Concrete Design by Anthony Adame

CONCRETE DESIGN PROJECT Anthony Adame Professor Aregawi Fall 2020 Concrete Design and Analysis of a One-Way Slab Structure

Beam Design: Beam Spanning from A1 to A2, Level 2, L Beam Tributary Length:

Dead Loads, weight per linear foot D1ROOF DROOF x Length 40 [psf] x 6’ D1ROOF = 240 [p/f] D2SLAB γ CONCRETE x Length x Slab Thickness 150 [pcf] x 6’ x (7”/12”) D2SLAB = 525 [p/f] D3BEAM γ CONCRETE x b x h 150 [pcf] x (10”/12”) x (16”/12”) D3BEAM = 166.67 [p/f] Live Loads, weight per linear food L1SNOW LSNOW x Length 30 [psf] x 6’

L1SNOW = 180 [p/f] Load Combinations and Factored Load per unit beam lenght DL = D1 + D2 + D3 DL = 240 + 525 +166.67 DL = 931.67 [p/f] LL = 180 [p/f] Combination 1 = 1.2 (DL) + 1.6 (LL) = 1406.004 [p/f] Combination 2 = 1.4 (DL) = 1304.338 [p/f] Load Combination 1 Controls, therefore wu = 1406.004 [p/f] Factored Moment Solving for Clear Span LN = Span – b/2 – b/2 = 30 ft – (10/12 [ft]) = 29.167 [ft] Moment = (wu x lN2)/ Moment Coefficient From ACI 6.5, Selected ACI Moment Coefficients

Max (+) Moment Coefficient for Beam is 14 Max (-) Moment Coefficient for Beam is 10 Max (+) Moment = = (wu x lN2)/ Moment Coefficient = (wu x lN2)/ 14 = 85.434 [kip -ft] Max (-) Moment =

= (wu x lN2)/ Moment Coefficient = (wu x lN2)/ 10 = 119.608 [kip -ft] Therefore, the controlling maximum moment will be 119.608 [kip -ft] Designing Flange Section with Tension Reinforcement, L Beam Clear Transverse Span= Sw = Short Span – b = 12 [ft] – 10/12 [ft] = 11.167 [ft] Flange thickness = Slab Thickness = 7 [in] Web Width = b = 10 [in] Effective Compression Controls Compression Flange Width, bE Combo 1 6.416 Combo 2 4.333 Combo 3 3.263 Control: Min(Combo) 3.264

Bottom Tensile Reinforcement Using No. 6 Bars, will mostly be exposed to weather due to roof conditions Diameter of bars: 0.75 [in], Area of bars: 0.44 [in] Total Area of Reinforcement, AS = No. of Bars x Area of bars = 3 x 0.44 = 1.23 [in2] Effective Depth, d For ACI 318, assuming exposed to weather (since on roof) and reinforcing is larger than bar No. 6 Clear Cover= CC = 2 [in] Using Stir Up with No. 4 Bar Diameter of No. 4 Bar = 0.5 [in] d = h- CC – DNO. 4 – 0.5D NO. 6

d= 13.125 [in] Tensile Force, T = Fy x AS = 79.2 [kip] Compressive Area = T / (0.85 x fC) = 31.059 [in2] Flange Area = bE *x h = 274.167 [in^2] Given that Flange Area is greater, the Whitney Stress Block is in the Flange Whitney Stress Block, a = bE / Compression Area = 0.793 [in] Checking Moment Capacity = φMn = 0.9 x As x fy x (d – (a/2)) = 907. 288 k – in = 75.607 k -ft Comparing to Max (+) Bending Moment = 75.607 < 85.434 Insufficent. Attempting 3 No. 7 (bar diameter of 0.875 [in] and area of 0.6 [in2]) generates a moment capacity of 101.427 [k-ft], which is insufficient. Attempting 3 No. 8 (bar diameter of 1 [in] and area of 0.79 [in2) generates a moment capacity of 131.052 [k-ft], which is sufficient as it is greater than the maximum moment 119.608 [k-ft]. Lastly designers must check for tension control. Then control is evaluated by ensure c/d is larger than 0.375 per ACI. c= a/B, where a is the Whitney Stress Block and B is given as 0.85 because fc is less than 4000 psi d = effective depth

c/d = 0.129, therefore sufficient Vertical Stirrup Shear Design Shears Stirrups are Necessary When VU < ΦVC where Φ is equal to 0.75 for shear design VU is given by ACI 6.5 Shear Coefficients, here the controlling coefficient is equal to 1.15

Solving for VU = VU = (coefficient x wu x LN)/2 VU = 23.580 [kip] Solving for ΦVC = VC = 2 x γ x Sqrt(fC) x b x d VC = 14.241 [kip] ΦVC = 10.681 [kip] 0.5 ΦVC = 5.34 [kip] Since VU > 0.5ΦVC stirrups are needed, additionally since VU > 0ΦVC designing for strength is required Solving for VS, REQUIRED = VS, REQUIRED = (VU - ΦVC )/0.75 VS, REQUIRED = 17.199 [kip]

