Prepared By ENGR. YOSHIAKI C. MIKAMI, BSCE RMP WALL FOOTING EXAMPLE: design a wall footing that carries a 250mm thick wall with a service load of π€π·πΏ = 60 ππ π and π€πΏπΏ = 35 ππ π. The bottom of the footing is 1.1m below the NGL where: π′π = 21 πππ , ππ¦ = 420 πππ, πΎπ πππ = 17 ππ π3, πΎππππ = 24 ππ π3, ππππππ€ = 150 ππ π2 Use temp=12mmØ bars and main=12mmØ a) b) c) d) e) f) Initial thickness Effective soil bearing pressure Area of footing Ultimate soil bearing pressure Solve for effective depth Design reinforcement DON’T PRACTICE UNTIL YOU GET IT RIGHT. PRACTICE WALL FOOTING SOLUTION: 1. Initial thickness π‘ = 20% ππ‘π. ππππππ πππ + 75ππ π‘ = 0.20 ∗ 1π + 0.075π π = π. ππππ 2. Effective soil bearing pressure ππππ = ππππππ€ − ΣπΎβ ππππ = 150 ππ 24 ππ ππππ = ππ π2 − πππ. πππ ππ΅ π3 ∗ 0.275π − 17 ππ π3 ∗ 0.825π 3. Required area of footing π΄ππ‘π = π’πππππ‘ππππ ππππ ππππ = 60ππ π+35ππ π ∗1π 129.375ππ/π2 π΄ππ‘π = 0.7343π2 = π΅πΏ = π΅(1π) π΅ = 0.7343π π ππ¦ π. ππ DON’T PRACTICE UNTIL YOU GET IT RIGHT. PRACTICE WALL FOOTING SOLUTION: 4. Ultimate bearing stress ππππ‘ππππ ππππ 1.2π·πΏ+1.6πΏπΏ ππ’ππ‘ = = π΄ππ‘π ππ’ππ‘ = π΄ππ‘π 1.2 60ππ π +1.6 35ππ π ∗1π 0.8π∗(1π) ππππ = πππ ππ΅ ππ 5. Footing depth Beam Shear: ππ’ππ‘ = ∅ππ ∅ππ = ∅0.17λ π′π ∗ ππ€ ∗ π ∅ππ = 0.75 ∗ 0.17(1) 21 π ππ2 ∗ 1000ππ ∗ π DON’T PRACTICE UNTIL YOU GET IT RIGHT. PRACTICE WALL FOOTING SOLUTION: 5. Footing depth ππ’ππ‘ = ππ’ππ‘ ∗ π΄ ππ’ππ‘ = 0.160 π 0.160 π ππ2 ∗ 1000ππ ∗ ππ2 ∗ 1000ππ ∗ 0.75 ∗ 0.17(1) 21 π ππ2 800ππ−250ππ 2 800ππ−250ππ 2 −π −π = ∗ 1000ππ ∗ π π = ππ. ππππ Note: as per provision 415.8, minimum d=150mm for footing on soil π = πππππ π‘ = 150ππ + 75ππ + 0.5 12ππ π‘ = 231ππ π ππ¦ πππππ DON’T PRACTICE UNTIL YOU GET IT RIGHT. PRACTICE WALL FOOTING SOLUTION: 5. Footing depth check: ππππ = 150 ππ ππππ = π΄ππ‘π = π΄ππ‘π = π2 − 24 πππ. ππ ππ΅ ππ ππ π3 ∗ 0.25π − 17 ππ π3 ∗ 0.85π 60ππ π+35ππ π ∗1π 129.55ππ/π2 0.7333π2 = π΅(1π) π΅ = 0.7333π π ππ¦ π. ππ ππππ€ = 250ππ − 75ππ − 0.5 ππ π πππ = ππππ¦π¦ DON’T PRACTICE UNTIL YOU GET IT RIGHT. PRACTICE WALL FOOTING SOLUTION: 6. Design reinforcement π€ = ππ’ππ‘ ∗ π΅ = 160 ππ π= 0.8π−0.25π 2 ππ’ππ‘ = π€π 2 2 = π2 ∗ 1π = 160 ππ π = 0.275π 160ππ π∗(0.275π)2 2 π΄πππ = π. ππππ΅ β π DON’T PRACTICE UNTIL YOU GET IT RIGHT. PRACTICE WALL FOOTING SOLUTION: 6. Design reinforcement ππ’ππ‘ = ∅ππ = 0.9 π′π