What was the weight of concrete in pounds per cubic foot?

Question

A batch of concrete consisted of $200$ lbs. fine aggregate, $350$ lbs coarse aggregate, $94$ lbs cement, and $5$ gallons water. The specific gravity of the sand and gravel may be taken as $2.65$ and that of the cement as $3.10$. What was the weight of concrete in place per cubic foot?

It shows here that the answer is $153$ lbs. But how?

Here is what I did: $$\frac{(94 \cdot 3.1)}{(550 \cdot 2.65)+(5 \cdot 7.481 \cdot 62.4)+(94 \cdot 3.1)} = \frac{291.4}{1457.5+41.71+291.4}$$ I was thinking about getting it proportions.

I also know that $density=\frac{mass}{volume}$. However, when I equate volume to be $1ft^3$, then the density of the concrete to be specific gravity $x$ water, I get an answer of 193.44 lbs. Please show steps.

Answer 1

I am puzzled as to why you have the weight of cement only in the numerator as well as adding it to the other weights in the denominator. To find the "weight per cubic foot" of concrete you need to divide the total weight of concrete by its volume. The total weight is 350+ 94+ 40= 484 pounds (for the water, 5 gallons is 40 pints and "a pint's a pound the world around"). Now use the specific gravities to determine the volume of each part and add them to find the total volume.

Answer 2

Note that

• volume $V=5\cdot 0.13+\frac{550}{2.65\cdot 62.4}+\frac{94}{3.10\cdot 62.4}=4.46 \,ft^3$

• weight $W=5\cdot 0.13\cdot 62.4+550+94=684.56 \,lbs$

thus

• $\frac W V=153.4 \,lbs/ft^3$

Answer 3

The mass of a batch is presumably about $200+350+94+5\times 8.34$ pounds

The volume of a batch is presumably about $\frac{200}{2.65 \times 8.34}+ \frac{350}{2.65 \times 8.34}+ \frac{94}{3.10 \times 8.34}+5$ US gallons or about $\frac{200}{2.65 \times 62.4}+ \frac{350}{2.65 \times 62.4}+ \frac{94}{3.10 \times 62.4}+ \frac{5}{7.48}$ cubic feet

So a batch weighs about $685.7$ pounds and has a volume of about $4.48$ cubic feet so dividing one by the other gives about $153$ pounds per cubic foot

This would be rather easier in metric units

Answer 4

Use direcly definition of density

$$ \bar {d }= \frac{\text{Total Weight}}{\text{Total Volume}} $$

$$=\dfrac {W_1+ W_2+W_3 +W_4}{ \dfrac{W_1}{d_1} + \dfrac{W_2}{d_2} + \dfrac{W_3}{d_3} +\dfrac{W_4}{d_4} }$$

that would give you after needed plug-ins as $ 153 \dfrac{\text{lbs}}{\text {ft^3}}.$