Equilibrium Requirements

Statics · FE Reference Handbook section

Statics

2 formulas

10 exam-style examples

~49 min

All Statics lectures

Learning objectives

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

This chapter section covers Equilibrium Requirements within Statics. 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 equilibrium requirements describes physically and when it applies.
State every one of the 2 relations 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 force in lbf or kN and distance in ft or m consistently.

Lecture

Why this section exists. Equilibrium Requirements is the part of Statics that lets you connect a determinate frame, truss or beam in equilibrium 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 one free body, three equilibrium equations, one unknown reported. 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 force in lbf or kN and distance in ft or m consistently. 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.

Crane lowering a steel plate girder onto bridge bearings while ironworkers guide it. — Photo 1. Where this shows up in practice: equilibrium requirements.
Capstone Studio instructional photograph

Statics — Equilibrium Requirements: 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 frame, truss or beam in equilibrium. 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 2 relations 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

Σ Fn	Quantity produced by "Σ Fn = 0" — read its definition and unit from the handbook line directly above the equation.
Σ Mn	Quantity produced by "Σ Mn = 0" — read its definition and unit from the handbook line directly above the equation.

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

Reactions on a simply supported beam

A 10 m simply supported beam carries a 40 kN point load 3 m from A and a 12 kN/m uniform load over the full span. Find both reactions.

Given

L = 10 m
P = 40 kN at 3 m from A
w = 12 kN/m over 10 m

Find

R_A and R_B

Start with the thinking

Replace the distributed load with its resultant at midspan.
Sum moments about A so R_A drops out of that equation.

Figure for Reactions on a simply supported beam

Step-by-step solution

UDL resultant
$W = w L = 12 (10) = 120 k N a t 5.00 m$
Moments about A — ΣM_A = 0: R_B(10) = 40(3) + 120(5)
Right side — 120 + 600 = 720 kN·m
Reaction
$R_{B} = 720/10 = 72.0 k N$
Vertical equilibrium
$R_{A} = 40 + 120 - 72.0 = 88.0 k N$

Answer: R_A = 88.0 kN, R_B = 72.0 kN

Why the other options are there

80/80 kN (point load position ignored)
R_A = 72, R_B = 88 (reactions swapped)

Reference: FE Reference Handbook — Statics — Equilibrium of rigid bodies

Example 2

Resultant of two concurrent forces

Forces of 300 N at 0° and 400 N at 90° act at a joint. What single force replaces them?

Given

F₁ = 300 N along x
F₂ = 400 N along y

Find

Magnitude and direction of the resultant

Start with the thinking

Perpendicular components add by Pythagoras.
Direction is measured from the +x axis.

Step-by-step solution

Magnitude
$R = (F x^{2} + F y^{2})$
Substitute
$R = (30 0^{2} + 40 0^{2}) = 250, 000$
Result
$R = 500 N$
Direction
$θ = arctan (400/300) = 53. 1^{\circ} ab o v e t h e x - a x i s$

Answer: R = 500 N at 53.1°

Why the other options are there

700 N (components added arithmetically)
36.9° (ratio inverted)

Reference: FE Reference Handbook — Statics — Resultants of force systems

Example 3

Support reactions on a simple beam — Equilibrium Requirements

A simply supported beam spans 34 ft and carries a 6 kip point load 14 ft from the left support. Find both reactions.

Given

L = 34 ft
P = 6 kip
a = 14 ft from A

Find

Reactions R_A and R_B

Start with the thinking

Sum moments about one support to isolate the other reaction.
Then use vertical equilibrium — never two moment equations.

Figure for Support reactions on a simple beam — Equilibrium Requirements

Step-by-step solution

Moment about A — ΣM_A = 0: R_B·L − P·a = 0
Substituting — R_B = P·a/L = 6(14)/34
Evaluate
$R_{B} = 2.47 k i p$
Vertical equilibrium — ΣF_y = 0: R_A + R_B − P = 0
Substituting
$R_{A} = 6 - 2.47 = 3.53 k i p$
Check — ΣM_B = R_A·L − P(L − a) = 0.00 ≈ 0 ✓

Answer: R_A = 3.53 kip, R_B = 2.47 kip

Why the other options are there

R_A = 2.47 kip (reactions swapped)
R_A = 3.00 kip (load assumed at midspan)

Reference: FE Reference Handbook — Statics → Equilibrium Requirements

Example 4

Support reactions on a simple beam — Equilibrium Requirements (2)

A simply supported beam spans 38 ft and carries a 9 kip point load 17 ft from the left support. Find both reactions.

