Design of Concrete Building Project

DESIGN OF CONCRETE STRUCTURES ARAC-634/ARCP-432 PROFESSOR HOWARD GIBBS CLASS PROJECT / SPRING 2018

WRITTEN REPORT JASON HALEC ERIC HARRIS 4-26-18

The project entails the design of a second floor area in a building classified as “Assembly”. The

area has multiple uses so the seats are moveable. Floor live loads are to be in accordance with American Society of Civil Engineers’ document, “Minimum Loads for Buildings and Other Structures”, otherwise known as ASCE 7. Dead loads consist of the weight of structural members only. The project calls for a concrete floor slab supported by concrete beams. The beams are supported by concrete columns and the columns are supported by spread footings. Concrete 28 day strength shall be 4,000 psi and the yield strength of the reinforcing is 60,000 psi. Allowable soil bearing capacity is 3,000 pounds per square foot. The cost of reinforced concrete (formwork, concrete, reinforcing, finishing, etc.) shall be $1,000 per cubic yard.

The project will be a team effort. Each student must be a part of a team and no team shall have less than two students nor more than four students. The project deliverables shall consist of design calculations and detail drawings for the following: ▪ Floor slab (thickness and reinforcement).

▪ Floor beams, assuming the slab assists the beams in supporting the loads (T-beams).

▪ Columns (tied); may be square or round.

▪ Spread footings.

▪ Overall cost of the structure.

Each team shall give a ten minute oral report on their solution, including visuals. Each member of the team must present a portion of the oral report. The oral reports must include the assumptions, design theories, and/or code provisions used in the preparation of your solutions.

TABLE OF CONTENTS A.

Section I: Design Specifications

B.

Section II: Slab Design

C.

Section III: Beam Design, Shear and Moment Reinforcing

D.

Section IV: Column Design, Longitudinal and Transverse Reinforcing

E.

Section V: Foundation Design

F.

Section VI: Cost Estimates

DESIGN SPECIFICATIONS A.

Section I: Design Specifications, Codes Used, Plan and Profile Views of Structure

1. Concrete: Assume normal weight concrete.

2. Soil: Assume soil weighs 110 pounds per cubic foot.

3. Slab thickness: Conform to the Code minimum thickness for one way slabs.

4. Beam Design: Beams may be designed as rectangular or T shaped. For either type, it must be demonstrated that all applicable Code provisions were followed.

5. Shear design: Stirrups spacing does not need to be adjusted; the initial spacing can be used for

the full length of the beam. Include in the report calculations indicating the section of the beam where shear reinforcing is no longer needed.

6. Columns: Design as a short column. Assume an eight (8) foot story height for elevation drawings. The “first floor” is a slab on grade that does not connect to the columns or footings.

7. Footings: Place the base of individual square footings at least four (4) feet below grade.

8. Connections: Anchor columns to footings using dowels.

9. Cost Estimates: Provide a cost estimate for the floor slab. Provide estimated cost for a single

DESIGN SPECIFICATIONS

beam and a total for all beams. Provide estimated cost for a single column and a total for all columns. Provide estimated cost for a single footing and a total for all footings. Provide a total estimated cost for the entire structure.

▪ Dead loads = weight of structural members ▪ Concrete Strength = 4,000 psi

▪ Reinforcing = 60,000 psi ▪ Soil Bearing Capacity = 3,000 psf

PRELIMINARY FLOOR PLAN 40’ X 50’ 1

2

3

4

5

6

50'-0" 10'-0"

10'-0"

10'-0"

10'-0"

A

20'-0"

B

40'-0"

20'-0"

Design 2nd floor Assembly

DESIGN SPECIFICATIONS

C

FIRST FLOOR PLAN SCALE 1/8" = 1' -0"

10'-0"

TABLE 4-1 MINIMUM UNIFORMLY DISTRIBUTED LIVE LOADS, L0 , AND MINIMUM CONCENTRATED LIVE LOADS Occupancy or Use

Uniform psi (kNim2)

Cone. lb(kN)

50 (2.4) 100 (4.79)

2,000 (8.9) 2,000 (8.9)

Apartments (see Residential) Access floor systems Office use Computer use Armories and drill rooms Assembly areas andtheaters Fixed seats (fastened to floor) Lobbies Movable seats Platforms (assembly) Stage floors Balconies (exterior) On one- and two-family residences only, and not exceeding 100 ft2 (9.3 m2 ) Bowling alleys, poolrooms, and similar recreational areas Catwalks for maintenance access Corridors First floor Other floors, same as occupancy served except as indicated

