--- title: "Steps of rectangular footing design" subject: "Design of RCC Structure" chapter: 11 key: "semester_7/design_of_rcc_structure:steps-of-rectangular-footing-design" source: https://www.bctengineer.com/q/semester_7/design_of_rcc_structure/steps-of-rectangular-footing-design site: BCT Engineering --- # Steps of rectangular footing design Design of RCC Structure · Chapter 11: Design of Footings - Asked in: 2065 Shrawan (old course), 7 marks - Video: [How to Design a Rectangular Footing with Concentric Loading - Part 1](https://www.youtube.com/watch?v=9gKOSwfTEeU) — PinoyCE - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_7/design_of_rcc_structure/steps-of-rectangular-footing-design - Whole chapter: https://www.bctengineer.com/md/semester_7/design_of_rcc_structure/chapter-11 ## Question Describe the steps for design of a rectangular RC footing. Why is shear reinforcement not provided in footing? ## Short answer Size footing for service load plus 10% self‑weight, keeping equal projections. Depth is governed by one‑way shear at d from column face and punching shear at d/2. Compute factored soil pressure, bending moments, and reinforcement (central band concentration in short direction). Shear reinforcement is omitted because depth is chosen so concrete shear capacity alone suffices, and stirrups are impractical in thick slabs cast on soil. ## Answer #### Steps for design of a rectangular RC footing (isolated) A rectangular footing is used when the column is rectangular, when there is a moment on the column, or when there is a space restriction on one side. 1. **Load and area**: take the column service load $P$ (add about 10% for the weight of footing and backfill) and the safe bearing capacity $q_a$. $$ A_{req} = \frac{1.1P}{q_a} $$ 2. **Dimensions**: choose $L \times B$ so that the projections beyond the column faces are equal in both directions, i.e. $L - a = B - c$ (where $a \times c$ is the column size), with $LB \ge A_{req}$. This gives about equal cantilever moments per metre width in both directions. If the column carries a moment $M$, check $q_{max} = \dfrac{1.1P}{LB}\left(1 + \dfrac{6e}{L}\right) \le q_a$ and $q_{min} \ge 0$ with $e = M/P$. 3. **Factored upward soil pressure**: $p_u = \dfrac{P_u}{LB}$ ($P_u = 1.5P$) (or the maximum pressure for a footing with moment). The self-weight is not included in bending and shear design. 4. **Depth** from the greater of: (a) one-way shear at a distance $d$ from the column face, and (b) punching shear at $d/2$ from the column face (permissible stress $k_s \cdot 0.25\sqrt{f_{ck}}$), and the bending requirement. Adopt $D$ and $d$. 5. **Bending moment** at the face of the column in each direction: $$ M_{u,L} = p_u B\frac{(L-a)^2}{8}, \qquad M_{u,B} = p_u L\frac{(B-c)^2}{8} $$ Find $A_{st}$ in each direction (minimum 0.12% of $bD$). 6. **Distribution of steel**: long-direction bars are spread uniformly over the width $B$. For short-direction bars, a fraction $\dfrac{2}{\beta + 1}$ of the total ($\beta = L/B$) is concentrated in a central band of width $B$; the rest is spread equally in the two outer portions (IS 456 cl. 34.3.1). 7. **Checks**: one-way shear $\tau_v \le \tau_c$ in both directions, punching shear, development length of the bars ($L_d \le$ available length from the face of the column), and bearing at the column-footing interface; provide dowels. 8. **Detail**: bars at the bottom, with 50 mm cover, plan and sections. #### Why shear reinforcement is not provided in a footing - A footing is a **thick member**, and its depth is made large enough that the shear stress is less than the permissible shear stress of concrete (for both one-way and punching shear). The concrete alone carries the shear, so stirrups are not needed. - Stirrups in a thick, heavy slab cast against soil are difficult to place and hold, and would cause congestion. - Bent-up bars or shear reinforcement are not efficient against punching in a thick footing, and the depth is easily increased at little cost. - Increasing the depth also gives rigidity, so the assumption of a uniform soil pressure is better satisfied. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. The required footing area is found from A. A = 1.1P/q_a (service load, 10% for self-weight) B. A = 1.5P/q_a C. A = P x q_a D. A = P/(1.1 q_a) Answer: A. About 10% of P is added for footing and backfill weight and divided by the safe bearing capacity. (Key point: Footing area) 2. Which pressure is used for bending and shear design? A. Factored net upward pressure p_u = P_u/(LB), excluding self-weight B. Service pressure including self-weight C. Zero pressure D. Pressure at the footing corner only Answer: A. Self-weight is balanced by soil reaction directly beneath it and does not cause bending or shear. (Key point: Design pressure) 3. In IS 456 cl. 34.3.1, what fraction of short-direction steel lies in the central band of width B? A. 2/(beta + 1) of the total B. 1/(beta + 1) C. All of it D. Half of it Answer: A. beta = L/B; the remaining steel is spread equally in the two outer portions. (Key point: Short-direction steel) 4. Depth of the footing is decided by checking A. One-way shear at d from the face and punching shear at d/2 B. Only bending C. Only bearing D. Only development length Answer: A. Shear usually governs; the depth is chosen so that tau_v <= tau_c and punching <= k_s x 0.25 sqrt(fck). (Key point: Depth criteria) 5. Why is shear reinforcement not provided in a footing? A. The footing is deep enough that concrete alone resists shear, and stirrups are hard to place in a thick slab B. Concrete has infinite shear strength C. Footings carry no shear D. Steel does not bond with concrete in footings Answer: A. Depth is increased until shear stress is below the permissible value, which is cheaper than stirrups. 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