Solving for VS, LIMIT = VS, LIMIT = 4 x x Sqrt(fC) x b x d VS, LIMIT = 28.482 [kip] Solving for AV = AV = 2 x No. 4AREA AV = 0.4 [in2] Checking Spacing, where minimum of the following four s controls Solving for s1 = s1 = (AV x fy) / (50 x b) s1 = 48 [in] Solving for s2 = s2 = (AV x fY) / (0.75 x Sqrt(fC) x b) s2 = 58.424 [in] Solving for s3 = s3 = d/2 s3 = 6.5 [in] Solving for s3 = s4 = 24 Therefore s = 6.5 [in] Solving for sSTRENGHT = sSTRENGHT = (AV x fY x d) / (VS, REQUIRED) sSTRENGHT = 18.141 [in] Minimum of sSTRENGHT and s controls Therefore s= 6.5 [in] Solving for VS, MAX = VS, MAX = 8 x Sqrt (fC) x b x d VS, MAX = 56.963 [kip] Solving for VS =

VS = (Av x fY x d) / s VS = 48 ksi ΦVS = 36 [kip] Solving for ΦVS + ΦVC = ΦVS + ΦVC = 46.681 [ksi] Given that VS, MAX or VS, MAX >VS > VS, REQUIRED Given that VU < ΦVS + ΦVC Sufficient Serviceability and Deflection Solving for Concrete Elasticity, EC EC = 33 x wC1.5 x Sqrt(fC) EC = 3321 [ksi] Steel Elasticity is known as Es ES = 29000 [ksi] Solving for the modular ration, n n = ES / EC n = 8.732 Solving for moment of deadload and live loads MD , ML, MT , with the controlling moment coefficient being 10, ( maximum moment per ACI moment coefficients), MD and wD = DLTOTAL x1 ft and wL = LLTOTAL x1 ft MD = (wD / LN2 )/10 MD = 79.256 [k-ft] ML = (wL / LN2 )/10 ML = 15.3125 [k-ft] MTOTAL = MD + ML MTOTAL = 94.569 [k-ft] Solving for reinforcement ratio, p p = As / (b × d) p = 0.0182

Solving for the neutral axis depth factor, k k = â&#x2C6;&#x161;2 Ă&#x2014; p Ă&#x2014; n + (p Ă&#x2014; n)2 â&#x2C6;&#x2019; p Ă&#x2014; n k = 0.42701 Solving for the cracked moment of inertia, ICRACKED ICRACKED =

b Ă&#x2014;(kĂ&#x2014;d)3 3

+ n Ă&#x2014; AS Ă&#x2014; (d â&#x2C6;&#x2019; k Ă&#x2014; d)2

ICRACKED = 1718.502 [in4] Solving for the gross moment of inertia, IGROSS IGROSS =

b Ă&#x2014;(h)3 12

IGROSS = 3413.333 [in4] Solving for the modulus of rupture, fR fR = 7.5 Ă&#x2014; â&#x2C6;&#x161;fc fR = 0.411 [ksi] Solving Cracking Moment, MCRACKED MCRACKED =

2Ă&#x2014;fR Ă&#x2014;IGROSS h

MCRACKED = 14.6133 [k-ft] Solving for Moment of Inertia for Dead Load Only, IE, D IE, D = (

đ?&#x2018;&#x20AC;đ??śđ?&#x2018;&#x2026;đ??´đ??śđ??žđ??¸đ??ˇ 3 ) đ?&#x2018;&#x20AC;đ??ˇ

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IE, D = 1729.125 [in4] Solving for Deflection due to only Dead Load, Î&#x201D;DD Î&#x201D;DD =

đ?&#x2018;¤ 5Ă&#x2014;( đ?&#x2018;&#x2018; )Ă&#x2014;(12Ă&#x2014;đ??żđ?&#x2018; )4 12

384 Ă&#x2014;đ??¸đ??ś Ă&#x2014;đ??źđ??¸

Î&#x201D;DD = 2.642 [in] Solving for Moment of Inertia for Dead and Live Load, IE,D,L IE,D,L = (

đ?&#x2018;&#x20AC;đ??śđ?&#x2018;&#x2026;đ??´đ??śđ??žđ??¸đ??ˇ 3 ) đ?&#x2018;&#x20AC;đ?&#x2018;&#x2021;đ?&#x2018;&#x201A;đ?&#x2018;&#x2021;đ??´đ??ż

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Ă&#x2014; đ??źđ??şđ?&#x2018;&#x2026;đ?&#x2018;&#x201A;đ?&#x2018;&#x2020;đ?&#x2018;&#x2020; + (1 â&#x2C6;&#x2019; (

đ?&#x2018;&#x20AC;đ?&#x2018;&#x2021;đ?&#x2018;&#x201A;đ?&#x2018;&#x2021;đ??´đ??ż

IE,D,L = 1724.756 [in4] Solving for Deflection due to only Live Load, Î&#x201D;DL

) ) Ă&#x2014; đ??źđ??śđ?&#x2018;&#x2026;đ??´đ??śđ??žđ??¸đ??ˇ

Î&#x201D;DL =

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Î&#x201D;DL = 3.160 [in] Solving for Total Deflection, Î&#x201D;D Î&#x201D;D = Î&#x201D;DL-Î&#x201D;DD Î&#x201D;D = 0.5184 [in] Per ACI for total allowed deflection:

Assuming roof is not likely to be damaged, deflection limit is LN / 240 = 1.45 which is greater than Î&#x201D;D, sufficient Per ACI for immediate deflections:

Both ends continuous, Allowed deflections = LN / 21 = 16.66 [in] for hminimum Given that h = 16 insufficient consider increasing or decrease span length.