NTS

DESIGN SPECIFICATIONS

60 (2.87) 100 (4.79) 100 (4.79) 100 (4.79) 150 (7.18) 100 (4.79) 60 (2.87) 75 (3.59) 40 (1.92)

100 (4.79)

Dining rooms and restaurants Dwellings (see Residential)

100 (4.79)

Elevator machine room grating (on area of 4 in.2 [2,580 mm2 ]) Finish light floor plate construction (on area of 1 in.2 [645 mm2 ]) Fire escapes On single-family dwellings only Fixed ladders

Gymnasiums-main floors and balconies Handrails, guardrails, and grab bars Hospitals Operating rooms, laboratories Patient rooms Corridors above first floor Hotels (see Residential) Libraries Reading rooms Stack rooms Corridors above first floor Manufacturing Light Heavy Marquees Office Buildings File and computer rooms shall be designed for heavier loads based on anticipated occupancy Lobbies and first-floor corridors Offices Corridors above first floor Penal Institutions Cell blocks Corridors Residential Dwellings (one- and two-family) Uninhabitable attics without storage Uninhabitable attics with storage Habitable attics and sleeping areas All other areas except stairs and balconies Hotels and multifamily houses Private rooms and corridors serving them Public rooms and corridors serving them Reviewing stands, grandstands, and bleachers

300 (1.33)

100 (4.79)

Dance halls and ballrooms Decks (patio and roof) Same as area served, or for the type of occupancy accommodated

Garages (passenger vehicles only) Trucks and buses Grandstands (see Stadiums and arenas, Bleachers)

Code Requirements to Follow

150 (7.18)

300 (1.33) 200 (0.89) 100 (4.79) 40 (1.92) See Section 4.4 40 (1.92)"¡h

100 (4.79) See Section 4.4 60 (2.87) 40 (1.92) 80 (3.83)

1,000 (4.45) 1,000 (4.45) 1,000 (4.45)

60 (2.87) 150 (7.18)' 80 (3.83)

1,000 (4.45) 1,000 (4.45) 1,000 (4.45)

125 (6.00) 250 (11.97)

2,000 (8.90) 3,000 (13.40)

75 (3.59)

100 (4.79) 50 (2.40) 80 (3.83) 40 (1.92) 100 (4.79)

10 (0.48) 20 (0.96) 30 (1.44) 40 (1.92) 40 (1.92) 100 (4.79) 100 (4.79)d

ASCE 7-05 Minimum Design Loads for Buildings and Other Structures

2,000 (8.90) 2,000 (8.90) 2,000 (8.90)

TABLE 4-1 MINIMUM UNIFORMLY DISTRIBUTED LIVE LOADS, L0, AND MINIMUM CONCENTRATED LIVE LOADS (continued) Occupancy or Use

Uniform psi (kN/m2)

Roofs Ordinary flat, pitched, and curved roofs Roofs used for promenade purposes Roofs used for roof gardens or assembly purposes Roofs used for other special purposes Awnings and canopies Fabric construction supported by a lightweight rigid skeleton structure All other construction Primary roof members, exposed to a work floor Single panel point of lower chord of roof trusses or any point along primary structural members supporting roofs over manufacturing, storage warehouses, and repair garages All other occupancies All roof surfaces subject to maintenance workers Schools Classrooms Corridors above first floor First-floor corridors

Cone. lb(kN)

20 (0.96)h 60 (2.87) 100 (4.79)

;

;

5 (0.24) nonreduceable 20 (0.96) 2,000 (8.9) 300 (1.33) 300 (1.33) 40 (1.92) 80 (3.83) 100 (4.79)

1,000 (4.45) 1,000 (4.45) 1,000 (4.45)

Sidewalks, vehicular driveways, and yards subject to trucking

250 (11.97)"

8,000 (35.60)f

Stadiums and arenas Bleachers Fixed seats (fastened to floor)

100 (4 . 7 9 / 60 (2 . 8 7 /

Stairs and exit ways One- and two-family residences on!y

100 (4.79) 40 (1.92)

200 (0.89)

Scuttles, skylight ribs, and accessible ceilings

20 (0.96)

Storage areas above ceilings

NTS

Storage warehouses (shall be designed for heavier loads if required for anticipated storage) Light Heavy Stores Retail First floor Upper floors Wholesale, all floors

