CELE Reinforced Concrete Footings, Bond and Development — Study Notes

Reinforced Concrete Footings, Bond and Development - Study Notes

Footings are critical load-bearing elements that transfer column and wall loads safely to the underlying soil. They must be designed to resist two distinct failure modes: (1) bearing failure (excessive settlement) controlled by soil pressure, and (2) structural failure (shear punching, beam shear, and flexure) governed by concrete strength and reinforcement. Additionally, the reinforcing bars embedded in footings must be developed—anchored adequately through bond—to mobilize their full yield strength. This chapter covers the complete design process: footing sizing from service loads, two-way punching shear checks, one-way beam shear verification, flexural analysis and bar placement, and critical development length calculations per NSCP 2015 (which adopts ACI 318 provisions). Mastery of these concepts is essential for the PRC Civil Engineer Licensure Examination.

1. Footing Sizing and Bearing Capacity

Footings must be sized from the service (working) load and the allowable soil bearing capacity $q_a$ (in kPa, typically provided by the geotechnical engineer or local code). The required plan area is:

A_{\text{req}} = \frac{P_{\text{service}}}{q_a}

For a square footing, the side length is $B = \sqrt{A_{\text{req}}}$ . Rectangular footings are sized similarly: $L \times B = A_{\text{req}}$ .

Critical Distinction: Sizing uses service (unfactored, working) loads. Once the plan dimensions are fixed, all subsequent structural design (shear, flexure, development) uses the factored design load $P_u = 1.4P_\text{DL} + 1.7P_\text{LL}$ (or per NSCP 2015 load factors).

Net Upward Design Pressure: With factored load and plan area $A_\text{footing}$ :

q_u = \frac{P_u}{A_\text{footing}}

This $q_u$ represents the upward pressure exerted by the soil on the footing base during design (ultimate) conditions and is used in shear and flexure calculations.

Practical Note (PRC Common Error): Candidates often confuse the sizing phase (service-load-based) with the design phase (factored-load-based). Always size first with service load and $q_a$ ; then apply load factors and use $q_u$ for structural checks.

Examples

Example 1.1: A reinforced concrete column transmits a service dead load of 800 kN and service live load of 500 kN to soil with allowable bearing capacity $q_a = 180$ kPa. Design a square footing plan area.

Solution

Step 1: Total service load.

P_{\text{service}} = 800 + 500 = 1300 \text{ kN}

Step 2: Required area.

A_{\text{req}} = \frac{P_{\text{service}}}{q_a} = \frac{1300}{180} = 7.222 \text{ m}^2

Step 3: Square side.

B = \sqrt{7.222} = 2.687 \text{ m}

Practical selection: Use $B = 2.70$ m (or 2.75 m to provide a small margin). Assume plan area = $2.70 \times 2.70 = 7.29$ m².

Step 4: Factored load (NSCP 2015).

P_u = 1.4(800) + 1.7(500) = 1120 + 850 = 1970 \text{ kN}

Step 5: Design pressure.

q_u = \frac{1970}{7.29} = 270 \text{ kPa}

This $q_u = 270$ kPa is used for all subsequent shear and flexure calculations.

Key points

Sizing area determined from service load and allowable soil bearing $q_a$ (working stress approach)
Once dimensions are set, switch to factored load $P_u$ and compute $q_u = P_u / A_\text{footing}$ for all structural design
Square footings: $B = \sqrt{A_\text{req}}$ ; rectangular: optimize based on column and site geometry
Footing thickness must be adequate for shear and development; typically 300–600 mm for low to medium loads
Plan dimensions are set at the base; top surface is usually level (for simple design) or sloped (for drainage)

2. Two-Way Punching Shear

Punching shear is a critical failure mode for footings. The column (or wall) tends to "punch" through the footing along a perimeter at a distance $d/2$ (effective depth) from the column faces. This is a two-way (or three-way) shear around the column.

Critical Perimeter for Square Column:

For a square interior column of side $c$ and footing effective depth $d$ , the critical perimeter is:

b_o = 4(c + d)

This perimeter is measured at $d/2$ from all four column faces, forming a square outline.

