Introduction
Designing a spread footing is a fundamental skill taught in Project Lead The Way (PLTW) courses, especially in the Engineering Design: Structures module (Unit 3.Plus, this activity challenges students to size a footing that safely supports a given load while satisfying code‑specified limits on bearing pressure, shear, and bending. That's why the “answers” to PLTW 3. 2‑9 are not just a set of numbers; they represent the logical process of translating loads, soil properties, and material strengths into a practical, constructible foundation. 2, Activity 9). In this article we walk through the entire problem, explain the engineering concepts behind each step, and provide a complete worked solution that can be used as a study guide or classroom reference.
1. What Is a Spread Footing?
A spread footing (also called a pad footing) is a shallow concrete slab that spreads a column or wall load over a larger area of soil. That said, its primary purpose is to keep soil bearing stress below the allowable limit and to prevent excessive settlement. Typical dimensions are rectangular or square, with a uniform thickness that resists bending and shear.
This is where a lot of people lose the thread Most people skip this — try not to..
Key design criteria:
| Criterion | Typical Requirement |
|---|---|
| Maximum bearing pressure | ≤ soil allowable pressure (often 2,000–3,000 lb/ft²) |
| Shear stress | ≤ concrete shear capacity (≈ 0.Now, 75√f′c) |
| Flexural stress | ≤ concrete flexural capacity (≈ 0. 6√f′c) |
| Minimum thickness | Often 6 in for residential, 8–12 in for heavier loads |
| Edge distance | ≥ 0. |
Understanding these limits is essential before any calculations begin.
2. PLTW 3.2‑9 Problem Statement
*A rectangular column with a cross‑section of 12 in × 12 in carries a factored axial load of 150 kips. Even so, concrete grade is 4,000 psi (f′c = 4 ksi) and steel grade is 60 ksi (fy = 60 ksi). On top of that, the column sits on a spread footing that must rest on a soil with an allowable bearing pressure of 2,500 lb/ft². Design the footing dimensions (length L, width B) and required thickness t such that bearing, shear, and flexural stresses are within limits. Assume a footing shape that is square for simplicity It's one of those things that adds up. Less friction, more output..
No fluff here — just what actually works.
The activity asks students to size the footing—determine L = B and t—while documenting the reasoning behind each decision And that's really what it comes down to..
3. Step‑by‑Step Design Procedure
3.1. Determine Required Footing Area from Bearing Capacity
The first limit to check is soil bearing pressure It's one of those things that adds up..
[ q_{allow}=2,500\ \text{lb/ft}^2 ]
[ P = 150\ \text{kips}=150,000\ \text{lb} ]
[ A_{min}= \frac{P}{q_{allow}} = \frac{150,000}{2,500}=60\ \text{ft}^2 ]
Since we are assuming a square footing, the side length L must satisfy
[ L^2 = A_{min} ;\Rightarrow; L = \sqrt{60}\approx 7.75\ \text{ft} ]
Round up to a convenient construction dimension, e.In practice, g. , 8 ft (96 in). This gives a safety margin for construction tolerances But it adds up..
3.2. Verify Shear Capacity
Concrete shear stress (τ) is approximated by
[ \tau = \frac{V}{b,t} ]
where
- V = shear force at the critical section (≈ 0.5 P for a uniformly loaded footing)
- b = width of the critical section (taken as footing width, L)
- t = footing thickness (unknown yet)
First compute V:
[ V = 0.5P = 0.5(150,000)=75,000\ \text{lb} ]
Concrete shear capacity (without steel) for f′c = 4 ksi:
[ \tau_{c} = 0.75\sqrt{f′c}=0.75\sqrt{4,000}=0.75\times 63.25\approx 47.4\ \text{psi} ]
Set τ ≤ τc:
[ \frac{75,000}{L,t} \le 47.4 ]
Convert L to inches (96 in):
[ \frac{75,000}{96,t} \le 47.4 ;\Rightarrow; t \ge \frac{75,000}{96 \times 47.4}= \frac{75,000}{4,550.4}\approx 16.
Thus, shear demands a thickness of ≈ 17 in Easy to understand, harder to ignore..
