Concrete structure design

This chapter covers the following:

Beam design
Slab design
Column design
Footing design

Beam design

Beam design focuses on analyzing and proportioning reinforced concrete members subjected mainly to bending and shear. You’ll use strain compatibility, internal force equilibrium, and strength reduction concepts to determine the required flexural reinforcement.

This section covers:

Singly and doubly reinforced beams
Rectangular and T-beam sections
Strength, ductility, and serviceability requirements consistent with reinforced concrete design principles

A diagram explaining the concrete compression stress block and strain distribution used in reinforced concrete beam design analysis. — Concrete beam design

$M_{u} = moment of moment diagram$

$z = lever arm = larger of ⎩ ⎨ ⎧ 0.9 \cdot d d - \frac{a}{2}$

$A_{s} = \frac{M _{u}}{ϕ f _{y} z}$

$A_{c} = \frac{A _{s} f _{y}}{0.85 f _{c}^{'}}$

If $A_{c} < A_{f}$ , the neutral axis (N.A.) is in the flange, so treat the section as a rectangular beam.

$R_{n} = \frac{M _{u}}{ϕ b d ^{2}}$

$ρ = 0.85 \cdot \frac{f _{c}^{'}}{f _{y}} (1 - 1 - \frac{2 \cdot R _{n}}{0.85 f _{c}^{'}})$

$ρ_{ma x} = 0.85 \cdot β \cdot \frac{f _{c}^{'}}{f _{y}} \cdot \frac{ϵ _{c u}}{ϵ _{c u} + ϵ _{t}}$

If $ρ < ρ_{ma x}$ , the beam is singly reinforced. Then:

$ρ_{min} = larger of ⎩ ⎨ ⎧ \frac{3 f _{c}^{'}}{f _{y}} \frac{200}{f _{y}}$

$ρ_{min} < ρ < ρ_{ma x}$

$A_{s} = ρ b d$

$a = \frac{A _{s} f _{y}}{0.85 f _{c}^{'} b}$

$c = \frac{a}{β _{1}}$

$ϵ_{t} = \frac{( d - c )}{c}, ϵ_{t} > 0.005$

Therefore $ϕ$ is OK, or calculate $ϕ$ :

$ϕ M_{n} = ϕ A_{s} f_{y} (d - \frac{a}{2}) > M_{u}$

But if $ρ > ρ_{ma x}$ , the beam is doubly reinforced. Then:

$A_{s} = A_{s}^{'} + ρ_{ma x} b d$

$A_{s}^{'} = (ρ - ρ_{ma x}) b d$

$A_{s} f_{y} = 0.85 f_{c}^{'} b_{1} c + A_{s}^{'} (\frac{c - d ^{'}}{c}) \cdot ϵ_{s} E_{s}$

Solve for $c$ and compute $a = β_{1} c$ .

Therefore $ϕ$ is OK, or calculate $ϕ$ .

$f_{s}^{'} = (\frac{A _{s}^{'}}{A _{s} - A _{s}^{'}}) f_{y}$

$ϕ M_{n} = ϕ [A_{s} f_{y} (d - \frac{a}{2}) + A_{s}^{'} f_{s}^{'} (d - d^{'})] > M_{u}$

If $A_{c} > A_{f}$ , the neutral axis (N.A.) is below the flange, so treat the section as a T-beam.

$b_{e ff} = smaller of ⎩ ⎨ ⎧ L /4 b_{w} + 16 t (beam spacing)$

$y = A_{f} \cdot (\frac{t}{2}) + (A_{c} - A_{f}) \cdot (\frac{t + \frac{( A _{c} - A _{f} )}{2 b _{w}}}{A _{c}})$

$z = lever arm = d - \overset{y}{ˉ}$

$A_{s} = \frac{M _{u}}{ϕ f _{y} z}$

$a = t + \frac{( A _{c} - A _{f} )}{b _{w}}, \frac{A _{s} f _{y}}{0.85 f _{c}^{'} b _{e ff}}$

$c = \frac{a}{β _{1}}$

$ϵ_{t} = \frac{( d - c )}{c}, ϵ_{t} > 0.005$

Therefore $ϕ$ is OK, or calculate $ϕ$ .

