As concrete dries, it shrinks-the higher the water content, the greater the shrinkage. If a concrete beam has a water content of w kg/m3, then it will shrink by a factor

S =

0.032w - 2.5

10,000

where S is the fraction of the original beam length that disappears due to shrinkage.

(a) A beam 12.027 m long is cast in concrete that contains 550 kg/m3 water. What is the shrinkage factor S?

How long will the beam be when it has dried? (Round your answer to three decimal places.)

(b) A beam is 12.047 m long when wet. We want it to shrink to 12.033 m, so the shrinkage factor should be

S = 0.0012.

What water content will provide this amount of shrinkage? (Round your answer to the nearest whole number.)

Question

Brylee

Mathematics

4 January, 07:27

As concrete dries, it shrinks-the higher the water content, the greater the shrinkage. If a concrete beam has a water content of w kg/m3, then it will shrink by a factor

S =

0.032w - 2.5

10,000

where S is the fraction of the original beam length that disappears due to shrinkage.

(a) A beam 12.027 m long is cast in concrete that contains 550 kg/m3 water. What is the shrinkage factor S?

How long will the beam be when it has dried? (Round your answer to three decimal places.)

(b) A beam is 12.047 m long when wet. We want it to shrink to 12.033 m, so the shrinkage factor should be

S = 0.0012.

What water content will provide this amount of shrinkage? (Round your answer to the nearest whole number.)

+5

Answers (1)

Know the Answer?

Harrison Charles · Answer 1

A.) Here, w = 550kg/m^3.

Therefore, S = (0.032 (550) - 2.5) / 10000 = 0.00151

After drying, the beam will be (1 - 0.00151) x 12.027 = 12.009

b.) 0.0012 = (0.032w - 2.5) / 10000

0.032w - 2.5 = 12

0.032w = 12 + 2.5 = 14.5

w = 14.5/0.032 = 453 kg/m^3

Therefore, the required water content is 453 kg/m^3