Welcome
to AffordableConcreteCutting.Net

*“We
Specialize in Cutting Doorways and Windows in Concrete Foundations”*

**Are You in New Castle ****New Hampshire****? Do You
Need Concrete Cutting?**

**We Are Your Local
Concrete Cutter**

**Call 603-622-4441**

**We Service New Castle
NH and all surrounding Cities & Towns**

*“No Travel Charges – Ever! Guaranteed!”*

Concrete
Cutting New Castle NH

Concrete
Cutter New Castle NH

Concrete
Coring New Castle NH

Core
Drilling New Castle NH

Concrete
Sawing New Castle NH

Concrete
Sawing New Castle New Hampshire

Concrete
Cutting New Castle New Hampshire

Concrete
Cutter New Castle New Hampshire

Concrete
Coring New Hampshire

Core
Driller New Castle NH

Core
Drilling New Castle New Hampshire

We
have already determined that: Substituting the above value of Ic in this equation, we have, after considerable reduction:
The above equation shows that we cannot select the percentage of steel at
random, since it evidently depends on the selected stresses for the steel and
concrete, and also on the ratio of their module. For example, consider a
high-grade concrete (1:2:4) whose modulus of elasticity is considered to be
2,900,000, and which has a limiting compressive stress of 2,700 pounds (c'),
which we may consider in conjunction with the limiting stress of 55,000 pounds
in the steel. The values of c, s, and r are therefore 2,700, 55,000, and 10
respectively. Substituting these values in Equation 18, we compute p = .012. What
percentage of steel would be required for ordinary stone concrete, with r = 15,
c =2,000, and s = 55,000 ANS 0.95 percent. The moment which resists the action
of the external forces is evidently measured by the product of the distance
from the center of gravity of the steel to the cancroids of compression of the
concrete, times the total compression of the concrete, or, otherwise, times the
tension in the steel. The compression in the concrete and the tension in the
steel are equal, and it is therefore only a matter of convenience to express
this product in terms of the tension in the steel. Therefore, adopting the
notation already mentioned, we may write the formula: But since the
computations are frequently made in terms of the dimensions of the concrete and
of the percentage of the reinforcing steel, it may be more convenient to write
the equation: From Equation 12 we have the total compression in the concrete.
Multiplying this by the distance from the steel to the cancroids of compression
(d - x), we have another equation for the moment: This equation is perfectly general,
except that it depends on the assumption as to the form of the stress-strain
diagram as described in Article 260. On the assumption that q = 'for ultimate
stresses in the concrete, the equation becomes: When the percentage of steel
used agrees with that computed from Equation 18, then Equations 20 and 22 will
give identically the same results; but when the percentage of steel is selected
arbitrarily, as is frequently done, then the proposed section should be tested
by both equations. When the percentage of steel is larger than that required by
Equation 18, the concrete will be compressed more than is intended before the
steel attains its normal tension. On the other hand, a lower percentage of
steel will require a higher unit-tension in the steel before the concrete
attains its normal compression. When the discrepancy between the percentage of
steel assumed and the true economical value is very great, the stress in the
steel (or the concrete) may become dangerously high when the stress in the
other element (on which the computation may have been made) is only normal.

268.
Example 1. What is the ultimate resisting moment of a concrete beam made of
1:3:5 concrete, which is 7 inches wide, 10 inches deep to the reinforcement,
and which uses 1.2 per cent of reinforcement? The concrete is supposed to have
a ratio for the module of elasticity (r) of 15. The ultimate strength of the
concrete is assumed as 2.000.

Answer.
From Table XIII, p = .012, and r = 15, Ic = .490; x =
.357 lcd = .175 d; d - x = .825 d. From Equation 22
we have: M0=>< 2,000 x .490 X 7 X10 x 8.25=330,137 inch-pounds. The total
compression in the concrete is the continued product of all the factors except
the last, and equals 40,017. But this equals the tension in the steel, whose
area = pbd = .012 x 7 X 10 .84 square inch. Therefore
the unit-stress in the steel would equal 40)017 ± .84 = 47,640 pounds per
square inch. This is considerably less than the usual ultimate of 55,000, and
shows that the percentage of steel is considerably in excess of the normal
value.

**Are You in New Castle ****New Hampshire****? Do You
Need Concrete Cutting?**

**We Are Your Local
Concrete Cutter**

**Call 603-622-4441**

**We Service New Castle
NH and all surrounding Cities & Towns**