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 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 8 bars of #9 (dia 1.128 in.) .

       Area of one bar of #9 = 1 in².

       A_s = 8-#9 bars. = 8 (1.00) = 8 in².

     Area of compression steel , A_s′ = 2-#8 = 2(0.79) = 1.58 in².

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 / β₁  = 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.

Problem 7-1 Solution of Indeterminate Structure - Propped cantilever

Problem 7-2 Solution of Indeterminate Structure - Continuous Beam