3.3. Verify Flexural (Bending) Capacity
The footing behaves like a slab supported on its edges. The maximum moment for a uniformly loaded square slab is:
[ M_{max}= \frac{q L^2}{16} ]
where (q = \frac{P}{A}) is the average pressure on the footing.
First compute q using the adopted size (8 ft × 8 ft = 64 ft²):
[ q = \frac{150,000}{64}=2,343.75\ \text{lb/ft}^2 ]
Convert q to lb/in² (divide by 144):
[ q = \frac{2,343.75}{144}=16.28\ \text{psi} ]
Now compute Mmax (in‑lb). Convert L to inches (96 in):
[ M_{max}= \frac{16.28 \times (96)^2}{16}= \frac{16.28 \times 9,216}{16}= \frac{150,100}{16}\approx 9,381\ \text{lb·in} ]
Flexural capacity of a rectangular concrete section (ignoring steel) is:
[ \phi M_n = \phi \cdot 0.9 f′c b t^2/6 ]
For a square footing, width b = L = 96 in. Using the ACI factor φ≈0.9 and simplifying:
[ \phi M_n = 0.9 \times 0.9 \times f′c \times \frac{b t^2}{6}=0.
Set (\phi M_n \ge M_{max}):
[ 0.81 \times 4,000 \times \frac{96 t^2}{6} \ge 9,381 ]
[ 0.81 \times 4,000 = 3,240 ]
[ \frac{96}{6}=16 ]
[ 3,240 \times 16 \times t^2 \ge 9,381 ;\Rightarrow; 51,840 t^2 \ge 9,381 ]
[ t^2 \ge \frac{9,381}{51,840}=0.181 ;\Rightarrow; t \ge 0.425\ \text{in} ]
The flexural check yields a tiny thickness, indicating bending is not governing for this load and footing size. In practice, minimum thickness requirements (often 6 in for residential, 8 in for commercial) dominate That's the part that actually makes a difference..
3.4. Apply Minimum Thickness & Construction Constraints
Code (ACI 318) typically requires a minimum thickness of 6 in for footings without heavy reinforcement, and 8 in when the footing is over 3 ft deep or when large moments are expected. Since our shear calculation demanded ~17 in, the controlling requirement is shear.
Thus, adopt:
- Length (L) = 8 ft (96 in)
- Width (B) = 8 ft (96 in)
- Thickness (t) = 18 in (round up to the nearest 2 in for bar placement)
3.5. Reinforcement Layout (Optional but Often Required)
Even though the activity focuses on sizing, PLTW expects a basic reinforcement plan:
-
Bottom steel: #5 bars at 12 in on center both ways.
Area per bar = 0.31 in² → total steel area ≈ 0.31 × (96/12)² = 0.31 × 64 = 19.8 in².
Steel ratio ρ = As/(b t) = 19.8/(96 × 18) ≈ 0.0115 (≈ 1.15 %). Below the 0.75 % minimum, so increase spacing to 9 in (ρ ≈ 0.015) And that's really what it comes down to.. -
Top steel (optional for temperature/shrinkage): #3 bars at 18 in on center.
All bars should be placed with a concrete cover of 2 in.
4. Full Set of Answers for PLTW 3.2‑9
| Item | Value | Reason |
|---|---|---|
| Required footing area | 60 ft² (minimum) | From bearing pressure limit |
| Adopted footing dimensions | 8 ft × 8 ft (64 ft²) | Rounded up for constructability |
| Average soil pressure | 2,344 lb/ft² | P/A with adopted area |
| Shear‑controlled thickness | 18 in | V/(L t) ≤ τc, rounded up |
| Bending‑controlled thickness | 0.5 % > 0.Because of that, 75 % minimum | |
| Top reinforcement | #3 @ 18 in O. This leads to both ways | Provides ρ ≈ 1. (optional) |
| Concrete cover | 2 in | Protects steel from corrosion |
| Final design | 8 ft × 8 ft × 18 in, #5 @ 9 in O.5 in (theoretical) | Flexural stress far below capacity |
| Minimum code thickness | 6 in (but overridden by shear) | ACI 318 |
| Bottom reinforcement | #5 @ 9 in O.C. C. C. |
5. Scientific Explanation Behind Each Check
5.1. Bearing Pressure
Soil behaves like a semi‑infinite elastic medium. The allowable bearing pressure is derived from field tests (plate load, Standard Penetration Test) and safety factors that account for variability in soil strength and load eccentricity. By ensuring P/A ≤ q_allow, we keep settlement within acceptable limits and avoid soil failure That's the part that actually makes a difference..
