Nominal flexural strength of a reinforced concrete beam

Compute the nominal flexural strength M_n of the reinforced concrete rectangular section of Figure 1.  

Given:

f_c′ = 5000 psi, f_y = 50,000 psi, b = 15 in., d = 22.5 in. and A_s= 4-#9 bars.

 Solution:

The calculation of the nominal flexural strength, M_n, is determined based on the ACI-318 guidelines for a singly reinforced section. The maximum usable strain at the extreme concrete fiber is assumed to be 0.003. It is also assumed that the steel has already yielded when the strength is reached (strain in concrete is 0.003). When the compressive strength of concrete (fc') exceeds 4000 psi, the value of β₁ is calculated as follows:

β₁ = 0.85 - 0.05 {(fc' - 4000)/1000} = 0.8

Given that tension steel consists of 4 bars of #9 (diameter d =1.128 in.) → A_s = 4-#9 bars. = 4 (1.00) = 4 in² (Area of a bar of #9 = 1 in²)

The internal forces acting on the section are calculated as:

C_c = 0.85 f_c′ a b = 0.85 (5) a (15) = 63.75a

T = A_s f_y = (4) (50) = 200kips

Applying static equilibrium, we get Cc = T→ 63.75 a =→ a = 3.14in.

Depth of neutral axis: 

x = a / β₁ = 3.14/0.8 = 3.93 in

C_c = 63.75 (3.14) = 200.18 kips

By straight line proportion we can calculate the strain in tension steel when the extreme concrete fiber has a compressive strain of 0.003.

ε_s = (d - x) (0.003) / x = (22.5 - 3.93) (0.003) / 3.93 = 0.014     

ε_y = f_y /E_s = 50 / 29000 = 0.00172

ε_s ≥ ε_y, this means that steel has yielded before crushing of concrete, which is a safe condition as it gives warning before failure.

The section is tension-controlled because the strain is more than 0.005, therefore the strength reduction factor will be taken as 0.9.

The nominal flexural strength is calculated as:

Moment = (force) (lever-arm) → M_n = T (d - a/2) = Cc (d - a/2) = 200.18 (22.5 - 3.14/2) = 4189.77 in-kips→ M_n = 349.15 ft-kips