Footing 1b

DESIGN OF SQUARE FOOTING 1-B

GIVEN: Dead Load, DL = 496.49 kN Live Load, LL = 260.3 kN Earthquake Load, E = 210.57 kN MDX = 10.73

kN-m MDY = 7.45

kN-m

MLX = 7.05

kN-m MLY = 2.39

kN-m

MEX = 157.77 kN-m MEY = 172.34 kN-m MATERIAL PROPERTIES Concrete compressive strength, f’c = 21 MPa Steel yield strength, fy = 275 MPa Normal weight concrete Unit weight of concrete, γc = 25

𝑘𝑁 𝑚^3

𝑘𝑁

Unit weight of soil, γs = 16.3 𝑚^3 Allowable soil bearing capacity, qa = 250 kPa

NSCP 2015 PROCEDURE PROVISIONS A. Footing Dimensions 1.) Assume thickness

422.8.2

COMPUTATION

t = 500 mm

Assume steel cover

Assume steel cover = 100 mm

Calculate Effective Depth, d

d = 500 mm – 100 mm = 400 mm

Calculate Effective Soil Bearing Capacity, qeff

𝑞𝑒𝑓𝑓 = 𝑞𝑎 − Σ𝛾ℎ 𝑞𝑒𝑓𝑓 = 250 − [(25)(0.5) + (16.3)(2 − 0.5)] 𝑞𝑒𝑓𝑓 = 213.05 𝑘𝑃𝑎 𝑞𝑒𝑓𝑓 =

Calculate Footing Base, B

𝑃𝑢𝑛𝑓𝑎𝑐𝑡𝑜𝑟𝑒𝑑 6𝑀𝑢𝑛𝑓𝑎𝑐𝑡𝑜𝑟𝑒𝑑 + 𝐵2 𝐵3

496.49 + 260.3 + 210.57 6( 175.55 + 182.18) + 𝐵2 𝐵3 B = 2.8432 m say B = 2.9 m 213.05 =

2.) Calculate the Ultimate Soil Bearing Capacity, qu Calculate ultimate load

Calculate ultimate moment along x

Calculate ultimate moment along x

𝑃𝑢 6𝑀𝑢𝑥 6𝑀𝑢𝑦 ± 3 ± 3 𝐵2 𝐵 𝐵 ; 𝑃𝑢 = 1.2(DL) + 1.0(LL) + 0.5(E) ; 𝑃𝑢 = 1.2(496.49) + 1.0(260.3) + 0.5(210.57) ; 𝑃𝑢 = 961.373 𝑘𝑁

𝑞𝑢 =

; 𝑀𝑢𝑥 = 1.2(𝑀𝐷𝑋 ) + 1.0(𝑀𝐿𝑋 ) + 0.5(𝑀𝐸𝑋 ) ; 𝑀𝑢𝑥 = 1.2(10.73) + 1(157.77) + 0.5(7.05) ; 𝑀𝑢𝑥 = 174.171 kN − m ; 𝑀𝑢𝑦 = 1.2(𝑀𝐷𝑦 ) + 1.0(𝑀𝐿𝑦 ) + 0.5(𝑀𝐸𝑦 ) ; 𝑀𝑢𝑦 = 1.2(7.45) + 1(172.34) + 0.5(2.39) ; 𝑀𝑢𝑦 = 182.475 kN − m 961.373 6(174.171) 6(182.475) ± ± 2.92 2.93 2.93 961.373 6(174.171) 6(182.475) 𝑞𝑢1 = + + 2.92 2.93 2.93 𝒒𝒖𝟏 = 𝟐𝟎𝟐. 𝟎𝟓𝟐 𝒌𝑷𝒂 961.373 6(174.171) 6(182.475) 𝑞𝑢2 = − + 2.92 2.93 2.93 𝒒𝒖𝟐 = 𝟏𝟏𝟔. 𝟑𝟓𝟔 𝒌𝑷𝒂 961.373 6(174.171) 6(182.475) 𝑞𝑢3 = + − 2.92 2.93 2.93 𝒒𝒖𝟑 = 𝟏𝟏𝟐. 𝟐𝟕 𝒌𝑷𝒂 961.373 6(174.171) 6(182.475) 𝑞𝑢4 = − − 2.92 2.93 2.93 𝒒𝒖𝟒 = 𝟐𝟔. 𝟓𝟕𝟒 𝒌𝑷𝒂 𝑞𝑢 =