ππ 2 π 1 − 0.59π 6.05π₯106 π β ππ = 0.9(21 π ππ2)(1000ππ) 169ππ 2 π(1 − 0.59π) π = π. πππππ π= ππ′π ππ¦ = 0.01128(21πππ) 420πππ π = π. πππππ π= π΄π ; ππ π΄π = πππ π΄π = 0.00056 1000ππ 169ππ π¨π = ππ. πππππ DON’T PRACTICE UNTIL YOU GET IT RIGHT. PRACTICE WALL FOOTING SOLUTION: 6. Design reinforcement π΄π _πππ = 0.0018π΄π = 0.0018π΅π‘ π΄π _πππ = 0.0018 1000ππ 250ππ π¨π_πππ = ππππππ > π¨π = πππ. ππππ ∴ πππ π¨π = ππππππ π = π΄π π (π )2 4 π π = 3.98 = 450ππ 2 π (12ππ)2 4 ππππ π π ππππππ = 1000ππ 3.97ππππ π π ππππππ = 251.3257ππ DON’T PRACTICE UNTIL YOU GET IT RIGHT. PRACTICE WALL FOOTING SOLUTION: 6. Design reinforcement 3 ∗ β = 3(250ππ) π πππ₯ = 450ππ π πππ₯ = 450ππ π πππ£π = 251.33ππ π ππ¦ 250ππ π΄π _π‘πππ = 0.0018 800ππ 250ππ π΄π _π‘πππ = 360ππ2 360ππ2 π=π 4 (12ππ)2 π = 3.1831 π ππ¦ π − ππππØ ππππ DON’T PRACTICE UNTIL YOU GET IT RIGHT. PRACTICE WALL FOOTING EXAMPLE: design a wall footing that carries a 325mm thick wall with a service load of π€π·πΏ = 200 ππ π and π€πΏπΏ = 180 ππ π and a moment load ππ·πΏ = 60 ππ π π and ππΏπΏ = 50 ππ π π. The bottom of the footing is 1.9m below the NGL where: π′π = 20.7 πππ, ππ¦ = 345 πππ , πΎπ πππ = 18 ππ π3 , πΎππππ = 23.5 ππ π3 , ππππππ€ = 185 ππ π2 Use temp=12mmØ bars and main=20mmØ DON’T PRACTICE UNTIL YOU GET IT RIGHT. PRACTICE WALL FOOTING SOLUTION: 1. Initial thickness π‘ = 20% ππ‘π. ππππππ πππ + 75ππ π‘ = 0.20 ∗ 1π + 0.075π π = π. ππππ 2. Effective soil bearing pressure ππππ = ππππππ€ − ΣπΎβ ππππ = 185 ππ ππππ = π2 − 23.5 πππ. ππ ππ΅ ππ ππ π3 ∗ 0.275π − 18 ππ π3 ∗ 1.625π 3. Required area of footing π 6π ππππ = + 2 πΏπ΅ 149.29 ππ πΏπ΅ π2 = (200ππ π+180ππ π)(1π) 6(60ππβπ π+50ππβπ π)(1π) + 1π π΅ (1π)π΅2 B = 3.73π π ππ¦ π. ππ DON’T PRACTICE UNTIL YOU GET IT RIGHT. PRACTICE WALL FOOTING SOLUTION: 4. Ultimate bearing stress ππππ‘ππππ ππππ 6(ππππ‘πππ ππππππ‘) ππ’ππ‘ = + 2 π΅πΏ ππ’ππ‘ = πΏπ΅ (1.2β200ππ π+1.6β180ππ π)(1π) ± 1π (3.8π) 6(1.2β60ππβπ π+1.6β50ππβπ π)(1π) (1π)(3.8π)2 ππππ_πππ = πππ. ππ ππ΅ ππππ_πππ = ππ. ππ ππ΅ ππ ππ 5. Footing depth Beam Shear: ππ’ππ‘ = ∅ππ ∅ππ = ∅0.17λ π′π ∗ ππ€ ∗ π ∅ππ = 0.75 ∗ 0.17(1) 20.7 π ππ2 ∗ 1000ππ ∗ π DON’T PRACTICE UNTIL YOU GET IT RIGHT. PRACTICE WALL FOOTING SOLUTION: 5. Footing depth DON’T PRACTICE UNTIL YOU GET IT RIGHT. PRACTICE WALL FOOTING SOLUTION: 5. Footing depth ππ’ππ‘ = ππ’ππ‘ ∗ π΄ ππ’ππ‘ = ππ΄ +0.2021π ππ2 2 ∗ 