Given

L = 38 ft
P = 9 kip
a = 17 ft from A

Find

Reactions R_A and R_B

Start with the thinking

Sum moments about one support to isolate the other reaction.
Then use vertical equilibrium — never two moment equations.

Figure for Support reactions on a simple beam — Equilibrium Requirements (2)

Step-by-step solution

Moment about A — ΣM_A = 0: R_B·L − P·a = 0
Substituting — R_B = P·a/L = 9(17)/38
Evaluate
$R_{B} = 4.03 k i p$
Vertical equilibrium — ΣF_y = 0: R_A + R_B − P = 0
Substituting
$R_{A} = 9 - 4.03 = 4.97 k i p$
Check — ΣM_B = R_A·L − P(L − a) = 0.00 ≈ 0 ✓

Answer: R_A = 4.97 kip, R_B = 4.03 kip

Why the other options are there

R_A = 4.03 kip (reactions swapped)
R_A = 4.50 kip (load assumed at midspan)

Reference: FE Reference Handbook — Statics → Equilibrium Requirements

Example 5

Support reactions on a simple beam — Equilibrium Requirements (3)

A simply supported beam spans 21 ft and carries a 27 kip point load 12 ft from the left support. Find both reactions.

Given

L = 21 ft
P = 27 kip
a = 12 ft from A

Find

Reactions R_A and R_B

Start with the thinking

Sum moments about one support to isolate the other reaction.
Then use vertical equilibrium — never two moment equations.

Figure for Support reactions on a simple beam — Equilibrium Requirements (3)

Step-by-step solution

Moment about A — ΣM_A = 0: R_B·L − P·a = 0
Substituting — R_B = P·a/L = 27(12)/21
Evaluate
$R_{B} = 15.43 k i p$
Vertical equilibrium — ΣF_y = 0: R_A + R_B − P = 0
Substituting
$R_{A} = 27 - 15.43 = 11.57 k i p$
Check — ΣM_B = R_A·L − P(L − a) = 0.00 ≈ 0 ✓

Answer: R_A = 11.57 kip, R_B = 15.43 kip

Why the other options are there

R_A = 15.43 kip (reactions swapped)
R_A = 13.50 kip (load assumed at midspan)

Reference: FE Reference Handbook — Statics → Equilibrium Requirements

Example 6

Support reactions on a simple beam — Equilibrium Requirements (4)

A simply supported beam spans 23 ft and carries a 16 kip point load 13 ft from the left support. Find both reactions.

Given

L = 23 ft
P = 16 kip
a = 13 ft from A

Find

Reactions R_A and R_B

Start with the thinking

Sum moments about one support to isolate the other reaction.
Then use vertical equilibrium — never two moment equations.

Figure for Support reactions on a simple beam — Equilibrium Requirements (4)

Step-by-step solution

Moment about A — ΣM_A = 0: R_B·L − P·a = 0
Substituting — R_B = P·a/L = 16(13)/23
Evaluate
$R_{B} = 9.04 k i p$
Vertical equilibrium — ΣF_y = 0: R_A + R_B − P = 0
Substituting
$R_{A} = 16 - 9.04 = 6.96 k i p$
Check — ΣM_B = R_A·L − P(L − a) = 0.00 ≈ 0 ✓

Answer: R_A = 6.96 kip, R_B = 9.04 kip

Why the other options are there

R_A = 9.04 kip (reactions swapped)
R_A = 8.00 kip (load assumed at midspan)

Reference: FE Reference Handbook — Statics → Equilibrium Requirements

Example 7

Support reactions on a simple beam — Equilibrium Requirements (5)

A simply supported beam spans 17 ft and carries a 18 kip point load 7 ft from the left support. Find both reactions.

Given

L = 17 ft
P = 18 kip
a = 7 ft from A

Find

Reactions R_A and R_B

Start with the thinking

Sum moments about one support to isolate the other reaction.
Then use vertical equilibrium — never two moment equations.

