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It
should be noted that the above equations apply only to symmetrical concrete
arch ribs which have their concrete abutments at the same level. Under such
conditions, Equation 53 is always satisfied when Equation 52 is satisfied. In
the very rare cases where concrete arches have to be designed on a different
basis, some of the simplifications given above cannot be utilized, and the work
becomes far more complicated. The solution of these very rare cases will not be
given here. Assume that Fig. 229 represents a portion of the cross-section of a
concrete arch at any point, the particular-portion having a total depth h equal
to the thickness of the concrete arch at that point, and a unit-width b, which
is presumably less than the total length of the concrete arch parallel to its
axis. Assume that there is reinforcement in both the top and bottom of this
section, and that the reinforcement is placed at a distance of Jo of the
thickness of the concrete arch from the top and the bottom, or from the
extrados and the intrados.

The moment of inertia of the plain concrete without
any reinforcement would evidently equal -- W. A transverse stress on such a
section will cause the bars on one side of the section (say the bottom) to be
in tension, while those in the top will be in compression. As already developed
in the treatment of concrete columns (Part III, Article 310), the steel which
is in compression will develop a compressive stress which is in proportion to
the ratio of the module of elasticity of the steel and the concrete; and we may
therefore consider that the area of steel in compression at the top (calling
its area A) is the equivalent of an area of concrete equal to the square
footage, in which r is as usual. The exact position of the neutral axis in a
section which is reinforced both in compression and in tension, depends upon
the percentage of steel which is used; but when the percentage is as large as
it usually is, the neutral axis is not far from the center of the section; and
since it very much simplifies the computations to consider it at the center of
the section, it will be so considered, and the moment of inertia of the steel
and concrete combined may be expressed by the equation. If, in any numerical
problem, it is considered preferable to place the steel so that the distance of
its center of gravity from the concrete surface of the concrete is greater or
less than 0.1 h, a corresponding change must be made in the second term of the
right-hand side of Equation 55.

Assume that p = .015 and that the thickness of
the concrete arch h equals 15 inches. For a unit-section 12 inches wide, the
area of the concrete would be uneven, which equals 180 square inches. Then
180X.015= 2.70=2A, since A is the area at either top or bottom. It should be
noted from Equation 55, that when, as is usual, the area of the steel in the
extrados and intrados remains constant, while the thickness of the concrete
arch varies, the increase in the moment of inertia is not strictly according to
the cube of the depth, but increases in accordance with two terms, one of which
varies as the cube of the depth, and the other as the square of the depth. To
illustrate the discrepancy, let us assume that the depth of the concrete arch
at the concrete abutment is 10 per cent greater than the depth at the crown; or
that, applying it to the above numerical case, the depth at the crown is 15
inches, and at the concrete abutment the depth is 16.5 inches. Using the
approximate rule that I varies as the cube of h, we find that 2 percent in excess
of the value found from computing the moment of inertia similarly on the assumption
that the depth at the concrete abutment is increased 50 percent, so that it
equals 22.5 inches, we find that the approximate rule will give a moment of
inertia which is nearly 8 per cent in excess of the actual.

**Are You in ****Harvard Massachusetts****? Do You
Need Concrete Cutting?**

**We Are Your Local Concrete
Cutting Company**

**Call ****508-283-3135**

**We Service all
surrounding Cities & Towns.**