125 (6.00) 250 (11.97)

100 (4.79) 75 (3.59) 125 (6.00)

Vehicle barriers

Code Requirements to Follow

DESIGN SPECIFICATIONS

g

1,000 (4.45) 1,000 (4.45) 1,000 (4.45)

See Section 4.4

Walkways and elevated platforms (other than exit ways)

60 (2.87)

Yards and terraces, pedestrian

100 (4.79)

"Floors in garages or portions of a building used for the storage of motor vehicles shall be designed for the uniformly distributed live loads of Table 4-1 or the following concentrated load: (I) for garages restricted to passenger vehicles accommodating not more than nine passengers, 3,000 lb (13.35 kN) acting on an area of 4.5 in. by 4.5 in. (114 mm by 114 mm) footprint of a jack; and (2) for mechanical parking structures without slab or deck that are used for storing passenger car only, 2,250 lb (10 kN) per wheel. hGarages accommodating trucks and buses shall be designed in accordance with an approved method, which contains provisions for truck and bus loadings.

'T he loading applies to stack room floors that support nonmobile, double-faced library book stacks subject to the following limitations: (I) The nominal book stack unit height shall not exceed 90 in. (2290 mm); (2) the nominal shelf depth shall not exceed 12 in. (305 mm) for each face; and (3) parallel rows of double-faced book stacks shall be separated by aisles not less than 36 in. (914 mm) wide. din addition to the vertical live loads, the design shall include horizontal swaying forces applied to each row of the seats as follows: 24 lb per linear ft of seat applied in a direction parallel to each row of seats and 10 lb per linear ft of seat applied in a direction perpendicular to each row of seats. The parallel and perpendicular horizontal swaying forces need not be applied simultaneously. eother uniform loads in accordance with an approved method, which contains provisions for truck loadings, shall also be considered where appropriate. f The concentrated wheel load shall be applied on an area of 4.5 in. by 4.5 in. (114 mm by 114 mm) footprint of a jack. gMinimum concentrated load on stair treads (on area of 4 in.2 [2,580 mm 2 ]) is 300 lb (1.33 kN). hWhere uniform roof live loads are reduced to less than 20 lb/ft2 (0.96 kN/m 2 ) in accordance with Section 4.9.1 and are applied to the design of structural members arranged so as to create continuity, the reduced roof live load shall be applied to adjacent spans or to alternate spans, whichever produces the greatest unfavorable effect. ;Roofs used for other special purposes shall be designed for appropriate loads as approved by the authority having jurisdiction.

I

TABLE 4-2 LIVE LOAD ELEMENT FACTOR, KLL Element

u

KLLa

Interior columns Exterior columns without cantilever slabs

4 4

Edge columns with cantilever slabs

3

Corner columns with cantilever slabs Edge beams without cantilever slabs Interior beams

2 2 2

All other members not identified including: Edge beams with cantilever slabs Cantilever beams One-way slabs Two-way slabs Members without provisions for continuous shear transfer normal to their span

1

In lieu of the preceding values, KLL is permitted to be calculated.

ASCE 7-05 Minimum Design Loads for Buildings and Other Structures

TABLE 2.1 TYPICAL DEAD LOADS FOR SOME COMMON BUILDING MATERIALS

NTS

Reinforced concrete Structural steel Plain concrete ivfovable steel partitions Plaster on concrete Suspended ceilings 5-Ply felt and gravel Hardwood flooring (7/8 in) 2 x 12 X 16 in double wood floors Wood studs with 1/2 in gypsum each side Clay brick wythes (4 in)

150 lb/cu ft 490 Lb/cu ft 145 lb/cq ft 4psf 5 psf 2psf 6 psf 4 psf 7 psf 8psf 39 psf

The approximate weights of some common building materials for roofs, wall, floors, and so on are presented in Table 2.1.

Code Requirements to Follow

DESIGN SPECIFICATIONS

ACI 318 - 11: Reinforced concrete beam design parameters

Reinforcement Ratio The amount of steel reinforcement in concrete members should be limited. Over-reinforcing (the placement of too much reinforcement) will not allow the steel to yield before the concrete crushes and there is a sudden failure.

