--- title: "Footing with load transfer check, 1000 kN" subject: "Design of RCC Structure" chapter: 11 key: "semester_7/design_of_rcc_structure:footing-with-load-transfer-check-1000" source: https://www.bctengineer.com/q/semester_7/design_of_rcc_structure/footing-with-load-transfer-check-1000 site: BCT Engineering --- # Footing with load transfer check, 1000 kN Design of RCC Structure · Chapter 11: Design of Footings - Asked in: 2080 Bhadra, 14 marks - Video: [Part 3 | Load transfer in footings |](https://www.youtube.com/watch?v=uJMQMyNYYbk) — Everything Civil - 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/footing-with-load-transfer-check-1000 - Whole chapter: https://www.bctengineer.com/md/semester_7/design_of_rcc_structure/chapter-11 ## Question Design an isolated footing supporting a square column 400 mm x 400 mm with 8-20 $\phi$ longitudinal reinforcement. The column is subjected to axial service load of 1000 kN. Consider base of footing is 1 m below the ground level. Take allowable bearing capacity of soil 100 kN/m$^2$, M15 concrete and Fe 415 steel. Also check transfer of load from column to footing. ## Short answer Use a 3.4 m x 3.4 m square footing, overall depth 500 mm (d=440 mm). Reinforcement: 16 mm bars @205 mm c/c both ways. One-way and two-way shear are safe; load transfer check shows that column bars as dowels can carry the excess 60 kN, so the connection is adequate. ## Answer #### Data and assumptions - Column 400 mm × 400 mm (long side $a=400$ mm along the footing length); footing concrete M15, steel Fe415; column concrete M20; column bars 8-20 mm. - Service load $P = 1000$ kN. - Allowable (safe) bearing capacity of soil = 100 kN/m² at 1.0 m depth. - Column concrete grade is not given; M20 is assumed for the column (footing is M15). - 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 Weight of footing and backfill is taken as 10% of the column load. $$ A_{req} = \frac{1.1\,P}{q_a} = \frac{1.1 \times 1000.0}{100} = 11.00\ \text{m}^2 $$ Provide a square footing $L \times B = 3.4 \times 3.4$ m, area = 11.56 m² $\ge$ 11.00 m². #### Factored soil pressure $P_u = 1.5 \times 1000.0 = 1500.0$ kN. $p_u = \dfrac{P_u}{BL} = \dfrac{1500.0}{3.4 \times 3.4} = 129.8$ kN/m². #### Bending moments and reinforcement Cantilever projections from the column face: along $L$: $(3400 - 400)/2 = 1500$ mm; along $B$: $(3400 - 400)/2 = 1500$ mm. $M_u = p_u \times (\text{width}) \times \dfrac{(\text{projection})^2}{2}$: $M_{u,L} = 496.3$ kN·m, $M_{u,B} = 496.3$ kN·m. Required depth for flexure: $M_{u,lim} = 0.138 f_{ck} b d^2$ is far larger than the moment for $d = 440$ mm, so the depth is governed by shear (one-way or punching). Trial **$D = 500$ mm, $d = 440$ 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 $B$) | 496.3 | 3331 | 16 mm @ 205 c/c | 3335 | | Along $B$ (width $L$) | 496.3 | 3331 | 16 mm @ 205 c/c | 3335 | #### Check for one-way shear (at distance $d$ from the column face) - Along $L$ side: $V_u = 467.6$ kN, $\tau_v = \dfrac{V_u}{B d} = 0.313$ MPa; $p_t = 0.223\%$, $\tau_c = 0.335$ MPa (Table 19). $\tau_v < \tau_c$, safe. - Along $B$ side: $V_u = 467.6$ kN, $\tau_v = 0.313$ MPa; $p_t = 0.223\%$, $\tau_c = 0.335$ MPa. Safe. #### Check for two-way (punching) shear (at $d/2$ from the column face) - Perimeter $b_o = 2[(a + d) + (c + d)] = 2[(400+440) + (400+440)] = 3360$ mm - $V_u = P_u - p_u (a+d)(c+d) = 1500.0 - 129.8 \times 0.840 \times 0.840 = 1408.4$ kN (average pressure; the effect of moment transfer is neglected). - $\tau_v = \dfrac{V_u}{b_o d} = 0.953$ MPa; $k_s = 0.5 + \beta_c = 1.00$, $\tau_c' = k_s \times 0.25\sqrt{f_{ck}} = 0.968$ MPa. $\tau_v < \tau_c'$, **safe**. #### Development length $L_d = \dfrac{0.87 f_y \phi}{4\tau_{bd}}$, $\tau_{bd} = 1.60$ MPa (M15, deformed bars): 16 mm bars (both directions) → $L_d = 903$ mm. Available length from column face (less 50 mm end cover): 1450 mm along $L$, 1450 mm along $B$. **Satisfied** in both directions. #### Bearing stress and transfer of load from column Column area $A_2 = 400\times400 = 0.160$ m². Bearing stress $= P_u/A_2 = 9.38$ MPa. Permissible bearing stress: column concrete $0.45 f_{ck} = 9.00$ MPa; footing concrete $0.45 f_{ck}\sqrt{A_1/A_2}$ with $\sqrt{A_1/A_2} \le 2$ gives up to 13.50 MPa. The lesser, 9.00 MPa, governs. Capacity of bearing = 1440 kN $<$ $P_u = 1500$ kN, so the excess 60 kN must be carried by dowels: $A_{s,dow} = \dfrac{60000}{0.87 f_y} = 166$ mm²; the column bars (8-20 mm, 2513 mm²) are continued into the footing as dowels - adequate. Dowel length in compression $L_{dc} = \dfrac{0.87 f_y\phi}{4 \times 1.25\,\tau_{bd}} = 903$ mm (and the same lap with the column bars). Available depth in footing = 500 - 50 - 32 = 418 mm $< L_{dc}$; dowels are bent at the bottom with 90° bends. #### Detailing ```text PLAN (3.4 m x 3.4 m) +------------------------------------+ | | | | | | | | | | | |--+---+---+---+---+---+---+---+---+-| | | | | | | | | | | | |--+---+---+---+---+---+---+---+---+-| | | | | +----------+ | | | | |--+---+---+-| column |--+---+---+-| | | | | +----------+ | | | | |--+---+---+---+---+---+---+---+---+-| | | | | | | | | | | | |--+---+---+---+---+---+---+---+---+-| | | | | | | | | | | | +------------------------------------+ | = bars along B, - = bars along L: 16 mm @ 205 c/c both ways ELEVATION | column | | | +---------------+----------+---------------+ | (dowels) | D = 500 mm | o o o o o o o o o o o o o | bottom mesh, 50 mm cover +------------------------------------------+ <---------------- 3400 mm -----------------> ``` - Bottom mesh: 16 mm @ 205 c/c along $L$ (lower layer), 16 mm @ 205 c/c along $B$ (upper layer), 50 mm clear cover; dowels 8-20 mm. **Answer:** Footing 3.4 m × 3.4 m × 500 mm thick; 16 mm @ 205 c/c along the length and 16 mm @ 205 c/c along the width. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. Column is M20 and footing M15. Which permissible bearing stress governs? A. 0.45 x 15 x 2 = 13.5 MPa from the footing B. The average of the two C. 0.45 x 20 = 9.00 MPa from the column concrete, as it is the lesser D. 0.25 sqrt(fck) Answer: C. The lesser of the column value (9.00 MPa) and the footing value (up to 13.50 MPa) governs the load transfer. (Key point: Governing permissible bearing stress) 2. Bearing capacity is 1440 kN against Pu = 1500 kN. How is the 60 kN excess handled? A. By dowels of area 60000/(0.87 fy) = 166 mm^2, met by the continued column bars B. By increasing the footing depth C. By using M25 in the footing D. No action is required Answer: A. The extra load is carried by steel across the interface; the 8-20 mm column bars extended into the footing exceed 166 mm^2. (Key point: Dowels for excess load) 3. Allowable punching shear stress for M15 concrete is: A. 0.25 sqrt(20) = 1.118 MPa B. 0.25 sqrt(15) = 0.968 MPa C. 0.45 x 15 = 6.75 MPa D. 0.335 MPa from Table 19 Answer: B. tau_c' = ks x 0.25 sqrt(fck) with ks = 1.0 and fck = 15 gives 0.968 MPa; tau_v = 0.953 MPa is just safe. (Key point: Punching shear strength with M15) 4. Why is the development length (903 mm for 16 mm bars) longer than with M20 (752 mm)? A. Fe415 is weaker here B. A 1 m deep base increases it C. Dowels need more length D. Lower grade M15 has lower design bond stress tau_bd = 1.60 MPa Answer: D. Ld = 0.87 fy phi/(4 tau_bd); weaker M15 concrete gives smaller tau_bd and so a longer Ld. 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