Beam Spanning from A1 to A2, Level 2, T Beam Discussion This beam follows the above beam design with the follow exceptions Tributary Area Length changes from 6 ft to 12 ft as displayed below

This then changes the load combination so that Dead Loads, weight per linear foot D1ROOF DROOF x Length 40 [psf] x 12’ D1ROOF = 4800 [p/f] D2SLAB γ CONCRETE x Length x Slab Thickness 150 [pcf] x 12’ x (7”/12”) D2SLAB = 1050 [p/f] D3BEAM remains = 166.67 [p/f] Live Loads, weight per linear food L1SNOW LSNOW x Length 30 [psf] x 12’ L1SNOW = 360 [p/f] Load Combinations and Factored Load per unit beam lenght DL = D1 + D2 + D3 DL = 240 + 525 +166.67 DL = 1696.67 [p/f] LL = 360 [p/f]

Combination 1 = 1.2 (DL) + 1.6 (LL) = 2612 [p/f] Combination 2 = 1.4 (DL) = 2375.3333 p/f] Load Combination 1 Controls, therefore wu = 2612 [p/f] Factored Moment Moment = (wu x LN2)/ Moment Coefficient From ACI 6.5, Selected ACI Moment Coefficients change to Max (+) Moment Coefficient for Beam is 11 Max (-) Moment Coefficient for Beam is 16 Max (+) Moment = = (wu x LN2)/ Moment Coefficient = (wu x LN2)/ 14 = 138.875 [kip -ft] Max (-) Moment = = (wu x lN2)/ Moment Coefficient = (wu x lN2)/ 10 = 202.001 [kip -ft] Therefore, the controlling maximum moment will be 202.001 [kip -ft] with the controlling moment coefficients used in deflection calculations was equal to 10. New Specification for T Beams require that be is controlled by minimum of Combo 1 = b x ( 2 x (Sw/2)) = 12 [ft] Combo 2 = b +2 ( 8 x h) = 22.166 [ft] Combo 3 = (2 x (LN/8)) = 8.125 [ft] Combination 3 controls, be =8.125 Additionally, new reinforcement specifications were necessary to achieve the necessary moment capacity. 4 No. 9 Bars were used. As therefore became 4 in2. The design moment capacity was found to be 224 kips, which is greater than maximum moment of 202 kips. Therefore, sufficient. For shear because the beam spans internally a new shear moment coefficient was needed. It is equal to 1.0. The maximum calculated shear was found to be 38 kip and the design shear was calculated to be 46 kips, therefore sufficient. The long-term beam deflection limit is the same as before 1.45 in and calculated design long-term was found to be

0.74 in, therefore sufficient. However, the immediate deflections for all beams were insufficient by 0.66, suggest increasing beam depth to h =17. Design of Beam A1 to A2, Level 1, L Beam It is necessary to repeat the design of an external perimeter L beam because of a difference in loading. The dead superimposed roof load is replaced with a dead superimposed floor load. Additionally, there is any dead load due to the wall resting on this beam. The live load is no longer governed by the live snow load but rather a live load of 50 psf per ASCE 7-10. This changes the limiting maximum moment to 135 k -ft. The design maximum moment was found to be 162 kips, using 3 No. 8 bars, therefore sufficient. Tension controlled, therefore sufficient. The shear calculated was found to be 26 kips, and with a shear coefficient of 1.15, the calculated shear was found to be 46 kip, therefore sufficient. Long term deflection was found to be 0.738, which was less than the maximum of 1.45, therefore sufficient. The immediate deflections for all beams were insufficient by 0.66, suggest increasing beam depth to h =17. Design of Beam A1 to A2, Level 1, T Beam It is necessary to repeat the design of an external perimeter T beam because of a difference in loading. The dead superimposed roof load is replaced with a dead superimposed floor load. The live load is no longer governed by the live snow load but rather a live load of 50 psf per ASCE 7-10. This changes the limiting maximum moment to 220 k -ft. The design maximum moment was found to be 224 kips, using 4 No. 9 bars, therefore sufficient. Tension controlled, therefore sufficient. The shear calculated was found to be 41 kips, and with a shear coefficient of 1.15, the calculated shear was found to be 46 kip, therefore sufficient. Long term deflection was found to be 1.23, which was less than the maximum of 1.45, therefore sufficient. However, the immediate deflections for all beams were insufficient by 0.66, suggest increasing beam depth to h =17.

Summary for Beam Design for L Beams Varaibles Fc Fy b h t,slab length of Trib Area Gamma Droof Dwall L Snow D1 (RooF) D2 (Wall) D3 (Slab) D4 (Beam) Dtotal L1 Combo 1

Roof L Beam A1 â&#x20AC;&#x201C; A2 3000.000 60000.000 10.000 16.000 7.000 6.000 150.000 40.000 10.000 30.000 240.000 0.000 525.000 166.667 931.667 180.000 1406.000

Level 1 L Beam A1 â&#x20AC;&#x201C; A2 3000.000 60000.000 10.000 16.000 7.000 6.000 150.000 30.000 10.000 50.000 180.000 60.000 525.000 166.667 931.667 300.000 1598.000

Units psi psi in in in ft pcf psf psf psf psf psf psf psf psf p/f

Combo 2 Control wu Span Length LN Max (+) Moment Max (-) Moment Controlling Moment Cc Sw Compression Flange Width Combo 1 Combo 2 Combo 3 Combo Control b,e Bar Diameter Bar Area Bars Count As Stirrup No.4 diameter Stirrup No. 4 area d FT Compression Area Flange Area Max of the two a Moment Capacity Moment Capacity beta c c/d Tension controlled Shear Coefficient Vu, from Coeff Vs,max Vc Phi Vc Vs,required Vs, limit Av s max smax 2 smax 3 smax 4 smax For Strenght s Vs Phi Vs Phs Vs and Phi Vc Vu Ec

