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 β₁ is calculated as given below;

 β₁  = 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 in².

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

     Area of compression steel , A_s′ = 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;

   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_c + C_s = 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.

     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 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. 

 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