Member Fixed-End Moments (Magnitudes)

Structural Analysis · FE Reference Handbook section

Structural Analysis

2 formulas

10 exam-style examples

~49 min

All Structural Analysis lectures

Learning objectives

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

This chapter section covers Member Fixed-End Moments (Magnitudes) 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 member fixed-end moments (magnitudes) 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 kip-ft consistently; EI in kip-in² needs a 1,728 factor.

Lecture

Why this section exists. Member Fixed-End Moments (Magnitudes) 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: member fixed-end moments (magnitudes).
Wikimedia Commons, CC BY-SA 4.0

Structural Analysis — Member Fixed-End Moments (Magnitudes): 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 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.

Handbook notes for this section

Definitions and conditions exactly as the handbook states them.

A B
a b
A B
Pa b2 P a2 b
L2 L2

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

Maximum shear and moment under a uniform load — Member Fixed-End Moments (Magnitudes)

A simply supported beam spans 26 ft under a uniform load of 4.50 kip/ft. Find the maximum shear and moment.

Given

w = 4.50 kip/ft
L = 26 ft

Find

V_max and M_max

Start with the thinking

Maximum shear sits at the supports, maximum moment at midspan.
wL²/8 is worth memorising.

Figure for Maximum shear and moment under a uniform load — Member Fixed-End Moments (Magnitudes)

Step-by-step solution

Reactions
$R = w L /2$
Substituting
$R = 4.50 (26) /2 = 58.50 k i p = V_{m} a x$
Midspan moment
$M = w L^{2} /8$
Substituting — M = 4.50(26)²/8 = 380.3 kip·ft

Answer: V_max = 58.50 kip; M_max = 380.3 kip·ft

Why the other options are there

1,521 kip·ft (cantilever formula used)
117.0 kip (total load reported as shear)

Reference: FE Reference Handbook — Structural Analysis → Member Fixed-End Moments (Magnitudes)

Example 2

Maximum shear and moment under a uniform load — Member Fixed-End Moments (Magnitudes) (2)

A simply supported beam spans 32 ft under a uniform load of 4.75 kip/ft. Find the maximum shear and moment.

Given

w = 4.75 kip/ft
L = 32 ft

Find

V_max and M_max

Start with the thinking

Maximum shear sits at the supports, maximum moment at midspan.
wL²/8 is worth memorising.

Figure for Maximum shear and moment under a uniform load — Member Fixed-End Moments (Magnitudes) (2)

Step-by-step solution

Reactions
$R = w L /2$
Substituting
$R = 4.75 (32) /2 = 76.00 k i p = V_{m} a x$
Midspan moment
$M = w L^{2} /8$
Substituting — M = 4.75(32)²/8 = 608.0 kip·ft

Answer: V_max = 76.00 kip; M_max = 608.0 kip·ft

Why the other options are there

2,432 kip·ft (cantilever formula used)
152.0 kip (total load reported as shear)

Reference: FE Reference Handbook — Structural Analysis → Member Fixed-End Moments (Magnitudes)

Example 3

Maximum shear and moment under a uniform load — Member Fixed-End Moments (Magnitudes) (3)

A simply supported beam spans 22 ft under a uniform load of 4.25 kip/ft. Find the maximum shear and moment.

Given

w = 4.25 kip/ft
L = 22 ft

Find

V_max and M_max

Start with the thinking

Maximum shear sits at the supports, maximum moment at midspan.
wL²/8 is worth memorising.

Figure for Maximum shear and moment under a uniform load — Member Fixed-End Moments (Magnitudes) (3)

Step-by-step solution

Reactions
$R = w L /2$
Substituting
$R = 4.25 (22) /2 = 46.75 k i p = V_{m} a x$
Midspan moment
$M = w L^{2} /8$
Substituting — M = 4.25(22)²/8 = 257.1 kip·ft

Answer: V_max = 46.75 kip; M_max = 257.1 kip·ft

Why the other options are there

1,029 kip·ft (cantilever formula used)
93.50 kip (total load reported as shear)

Reference: FE Reference Handbook — Structural Analysis → Member Fixed-End Moments (Magnitudes)

Example 4

Maximum shear and moment under a uniform load — Member Fixed-End Moments (Magnitudes) (4)

A simply supported beam spans 36 ft under a uniform load of 5.00 kip/ft. Find the maximum shear and moment.