1304.333 1406.000 1.406 30.000 29.167 85.434 119.608 119.608 2.000 11.167

1304.333 1598.000 1.598 30.000 29.167 97.101 135.941 135.941 2.000 11.167

p/f p/f k/f ft ft k-ft k-ft k-ft in ft

6.416 4.333 3.263 3.263 39.156 1.000 0.790 3.000 2.370 0.500 0.200 13.000 142.200 55.765 274.092 274.092 1.424 1572.608 131.051 0.850 1.675 0.129 Yes 1.150 23.580 56.963 14.241 10.681 17.199 28.482 0.400 48.000 58.424 6.500 24.000 6.500 18.141 48.000 36.000 46.681 23.580 3321.000

6.416 4.333 3.263 3.263 39.156 1.128 1.000 3.000 3.000 0.500 0.200 12.936 180.000 70.588 274.092 274.092 1.803 1949.610 162.467 0.850 2.121 0.164 Yes 1.150 26.800 56.683 14.171 10.628 21.562 28.341 0.400 48.000 58.424 6.468 24.000 6.468 14.398 48.000 36.000 46.628 26.800 3321.000

ft ft ft ft in in in^2 in^2 in in in kip in^2 in^2 in k - in k -ft in

kip kip kip kip kip kip in^2 in in in in in in kip kip kip kip ksi

Es n wDL wL1 Ln MD ML MT h d As b rho rho n k Icr Ig fr MC Ie Delta dd Ie,2 Delta dl Deflection

29000.000 8.732 931.670 180.000 29.167 79256.649 15312.500 94569.149 16.000 13.000 2.370 10.000 0.018 0.159 0.427 1718.502 3413.333 411.000 14.613 1729.125 2.642 1724.755 3.160 0.518

29000.000 8.732 931.670 300.000 29.167 79256.649 25520.833 104777.483 16.000 13.000 3.000 10.000 0.023 0.202 0.465 2003.517 3413.333 411.000 14.613 2012.354 2.270 2007.342 3.008 0.738

Roof T Beam 3000.000 60000.000 10.000 16.000 7.000 12.000 150.000 40.000 40.000 10.000 30.000 50.000 480.000 1050.000 166.667 1696.667 360.000 2612.000 2375.333 2612.000 2.612 30.000

Level 1 T Beam 3000.000 60000.000 10.000 16.000 7.000 12.000 150.000 40.000 40.000 10.000 30.000 50.000 360.000 1050.000 166.667 1576.667 600.000 2852.000 2207.333 2852.000 2.852 30.000

ksi p/f p/f ft p-ft p-ft p-ft in in in^2 in

in^4 in^4 psi k-ft in^4 in in^4 in in

Summary for Beam Design for T Beams Varaibles Fc Fy b h t,slab length of Trib Area Gamma Droof Dfloor Dwall Lsnow Lfloor D1 D2 D3 Dtotal L1 Combo 1 Combo 2 Control wu Span Length

Units psi psi in in in ft pcf psf psf psf psf psf psf psf psf psf psf p/f p/f p/f k/f ft, short

LN Max (-) Moment Max (+) Moment Controlling Moment Cc Sw Compression Flange Width Combo 1 Combo 2 Combo 3 Combo Control b,e Bar Diameter Bar Area Bars Count As Stirrup No.4 diameter Stirrup No. 4 area d FT Compression Area Flange Area Max of the two a Moment Capacity Moment Capacity beta c c/d Shear Coefficient Vu, from Coeff Vs,max Vc Phi Vc Vs,required Vs, limit Av s max smax 2 smax 3 smax 4 smax For Strenght s Vs Phi Vs Phs Vs and Phi Vc Vu Ec

29.167 202.001 138.876 202.001 2.000 11.167

29.167 220.562 151.636 220.562 2.000 11.167

ft k-ft k-ft k-ft in ft

12.000 22.167 8.125 8.125 97.500 1.128 1.000 4.000 4.000 0.500 0.200 12.936 240.000 94.118 682.500 682.500 0.965 2689.923 224.160 0.850 1.136 0.088 1.000 38.092 56.683 14.171 10.628 36.618 28.341 0.400 48.000 58.424 6.468 24.000 6.468 8.478 48.000 36.000 46.628 38.092 3321.000

12.000 22.167 8.125 8.125 97.500 1.128 1.000 4.000 4.000 0.500 0.200 12.936 240.000 94.118 682.500 682.500 0.965 2689.923 224.160 0.850 1.136 0.088 1.000 41.592 56.683 14.171 10.628 41.285 28.341 0.400 48.000 58.424 6.468 24.000 6.468 7.520 48.000 36.000 46.628 41.592 3321.000

ft ft ft ft in in in^2 in^2 in in in kip in^2 in^2 in k - in k -ft in

kip kip kip kip kip kip in^2 in in in in in in kip kip kip kip ksi

Es n wDL wL1 Ln MD ML MT h d As b rho rho n k Icr Ig fr MC Ie Delta dd Ie,2 Delta dl Deflection

29000.000 8.732 1696.667 360.000 29.167 144334.519 30625.000 174959.519 16.000 13.000 4.000 10.000 0.031 0.269 0.512 2388.695 3413.333 411.000 14.613 2389.758 3.481 2389.292 4.220 0.739