The reinforcement ratio in concrete beam design is th following fraction: The reinforcement ratio , ρ, must be less than a value determined with a concrete strain of 0.003 and tensile strain of 0.004 (minimum). When the strain in the reinforcement is 0.005 or greater, the section is tension controlled. (For smaller strains the resistance factor reduces to 0.65 because the stress is less than the yield stress in the steel.)

Maximum Reinforcement Based on the limiting strain of 0.005 in the steel, x(or c) = 0.375d so α = β1 (0.375d) to find As-max The values of β1 are presented in the following Table: Table A.7 STEEL GRADES / YIELD STRESSES & RHO

DESIGN SPECIFICATIONS

ACI 318 - 11: Reinforced concrete beam design parameters

Minimum Reinforcement Minimum reinforcement is provided even if the concrete can resist the tension, in order to control cracking.

Minimum required reinforcement:

but not less than where: fy is the yield strength in psi bw is the width of the web of a concrete T-Beam cross section d is the effective depth from the top of a reinforced concrete beam to the centroid of the tensile steel Cover for Reinforcement Cover of concrete over/under the reinforcement must be provided to protect the steel from corrosion. For indoor exposure, 1.5 inch is typical for beams and columns, 0.75 inch is typical for slabs, and for concrete cast against soil, 3 inch minimum is required.

Bar Spacing Minimum bar spacings are specified to allow proper consolidation of concrete around the reinforcement. The minimum spacing is the maximum of 1 in, a bar diameter, or 1.33 times the maximum aggregate size.

Effective width beff In case of T-Beams or Gamma-Beams, the effective slab can be calculated as follows: For interior T-sections, beff is the smallest of:

L/4, bw + 16t, or center to center of beams For exterior T-sections, bE is the smallest of

bw + L/12, bw + 6t, or bw + ½(clear distance to next beam) When the web is in tension the minimum reinforcement required is the same as for rectangular

DESIGN SPECIFICATIONS

sections with the web width (bw) in place of b.

When the flange is in tension (negative bending), the minimum reinforcement required is the greater value of

where: fy is the yield strength in psi bw is the width of the web of a concrete T-Beam cross section beff is the effective flange width

Stress and Strain Distribution For R/C Beams

DESIGN SPECIFICATIONS A.7 Table – Areas of Steel bars

SLAB DESIGN B.

Section II: Slab Design Calculations, Drawings to include Plan and Section Views with Details of Reinforcing

ONE-WAY SLAB – ANALYSIS (SECTION) 1.

2.

3.

4.

Solve for Reaction; Reaction becomes load beam Determine the weight of slab Solve for one beam and double the load Draw the shear and moment for slab on beam

Assume 6” in slab

SLAB DESIGN

Analysis of One-way Slab

ONE-WAY SLAB ANALYSIS Assume Simply – supported beam design Span length = 10 feet f’c = 4,000 psi fy = 60, 000 psi LL = 100 psf (Table 4.1 – Assembly, Movable seats and Tables)

ANALYSIS MEMBER

DL = Weight of Structure, Reinforced Concrete = 150 pcf h = L/20

h

SLAB DESIGN

12” 10’ – 0”

6”

5”

D = 5” (Assume ½” Diameter Bars with ¾” Cover)

ONE-WAY SLAB ANALYSIS- CALCULATIONS Solving for factored moment h = L/20 = (10’) (12”/1’) / 20 = 6” And Shrinkage/Temp of Steel

Wu = 1.2 DL + 1.6 LL = 1.2 (75) + 1.6 (100) Wu = 90 + 160 = 250 plf Mu = wl²/ 8 (FOR UNIFORM LOAD) = (250) (10²) / 8 = 3125 ft – lb.

Mu / Φ bd² = 3125 (12) / (0.90) (12) (5)² = 3125 (12) / 270 = = 37500 / 270 = 138.89 psi (Per table A.3 = 138.89 psi = ρ = 0.00225) Use ρ min = 0.0033

Use

As = ρ * b * d = (0.0033) (12)(5) = 0.198 #3 @ 6 ½ “ , As = 0.20 #4 @ 12” , As = 0.20

SLAB DESIGN

Use

#4 Diameter = 0.5”, d (actual) = 5” Shrinkage and Temp. of Steel As = (0.0018) b*h = (0.0018) (12) (6) = 0.1326 in², USE # 3 @ 10, As = 0.13in²/ft 0.0018 is ρ min Per Table A.12

ONE-WAY SLAB ANALYSIS- CALCULATIONS Wu = 250 plf slab

1. Ra = Rb = wl / 2 = (250) (10) / 2 = 1250 plf 2. Va = Vb = Ra = Rb = wl / 2 = 1250 plf 10’ – 0”

3. Mmax = wl² / 8 = 3125 ft. - lb.

Ra

Rb

FBD 5’ – 0”

Solving for Reactions, Shear, and Moment

5’ – 0”

Va – 1250 lb.