1000ππ ∗ 3800ππ−325ππ 2 −π ππ΄ −0.0758π ππ2 0.2021π ππ2−0.0758π ππ2 = 1737.5ππ+325ππ+π 3800ππ ππ΄ − 0.0758 π ππ2 = 3.327π₯10−5 π ππ3 2062.5ππ ππ΄ = 3.327π₯10−5 π ππ3 π + 0.144 π ππ2 +π 3.327π₯10−5 π ππ3 π +0.144π ππ2 +0.2021π ππ2 β 1000ππ β 2 1737.5ππ − π = 0.75 ∗ 0.17(1) 20.7 π ππ2 β 1000ππ βπ π = πππ. ππ ππ DON’T PRACTICE UNTIL YOU GET IT RIGHT. PRACTICE WALL FOOTING SOLUTION: 5. Footing depth check: π‘πππ€ = 411.58ππ + 75ππ + 0.5 20ππ π‘πππ€ = 496.58 ππ π ππ¦ πππ ππ ππππ€ = 500ππ − 75ππ − 0.5 20ππ π πππ = πππ ππ ππππ = 185 ππ ππππ = π2 − 23.5 πππ. ππ ππ΅ ππ ππππ = π πΏπ΅ 148.05 ππ + ππ π3 (0.5π) − 18 ππ π3 (1.4π) 6π πΏπ΅2 π2 = (200ππ π+180ππ π)(1π) 6(60ππβπ π+50ππβπ π)(1π) + 1π π΅ (1π)π΅2 B = 3.75π π ππ¦ π. ππ DON’T PRACTICE UNTIL YOU GET IT RIGHT. PRACTICE WALL FOOTING SOLUTION: 6. Design reinforcement ππ΅ −0.0758π ππ2 0.2021π ππ2−0.0758π ππ2 = 1737.5ππ+325ππ 3800ππ ππ΅ = 0.144 π ππ2 1737.5 ππ 1000ππ + ππ2 2 1 2 (0.2021 π ππ2 − 0.144 π ππ2)(1737.5ππ)( 1737.5ππ )(1000ππ) 2 3 π΄πππ = πππ. ππππππ π΅ β ππ ππ’ππ‘ = 0.144 π 1737.5ππ DON’T PRACTICE UNTIL YOU GET IT RIGHT. PRACTICE WALL FOOTING SOLUTION: 6. Design reinforcement ππ’ππ‘ = ∅ππ = 0.9 π′π ππ 2 π 1 − 0.59π 275.83π₯106 π β ππ = 0.9(20.7 π ππ2)(1000ππ) 415ππ 2 π(1 − 0.59π) π = π. πππππ π= ππ′π ππ¦ = 0.09084(20.7πππ) 345πππ π = π. πππππ π= π΄π ; ππ π΄π = πππ π΄π = 0.00545 1000ππ 415ππ π¨π = ππππ. ππ πππ DON’T PRACTICE UNTIL YOU GET IT RIGHT. PRACTICE WALL FOOTING SOLUTION: 6. Design reinforcement π΄π _πππ = 0.002π΄π = 0.002π΅π‘ π΄π _πππ = 0.002 1000ππ 500ππ π¨π_πππ = πππππππ > π¨π = ππππ. πππππ ∴ πππ π¨π = ππππ. πππππ π = π΄π π (π )2 4 π π = 7.2 = 2261.75ππ 2 π (20ππ)2 4 ππππ π π ππππππ = 1000ππ 7.2ππππ π π ππππππ = 138.8ππ DON’T PRACTICE UNTIL YOU GET IT RIGHT. PRACTICE WALL FOOTING SOLUTION: 6. Design reinforcement 3 ∗ β = 3(475ππ) π πππ₯ = 450ππ π πππ₯ = 450ππ π πππ£π = 138.8ππ π ππ¦ 125ππ π΄π _π‘πππ = 0.002 3800ππ 500ππ π΄π _π‘πππ = 3800ππ2 3800ππ2 π=π 4 (12ππ)2 π = 33.6 π ππ¦ ππ − ππππØ ππππ DON’T PRACTICE UNTIL YOU GET IT RIGHT. PRACTICE RECTANGULAR FOOTING EXAMPLE: a rectangular column footing is to support a 450mmx450mm square tied column that carries a dead load of 950 kN and live load of 550 kN. The column is reinforced with a 6-20mmØ bars. The base of the footing is 1.8m below the natural grade line where the allowable soil pressure