--- title: "Isolated footing 500x500 with moment, M25" subject: "Design of RCC Structure" chapter: 11 key: "semester_7/design_of_rcc_structure:isolated-footing-500x500-with-moment-m25" source: https://www.bctengineer.com/q/semester_7/design_of_rcc_structure/isolated-footing-500x500-with-moment-m25 site: BCT Engineering --- # Isolated footing 500x500 with moment, M25 Design of RCC Structure · Chapter 11: Design of Footings - Asked in: 2071 Chaitra, 14 marks - Video: [RCC 1 - Design of Isolated Column Footing](https://www.youtube.com/watch?v=epm-i1SXxIg) — Learn with Shruti - 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/isolated-footing-500x500-with-moment-m25 - Whole chapter: https://www.bctengineer.com/md/semester_7/design_of_rcc_structure/chapter-11 ## Question Design an isolated footing to carry a column load of 1300 kN and BM of 100 kN-m from both axes of column. Column is 500 mm x 500 mm in size with 25 mm diameter longitudinal steel. The bearing capacity of soil is 220 kN/m$^2$. Consider depth of foundation as 1.70 m. Take unit weight of soil as 18.5 kN/m$^3$. Use M25 grade concrete and Fe415 steel. ## Short answer Square footing 3.1 m × 3.1 m, overall depth 650 mm (effective 590 mm). Main reinforcement 16 mm @ 185 mm c/c both ways; one-way and punching shear are within limits. ## Answer #### Data and assumptions - Column 500 mm × 500 mm (long side $a=500$ mm along the footing length); footing concrete M25, steel Fe415; column concrete M25; column bars 8-25 mm. - Service load $P = 1300$ kN, service moment $M = 100$ kN·m (about the other axis also 100 kN·m). - Allowable (safe) bearing capacity of soil = 220 kN/m² at 1.7 m depth, soil unit weight 18.5 kN/m³. - Loads are service loads; moments of 100 kN·m act about both axes, so a square footing is used. - Clear cover 50 mm for the footing (IS 456 cl. 26.4.2.2); effective depth $d = D - 60$ mm (cover + half bar). Load factor 1.5. #### Size of footing Net safe bearing capacity = $q - \gamma D_f = 220 - 18.5 \times 1.7 = 188.55$ kN/m² (this takes care of the weight of footing and backfill). $$ A_{req} = \frac{P}{q_{net}} = \frac{1300.0}{188.55} = 6.89\ \text{m}^2 $$ Eccentricity $e = M/P = 100.0/1300.0 = 0.077$ m (and 0.077 m in the other direction), which is less than $L/6$, so there is no tension under the base. The size is chosen so that $$ q_{max} = \frac{P}{BL}\left(1 + \frac{6e}{L} + \frac{6e_y}{B}\right) \le q_a $$ Try $L \times B = 3.1 \times 3.1$ m ($A = 9.61$ m²): $q_{max} = 175.6$ kN/m² $\le 188.6$ and $q_{min} = 95.0$ kN/m² $> 0$. **Safe.** #### Factored soil pressure $P_u = 1.5 \times 1300.0 = 1950.0$ kN, $M_u = 150.0$ kN·m. $p_u = \dfrac{P_u}{BL}\left(1 \pm \dfrac{6e}{L} \pm \dfrac{6e_y}{B}\right)$: $p_{u,max} = 263.3$ kN/m², $p_{u,min} = 142.5$ kN/m² (average 202.9 kN/m²). Self-weight does not cause bending in the footing, so it is excluded. #### Bending moments and reinforcement Cantilever projections from the column face: along $L$: $(3100 - 500)/2 = 1300$ mm; along $B$: $(3100 - 500)/2 = 1300$ mm. With the maximum pressure assumed over the cantilever (conservative): $M_{u,L} = 689.8$ kN·m, $M_{u,B} = 689.8$ kN·m. Required depth for flexure: $M_{u,lim} = 0.138 f_{ck} b d^2$ is far larger than the moment for $d = 590$ mm, so the depth is governed by shear (one-way or punching). Trial **$D = 650$ mm, $d = 590$ mm**, found by the shear checks below. $A_{st} = \dfrac{0.5 f_{ck}}{f_y}\left[1 - \sqrt{1 - \dfrac{4.6 M_u}{f_{ck} b d^2}}\right] b d$; minimum steel = 0.12% of $bD$. | Direction | $M_u$ (kN·m) | $A_{st}$ req (mm²) | Provide | $A_{st}$ prov (mm²) | |---|---|---|---|---| | Along $L$ (width $B$) | 689.8 | 3341 | 16 mm @ 185 c/c | 3369 | | Along $B$ (width $L$) | 689.8 | 3341 | 16 mm @ 185 c/c | 3369 | #### Check for one-way shear (at distance $d$ from the column face) - Along $L$ side: $V_u = 579.6$ kN, $\tau_v = \dfrac{V_u}{B d} = 0.317$ MPa; $p_t = 0.184\%$, $\tau_c = 0.319$ MPa (IS 456 Table 19, SP 16 formula). $\tau_v < \tau_c$, safe. - Along $B$ side: $V_u = 579.6$ kN, $\tau_v = 0.317$ MPa; $p_t = 0.184\%$, $\tau_c = 0.319$ MPa. Safe. #### Check for two-way (punching) shear (at $d/2$ from the column face) - Perimeter $b_o = 2[(a + d) + (c + d)] = 2[(500+590) + (500+590)] = 4360$ mm - $V_u = P_u - p_u (a+d)(c+d) = 1950.0 - 202.9 \times 1.090 \times 1.090 = 1708.9$ kN (average pressure; the effect of moment transfer is neglected). - $\tau_v = \dfrac{V_u}{b_o d} = 0.664$ MPa; $k_s = 0.5 + \beta_c = 1.00$, $\tau_c' = k_s \times 0.25\sqrt{f_{ck}} = 1.250$ MPa. $\tau_v < \tau_c'$, **safe**. #### Development length $L_d = \dfrac{0.87 f_y \phi}{4\tau_{bd}}$, $\tau_{bd} = 2.24$ MPa (M25, deformed bars): 16 mm bars (both directions) → $L_d = 645$ mm. Available length from column face (less 50 mm end cover): 1250 mm along $L$, 1250 mm along $B$. **Satisfied** in both directions. #### Bearing stress and transfer of load from column Column area $A_2 = 500\times500 = 0.250$ m². Bearing stress $= P_u/A_2 = 7.80$ MPa. Permissible bearing stress: column concrete $0.45 f_{ck} = 11.25$ MPa; footing concrete $0.45 f_{ck}\sqrt{A_1/A_2}$ with $\sqrt{A_1/A_2} \le 2$ gives up to 22.50 MPa. The lesser, 11.25 MPa, governs. Capacity of bearing = 2812 kN $\ge$ $P_u = 1950$ kN, so the whole load can be transferred by bearing. Provide minimum dowels of 0.5% of column area = 1250 mm²: use 8 bars of 25 mm (3927 mm²) as dowels, same as column bars. Dowel length in compression $L_{dc} = \dfrac{0.87 f_y\phi}{4 \times 1.25\,\tau_{bd}} = 806$ mm (and the same lap with the column bars). Available depth in footing = 650 - 50 - 32 = 568 mm $< L_{dc}$; dowels are bent at the bottom with 90° bends. #### Detailing ```text PLAN ELEVATION +------------------+ col | # # # # # # | | | | # +------+ # # | _|_|__ | # | col | # # | |_____| <- D = 650 mm | # +------+ # # | bars both ways | # # # # # # | +------------------+ L = 3.1 m, B = 3.1 m ``` - Bottom mesh: 16 mm @ 185 c/c along $L$ (lower layer), 16 mm @ 185 c/c along $B$ (upper layer), 50 mm clear cover; dowels 8-25 mm. **Answer:** Footing 3.1 m × 3.1 m × 650 mm thick; 16 mm @ 185 c/c along the length and 16 mm @ 185 c/c along the width. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. For q = 220 kN/m^2, depth 1.7 m and soil unit weight 18.5 kN/m^3, how is the net safe bearing capacity found? A. 220 + 18.5 x 1.7 = 251.5 kN/m^2 B. 220 x 1.7 = 374 kN/m^2 C. 220 - 18.5 x 1.7 = 188.55 kN/m^2 D. 220 / 1.5 = 146.7 kN/m^2 Answer: C. The overburden pressure gamma x D_f is subtracted, which covers the weight of footing and backfill. (Key point: Net bearing capacity) 2. Why is a square footing chosen in this problem? A. The column is square B. Equal moments of 100 kN-m act about both axes, so the same size is needed in both directions C. Square footings need no shear check D. It saves reinforcement in the short direction Answer: B. With moments about both axes, the pressure varies in both directions, and equal moments make a square footing appropriate. (Key point: Biaxial moment and footing shape) 3. Where is the critical section for two-way (punching) shear taken, and what is compared? A. At d from the column face; tau_v compared with tau_c from Table 19 B. At the column face; tau_v compared with 0.45 f_ck C. At d/2 from the column face; tau_v compared with k_s x 0.25 sqrt(f_ck) D. At the footing edge; tau_v compared with 0.12% Answer: C. Punching shear is checked on a perimeter at d/2 from the column face against k_s x 0.25 sqrt(f_ck) = 1.25 MPa. (Key point: Punching shear check) 4. The required dowel compression length is 806 mm but the available depth is only 568 mm. What is done? A. The footing depth is ignored B. Dowels are bent at the bottom with 90 degree bends C. Dowels are cut short to 568 mm D. Dowel diameter is reduced Answer: B. When the straight length is insufficient, the dowels are bent at the bottom of the footing to develop the required anchorage. 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