All-In-One End of Chapter Problem 104-003

[DanHalen]

First let me state that I'm using the All-In-One Civil Engineering PE Breadth & Depth Exam Guide, 2nd Ed. I am reviewing the end of chapter problems and run across a little snag on Prob. 104-003. The problem states:

The table below shows characteristics of a concrete mix. If the total unit weight of the concrete is 142 lb/ft3, what is the air content (percent)?

Component

Weight (lbs)

SG

Volume(ft³)

Cement

450

3.15

2.2894

Sand

870

2.65

5.2612

Stone Aggregate

1106

2.50

*6.0897*

Water

205

1.0

3.2853

Total

2206

16.9256

How do they get 6.0897? When you run the numbers for V = (1106)/(2.50*62.4) = 7.0897 not 6.0897. Is this a typo?

 

[civilized_naah]

Yes, you are right. The correct numbers are:

Volume of coarse aggregate = 7.0897 cu ft

Total weight = 2631 lb

Total volume = 2631/142 = 18.5282 cu ft

Volume of cement + FA + CA + water = 17.9256 cu ft

Air volume = 0.6025 cu ft

Air content = 0.6025/18.5282 = 3.25%