Given

w = 5.00 kip/ft
L = 36 ft

Find

V_max and M_max

Start with the thinking

Maximum shear sits at the supports, maximum moment at midspan.
wL²/8 is worth memorising.

Figure for Maximum shear and moment under a uniform load — Member Fixed-End Moments (Magnitudes) (4)

Step-by-step solution

Reactions
$R = w L /2$
Substituting
$R = 5.00 (36) /2 = 90.00 k i p = V_{m} a x$
Midspan moment
$M = w L^{2} /8$
Substituting — M = 5.00(36)²/8 = 810.0 kip·ft

Answer: V_max = 90.00 kip; M_max = 810.0 kip·ft

Why the other options are there

3,240 kip·ft (cantilever formula used)
180.0 kip (total load reported as shear)

Reference: FE Reference Handbook — Structural Analysis → Member Fixed-End Moments (Magnitudes)

Example 5

Maximum shear and moment under a uniform load — Member Fixed-End Moments (Magnitudes) (5)

A simply supported beam spans 18 ft under a uniform load of 4.25 kip/ft. Find the maximum shear and moment.

Given

w = 4.25 kip/ft
L = 18 ft

Find

V_max and M_max

Start with the thinking

Maximum shear sits at the supports, maximum moment at midspan.
wL²/8 is worth memorising.

Figure for Maximum shear and moment under a uniform load — Member Fixed-End Moments (Magnitudes) (5)

Step-by-step solution

Reactions
$R = w L /2$
Substituting
$R = 4.25 (18) /2 = 38.25 k i p = V_{m} a x$
Midspan moment
$M = w L^{2} /8$
Substituting — M = 4.25(18)²/8 = 172.1 kip·ft

Answer: V_max = 38.25 kip; M_max = 172.1 kip·ft

Why the other options are there

688.5 kip·ft (cantilever formula used)
76.50 kip (total load reported as shear)

Reference: FE Reference Handbook — Structural Analysis → Member Fixed-End Moments (Magnitudes)

Example 6

Maximum shear and moment under a uniform load — Member Fixed-End Moments (Magnitudes) (6)

A simply supported beam spans 29 ft under a uniform load of 1.50 kip/ft. Find the maximum shear and moment.

Given

w = 1.50 kip/ft
L = 29 ft

Find

V_max and M_max

Start with the thinking

Maximum shear sits at the supports, maximum moment at midspan.
wL²/8 is worth memorising.

Figure for Maximum shear and moment under a uniform load — Member Fixed-End Moments (Magnitudes) (6)

Step-by-step solution

Reactions
$R = w L /2$
Substituting
$R = 1.50 (29) /2 = 21.75 k i p = V_{m} a x$
Midspan moment
$M = w L^{2} /8$
Substituting — M = 1.50(29)²/8 = 157.7 kip·ft

Answer: V_max = 21.75 kip; M_max = 157.7 kip·ft

Why the other options are there

630.8 kip·ft (cantilever formula used)
43.50 kip (total load reported as shear)

Reference: FE Reference Handbook — Structural Analysis → Member Fixed-End Moments (Magnitudes)

Example 7

Maximum shear and moment under a uniform load — Member Fixed-End Moments (Magnitudes) (7)

A simply supported beam spans 30 ft under a uniform load of 2.75 kip/ft. Find the maximum shear and moment.

Given

w = 2.75 kip/ft
L = 30 ft

Find

V_max and M_max

Start with the thinking

Maximum shear sits at the supports, maximum moment at midspan.
wL²/8 is worth memorising.

