Solved example on strength of doubly reinforced concrete beam

Problem 9-3

Compute the nominal flexural strength Mn of the reinforced concrete rectangular section given below in figure 9-3(a).  Take fc′ = 5 ksi,   beam width b = 14 in., effective depth d = 21 in. The beam has tension steel of As = 4-#10 bars placed in one layer  fy = 60 ksi, and compression steel of  As′ = 2-#7 bars, fs′ = 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 Mn 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  fc′  grerater than 4000 psi the value  of β1 is calculated as given below;

β1  = 0.85 - 0.05 {( fc′ -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 in2.

As = 4-#10 bars. = 4 (1.27) = 5.08 in2.

Area of compression steel , As′ = 2-#7 = 2(0.6) = 1.2 in2.

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

Cc = 0.85 fcba = 0.85 (5) (14) a = 59.5 a

Cs = (fs′ - 0.85 fc′) As′ = (60 - 0.85*5) 1.2 = 66.9 kips

T =  As fy  =  (5.08) (60) = 304.8 kips

Applying static equilibrium, we get  Cc + Cs = T;

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 / β1  = 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.

εs′ = ( x - d′ ) (0.003) / x = (4.997 - 2.5) (0.003) / 4.997 =  0.0015

yield strain of steel, εy = fy /Es = 60 / 29000 = 0.00207

εs′  is less  than  εy , 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 fs′ in compression rebar at the time of crushing of concrete.

fs′ = εs′ * Es ={( x - d′ ) (0.003) / x}*29000

Cs = (fs′ - 0.85 fc′) As′ =[{( x - 2.5 ) (0.003) / x}*29000 - 0.85 * 5] *1.2

Cc = 0.85 fcba = 0.85 *5 *14 *(0.8*x)

T =  As fy  = 5.08*60 = 304.8 kip

Apply equilibrium of froces  at the section;

Cc + Cs = T;

substituting the values of Cc, Cs, T in the above equation and simplifying we get;

47.6 x2 - 205.5 x - 261 = 0;

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

therefore  x = 5.34 in.

a =  β1 x = 0.8 * 5.34 = 4.27 in.

εs′ = (5.34 - 2.5) (0.003) / 5.34  = 0.0016

fs′ = 0.0016*29000 = 46.4 ksi

Cc = = 0.85 *5 *14 *(4.27) = 254.065 kip

Cs = (46.4 - 0.85*5)(1.2) = 50.58 kip

Cc + Cs = 254.065 + 50.58 = 304.65 kip    (equal to T; confirms calculation)

Nominal Flexural Strength

Mn = Cc (d - a/2)  + Cs (d - d′ )

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

Mn = 477.39 ft-kip

Strain in the Tension Steel

Using strain diagram we can calculate the strain in tension steel

εs = 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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