Nominal Punching Shear Strength:

The critical equation for square/compact columns (per NSCP 2015 / ACI 318) is:

V_c = 0.33 \, \lambda \sqrt{f'_c} \, b_o d

where:

$f'_c$ = concrete compressive strength (MPa)
$\lambda$ = 1.0 (normal-weight concrete); 0.75 (lightweight); typically 1.0 in design
$b_o$ = critical perimeter (mm)
$d$ = effective depth (mm)
$V_c$ is in Newtons; divide by 1000 for kN

The design shear strength is $\phi V_c$ with $\phi = 0.75$ for shear (NSCP 2015).

Punching Demand (Factored Shear Force):

The upward soil pressure acts over the entire footing area. The punching demand is the total upward pressure outside the critical perimeter:

V_u = q_u \left[ A_{\text{footing}} - (c + d)^2 \right]

Note: $(c + d)^2$ is the area inside the critical perimeter where shear stress is not computed.

Design Check:

\phi V_c \ge V_u \quad \Rightarrow \quad 0.75 V_c \ge V_u

If this check fails, either (a) increase footing thickness $d$ , (b) increase concrete strength $f'_c$ , or (c) add punching shear reinforcement (stirrups or shear studs).

NSCP 2015 Alternative Equations (for reference):

ACI 318 / NSCP also permits two alternative expressions involving the aspect ratio $\beta_c$ (column perimeter / column width) and the position $\alpha_s$ . These typically yield higher $V_c$ values for rectangular columns and are not governing for square columns.

Examples

Example 2.1: Check two-way punching shear for the footing from Example 1.1. Given: square footing $B = 2.70$ m, square column $c = 0.40$ m (400 mm), footing effective depth $d = 0.50$ m (500 mm), concrete strength $f'_c = 28$ MPa, factored load $P_u = 1970$ kN, $\lambda = 1.0$ (normal-weight).

Solution

Step 1: Compute design pressure.

q_u = \frac{P_u}{B^2} = \frac{1970}{2.70^2} = \frac{1970}{7.29} = 270.1 \text{ kPa} = 0.2701 \text{ N/mm}^2

Step 2: Critical perimeter.

b_o = 4(c + d) = 4(400 + 500) = 4(900) = 3600 \text{ mm}

Step 3: Nominal punching shear strength.

V_c = 0.33 \times 1.0 \times \sqrt{28} \times 3600 \times 500

= 0.33 \times 5.292 \times 1\,800\,000 = 3\,143\,520 \text{ N} = 3143.5 \text{ kN}

Step 4: Design strength.

\phi V_c = 0.75 \times 3143.5 = 2357.6 \text{ kN}

Step 5: Punching demand.

V_u = q_u \left[ A_\text{footing} - (c+d)^2 \right] = 270.1 \left[ 7.29 - (0.90)^2 \right]

= 270.1 \times (7.29 - 0.81) = 270.1 \times 6.48 = 1750.2 \text{ kN}

Step 6: Check.

\phi V_c = 2357.6 \text{ kN} \ge V_u = 1750.2 \text{ kN} \quad \checkmark

Result: Punching shear is adequate. No additional shear reinforcement is required for punching.

Example 2.2 (PRC-style): A rectangular footing is $4.0 \times 3.0$ m, $d = 0.60$ m, supports a square column $0.50 \times 0.50$ m, $f'_c = 35$ MPa, $P_u = 2800$ kN. Is punching shear OK?

Solution

Step 1: Footing area and design pressure.

A = 4.0 \times 3.0 = 12.0 \text{ m}^2 = 12\,000 \text{ mm}^2

q_u = \frac{2800}{12.0} = 233.33 \text{ kPa}

Step 2: Critical perimeter (square column).

b_o = 4(500 + 600) = 4(1100) = 4400 \text{ mm}

Step 3: Shear strength.

V_c = 0.33 \times 1.0 \times \sqrt{35} \times 4400 \times 600

= 0.33 \times 5.916 \times 2\,640\,000 = 5\,156\,000 \text{ N} = 5156 \text{ kN}

\phi V_c = 0.75 \times 5156 = 3867 \text{ kN}

Step 4: Demand.

V_u = 233.33 \left[ 12.0 - (1.1)^2 \right] = 233.33 \times (12.0 - 1.21) = 233.33 \times 10.79 = 2520 \text{ kN}

Step 5: Check.

