Concrete

Materials Science · FE Reference Handbook section

Materials Science

0 formulas

10 exam-style examples

~45 min

All Materials Science lectures

Learning objectives

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

This chapter section covers Concrete within Materials Science. 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 concrete describes physically and when it applies.
State every one of the 0 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: percent versus fraction in composition and strain.

Lecture

Why this section exists. Concrete is the part of Materials Science that lets you connect a steel, concrete or polymer specimen under test 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 a definition, a phase-diagram read, or a one-line property calculation. 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. percent versus fraction in composition and strain. 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.

Reinforcing steel and formwork placed in an excavated bridge footing. — Photo 1. Where this shows up in practice: concrete.
Wikimedia Commons, public domain

Materials Science — Concrete: 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 steel, concrete or polymer specimen under test. 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 0 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.

AVERAGE 28-DAY COMPRESSIVE
8,000
6,000
NO ADDED AIR
STRENGTH, PSI
4,000
RECOMMENDED
2,000 PERCENT
ENTRAINED AIR
1,000
0.40 0.60 0.80 1.00
W/C BY WEIGHT
Concrete strength decreases with increases in water-cement
ratio for concrete with and without entrained air.
Concrete Manual, 8th ed., U.S. Bureau of Reclamation, 1975.
Water-cement (W/C) ratio is the primary factor affecting the strength of concrete. The figure above shows how W/C expressed
as a ratio of weight of water and cement by weight of concrete mix affects the compressive strength of both air-entrained and
non-air-entrained concrete.
6,000 IN AIR AFTER 28 DAYS CONTINUOUSLY MOIST CURED
COMPRESSIVE STRENGTH, PSI
IN AIR AFTER 14 DAYS
5,000 IN AIR AFTER 7 DAYS
IN AIR AFTER 3 DAYS
4,000

Core formulas for this FE topic

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

This section is conceptual; there are no equations to memorise.

Worked exam-style examples

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

Example 1

Water–cement ratio and mix water — Concrete

A mix uses 668 lb of cement per cubic yard at w/c = 0.59. How much mix water is needed per cubic yard, in pounds and gallons?

Given

Cement = 668 lb/yd³
w/c = 0.59
Water = 8.34 lb/gal

Find

Water content

Start with the thinking

The w/c ratio is by mass, not volume.
Lower w/c means higher strength and lower workability.

Step-by-step solution

Definition
$w / c = ma sso f w a t er / ma sso f ce m e n t$
Rearrange
$w a t er = (w / c) \times ce m e n t$
Substituting
$w a t er = 0.59 \times 668 = 394.1 l b / y d^{3}$
Convert
$394.1/8.34 = 47.26 g a l / y d^{3}$
Trend check
$a tw / c = 0.59 t h ee x p ec t e d 28 - d a y s t r e n g t hi so n t h eor d er o f 39 p s i$

Answer: ≈ 394.1 lb (47.3 gal) per yd³

Why the other options are there

1,132 lb (ratio inverted)
14.6 lb (volume conversion misapplied)

Reference: FE Reference Handbook — Materials Science → Concrete

Example 2

Water–cement ratio and mix water — Concrete (2)

A mix uses 649 lb of cement per cubic yard at w/c = 0.60. How much mix water is needed per cubic yard, in pounds and gallons?

Given

Cement = 649 lb/yd³
w/c = 0.60
Water = 8.34 lb/gal

Find

Water content

Start with the thinking

The w/c ratio is by mass, not volume.
Lower w/c means higher strength and lower workability.

Step-by-step solution

Definition
$w / c = ma sso f w a t er / ma sso f ce m e n t$
Rearrange
$w a t er = (w / c) \times ce m e n t$
Substituting
$w a t er = 0.60 \times 649 = 389.4 l b / y d^{3}$
Convert
$389.4/8.34 = 46.69 g a l / y d^{3}$
Trend check
$a tw / c = 0.60 t h ee x p ec t e d 28 - d a y s t r e n g t hi so n t h eor d er o f 34 p s i$

Answer: ≈ 389.4 lb (46.7 gal) per yd³

Why the other options are there

1,082 lb (ratio inverted)
14.4 lb (volume conversion misapplied)

Reference: FE Reference Handbook — Materials Science → Concrete

Example 3

Water–cement ratio and mix water — Concrete (3)

A mix uses 630 lb of cement per cubic yard at w/c = 0.57. How much mix water is needed per cubic yard, in pounds and gallons?

Given

Cement = 630 lb/yd³
w/c = 0.57
Water = 8.34 lb/gal

Find

Water content

Start with the thinking

The w/c ratio is by mass, not volume.
Lower w/c means higher strength and lower workability.

