.5 kip·ft (simple-span value)
250.0 kip·ft (cantilever value)

Reference: FE Reference Handbook — Structural Analysis → Plane Frame

Example 3

Fixed-end moments for a loaded member — Plane Frame (3)

A member of a rigid frame spans 19 ft and carries 1.75 kip/ft. What are the fixed-end moments, and what is the midspan moment once they are applied?

Given

w = 1.75 kip/ft
L = 19 ft
Both ends fixed

Find

FEM and midspan moment

Start with the thinking

Fixed-end moment for a uniform load is wL²/12.
The midspan value is wL²/24 with fixed ends.

Step-by-step solution

Fixed-end moment
$F E M = w L^{2} /12$
Substituting — FEM = 1.75(19)²/12 = 52.6 kip·ft at each end
Midspan
$M_{m} i d = w L^{2} /24$
Substituting — M_mid = 1.75(19)²/24 = 26.3 kip·ft
Check — simple-span value wL²/8 = 79.0 kip·ft is the sum of the end and mid effects ✓

Answer: FEM = 52.6 kip·ft; midspan = 26.3 kip·ft

Why the other options are there

79.0 kip·ft (simple-span value)
315.9 kip·ft (cantilever value)

Reference: FE Reference Handbook — Structural Analysis → Plane Frame

Example 4

Fixed-end moments for a loaded member — Plane Frame (4)

A member of a rigid frame spans 16 ft and carries 1.50 kip/ft. What are the fixed-end moments, and what is the midspan moment once they are applied?

Given

w = 1.50 kip/ft
L = 16 ft
Both ends fixed

Find

FEM and midspan moment

Start with the thinking

Fixed-end moment for a uniform load is wL²/12.
The midspan value is wL²/24 with fixed ends.

Step-by-step solution

Fixed-end moment
$F E M = w L^{2} /12$
Substituting — FEM = 1.50(16)²/12 = 32.0 kip·ft at each end
Midspan
$M_{m} i d = w L^{2} /24$
Substituting — M_mid = 1.50(16)²/24 = 16.0 kip·ft
Check — simple-span value wL²/8 = 48.0 kip·ft is the sum of the end and mid effects ✓

Answer: FEM = 32.0 kip·ft; midspan = 16.0 kip·ft

Why the other options are there

48.0 kip·ft (simple-span value)
192.0 kip·ft (cantilever value)

Reference: FE Reference Handbook — Structural Analysis → Plane Frame

Example 5

Fixed-end moments for a loaded member — Plane Frame (5)

A member of a rigid frame spans 26 ft and carries 1.50 kip/ft. What are the fixed-end moments, and what is the midspan moment once they are applied?

Given

w = 1.50 kip/ft
L = 26 ft
Both ends fixed

Find

FEM and midspan moment

Start with the thinking

Fixed-end moment for a uniform load is wL²/12.
The midspan value is wL²/24 with fixed ends.

Step-by-step solution

Fixed-end moment
$F E M = w L^{2} /12$
Substituting — FEM = 1.50(26)²/12 = 84.5 kip·ft at each end
Midspan
$M_{m} i d = w L^{2} /24$
Substituting — M_mid = 1.50(26)²/24 = 42.3 kip·ft
Check — simple-span value wL²/8 = 126.8 kip·ft is the sum of the end and mid effects ✓

Answer: FEM = 84.5 kip·ft; midspan = 42.3 kip·ft

Why the other options are there

126.8 kip·ft (simple-span value)
507.0 kip·ft (cantilever value)

Reference: FE Reference Handbook — Structural Analysis → Plane Frame

Example 6

Fixed-end moments for a loaded member — Plane Frame (6)

A member of a rigid frame spans 26 ft and carries 1.75 kip/ft. What are the fixed-end moments, and what is the midspan moment once they are applied?

Given

w = 1.75 kip/ft
L = 26 ft
Both ends fixed

Find

FEM and midspan moment

Start with the thinking

Fixed-end moment for a uniform load is wL²/12.
The midspan value is wL²/24 with fixed ends.

Step-by-step solution

Fixed-end moment
$F E M = w L^{2} /12$
Substituting — FEM = 1.75(26)²/12 = 98.6 kip·ft at each end
Midspan
$M_{m} i d = w L^{2} /24$
Substituting — M_mid = 1.75(26)²/24 = 49.3 kip·ft
Check — simple-span value wL²/8 = 147.9 kip·ft is the sum of the end and mid effects ✓

Answer: FEM = 98.6 kip·ft; midspan = 49.3 kip·ft

Why the other options are there

147.9 kip·ft (simple-span value)
591.5 kip·ft (cantilever value)

Reference: FE Reference Handbook — Structural Analysis → Plane Frame

Example 7

Fixed-end moments for a loaded member — Plane Frame (7)

A member of a rigid frame spans 17 ft and carries 3.50 kip/ft. What are the fixed-end moments, and what is the midspan moment once they are applied?

Given

w = 3.50 kip/ft
L = 17 ft
Both ends fixed

Find

FEM and midspan moment

Start with the thinking

Fixed-end moment for a uniform load is wL²/12.
The midspan value is wL²/24 with fixed ends.