$ϕ M_{n} = ϕ 0.85 f_{c}^{'} [t (b_{e ff} - b_{w}) (d - \frac{t}{2}) + a b_{w} (d - \frac{a}{2})] > M_{u}$

Shear design

$ϕ V_{n} \geq V_{u}$

Nominal shear strength:

$V_{n} = V_{c} + V_{s}$

$V_{c} = 2 λ f_{c}^{'} b_{w} d$

Where:

$λ = 1.0$ for normal weight concrete (NWC)
$λ = 0.75$ for lightweight concrete

$V_{s} = \frac{A _{v} f _{y} d}{s} (may not exceed 8 b_{w} d / f_{c}^{'})$

Required and maximum-permitted stirrup spacing $s$

If:

$V_{u} \leq \frac{ϕ V _{c}}{2} \Rightarrow No stirrups required$
If:

$V_{u} > \frac{ϕ V _{c}}{2} \Rightarrow Use the following table (A_{v} given)$

Please see Stirrup spacing table from FE Handbook.

Slab design

Slab design involves selecting slab thickness and reinforcement to resist flexural moments and control cracking under dead and live loads. This section covers minimum thickness requirements, self-weight calculations, and flexural reinforcement design using strength-based methods. It also includes temperature and shrinkage reinforcement for durability and long-term performance.

Select minimum slab thickness according to guidelines $(t)$

Calculate self weight from slab thickness $t$ .

$self wt = (\frac{t}{12}) \cdot 150$

Calculate moment from dead load and live load combination:

$ρ_{max} = 0.85 \cdot β \cdot \frac{f _{c}^{'}}{f _{y}} \cdot \frac{ϵ _{u}}{ϵ _{u} + ϵ _{t}}$

Consider $ρ = ρ_{max}$ to calculate $d$ :

$d_{req} = \frac{M _{max}}{ϕρ f _{y} b ( 1 - 0.59 \frac{ρ f _{y}}{f _{c}^{'}} )}$

Adjust $d$ :

If $d_{pro} > d_{req}$

$R_{n} = \frac{M _{u}}{ϕ b d ^{2}}$

$ρ = 0.85 \cdot \frac{f _{c}^{'}}{f _{y}} (1 - 1 - \frac{2 \cdot R _{n}}{0.85 f _{c}^{'}})$

$A_{s} = ρ b d$

Check:

$A_{s} > A_{s, min}$

Select temperature and shrinkage steel

Column design

Column design addresses the strength and stability of reinforced concrete members subjected to axial load, with or without bending. This section covers axial load capacity, longitudinal reinforcement limits, and confinement requirements using ties or spirals. Interaction relationships and strength reduction factors are used to verify adequacy, with attention to reinforcement detailing and spacing requirements.

$ϕ P_{n} = α ϕ [0.85 f_{c}^{'} (A_{g} - A_{s t}) + f_{y} A_{s t}]$

Use:

$A_{s t} = s \cdot A_{g}; s = (0.01 \sim 0.08)$

If longitudinal bar $#10$ or smaller, then use tie $#3$ .

Else use tie $#4$ .

Spacing of tie bar = smallest of:

$48 \times tie bar dia$
$16 \times longitudinal bar dia$
$least dimension$

Select arrangement as below so that clear spacing between longitudinal bars $> 1"$ or dia of tie bar.

$R_{n} = \frac{M _{n}}{ϕ f _{c}^{'} A _{g} h}$

$ϕ P_{n x} = ϕ K_{n x} f_{c}^{'} A_{g}$

$ϕ P_{n o} = α ϕ [0.85 f_{c}^{'} (1 - s) + s f_{y}] A_{g}$

$\frac{1}{ϕ P _{n}} = \frac{1}{ϕ P _{n x}} + \frac{1}{ϕ P _{n y}} - \frac{1}{ϕ P _{n o}}$

$ϕ P_{n} > P_{u} so OK$

A set of reinforced concrete column detailing diagrams showing proper lateral tie and hoop configurations with spacing limits for confinement. — Concrete lateral ties for column

Footing design

Footing design focuses on transferring loads from columns or walls to supporting soil without exceeding allowable bearing pressures. This section covers required footing area, factored soil pressures, and effective depth selection. Flexural, one-way shear, and punching shear checks are used to confirm adequate strength and structural integrity for isolated and combined footings.

$A = larger of ⎩ ⎨ ⎧ \frac{D L + LL}{BC} \frac{D L + LL + EQ L}{1.33 \cdot BC} \frac{D L + LL + W L}{1.33 \cdot BC}$

$q_{u} = \frac{1.2 \cdot D L + 1.6 \cdot LL}{A}$

Adjust $d = 2 A$

A schematic showing the geometry and critical section for punching shear around a concentrated column load on a concrete footing. — Concrete footing beam and punching shear

Check beam shear:

$q_{u} \cdot w < 2 λ f_{c}^{'} \cdot b d$

Check punching shear:

$q_{u} (A - A_{o}) < 4 λ f_{c}^{'} \cdot P_{o} d$

So, $d$ is OK.

