Truck Loading Example Calcs

8/12/2019 Truck Loading Example Calcs

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5/89

ASSUMPTI ONS 2. 0 MG LESLI E LANE RESERVOI R

DATE: 9/ 22/ 2000 ESCONDI DO, CALI FORNI A

LeslieLane.wpf Page 1

I. ASSUMPTIONS

Basic Tank Data:

Tank inside diameter = 107.0 ft

Water depth near wall = 30.00 ft (above floor)

Maximum water depth for overload = 31.00 ft (above floor)

Roof placed on top of wall

Distance from floor to top of wall = 31.00 ft

Center-to-center distance between interior roof supports = 21.00 feet

Superimposed roof load = 50.00 psf live load + 200.0 psf soil load+ H2O truck loading

The corewall is of uniform thickness and poured in place.

Poured in place wall thickness = t wall = 10.00 in

Wall edge condition: neoprene pad supports

Prestressing Information:Circumferential prestressing system: 0.3750" diameter galvanized strand.

Circumferential prestressing machine tolerance = 1.500%

Minimum ultimate strength of circumferential prestressing steel = 240.0 ksi

Prestressing stress loss = 25.00 ksi

Final shotcrete cover depth over prestressing = 1.500 in

Center-to-center distance between layers of prestressing = 0.7500 in

Minimum final circumferential compression for differential temperature and dryness:

Above grade = 200.0 psi

Below grade = 50.0 psiPrestressing will taper from 200.0 psi to 50.00 psi over 6.000 ft beginning at theminimum backfill elevation

Vertical prestressing system: Dywidag threadbars

Minimum final vertical compression in corewall = 200.0 psi

Minimum ultimate strength of vertical prestressing steel = 150.0 ksi

Vertical prestressing stress tolerance = 1.500%

Maximum spacing of vertical prestressing = 50.00 inSoil Information:

Backfill height = h backfill = 33.54 ft max. to 31.00 ft. min. (above floor)

Static equivalent liquid backfill pressure = 1192. psf (trapezoidal distribution)Seismic equivalent liquid backfill pressure = 36.00 psf (inverted triangular distribution)

Backfill soil density = 130.0 pcf

Soil bearing capacity = 6000 psf (Gross) for Columns= 7000 psf (Gross) for Walls


6/89

ASSUMPTI ONS 2. 0 MG LESLI E LANE RESERVOI R



Misc. Information:

Concrete strength: core-wall = 4000 psishotcrete = 4000 psicolumns = 4000 psiroof slab = 4000 psifootings = 3500 psi

floor slab = 3500 psisump = 3500 psiSlopes: center of floor to wall = 1.500% ( + = down, - = up)

center of roof to wall = 1.500% ( + = down, - = up)

Thickness of floor slab = 6.000 in

Steel reinforcement required in floor slab for temperature and drying shrinkage = 0.5000%

Steel reinforcement required in wall footing for temperature and drying shrinkage = 0.5000%

Friction factor between floor and subgrade = 1.500UBC 97 SEISMIC DESIGN

Effective seismic horizontal acceleration = C v x IRhoriz x T

Effective seismic vertical acceleration = C v x I x B R ver x T

T=[(4 x ( WS +WR +WI ) ) / g r k ] 1/ 2

Wher e, W S = Wei ght of wal l AWWA D110-95, Eq 4-8

= 1610. k ( See Sei smi c Cabl e Desi gn)

WR = Wei ght of r oof

= 3662. k ( See Sei smi c Cabl e Desi gn)

WI = Wei ght of wat er

= 1. 686e+04 k ( See Sei smi c Cabl e Desi gn)

g = accel er at i on due t o gr avi t y

= 32. 20 f t / sec 2

r = r adi us of t ank at mi d- hei ght of wal l

= 53. 50 f t

k = spr i ng st i f f ness

Wher e, k = 144 [ A x E Cos 2" + 2 x G x W x L p] AWWA D110-95, Eq 4-9 L c x S c t x S p

A = ar ea of st r and

= 0. 8900 i n 2

E = modul us of el ast i ci t y

= 26 x 10 6 psi

" = angl e of st r and

= 45 0

Lc = l engt h of st r and

= 27. 13 i n


7/89

REFERENCES 2.0 MG LESLIE LANE RESERVOIR

REVISED: 11/02/2000 ESCONDIDO, CALIFORNIA


Sc = spaci ng bet ween st ands

= 46. 19 i n

G = shear modul us

= 52. 00 psi

W = wi dt h of bear i ng pad

= 6. 000 i n

LP = l engt h of bear i ng pad

= 1. 000 i n

t = t hi ckness of bear i ng pad

= 1. 250 i n

SP = spaci ng of bear i ng pad

= 1. 000 i n

k= 144 [ 0. 08900 i n 2 x 26 x 10 6 psi x Cos 2 45 0 + 2 x 52. 00 psi x 6. 000 i n x 1. 000 i n ]

27. 13 i n x 46. 19 i n 1. 000 i n x 1. 000 i nk= 222. 8 k

T=[ ( 4 x ( 1610. k + 3662. k + 1. 686e+04 k) ] 1/ 232. 2 f t / s 2 x 53. 50 f t x 222. 8 k

T= 0. 8512

Soil Profile = S BSeismic Source Type = A/B

Z = 0.40Ca = 0.40Cv = 0.60I = 1.25R

hor= 4.500

Rvert = 3.00Nv = 1.0Na = 1.0B = 2/3

Uo = 0.1958 g

Vo = 0.1958 gAWWA SEISMIC DESIGN

Effective seismic horizontal acceleration = (Z H x I x C)/R HEffective seismic vertical acceleration = (Z V x I x C)/R VWhere, Z H = 0.4000ZV = 0.2667I = 1.000

C = 2.750 (conservative)RH = 4.500RV = 3.000

Uo = 0.2444 gVo = 0.2444 gAdopt: Effective seismic horizontal acceleration = 0.2444g

Adopt: Effective seismic vertical acceleration = 0.2444g

Seismic spectral velocity = 1.250 ft/sec (at a convective period of 6.792 sec.)

II. REFERENCES


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REFERENCES 2.0 MG LESLIE LANE RESERVOIR

DATE: 9/ 22/ 2000 ESCONDIDO, CALIFORNIA


A. Heins, C.P., and Firmage, D.A., Design of Modern Steel Highway Bridges . JohnWiley and Sons, Inc., New York, New York (1979).

B. Park, R., and Gamble, W.L., Reinforced Concrete Slabs. John Wiley and Sons,Inc., New York, New York (1980).

C. Nuclear Reactors and Earthquakes , Chap. 6 and Appendix F. U.S. NuclearRegulatory Commission publication, Division of Technical Information, TID-7024,

National Technical Information Service (1963).D. Wang, Chu-Kia, and Salmon, Charles G., Reinforced Concrete Design, third edition.

Harper and Row, Publishers, Inc., New York, New York (1979).

E. Peck, Ralph B., Hanson, Walter E., and Thornburn, Thomas H., FoundationEngineering, second edition. John Wiley and Sons, Inc., New York, New York(1974).

F. Manual of Steel Construction, Eighth Edition, Chap. 5. American Institute ofSteel Construction, Inc., Chicago, Illinois (1980).

G. Post-Tensioning Manual, Fifth Edition, Chap. 3. Post-Tensioning Institute,Phoenix, Arizona (1990).

H. ACI Manual of Concrete Practice, Part II. American Concrete Institute, Detroit,Michigan (1983).

I. Jacobsen, Lydik S., "Impulsive Hydrodynamics of Fluid Inside a Cylindrical Tankand of Fluid Surrounding a Cylindrical Pier", Bulletin of the SeismologicalSociety of America, 39:189 (1949).

J. Handbook of Molded and Extruded Rubber, Second Edition. Goodyear Tire and RubberCompany, Akron, Ohio (1959).

K. AWWA Standard for Wire-Wound Circular Prestressed-Concrete Water Tanks, AmericanWater Works Association, Denver, Colorado (1995).


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NOMENCLATURE 2.0 MG LESLIE LANE RESERVOIR

DATE: 9/22/2000 ESCONDIDO, CALIFORNIA


III . TANK DESIGN NOMENCLATURE

a side of square area loaded by tire (ft).

a height of Whitney's stress block (in).