Step-by-step solution

Fixed-end moment
$F E M = w L^{2} /12$
Substituting — FEM = 3.50(17)²/12 = 84.3 kip·ft at each end
Midspan
$M_{m} i d = w L^{2} /24$
Substituting — M_mid = 3.50(17)²/24 = 42.1 kip·ft
Check — simple-span value wL²/8 = 126.4 kip·ft is the sum of the end and mid effects ✓

Answer: FEM = 84.3 kip·ft; midspan = 42.1 kip·ft

Why the other options are there

126.4 kip·ft (simple-span value)
505.8 kip·ft (cantilever value)

Reference: FE Reference Handbook — Structural Analysis → Plane Frame

Example 8

Fixed-end moments for a loaded member — Plane Frame (8)

A member of a rigid frame spans 21 ft and carries 1.25 kip/ft. What are the fixed-end moments, and what is the midspan moment once they are applied?

Given

w = 1.25 kip/ft
L = 21 ft
Both ends fixed

Find

FEM and midspan moment

Start with the thinking

Fixed-end moment for a uniform load is wL²/12.
The midspan value is wL²/24 with fixed ends.

Step-by-step solution

Fixed-end moment
$F E M = w L^{2} /12$
Substituting — FEM = 1.25(21)²/12 = 45.9 kip·ft at each end
Midspan
$M_{m} i d = w L^{2} /24$
Substituting — M_mid = 1.25(21)²/24 = 23.0 kip·ft
Check — simple-span value wL²/8 = 68.9 kip·ft is the sum of the end and mid effects ✓

Answer: FEM = 45.9 kip·ft; midspan = 23.0 kip·ft

Why the other options are there

68.9 kip·ft (simple-span value)
275.6 kip·ft (cantilever value)

Reference: FE Reference Handbook — Structural Analysis → Plane Frame

Example 9

Fixed-end moments for a loaded member — Plane Frame (9)

A member of a rigid frame spans 22 ft and carries 4.00 kip/ft. What are the fixed-end moments, and what is the midspan moment once they are applied?

Given

w = 4.00 kip/ft
L = 22 ft
Both ends fixed

Find

FEM and midspan moment

Start with the thinking

Fixed-end moment for a uniform load is wL²/12.
The midspan value is wL²/24 with fixed ends.

Step-by-step solution

Fixed-end moment
$F E M = w L^{2} /12$
Substituting — FEM = 4.00(22)²/12 = 161.3 kip·ft at each end
Midspan
$M_{m} i d = w L^{2} /24$
Substituting — M_mid = 4.00(22)²/24 = 80.7 kip·ft
Check — simple-span value wL²/8 = 242.0 kip·ft is the sum of the end and mid effects ✓

Answer: FEM = 161.3 kip·ft; midspan = 80.7 kip·ft

Why the other options are there

242.0 kip·ft (simple-span value)
968.0 kip·ft (cantilever value)

Reference: FE Reference Handbook — Structural Analysis → Plane Frame

Example 10

Fixed-end moments for a loaded member — Plane Frame (10)

A member of a rigid frame spans 23 ft and carries 3.75 kip/ft. What are the fixed-end moments, and what is the midspan moment once they are applied?

Given

w = 3.75 kip/ft
L = 23 ft
Both ends fixed

Find

FEM and midspan moment

Start with the thinking

Fixed-end moment for a uniform load is wL²/12.
The midspan value is wL²/24 with fixed ends.

Step-by-step solution

Fixed-end moment
$F E M = w L^{2} /12$
Substituting — FEM = 3.75(23)²/12 = 165.3 kip·ft at each end
Midspan
$M_{m} i d = w L^{2} /24$
Substituting — M_mid = 3.75(23)²/24 = 82.7 kip·ft
Check — simple-span value wL²/8 = 248.0 kip·ft is the sum of the end and mid effects ✓

Answer: FEM = 165.3 kip·ft; midspan = 82.7 kip·ft

Why the other options are there

248.0 kip·ft (simple-span value)
991.9 kip·ft (cantilever value)

Reference: FE Reference Handbook — Structural Analysis → Plane Frame

Self-check

Answer these without notes before moving on.

Without looking, state the relation on this page whose left-hand side is the quantity most often requested, and name every symbol in it.
Which assumption, if violated, makes the main relation of this section invalid?
Given a determinate or one-degree indeterminate structure, what is the first quantity you would compute, and why that one first?
Which unit conversion in this subject most often produces a wrong answer choice, and what is its numerical factor?
Rework Example 1 above from the givens alone, without reading the solution lines.

Chapter summary

Plane Frame contains 1 relation; you must be able to find this page in under 15 seconds.
Exam style: reactions, an influence-line ordinate, or one deflection.
Unit rule: keep kip-ft consistently; EI in kip-in² needs a 1,728 factor.
Work the 10 examples until the solution path, not the answer, is automatic.

Common traps in this section

keep kip-ft consistently; EI in kip-in² needs a 1,728 factor
Answering the intermediate quantity instead of the quantity requested.
Rounding intermediate values before the final step.
Using a relation from an adjacent handbook section that shares a symbol.
Skipping the sketch — most lost points on this page start with a misread geometry.

Plane Frame

Learning objectives

Lecture

Theory, developed

Notation used in this section

Handbook notes for this section

Core formulas for this FE topic

3m+r=3j+cStableandstaticallydeterminate

Worked exam-style examples

Self-check

Chapter summary

Common traps in this section