29000.000 8.732 1576.667 600.000 29.167 134126.157 51041.667 185167.824 16.000 13.000 4.000 10.000 0.031 0.269 0.512 2388.695 3413.333 411.000 14.613 2390.020 3.234 2389.199 4.467 1.232

ksi p/f p/f ft p-ft p-ft p-ft in in in^2 in

in^4 in^4 psi k-ft in^4 in in^4 in in

Slab Design Slab A1, A2, B2, B1, Level 1 Concrete conditions will be the same used in beam design: f,c = 3 [ksi] f,y = 60 [ksi] Total depth of the slab was given as 7 [in], for the calculation the slab will be considered to be continuous on both ends h = 7 i[n] For reinforcement, assume to use bar No. 4 with a diameter of 0.5 in and an area of 0.2 [in2]. Assume that the slab will not be exposed to weather or ground, so cc = 0.75 in. The effective depth this then equal to d = h â&#x20AC;&#x201C; cc â&#x20AC;&#x201C; (bar diameter /2) d = 6 [in]

The slab will be analyzed using 12’ strip width, b. Load Combination Dead Loads D1 = Gamma Concrete x Slab Thickness x Strip Width D1 = 150 pcf x h/12 x b/12 D1 = 87.5 [p/f] D2 = Dead Floor Load x Strip Width D2 = 30 psf x b/12 D2 = 30 [p/f] DT = 117.5 [p/f] Live Load LL = 50 psf x b/12 LL = 50 [p/f] Factored Load Combinations Combination 1 = 1.2 (DL) + 1.6 (LL) = 164.5 [p/f] Combination 2 = 1.4 (DL) = 221 [p/f] Load Combination 1 Controls, therefore wu = 0.221 [kip/f] Factored Moment Solving for Clear Span LN = Span – b/2 – b/2 = 12 ft – (10/12 [ft]) = 11.17 [ft] Moment = (wu x LN2)/ Moment Coefficient From ACI 6.5, Selected ACI Moment Coefficients

Moment Coefficient for Slab Support on Outer Edge is 24 = (wu x LN2)/ Moment Coefficient = (wu x LN2)/ 24 = 85.434 [kip -ft] Beam Spacing = 12 [ft], short span length Beam Width = 10 [in] = 0.83[ft] Solving for Ln= Ln= Beam Spacing â&#x20AC;&#x201C; Beam Width Ln = 11.17 [ft] Beta is given as 0.85 as Fc is less than 4000 [psi] Assuming that No. 4 bars will be used for reinforcement Area No. 4 = 0.2 [in] Solving for rho REQUIRED given rectangular cross-section= rho REQUIRED =

0.85 Ă&#x2014;đ?&#x2018;&#x201C;đ?&#x2018;? đ?&#x2018;&#x201C;đ?&#x2018;Ś

2 Ă&#x2014;đ?&#x2018;&#x20AC;

đ?&#x2018;˘ ) Ă&#x2014; (1 â&#x2C6;&#x2019; â&#x2C6;&#x161;1 â&#x2C6;&#x2019; 0.85 Ă&#x2014;đ?&#x2018;&#x201C;đ?&#x2018;?Ă&#x2014;0.90Ă&#x2014;đ?&#x2018;?Ă&#x2014;đ?&#x2018;&#x2018; 2

rho REQUIRED = 0.0006 rho MINIMUM is given as 0.0018 by ACI Using rho REQUIRED to solve for As REQUIRED = As REQUIRED = rho REQUIRED x b x d As REQUIRED = 0.04 [in2] Solving for As MINIMUM = As MINIMUM = rho REQUIRED x b x d As MINIMUM = 0.04 [in2]

Controlling Area of reinforcement is the greater of the, so As MINIMUM controls Next a designer must check for bar spacing which is controlled by check the smallest of three combinations: one of which will require solving for the stress block, a Solving for a = đ??´đ?&#x2018; đ?&#x2018;Ľ đ?&#x2018;&#x201C;đ?&#x2018;Ś

a = 0.85 đ?&#x2018;Ľ đ?&#x2018;&#x201C;đ?&#x2018;? đ?&#x2018;Ľ đ?&#x2018;? a = 0.30 [in] s combinations Combination 1 đ?&#x2018;?đ?&#x2018;Ľđ?&#x2018;&#x17D;

s = đ??´đ?&#x2018; ,đ?&#x2018;?đ?&#x2018;&#x153;đ?&#x2018;&#x203A;đ?&#x2018;Ąđ?&#x2018;&#x;đ?&#x2018;&#x153;đ?&#x2018;&#x2122;đ?&#x2018;&#x2122;đ?&#x2018;&#x2013;đ?&#x2018;&#x203A;đ?&#x2018;&#x201D; s = 15.87 in Combination 2 s=3xh s = 21 [ in] Combination 3 s = 18 [in] per ACI Combination 1 controls as it in the minimum. Ensure the slab is tension controlled: c=a/B c = 0.30 / 0.85 c /d = 0.058 therefore, tension controlled and sufficient. Lastly is to design the spacing of the slabâ&#x20AC;&#x2122;s reinforcement:

Ln = 11.17 ft Ln / 4 = 2.7923 [ft] = 33.51 [in] Ln / 3 = 3.723 [ft] = 44.68 [in] Top Reinforcement Length be 2(Ln/3) + Beam Length =8.44 [ft] = 101.36 [in] In order to solve for all slab spans a similar process was followed as above. There main difference were the moment coefficients that was depend on the slabâ&#x20AC;&#x2122;s location relative to the perimeter. See summary below. Level 1

Slab 1

Slab 2

Slab 3

Location

Support 1

Span 1-2

Support 2

Span 2-3

Support 3

Span 3-4

Support 4

1/K

Mu, (k-ft)

