Concrete structure design

This chapter covers the following:

Beam design
Slab design
Column design
Footing design

Exam tip: these concrete design formulas use US customary units (psi, in, lb) - any $f_{c}^{'}$ term assumes $f_{c}^{'}$ is in psi. Watch for mixed units within a single problem too - $f_{c}^{'}$ and $f_{y}$ are usually given in psi while $M_{u}$ is often given in ft-kip or in-kip, so convert everything to consistent units (typically psi and inches) before plugging into a formula. You don’t need to memorize these formulas: the NCEES FE Reference Handbook lists each one under concrete design, so focus on locating the right equation quickly rather than recalling it from memory.

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

$M_{u} = moment of moment diagram$

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

Using the smaller of the two values keeps the first-pass steel area estimate conservative, since the actual stress-block depth $a$ isn’t known yet.

$A_{s} = \frac{M _{u}}{ϕ f _{y} z}$

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

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

The $z$ -based $A_{s}$ above is only a first-pass estimate, used to check whether the neutral axis falls in the flange (rectangular section) or below it (T-beam). For a rectangular section, the steps below then size the final steel.

Step 1: Classify the section as singly or doubly reinforced.

$R_{n} = \frac{M _{u}}{ϕ b d ^{2}}$

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

Compare $ρ$ to the maximum ratio $ρ_{ma x}$ , which uses the concrete’s crushing strain $ϵ_{c u}$ and the tension steel strain $ϵ_{t}$ to keep the section ductile:

$ρ_{ma x} = 0.85 \cdot β_{1} \cdot \frac{f _{c}^{'}}{f _{y}} \cdot \frac{ϵ _{c u}}{ϵ _{c u} + ϵ _{t}}$

Step 2: Singly reinforced design path.

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

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

$ρ_{min} < ρ < ρ_{ma x}$

$A_{s} = ρ b d$

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

$c = \frac{a}{β _{1}}$

Here $β_{1}$ is the ACI stress-block depth factor - $0.85$ for $f_{c}^{'} \leq 4, 000$ psi, decreasing by $0.05$ for each $1, 000$ psi above that (down to a floor of $0.65$ ).

$ϵ_{t} = 0.003 (\frac{d _{t} - c}{c})$

Here $d_{t}$ is the depth to the extreme layer of tension steel, equal to $d$ when the steel is in a single layer. If $ϵ_{t} \geq 0.005$ , the section is tension-controlled and $ϕ = 0.90$ . If $ϵ_{t} < 0.005$ , the section is in the transition zone and $ϕ$ must be recalculated using ACI’s linear transition equation between $ϕ = 0.65$ (compression-controlled) and $ϕ = 0.90$ (tension-controlled).

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

Step 3: Doubly reinforced design path.

But if $ρ > ρ_{ma x}$ , the beam is doubly reinforced - compression steel $A_{s}^{'}$ is added because a singly reinforced section can’t carry the moment within the ductility limit. Then:

$A_{s} = A_{s}^{'} + ρ_{ma x} b d$

$A_{s}^{'} = (ρ - ρ_{ma x}) b d$

$A_{s} f_{y} = 0.85 f_{c}^{'} ba + A_{s}^{'} f_{s}^{'}$

$a = β_{1} c, f_{s}^{'} = E_{s} (0.003) (\frac{c - d ^{'}}{c}) \leq f_{y}$

Solve for $c$ , then check $ϕ$ using the same tension-controlled rule described above.

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

Example: Singly reinforced beam

Given $b = 12"$ , $d = 20"$ , $f_{c}^{'} = 4, 000$ psi, $f_{y} = 60, 000$ psi, and $M_{u} = 150$ ft-kip $= 1, 800$ in-kip. Using $z = 0.9 d = 18"$ : $A_{s} = \frac{M _{u}}{ϕ f _{y} z} = \frac{1 , 800}{0.9 ( 60 ) ( 18 )} = 1.85 in^{2}$ . Refining with $a = \frac{A _{s} f _{y}}{0.85 f _{c}^{'} b} = 2.72"$ gives $z = d - \frac{a}{2} = 18.64"$ , so $A_{s} \approx 1.79 in^{2}$ .

Answer: $A_{s} \approx 1.79 in^{2}$

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

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

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

$z = lever arm = d - \overset{y}{ˉ}$

$A_{s} = \frac{M _{u}}{ϕ f _{y} z}$

$a = t + \frac{( A _{c} - A _{f} )}{b _{w}}$

This geometric expression for $a$ gives a quick initial estimate directly from the flange and web areas. Once $A_{s}$ has been computed from $z$ , refine $a$ using the force-equilibrium expression below, and carry that refined value forward into the remaining checks:

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

$c = \frac{a}{β _{1}}$

$ϵ_{t} = 0.003 (\frac{d _{t} - c}{c})$

Check $ϕ$ using the same tension-controlled rule described earlier in this section.