A b bearing area of vertical prestressing anchorage (in2).

A' b maximum area of the portion of the concrete anchorage surface for the verticalprestressing that is geometrically similar to and concentric with the area of theanchorage (in 2).

a bay plan area of one interior roof bay (ft2).

a bar area of reinforcing bar or prestressing tendon (in2).

alpha c ratio of the flexural stiffness of column to flexural stiffness of slab (1).

alpha ec ratio of flexural stiffness of equivalent column to flexural stiffness of slab(1).

alpha min minimum alpha c to satisfy 1994 UBC, Section (1913.6.10(a)) (1).

A c area of concrete in compression for minimum eccentricity (in2), area of core

concrete column (in 2).

A c,b area of concrete in compression for balanced condition (in 2). A c,max area of concrete in compression for maximum eccentricity (in

2).

A footing area of column footing (ft2).

A g gross cross-sectional area (in2).

Ag column gross cross-sectional area of column strip (in2).

Ag middle gross cross-sectional area of middle strip (in2).

A int area of tank roof supported entirely by interior columns (ft2).

A long plan area one exterior roof bay at the longest end span (ft2).

A roof area of tank roof (ft2).

A short plan area one exterior roof bay at the shortest end span (ft 2). As area of reinforcing (in 2).

As drag required area of reinforcing from ACI drag resistance formula (in2).

As max maximum amount of reinforcing allowed (in2).

As min minimum amount of reinforcing allowed (in2).

A shear web area of structural tubing resisting shear (in2).

As seismic required area of reinforcing for seismic loads (in2).

As temp required area of reinforcing for temperature effects (in2).

A strand area of circumferential prestressing strand (in2).

As water required area of reinforcing for water loads (in 2).


10/89




A t tributary area of roof (kips/lineal foot of wall).

A v required area of structural tubing for roof shear connection (in2).

b width of compression area of concrete section (in).

BD downward drag of backfill on tank wall (plf).

Bd ratio of maximum factored axial dead load to maximum total factored axial load(1).

beta 1 factor defined in 1994 UBC, Section (1910.2.7.3).

beta a ratio of distributed dead load to distributed live load on roof (1).

b f width of flange of structural tubing (in).

bw width of compression area for section in bending (in).

C stiffness constant (in 4).

Cc compressive force in concrete at minimum eccentricity (kips).

Cc,b compressive force in concrete for balanced condition (kips).

Cc,max compressive force in concrete at maximum eccentricity (kips).Cm column design constant (1).

Cs initial initial compressive stress in tank wall (psi).

Cs b,e compressive stress due to seismic backfill pressure

Cs b,s compressive stress in wall due to static backfill pressure (psi).

Cs p compressive stress in wall due to circumferential prestressing (psi).

d f depth to tensile reinforcing for flexure (in).

dv depth to tensile reinforcing for shear (in).

dbar diameter of reinforcing bar (in).

d column diameter of column (in).delta initial deflection of column (in).

delta comp compressive deflection of bearing pad (%).

delta b design moment magnification factor (1).

delta radial maximum inward wall movement (in).

delta s positive moment magnification factor (1).

DL dead load (units as noted).

d tank diameter of tank at mid height of wall (ft).

e eccentricity of column axial load (in).

E Modulus of Elasticity (psi).Ebar Modulus of Elasticity of prestressing tendon (psi).

e b eccentricity of the axial load on the column which corresponds to the point oftransition between the compression controls region and the tension controlsregion (in).

e initial radial strain in bearing pad due to prestressing (1).

e max maximum radial strain in bearing pad (1).

e min eccentricity of the axial load on the column which corresponds to the point oftransition between the maximum axial compressive strength controls region and thecompression controls region (in).


11/89




e temp radial strain in bearing pad due to temperature effects (1).

EL earthquake load (units as noted).

fa allowable tensile force in reinforcing (psi).

F friction factor between floor and subgrade (1).

f'c compressive stress of concrete (psi).

f cp permissible compressive stress in concrete at vertical prestressing anchorage(psi).

Fs force in reinforcing steel (kips).

Fv allowable shear stress in structural tubing

fy yield strength of floor reinforcing (psi).

Fy summation of forces in the "Y" direction (lb).

fy yield strength of reinforcing or structural steel (psi).

gammabackfill weight density of soil backfill material (pcf).

gammaconcrete weight density of concrete (pcf).

gammaequivalent backfill equivalent liquid density of soil backfill material (pcf).gammaf fraction of column moment transferred to roof slab through flexure (1).

gammafluid weight density of fluid (pcf).

gammav fraction of column moment transferred to roof slab through shear (1).

hbackfill height of backfill above wall footing (ft).

h column side dimension of equivalent square column (in).

Hc hoop force in wall from convective component of water during seismic excitation(k/ft).

ho elevation of center of impulsive water mass above top of wall footing (ft).

Hi hoop force in wall from impulsive component of water during seismic excitation(k/ft).

Hv hoop force in wall from vertical component of water during seismic excitation(k/ft).

hwall height of tank wall (ft).

hwater height of normal fluid level above tank floor at wall (ft).

I gross moment of inertia of gross cross section (in4).

J c section property analogous to the polar moment of inertia

k stability factor for frame member (1).

K column flexural stiffness of column (k-ft).

K ec flexural stiffness of equivalent column (k-ft).

K slab stiffness of slab section (k-ft).


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K t torsional stiffness of slab section (k-ft).

L2 longer span length from center to center of supports in direction transverse fromdirection in which moments are being determined (ft).

L2' shorter span length from center to center of supports in direction transversefrom direction in which moments are being determined (ft).

Lcolumn length of column from top of footing to bottom of drop panel (ft).

Ld basic tensile development length of reinforcing steel (in).

Ldb basic development length for reinforcing in compression (in).

Ldevelopment required development length of column reinforcing in footing (in).

Ldh modified tensile or compression development length for standard hooks (in).

Ldrop length of side of square drop panel (ft).

Le,max maximum exterior span length from center of column to center of wall (ft).

Le,min minimum exterior span length from center of column to center of wall (ft).

Lfooting length of side of square column footing (ft).

Lhb

basic development length for standard hook in tension (in).

Li,max maximum interior span length from center to center of column supports (ft).

LL live load (units as noted).

Ln length of longer clear span in direction in which moments are being determined(ft).

Ln' length of shorter clear span in direction in which moments are being determined(ft).

Ln,e maximum exterior clear span (ft).

Ln,i maximum interior clear span (ft).

Ls maximum dimension of floor slab (ft).

Lsplice

required length of splice for vertical column reinforcing (in).

M2b required bending capacity of column due to distributed loads on roof, from ACI318R-89, Section 13.6.3.3 (k-ft).

Mc total required bending capacity of column due to all loads (k-ft).

Md required bending capacity of column due to initial deflection (k-ft).

Mf moment transferred from column to roof slab through flexure (k-ft).

Mfooting design bending moment in wall footing (lb-ft/ft).

Mi moment in column due to inertia of the column under seismic loads (k-ft).

Mn nominal bending capacity of section (k-ft/ft).

Mn,b nominal bending capacity of the column at balanced condition (k-ft).

Mn,min nominal bending capacity of the column at minimum eccentricity (k-ft).Mo nominal bending capacity of the column at maximum eccentricity (k-ft).

Mo,c total factored static moment in roof slab in a cantilever span (k-ft).

Mo,e total factored static moment in roof slab in an exterior span (k-ft).


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Mo,i total factored static moment in roof slab in an interior span (k-ft).

Mu factored design moment (k-ft/ft).

MUS minimum ultimate stress (psi).

Mv moment transferred from column to roof slab through shear (k-ft).

Mw moment in column due to inertia of the water around the column under seismicloads (k-ft).

n number of seismic cables required per set (ea).

Nab neutral axis location for column at balanced condition (in).

Namax neutral axis location for column at maximum eccentricity (in).

Namin neutral axis location for column at minimum eccentricity (in).

overhang distance from the exterior face of the corewall to the edge of the roof slab(in).

Pc critical buckling load on column (k-ft).

Pdt prestressing force in tank wall for differential temperature and dryness effects(k/ft).