\phi V_c = 3867 \text{ kN} \ge V_u = 2520 \text{ kN} \quad \checkmark

Result: Punching is adequate.

Key points

Critical perimeter for punching: $b_o = 4(c + d)$ measured at $d/2$ from column face
Governing shear strength for square columns: $V_c = 0.33 \lambda \sqrt{f'_c} b_o d$
Punching demand: $V_u = q_u [A_\text{footing} - (c+d)^2]$ (pressure outside the critical perimeter)
Design check: $\phi V_c \ge V_u$ with $\phi = 0.75$
If punching shear is insufficient, increase footing thickness, $f'_c$ , or provide shear reinforcement
Common board-exam error: Using the gross footing area instead of the area outside the critical perimeter in the $V_u$ calculation

3. One-Way (Beam) Shear

In addition to punching shear, the footing must be checked for one-way (beam) shear along any vertical plane at a distance $d$ from the column face across the full width of the footing.

Critical Section for Beam Shear:

The critical section is at distance $d$ from the face of the column (not from the center). For a square column of side $c$ on a footing of width $B$ , the cantilevered length is:

\ell_{\text{cant}} = \frac{B - c}{2}

The one-way shear check is performed at distance $d$ from the column face:

d_\text{section} = \ell_{\text{cant}} - d = \frac{B - c}{2} - d

One-Way Shear Force:

V_u = q_u \times B \times d_\text{section}

(Here $B$ is the footing width perpendicular to the shear plane; for a square footing, both directions have the same $V_u$ by symmetry.)

Nominal One-Way Shear Strength:

V_c = 0.17 \, \lambda \sqrt{f'_c} \, B d

Design strength: $\phi V_c = 0.75 V_c$ (shear, $\phi = 0.75$ ).

Design Check:

\phi V_c \ge V_u

Relationship Between the Two Shear Checks:

For typical footings with moderate overhang, punching shear is usually more critical (less margin). One-way shear typically governs only for very thin, heavily overhanging footings. It is still checked to satisfy code requirements.

Examples

Example 3.1: Check one-way shear for the footing in Example 2.1 ( $B = 2.70$ m, $c = 0.40$ m, $d = 0.50$ m, $f'_c = 28$ MPa, $q_u = 270.1$ kPa).

Solution

Step 1: Cantilever length to shear section.

\ell_{\text{cant}} = \frac{B - c}{2} = \frac{2.70 - 0.40}{2} = \frac{2.30}{2} = 1.15 \text{ m}

d_{\text{section}} = 1.15 - 0.50 = 0.65 \text{ m}

Step 2: One-way shear force.

V_u = q_u \times B \times d_{\text{section}} = 270.1 \times 2.70 \times 0.65 = 475.1 \text{ kN}

Step 3: One-way shear strength.

V_c = 0.17 \times 1.0 \times \sqrt{28} \times 2700 \times 500 = 0.17 \times 5.292 \times 1\,350\,000

= 1\,212\,000 \text{ N} = 1212 \text{ kN}

\phi V_c = 0.75 \times 1212 = 909 \text{ kN}

Step 4: Check.

\phi V_c = 909 \text{ kN} \ge V_u = 475.1 \text{ kN} \quad \checkmark

Result: One-way shear is adequate.

Key points

One-way shear critical section is at distance $d$ from the column face
One-way shear strength: $V_c = 0.17 \lambda \sqrt{f'_c} B d$ (lower coefficient than punching)
Design check: $\phi V_c \ge V_u$ with $\phi = 0.75$
For typical footings, punching shear is usually more critical than one-way shear
Always perform both checks; omitting one is a common exam error

4. Flexural Design

After confirming that shear capacity is adequate, the footing must be designed for flexure (bending moment). The critical section for moment in a footing is at the face of the column.

Critical Section for Moment:

The footing acts as a cantilever beam extending from the column face to the edge. At the column face, the cantilevered slab width is $\frac{B - c}{2}$ (for a square footing and square column).

Flexural Demand (Factored Moment):

The upward soil pressure over the cantilevered length $\ell = \frac{B - c}{2}$ creates a moment:

M_u = q_u \times B \times \frac{\ell^2}{2} = q_u \times B \times \frac{1}{2} \left( \frac{B - c}{2} \right)^2

Simplifying:

M_u = \frac{q_u \times B \times (B - c)^2}{8}

For footings extending in both directions symmetrically, apply this calculation in each direction.