Shear

0

Mmax = 3125 ft – lb.

Vb – 1250 lb.

SLAB DESIGN Moment

0

BEAM DESIGN C.

Section III: Beam Design Calculations, Drawings to include Plan and Section Views with Details of Shear and Moment Reinforcing

T-BEAM DESIGN - ANALYSIS 1. Solve for b & d; solve for T-beam 2. Mu doesn’t change until the weight of the beam changes 3. Wu = 1.2 DL + Reaction of slab on beam (Uniform load on beams)

6”

Analysis of T - Beam Assume 12” column

12” Assumed

DL = 150 pcf * length of the beam Ra + B slab = 2500 plf (each beam supports two slabs)

BEAM DESIGN

T-BEAM DESIGN ANALYSIS

10’ – 0” hf h Analysis of T - Beam As

hf = 6” bw = 12” h = 18” d = 15 ½”

d 2 ½”

bw

10’

BEAM DESIGN

9’ 12”

12”

T-BEAM DESIGN ANALYSIS Wu = 250 plf

1. Wu = 1.2 (150) (20’) + 2500 = 6100 plf

slab

2. Ra = Rb = wl / 2 = (6100)(20)² / 2 = 61000 ft. – lbs.

3. Va = Vb = Ra = Rb = 61 kips

20’ – 0”

4. Mmax = wl² / 8 = 61 k/ft (20 ft)² / 8 = 3050 ft-k

Ra

Rb

FBD Solving for Reactions, Shear, and Moment

10’ – 0”

10’ – 0”

Va – 61 k

Shear

0

Mmax = 3050 ft – lb.

Vb – 61 k

BEAM DESIGN Moment

0

T-BEAM DESIGN CALCULATIONS Solving for bf, As, and Trial Steel

bf = 16 hf + bw = 16 (6) + 12 = 108“ = span/4 = (20 ft) (12 in/ ft ) = 60” = clear span + bw = 108” + 12” = 120” As min (for check) = 3 √f’c * bwd = 3 √4000/60000 (12) (15.5) = 0.59 in² or……

200 bwd/ fy = 0.62in² z = 0.9d = 0.9 (15.5) = 13.95 in or.…. d – hf / 2 = 15.5 – 6/2 = 12.5 in

USE

Trial Steel Mn = Mn /Φ = 3050 ft-k / 0.9 = 3388.89 ft.-k

As = Mn / fy*z = 3388.89 (1000) / (60000) (13.95) = 4.05 in²

BEAM DESIGN

T-BEAM DESIGN CALCULATIONS Solving for As

As= 4.05in² Ac = As*fy/ 0.85 f’c = (4.05) (60000)/ (0.85) (4000) = 71.47 in a = Ac/bf = 71.47 / 60 = 1.19 in z = 15.5 – 1.192 = 14.91 in As = (3388.89) (1000) / (60000) (14.91) = 3.79 in² Ac = (3.79) (60000) / (0.85) (4000) = 66.88 in² a = Ac/bf = 66.88 / 60 = 1.11 in z = 15.5 – 1.11 / 2 = 14.95 in As = (3388.89) (1000) / (60000) (14.95) = 3.77 in²

USE

Ac = (3.77) ( 60000) / (0.85) (4000) = 66.53 in²

BEAM DESIGN

a = Ac/bf = 66.53 / 60 = 1.11 in z = 15.5 – 1.11 / 2 = 14.95 in Check: 3.77 in > 0.59” in, Φ = 0.90 (from previous calc’s) Use 3 # 10 , As = 3.79 in² (Web with min = 10.44” OK)

T-BEAM DESIGN CALCULATIONS Solving for Development Length

#10 bars, from Table Areas of Steel , db = 1.27in Steel bars are not coated fy = 4,000 psi f’c = 60,000 psi As = 3.44in² (required steel) ψt * ψe * ψs* λ = each value = 1.0 Ktr = 0 C= 3” side cover C= 3/2 = 1.5 half of the bar spacing (min value)

Development Length, ld Calculations

ld = (3/40) fy/ √f’c { ψt * ψe * ψs* λ / C + ktr / db } db C + ktr / db = 1.5” + 0 / 1.27 = 1.181

= (3/40) (60,000/ √4000) {(1)(1)(1)(1) / 1.181)} 1.27 = 76.51 in.