Pdynamic prestressing force in the tank wall for hydrodynamic hoop forces (k/ft).PL point load on roof slab (lbs).

Pn,b nominal axial capacity of column at balanced condition (kips).

Pol prestressing force for hydrostatic overload (k/ft).

Ppad,static static vertical load on bearing pad (plf).

Ppad,seismic seismic vertical load on bearing pad (plf).

Pservice total unfactored service load on column footing (kips).

Psl prestressing force for hydrostatic service load (k/ft).

Ps,static static soil load on wall footing (psf).

Ptotal total prestressing force required (kips).Pu factored axial demand on column (kips).

Pu,static static structure load on wall footing (psf).

Pv required vertical prestressing force (k/ft).

Pw,static static water load on wall footing (psf).

r radius of gyration of column (in).

rho reinforcing ratio (1).

rho b reinforcing ratio which produces a balanced strain condition (1).

rho max maximum reinforcing ratio (1).

rho min minimum reinforcing ratio (1).rho s column spiral reinforcing ratio (1).

rho water density of fluid in tank (pcf).

Ru coefficient of resistance (psi).

s length of prospective failure plane (ft).

S shape factor of bearing pad (1).

stress allow allowable compressive stress in bearing pad (psi).

stress pad compressive stress in bearing pad (psi).


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Ss strain in reinforcing steel (in/in).

Sy yield strain of reinforcing steel (in/in).

t drop thickness of drop panel slab (in).

t f thickness of flange of structural tubing (in).

t flash coat thickness of initial coating of shotcrete placed over corewall prior to wrapping(in).

t floor thickness of floor slab (in).

t footing thickness of footing (in).

t pad thickness of bearing pad (in).

t roof thickness of roof slab (in).

t roof pad thickness of bearing pad between wall and roof (in).

t wall thickness of the tank wall (in).

t wall,average average thickness of the tank wall (in).

t wall,base thickness of the cast-in-place corewall at the base (in).

t wall,mid thickness of the tank wall at midheight (in).

t wall,top thickness of the cast-in-place corewall at the top (in).

U total factored load on tank roof (units as noted).

uo horizontal acceleration of ground due to seismic excitation (ft/sec2).

v roof shear per lineal foot of tank wall (kips/ft).

V total roof shear (kips).

v1,2 maximum and minimum shear stress in roof slab due to column moment transfer(psi).

Vc nominal capacity of concrete section in shear neglecting reinforcing (kips).

Vce,circle nominal shear capacity of roof slab in exterior bay under circular twoway actionneglecting reinforcing (kips).

Vce,square nominal shear capacity of roof slab in exterior bay under square twoway actionneglecting reinforcing (kips).

Vce,straight nominal shear capacity of roof slab in exterior bay under oneway actionneglecting reinforcing (kips).

Vci,circle nominal shear capacity of roof slab in interior bay under circular twoway actionneglecting reinforcing (kips).

Vci,square nominal shear capacity of roof slab in interior bay under square twoway actionneglecting reinforcing (kips).

Vci,straight nominal shear capacity of roof slab in interior bay under oneway actionneglecting reinforcing (kips).

v connection roof shear per connection (kips).Vconvective base shear due to weight of fluid acting convectively on the tank wall (kips).


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v c,square nominal shear capacity of footing under square twoway action neglectingreinforcing (psi).

v c,straight nominal shear capacity of footing under oneway action neglecting reinforcing(psi).

Vfooting design shear in wall footing (lb).

Vimpulsive base shear due to weight of fluid acting impulsively on the tank wall (kips).Vn nominal shear capacity of section (kips/ft).

vo vertical acceleration of ground due to seismic excitation (ft/sec2).

Vs volume of spiral reinforcing (in3/in).

v set required capacity of one set of seismic cables (ft).

Vs,square nominal shear capacity of reinforcing under square twoway action (kips)

Vs,straight nominal shear capacity of reinforcing under oneway action (kips)

Vu factored design shear (kips/ft).

Vue,circle factored shear force on roof slab in exterior bay due to circular twoway action(kips).

Vue,square factored shear force on roof slab in exterior bay due to square twoway action(kips).

Vue,straight factored shear force on roof slab in exterior bay due to oneway action (kips).

Vui,circle factored shear force on roof slab in interior bay due to circular twoway action(kips).

Vue,square factored shear force on roof slab in interior bay due to square twoway action(kips).

Vui,straight factored shear force on roof slab in interior bay due to oneway action (kips).

vu,square factored shear stress on footing due to square twoway action (psi).

Vu,square factored shear force on footing due to square twoway action (kips).

vu,straight factored shear stress on footing due to oneway action (psi).Vu,straight factored shear force on footing due to oneway action (kips).

w circular frequency of oscillation of fluid surface (rad/sec).

wpad total width of bearing pads (in).

Wd factored distributed dead load on roof for column design (psf).

Wl factored distributed live load on roof for column design (psf).

Wcolumn effective weight of columns which is transferred to the roof diaphragm (kips).

Wdead total dead weight on tank roof (kips).

Wdrop total weight of drop panels (kips).

Wlive total live weight on tank roof (kips).Wroof total weight resisted by roof diaphragm (kips).

Wslab weight of roof slab excluding drop panels (kips).

Wwall total weight of tank wall (kips).

Wwater total weight of tank contents (kips).


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X smaller dimension of rectangular area (in).

Xc distance from column centerline to centroid of compression area at minimumeccentricity (in).

Xc,b distance from column centerline to centroid of compression area for balancedcondition (in).

Xc,max distance from column centerline to centroid of compression area at maximumeccentricity (in).

Xroof location of centroid of roof mass above top of wall footing (ft).

Xs sum of moment arms for reinforcing in compression at minimum eccentricity (in).

Xs,b sum of moment arms for reinforcing in compression for balanced condition (in).

Xs,max sum of moment arms for reinforcing in compression at maximum eccentricity (in).

Xwall location of centroid of wall mass above top of wall footing (ft).

Y larger dimension of rectangular area (in).

Ysoil relative location of center of soil pressure from toe of wall footing (1).

Ywall relative location of center of bearing pad from toe of wall footing (1).

Ywater relative location of center of water pressure from toe of wall footing (1).

Y" distance from the base of the wall to the point of investigation (ft).

/0 shear angle of bearing pad (1).

/0 strength reduction factor for concrete design (1).

/0h angular amplitude of free oscillation at the water surface

/0Mo(max) maximum nominal bending capacity of the column with no axial load and reduced bythe ACI capacity reduction factor (k-ft).

/0P n(max) maximum nominal axial capacity of the column with no bending moment and reducedby the ACI capacity reduction factor (kips).


17/89

FLAT SLAB 2.0 MG LESLIE LANE RESERVOIR

REVISED: 02/02/2001 ESCONDIDO, CALIFORNIA


IV. FLAT SLAB DESIGN

The flat slab roof and column designs satisfy as a minimum the requirements of the ACI 318-95and ACI 350-89. The Direct Design Method as outlined in the ACI 318-95, Chapter 19, isutilized for these designs.

A. Assumptions

Live Loads - Case [1]:

snow = 0.0000 psf

peagravel = 0.0000 psf

soil = 100.0 psf

people = 0.0000 psf

other misc. = 50.00 psfLive Load = 150.00 psf

Superimposed Dead Loads - Case [1]:

peagravel = 0.0000 psfsoil = 100.0 psf

other misc. = 0.0000 psfDead load = 100.0 psf

Roof slab thickness - t roof = 12.00 in

Drop panel thickness - t drop = 7.500 in

Drop panel size - L drop = 7.750 ft x 7.750 ft

Column diameters - d column = 24.00 in (Interior)

= 24.00 in (Exterior)

Equivalent square column sizes - h column = 21.27 in (Interior)= 21.27 in (Exterior)

Concrete weight = 150.0 pcf

All reinforcing bars are Grade 60.