Figure for Maximum shear and moment under a uniform load — Member Fixed-End Moments (Magnitudes) (7)

Step-by-step solution

Reactions
$R = w L /2$
Substituting
$R = 2.75 (30) /2 = 41.25 k i p = V_{m} a x$
Midspan moment
$M = w L^{2} /8$
Substituting — M = 2.75(30)²/8 = 309.4 kip·ft

Answer: V_max = 41.25 kip; M_max = 309.4 kip·ft

Why the other options are there

1,238 kip·ft (cantilever formula used)
82.50 kip (total load reported as shear)

Reference: FE Reference Handbook — Structural Analysis → Member Fixed-End Moments (Magnitudes)

Example 8

Maximum shear and moment under a uniform load — Member Fixed-End Moments (Magnitudes) (8)

A simply supported beam spans 43 ft under a uniform load of 2.50 kip/ft. Find the maximum shear and moment.

Given

w = 2.50 kip/ft
L = 43 ft

Find

V_max and M_max

Start with the thinking

Maximum shear sits at the supports, maximum moment at midspan.
wL²/8 is worth memorising.

Figure for Maximum shear and moment under a uniform load — Member Fixed-End Moments (Magnitudes) (8)

Step-by-step solution

Reactions
$R = w L /2$
Substituting
$R = 2.50 (43) /2 = 53.75 k i p = V_{m} a x$
Midspan moment
$M = w L^{2} /8$
Substituting — M = 2.50(43)²/8 = 577.8 kip·ft

Answer: V_max = 53.75 kip; M_max = 577.8 kip·ft

Why the other options are there

2,311 kip·ft (cantilever formula used)
107.5 kip (total load reported as shear)

Reference: FE Reference Handbook — Structural Analysis → Member Fixed-End Moments (Magnitudes)

Example 9

Maximum shear and moment under a uniform load — Member Fixed-End Moments (Magnitudes) (9)

A simply supported beam spans 19 ft under a uniform load of 2.50 kip/ft. Find the maximum shear and moment.

Given

w = 2.50 kip/ft
L = 19 ft

Find

V_max and M_max

Start with the thinking

Maximum shear sits at the supports, maximum moment at midspan.
wL²/8 is worth memorising.

Figure for Maximum shear and moment under a uniform load — Member Fixed-End Moments (Magnitudes) (9)

Step-by-step solution

Reactions
$R = w L /2$
Substituting
$R = 2.50 (19) /2 = 23.75 k i p = V_{m} a x$
Midspan moment
$M = w L^{2} /8$
Substituting — M = 2.50(19)²/8 = 112.8 kip·ft

Answer: V_max = 23.75 kip; M_max = 112.8 kip·ft

Why the other options are there

451.3 kip·ft (cantilever formula used)
47.50 kip (total load reported as shear)

Reference: FE Reference Handbook — Structural Analysis → Member Fixed-End Moments (Magnitudes)

Example 10

Maximum shear and moment under a uniform load — Member Fixed-End Moments (Magnitudes) (10)

A simply supported beam spans 36 ft under a uniform load of 5.00 kip/ft. Find the maximum shear and moment.

Given

w = 5.00 kip/ft
L = 36 ft

Find

V_max and M_max

Start with the thinking

Maximum shear sits at the supports, maximum moment at midspan.
wL²/8 is worth memorising.

Figure for Maximum shear and moment under a uniform load — Member Fixed-End Moments (Magnitudes) (10)

Step-by-step solution

Reactions
$R = w L /2$
Substituting
$R = 5.00 (36) /2 = 90.00 k i p = V_{m} a x$
Midspan moment
$M = w L^{2} /8$
Substituting — M = 5.00(36)²/8 = 810.0 kip·ft

Answer: V_max = 90.00 kip; M_max = 810.0 kip·ft

Why the other options are there

3,240 kip·ft (cantilever formula used)
180.0 kip (total load reported as shear)

Reference: FE Reference Handbook — Structural Analysis → Member Fixed-End Moments (Magnitudes)

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

Member Fixed-End Moments (Magnitudes) contains 2 relations; 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.