Worked Example: Concrete Beam Flexure Basics

This worked example demonstrates how to find the nominal moment capacity ( $M_n$ ) and the design moment capacity ( $\phi M_n$ ) of a singly reinforced concrete beam.

1. Problem Statement

A simply supported rectangular reinforced concrete beam has a width $b = 300 \text{ mm}$ and an effective depth $d = 500 \text{ mm}$ . It is reinforced with tension steel having an area $A_s = 1500 \text{ mm}^2$ . The concrete strength is $f_c' = 28 \text{ MPa}$ and the steel yield strength is $f_y = 420 \text{ MPa}$ . Calculate the nominal moment capacity ( $M_n$ ) and the design moment capacity ( $\phi M_n$ ) assuming the section is tension-controlled ( $\phi = 0.90$ ).

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2. Given Data

Beam Width (b)

Effective Depth (d)

Steel Area (A_s)

mm²

Concrete Strength (f_c')

MPa

Steel Yield Strength (f_y)

MPa

Strength Reduction Factor (φ)

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3. Formula and Calculation

Calculate the depth of the equivalent stress block ( $a$ ): $a = \frac{A_s f_y}{0.85 f_c' b}$ For initial values: $a = \frac{1500 \times 420}{0.85 \times 28 \times 300} = 88.24 \text{ mm}$

Calculate the nominal moment capacity ( $M_n$ ):

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(Note: The result $M_n$ is in $\text{N·mm}$ . To convert to $\text{kN·m}$ , divide by $1,000,000$ )

Calculate the design moment capacity ( $\phi M_n$ ): $\phi M_n = \phi \times M_n$ For initial values: $0.90 \times 287,204,400 \text{ N·mm} = 258,483,960 \text{ N·mm} = 258.5 \text{ kN·m}$

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