Concrete

Materials Science · FE Reference Handbook section

Materials Science

0 formulas

10 exam-style examples

~45 min

All Materials Science lectures

Handbook notes for this section

Definitions and conditions exactly as the handbook states them.

Concrete strength decreases with increases in water-cement
ratio for concrete with and without entrained air.
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
6,000 IN AIR AFTER 28 DAYS CONTINUOUSLY MOIST CURED

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

$Cement = 668 lb/yd^{3}$
$w/c = 0.59$
$Water = 8.34 lb/gal$

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
$w/c = mass of water / mass of cement$
Rearrange
$water = (w/c) \times cement$
Substituting
$water = 0.59 \times 668 = 394.1 lb/yd^{3}$
Convert
$394.1/8.34 = 47.26 gal/yd^{3}$
Trend check
$at w/c = 0.59 the expected 28-day strength is on the order of 39 psi$

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

$Cement = 649 lb/yd^{3}$
$w/c = 0.60$
$Water = 8.34 lb/gal$

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
$w/c = mass of water / mass of cement$
Rearrange
$water = (w/c) \times cement$
Substituting
$water = 0.60 \times 649 = 389.4 lb/yd^{3}$
Convert
$389.4/8.34 = 46.69 gal/yd^{3}$
Trend check
$at w/c = 0.60 the expected 28-day strength is on the order of 34 psi$

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

$Cement = 630 lb/yd^{3}$
$w/c = 0.57$
$Water = 8.34 lb/gal$

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
$w/c = mass of water / mass of cement$
Rearrange
$water = (w/c) \times cement$
Substituting
$water = 0.57 \times 630 = 359.1 lb/yd^{3}$
Convert
$359.1/8.34 = 43.06 gal/yd^{3}$
Trend check
$at w/c = 0.57 the expected 28-day strength is on the order of 52 psi$

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

$Cement = 716 lb/yd^{3}$
$w/c = 0.58$
$Water = 8.34 lb/gal$

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
$w/c = mass of water / mass of cement$
Rearrange
$water = (w/c) \times cement$
Substituting
$water = 0.58 \times 716 = 415.3 lb/yd^{3}$
Convert
$415.3/8.34 = 49.79 gal/yd^{3}$
Trend check
$at w/c = 0.58 the expected 28-day strength is on the order of 45 psi$

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

$Cement = 653 lb/yd^{3}$
$w/c = 0.54$
$Water = 8.34 lb/gal$

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
$w/c = mass of water / mass of cement$
Rearrange
$water = (w/c) \times cement$
Substituting
$water = 0.54 \times 653 = 352.6 lb/yd^{3}$
Convert
$352.6/8.34 = 42.28 gal/yd^{3}$
Trend check
$at w/c = 0.54 the expected 28-day strength is on the order of 79 psi$

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

$Cement = 519 lb/yd^{3}$
$w/c = 0.46$
$Water = 8.34 lb/gal$

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
$w/c = mass of water / mass of cement$
Rearrange
$water = (w/c) \times cement$
Substituting
$water = 0.46 \times 519 = 238.7 lb/yd^{3}$
Convert
$238.7/8.34 = 28.63 gal/yd^{3}$
Trend check
$at w/c = 0.46 the expected 28-day strength is on the order of 238.0 psi$

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

$Cement = 658 lb/yd^{3}$
$w/c = 0.54$
$Water = 8.34 lb/gal$

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
$w/c = mass of water / mass of cement$
Rearrange
$water = (w/c) \times cement$
Substituting
$water = 0.54 \times 658 = 355.3 lb/yd^{3}$
Convert
$355.3/8.34 = 42.60 gal/yd^{3}$
Trend check
$at w/c = 0.54 the expected 28-day strength is on the order of 79 psi$

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

$Cement = 650 lb/yd^{3}$
$w/c = 0.42$
$Water = 8.34 lb/gal$

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
$w/c = mass of water / mass of cement$
Rearrange
$water = (w/c) \times cement$
Substituting
$water = 0.42 \times 650 = 273.0 lb/yd^{3}$
Convert
$273.0/8.34 = 32.73 gal/yd^{3}$
Trend check
$at w/c = 0.42 the expected 28-day strength is on the order of 414.5 psi$

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

$Cement = 506 lb/yd^{3}$
$w/c = 0.59$
$Water = 8.34 lb/gal$

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
$w/c = mass of water / mass of cement$
Rearrange
$water = (w/c) \times cement$
Substituting
$water = 0.59 \times 506 = 298.5 lb/yd^{3}$
Convert
$298.5/8.34 = 35.80 gal/yd^{3}$
Trend check
$at w/c = 0.59 the expected 28-day strength is on the order of 39 psi$

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

$Cement = 578 lb/yd^{3}$
$w/c = 0.41$
$Water = 8.34 lb/gal$

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
$w/c = mass of water / mass of cement$
Rearrange
$water = (w/c) \times cement$
Substituting
$water = 0.41 \times 578 = 237.0 lb/yd^{3}$
Convert
$237.0/8.34 = 28.41 gal/yd^{3}$
Trend check
$at w/c = 0.41 the expected 28-day strength is on the order of 476.1 psi$

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