Required Flexural Reinforcement:

Using the standard beam design formula, the required tension steel area is:

A_s = \frac{M_u}{\phi f_y \left( d - \frac{a}{2} \right)}

where $a = \frac{A_s f_y}{0.85 f'_c B}$ (depth of compression block).

Alternatively, use the design coefficient $R_u = \frac{M_u}{B d^2}$ and solve iteratively or use ACI tables.

For simplicity in hand calculations (as in PRC exams), assume $a \approx d/5$ or use the approximate formula:

A_s \approx \frac{M_u}{0.9 f_y d}

Minimum Reinforcement:

Per ACI 318 / NSCP 2015, the minimum reinforcement ratio is:

\rho_{\min} = \frac{1.4}{f_y} = \frac{1.4}{415} = 0.00337

So $A_{s,\min} = \rho_{\min} \times B \times d = 0.00337 B d$ .

Bar Layout (Reinforcement Distribution):

For a square footing, bars run in both directions. Typically, place the required $A_s$ near the tension face (bottom). The bars should be:

Distributed uniformly across the width $B$
Spaced not more than $d$ or $300$ mm apart (NSCP/ACI)
Adequately covered and developed (covered in Bond and Development section)

Common Practice: In footings, reinforce both the bottom (tension) in both orthogonal directions. Some designs also use top bars (negative reinforcement) near the column to control crack control if loads are eccentric.

Examples

Example 4.1: Design flexural reinforcement for the footing in Example 2.1. Given: $B = 2.70$ m, $c = 0.40$ m, $d = 0.50$ m, $f'_c = 28$ MPa, $f_y = 415$ MPa, $q_u = 270.1$ kPa.

Solution

Step 1: Cantilever length.

\ell = \frac{B - c}{2} = \frac{2.70 - 0.40}{2} = 1.15 \text{ m}

Step 2: Factored moment at column face.

M_u = q_u \times B \times \frac{\ell^2}{2} = 270.1 \times 2.70 \times \frac{(1.15)^2}{2}

= 270.1 \times 2.70 \times \frac{1.3225}{2} = 270.1 \times 2.70 \times 0.66125

= 483.0 \text{ kN}\cdot\text{m}

Step 3: Approximate required area (using $A_s \approx M_u / (0.9 f_y d)$ ).

A_s \approx \frac{483.0 \times 10^6}{0.9 \times 415 \times 500} = \frac{483\,000\,000}{186\,750} = 2586 \text{ mm}^2

Step 4: Check minimum reinforcement.

A_{s,\min} = 0.00337 \times 2700 \times 500 = 4544 \text{ mm}^2

Since $A_{s,\min} > A_s$ (from moment), use minimum reinforcement:

A_s = 4544 \text{ mm}^2

Step 5: Bar selection and spacing. Use $20$ mm diameter bars ( $A_b = 314$ mm² each):

N = \frac{4544}{314} \approx 14.5 \text{ bars} \quad \Rightarrow \text{ use } 15 \text{ bars}

Spacing: $s = \frac{B}{N-1} = \frac{2700}{14} \approx 193 \text{ mm}$ (acceptable, $< 300$ mm).

Alternatively, use $16$ mm bars ( $A_b = 201$ mm²):

N = \frac{4544}{201} \approx 22.6 \text{ bars} \quad \Rightarrow \text{ use } 23 \text{ bars}, \ s \approx 121 \text{ mm}

Result: Provide $15$ bars at $\approx 190$ mm c/c (or $23$ bars at $\approx 120$ mm c/c if using smaller diameter). Bars run in both principal directions.

Key points

Critical flexural section is at the face of the column
Moment: $M_u = q_u \times B \times \ell^2 / 2$ where $\ell = (B - c) / 2$ is the cantilever length
Simplification: $M_u = q_u B (B - c)^2 / 8$
Required area: $A_s = M_u / [0.9 f_y d]$ (approximate); iterate for exact calculation
Minimum reinforcement: $\rho_{\min} = 1.4 / f_y \approx 0.0034$ (per ACI 318)
Bar spacing: $\le d$ or $300$ mm (typically 150–250 mm in practice)
All bars must be developed; development length is calculated separately (see Bond and Development section)

5. Bond and Development Length

Reinforcing bars embedded in concrete develop their yield strength through bond stress (adhesion and friction) along the bar surface. The bar must be long enough to transfer the full yield force to the concrete. This embedment length is called the development length $\ell_d$ .