1.148

1.968

2.756

2.505

1.722

2.505

2.756

1.968

1.148

Mu, (k-in)

13.779

23.621

33.069

30.063

20.668

30.063

33.069

23.621

13.779

2.55

0.014

0.024

0.034

0.031

0.021

0.031

0.034

0.024

0.014

rho, required

0.0006

0.0010

0.0014

0.0013

0.0009

0.0013

0.0014

0.0010

0.0006

rho, min

0.0018

As_required, in ^2 As_min, (in^2)

0.04

0.07

0.10

0.09

0.06

0.09

0.10

0.07

0.04

0.151

As, controlling (in^2 ) Bar Used (#)

0.151

S,1 (in)

15.87

S, 2 (in)

S,3 (in)

S, controlling, (in) S, (round up), (in)

15.87

a, (in)

0.30

c (in)

0.349

c/d

0.0581

tension control

0.375

max, (ten or c/d)

0.3750

tension controlled?

yes

Roof Slab Designs In order to design the slab for the roof the same methodology was followed for the Level 1 Span. Again No 4. Bars were used. However, due to a difference in loads, a different wu (load combination factor) calculation was necessary shown below.

Load Combination Dead Loads D1 = Gamma Concrete x Slab Thickness x Strip Width D1 = 150 pcf x h/12 x b/12 D1 = 87.5 [p/f]

D2 = Dead Roof Load x Strip Width D2 = 40 psf x b/12 D2 = 40 [p/f]

DT = 127.5 [p/f]

Snow Load LL = 30 psf x b/12 LL = 30 [p/f]

Factored Load Combinations Combination 1 = 1.2 (DL) + 1.6 (LL) = 201 [p/f] Combination 2 = 1.4 (DL) = 178.5 [p/f] Load Combination 1 Controls, therefore wu = 0.201 [kip/f]. All other variables remain the same. See below for summary outputs. Roof

Slab 1

Slab 2

Slab 3

Location

Support 1

Span 1-2

Support 2

Span 2-3

Support 3

Span 3-4

Support 4

1/K

Mu, (k-ft)

1.044

1.790

2.506

2.279

1.566

2.279

2.506

1.790

1.044

Mu, (k-in)

12.532

21.483

30.076

27.342

18.798

27.342

30.076

21.483

12.532

2.55

0.013

0.022

0.031

0.028

0.019

0.028

0.031

0.022

0.013

0.0005

0.0009

0.0013

0.0012

0.0008

0.0012

0.0013

0.0009

0.0005

rho, required

rho, min

0.0018

As_required, in

0.04

0.07

0.09

0.06

0.09

0.07

0.04

As_min, (in^2)

0.151

As, controlling

0.151

Bar Used (#)

S,1 (in)

15.87

S, 2 (in)

S,3 (in)

S, controlling,

15.87

S, (round up), (in)

a, (in)

0.30

c (in)

0.349

c/d

0.0581

tension control

0.375

max, (ten or c/d)

0.3750

tension

yes

(in^2 )

(in)

controlled?

Spacing values remain the same as Level 1 Slab.

Column Design Given the structure geometry, only two columns need to be designed. It is known that the Level 1 will control as they upload the Level 1 Slab and the Roof Slap. From the Level 1 columns and exterior and interior beam will need to designed to find a controlling column design for the entire structure. Corner Beam A1 Level 1 Tributary Area

Calculating Axial Load from bottom of column A1 Dead Loads

Pcolumn = height of column x column h x column b x gammaconcrete Pcolumn = 2.708 [kips]

PBeam AB = beam depth x beam width x AB trib Length x gammaconcrete PBeam AB = 1 [kips] PBeam AB total = PBeam AB x Number of Beams Supported PBeam AB total = 2 [kips]

PBeam 1,2 = beam depth x beam width x 1,2 trib Length x gammaconcrete PBeam 1,2= 2.5 [kips] PBeam 1,2 total = PBeam 1,2 x Number of Beams Supported PBeam 1,2 total = 5 [kips]

PSlab = slab thickness x AB trib Length x 1,2 trib Length x gammaconcrete PSlab = 7.875 [kips] PSlab Total = PSLAB x Number of Slabs Supported PSlab Total = 15.75 [kips[

PSuperimposed Dead Roof Load = AB trib Length x 1,2 trib Length x DRoof PSuperimposed Dead Roof Load = 2.7 kips

PSuperimposed Dead Floor Load = AB trib Length x 1,2 trib Length x DFloor PSuperimposed Dead Floor Load = 4.5 [kip]

Totaling all the above Dead Loads gives DL = 31.758 [kips]

For the Live Floor Loads,

PLive Snow Load = AB trib Length x 1,2 trib Length x LSnow PLive Snow Load = 4.5 [kip] For the Live Snow Loads, PLive Floor Load = AB trib Length x 1,2 trib Length x LFloor PLive Floor Load = 2.7 [kip]

Axial Loading Is Controlled by the maximum of the three combinations:

Combo 1: 44.461 [kips] Combo 2: 46.66 [kips] Combo 3: 46.93[kips] Combination 3 controls, therefore Pu = 46.93 [kips] For moment imposed on the column, the negative moment calculated in the section Beam Design will be used for simplification: Mu = 135.941[k-ft] Now a column interaction chart will be used to solve for sufficient reinforcement. First we will calculate Îł, given the b and h of the column are 10 [in] by 10 [in]. Assuming a clear clear of 1.50 [in] per ACI and effective cover 2.5 [in] will be assumed to account for the diameter of the unknown reinforcement