is 210 kPa. Using π′π = 20.7 πππ, ππ¦ = 420 πππ, πΎπ πππ = 17 ππ π3, πΎππππ = 23.5 ππ π3 DON’T PRACTICE UNTIL YOU GET IT RIGHT. PRACTICE RECTANGULAR FOOTING SOLUTION: 1. Initial thickness π‘ = 20% ππ‘π. ππππππ πππ + 75ππ π‘ = 0.20 ∗ 2200ππ + 75ππ π = πππππ 2. Effective soil bearing pressure ππππ = ππππππ€ − ΣπΎβ ππππ = 210 ππ ππππ = π2 − 23.5 πππ. ππ ππ΅ ππ ππ π3 (0.515π) − 17 ππ π3 (1.285π) 3. Required area of footing π 6π ππππ = + 2 π΅πΏ 176.05 ππ π΅πΏ π2 = 950ππ+550ππ 2.2π πΏ L = 3.87π π ππ¦ π. ππ DON’T PRACTICE UNTIL YOU GET IT RIGHT. PRACTICE RECTANGULAR FOOTING SOLUTION: 4. Ultimate bearing stress ππππ‘ππππ ππππ ππ’ππ‘ = ππ’ππ‘ = π΅πΏ 1.2β950ππ+1.6β550ππ 2.2π (3.9π) ππππ = πππ. ππ ππ΅ ππ 5. Footing depth Beam Shear: a) along short span (2.2m) ππ’ππ‘ = ∅ππ ∅ππ = ∅0.17λ π′π ∗ ππ€ ∗ π ∅ππ = 0.75 ∗ 0.17(1) 20.7 π ππ2 ∗ 2200ππ ∗ π DON’T PRACTICE UNTIL YOU GET IT RIGHT. PRACTICE RECTANGULAR FOOTING SOLUTION: 5. Footing depth ππ’ππ‘ = ππ’ππ‘ ∗ π΄ ππ’ππ‘ = 0.23543 π 0.23543 π ππ2 ∗ 2200ππ ∗ ππ2 ∗ 2200ππ ∗ 0.75 ∗ 0.17(1) 20.7 π ππ2 3900ππ−450ππ 2 3900ππ−450ππ 2 −π −π = ∗ 2200ππ ∗ π π πππππ ππππ = πππ. ππ ππ DON’T PRACTICE UNTIL YOU GET IT RIGHT. PRACTICE RECTANGULAR FOOTING SOLUTION: 5. Footing depth b) along long span (3.9m) ππ’ππ‘ = ∅ππ ∅ππ = 0.75 ∗ 0.17(1) 20.7 π ππ2 ∗ 3900ππ ∗ π DON’T PRACTICE UNTIL YOU GET IT RIGHT. PRACTICE RECTANGULAR FOOTING SOLUTION: 5. Footing depth ππ’ππ‘ = ππ’ππ‘ ∗ π΄ ππ’ππ‘ = 0.23543 π 0.23543 π ππ2 ∗ 3900ππ ∗ ππ2 ∗ 3900ππ ∗ 0.75 ∗ 0.17(1) 20.7 π ππ2 2200ππ−450ππ 2 2200ππ−450ππ 2 −π −π = ∗ 3900ππ ∗ π π ππππ ππππ = πππ. ππ ππ DON’T PRACTICE UNTIL YOU GET IT RIGHT. PRACTICE RECTANGULAR FOOTING SOLUTION: 5. Footing depth Punching Shear: ∅ππ = ∅0.33λ π′π ∗ ππ ∗ π ∅ππ = 0.75 ∗ 0.33(1) 20.7 π ππ2 4 450ππ + π (π) DON’T PRACTICE UNTIL YOU GET IT RIGHT. PRACTICE RECTANGULAR FOOTING SOLUTION: 5. Footing depth ππ’ππ‘ = ππ’ππ‘ ∗ π΄ ππ’ππ‘ = 0.23543 π 0.23543 π ππ2 ππ2 2200ππ β 3900ππ − 450ππ + π 2200ππ β 3900ππ − 450ππ + π 0.75 ∗ 0.33(1) 20.7 π ππ2 2 2 = 4 450ππ + π (π) π πππππ = πππ. ππ ππ ππππ£π = πππ₯πππ’π(ππ βπππ‘ π πππ , πππππ π πππ , πππ’ππβ ) ππππ£π = πππ₯πππ’π(497.99ππ, 252.60ππ, 450.78ππ) π ππππ = πππ. ππππ DON’T PRACTICE UNTIL YOU GET IT RIGHT. PRACTICE RECTANGULAR FOOTING SOLUTION: 5. Footing depth check: π‘πππ€ = 497.99ππ + 75ππ + 1.5 20ππ π‘πππ€ = 602.99 ππ π ππ¦ πππ ππ ππππ€ = 625ππ − 75ππ − 1.5 20ππ π πππ = πππ ππ ππππ = 210 ππ ππππ = π2 − 23.5 πππ. ππ ππ΅ ππ ππππ = π π΅πΏ 175.34 ππ π2 = ππ π3 (0.625π) − 17 ππ π3 (1.175π) 950ππ+550ππ 2.2π πΏ L = 3.89π π ππ¦ π. ππ DON’T PRACTICE UNTIL YOU GET IT RIGHT. PRACTICE RECTANGULAR FOOTING SOLUTION: 6. Design reinforcement (along long span) ππ’ππ‘ = 0.23543 π ππ2 1725ππ 1725ππ 2 2200ππ π΄πππ = πππ. ππππππ π΅ β ππ DON’T PRACTICE UNTIL YOU GET IT RIGHT. PRACTICE RECTANGULAR FOOTING SOLUTION: 6. Design reinforcement (along long span) ππ’ππ‘ = ∅ππ = 0.9 π′π ππ 2 π 1 − 0.59π 770.61π₯106 π β ππ = 0.9(20.7 π ππ2)(2200ππ) 520ππ 2 π(1 − 0.59π) π = π. πππππ π= ππ′π ππ¦ = 0.07265(20.7πππ) 420πππ π = π. πππππ π= π΄π ; ππ π΄π = πππ π΄π = 0.00358 2200ππ 520ππ π¨π = ππππ. ππ πππ DON’T PRACTICE UNTIL YOU GET IT RIGHT. PRACTICE RECTANGULAR FOOTING SOLUTION: 6. Design reinforcement (along long span) π΄π _πππ = 0.0018π΄π = 0.0018π΅π‘ π΄π _πππ = 0.0018 2200ππ 625ππ π¨π_πππ = πππππππ < π¨π = ππππ. πππππ ∴ πππ π¨π = ππππ. ππ πππ π = π΄π π (π )2 4 π = 4096.21ππ 2 π (20ππ)2 4 π = 13.04 π ππ¦ ππ − ππππØ ππππ πππππ ππππ ππππ DON’T PRACTICE UNTIL YOU GET IT RIGHT. PRACTICE RECTANGULAR FOOTING SOLUTION: 6. Design reinforcement (along short span) ππ’ππ‘ = 0.23543 π ππ2 875ππ 875ππ 2 3900ππ π΄πππ = πππ. ππππππ π΅ β ππ DON’T PRACTICE UNTIL YOU GET IT RIGHT. PRACTICE RECTANGULAR FOOTING SOLUTION: 6. Design reinforcement (along short span) ππ’ππ‘ = ∅ππ = 0.9 π′π ππ 2 π 1 − 0.59π 351.49π₯106 π β ππ = 0.9(20.7 π ππ2)(3900ππ) 520ππ 2 π(1 − 0.59π) π = π. πππππ π= ππ′π ππ¦ = 0.01808(20.7πππ) 420πππ π = π. πππππ π= π΄π ; ππ π΄π = πππ π΄π = 0.00089 3900ππ 520ππ π¨π = ππππ. ππ πππ DON’T PRACTICE UNTIL YOU GET IT RIGHT. PRACTICE RECTANGULAR FOOTING SOLUTION: 6. Design reinforcement (along short span) π΄π _πππ = 0.0018π΄π = 0.0018πΏπ‘ π΄π _πππ = 0.0018 3900ππ 625ππ π¨π_πππ = ππππ. π πππ > π¨π = ππππ. ππ πππ ∴ πππ π¨π = ππππ. ππππ π = π΄π π (π )2 4 π = 4387.5ππ 2 π (20ππ)2 4 π = 13.97 π ππ¦ ππ − ππππØ ππππ πππππ πππππ ππππ DON’T PRACTICE UNTIL YOU GET IT RIGHT. PRACTICE RECTANGULAR FOOTING SOLUTION: 7. Band Width ππππ ππππ‘πππ ππππ π€πππ‘β ππππ‘ππ = 2 ; π½+1 = 2 1.77+1 π½= πΏ π΅ π½ = 3.9π 2.2π = 1.77 ππππ ππππ‘πππ ππππ π€πππ‘β 13 ππππ€ = 9.39 π ππ¦ ππ ππππ πππππ‘ = 14 − 10 = 4 ππππ (πππ‘π: πππ 1 πππ π π¦ππππ‘ππ¦) DON’T PRACTICE UNTIL YOU GET IT RIGHT. PRACTICE