$M_{u} = \frac{1}{8} q_{u} (b - a)^{2}$

$R_{n} = \frac{M _{u}}{ϕ b d ^{2}}$

$ρ = 0.85 \cdot \frac{f _{c}^{'}}{f _{y}} (1 - 1 - \frac{2 \cdot R _{n}}{0.85 f _{c}^{'}})$

$ρ_{m i n} = max (\frac{3 f _{c}^{'}}{f _{y}}, \frac{200}{f _{y}})$

$A_{s} = max (ρ_{m i n} b d, ρ b d)$

A schematic illustrating the analysis and design of a combined footing under two column loads, including resultant location, footing dimensions, and soil pressure distribution. — Concrete combined footing design

This chapter covers the following:

Beam design
Slab design
Column design
Footing design

Beam design

This section covers:

Singly and doubly reinforced beams
Rectangular and T-beam sections
Strength, ductility, and serviceability requirements consistent with reinforced concrete design principles

$M_{u} = moment of moment diagram$

$z = lever arm = larger of ⎩ ⎨ ⎧ 0.9 \cdot d d - \frac{a}{2}$

$A_{s} = \frac{M _{u}}{ϕ f _{y} z}$

$A_{c} = \frac{A _{s} f _{y}}{0.85 f _{c}^{'}}$

If $A_{c} < A_{f}$ , the neutral axis (N.A.) is in the flange, so treat the section as a rectangular beam.

$R_{n} = \frac{M _{u}}{ϕ b d ^{2}}$

$ρ = 0.85 \cdot \frac{f _{c}^{'}}{f _{y}} (1 - 1 - \frac{2 \cdot R _{n}}{0.85 f _{c}^{'}})$

$ρ_{ma x} = 0.85 \cdot β \cdot \frac{f _{c}^{'}}{f _{y}} \cdot \frac{ϵ _{c u}}{ϵ _{c u} + ϵ _{t}}$

If $ρ < ρ_{ma x}$ , the beam is singly reinforced. Then:

$ρ_{min} = larger of ⎩ ⎨ ⎧ \frac{3 f _{c}^{'}}{f _{y}} \frac{200}{f _{y}}$

$ρ_{min} < ρ < ρ_{ma x}$

$A_{s} = ρ b d$

$a = \frac{A _{s} f _{y}}{0.85 f _{c}^{'} b}$

$c = \frac{a}{β _{1}}$

$ϵ_{t} = \frac{( d - c )}{c}, ϵ_{t} > 0.005$

Therefore $ϕ$ is OK, or calculate $ϕ$ :

$ϕ M_{n} = ϕ A_{s} f_{y} (d - \frac{a}{2}) > M_{u}$

But if $ρ > ρ_{ma x}$ , the beam is doubly reinforced. Then:

$A_{s} = A_{s}^{'} + ρ_{ma x} b d$

$A_{s}^{'} = (ρ - ρ_{ma x}) b d$

$A_{s} f_{y} = 0.85 f_{c}^{'} b_{1} c + A_{s}^{'} (\frac{c - d ^{'}}{c}) \cdot ϵ_{s} E_{s}$

Solve for $c$ and compute $a = β_{1} c$ .

Therefore $ϕ$ is OK, or calculate $ϕ$ .

$f_{s}^{'} = (\frac{A _{s}^{'}}{A _{s} - A _{s}^{'}}) f_{y}$

$ϕ M_{n} = ϕ [A_{s} f_{y} (d - \frac{a}{2}) + A_{s}^{'} f_{s}^{'} (d - d^{'})] > M_{u}$

If $A_{c} > A_{f}$ , the neutral axis (N.A.) is below the flange, so treat the section as a T-beam.

$b_{e ff} = smaller of ⎩ ⎨ ⎧ L /4 b_{w} + 16 t (beam spacing)$

$y = A_{f} \cdot (\frac{t}{2}) + (A_{c} - A_{f}) \cdot (\frac{t + \frac{( A _{c} - A _{f} )}{2 b _{w}}}{A _{c}})$

$z = lever arm = d - \overset{y}{ˉ}$

$A_{s} = \frac{M _{u}}{ϕ f _{y} z}$

$a = t + \frac{( A _{c} - A _{f} )}{b _{w}}, \frac{A _{s} f _{y}}{0.85 f _{c}^{'} b _{e ff}}$

$c = \frac{a}{β _{1}}$

$ϵ_{t} = \frac{( d - c )}{c}, ϵ_{t} > 0.005$

Therefore $ϕ$ is OK, or calculate $ϕ$ .

