**Solved example on strength of doubly reinforced concrete beam **

**Problem 9-3**

Compute the nominal flexural strength M_{n} of the
reinforced concrete rectangular section given below in figure
9-3(a). Take *f*_{c}′ = 5 ksi, * *beam width* b* = 14 in.,
effective depth* d* = 21 in. The beam has tension steel of *A*_{s}
= 4-#10 bars placed in one layer *f*_{y} =
60 ksi, and compression steel of *A*_{s}′
= 2-#7 bars, *f*_{s}′ = 60 ksi, effective cover *d*′ of
2.5 in.

Figure 9-3

**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.

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 4 bars of #10 (dia 1.27 in.) .

Area of one bar of #10 = 1.27
in^{2}.

* A*_{s} = 4-#10
bars. = 4 (1.27) = 5.08
in^{2}.

Area of compression steel , *A*_{s}′
= 2-#7 = 2(0.6) = 1.2
in^{2}.

The internal forces acting on the section shown in figure 9-3(c) are calculated as given below;

C_{c} = 0.85 *f*_{c}′ *ba* = 0.85
(5) (14) a = 59.5 *a*

C_{s} = (*f*_{s}′ - 0.85 *f*_{c}′) *
A*_{s}′ = (60 - 0.85*5) 1.2 = 66.9 kips

T = *A*_{s} *f*_{y}
= (5.08) (60) = 304.8 kips

Applying static equilibrium, we get C

59.5 *a* +
66.9 = 304.8

therefore depth of stress block, *a* = (304.8- 66.9)/59.5 =
3.998 in

Depth of neutral axis *x* = *a */ β_{1}
= 3.998/0.8 = 4.997 in

By straight line proportion (figure 9-3(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 =
(4.997 - 2.5) (0.003) / 4.997 = 0.0015

yield strain of steel, ε_{y} = *f*_{y}
/E_{s} = 60 / 29000 = 0.00207

*ε*_{s}′ is
less than ε_{y} ,
this means that compression steel has not yielded before crushing of concrete.
hence the assumption is not confirmed and therefore the calculated value of *x* is
not valid.

In such cases when compression steel has not yielded we have to find the depth of neutral axis by considering the equilibrium of forces acting on the section;

Force in compression steel will be calculated on the basis of
actual stress *f*_{s}′ in compression rebar
at the time of crushing of concrete.

* f*_{s}′ = *ε*_{s}′
* E_{s} ={( x - d′ ) (0.003) / x}*29000

C_{s} = (*f*_{s}′ - 0.85 *f*_{c}′) *
A*_{s}′ =[{( x - 2.5 ) (0.003) / x}*29000
- 0.85 ** 5*] *1.2

C_{c} = 0.85 *f*_{c}′ *ba
= *0.85 *5 *14 *(0.8*x)

T = *A*_{s} *f*_{y}
= 5.08*60 = 304.8 kip

Apply equilibrium of froces at the section;

C_{c} + C_{s} = T;

substituting the values of C_{c}, C_{s}, T in the above equation and simplifying
we get;

47.6 x^{2}
- 205.5 x
- 261 = 0;

the above quadratic equation is solved to get the positive value of x;

therefore x = 5.34 in.

a
= β_{1} x
= 0.8 * 5.34 = 4.27 in.

* ε*_{s}′ = (5.34 - 2.5) (0.003) /
5.34 = 0.0016

*f*_{s}′ = 0.0016*29000 = 46.4 ksi

C_{c} = *= *0.85 *5 *14 *(4.27) = 254.065 kip

C_{s} = (46.4 - 0.85*5)(1.2) = 50.58 kip

C_{c} + C_{s} = 254.065 + 50.58 = 304.65 kip
(equal to T; confirms calculation)

Nominal Flexural Strength

M_{n} = C_{c} (d - a/2) + C_{s}
(d - d′ )

= 254.065 (21 - 4.27/2) + 50.58*(21 - 2.5) = 5728.66 in-kip

**M**_{n}**
= 477.39 ft-kip**

Strain in the Tension Steel

Using strain diagram we can calculate the strain in tension steel

ε_{s} = 0.003(21 - 5.34)
/ 5.34 = 0.0088

The strain in the tension steel is more than yield strain and also greater than 0.005. Hence the section is tension-controlled. Therefore the strength reduction factor will be equal to 0.9

You can also use our reinforced concrete calculator to solve this problem or visit other solved examples on strength of Reinforced concrete beam Prob 9-1 and Prob 9-2

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