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

Shear design

$ϕ V_{n} \geq V_{u}$

The critical section for one-way shear is taken at a distance $d$ from the face of the support - sections closer to the support are strengthened by direct compression strut action, so $V_{u}$ is evaluated at that location rather than right at the support face.

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 all-lightweight concrete and $0.85$ for sand-lightweight concrete

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

Required and maximum-permitted stirrup spacing $s$

If:

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

$V_{u} > \frac{ϕ V _{c}}{2} \Rightarrow stirrups are required$

Where $\frac{ϕ V _{c}}{2} < V_{u} \leq ϕ V_{c}$ , only minimum stirrups are needed: the required spacing is the smaller of $s = \frac{A _{v} f _{y}}{0.75 f _{c}^{'} b _{w}}$ and $s = \frac{A _{v} f _{y}}{50 b _{w}}$ . Where $V_{u} > ϕ V_{c}$ , the stirrups must carry $V_{s} = \frac{V _{u}}{ϕ} - V_{c}$ , so the required spacing is $s = \frac{A _{v} f _{y} d}{V _{s}}$ .

As a quick reference, maximum stirrup spacing is the smaller of $d /2$ or $24"$ , tightened to the smaller of $d /4$ or $12"$ when $V_{s}$ exceeds $4 f_{c}^{'} b_{w} d$ .

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 the factored moment from the dead and live load combination, here for a simply supported one-way slab strip of span $L$ :

$w_{u} = 1.2 w_{D} + 1.6 w_{L}, M_{max} = \frac{w _{u} L ^{2}}{8}$

Then find the maximum reinforcement ratio:

$ρ_{max} = 0.85 \cdot β_{1} \cdot \frac{f _{c}^{'}}{f _{y}} \cdot \frac{ϵ _{c u}}{ϵ _{c u} + ϵ _{t}}$

Set $ρ = ρ_{max}$ to calculate the required depth $d$ :

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

Adjust $d$ :

If $d_{pro} > d_{req}$ , compute $ρ$ from $M_{u}$ using the same $R_{n} \to ρ$ relationship shown in the beam design section above (with the slab’s own $b$ , $d_{p ro}$ , and $ϕ$ ), then find the reinforcement:

$A_{s} = ρ b d$

Check:

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

Select temperature and shrinkage steel

For Grade 60 deformed bars, ACI 318 requires temperature and shrinkage reinforcement, placed perpendicular to the main steel, of at least

$A_{s, temp} = 0.0018 A_{g} = 0.0018 bh$

where $h$ is the total slab thickness, at a spacing no greater than the smaller of $5 h$ or $18"$ . The same $0.0018 A_{g}$ is also $A_{s, min}$ for the main flexural steel of a one-way slab, in place of the beam limits $3 f_{c}^{'} / f_{y}$ and $200/ f_{y}$ .

Example: One-way slab reinforcement

Given $b = 12"$ (per foot of width), $d_{p ro} = 5"$ , $f_{c}^{'} = 4, 000$ psi, $f_{y} = 60, 000$ psi, and $M_{u} = 96$ in-kip per foot of width. $R_{n} = \frac{M _{u}}{ϕ b d ^{2}} = 0.356$ ksi, so $ρ = 0.0063$ and $A_{s} = ρ b d = 0.38 in^{2}$ per foot, which exceeds $A_{s, min}$ (for example, $0.0018 bh = 0.13 in^{2}$ per foot for a 6 in slab).

Answer: $A_{s} \approx 0.38 in^{2}$ per foot of slab width.

Column design

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

Here $α$ is a reduction factor that accounts for accidental eccentricity in axially loaded columns - $0.80$ for tied columns and $0.85$ for spiral columns.

Use:

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

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.

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

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

Here $K_{n x}$ is a nondimensional axial-load coefficient for bending about the x-axis, read from an interaction chart for the given $R_{n}$ and reinforcement ratio (an equivalent $ϕ P_{n y}$ and $K_{n y}$ apply for bending about the y-axis).

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

(Here $ρ_{g}$ is the same steel ratio defined above, and $ϕ P_{n o}$ is the pure axial capacity with no eccentricity, so $α$ does not apply.)

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

$ϕ P_{n} > P_{u} so OK$

Example: Tied column axial capacity

Given $A_{g} = 256 in^{2}$ , $f_{c}^{'} = 4, 000$ psi, $f_{y} = 60, 000$ psi, $ρ_{g} = 0.02$ , $α = 0.80$ , $ϕ = 0.65$ (ignoring bending). $A_{s t} = ρ_{g} A_{g} = 5.12 in^{2}$ , so $ϕ P_{n} = α ϕ [0.85 f_{c}^{'} (A_{g} - A_{s t}) + f_{y} A_{s t}] \approx 603 kip$ .