Why Development Matters in Footings:

In a footing, bars placed near the bottom are anchored by embedment into the footing depth. If $\ell_d$ is too short, the bars cannot reach their yield force, and the footing will fail prematurely (bar slip). The development length depends on:

Bar diameter $d_b$
Concrete strength $f'_c$
Yield strength $f_y$
Bar position (top/bottom, confined/unconfined)
Cover and spacing

Simplified Development Length Formula (NSCP 2015 / ACI 318):

For tension bars (normal-weight concrete, $\lambda = 1.0$ ):

\ell_d = \frac{f_y \psi_t \psi_e}{K \lambda \sqrt{f'_c}} d_b

where:

$f_y$ = yield strength of bar (MPa), typically 415 or 500 for Philippine steel
$d_b$ = bar diameter (mm)
$f'_c$ = concrete strength (MPa)
$K$ = coefficient depending on bar size:
- $K = 2.1$ for bars $\le 20$ mm diameter ("small bars")
- $K = 1.7$ for bars $> 20$ mm diameter ("large bars")
$\psi_t$ = top-bar factor:
- $\psi_t = 1.3$ for top bars (cast with fresh concrete above)
- $\psi_t = 1.0$ for bottom bars
$\psi_e$ = epoxy-coating factor:
- $\psi_e = 1.5$ for epoxy-coated bars (rare in footings)
- $\psi_e = 1.0$ for uncoated bars (normal)
$\lambda$ = 1.0 for normal-weight concrete

Minimum Development Length:

Regardless of calculation, $\ell_d \ge 300$ mm per NSCP 2015 / ACI 318.

Modified (Favorable) Conditions:

The above formula applies when bars have adequate cover and spacing:

Clear cover $\ge d_b$ and
Clear spacing between bars $\ge d_b$ and
Minimum stirrups present

If these conditions are not met, the coefficients $K$ are reduced (less-favorable conditions):

$K = 1.4$ for small bars ( $\le 20$ mm)
$K = 1.1$ for large bars ($> 20$ mm)

Hooks and Splices:

Hooks (90° or 180°) reduce the required development length. A standard 180° hook can develop a bar in approximately $0.7 \ell_d$ (straight embedment). Hooks are used when straight length is constrained.
Mechanical anchorages (plates, caps) can also reduce $\ell_d$ .
Splices (lap, welded, mechanical) must develop both bars; lap-splice length typically $= 1.3 \ell_d$ or per code tables.

Application to Footings:

In a footing, bars are placed near the bottom (tension face). The development length is measured from the critical section (at the column face, where moment is maximum and bars reach peak stress) to the end of the bar. Typically:

\text{Available length} = \text{half footing width} - \text{cover at edge}

For example, if $B = 2.70$ m and cover = 75 mm:

\text{Available length} = \frac{2.70}{2} - 0.075 = 1.35 - 0.075 = 1.275 \text{ m} = 1275 \text{ mm}

If the calculated $\ell_d > 1275$ mm, the straight bar cannot fit; use hooks or reduce bar diameter.

Common Board-Exam Pitfall:

Forgetting to add $\psi_t$ for top bars (very common; $\psi_t = 1.3$ increases $\ell_d$ by 30%)
Using $K = 2.1$ for all bars (only for $\le 20$ mm)
Omitting the 300 mm minimum
Calculating available length incorrectly (not accounting for cover or bar bend radius)

Examples

Example 5.1: Compute development length for a $20$ mm bottom bar in the footing of Example 4.1. Given: $f_y = 415$ MPa, $f'_c = 28$ MPa, bottom bar ( $\psi_t = 1.0$ ), uncoated ( $\psi_e = 1.0$ ), adequate cover and spacing (favorable conditions), normal-weight concrete ( $\lambda = 1.0$ ).