Figure for Support reactions on a simple beam — Equilibrium Requirements (5)

Step-by-step solution

Moment about A — ΣM_A = 0: R_B·L − P·a = 0
Substituting — R_B = P·a/L = 18(7)/17
Evaluate
$R_{B} = 7.41 k i p$
Vertical equilibrium — ΣF_y = 0: R_A + R_B − P = 0
Substituting
$R_{A} = 18 - 7.41 = 10.59 k i p$
Check — ΣM_B = R_A·L − P(L − a) = 0.00 ≈ 0 ✓

Answer: R_A = 10.59 kip, R_B = 7.41 kip

Why the other options are there

R_A = 7.41 kip (reactions swapped)
R_A = 9.00 kip (load assumed at midspan)

Reference: FE Reference Handbook — Statics → Equilibrium Requirements

Example 8

Support reactions on a simple beam — Equilibrium Requirements (6)

A simply supported beam spans 17 ft and carries a 19 kip point load 9 ft from the left support. Find both reactions.

Given

L = 17 ft
P = 19 kip
a = 9 ft from A

Find

Reactions R_A and R_B

Start with the thinking

Sum moments about one support to isolate the other reaction.
Then use vertical equilibrium — never two moment equations.

Figure for Support reactions on a simple beam — Equilibrium Requirements (6)

Step-by-step solution

Moment about A — ΣM_A = 0: R_B·L − P·a = 0
Substituting — R_B = P·a/L = 19(9)/17
Evaluate
$R_{B} = 10.06 k i p$
Vertical equilibrium — ΣF_y = 0: R_A + R_B − P = 0
Substituting
$R_{A} = 19 - 10.06 = 8.94 k i p$
Check — ΣM_B = R_A·L − P(L − a) = 0.00 ≈ 0 ✓

Answer: R_A = 8.94 kip, R_B = 10.06 kip

Why the other options are there

R_A = 10.06 kip (reactions swapped)
R_A = 9.50 kip (load assumed at midspan)

Reference: FE Reference Handbook — Statics → Equilibrium Requirements

Example 9

Support reactions on a simple beam — Equilibrium Requirements (7)

A simply supported beam spans 39 ft and carries a 19 kip point load 16 ft from the left support. Find both reactions.

Given

L = 39 ft
P = 19 kip
a = 16 ft from A

Find

Reactions R_A and R_B

Start with the thinking

Sum moments about one support to isolate the other reaction.
Then use vertical equilibrium — never two moment equations.

Figure for Support reactions on a simple beam — Equilibrium Requirements (7)

Step-by-step solution

Moment about A — ΣM_A = 0: R_B·L − P·a = 0
Substituting — R_B = P·a/L = 19(16)/39
Evaluate
$R_{B} = 7.79 k i p$
Vertical equilibrium — ΣF_y = 0: R_A + R_B − P = 0
Substituting
$R_{A} = 19 - 7.79 = 11.21 k i p$
Check — ΣM_B = R_A·L − P(L − a) = 0.00 ≈ 0 ✓

Answer: R_A = 11.21 kip, R_B = 7.79 kip

Why the other options are there

R_A = 7.79 kip (reactions swapped)
R_A = 9.50 kip (load assumed at midspan)

Reference: FE Reference Handbook — Statics → Equilibrium Requirements

Example 10

Support reactions on a simple beam — Equilibrium Requirements (8)

A simply supported beam spans 30 ft and carries a 13 kip point load 11 ft from the left support. Find both reactions.

Given

L = 30 ft
P = 13 kip
a = 11 ft from A

Find

Reactions R_A and R_B

Start with the thinking

Sum moments about one support to isolate the other reaction.
Then use vertical equilibrium — never two moment equations.

Figure for Support reactions on a simple beam — Equilibrium Requirements (8)

Step-by-step solution

Moment about A — ΣM_A = 0: R_B·L − P·a = 0
Substituting — R_B = P·a/L = 13(11)/30
Evaluate
$R_{B} = 4.77 k i p$
Vertical equilibrium — ΣF_y = 0: R_A + R_B − P = 0
Substituting
$R_{A} = 13 - 4.77 = 8.23 k i p$
Check — ΣM_B = R_A·L − P(L − a) = 0.00 ≈ 0 ✓

Answer: R_A = 8.23 kip, R_B = 4.77 kip

Why the other options are there

R_A = 4.77 kip (reactions swapped)
R_A = 6.50 kip (load assumed at midspan)

Reference: FE Reference Handbook — Statics → Equilibrium Requirements

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 frame, truss or beam in equilibrium, 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

Equilibrium Requirements contains 2 relations; you must be able to find this page in under 15 seconds.
Exam style: one free body, three equilibrium equations, one unknown reported.
Unit rule: keep force in lbf or kN and distance in ft or m consistently.
Work the 10 examples until the solution path, not the answer, is automatic.

Common traps in this section

keep force in lbf or kN and distance in ft or m consistently
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.