Maximum span from center-to-center of supports:

interior bay - L i,max = 21.00 ft

exterior bay - L e,max = 15.70 ft (to centerline of wall)

Minimum span from center-to-center of supports:

exterior bay - L e,min = 10.16 ft (to centerline of wall)


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FLAT SLAB 2. 0 MG LESLI E LANE RESERVOI R

REVISED: 02/02/2001 ESCONDI DO, CALI FORNI A


B. Deflection Requirements

The minimum thickness of the two-way flat slab with drop panels will be investigated forinterior and exterior spans as computed from ACI 318-95, Section 9.5.3.2.

basic minimum thickness for exterior span:

t roof(minimum) = L n,e /33 ACI 318-95, Table 9.5(c)

where, L n,e = L e,max - (h column )/2

Ln,e = 15.70 ft - (21.27 in x 1 ft) = 14.81 ft 2 x 12 in

t roof(minimum) = 14.81 ft x 12 in/ft = 5.387 in 33

basic minimum thickness for interior span:

t roof(minimum) = L n,i /36 ACI 318-95, Table 9.5 (c)

where, L n,i = L i,max - h column

Ln,i = 21.00 ft - (21.27 in x 1 ft) = 19.23 ft 12 in

t roof(minimum) = 19.23 ft x 12 in/ft = 6.409 in 36

In order to reduce the reinforcing requirements and provide a more ductile slab, athicker slab will be provided.

ADOPT: t roof = 12.00 in


19/89


REVI SED: 02/ 02/ 2001 ESCONDI DO, CALI FORNI A


C. Factored Loads - Load Case [1]

The required capacity of the slab will be calculated in accordance with ACI 350-89,Sect.(2.6.5) and ACI 318-95, Section (9.2).

The shear load will be carried by the concrete only and does not require theEnvironmental Durability Factor per ACI 350.

U1 = 1.40 x DL + 1.70 x LL ACI 350R-89, Sect. (2.6.5)

U2 = 1.05 x DL + 1.28 x LL + 1.40 x EL ACI 350R-89, Sect. (2.6.5)

U3 = 0.90 x DL + 1.43 x EL ACI 350R-89, Sect. (2.6.5)

1. Dead load:

DL = t roof x 150 pcf + superimposed dead load

= (12.00 in/12 in/ft) x 150.0 pcf + 100.0 psf

= 250.0 psf

ADOPT: DL = 250.0 psf

DLdrop = additional dead load at drop panels

= t drop x 150.0 pcf

= (7.500 in/12 in/ft) x 150.0 pcf

= 93.75 psfADOPT: DL drop = 93.75 psf

2. Earthquake load:

EL = v o x (DL + LL soil )

EL = 0.2444 x (250.0 + 100.0) psf = 85.54 psfADOPT: EL = 85.54 psf

3. ACI Factored Load

U1shear = 1.40 x 250.0 psf + 1.70 x 150.0 psf = 605.0 psf

U2shear = 1.05 x 250.0 psf + 1.28 x 150.0 psf + 1.40 x 85.54 psf = 574.3 psf

U3shear = 0.90 x 250.0 psf + 1.43 x 85.54 psf = 347.3 psf

ADOPT: U = 605.0 psf (load case U1)

U1drop,shear = 1.40 x 93.75 psf = 131.3 psf

U2drop,shear = 1.05 x 93.75 psf + 1.40 x (0.2444 x 93.75 psf) = 130.5 psf

U3drop,shear = 0.90 x 93.75 psf + 1.43 x (0.2444 x 93.75 psf) = 117.1 psfADOPT: U drop = 131.3 psf (load case U1 drop )


20/89




4. Vehicle Load

The vehi cl e l oads wi l l be cal cul at ed based on t he l oad f act or s f r om ACI 318- 95. Foa movi ng vehi cl e, t he st at i c l oad wi l l al so be scal ed wi t h an i mpact f act or ( I ) .Fr om Ref er ence 1, t he i mpact f act ors wi l l be 1. 0 and 1. 3 f or st at i onar y and movi ngvehi cl es r espect i vel y.

Vehi cl e #1: 3. 200e+04 l bs

I = 1. 300 Ref. 1, EQ. (3.1)

VL1 u1 = 1. 7 x I x vehi cl e wei ght

= 1. 7 x 1. 300 x 3. 200e+04 l bs = 7. 072e+04 l bs

VL1 u2 = 1. 28 x vehi cl e wei ght + 1. 40 x ( v o x vehi cl e wei ght )

= 1. 28 x 3. 200e+04 l bs + 1. 40 x ( 0. 2444 x 3. 200e+04 l bs)

= 5. 191e+04 l bs

ADOPT: VL = VL1 = 7.072e+04 lbs

5. Wheel Load

Rear wheel l oads wi l l be di st r i but ed as per AASHTO desi gn vehi cl e l oads ( see Fi gur e3. 1, Ref er ence 1) .

VL1 rear = 0. 4 x VL1

= 0. 4 x 7. 072e+04 l bs

= 2. 829e+04 l bs


21/89




D. Shear Requirements - Load Case [1]

The shear st r engt h under wi de- beam act i on i nsi de t he dr op panel and under t wo- way act i oni nsi de and out si de the dr op panel wi l l be i nvest i gat ed.

MECHANISM: An i nt er i or bay wi t h a st r ai ght pot ent i al f ai l ur e sur f ace al ong t he f ul lwi dt h of t he bay and l ocat ed a di st ance d f r om t he edge of t he equi val entsquar e col umn

Vu i , s t r ai ght = U x L i , max x s 1 + U drop x L drop x s' 1 + V1

wher e, s 1 = ( L i , max ) / 2 - ( h col umn / 2 + d) ACI 318-95 Sect. (11.1.3.1)

s ' 1 = ( L drop ) / 2 - ( h col umn / 2 + d)

d = t r oof + t drop - cover - 1. 5 x d bar

= 12. 00 i n + 7. 500 i n - 2. 000 i n - 1. 5 x 0. 8750 i n

= 16. 19 i n

s 1 = 21. 00 f t / 2 - ( 21. 27 i n/ 2 + 16. 19 i n) / 12 = 8. 265 f t

s' 1 = 7. 750 f t / 2 - ( 21. 27 i n/ 2 + 16. 19 i n) / 12 = 1. 640 f tVu i , s t r ai ght = ( 605. 0psf x 21. 00f t x 8. 265f t + 131. 1psf x 7. 750f t x 1. 640f t + 7. 072e+04

l bs) / ( 1000l bs/ k)

= 177. 4 k

Vc i , s t r ai ght = 2 x %( f ' c) x b w x ( d - t drop ) ACI 318-95, EQ. (11-3)

wher e, b w = L i , max = 21. 00 f t

Vc i , s t r ai ght = ( 2 x %( 4000psi ) x 21. 00f t x ( 16. 19i n - 7. 500i n) x 12i n/ f t ) / ( 1000l b/ k)

= 276. 9 k

/OVc = 0. 8500 x 276. 9 k = 235. 4 k > Vu i , s t r ai ght therefore, O.K.

MECHANISM: An ext er i or bay wi t h a st r ai ght pot ent i al f ai l ur e sur f ace al ong t he f ul lwi dt h of t he bay and l ocat ed a di st ance d f r om t he edge of t he equi val entsquar e col umn

The shear f or ce i n an ext er i or bay wi l l be scal ed by an addi t i onal f act or f or endef f ect s

Vu e, s t r a ight = 1. 15 x [ U x L i , max x s 2 + U drop x L drop x s' 1] + V1

wher e, s 2 = L e, max / 2 - ( h col umn / 2 + d) ACI 318-95, Sect. (11.1.3.1)

d = 16. 19 i n

s 2 = 15. 70 f t / 2 - ( 21. 27 i n/ 2 + 16. 19 i n) / 12 = 5. 615 f t

Vu e, s t r a ight = ( 1. 15 x [ 605. 0 psf x 21. 00 f t x 5. 615 f t + 131. 3 psf x 7. 750 f t x1. 640 f t ] + 7. 072e+04 l bs) / ( 1000 l bs/ k)

= 154. 7 k

Vc e, s t r a ight = 2 x %( f ' c) x b w x ( d - t drop ) ACI 318-89, EQ. (11-3)

b w = L i , max = 21. 00 f t

Vc e, s t r a ight = ( 2 x %( 4000psi ) x 21. 00f t x ( 16. 19i n - 7. 500i n) x 12i n/ f t ) / ( 1000l b/ k)

= 276. 9 k

/OVc = 0. 8500 x 276. 9 k = 235. 4 k > Vu e , s t r ai ght therefore, O.K.