Step-by-step solution

Definition
$w / c = ma sso f w a t er / ma sso f ce m e n t$
Rearrange
$w a t er = (w / c) \times ce m e n t$
Substituting
$w a t er = 0.57 \times 630 = 359.1 l b / y d^{3}$
Convert
$359.1/8.34 = 43.06 g a l / y d^{3}$
Trend check
$a tw / c = 0.57 t h ee x p ec t e d 28 - d a y s t r e n g t hi so n t h eor d er o f 52 p s i$

Answer: ≈ 359.1 lb (43.1 gal) per yd³

Why the other options are there

1,105 lb (ratio inverted)
13.3 lb (volume conversion misapplied)

Reference: FE Reference Handbook — Materials Science → Concrete

Example 4

Water–cement ratio and mix water — Concrete (4)

A mix uses 716 lb of cement per cubic yard at w/c = 0.58. How much mix water is needed per cubic yard, in pounds and gallons?

Given

Cement = 716 lb/yd³
w/c = 0.58
Water = 8.34 lb/gal

Find

Water content

Start with the thinking

The w/c ratio is by mass, not volume.
Lower w/c means higher strength and lower workability.

Step-by-step solution

Definition
$w / c = ma sso f w a t er / ma sso f ce m e n t$
Rearrange
$w a t er = (w / c) \times ce m e n t$
Substituting
$w a t er = 0.58 \times 716 = 415.3 l b / y d^{3}$
Convert
$415.3/8.34 = 49.79 g a l / y d^{3}$
Trend check
$a tw / c = 0.58 t h ee x p ec t e d 28 - d a y s t r e n g t hi so n t h eor d er o f 45 p s i$

Answer: ≈ 415.3 lb (49.8 gal) per yd³

Why the other options are there

1,234 lb (ratio inverted)
15.4 lb (volume conversion misapplied)

Reference: FE Reference Handbook — Materials Science → Concrete

Example 5

Water–cement ratio and mix water — Concrete (5)

A mix uses 653 lb of cement per cubic yard at w/c = 0.54. How much mix water is needed per cubic yard, in pounds and gallons?

Given

Cement = 653 lb/yd³
w/c = 0.54
Water = 8.34 lb/gal

Find

Water content

Start with the thinking

The w/c ratio is by mass, not volume.
Lower w/c means higher strength and lower workability.

Step-by-step solution

Definition
$w / c = ma sso f w a t er / ma sso f ce m e n t$
Rearrange
$w a t er = (w / c) \times ce m e n t$
Substituting
$w a t er = 0.54 \times 653 = 352.6 l b / y d^{3}$
Convert
$352.6/8.34 = 42.28 g a l / y d^{3}$
Trend check
$a tw / c = 0.54 t h ee x p ec t e d 28 - d a y s t r e n g t hi so n t h eor d er o f 79 p s i$

Answer: ≈ 352.6 lb (42.3 gal) per yd³

Why the other options are there

1,209 lb (ratio inverted)
13.1 lb (volume conversion misapplied)

Reference: FE Reference Handbook — Materials Science → Concrete

Example 6

Water–cement ratio and mix water — Concrete (6)

A mix uses 519 lb of cement per cubic yard at w/c = 0.46. How much mix water is needed per cubic yard, in pounds and gallons?

Given

Cement = 519 lb/yd³
w/c = 0.46
Water = 8.34 lb/gal

Find

Water content

Start with the thinking

The w/c ratio is by mass, not volume.
Lower w/c means higher strength and lower workability.

Step-by-step solution

Definition
$w / c = ma sso f w a t er / ma sso f ce m e n t$
Rearrange
$w a t er = (w / c) \times ce m e n t$
Substituting
$w a t er = 0.46 \times 519 = 238.7 l b / y d^{3}$
Convert
$238.7/8.34 = 28.63 g a l / y d^{3}$
Trend check
$a tw / c = 0.46 t h ee x p ec t e d 28 - d a y s t r e n g t hi so n t h eor d er o f 238.0 p s i$

Answer: ≈ 238.7 lb (28.6 gal) per yd³

Why the other options are there

1,128 lb (ratio inverted)
8.8 lb (volume conversion misapplied)

Reference: FE Reference Handbook — Materials Science → Concrete

Example 7

Water–cement ratio and mix water — Concrete (7)

A mix uses 658 lb of cement per cubic yard at w/c = 0.54. How much mix water is needed per cubic yard, in pounds and gallons?

Given

Cement = 658 lb/yd³
w/c = 0.54
Water = 8.34 lb/gal

Find

Water content

Start with the thinking

The w/c ratio is by mass, not volume.
Lower w/c means higher strength and lower workability.