Answer: $ϕ P_{n} \approx 603 kip$ , checked against the factored axial demand $P_{u}$ .

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

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

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

Assume a trial depth $d$ , then verify it against the one-way and punching-shear checks below - these confirm the trial value or signal that $d$ needs to be increased.

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:

The critical section for one-way (beam) shear is taken at a distance $d$ from the face of the column, across width $w$ :

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

Check punching shear:

The critical section for punching (two-way) shear runs around a perimeter $P_{o}$ located $d /2$ from the face of the column, enclosing the area $A_{o}$ shown in the figure above:

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

Both checks use $ϕ = 0.75$ , the resistance factor for shear. If both pass, the trial depth $d$ is adequate for shear and you can proceed to flexural design; if either fails, increase $d$ and check again.

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

Compute $ρ$ from $M_{u}$ using the same $R_{n} \to ρ$ relationship shown in the beam design section above (with the footing’s own $b$ , $d$ , and $ϕ$ ), then compare it to the minimum reinforcement ratio:

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

$A_{s} = max (ρ_{m i n} b d, ρ b d)$

Example: Isolated footing area

Given $D L = 100$ kip, $LL = 60$ kip, and $BC = 4$ ksf, with no wind or seismic load. $A = \frac{D L + LL}{BC} = \frac{160}{4} = 40 ft^{2}$ .

Answer: the footing needs a plan area of at least $40 ft^{2}$ , for example a $6. 3^{'} \times 6. 3^{'}$ square footing.

Combined footings extend these same flexural, one-way shear, and punching shear checks to a base supporting two or more columns, sizing and positioning the footing so the resultant of the column loads keeps the soil pressure uniform.

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

$M_{u} = moment of moment diagram$

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

Using the smaller of the two values keeps the first-pass steel area estimate conservative, since the actual stress-block depth $a$ isn’t known yet.

$A_{s} = \frac{M _{u}}{ϕ f _{y} z}$

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

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

Step 1: Classify the section as singly or doubly reinforced.

$R_{n} = \frac{M _{u}}{ϕ b d ^{2}}$

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

Compare $ρ$ to the maximum ratio $ρ_{ma x}$ , which uses the concrete’s crushing strain $ϵ_{c u}$ and the tension steel strain $ϵ_{t}$ to keep the section ductile:

$ρ_{ma x} = 0.85 \cdot β_{1} \cdot \frac{f _{c}^{'}}{f _{y}} \cdot \frac{ϵ _{c u}}{ϵ _{c u} + ϵ _{t}}$

Step 2: Singly reinforced design path.

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

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

$ρ_{min} < ρ < ρ_{ma x}$

$A_{s} = ρ b d$

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

$c = \frac{a}{β _{1}}$

Here $β_{1}$ is the ACI stress-block depth factor - $0.85$ for $f_{c}^{'} \leq 4, 000$ psi, decreasing by $0.05$ for each $1, 000$ psi above that (down to a floor of $0.65$ ).

$ϵ_{t} = 0.003 (\frac{d _{t} - c}{c})$

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

Step 3: Doubly reinforced design path.

But if $ρ > ρ_{ma x}$ , the beam is doubly reinforced - compression steel $A_{s}^{'}$ is added because a singly reinforced section can’t carry the moment within the ductility limit. Then:

$A_{s} = A_{s}^{'} + ρ_{ma x} b d$

$A_{s}^{'} = (ρ - ρ_{ma x}) b d$

$A_{s} f_{y} = 0.85 f_{c}^{'} ba + A_{s}^{'} f_{s}^{'}$

$a = β_{1} c, f_{s}^{'} = E_{s} (0.003) (\frac{c - d ^{'}}{c}) \leq f_{y}$

Solve for $c$ , then check $ϕ$ using the same tension-controlled rule described above.

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

Example: Singly reinforced beam

Given $b = 12"$ , $d = 20"$ , $f_{c}^{'} = 4, 000$ psi, $f_{y} = 60, 000$ psi, and $M_{u} = 150$ ft-kip $= 1, 800$ in-kip. Using $z = 0.9 d = 18"$ : $A_{s} = \frac{M _{u}}{ϕ f _{y} z} = \frac{1 , 800}{0.9 ( 60 ) ( 18 )} = 1.85 in^{2}$ . Refining with $a = \frac{A _{s} f _{y}}{0.85 f _{c}^{'} b} = 2.72"$ gives $z = d - \frac{a}{2} = 18.64"$ , so $A_{s} \approx 1.79 in^{2}$ .