BEAM DESIGN

Ratio = Areq / Aprov = 3.44in² / 3.79in² = 0.91 Ld = 76.51 (Arequired/ Aprovided) Ld = 0.91 (76.51) Ld = 69.62” Ld = 70” Use 70” – Development Length

T-BEAM DESIGN CALCULATIONS Solving for Stirrups And determine spacing

1.

2.

3.

4.

Calculate Theoretical stirrup spacing Determine max spacing to provide min area of shear reinforcement Compute max spacing; d/4 ≤ 12 in, if Vs > 4 √ f’c *bwd Determine min practical spacing

Vu= Vu at left end = 61 k = 61,000 lb.

Vu = at a distance d from face of support = - Wu (d) + Ra = -(6100) (15.5/12 in) + 61000 lb. = 53,120.83 lb. ΦVc = Φ 2 λ (√ f’c) bwd = (0.75) (2) (1) (√4000) (12) (15.5) = 17,645.51 lb.

Vu = Φ Vc + Φ Vcs ΦVs = Vu – ΦVc = 53120.83 – 17645.51 = 35,475.32 lb.

Vs = 35,475.32 / 0.75 = 47,300.43 lb.

S = Av*fy*d / Vs = (2) (0.11) (60000) ( 15.5) / 47,300.43 = 4.33 in (Theoretical Spacing)

BEAM DESIGN

S = Av*fy / 0.75 √ f’c * bw = 2 (0.11) ( 60000) / 0.75 √4000 (12) = 23.19 in (Max Spacing) ≤ S = Av*fy / 50 *bw = 2 (0.11) (60000) / (50) (12) = 22 in Vs = 47300.43 < 4 √ f’c *bwd = 4 (√4000) (12) (15.5) = 47,058 lb.

S = d/4 = 15.5 = 3.875 ≤ 12in ≈ 4in = S

COLUMN DESIGN D.

Section IV: Column Design Calculations, Drawings to include Plan and Section Views with Details of Longitudinal and Transverse Reinforcing

COLUMN DESIGN - ANALYSIS 1. Design Rectangular / Square column with ties 2. Provided for no more than 4” inches for the ties above/ below 3. Draw cross section

Analysis of Columns

d

61000 lb. b

Rb2

61000 lb.

Rb1

18”

COLUMN DESIGN

8’ – 0” 6’ – 6”

COLUMN DESIGN CALCULATIONS Solving for Tied Columns φ for tied columns = 0.65 Rb1 = 61000 lb.

Rb2 = 61000 lb.

Ru = 122000 lb. = 122 k Ru = Pu Ast = Ag (0.02)

ΦPn = Φ (0.80) [ 0.85 f’c (Ag – Ast) + fy Ast ] 144 = (0.65) (0.80) [ (0.85) (4) (Ag – 0.02Ag) + (60) (0.02Ag) 144 = 0.52 [3.4 (Ag – 0.002Ag) + 1.2 Ag 144 = 0.52 [ 3.4Ag – 0.068Ag) + 1.2 Ag]

COLUMN DESIGN

144 = 0.52 [Ag (3.4 – 0.068) + 1.2 Ag] 144 = 0.52 (3.332Ag + 1.2Ag) 144 = 1.733Ag + 0.624Ag 144 = 2.351Ag 144/2.351 = Ag = 61.09 in2 8 x 8 = 64.00, USE 8” X 8” Sq. Column

COLUMN DESIGN CALCULATIONS Solving for Bars & Ties

Design of Longitudinal Bars Ag = 64 in² ΦPn = Φ (0.80) [ 0.85 f’c (Ag – Ast) + fy Ast ] 144 = (0.65) (0.80) [ (0.85) (4) (64 – Ast) + (60)Ast ] 144 = 0.52 [3.4 (64 – Ast) + 60Ast ] 144 = 0.52 [ 217.6 – 3.4Ast) + 60Ast ]

144 = 0.52 [ 217.6 + 56.6Ast ] 144 = 113.15 + 29.43 144 – 113.15 / 29.43 = Ast = 1.05 in² = 4 #5, Ast = 1.23 in²

USE

Design of Ties

COLUMN DESIGN

a.) 16 (0.625) = 10”

#5 Dia = 0.625

b) 48 (0.375) = 18”

#3 Dia = 0.375

c) least dimension = 8”

USE #3 ties @ 8”

Check: STL % = 1.23 / (8) (8) = 0.019 > 0.010 OK

FOUNDATION DESIGN E.