Solution

Step 1: Identify bar size and coefficient. Since $d_b = 20$ mm (at the boundary), use $K = 2.1$ (small bar, favorable).

Step 2: Apply development formula.

\ell_d = \frac{f_y \psi_t \psi_e}{K \lambda \sqrt{f'_c}} d_b = \frac{415 \times 1.0 \times 1.0}{2.1 \times 1.0 \times \sqrt{28}} \times 20

= \frac{415}{2.1 \times 5.292} \times 20 = \frac{415}{11.113} \times 20 = 37.35 \times 20 = 747 \text{ mm}

Step 3: Check minimum.

\ell_d = 747 \text{ mm} \ge 300 \text{ mm} \quad \checkmark

Step 4: Check available embedment (from column face to footing edge). Available = half width – cover = $1.35 - 0.075 = 1.275$ m = $1275$ mm.

Since available ($1275$ mm) $\gg$ required ($747$ mm), development is easily satisfied. No hooks needed.

Example 5.2: Compute $\ell_d$ for a $25$ mm top bar (same footing conditions, but positioned as a top bar). $\psi_t = 1.3$ (top bar), others as in Example 5.1.

Solution

Step 1: Bar size and coefficient. Since $d_b = 25$ mm $> 20$ mm, use $K = 1.7$ (large bar, favorable).

Step 2: Top-bar development length.

\ell_d = \frac{415 \times 1.3 \times 1.0}{1.7 \times 1.0 \times \sqrt{28}} \times 25

= \frac{539.5}{1.7 \times 5.292} \times 25 = \frac{539.5}{8.996} \times 25 = 60.0 \times 25 = 1500 \text{ mm}

Step 3: Check.

\ell_d = 1500 \text{ mm} \ge 300 \text{ mm} \quad \checkmark

Step 4: Available embedment. If the top bar extends from the column face inward (into the footing thickness), the available length is typically the footing thickness (if embedded fully). Assuming footing thickness $h = 0.60$ m (typical):

\text{Available} \approx 600 - \text{cover} = 600 - 50 = 550 \text{ mm}

Since required ($1500$ mm) $> $ available ($550$ mm), straight embedment is insufficient. Use a 180° hook (reduces length to $\approx 0.7 \times 1500 = 1050$ mm, still tight) or bend the bar (form an L or anchor it differently).

Example 5.3 (Unfavorable conditions): Recompute $\ell_d$ for the $20$ mm bottom bar from Example 5.1, but assuming inadequate spacing/cover (unfavorable condition). All other parameters unchanged.

Solution

Step 1: Unfavorable conditions apply. Use reduced coefficient $K = 1.4$ (instead of 2.1).

Step 2: Unfavorable development length.

\ell_d = \frac{415 \times 1.0 \times 1.0}{1.4 \times 1.0 \times \sqrt{28}} \times 20

= \frac{415}{1.4 \times 5.292} \times 20 = \frac{415}{7.409} \times 20 = 56.0 \times 20 = 1120 \text{ mm}

Compare to favorable ($747$ mm): unfavorable is 50% longer (1120 vs 747 mm).

Step 3: Check availability. Available $= 1275$ mm (same as before).

Since $1120 < 1275$ mm, development is still OK, but the margin is tighter. Design principle: Provide adequate spacing and cover to benefit from the favorable (shorter) $\ell_d$ .

Key points

Development length $\ell_d$ ensures bar reaches full yield force via bond
Simplified formula: $\ell_d = [f_y \psi_t \psi_e / (K \lambda \sqrt{f'_c})] d_b$
Coefficient $K$ : 2.1 for bars $\le 20$ mm; 1.7 for larger bars (favorable conditions with adequate cover/spacing)
Unfavorable conditions reduce $K$ to 1.4 and 1.1 respectively
Minimum: $\ell_d \ge 300$ mm always
Top-bar factor $\psi_t = 1.3$ (increases $\ell_d$ ); bottom-bar factor $\psi_t = 1.0$
Epoxy factor $\psi_e = 1.0$ (normal uncoated); $\psi_e = 1.5$ (epoxy-coated, rare)
Available embedment is typically half the footing width minus cover at the edge
Hooks can reduce required length; splices require longer lap lengths
Critical section for development in footings is typically at the column face (maximum moment)

6. Practical Design Workflow and Common Pitfalls

A systematic approach ensures accurate footing design and helps avoid exam errors:

Step-by-Step Workflow:

Phase 1: Sizing (Service Loads)

Obtain column load (D + L) and allowable soil bearing $q_a$ from geotechnical report.
Compute required area: $A_{\text{req}} = P_{\text{service}} / q_a$ .
Select plan dimensions $B \times L$ (typically square, $B = \sqrt{A}$ ).
Choose footing thickness $h$ (typically 0.3–0.6 m for light to medium loads; 0.8–1.2 m for heavy industrial loads). Rule of thumb: $h \approx (B - c) / 4$ to $(B - c) / 2$ for moderate overhang.

Phase 2: Structural Design (Factored Loads)

Apply load factors: $P_u = 1.4 P_{\text{DL}} + 1.7 P_{\text{LL}}$ (NSCP 2015).
Compute design pressure: $q_u = P_u / A$ .
Two-way shear check: Compute $b_o = 4(c + d)$ , $V_c = 0.33\lambda\sqrt{f'_c}b_o d$ , demand $V_u = q_u[A - (c+d)^2]$ . Verify $\phi V_c \ge V_u$ ( $\phi = 0.75$ ).
One-way shear check: Compute $V_u$ at distance $d$ from column face, $V_c = 0.17\lambda\sqrt{f'_c}Bd$ . Verify $\phi V_c \ge V_u$ .
Flexure design: Compute $M_u = q_u B (B-c)^2 / 8$ . Find $A_s$ using $A_s = M_u / (0.9 f_y d)$ . Compare to $A_{s,\min} = 0.00337 B d$ ; use the larger.
Select bar size and spacing. Verify spacing $\le 300$ mm.
Development length: For each bar, compute $\ell_d$ using the formula. Check that available embedment $\ge \ell_d$ . If not, use hooks or reduce bar diameter.

Phase 3: Detailing

Show bar layout (plan and section views).
Specify cover (typically 40–50 mm for interior footings, 75 mm for exposed surfaces).
Provide clear dimensions, bar size, spacing, and embedment lengths.
Note any hooks or bends.

Common Pitfalls (PRC Board Exams):

Pitfall	Impact	Prevention
Confusing service and factored loads	Wrong footing size or design pressure	Always label: sizing uses service $P$ ; design uses $P_u$
Using gross area for punching demand	Overestimating $V_u$ , incorrect conclusion	Subtract $(c+d)^2$ from footing area; this is the area outside the critical perimeter
Omitting one-way shear check	Missing a potential failure mode	Always check both punching and beam shear; list both in solutions
Wrong critical section for moment	Using wrong $\ell$ in $M_u$ calculation	Moment is at the column face, not at centerline or edge
Applying $\ell_d$ from edge instead of from critical section	Underestimating required embedment	In footings, development is measured from the column face (critical section) toward the edge
Forgetting $\psi_t = 1.3$ for top bars	Underestimating $\ell_d$ for top bars (can cause exam errors)	Top bar factor always 1.3; bottom bar always 1.0; check bar position
Using $K = 2.1$ for all bars	Wrong $\ell_d$ for bars $> 20$ mm	Split: $K = 2.1$ for $\le 20$ mm; $K = 1.7$ for $> 20$ mm
Minimum $\ell_d$ of 300 mm	Can be critical for small bars in thick footings	Always take max( $\ell_d$ calc, 300 mm)
Inadequate available length	Bars cannot develop; foundation fails	Always compute available embedment and compare to required $\ell_d$
Ignoring unfavorable conditions	Underestimating $\ell_d$ when cover/spacing are tight	Check code requirements for cover and spacing; if not met, use unfavorable $K$

Exam Strategy:

Label all quantities clearly. Examiners grade step-by-step; unclear notation loses points.
Show load factor application. Write " $P_u = 1.4(800) + 1.7(500) = ...$ " explicitly.
Two-way AND one-way shear. Never omit either check, even if one is clearly adequate.
Unit consistency. Keep dimensions in mm (or all in m), convert stresses to MPa, forces to kN. Show conversions: " $b_o = 3600$ mm $= 3.6$ m."
Development length table or formula. Write the formula and plug in values step-by-step, showing the coefficient choice ( $K = 2.1$ or $1.7$, $\psi_t$ , etc.).
Available vs. required. Explicitly compare available embedment to $\ell_d$ ; state whether hooks are needed.