5.2. Shear
Shear in a footing occurs along a critical perimeter located at a distance d (approximately 0.5 t) from the column face. The shear stress distribution is highest near the column edge because the load is transferred over a short lever arm. The equation τ = V/(b t) assumes a uniform shear flow across the footing width, which is conservative for rectangular footings And that's really what it comes down to. That's the whole idea..
5.3. Flexure
The footing slab acts like a simply supported beam in two directions. The maximum moment for a uniformly loaded square slab is derived from plate theory (M = qL²/16). Day to day, concrete’s flexural capacity is governed by the modulus of rupture (≈ 0. Because of that, 6√f′c). In most spread footings, bending is rarely controlling because the footing width is large relative to the load magnitude, which is why shear often dictates thickness Small thing, real impact..
5.4. Reinforcement Ratio
The steel ratio ρ ensures that the concrete does not crack under service loads and that the steel yields before concrete crushes. 15 %) for footings, but many designers adopt 0.On top of that, 75 % as a practical lower bound for durability. 0015 (0.ACI 318 recommends a minimum ρ of 0.Adjusting bar spacing changes ρ directly.
Quick note before moving on.
6. Frequently Asked Questions (FAQ)
Q1: Why do we assume a square footing when the problem allows any shape?
A: A square shape simplifies calculations and provides symmetric load distribution, which is ideal for teaching concepts. In practice, rectangular footings are used when column spacing or site constraints dictate different dimensions.
Q2: Can we reduce the thickness by using higher‑strength concrete?
A: Yes. Increasing f′c raises both shear and flexural capacities (τc ∝ √f′c, M_n ∝ f′c). Here's one way to look at it: using 6,000 psi concrete would lower the required thickness from 18 in to roughly 14 in for the same shear demand That alone is useful..
Q3: What if the soil bearing capacity is lower, say 1,500 lb/ft²?
A: The required area would increase to (A = P/q_{allow}=150,000/1,500=100 ft²). The footing side would become √100 = 10 ft, and the shear thickness would change accordingly (larger L reduces shear stress, potentially allowing a thinner footing) Small thing, real impact. Practical, not theoretical..
Q4: Do we need to check punching shear under the column?
A: For spread footways, punching shear is typically evaluated at a distance d/2 from the column face. In this activity, the shear check performed on the entire footing perimeter is sufficient because the column is small (12 in) relative to the footing size And that's really what it comes down to..
Q5: How does eccentric loading affect the design?
A: Eccentricity shifts the resultant load away from the centroid, creating a non‑uniform pressure distribution. The design would require a larger footing area on the side of higher pressure and possibly a different shape to keep stresses within limits But it adds up..
7. Practical Tips for Students
- Round early, round often – Use construction‑friendly dimensions (whole inches, ½‑foot increments) to avoid confusion later.
- Keep a units sheet – Converting between kip, lb, ft, and in is a common source of error. Write every quantity with its unit before plugging into formulas.
- Check the most restrictive condition first – Bearing pressure gives a minimum area; shear usually provides a minimum thickness. Flexure is often non‑governing but still worth verifying.
- Document assumptions – State soil bearing capacity, concrete grade, safety factors, and any simplifications (uniform load, square footing). This mirrors real‑world engineering reports.
- Use a spreadsheet – Small changes in load or soil properties can be explored quickly, helping you see the sensitivity of each design parameter.
8. Conclusion
The PLTW 3.On the flip side, 2‑9 “Sizing a Spread Footing” activity encapsulates the core workflow of foundation design: start with soil bearing limits, verify shear, confirm flexure, and finish with reinforcement detailing. By following the systematic approach outlined above, students can confidently arrive at a safe, code‑compliant footing—in this case an 8 ft × 8 ft × 18 in concrete pad reinforced with #5 bars at 9 in on center. Mastery of these steps not only earns the correct answer for the PLTW worksheet but also builds a solid foundation (pun intended) for future structural engineering challenges.