3.) Check on one-way shear

𝑞𝑢1 − 𝑞𝑢3 𝑞′𝑢1 = 𝑞𝑢1 − ( )(0.85) 𝐵 202.052 − 112.27 𝑞′𝑢1 = 202.052 − ( )(0.85) 2.9 𝒒′𝒖𝟏 = 𝟏𝟕𝟓. 𝟕𝟒 𝒌𝑷𝒂 𝑞𝑢2 − 𝑞𝑢4 𝑞′𝑢2 = 𝑞𝑢2 − ( )(0.85) 𝐵 116.356 − 26.574 𝑞′𝑢2 = 116.356 − ( )(0.85) 2.9 𝒒′𝒖𝟐 = 𝟗𝟎. 𝟎𝟒 𝒌𝑷𝒂 𝐵−𝐶 𝑞′𝑢1 + 𝑞′𝑢2 + 𝑞𝑢1 + 𝑞𝑢2 − 𝑑)(𝐵)( ) 2 4 2.9 − 0.4 175.74 + 90.04 + 202.05 + 116.35 𝑉𝑢 = ( − 0.4)(2.9)( ) 2 4 𝑽𝒖 = 𝟑𝟔𝟎. 𝟎𝟎 𝒌𝑷𝒂 𝑉𝑢 = (

Eq. (411-1)

Calculate Nominal Shear load

𝑉𝑛 = 0.17(𝜆)√𝑓′𝑐𝑏𝑤𝑑 𝑉𝑛 = 0.17(1)√21(2900)(400)/1000

𝑽𝒏 = 𝟗𝟎𝟑. 𝟔𝟖 𝒌𝑵 Use 0.75 reduction factor 411.6.1.6

𝜙𝑉𝑛 = 0.75(903.68) 𝝓𝑽𝒏 = 𝟔𝟕𝟕. 𝟕𝟔 𝒌𝑵 𝑉𝑢 < 𝜙𝑉𝑛 𝟑𝟔𝟎. 𝟎𝟎 𝒌𝑵 < 𝟔𝟕𝟕. 𝟕𝟔𝒌𝑵, ok!

4.) Check on two-way shear

Calculate the Ultimate Shear Capacity, Vu

𝑞1 + 𝑞2 +𝑞3 + 𝑞4 4 202.05 + 116.36 + 112.27 + 26.57 𝑞𝑎𝑣𝑒 = 4 𝒒𝒂𝒗𝒆 = 𝟏𝟏𝟒. 𝟑𝟏 𝒌𝑷𝒂 𝑞𝑎𝑣𝑒 =

𝑉𝑢 = 𝑞𝑎𝑣𝑒 (𝐵2 − (𝑐 + 𝑑)2 ) 𝑉𝑢 = 114.31(2.92 − (0.4 + 0.4)2 ) 𝑽𝒖 = 𝟖𝟖𝟖. 𝟐𝟏 𝒌𝑵 𝑉𝑛 = 0.33𝜆√𝑓′𝑐 𝑏𝑜 𝑑 𝑉𝑛 = 0. .33(1)√21(4)(400 + 400)/1000 𝑽𝒏 = 𝟏𝟗𝟑𝟓. 𝟔𝟖 𝒌𝑵

Eq. (411-39)

Use 0.75 reduction factor 411.6.1.6 Compare Ultimate Shear to the Reduced Nominal Shear

𝜙𝑉𝑛 = 0.75(1935.68) 𝝓𝑽𝒏 = 𝟏𝟒𝟓𝟏. 𝟕𝟔 𝒌𝑵 𝑉𝑢 < 𝜙𝑉𝑛 𝟖𝟖𝟖. 𝟐𝟏 𝒌𝑵 < 𝟏𝟒𝟓𝟏. 𝟕𝟔𝒌𝑵, ok! Therefore, The footing dimension will be 2.9mx2.9mx0.5m

B. Main Reinforcement 2

412.3.3

Calculate Nominal moment load

𝑀𝑢 = 14.313(2.9)[(2.9 − 0.4)8 ] 𝑀𝑢 = 𝟐𝟓𝟖. 𝟗𝟗𝟏 𝒌𝑵

Calculate for the value of w by equating the value of Mu to ϕMn

𝑀𝑢 = 𝜙𝑀𝑛 𝑀𝑢 = 𝜙(𝜆)√𝑓 ′ 𝑐(𝑏𝑤)(𝑑)2 (𝑤(1 − 0.59𝑤)) 258.991(106 ) = 0.9(1.0)√21(2900)(400)2 (𝑤(1 − 0.59𝑤)) w= 0.030066112