Step-by-step solution

Definition
$w / c = ma sso f w a t er / ma sso f ce m e n t$
Rearrange
$w a t er = (w / c) \times ce m e n t$
Substituting
$w a t er = 0.54 \times 658 = 355.3 l b / y d^{3}$
Convert
$355.3/8.34 = 42.60 g a l / y d^{3}$
Trend check
$a tw / c = 0.54 t h ee x p ec t e d 28 - d a y s t r e n g t hi so n t h eor d er o f 79 p s i$

Answer: ≈ 355.3 lb (42.6 gal) per yd³

Why the other options are there

1,219 lb (ratio inverted)
13.2 lb (volume conversion misapplied)

Reference: FE Reference Handbook — Materials Science → Concrete

Example 8

Water–cement ratio and mix water — Concrete (8)

A mix uses 650 lb of cement per cubic yard at w/c = 0.42. How much mix water is needed per cubic yard, in pounds and gallons?

Given

Cement = 650 lb/yd³
w/c = 0.42
Water = 8.34 lb/gal

Find

Water content

Start with the thinking

The w/c ratio is by mass, not volume.
Lower w/c means higher strength and lower workability.

Step-by-step solution

Definition
$w / c = ma sso f w a t er / ma sso f ce m e n t$
Rearrange
$w a t er = (w / c) \times ce m e n t$
Substituting
$w a t er = 0.42 \times 650 = 273.0 l b / y d^{3}$
Convert
$273.0/8.34 = 32.73 g a l / y d^{3}$
Trend check
$a tw / c = 0.42 t h ee x p ec t e d 28 - d a y s t r e n g t hi so n t h eor d er o f 414.5 p s i$

Answer: ≈ 273.0 lb (32.7 gal) per yd³

Why the other options are there

1,548 lb (ratio inverted)
10.1 lb (volume conversion misapplied)

Reference: FE Reference Handbook — Materials Science → Concrete

Example 9

Water–cement ratio and mix water — Concrete (9)

A mix uses 506 lb of cement per cubic yard at w/c = 0.59. How much mix water is needed per cubic yard, in pounds and gallons?

Given

Cement = 506 lb/yd³
w/c = 0.59
Water = 8.34 lb/gal

Find

Water content

Start with the thinking

The w/c ratio is by mass, not volume.
Lower w/c means higher strength and lower workability.

Step-by-step solution

Definition
$w / c = ma sso f w a t er / ma sso f ce m e n t$
Rearrange
$w a t er = (w / c) \times ce m e n t$
Substituting
$w a t er = 0.59 \times 506 = 298.5 l b / y d^{3}$
Convert
$298.5/8.34 = 35.80 g a l / y d^{3}$
Trend check
$a tw / c = 0.59 t h ee x p ec t e d 28 - d a y s t r e n g t hi so n t h eor d er o f 39 p s i$

Answer: ≈ 298.5 lb (35.8 gal) per yd³

Why the other options are there

857.6 lb (ratio inverted)
11.1 lb (volume conversion misapplied)

Reference: FE Reference Handbook — Materials Science → Concrete

Example 10

Water–cement ratio and mix water — Concrete (10)

A mix uses 578 lb of cement per cubic yard at w/c = 0.41. How much mix water is needed per cubic yard, in pounds and gallons?

Given

Cement = 578 lb/yd³
w/c = 0.41
Water = 8.34 lb/gal

Find

Water content

Start with the thinking

The w/c ratio is by mass, not volume.
Lower w/c means higher strength and lower workability.

Step-by-step solution

Definition
$w / c = ma sso f w a t er / ma sso f ce m e n t$
Rearrange
$w a t er = (w / c) \times ce m e n t$
Substituting
$w a t er = 0.41 \times 578 = 237.0 l b / y d^{3}$
Convert
$237.0/8.34 = 28.41 g a l / y d^{3}$
Trend check
$a tw / c = 0.41 t h ee x p ec t e d 28 - d a y s t r e n g t hi so n t h eor d er o f 476.1 p s i$

Answer: ≈ 237.0 lb (28.4 gal) per yd³

Why the other options are there

1,410 lb (ratio inverted)
8.8 lb (volume conversion misapplied)

Reference: FE Reference Handbook — Materials Science → Concrete

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 steel, concrete or polymer specimen under test, 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

Concrete contains 0 relations; you must be able to find this page in under 15 seconds.
Exam style: a definition, a phase-diagram read, or a one-line property calculation.
Unit rule: percent versus fraction in composition and strain.
Work the 10 examples until the solution path, not the answer, is automatic.

Common traps in this section

percent versus fraction in composition and strain
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.