Solving for Îł,

γh = h – (2.5*2) γh = 7 γ = γh / y γ = 0.5 From I can tell there exist no interaction chart with a γ less than 0.6 Additionally, it was discovered that 10’ by 10’ with a fc equal to 3000 psi can not carry the moment sufficient. There a 10’ by 10’ column is insufficient. In order to main a concrete strength of 3000 psi, I suggest a 12’ by 12’ column be used instead. Recalculated γ = 0.58, used γ = 0.6 None of the interaction diagrams in our textbook have a chart for a concrete strength equal to 3000 psi. An interaction diagram for 3000 psi was found here: http://site.iugaza.edu.ps/sshihada/files/2017/09/INTERACTION-DIAGRAMS.pdf It should be noted that traditionally the x and y axis for a moment diagram are equal to

However, this particular chart uses:

Where Pne = Moment Solving gives: Kn = 0.1086 Rn = 0.3146 Next locate on interaction diagram, seen below with the lower blue line

Rho can be estimated to be 0.062. Per Section 10.6.11 rho must be between 0.01 and 0.08, so it is sufficient. To find the total area of reinforcement solve for As where Ag is 144 in (12 by 12) As = rho x Gross Area As = 8.928 [in2] Per the interaction chart picture, this column design has a total of 8 bars. Taking As and dividing it by the number of bars will indicate which bar should be useds. As = 8.928 / 8 As Individual = 1.116 [in2] Therefore 8 No. 10 Bars should be used. Per ACI Section 25.7.2. No 4. Bars should be used for ties since No. 10 bars were selected for the longitudinal reinforcement. Next, we must check how far to distance the column ties. Per ACI:

S1 = 20.32 [in]

S2 = 24 [in] S3 =12 [in] Combination 3 controls, so s = 12 in By inspection the spacing between the longitudinal reinforcement is less than 6, estimated 5.26 [in] so the following configuration will be needed:

Column Design B2, Level 1 Tributary Area

This column will carry a greater load due to its larger tributary area. The steps used are repeated from the same process. It should be noted in order to carry the moment and the axial load sufficiently a 14 in by 14 in beam will be used. The rho found can be seen in the above interaction diagram as the top orange line. No 11 bars should be used for the longitudinal reinforcement with No 4. ties spaced at 14 in. Below is a summary of the two column designs, with the loading corrected for the new recommended dimensions. The interior column controls. Sufficient. Level 1 slab thickness column height column l column w beam depth h beam width b beam AB trib length beam 12 trib lehgt

Column A1 7.000 26.000 12.000 12.000 16.000 10.000 6.000 15.000

Column B2 7.000 26.000 14.000 14.000 16.000 10.000 12.000 30.000

in ft in in in in ft ft

Pcol P, Beam AB 2 * P, Beam AB P, Beam 12 2* P, Beam 13 P * Slab 2*P, Slab PSuperImposed Dead Roof PSuperImposed Dead Floor Plivesnow Plive floor PD PL Psnow Unfactored P Combo 1 Combo 2 Combo 3 Controlling Pu Pn b h Pu/bh Mn Mn bh^2 Mn/bh^2 ksi yh y Kn y Rn x rho As Min Bar Area s1 s2 s2 s, control

3.900 1.000 2.000 2.500 5.000 7.875 15.750 3.600 2.700 2.700 4.500 32.950 4.500 2.700 40.15 46.130 48.090 48.360 48.360 48.360 12.000 12.000 0.336 135.941 1631.292 1728.000 0.944 7.000 0.583 0.112 0.315 0.062 8.928 1.116 20.320 24.000 12.000 12.000

5.308 2.000 4.000 5.000 10.000 31.500 63.000 14.400 10.800 10.800 18.000 107.508 18.000 10.800 136.308 150.512 163.210 164.290 164.290 164.290 14.000 14.000 0.838 220.562 2646.744 2744.000 0.965 9.000 0.643 0.279 0.322 0.062 12.152 1.519 22.560 24.000 14.000 14.000

Foundation Design Given the structure geometry, only two columns need to be designed. Footing for Columns A1

kips kips kips kips kips kips kips kips kips kips kips kips kips kips kips kips kips kips kips in in ksi k-ft k-in in^3 ksi

in^2 in^2 in in in in

From column design, the unfactored and factored service load of the column was calculated: P = 40.15 [kip] Pu = 48.36 [kip] The allowable soil bearing pressure was given as q allow = 3000 [psf] The clear cover per ACI Section 7.7 is 3 because it the column footing will be in the below ground Cc = 3 [in] Solving for the dimension of the footing, Areq = Areq = đ?&#x2018;&#x17E;

đ?&#x2018;&#x192; đ?&#x2018;&#x17D;đ?&#x2018;&#x2122;đ?&#x2018;&#x2122;đ?&#x2018;&#x153;đ?&#x2018;¤

Areq = 13.383 [ft2] Assuming the footing will be a square, the length (L) will be equal to the width (B) and since a square column c1 = c2 = c12 [in] or 1 [ft] Solving for B and L B = â&#x2C6;&#x161;đ??´đ?&#x2018;&#x;đ?&#x2018;&#x2019;đ?&#x2018;&#x17E; B = L = 3.65 [ft] Trial and Error is then used to determine a Effective Depth (d) that will satisfy Vc > Vu It was discovered that the minimum d that will be sufficient is 5 in. d = 5 [in] Solving for qu = qu =