$ϕ M_{n} = ϕ 0.85 f_{c}^{'} [t (b_{e ff} - b_{w}) (d - \frac{t}{2}) + a b_{w} (d - \frac{a}{2})] > M_{u}$

Shear design

$ϕ V_{n} \geq V_{u}$

Nominal shear strength:

$V_{n} = V_{c} + V_{s}$

$V_{c} = 2 λ f_{c}^{'} b_{w} d$

Where:

$λ = 1.0$ for normal weight concrete (NWC)
$λ = 0.75$ for lightweight concrete

$V_{s} = \frac{A _{v} f _{y} d}{s} (may not exceed 8 b_{w} d / f_{c}^{'})$

Required and maximum-permitted stirrup spacing $s$

If:

$V_{u} \leq \frac{ϕ V _{c}}{2} \Rightarrow No stirrups required$
If:

$V_{u} > \frac{ϕ V _{c}}{2} \Rightarrow Use the following table (A_{v} given)$

Please see Stirrup spacing table from FE Handbook.

Slab design

Select minimum slab thickness according to guidelines $(t)$

Calculate self weight from slab thickness $t$ .

$self wt = (\frac{t}{12}) \cdot 150$

Calculate moment from dead load and live load combination:

$ρ_{max} = 0.85 \cdot β \cdot \frac{f _{c}^{'}}{f _{y}} \cdot \frac{ϵ _{u}}{ϵ _{u} + ϵ _{t}}$

Consider $ρ = ρ_{max}$ to calculate $d$ :

$d_{req} = \frac{M _{max}}{ϕρ f _{y} b ( 1 - 0.59 \frac{ρ f _{y}}{f _{c}^{'}} )}$

Adjust $d$ :

If $d_{pro} > d_{req}$

$R_{n} = \frac{M _{u}}{ϕ b d ^{2}}$

$ρ = 0.85 \cdot \frac{f _{c}^{'}}{f _{y}} (1 - 1 - \frac{2 \cdot R _{n}}{0.85 f _{c}^{'}})$

$A_{s} = ρ b d$

Check:

$A_{s} > A_{s, min}$

Select temperature and shrinkage steel

Column design

$ϕ P_{n} = α ϕ [0.85 f_{c}^{'} (A_{g} - A_{s t}) + f_{y} A_{s t}]$

Use:

$A_{s t} = s \cdot A_{g}; s = (0.01 \sim 0.08)$

If longitudinal bar $#10$ or smaller, then use tie $#3$ .

Else use tie $#4$ .

Spacing of tie bar = smallest of:

$48 \times tie bar dia$
$16 \times longitudinal bar dia$
$least dimension$

Select arrangement as below so that clear spacing between longitudinal bars $> 1"$ or dia of tie bar.

$R_{n} = \frac{M _{n}}{ϕ f _{c}^{'} A _{g} h}$

$ϕ P_{n x} = ϕ K_{n x} f_{c}^{'} A_{g}$

$ϕ P_{n o} = α ϕ [0.85 f_{c}^{'} (1 - s) + s f_{y}] A_{g}$

$\frac{1}{ϕ P _{n}} = \frac{1}{ϕ P _{n x}} + \frac{1}{ϕ P _{n y}} - \frac{1}{ϕ P _{n o}}$

$ϕ P_{n} > P_{u} so OK$

Footing design

$A = larger of ⎩ ⎨ ⎧ \frac{D L + LL}{BC} \frac{D L + LL + EQ L}{1.33 \cdot BC} \frac{D L + LL + W L}{1.33 \cdot BC}$

$q_{u} = \frac{1.2 \cdot D L + 1.6 \cdot LL}{A}$

Adjust $d = 2 A$

Check beam shear:

$q_{u} \cdot w < 2 λ f_{c}^{'} \cdot b d$

Check punching shear:

$q_{u} (A - A_{o}) < 4 λ f_{c}^{'} \cdot P_{o} d$

So, $d$ is OK.

$M_{u} = \frac{1}{8} q_{u} (b - a)^{2}$

$R_{n} = \frac{M _{u}}{ϕ b d ^{2}}$

$ρ = 0.85 \cdot \frac{f _{c}^{'}}{f _{y}} (1 - 1 - \frac{2 \cdot R _{n}}{0.85 f _{c}^{'}})$

$ρ_{m i n} = max (\frac{3 f _{c}^{'}}{f _{y}}, \frac{200}{f _{y}})$

$A_{s} = max (ρ_{m i n} b d, ρ b d)$