22/89




MECHANISM: An i nt er i or bay wi t h a ci r cul ar pot ent i al f ai l ur e sur f ace l ocat ed a di st anced/ 2 f r om t he f ace of t he col umn:

Vu i , c i r c l e = U x [ L i , max 2 - x ( s 3/ 2) 2] + U drop x [ L drop 2 - B x ( s 3/ 2) 2] + V1 ACI 318-95, Sect. (11.12)

wher e, s 3 = d col umn + d

= ( 24. 00 i n + 16. 19 i n) / ( 12 i n/ f t ) = 3. 349 f t

Vu i , c i r c l e = ( 605. 0 psf x [( 21. 00 f t ) 2 - B x (3. 349 f t / 2) 2] + 131. 1 psf x [ ( 7. 750 f t ) - B x (3. 349 f t / 2) 2] + 7. 072e+04l bs) / ( 1000 l bs/ k)

= 338. 9 k

Vc i , c i r c l e = 4 x %( f ' c) x b o x d ACI 318-95, EQ. (11-38)

wher e, b o = B x s 3 = x 3. 349 f t = 10. 52 f t

Vc i , c i r c l e = 4 x %( 4000 psi ) x 10. 52 f t x 16. 19 i n x 12 i n/ f t / ( 1000 l bs/ k)

= 517. 0 k

/OVc = 0. 8500 x 517. 0 k = 439. 5 k > Vu i , c i r c l e therefore, O.K.

MECHANISM: An ext er i or bay wi t h a ci r cul ar pot ent i al f ai l ur e sur f ace l ocat ed a di st anced/ 2 f r om t he f ace of an edge col umn:

Vu e, c i r c l e = 1. 15 x [ U x [ L i , max x ( L i , max + L e, max ) / 2 - B x ( s 3/ 2) 2] + U drop x[ Ldrop 2 - B x ( s 3/ 2) 2] ] + V1 ACI 318-95, Sect. (11.12)

Vu e, c i r c l e = ( 1. 15 x [ 605. 0 psf x [ 21. 00 f t x(21. 00 f t + 16. 19 f t ) / 2 - B x(3. 349 f t / 2) + 131. 3 psf x [ ( 7. 750 f t ) 2 - B x (3. 349 f t / 2) 2] ] + 7. 072e+04 l bs) / ( 1000

l bs / k)

= 340. 4 k

Vc e, c i r c l e = 4 x %( f ' c) x b o x d ACI 318-95, EQ. (11-38)wher e, b o = B x s 4 = B x 3. 349 f t = 10. 52 f t

Vc e, c i r c l e = 4 x %( 4000 psi ) x 10. 52 f t x 16. 19 i n x 12 i n/ f t / ( 1000 l bs/ k)

= 517. 0 k

/OVc = 0. 8500 x 517. 0 k = 439. 5 k > Vu e, c i r cl e therefore, O.K.


23/89




MECHANISM: An i nt er i or bay wi t h a squar e pot ent i al f ai l ur e sur f ace l ocat ed a di st ance( d - t drop ) / 2 f r om t he edge of t he dr op panel

Vu i , square = U x ( L i , max 2 - ( L drop + d - t drop ) 2) + V1 ACI 318-95, Sect. (11.12)

( 605. 0 psf x ( ( 21. 00 f t ) 2 - ( 7. 750 f t + ( 16. 19 i n - 7. 500 i n) / ( 12 i n/ f t ) ) 2] + 7. 072e+04 l bs) x 0. 001 k/ l b

= 294. 1 k

Vc i , square = 4 x %( f ' c) x b o x ( d - t drop ) ACI 318-95, EQ. (11-38)

wher e, b o = 4 x ( L drop + ( d - t drop ) / 12)

= 4 x ( 7. 750 f t + ( 16. 19 i n - 7. 500 i n) / ( 12 i n/ f t ) )

= 33. 90 f t

Vc i , square = 4 x %( 4000 psi ) x 33. 90 f t x ( 16. 19 i n - 7. 500 i n) x 12 i n/ f t / ( 1000 l bs/ k)

= 893. 9 k

/OVc = 0. 8500 x 893. 9 k = 759. 9 k > Vu i , square therefore, O.K.

MECHANISM: An exter i or bay wi t h a squar e pot ent i al f ai l ur e sur f ace l ocat ed a di st ance( d - t drop ) / 2 f r om t he edge of t he dr op panel

Vu e, square = 1. 15 x U x [ L i , max x ( L e, max + L i , max ) / 2 - ( L drop + d - t drop ) 2) + V1 ACI 318-95, Sect. (11.12)

= ( ( 1. 15 x 605. 0 psf x [ 21. 00 f t x ( 16. 19 f t + 21. 00 f t ) / 2 - ( 7. 750 f t +( 16. 19 i n - 7. 500 i n) / ( 12 i n/ f t ) ) 2] ) + 7. 072e+04 l bs) / ( 1000 l bs/ k)

= 288. 9 k

Vc e, square = 4 x %( f ' c) x b o x ( d - t drop ) ACI 318-95, EQ. (11-38)

wher e, b o = 4 x ( L drop + ( d - t drop ) / 12)

= 4 x ( 7. 750 f t + ( 16. 19 i n - 7. 500 i n) / ( 12 i n/ f t ) )

= 33. 90 f t

Vc e, square = 4 x %( 4000 psi ) x 33. 90 f t x ( 16. 19 i n - 7. 500 i n) x 12 i n/ f t / ( 1000 l bs/ k)

= 893. 9 k

/OVc e, square = 0. 8500 x 893. 9 k = 759. 9 k > Vu e, square therefore O.K.


24/89




E. Moments in Roof Slab - Load Case [1]

The moments in the roof slab will be due to distributed loads and point loads. Thedesign for bending due to distributed loads will be based on the ACI 318-95 Direct DesignMethod and ACI 350-89, Section 2.6.5.a. The moments caused by point loads (if required)will be calculated separately and superimposed upon the distributed load moments.

1. Factored load moments due to distributed loads:

U1moment = 1.30 x U1 shear = 1.30 x 605.0 psf = 786.5 psf



ADOPT: U = 786.5 psf (load case 1)

U1drop,moment = 1.30 x U1 drop,shear = 1.30 x 131.3 psf = 170.6 psf



ADOPT: U drop = 170.6 psf (load case 1)

Mo,i = total factored static moment in an interior span per ACI 350-89,Sect.2.6.5.a.

Mo,i = (U x L 2 x L n,i2)/8 + U drop x L drop x (L drop - h column )

2/8ACI 318-95, EQ. (13-3)

where, L 2 = L i,max = 21.00 ft CRSI Handbook-82, Sect. 10

Mo,i = 786.5 psf x 21.00 ft x (19.23 ft)2/(8 x 1000 lb/k) + 170.6 psf x

7.750 ft x (7.750 ft - 21.27 in/12 in/ft) 2/(8 x 1000 lb/k)

= 769.2 k-ft

Mo,e = total factored static moment in an exterior spanMo,e = (U x L 2 x L n,e

2)/8 + U drop x L drop x (L drop - h column )2/8

ACI 318-95, EQ. (13-3)where, L 2 = L i,max = 21.00 ft CRSI Handbook-82, Sect. 10

Mo,e = (786.5 psf x 21.00 ft x (14.81 ft)2/(8 x 1000 lb/k) + 170.6 psf

x 7.750 ft x(7.750 ft - 21.27 in/12 in/ft) 2/(8 x 1000 lb/k)

= 459.0 k-ft


25/89




2. Service load moments due to distributed loads:

Mo,i = total service load static moment in an interior span

Mo,i = ((DL + LL) x L 2 x L n,i2)/8 + DL drop x L drop x (L drop - h column )

2/8ACI 318-95, EQ. (13-3)

where, L 2 = L i,max = 21.00 ft CRSI Handbook-82, Sect. 10Mo,i = (250.0 psf + 150.0 psf) x 21.00 ft x (19.23 ft)

2/(8 x 1000 lb/k) + 93.75 psf x 7.750 ft x (7.750 ft - 21.27 in/12 in/ft) 2/(8 x 1000 lb/k)

= 391.4 k-ft

Mo,e = total service load static moment in an exterior span

Mo,e = ((DL + LL) x L 2 x L n,e2)/8 + DL drop x L drop x (L drop - h column )

2/8ACI 318-95, EQ. (13-3)

where, L 2 = L i,max = 21.00 ft CRSI Handbook-82, Sect. 10

Mo,e = (250.0 psf + 150.0 psf) x 21.00 ft x (14.81 ft)2/(8 x 1000 lb/k)

+ 93.75 psf x 7.750 ft x (7.750 ft - 21.27 in/12 in/ft) 2/(8 x 1000 lb/k)

= 233.5 k-ft

Longitudinal Distribution of Total Factored Static Moment:

The total factored static moment in an INTERIOR span shall be distributedlongitudinally based on ACI 318-95, Sect. (13.6.3.2).