RECTANGULAR FOOTING SOLUTION: 8. Development Length ππ ππππ = 9 10 β ππ¦ π′π β πΌπ½πΎλ π+ππ‘π ππ ; π+ππ‘π ππ ≤ 2.5 20ππ = 85ππ 2 2200ππ−2 75ππ −20ππ π2 = = 145ππ 15−1 9 414πππ 1 1 (1)(1) = 20ππ β β 85ππ+0 10 20.7πππ 20ππ π1 = 75ππ + ππ ππ = 655.16ππ πππ π’ππππππ = 2200ππ−450ππ 2 = 875ππ > ππ ∴ πππππ’ππ‘π DON’T PRACTICE UNTIL YOU GET IT RIGHT. PRACTICE RECTANGULAR FOOTING EXAMPLE: design the footing for a rectangular column 400mmx600mm which is to carry ππ·πΏ = 780 ππ, ππ·πΏ = 340 ππ β π, ππΏπΏ = 650 ππ, ππΏπΏ = 280 ππ β π. The footing dimension is limited to 3 m. the bottom of footing is 1.5m below the NGL with an allowable bearing capacity of ππππππ€ = 160 ππ π2. Using π′π = 27.5 πππ, ππ¦ = 275 πππ, πΎπ πππ = 16 ππ π3, πΎππππ = 23.5 ππ π3 , 25mmØ bars DON’T PRACTICE UNTIL YOU GET IT RIGHT. PRACTICE RECTANGULAR FOOTING SOLUTION: 1. Initial thickness π‘ = 20% ππ‘π. ππππππ πππ + 75ππ π‘ = 0.20 ∗ 3000ππ + 75ππ π = πππππ 2. Effective soil bearing pressure ππππ = ππππππ€ − ΣπΎβ ππππ = 160 ππ ππππ = π2 − 23.5 πππ. ππ ππ΅ ππ ππ π3 (0.675π) − 16 ππ π3 (0.825π) 3. Required area of footing π 6π ππππ = + 2 π΅πΏ 130.94 ππ π΅πΏ π2 = 780ππ+650ππ 3π πΏ + 6(340ππβπ+280ππβπ) (3π)πΏ2 L = 5.396π π ππ¦ π. ππ DON’T PRACTICE UNTIL YOU GET IT RIGHT. PRACTICE RECTANGULAR FOOTING SOLUTION: 4. Ultimate bearing stress ππππ‘ππππ ππππ 6(ππππ‘πππ ππππππ‘) ππ’ππ‘ = + 2 ππ’ππ‘ = π΅πΏ 1.2β780ππ+1.6β650ππ 3π (5.4π) ππππ_πππ = πππ. ππ ππ΅ ππππ_πππ = ππ. ππ ππ΅ ± πΏπ΅ 6(1.2β340ππβπ+1.6β280ππβπ) (3π)(5.4π)2 ππ ππ 5. Footing depth Beam Shear: a) along short span (3m) ππ’ππ‘ = ∅ππ ∅ππ = ∅0.17λ π′π ∗ ππ€ ∗ π ∅ππ = 0.75 ∗ 0.17(1) 27.5 π ππ2 ∗ 3000ππ ∗ π DON’T PRACTICE UNTIL YOU GET IT RIGHT. PRACTICE RECTANGULAR FOOTING SOLUTION: 5. Footing depth DON’T PRACTICE UNTIL YOU GET IT RIGHT. PRACTICE RECTANGULAR FOOTING SOLUTION: 5. Footing depth ππ’ππ‘ = ππ’ππ‘ ∗ π΄ ππ’ππ‘ = ππ΄ +0.18069π ππ2 2 ∗ 3000ππ ∗ 2400ππ − π ππ΄ −0.06326π ππ2 0.18069π ππ2−0.06326π ππ2 = 2400ππ+600ππ+π 5400ππ ππ΄ − 0.06326 π ππ2 = 2.17π₯10−5 π ππ3 3000ππ ππ΄ = 2.17π₯10−5 π ππ3 π + 0.12849 π ππ2 +π 2.17π₯10−5 π ππ3 π +0.12849π ππ2 +0.18069π ππ2 β 3000ππ β 2 2400ππ − π = 0.75 ∗ 0.17(1) 27.5 π ππ2 β 3000ππ βπ π πππππ ππππ = πππ. ππ ππ DON’T PRACTICE UNTIL YOU GET IT RIGHT. PRACTICE RECTANGULAR FOOTING SOLUTION: 5. Footing depth b) along long span (5.4m) ππ’ππ‘ = ∅ππ ∅ππ = 0.75 ∗ 0.17(1) 27.5 π ππ2 ∗ 5400ππ ∗ π DON’T PRACTICE UNTIL YOU GET IT RIGHT. PRACTICE RECTANGULAR FOOTING SOLUTION: 5. Footing depth ππ’ππ‘ = ππ’ππ‘ ∗ π΄ ππ’ππ‘ = 0.18069π ππ2+0.06326π ππ2 2 ∗ 5400ππ ∗ 1300ππ − π 0.18069π ππ2+0.06326π ππ2 ∗ 5400ππ ∗ 2 0.75 ∗ 0.17(1) 27.5 π ππ2 β 5400ππ 1300ππ − π = βπ π ππππ ππππ = πππ. ππ ππ DON’T PRACTICE UNTIL YOU GET IT RIGHT. PRACTICE RECTANGULAR FOOTING SOLUTION: 5. Footing depth Punching Shear: ∅ππ = ∅0.33λ π′π ∗ ππ ∗ π ∅ππ = 0.75 ∗ 0.33(1) 27.5 π 2 600ππ + π + 400ππ + π (π) ππ2 DON’T PRACTICE UNTIL YOU GET IT RIGHT. PRACTICE RECTANGULAR FOOTING SOLUTION: 5. Footing depth ππ΅ −0.06326π ππ2 0.18069π ππ2−0.06326π ππ2 = 2400ππ−0.5π 5400ππ ππΆ −0.06326π ππ2 0.18069π ππ2−0.06326π ππ2 2400ππ+600ππ+0.5π = 5400ππ DON’T PRACTICE UNTIL YOU GET IT RIGHT. PRACTICE RECTANGULAR FOOTING SOLUTION: 5. Footing depth ππ΅ = 2.17π₯10−5 π ππ΅ = ππ3 2400ππ − 0.5π + 0.06326 −1.085π₯10−5 π ππ3 π + 0.11534 π ππ2 ππΆ = 2.17π₯10−5 π ππΆ = ππ3 3000ππ + 0.5π + 0.06326 1.085π₯10−5 π ππ3 π + 0.12849 π ππ2 π π ππ2 ππ2 ππ’ππ‘ = ππ’ππ‘ ∗ π΄ ππ’ππ‘ = ππ΅ +ππΆ 2 0.18069π ππ2+0.06326π ππ2 2 3000ππ β 5400ππ − (600ππ + π) β (400ππ + π) DON’T PRACTICE UNTIL YOU GET IT RIGHT. PRACTICE RECTANGULAR FOOTING SOLUTION: 5. Footing depth 0.18069π ππ2+0.06326π ππ2 2 3000ππ β 5400ππ − −1.085π₯10−5 π ππ3 π +0.11534π ππ2 + 1.085π₯10−5 π ππ3 π +0.12849π ππ2 2 (600ππ + π) β (400ππ + π) = 0.75 ∗ 0.33(1) 27.5 π 2 600ππ + π + 400ππ + π (π) ππ2 π πππππ = πππ. ππ ππ ππππ£π = πππ₯πππ’π(ππ βπππ‘ π πππ , πππππ π πππ , πππ’ππβ ) ππππ£π = πππ₯πππ’π(462.51ππ, 200.57ππ, 401.36ππ) π ππππ = πππ. ππ ππ DON’T PRACTICE UNTIL YOU GET IT RIGHT. PRACTICE RECTANGULAR FOOTING SOLUTION: 5. Footing depth check: π‘πππ€ = 462.51ππ + 75ππ + 1.5 25ππ π‘πππ€ = 575.01 ππ π ππ¦ πππ ππ ππππ€ = 600ππ − 75ππ − 1.5 25ππ π πππ = πππ. πππ ππππ = 160 ππ − 23.5 ππ ππππ = ππ π2 πππ. π ππ΅ 6π π΅πΏ2 780ππ+650ππ 131.5 ππ π2 = 3π πΏ ππππ = π π΅πΏ π3 (0.6π) − 16 ππ π3 (0.9π) + + 6(340ππβπ+280ππβπ) (3π)πΏ2 L = 5.38π π ππ¦ π. ππ DON’T PRACTICE UNTIL YOU GET IT RIGHT. PRACTICE RECTANGULAR FOOTING SOLUTION: 6. Design reinforcement (along long span) ππ· −0.06326π ππ2 0.18069π ππ2−0.06326π ππ2 = 2400ππ+600ππ 5400ππ ππ· = 0.12849 π ππ2 2400ππ 3000ππ + ππ2 2 1 2 (0.18069 π ππ2 − 0.12849 π ππ2)(2400ππ)( 2400ππ )(3000ππ) 2 3 π΄πππ = ππππ. ππππππ π΅ β ππ ππ’ππ‘ = 0.12849 π 2400ππ DON’T PRACTICE UNTIL YOU GET IT RIGHT. PRACTICE RECTANGULAR FOOTING SOLUTION: 6. Design reinforcement (along long span) ππ’ππ‘ = ∅ππ = 0.9 π′π ππ 2 π 1 − 0.59π 1410.83π₯106 π β ππ = 0.9(27.5 π ππ2)(3000ππ) 487.5ππ 2 π(1 − 0.59π) π = π. πππππ π= ππ′π ππ¦ = 0.08413(27.5πππ) 275πππ π = π. πππππ π= π΄π ; ππ π΄π = πππ π΄π = 0.00841 3000ππ 487.5ππ π¨π = πππππ. πππππ DON’T PRACTICE UNTIL YOU GET IT RIGHT. PRACTICE RECTANGULAR FOOTING SOLUTION: 6. Design reinforcement (along long span) π΄π _πππ = 0.002π΄π = 0.002π΅π‘ π΄π _πππ = 0.002 3000ππ 600ππ π¨π_πππ = πππππππ < π¨π = πππππ. πππππ ∴ πππ π¨π = πππππ. πππππ π = π΄π π (π )2 4 π = 12299.63ππ 2 π (25ππ)2 4 π = 25.06 π ππ¦ ππ − ππππØ ππππ πππππ ππππ ππππ DON’T PRACTICE UNTIL YOU GET IT RIGHT. PRACTICE RECTANGULAR FOOTING SOLUTION: 6. Design reinforcement (along short span) ππ’ππ‘ = 0.06326π ππ2+0.18069π ππ2 2 5400ππ 1300ππ 1300ππ 2 π΄πππ = πππ. ππππππ π΅ β ππ DON’T PRACTICE UNTIL YOU GET IT RIGHT. PRACTICE RECTANGULAR FOOTING SOLUTION: 6. Design reinforcement (along short span) ππ’ππ‘ = ∅ππ = 0.9 π′π ππ 2 π 1 − 0.59π 556.57π₯106 π β ππ = 0.9(27.5 π ππ2)(5400ππ) 487.5ππ 2 π(1 − 0.59π) π = π. πππππ π= ππ′π ππ¦ = 0.01771(27.5πππ) 275πππ π = π. πππππ π= π΄π ; ππ π΄π = πππ π΄π = 0.00177 5400ππ 487.5ππ π¨π = ππππ. ππ πππ DON’T PRACTICE UNTIL YOU GET IT RIGHT. PRACTICE RECTANGULAR FOOTING SOLUTION: 6. Design reinforcement (along short span) π΄π _πππ = 0.002π΄π = 0.002πΏπ‘ π΄π _πππ = 0.002 5400ππ 600ππ π¨π_πππ = πππππππ > π¨π = ππππ. πππππ ∴ πππ π¨π = πππππππ π = π΄π π (π )2 4 π = 6480ππ 2 π (25ππ)2 4 π = 13.2 π ππ¦ ππ − ππππØ ππππ πππππ πππππ ππππ DON’T PRACTICE UNTIL YOU GET IT RIGHT. PRACTICE RECTANGULAR FOOTING SOLUTION: 7. Band Width ππππ ππππ‘πππ ππππ π€πππ‘β ππππ‘ππ = 2 ; π½+1 = 2 1.8+1 π½= πΏ π΅ π½ = 5.4π 3π = 1.8 ππππ ππππ‘πππ ππππ π€πππ‘β 13 ππππ€ = 9.29 π ππ¦ ππ ππππ πππππ‘ = 14 − 10 = 4 ππππ (πππ‘π: πππ 1 πππ π π¦ππππ‘ππ¦) DON’T PRACTICE UNTIL YOU GET IT RIGHT. PRACTICE RECTANGULAR FOOTING SOLUTION: 8. Development Length ππ ππππ = 9 10 β ππ¦ π′π β πΌπ½πΎλ π+ππ‘π ππ ; π+ππ‘π ππ ≤ 2.5 25ππ = 87.5ππ 2 3000ππ−2 75ππ −25ππ π2 = = 100.89ππ 29−1 9 275πππ 1 1 (1)(1) = 25ππ β β 87.5ππ+0 10 27.5πππ 25ππ π1 = 75ππ + ππ ππ = 471.96ππ πππ π’ππππππ = 3000ππ−400ππ 2 = 1300ππ > ππ ∴ πππππ’ππ‘π DON’T PRACTICE UNTIL YOU GET IT RIGHT. PRACTICE