Answer: $A_{s} \approx 1.79 in^{2}$

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

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

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

$z = lever arm = d - \overset{y}{ˉ}$

$A_{s} = \frac{M _{u}}{ϕ f _{y} z}$

$a = t + \frac{( A _{c} - A _{f} )}{b _{w}}$

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

$c = \frac{a}{β _{1}}$

$ϵ_{t} = 0.003 (\frac{d _{t} - c}{c})$

Check $ϕ$ using the same tension-controlled rule described earlier in this section.

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

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 all-lightweight concrete and $0.85$ for sand-lightweight concrete

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

Required and maximum-permitted stirrup spacing $s$

If:

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

$V_{u} > \frac{ϕ V _{c}}{2} \Rightarrow stirrups are required$

As a quick reference, maximum stirrup spacing is the smaller of $d /2$ or $24"$ , tightened to the smaller of $d /4$ or $12"$ when $V_{s}$ exceeds $4 f_{c}^{'} b_{w} d$ .

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 the factored moment from the dead and live load combination, here for a simply supported one-way slab strip of span $L$ :

$w_{u} = 1.2 w_{D} + 1.6 w_{L}, M_{max} = \frac{w _{u} L ^{2}}{8}$

Then find the maximum reinforcement ratio:

$ρ_{max} = 0.85 \cdot β_{1} \cdot \frac{f _{c}^{'}}{f _{y}} \cdot \frac{ϵ _{c u}}{ϵ _{c u} + ϵ _{t}}$

Set $ρ = ρ_{max}$ to calculate the required depth $d$ :

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

Adjust $d$ :

$A_{s} = ρ b d$

Check:

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

Select temperature and shrinkage steel

For Grade 60 deformed bars, ACI 318 requires temperature and shrinkage reinforcement, placed perpendicular to the main steel, of at least

$A_{s, temp} = 0.0018 A_{g} = 0.0018 bh$

Example: One-way slab reinforcement

Given $b = 12"$ (per foot of width), $d_{p ro} = 5"$ , $f_{c}^{'} = 4, 000$ psi, $f_{y} = 60, 000$ psi, and $M_{u} = 96$ in-kip per foot of width. $R_{n} = \frac{M _{u}}{ϕ b d ^{2}} = 0.356$ ksi, so $ρ = 0.0063$ and $A_{s} = ρ b d = 0.38 in^{2}$ per foot, which exceeds $A_{s, min}$ (for example, $0.0018 bh = 0.13 in^{2}$ per foot for a 6 in slab).

Answer: $A_{s} \approx 0.38 in^{2}$ per foot of slab width.

Column design

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

Here $α$ is a reduction factor that accounts for accidental eccentricity in axially loaded columns - $0.80$ for tied columns and $0.85$ for spiral columns.

Use:

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

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.

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

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

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

(Here $ρ_{g}$ is the same steel ratio defined above, and $ϕ P_{n o}$ is the pure axial capacity with no eccentricity, so $α$ does not apply.)

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

$ϕ P_{n} > P_{u} so OK$

Example: Tied column axial capacity

Given $A_{g} = 256 in^{2}$ , $f_{c}^{'} = 4, 000$ psi, $f_{y} = 60, 000$ psi, $ρ_{g} = 0.02$ , $α = 0.80$ , $ϕ = 0.65$ (ignoring bending). $A_{s t} = ρ_{g} A_{g} = 5.12 in^{2}$ , so $ϕ P_{n} = α ϕ [0.85 f_{c}^{'} (A_{g} - A_{s t}) + f_{y} A_{s t}] \approx 603 kip$ .

Answer: $ϕ P_{n} \approx 603 kip$ , checked against the factored axial demand $P_{u}$ .

Footing design

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

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

Assume a trial depth $d$ , then verify it against the one-way and punching-shear checks below - these confirm the trial value or signal that $d$ needs to be increased.

Check beam shear:

The critical section for one-way (beam) shear is taken at a distance $d$ from the face of the column, across width $w$ :

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

Check punching shear:

The critical section for punching (two-way) shear runs around a perimeter $P_{o}$ located $d /2$ from the face of the column, enclosing the area $A_{o}$ shown in the figure above:

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

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

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

$A_{s} = max (ρ_{m i n} b d, ρ b d)$

Example: Isolated footing area

Given $D L = 100$ kip, $LL = 60$ kip, and $BC = 4$ ksf, with no wind or seismic load. $A = \frac{D L + LL}{BC} = \frac{160}{4} = 40 ft^{2}$ .

Answer: the footing needs a plan area of at least $40 ft^{2}$ , for example a $6. 3^{'} \times 6. 3^{'}$ square footing.