--- title: "Footing for 1600 kN with shear check" subject: "Design of RCC Structure" chapter: 11 key: "semester_7/design_of_rcc_structure:footing-for-1600-kn-with-shear" source: https://www.bctengineer.com/q/semester_7/design_of_rcc_structure/footing-for-1600-kn-with-shear site: BCT Engineering --- # Footing for 1600 kN with shear check Design of RCC Structure · Chapter 11: Design of Footings - Asked in: 2076 Asoj, 10 marks - Video: [One way shear in footing | Beam shear in footing | Critical section for one way shear in footing](https://www.youtube.com/watch?v=FIm56UO6ekU) — Civil Engineering Mastery - 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-for-1600-kn-with-shear - Whole chapter: https://www.bctengineer.com/md/semester_7/design_of_rcc_structure/chapter-11 ## Question Design an isolated footing to carry an axial factored load of 1600 kN. The column is 400 mm by 400 mm in size with 20 mm dia longitudinal bars. The bearing capacity of soil is 180 kN/m$^2$. For footing, adopt M20 concrete and 415 grade HSD bars. Check for shear is necessary. ## Short answer Service load = 1600/1.5 = 1066.7 kN; area = 1.1*1066.7/180 = 6.52 m², provide 2.6 m square. Factored pressure 236.7 kN/m²; depth D=550 mm (d=490 mm). Reinforcement 12 mm @ 130 c/c both ways; shear safe. Bearing stress 10 MPa > 9 MPa, excess 160 kN carried by column dowels. ## Answer #### Data and assumptions - Column 400 mm × 400 mm (long side $a=400$ mm along the footing length); footing concrete M20, steel Fe415; column concrete M20; column bars 8-20 mm. - Given loads are factored: $P_u = 1600$ kN. Service loads = factored / 1.5: $P = 1066.7$ kN. - Allowable (safe) bearing capacity of soil = 180 kN/m². - The load of 1600 kN is a factored load. - 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. $$ A_{req} = \frac{1.1\,P}{q_a} = \frac{1.1 \times 1066.7}{180} = 6.52\ \text{m}^2 $$ Provide a square footing $L \times B = 2.6 \times 2.6$ m, area = 6.76 m² $\ge$ 6.52 m². #### Factored soil pressure $P_u = 1.5 \times 1066.7 = 1600.0$ kN. $p_u = \dfrac{P_u}{BL} = \dfrac{1600.0}{2.6 \times 2.6} = 236.7$ kN/m². #### Bending moments and reinforcement Cantilever projections from the column face: along $L$: $(2600 - 400)/2 = 1100$ mm; along $B$: $(2600 - 400)/2 = 1100$ mm. $M_u = p_u \times (\text{width}) \times \dfrac{(\text{projection})^2}{2}$: $M_{u,L} = 372.3$ kN·m, $M_{u,B} = 372.3$ kN·m. Required depth for flexure: $M_{u,lim} = 0.138 f_{ck} b d^2$ is far larger than the moment for $d = 490$ mm, so the depth is governed by shear (one-way or punching). Trial **$D = 550$ mm, $d = 490$ 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 $B$) | 372.3 | 2183 | 12 mm @ 130 c/c | 2262 | | Along $B$ (width $L$) | 372.3 | 2183 | 12 mm @ 130 c/c | 2262 | #### Check for one-way shear (at distance $d$ from the column face) - Along $L$ side: $V_u = 375.4$ kN, $\tau_v = \dfrac{V_u}{B d} = 0.295$ MPa; $p_t = 0.178\%$, $\tau_c = 0.310$ MPa (IS 456 Table 19, SP 16 formula). $\tau_v < \tau_c$, safe. - Along $B$ side: $V_u = 375.4$ kN, $\tau_v = 0.295$ MPa; $p_t = 0.178\%$, $\tau_c = 0.310$ MPa. Safe. #### Check for two-way (punching) shear (at $d/2$ from the column face) - Perimeter $b_o = 2[(a + d) + (c + d)] = 2[(400+490) + (400+490)] = 3560$ mm - $V_u = P_u - p_u (a+d)(c+d) = 1600.0 - 236.7 \times 0.890 \times 0.890 = 1412.5$ kN (average pressure; the effect of moment transfer is neglected). - $\tau_v = \dfrac{V_u}{b_o d} = 0.810$ MPa; $k_s = 0.5 + \beta_c = 1.00$, $\tau_c' = k_s \times 0.25\sqrt{f_{ck}} = 1.118$ MPa. $\tau_v < \tau_c'$, **safe**. #### Development length $L_d = \dfrac{0.87 f_y \phi}{4\tau_{bd}}$, $\tau_{bd} = 1.92$ MPa (M20, deformed bars): 12 mm bars (both directions) → $L_d = 564$ mm. Available length from column face (less 50 mm end cover): 1050 mm along $L$, 1050 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 $= P_u/A_2 = 10.00$ 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 18.00 MPa. The lesser, 9.00 MPa, governs. Capacity of bearing = 1440 kN $<$ $P_u = 1600$ kN, so the excess 160 kN must be carried by dowels: $A_{s,dow} = \dfrac{160000}{0.87 f_y} = 443$ mm²; the column bars (8-20 mm, 2513 mm²) are continued into the footing as dowels, which is adequate. Dowel length in compression $L_{dc} = \dfrac{0.87 f_y\phi}{4 \times 1.25\,\tau_{bd}} = 752$ mm (and the same lap with the column bars). Available depth in footing = 550 - 50 - 24 = 476 mm $< L_{dc}$; dowels are bent at the bottom with 90° bends. #### Detailing ```text PLAN ELEVATION +------------------+ col | # # # # # # | | | | # +------+ # # | _|_|__ | # | col | # # | |_____| <- D = 550 mm | # +------+ # # | bars both ways | # # # # # # | +------------------+ L = 2.6 m, B = 2.6 m ``` - Bottom mesh: 12 mm @ 130 c/c along $L$ (lower layer), 12 mm @ 130 c/c along $B$ (upper layer), 50 mm clear cover; dowels 8-20 mm. **Answer:** Footing 2.6 m × 2.6 m × 550 mm thick; 12 mm @ 130 c/c along the length and 12 mm @ 130 c/c along the width. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. The 1600 kN given is a factored load. How is the footing area found? A. Use 1600 kN directly with q_a B. Convert to service load 1600/1.5 = 1066.7 kN and use 1.1P/q_a = 6.52 m^2 C. Multiply 1600 by 1.5 first D. Use 0.9 x 1600 kN Answer: B. Soil bearing capacity is a service-level value, so the factored load is divided by 1.5 before sizing the footing. (Key point: Factored vs service load) 2. What is compared in the one-way shear check at distance d from the column face? A. tau_v = V_u/(B d) = 0.295 MPa against tau_c = 0.310 MPa from Table 19 for p_t = 0.178% B. tau_v against 0.25 sqrt(f_ck) C. The bearing stress against 0.45 f_ck D. V_u against the bending moment Answer: A. Nominal shear stress is compared with the design shear strength tau_c, which depends on percentage tension steel; here 0.295 < 0.310 so safe. (Key point: One-way shear) 3. Why is 9.00 MPa (0.45 f_ck of the column) the permissible bearing stress rather than 18 MPa? A. Footing concrete is weaker than column concrete B. The lesser of column-concrete stress and footing-concrete stress (up to 0.45 f_ck sqrt(A1/A2), max 2 times) governs C. 18 MPa applies only to Fe500 steel D. Code gives no upper limit Answer: B. The permissible bearing stress is taken as the lesser of the two values; the column concrete's 9.00 MPa is less than the footing's 18.00 MPa. 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