đ?&#x2018;&#x192;đ?&#x2018;˘ đ??´đ?&#x2018;&#x;đ?&#x2018;&#x2019;đ?&#x2018;&#x17E;

qu = 3.61 [k/ft2] Given that the footing is a column the punching shear perimeter will be, bo = bo = 4 x (c+d) bo = 68 [in] Solving for Vu = Vu = Pu - qu x (c +d) x (c+d)

Vu = 47.130 [kips] Solving for Vc = Vc = 4 x 0.9 x d x bo x â&#x2C6;&#x161;đ?&#x2018;&#x201C;đ?&#x2018;? Vc = 55.867 [kips] Vc > Vu , therefore sufficient. Next, solving for area of necessary reinforcement to find spacing of the reinforcement Solving for the factored moment at the critical section, Mu đ??ż

đ?&#x2018;? 2

Mu = qu x B x 0.5 x (2 â&#x2C6;&#x2019; 2)

Mu = 11.67 [k -ft] = 140.12 [ k -in] Solving for reinforcement ratio rho REQUIRED, rho REQUIRED =

0.85 Ă&#x2014;đ?&#x2018;&#x201C;đ?&#x2018;? đ?&#x2018;&#x201C;đ?&#x2018;Ś

2 Ă&#x2014;đ?&#x2018;&#x20AC;

Ă&#x2014; (1 â&#x2C6;&#x2019; â&#x2C6;&#x161;1 â&#x2C6;&#x2019; 0.90 Ă&#x2014;đ?&#x2018;?Ă&#x2014;đ?&#x2018;&#x2018;2 đ?&#x2018;Ľ đ?&#x2018;˘0.85 đ?&#x2018;Ľ đ?&#x2018;&#x201C;đ?&#x2018;?)

rho REQUIRED = 0.00243 Solving for Whitney Stress Block, a = a=

đ?&#x2018;&#x;â&#x201E;&#x17D;đ?&#x2018;&#x153;,đ?&#x2018;&#x;đ?&#x2018;&#x2019;đ?&#x2018;&#x17E; đ?&#x2018;Ľ đ?&#x2018;&#x2018; đ?&#x2018;Ľ đ?&#x2018;&#x201C;đ?&#x2018;? 0.85 đ?&#x2018;Ľ đ?&#x2018;&#x201C;đ?&#x2018;?

a = 0.286 [in] Beta is given as 0.85 since fc is less than 4000 psi. Solving for c, c=a/B c = 0.3368 [in] Solving for strain of rebar, et et = 0.0003 x

đ?&#x2018;&#x2018;â&#x2C6;&#x2019;đ?&#x2018;? đ?&#x2018;?

et = 0.0414 Given than et = 0.0414 is greater than 0.005, per ACI tension controlled. Sufficient. Solving for area of reinforcement given moment, As REQUIRED = As REQUIRED = rho REQUIRED x B x d As REQUIRED = 0.534 [in2] Checking to see if this is greater than As MINIMUM

As MINIMUM = 0.0018 x B x h Given that we know d and assume that No. 4 Bars will be used for reinforcing h = d + cc + dno.4 + (dno.4/2) h = 8.75 in therefore, As MINIMUM = 0.691 [in2] and it controls. As = 0.691 [in2] Spacing can now be found s, 2

đ?&#x2018;&#x2018; đ??ľ đ?&#x2018;Ľ đ?&#x2018;?đ?&#x2018;&#x2013; đ?&#x2018;Ľ đ?&#x2018; đ?&#x2018;&#x153;4. 4

đ??´đ?&#x2018;

s = 12.460 [in] Spacing Requirement Check Combination 1: s Combination 2: 18 in Combination 3: 3h = 26.25 Combination 1 controls for minimum spacing Now to check one way shear such that Vc > Vu x = (L/2) â&#x20AC;&#x201C; (c/2) â&#x20AC;&#x201C; d x = 0.912 [ft] Vu = qu x B x (x) Vu = 12.062 [ft] Vc = 2 x 0.75 x d x B x SQRT(Fc, psi) Vc = 18.033 [kips] Vc > Vu, so sufficient. Footing Design for B2 The same methodology for the second footing was employed. Notable changed when plugging into the excel was a new factored and unfactored load, new dimensions as the column was suggested to be 14 by 14 inches, and d found by trial and error. See a summary below:

Variables

Footing A1

Footing B2

12.000

14.000

12.000

14.000

1.000

1.167

Unfactored Load, P

40.150

136.308

kips

Factored Load, Pu

48.360

164.290

kips

3000.000

psi

60000.000

psi

3000.000

psf

3.000

13.383

45.436

ft2

3.658

6.741

43.900

80.887

3.658

6.741

43.900

80.887

3.613

3.616

k/ft^2

0.025

k/in^2

5.000

11.000

d ^2

25.000

121.000

68.000

100.000

47.130

164.064

kips

Vc=

55.868

180.748

kips

Larger

55.868

180.748

Sufficient

11676.964

94656.020

11.677

94.656

kip ft

140.124

1135.872

kip in

0.043

0.057

0.052

rho

0.002

0.286

0.571

0.850

0.337

0.672

0.042

0.046

check against

0.005

q allow

p-ft

Tension Controlled

0.042

0.046

Yes

8.750

14.750

As Min

0.691

2.148

in^2

As, required

0.534

1.963

in^2

no 4. diameter

0.500

As control

0.691

2.148

in^2

12.460

7.392

18.000

26.250

44.250

s, controlling

12.460

7.392

0.705

2.189

0.912

1.870

12.062

45.585

kips

18.034

73.101

Sufficient

Drawings