**Calculation of Flexural Strength of doubly Reinforced Concrete beam**

**Problem 9-2**

**
Compute the nominal flexural strength M _{n} of the
reinforced concrete rectangular section given below in figure
9-2(a). Take f_{c}′ = 5000
psi, width b = 13 in.,
effective depth d = 25 in. The beam has tension steel of A_{s}
= 8-#9 bars placed in two layers with a spacing of 1 in. between
them, f_{y} =
40,000 psi, and compression steel of A_{s}′
= 2-#8 bars, f_{s}′
= 40,000 psi, effective cover d′ of 3 in. **

Figure 9-2

**Solution:**

The given section is doubly reinforced
with steel in tension as well as compression zone of the section. The computation of nominal flexural
strength M_{n} is based on the guidelines of ACI-318. The
maximum value of usable strain at the extreme concrete fiber
is assumed to be 0.003.In the case of two layers of tension
steel, the effective depth is taken from the centroid of tension
steel to the top edge of the section.

For *f*_{c}′ grerater than
4000 psi the value of β_{1} is calculated as given
below;

β_{1} = 0.85 - 0.05 {( *f*_{c}′
-4000)/1000} = 0.8

Assume that the compression steel has yielded when the strength is reached (strain in concrete is 0.003).

Given that tension steel consists of 8 bars of #9 (dia 1.128 in.) .

Area of one bar of #9 = 1
in^{2}.

* A*_{s} = 8-#9
bars. = 8 (1.00) = 8
in^{2}.

Area of compression steel , *A*_{s}′
= 2-#8 = 2(0.79) = 1.58
in^{2}.

The internal forces acting on the section shown in figure 9-2(c) are calculated as given below;

C_{c} = 0.85 *f*_{c}′ *ba* = 0.85
(5) (13) a = 55.25 *a*

C_{s} = (*f*_{s}′ - 0.85 *f*_{c}′) *
A*_{s}′ = (40 - 0.85*5) 1.58 = 56.485 kips

T = *A*_{s} *f*_{y}
= (8) (40) = 320 kips

Applying static equilibrium, we get C_{c} + C_{s} = T;

55.25 *a* +
56.485 = 320

therefore depth of stress block, *a* = (320
- 56.485)/55.25 = 4.77 in

Depth of neutral axis *x* = *a */ β_{1}
= 4.77/0.8 = 5.96 in

C_{c} = 55.25 (4.77) = 263.54 kips

By straight line proportion (figure 9-2(b)) we can calculate the strain in compression steel when the extreme concrete fiber has a compressive strain of 0.003.

*ε*_{s}′ = ( x - d′ ) (0.003) / x =
(5.96 - 3) (0.003) / 5.96 = 0.00149

yield strain of steel, ε_{y} = *f*_{y}
/E_{s} = 40 / 29000 = 0.00138

*ε*_{s}′ is greater than ε_{y} ,
this means that compression steel has yielded before crushing of concrete,
hence the assumption is verified and the value of *x* is
valid.

If Compression steel does not yield; then ?? see Example 9-3

Now we determine the strain in the two layers of tension steel.

Using strain diagram we can calculate the strain in tension steel

Lower most layer of tension steel

ε_{s1} = 0.003(25 + 0.5 + 1.28/2 - 5.96)
/ 5.96 = 0.01

Second layer of tension steel

ε_{s2} = 0.003(25 - 0.5 - 1.28/2 - 5.96) / 5.96 =
0.009

The strain in both the layers of tension steel is more than yield strain and also greater than 0.005. Hence the section is tension-controlled.

The nominal flexural strength is computed as given below;
M_{n} = C_{c} (d - a/2) + C_{s}
(d - d′ )

= 263.54 (25 - 4.77/2) + 56.485 (25 - 3) = 7202.63 in-kips

**M**_{n}** = 600.22 ft-kips **

see another problem 9-1 and problem 9-3 on Reinforced Concrete

You can also use our
**Reinforced concrete calculator** for quickly finding the strength of reinforced concrete sections.

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