Calculate the ratio of reinforcement

𝑤𝑓′𝑐 𝑓𝑦 0.030066112(21) ρ= 275 ρ=

ρ = 0.002295958 410.6.1 Eq. (410-3)

Calculate the minimum reinforcement

1.4 𝑓𝑦 1.4 = 275 = 0.00591

ρ𝑚𝑖𝑛 = ρ𝑚𝑖𝑛 ρ𝑚𝑖𝑛

Use ρ𝑚𝑖𝑛 Calculate the Area of reinforcement

𝐴𝑠 = ρbwd 𝐴𝑠 = 0.00591(2900)(400) 𝐴𝑠 = 5905.455𝑚𝑚2

Determine the number of 20mm RSB

N=𝜋 4

As (𝑑𝑏)2

=

5905.455 𝜋 (20)2 4

=19 bars

bw − c 2900 − 400 l𝑑𝑝𝑟𝑜𝑣𝑖𝑑𝑒𝑑 = ( − 𝑐𝑐) = ( − 100) = 1150𝑚𝑚 2 2 Eq. (412-1)

412.3.4.1

412.3.4.2 412.3.4.3

Calculate the required development length Where: φ𝑡 = bar location; not more than 300mm of fresh concrete below horizontal reinforcement φ𝑒 =coating factor; uncoated φ𝑠 =bar size facto

1 fy φ𝑡 φ𝑒 φ𝑠 l𝑑𝑟𝑒𝑞𝑑 = ( ) ( ) (𝑑𝑏) )( 𝐶𝑏 + 𝑘𝑡𝑟 1.1 𝜆√𝑓 ′ 𝑐 𝑑𝑏 𝐶𝑏+𝑘𝑡𝑟 100+0 where: = =5>2.5 𝑑𝑏

1 275 1.0(1.0)(0.8) l𝑑𝑟𝑒𝑞𝑑 = ( ) ( )( ) (20) 1.1 1√21 2.5 l𝑑𝑟𝑒𝑞𝑑 = 349.15𝑚𝑚 𝑙𝑑𝑟𝑒𝑞𝑑 < 𝑙𝑑𝑝𝑟𝑜𝑣𝑖𝑑𝑒𝑑 𝟑𝟒𝟗. 𝟏𝟓𝒎𝒎 < 𝟏𝟏𝟓𝟎𝒎𝒎, ok!

C. Load Transfer From Column to Footing 𝐵𝑢 = Pu = 1051.041kN 𝐴 8.41 √ 2 =√ = 7.25 > 2 𝐴1 0.16 𝐵𝑛 = 2(0.85)(f ′ c)A1 𝐵𝑛 = 2(0.85)(f ′ c)1600 𝑩𝒏 = 𝟓𝟕𝟏𝟐𝐤𝐍 𝜙𝐵𝑛 = 0.65(5712) = 3712.8𝑘𝑁 𝐵𝑢 < 𝜙𝐵𝑛 𝟏𝟎𝟓𝟏. 𝟎𝟒𝒌𝑵 < 𝟑𝟕𝟏𝟐. 𝟖𝒌𝑵, ok!

20

Provide Minimum Reinforcement (dowels) 𝐴𝑠𝑚𝑖𝑛 = 0.005(Ag) 𝐴𝑠𝑚𝑖𝑛 = 0.005(160000) 𝑨𝒔𝒎𝒊𝒏 = 𝟖𝟎𝟎𝒎𝒎𝟐 N=𝜋 4

As (𝑑𝑏)2

=𝜋 4

800 (16)2

=4 bars

0.24fyφr 0.24(275)(1.0) ) (db) = ( ) (16) = 230.44𝑚𝑚 ′ 1.0(√21) 𝜆√𝑓 𝑐 l𝑑𝑐 = 0.043(𝑓𝑦)(φr)(db) = 0.043)(275)(1.0)(16) = 189.20𝑚𝑚 l𝑑𝑐 = 200𝑚𝑚 l𝑑𝑐 = (

use l𝑑𝑐 = 230.44𝑚𝑚