EGB375 · Design of Concrete Structures
Shallow (Isolated Pad) Footing Design
An isolated pad footing spreads a single column load onto the soil, and in EGB375 Design of Concrete Structures at Queensland University of Technology it is designed to AS3600-2018 with four checks in two families: service (unfactored) loads fix the plan size so the soil is not overstressed, while factored ultimate loads verify the concrete strength. This chapter builds the workflow end to end — net allowable bearing pressure and pad size, the ultimate net soil pressure, punching (two-way) shear on a perimeter dom/2 beyond the column, flexure on the cantilevering pad, and one-way (flexural) shear at dv — reusing the slab-flexure and shear methods from earlier in the unit.
What this chapter covers
- 01Why a footing is a thick upside-down cantilever, and the service-vs-ultimate split
- 02Net allowable soil pressure qa,net = qa - 25D and the size (bearing) check
- 03Ultimate net soil pressure Pnet = (1.2 NG + 1.5 NQ)/Apad
- 04Punching (two-way) shear at dom/2: perimeter u, Vuo = u dom fcv and phi = 0.7
- 05The punching stress fcv = 0.17(1 + 2/beta_h) sqrt(f'c) and its 0.34 sqrt(f'c) cap
- 06Flexure at 0.7 a_sup: the cantilever moment M* and the design steel
- 07Minimum and crack-control steel (which usually governs a thick pad)
- 08One-way (flexural) shear at dv = max(0.72D, 0.9d) with phi = 0.75 and no stirrups
- 09How the two shear checks differ (dom vs dv, phi = 0.7 vs 0.75)
- 10Deepening the pad as the reliable fix when a strength check fails
Worked example: full isolated pad footing design (four checks)
- +1Net allowable pressure. qa,net = qa − 25D = 200 − 25×0.50 = 187.5 kPa.
- +2Size (Check 1). A_req = (NG+NQ)/qa,net = 1000/187.5 = 5.33 m^2 → side √5.33 = 2.31 m → provide a 2.4 m square pad (A_pad = 5.76 m^2). Service pressure = 1000/5.76 = 173.6 kPa ≤ 187.5, so bearing is OK.
- +1Ultimate net pressure. Pnet = (1.2×650 + 1.5×350)/5.76 = 1305/5.76 = 226.6 kPa.
- +1Effective depths. dx = 500 − 50 − 20/2 = 440 mm; dy = 500 − 50 − 20 − 20/2 = 420 mm; dom = ½(440+420) = 430 mm.
- +2Punching demand (Check 2). Critical side = c + dom = 400 + 430 = 830 mm → u = 4×830 = 3320 mm; A_crit = 0.830^2 = 0.689 m^2. V*_punch = Pnet(A_pad − A_crit) = 226.6×(5.76 − 0.689) = 1149 kN.
- +2Punching capacity. Square column so beta_h = 1: fcv = 0.17(1+2)√32 = 2.89 MPa, but the cap 0.34√32 = 1.923 MPa governs. Vuo = u·dom·fcv = 3320×430×1.923 = 2.746×10^6 N = 2746 kN; phi·Vuo = 0.7×2746 = 1922 kN ≥ 1149, so punching is OK.
- +2Cantilever moment (Check 3). a_sup = ½×400 = 200 mm; critical section at 0.7×200 = 140 mm from the centre, so L_cant = 1200 − 140 = 1.060 m. M* = Pnet·L_cant^2/2 = 226.6×1.060^2/2 = 127.3 kN·m/m.
- +2Flexural and minimum steel. Solving the stress block (alpha2 = 0.802, gamma = 0.89, phi = 0.85, d = 420 mm) needs Ast ≈ 726 mm^2/m. But fct.f = 0.6√32 = 3.39 MPa gives rho_min = 0.19(500/420)^2×3.39/500 → Ast,min = 768 mm^2/m, and rho_crack = 0.75×0.0035×(500/420) → Ast,crack = 1313 mm^2/m (largest, so crack steel governs). Provide N20 @ 220 mm each way (1409 mm^2/m ≥ 1313).
- +1One-way shear (Check 4). dv = max(0.72×500, 0.9×420) = max(360, 378) = 378 mm. Critical section at 200 + 378 = 578 mm from centre, so L_out = 1200 − 578 = 622 mm. V* = 226.6×0.622 = 140.9 kN/m; coexisting M* = 226.6×0.622^2/2 = 43.8 kN·m/m.
- +2Shear capacity. ex = [43.8×10^6/378 + 140.9×10^3]/(2×200000×1409) = 4.56×10^-4. With kdg = 32/(16+20) = 0.889, kv = [0.4/(1+1500×0.000456)]·[1300/(1000+0.889×378)] = 0.238×0.973 = 0.231. Vuc = 0.231×1000×378×√32 = 494 kN/m; phi·Vuc = 0.75×494 = 371 kN/m ≥ 140.9, so one-way shear is OK.