Section V: Foundation Design Calculations, Drawings to include Plan and Section Views with Details of Reinforcing

SPREAD FOOTING DESIGN - ANALYSIS 1.

2.

3.

4.

Analysis of Spread Footing

1.

Determine Footing Sizes Determine Footing Depth Design Column Dowels Draw Plan and Section

Column loads: • Live load: 110 psf • Dead load: 150 pcf • Footing uplift: 0 kips

• Column size: 1 ft. x 1 ft.

2.

Soil information: • Allowable soil bearing capacity: 3000 psf • Soil cover above footing: 1 ft

FOUNDATION DESIGN

• Unit weight of soil: 110 pcf 3.

Materials used: • Concrete strength at 28 day = 3000 psi • Yield strength of rebars = 60 ksi Design code: ACI 318-05

SPREAD FOOTING DESIGN CALCULATIONS Solving for footing size

Weight of soil = ꙋsoil = 110 lb / ft3 Soil bearing Capacity = 3,000 psf Weight of Concrete = ꙋcon = 150 lb / ft3 qe = qa – (qsoil – qftg) = 3000 – (24/12)150 – (24/12)100 = 3000 – 300 – 220 = 3000 – 520 = 2480 psf a = 122.77 / 2.48 = 49.5 ft2 ≈ 7’ x 7’ USE 7’ X 7’ Spread Footings

FOUNDATION DESIGN

SPREAD FOOTING DESIGN CALCULATIONS Solving for footing depth

The Factored footing pressure can be calculated as Qu = 1.2 x 150 + 1.6 x 100 = 340 / footing area (56.3 ft²) = 6.03 psf Check punching Shear Assume the reinforcements are #10 bars, , the effective depth D = 15.5 – 2.5” cover – 1.27 (one bar size) = 11.73 = 11inches” The punch shear stress can be calculated as Vu = 60.3 (7.5² - (1 +11”)² *1000{4 x 11” x (1 + 11”) x 144} = 163.3 psi The shear Strength of concrete is ΦVc = 0.75 (4) (3000)1/2 = 164 psi OK.

FOUNDATION DESIGN

SPREAD FOOTING DESIGN - ANALYSIS DESIGNING COLUMN DOWELS Solving for column dowels

1. The bearing capacity of concrete at column base is = (0.65)(0.85)(4)(7)(7) = 108.29 kips 2. The factor column load is Pu = 122.77 kips ( load combination from calculations above) 3. The required area of column dowels is As = (122.77 â&#x20AC;&#x201C; 108.29) / 60 = .241 in2 4. The minimum dowel area is As,min = (0.0005)(14 in)(14 in) = 0.098 in2

FOUNDATION DESIGN

Use 4 -#5 dowels As = 1.23in2 PLAN VIEW

ELEVATION

ASSEMBLY DESIGN DRAWINGS - 1

ASSEMBLY DESIGN

ASSEMBLY DESIGN DRAWINGS - 2

ASSEMBLY DESIGN

ASSEMBLY DESIGN DRAWINGS - 3

ASSEMBLY DESIGN

COST ESTIMATES F.

Section VI: Cost Estimates

COST ESTIMATION (SLAB, BEAM, COLUMN)

Cost Estimation using BIM quantity take-offs

COST ESTIMATION

FOOTING COST = $ 13.61 yd³ x $1000 / yd³ = 367.50 x 18 (# of footing) = $6,615

Grand Total

$54,759 + footing estimate*

COST ESTIMATION (BREAKDOWN) Cost Estimation of Project

The cost of reinforced concrete (formwork, concrete, reinforcing, finishing, etc.) shall be $1,000 per cubic yard (yd³) • SLAB COST = 38. 722 yd³ x $1000 / yd³= $38,722 • BEAM COST = 14.111 yd³ x $1000 / yd³ = $14,111 • COLUMN COST = 0.107 yd³ x $1000 / yd³ = 107 lb x 18 (# of columns) = $1,926 • FOOTING COST* = $ 13.61 yd³ x $1000 / yd³ = 367.50 x 18 (# of footings) = $6,615 • TOTAL COST = $ 54,759 + $ 6,615 = $61, 374 (+ Design Contingency)

COST ESTIMATION

SLAB DESIGN NOTES

Initial footing calculations indicated combined footings were required to support

the loads of the structure. A 7-foot square footing is determined when the project defined 3000 psf soil bearing capacity was compared to the calculated loads.