Based on the ACI 318-95, Sect. (13.6.3.3), the total factored static moment inan EXTERIOR span shall be distributed longitudinally based on the supportcondition at the exterior edge of the slab. Due to the fact that the length ofthe end span varies depending on which column line is analyzed, the supportcondition will fall under one of two categories. For moderate to long end spans,the support condition for the exterior span is most accurately represented byColumn (1) of Sect. 13.6.3.3. For short end spans, the support condition for thefirst interior span is most accurately represented by Column (3) of Sect.13.6.3.3. Both cases will be completely investigated in the calculations whichfollow.

Transverse Distribution of Factored Moment:

The transverse distribution of the negative moment in interior spans, negativemoment in exterior spans, and positive moment shall be as prescribed by the ACI318-95, Sect. (13.6.4).


26/89

FLAT SLAB 2. 0 MG LESLI E LANE



3. Moments due to vehicle load:

The moment s due t o t he wheel l oad on t he sl ab wi l l be cal cul at ed based on t hedi scussi on i n Ref er ence 2, Sect . 3. 5. 2, concer ni ng concent r at ed l oads on sl abs.Ref er ence 2 appl i es speci f i cal l y t o one l oad appl i ed t o t he sl ab. I t i s assumed t hatt he one wheel l oad i s appl i ed at t he poi nt whi ch causes maxi mum moment i n t he sl ab

f or posi t i ve or negat i ve moment s. These poi nt s ar e 0. 50L and 0. 25L r espect i vel y.Based on t hi s assumpt i on, t he second wheel shoul d be at l east 7 f t f r om t he f i r st .

The f act t hat t he second wheel wi l l be l ocat ed no l ess t han 7 f t f r omt he poi nt whi chcauses maxi mumposi t i ve moment wi l l ef f ect i vel y r educe t he posi t i ve moment caused byt he second wheel t o aproxi mat el y one t hi r d of t hat caused by t he wheel whi ch i sl ocat ed at 0. 50L.

The negat i ve moment wi l l not be ef f ect ed near l y as much by t he l ocat i on of t he wheell oad. Theref or e, t he conser vat i ve way t o anal i ze f or t he negat i ve moment caused byt wo wheel l oads wi l l be t o doubl e t he moment caused by one wheel l oad.

Mv, c , i , neg = negat i ve moment caused by vehi cl e l oad on col umn st r i p i n an i nt eri or span

Mv,m, i , neg = negat i ve moment caused by vehi cl e l oad on mi ddl e st r i p i n an i nt eri or span

Mv, i , p os = posi t i ve moment caused by vehi cl e l oad on any st r i p i n an i nt er i or span

Mv,c , e ,neg = negat i ve moment caused by vehi cl e l oad on col umn st r i p i n an ext er i or span

Mv, m, e, neg = negat i ve moment caused by vehi cl e l oad on col umn st r i p i n an ext er i or span

Mv,e, pos = posi t i ve moment caused by vehi cl e l oad on col umn st r i p i n an ext er i or span

a. Vehicle #1

a1 = %( VL1 r ear / ( t i re pressure) )a1 = %( 2. 829e+04 l bs/ 40. 00 psi ) x 1 f t / 12 i n = 2. 216 f tFor an I nt er i or span:

a1/ L i , max = 2. 216 f t / 21. 00 f t = 0. 1055

assumi ng al pha 1 = bet a t = 0,M1 v, c , i , n eg / VL1 rear = 0. 360 Ref. 2, Fig. (3.44)

M1 v, c , i , n eg = 2 x ( M1 v, c , i , neg / VL1 r ear x VL1 rear ) x L 2/ 2

= 2 x ( 0. 360 x 2. 829e+04 l b) x 21. 00 f t / 2 x 0. 001 k/ l b

= 213. 9 k- f t

assumi ng al pha 1 = bet a t = i nf i ni t y,

M1 v,m, i , neg / VL1 rear = 0. 231 Ref. 2, Fig. (3.44)

M1 v,m, i , neg = 2 x ( M1 v,m, i , neg / VL1 r ear x VL1 rear ) x L 2/ 2

= 2 x ( 0. 231 x 2. 829e+04 l b) x 21. 00 f t / 2 x 0. 001 k/ l b= 137. 2 k- f t


27/89




assumi ng al pha 1 = bet a t = 0,

M1 v, i , pos / VL1 rear = 0. 210 Ref. 2, Fig. (3.43)

M1 v, i , pos = 1. 333 x ( M1 v, i , pos / VL1 r ear x VL1 rear ) x L 2/ 2

= 1. 333 x ( 0. 210 x 2. 829e+04 l b) x 21. 00 f t / 2 x 0. 001 k/ l b

= 83. 15 k- f t

For an Ext er i or span:

a1/ L e, max = 2. 216 f t / 15. 70 = 0. 1412


M1 v,c , e ,neg / VL1 rear = 0. 350 Ref. 2, Fig. (3.44)

M1 v,c , e ,neg = 2 x ( M1 v,c , e, neg / VL1 r ear x VL1 rear ) x L 2/ 2

= 2 x ( 0. 350 x 2. 829e+04 l b) x 21. 00 f t / 2 x 0. 001 k/ l b

= 207. 9 k- f t

assumi ng al pha 1 = bet a t = i nf i ni t y,

M1 v, m, e, neg / VL1 rear = 0. 230 Ref. 2, Fig. (3.44)

M1 v, m, e, neg = 2 x ( M1 v, m, e, neg / VL1 r ear x VL1 rear ) x L 2/ 2

= 2 x ( 0. 230 x 2. 829e+04 l b) x 21. 00 f t / 2 x 0. 001 k/ l b

= 136. 6 k- f t


M1 v,e, pos / VL1 rear = 0. 185 Ref. 2, Fig. (3.43)

M1 v,e, pos = 1. 333 x ( M1 v,e, pos / VL1 rear x VL1 rear ) x L 2/ 2

= 1. 333 x ( 0. 185 x 2. 829e+04 l b) x 21. 00 f t / 2 x 0. 001 k/ l b

= 73. 25 k- f t

b. TABLE 1. Summary of Moments due to vehicle load:

Ser vi ce Load Locat i on Fact or ed Load Moment ( k- f t ) Moment

Mv,c , e ,neg M1 v,c , e ,neg x 1. 3 = 266. 4 M1 v,c , e, neg / 1. 7 = 122. 3

Mv, m, e, neg M1 v, m. e, neg x 1. 3 = 173. 8 M1 v, m. e, neg / 1. 7 = 80. 35

Mv,e, pos M1 v,e, pos x 1. 3 = 87. 50 M1 v,e, pos / 1. 7 = 43. 09

Mv, c , i , n eg M1 v, c , i , n eg x 1. 3 = 278. 1 M1 v, c , i , neg / 1. 7 = 125. 8

Mv,m, i , neg M1 v,m, i , neg x 1. 3 = 178. 1 M1 v,m, i , neg / 1. 7 = 80. 71

Mv, i , pos M1 v, i , pos x 1. 3 = 108. 1 M1 v, i , pos / 1. 7 = 48. 92


28/89




F. Moment Summary

TABLE 2. Summary of Moments in Slab (worst case) Fact or ed Load Servi ce Load

Locat i on Equat i on f or Moment Moment MomentExt er i or Col umn 0. 75 x 0. 75 x M o ,e , exte r i o r + M v,c , e ,neg = 528. 5 f t - k 253. 6 f t - kSt r i p - Top or 0. 26 x 1. 00 x M o, i , ext e r i or + M v,c , e ,neg = 470. 3 f t - k 224. 0 f t - k