Key terms
- Net allowable pressure qa,net
- The soil bearing pressure left for the column load after the pad self-weight is subtracted: qa,net = qa - 25D (kPa), with D the pad thickness in metres and 25 kN/m^3 the unit weight of reinforced concrete. It is used with service (unfactored) loads to size the pad.
- Ultimate net soil pressure Pnet
- The factored net upward soil pressure used for all strength checks, Pnet = (1.2 NG + 1.5 NQ)/Apad (kPa). The pad self-weight is not included because it is carried straight into the soil and does not bend or punch the pad.
- Punching (two-way) shear
- The check that the column does not punch a cone through the pad. The critical perimeter u is set dom/2 beyond the column face; the demand is V*_punch = Pnet(Apad - Acrit) and the capacity is Vuo = u dom fcv, with the check V*_punch <= phi Vuo and phi = 0.7.
- Punching stress fcv
- The concrete punching-shear stress, fcv = 0.17(1 + 2/beta_h) sqrt(f'c), capped at 0.34 sqrt(f'c) (MPa), where beta_h is the ratio of the long to short column side. For a square or stocky column (beta_h < 2) the 0.34 sqrt(f'c) cap governs.
- Mean effective depth dom
- The average of the two bar-layer effective depths, dom = 0.5(dx + dy) in mm, used in the punching check. It differs from dv, the one-way shear depth.
- Flexural critical section (0.7 a_sup)
- The section at which the cantilever moment is taken, located 0.7 a_sup from the column centreline, where a_sup is half the column width. The design moment is Pnet times the cantilever span squared over two, per metre of width.
- Minimum and crack-control steel
- The tension steel must be the largest of the bending requirement rho_BM, the minimum-strength steel rho_min = 0.19(D/d)^2 fct.f/fsy (with fct.f = 0.6 sqrt(f'c)) and the crack-control steel rho_crack = 0.75 x 0.0035 (D/d). In a thick pad the crack-control minimum usually governs.
- One-way (flexural) shear
- The beam-type shear check on a full-width plane at dv = max(0.72D, 0.9d) from the column face. With no stirrups, capacity is Vuc = kv bv dv sqrt(f'c); the check is V* <= phi Vuc with phi = 0.75 (distinct from the punching phi = 0.7).
Shallow (Isolated Pad) Footing Design FAQ
Why are service loads used to size the footing but factored loads used for the strength checks?
Bearing is a serviceability condition: the pad must not overstress the soil under the loads it actually sees day to day, which are the unfactored service loads NG + NQ. Once the plan area is fixed, the concrete itself (punching, flexure, one-way shear) is a strength condition, so it is checked against the factored ultimate pressure Pnet = (1.2 NG + 1.5 NQ)/Apad. Mixing the two up, for example sizing on factored loads, gives an oversized pad and is a common error.
What is the difference between the two shear checks on a footing?
A footing carries two distinct shear checks. Punching (two-way) shear guards against the column punching a cone through the pad; it acts on a perimeter dom/2 beyond the column, uses the mean depth dom, and takes phi = 0.7. One-way (flexural) shear treats the pad as a wide beam and is checked on a full-width plane at dv = max(0.72D, 0.9d) from the face, using phi = 0.75. Using the wrong effective depth or the wrong phi in either is a classic lost mark.
Can AI help me with shallow footing design in EGB375?
Yes, for learning and rehearsal. Sia is an AI tutor that explains the footing workflow step by step: net allowable pressure and sizing, then the ultimate pressure and the punching, flexure and one-way-shear checks, and it can check your reasoning on practice pads and highlight where the crack-control minimum governs. It supports your study rather than replacing it: it will not sit the closed-book exam for you, and no tool can promise a particular mark or grade. Always confirm code clauses against AS3600 and your unit materials.
Studying with AI? Sia — free AI civil engineering tutor works through EGB375 step by step.
Exam move
Treat the footing as a fixed four-check routine, not a set of loose formulae. First size the pad on service loads: qa,net = qa - 25D, then A_req = (NG + NQ)/qa,net, and pick a square pad whose service pressure sits at or below qa,net. Then switch to the factored net pressure Pnet = (1.2 NG + 1.5 NQ)/Apad and run the three strength checks in order: punching (two-way) shear on the perimeter dom/2 beyond the column with phi = 0.7 and the fcv cap of 0.34 sqrt(f'c); flexure at 0.7 a_sup, taking the largest of rho_BM, rho_min and rho_crack for the steel; and one-way (flexural) shear at dv from the face with phi = 0.75. Rehearse one full pad end to end until the order and the two different phi values are automatic, keep service and ultimate loads strictly apart, and remember that deepening the pad is the reliable fix when a strength check fails. Because the final exam is closed-book and paper-based, condense the whole workflow, the AS3600 factors and one worked pad onto your permitted note sheets. The exam is marked out of 100 and worth 50% of the unit; once you confirm its duration on the QUT exam timetable, divide those minutes by 100 to set a consistent minutes-per-mark rate and spend on a footing question in proportion to its marks.
