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Compute the nominal flexural strength M_{n} of the reinforced concrete rectangular section of Figure 1.

**Given:**

*f _{c}*′ = 5000 psi,

** 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 β_{1} is calculated as follows:

*β _{1} *

Given that tension steel consists of 4 bars of #9 (diameter d =1.128 in.) *→ A*_{s} = 4-#9 bars. = 4 (1.00) = 4 in^{2} (Area of a bar of #9 = 1 in^{2})

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 / β _{1} = 3.14/0.8 = 3.93 in*

*C _{c} *

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} *

*ε _{y}*

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 *

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