Ext er i or Col umn 0. 63 x 0. 60 x M o ,e , exte r i o r + M v,e, pos = 268. 7 f t - k 131. 3 f t - k St r i p - Bot t om

Ext er i or Mi ddl e 0. 75 x 0. 25 x M o ,e , exte r i o r + M v, m, e, neg = 263. 7 f t - k 124. 2 f t - k St r i p - Top or 0. 26 x 0. 00 x M o, i , ext e r i or + M v, m, e, neg = 177. 6 f t - k 80. 37 f t - k

Ext er i or Mi ddl e 0. 63 x 0. 40 x M o ,e , exte r i o r + M v,e, pos = 210. 9 f t - k 101. 9 f t - kSt r i p - Bot t om

Fi r st I nt er i or 0. 70 x 0. 75 x M o, i , i nt er i or + M v, c , i , n eg = 681. 8 f t - k 331. 3 f t - k Col umn St r i p - Top 0. 65 x 0. 75 x M o, i , i nt er i or + M v, c , i , n eg = 653. 0 f t - k 316. 6 f t - k

Fi r st I nt er i or 0. 52 x 0. 60 x M o, i , i nt er i or + M v, i , pos = 348. 1 f t - k 171. 0 f t - k Col umn St r i p - Bot t om

Fi r st I nt er i or 0. 70 x 0. 25 x M o, i , i nt er i or + M v,m, i , neg = 313. 0 f t - k 149. 2 f t - k Mi ddl e St r i p - Top or 0. 65 x 0. 25 x M o, i , i nt er i or + M v,m, i , neg = 303. 4 f t - k 144. 3 f t - k

Fi r st I nt er i or 0. 52 x 0. 40 x M o, i , i nt er i or + M v, i , pos = 268. 1 f t - k 130. 3 f t - k Mi ddl e St r i p - Bot t om

I nt er i or Col umn 0. 65 x 0. 75 x M o, i , i nt er i or + M v, c , i , n eg = 653. 0 f t - k 316. 6 f t - k St r i p - Top

I nt er i or Col umn 0. 35 x 0. 60 x M o, i , i nt er i or + M v, i , pos = 269. 6 f t - k 131. 1 f t - k

St r i p - Bot t omI nt er i or Mi ddl e 0. 65 x 0. 25 x M o, i , i nt er i or + M v,m, i , neg = 303. 4 f t - k 144. 3 f t - k

St r i p - Top

I nt er i or Mi ddl e 0. 35 x 0. 40 x M o, i , i nt er i or + M v, i , pos = 215. 8 f t - k 103. 7 f t - k St r i p - Bot t om


29/89




G. Flexural Design

A. Method for determining required reinforcing:

The required area of tension reinforcing within each strip was determined by thefollowing procedure:

Mu # /0Mn

/0Mn = / 0 x As x fy x (d - a/2)

where, As = rho x b x d

a = As x fy = rho x d x fy 0.85 x f'c x b 0.85 x f'c

/0Mu = / 0 x rho x b x d x fy x (d - rho x d x fy ) Ref. 4, EQ. (3.6.3) 2 x 0.85 x f'c

by defining the following:

m = fy Ref. 4, EQ. (3.6.4a) 0.85 x f'c

and substituting Ref. 4, EQ. (3.6.3) and (3.6.4a) into the above,

Ru = rho x fy x (1 - 0.5 x rho x m)

rho = 1(1 - %(1 - 2 x m x R u )) Ref. 4, EQ. (3.6.5) m fy

As required = rho x b x d

B. Maximum reinforcement ratio:

rho max = 0.75 x rho b ACI 318-95, Sect. (10.3.3)

where, rho b = the reinforcing ratio producing a balanced strain condition

rho b = 0.85 x f'c x beta 1 x 87000 Ref. 4, EQ. (3.5.4) fy 87000 + fy

rho max = 0.75 x 0.85 x 4000 psi x 0.8500 x 87000 = 0.0213860000 psi 87000 + 60000

C. Minimum reinforcement ratio:

rho min = 0.0018 ACI 318-95, Sect. (7.12.2.1)

As min = rho min x Ag

for column strips reinforcing at drop panels, Ag = b x t roof + L drop x t drop

Ag = 126.0 in x 12.00 in + 7.750 ft x 12 in/ft x 7.500 in = 2210. sqin As min = 0.0018 x 2210. sqin = 3.977 sqin

for all other strips, Ag = b x t roof

Ag = 126.0 in x 12.00 in = 1512. sqin

As min = 0.0018 x 1512. sqin = 2.722 sqin

D. Calculation of Z values:

At service loads, we assume a linear stress strain relationship. In addition, fora linear stress strain relationship, the location of the neutral axis coincides with


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the location of the center of gravity of the section. Therefore, solving for thedistance to the neutral axis, x,

(A c)(x) = (A t )(d - x) + (A t ')(d' - x)

where: A c = 1/2 (b)(x)

b = width of section

x = distance to neutral axis A t = equivalent concrete area of tension steel

= (n)(A s )

A t '= equivalent concrete area of additional tension steel (theadditional steel is typically in compression, however hereit is in tension)

= (n)(A s ')

n = modular ratio = E s /E c = 8.044

manipulating the equation

(1/2)(b)(x) 2 + n(A s + A s ')(x) - n[(A s )(d) + (A s ')(d')] = 0

thereforex = -n(A s + A s ') + %[(n(A s + A s '))

2 + (2)(b)(n)[(A s )(d) + (A s ')(d')]] b

The service load moments is:

Mservice = (j)(A s )(f s ) + (j')(A s ')(f s ')

where, j = moment arm for tension steel = d - x/3j'= moment arm for additional tension steel = d' - x/3

from strain compatibility,

f s f s '

=d - x d' - x

therefore,

f s ' = f s (d' - x)/(d - x)

substituting and solving for f s ,

f s = M service /[(d - x/3)(A s ) + (d' - x/3)[(d' - x)/(d - x)](A s ')]

Z = f s 3%[(2)(d c) 2(spacing)] ACI 318-95, EQ. (10-5)

Z # 115 k/in ACI 350-89, Sect. (2.6.6)


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COLUMN DESI GN 2. 0 MG LESLI E LANE RESERVOI R



E. Reinforcing

TABLE 2. Reinforcement Parameters and Provisions

Location b d As bar # of As (n) (in) rho required size bars provided spacing

Exterior Column 93.00 16.19 0.005043 7.593 7 16 9.600 7.875Strip - Top

Exterior Column 126.0 8.688 0.006672 7.304 7 16 9.600 7.875Strip - Bottom

Exterior Middle 126.0 8.875 0.006247 6.988 6 16 7.040 7.875Strip - Top

Exterior Middle 126.0 8.875 0.004937 5.521 6 16 7.040 7.875Strip - Bottom

First Interior 93.00 16.19 0.006602 9.939 7 20 12.00 6.300Column Strip -Top

First Interior 126.0 8.688 0.008820 9.655 7 20 12.00 6.300

Column Strip -Bottom

First Interior 126.0 8.875 0.007505 8.393 6 20 8.800 6.300Middle Strip -Top

First Interior 126.0 8.875 0.006359 7.111 6 20 8.800 6.300Middle Strip -Bottom

Interior Column 93.00 16.19 0.006305 9.492 7 20 12.00 6.300Strip - Top

Interior Column 126.0 8.688 0.006696 7.330 7 20 12.00 6.300Strip - Bottom

Interior Middle 126.0 8.875 0.007258 8.116 6 20 8.800 6.300Strip - Top

Interior Middle 126.0 8.875 0.005057 5.655 6 20 8.800 6.300Strip - Bottom


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F. Check for flexural crack control

TABLE 3. Reinforcement Parameters and Provisions

Location b d d c* d' A s A s ' x fs Z (in) (in) (in) (in) spacing provided provided (in) (ksi) (k/in)Exterior Column 93.00 16.19 2.438 10.81 7.875 9.600 12.00 5.399 15.83 71.85Strip - Top

Exterior Column 126.0 8.688 2.438 3.313 7.875 9.600 6.000 2.770 20.79 94.38Strip - Bottom