Isolated square footings of that size will not fit within a column spacing of 10feet. An isolated square footing of 4'x'4', for example, requires a qe to be at least 7670 psf. 48ft2 = 122.77k / 7.67 psf. In light of seeing this problem, we continued with the design as-is for demonstration purposes.

DESIGN CONCLUSION

RESOURCES 1. Jack C. McCormac, Russell H. Brown, DESIGN OF REINFORCED CONCRETE 8th Edition, ACI 318 â&#x20AC;&#x201C; 08 Code Edition, John Wiley and Sons, Inc., 2009.

2. AISC. AISC MANUAL OF STEEL CONSTRUCTION, 13rd Edition, American Institute of Steel Construction, 2008.

3. James Ambrose, Patrick Tripeny, SIMPLIFIED DESIGN of STEEL STRUCTURES, 8th Edition, John Wiley and Sons, Inc., 2007.

4. William T. Segui, LRFD STEEL DESIGN, 5th Edition, Cengage Learning, Thomson Publishing, 2012.

5. American Society of Civil Engineers. MINIMUM DESIGN LOADS FOR BUILDINGS AND OTHER STRUCTURES, ASCE, Pgs. 12-13, 2006.

Any photos retrieved from: Google Images

PROJECT REFERENCES

TERMINOLOGY US CUSTOMARY UNITS

=depthof equivalent.rectangular stress block, in.

a

Ag =gross area of column, m2 As =areaof tension reinforcement, A/ =areaof compression reinfuzcement.m2 As, =total area of longitudinalreinforcement,in2 A,. = areaof shearreinforcement wi1hln a distances, in. b = width of compression face of member, in. be =effective compression flange width, in. b.., =web width, in.

B1 =ratio of depth of rectangular stress block, a, to depth to neutral axis, c

=ratio of totalreinforcement areato cross-sectional

areaofcolumn=Asr/Ag s =spacing cJfsheartiesmeasured alonglongitudinal axisofmem.bel; in.

Vc = nominalshear strength provided by concrete, lb strength at seeti.on, lb Vn nominal Ď&#x2020;Vs=design shearstrengthat section, lb _ = nominal shear strength provided by reinforcement, lb Vu =factored shear forceat section, lb

c

= distance:fromextreme-compressionfiber to neutral axis.

BARSIZE

d

=distance fromextreme compression fiber to centroid of

#3

0.37S

0.11

#4

0.500

0.20

in.

DIAMETER.IN.

AREA. IN2.

WEIGHT, LB/FF

0.376

nonprestressedtension reinforcement,in. d, = distanceftomextreme compression fiber to extreme tensionsteel,in.

Ee = modulus of elasticity= 33w!-5ffc ,psi

#5

0.625

0.31

1.043

#6

0.750

0.44

1.502

ÂŁ1

#7

0.875

0.60

2.044

#8

l.(X)()

0.79

2.670

#9

1.128

1.00

3.400

#10

1.270

1.27

4.303

#11

1.410

1.56

5.313

#14

1.693

2.2S

1.650

#18

2.251

4.00

=net tensilestrain in extreme tension steel st nominal strength

1c =compressive strength of concrete, psi J;, =yieldstrength of steel reinforcement.psi h_r=T-beam flange thickness, in.

Mc =factored column moment, includingslenderness effect, in.-lb

TERMS/ FORMULAS

p

M,. = nominal moment strength at section, in.-lb = d e s i g n moment strength at section, in.-lb M. =factored moment at section, in.-lb P,. = nominal axial load strength at given eccentricity, lb l{IP,.= design axialload strength at given eccentricity, lb Pu

=factored axial foice at section, lb

LOADFACTORSFORREQUIREDSTRENGTH

U= 1.4D U = 1.2D + 1.6L

0.668

13.60

DESIGN OF CONCRETE STRUCTURES ARAC-634/ARCP-432 PROFESSOR HOWARD GIBBS CLASS PROJECT / SPRING 2018

THANK YOU! QUESTION? COMMENTS? PROVISIONS?