Key points

Six-step design sequence: sizing → load factors → punching → one-way → flexure → development
Always distinguish service loads (sizing) from factored loads (design)
Perform both punching and beam shear checks; omitting one is a common error
Moment critical section is at the column face; cantilevered length = $(B - c) / 2$
Development length must account for bar position ( $\psi_t = 1.3$ for top bars, 1.0 for bottom)
Available embedment must be checked against required $\ell_d$ ; use hooks if needed
Clear documentation, labeled diagrams, and step-by-step calculations are essential for exam success

Diagrams

Footing Design Workflow: Service Load Sizing to Development Check

Systematic flowchart showing the six-phase footing design process: (1) service-load sizing, (2) factored-load setup, (3) two-way punching shear check, (4) one-way beam shear check, (5) flexural design and reinforcement selection, and (6) development length verification. Yellow diamonds indicate critical decision points where design iteration may be needed.

Development Length Decision Tree: Selecting Coefficient K and Factor ψ_t

Decision tree for computing development length, highlighting the critical choices: bar-size coefficient K (2.1 vs 1.7), favorable vs unfavorable conditions, and bar position factor ψ_t (1.0 vs 1.3). Guides students step-by-step to the correct formula and handling of marginal cases.

Footing Design: Core Concepts and Relationships

Mind map organizing footing design into five main branches: sizing (service-load based), two-way and one-way shear checks (with formulas), flexural design (with critical section and moment formula), development (bond and anchorage), and detailing (practical layout and documentation). Shows hierarchical relationships and key parameters.

Shear Adequacy Assessment: Pass/Fail States and Remediation

State diagram showing the shear check logic: compute nominal strength, compare design demand to factored capacity, and if inadequate, decide on remediation (increase thickness, strength, or add reinforcement) and recalculate. Illustrates the iterative nature of design.

In summary

Reinforced concrete footings transmit column loads safely to the soil. Design involves (1) sizing the plan area from service loads and allowable soil bearing, (2) checking shear via two criteria—punching (two-way, critical perimeter $b_o = 4(c + d)$ , $V_c = 0.33\lambda\sqrt{f'_c}b_o d$ ) and beam (one-way, $V_c = 0.17\lambda\sqrt{f'_c}Bd$ ), (3) designing flexure at the column face ( $M_u = q_u B \ell^2 / 2$ , where $\ell = (B-c)/2$ ), and (4) verifying development length ( $\ell_d$ per NSCP 2015/ACI 318) so bars reach yield. Key distinctions: sizing uses service loads and allowable soil pressure; structural design uses factored loads and $q_u$ . Punching demand excludes the area inside the critical perimeter. Development length depends on bar size ( $K = 2.1$ for $\le 20$ mm, $1.7$ for larger), position ( $\psi_t = 1.3$ for top bars, $1.0$ for bottom), and whether cover/spacing are adequate. The workflow is systematic: size → load factors → punching → beam shear → flexure → development. Common exam errors include confusing service and factored loads, omitting one shear check, applying the wrong bar coefficient for development, and forgetting the top-bar factor. Mastery of these concepts, paired with careful step-by-step calculation and clear documentation, is essential for the PRC Civil Engineer Licensure Examination.

Reinforced Concrete Footings, Bond and Development is chapter 6 of 7 in the Reinforced & Prestressed Concrete part of Super Tutor's CELE review. The Civil Engineer Licensure Examination is given by Professional Regulation Commission (PRC) — Board of Civil Engineering; its dates, format and passing rules are on the CELE page. See the CELE guide.

CELE Reinforced & Prestressed Concrete — Reinforced Concrete Footings, Bond and DevelopmentStudy Notes

Reinforced Concrete Footings, Bond and Development - Study Notes

1. Footing Sizing and Bearing Capacity

2. Two-Way Punching Shear

3. One-Way (Beam) Shear

4. Flexural Design

5. Bond and Development Length

6. Practical Design Workflow and Common Pitfalls

Diagrams

Study Reinforced Concrete Footings, Bond and Development in the Super Tutor app