Exterior Middle 126.0 8.875 2.375 3.125 7.875 7.040 8.800 2.526 25.50 113.8Strip - Top

Exterior Middle 126.0 8.875 2.375 3.125 7.875 7.040 4.400 2.474 21.21 94.62Strip - Bottom

First Interior 93.00 16.19 2.438 10.81 6.300 12.00 12.00 5.693 17.77 74.91Column Strip -Top

First Interior 126.0 8.688 2.438 3.313 6. 300 12. 00 6. 000 2.994 22.06 92.96Column StripBottom

First Interior 126.0 8.875 2.375 3.125 6.300 8.800 8.800 2.716 25.07 103.8Middle Strip -Top

First Interior 126.0 8.875 2.375 3.125 6.300 8.800 4.400 2.684 22.05 91.34Middle Strip -Bottom

Interior Column 93.00 16.19 2.438 10.81 6.300 12.00 12.00 5.693 16.99 71.59Strip - Top

Interior Column 126.0 8.688 2.438 3.313 6.300 12.00 6.000 2.994 16.91 71.26Strip - Bottom

Interior Middle 126.0 8.875 2.375 3.125 6.300 8.800 8.800 2.716 24.25 100.4Strip - Top

Interior Middle 126.0 8.875 2.375 3.125 6.300 8.800 4.400 2.684 17.55 72.68Strip - Bottom


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H. Flexural Capacity

The capacity of each design strip can be calculated based on the formula below:

Capacity = / 0 x T x (d - a/2)

where, T = tensile force in reinforcing = As provided x fy

d = depth to reinforcing from Table 2

a = height of compressive stress block in concrete

= As provided x fy0.85 x f'c x b

Capacity = / 0 x As provided x fy x (d - ( As provided x fy))

1.7 x f'c x b

TABLE 3. Flexural Capacity of Roof Design Strips in Bending

Location Capacity (ft-k)Exterior Column Strip - Top 660.0

Exterior Column Strip - Bottom 364.3

Exterior Middle Strip - Top 265.5

Exterior Middle Strip - Bottom 265.5

First Interior Column Strip - Top 812.6

First Interior Column Strip - Bottom 423.8

First Interior Middle Strip - Top 327.1

First Interior Middle Strip - Bottom 327.1

Interior Column Strip - Top 812.6Interior Column Strip - Bottom 423.8

Interior Middle Strip - Top 327.1

Interior Middle Strip - Bottom 327.1

V. COLUMN DESIGN


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A. Stiffness Calculations

1. Slab Stiffness

E = 57,000 x %(4000) = 3.605e+06 psi

I gross = 1/12 x L i,max x t roof3

= 1/12 x 21.00 ft x 12 in/ft x (12.00 in)3

= 3.629e+04 in4

K slab = 4 x E x I gross /L i,max = 4 x 3.605e+06 psi x 3.629e+04 in 4 = 1.730e+05 ft-k 21.00 ft x (12 in/ft) 2 x 1000 lb/k

2. Torsional Stiffness of Roof Slab at Columns

C = (1 - 0.63 x X) x X 3 x Y ACI 318-95, Sect.(13.0) Y 3

where, for roof slab: X = t slab = 12.00 in

Y = h column = 21.27 in

for drop panel: X = minimum(C1,C2)

Y = maximum(C1,C2)

where,

C1 = t slab + t drop = 12.00 in + 7.500 in = 19.50 in

C2 = h column = 21.27 in

Therefore,

X = 19.50 in

Y = 21.27 in

Cslab = (1 - 0.63 x 12.00 in) x (12.00 in)3 x 21.27 in = 7897. in 4

21.27 in 3

Cdrop = (1 - 0.63 x 19.50 in) x (19.50 in)3 x 21.27 in = 2.221e+04 in 4

21.27 in 3

K t = torsional stiffness of slab section

= [L i,max x ((1 - Y/L i,max )5 + (1 - L drop /L i,max )

5 x (1/C slab - 1/C drop ))]-1

20 E x C drop E

= [21.00 x ( (1 - 21.27 in x 1 ft/12 in) 5 + (1 - 7.750 ft) 5 x ( 1 - 1 ))] 20 21.00 ft 21.00 ft 7897. 2.221e+04

3.605e+06 psi x 2.221e+04 in 4 3.605e+06 psi

= 9.245e+10 in 2-lb/ft x 1 ft 2/144 in 2 x 0.001 k/lb = 6.420e+05 ft-k


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3. Column Stiffness

E = 57000 x %(fc) = 57000 x %(4000 psi) = 3.605e+06 psiLcolumn = distance from top of the floor to top of wall + t roof pad - t drop -t column footing

= 31.00 ft + (0.5000 in - 7.500 in - 15.00 in)/12 in/ft = 29.17 ft

I gross = B/4 x (d column /2)4

= B/4 x (24.00/2) 4 = 1.629e+04 in 4

K column = 4 x E x I gross /L column = 4 x 3.605e+06 psi x 1.629e+04 in 4

29.17 ft x (12 in/ft) 2 x 1000 lb/k = 5.592e+04 ft-k

K ec = (K column-1 + K t

-1 ) -1

= ((5.592e+04 ft-k) -1 + (6.420e+05 ft-k) -1 ) -1 = 5.144e+04 ft-kalpha ec = K ec /K slab = (5.144e+04)/(1.730e+05) = 0.2972

B. Column Loads

Length designations are as follows:Ln = longer clear span in direction moments are being determinedL2 = transverse spanLn'= shorter clear span in direction moments are being determinedL2'= shorter transverse span

r = radius of gyration = 0.25 x d column The column will be assumed to be braced against sidesway. Under seismic loading, theroof-wall shear connection allows 0.1250 in of movement in any direction which will betreated as an initial displacement of the column-roof joint. All moments caused byseismic loading would logically be multiplied by delta s and all static moments would bemultiplied by delta b (ACI 318-95, EQ. (10-9)). However, due to the small magnitude ofthe lateral deflection (0.1250 in) and for simplicity, all calculated moments will beconservatively magnified by delta b.k = 1.00 (assuming a braced frame)for interior columns,

A bay = L i,max2

for exterior columns, A bay = L i,max x (L e + L i,max )/2

Slenderness = k x L column /rSince Slenderness > 22, moment magnification will be calculated for each case.Since Slenderness < 100, approximate methods may be used to evaluate slenderness

effects (ACI 318-95, Sect. (10.11.5)).column weight = B x (d column /2) 2 x L column x 150.0 pcfcolumn weight + drop panel weight = column weight + [(L drop )

2 x t drop x 150.00 pcf]


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TABLE 4. Column Design Variables Consistent for All Load Cases

exterior exteriorcolumn at column at

interior longest shortest

column end span end spanLn (ft) 19.23 19.23 19.23L2 (ft) 21.00 21.00 21.00Ln' (ft) 19.23 11.92 6.773L2' (ft) 21.00 21.00 21.00d column (in) 24.00 24.00 24.00r gyration (in) 6.000 6.000 6.000k 1.000 1.000 1.000Li,max (ft) 21.00 21.00 21.00Le (ft) 0 12.81 7.659

A bay (ft2

) 441.0 355.0 300.9Lcolumn (ft) 29.17 29.17 29.17sl ender ness 58. 33 58. 33 58. 33

column weight (lbs) 1.374e+04 1.374e+04 1.374e+04column + drop (lbs) 1.938+04 1.938e+04 1.938e+04

Dead load:

DL = distributed dead load on column

For soil and gravel treated as Live Loads,

DL = roof weight + other dead load

For soil and gravel treated as Dead Loads,DL = roof weight + other dead load + soil dead load + gravel dead load

Live load:

LL = distributed live load on column

For soil and gravel treated as Live Loads,

LL = live loads + soil live load + gravel live load

For soil and gravel treated as Dead Loads,

LL = live loads

Earthquake load - for SEISMIC cases only:

EL = distributed earthquake load on column = v o x DL

Factored loads:

for static cases,

Wd = 1.3 x 1.4 x DL

WL = 1.3 x 1.7 x LL

for seismic cases,

Wd = 1.3 x 1.05 x DL


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WL = 1.3 x 1.28 x LL + 1.3 x 1.40 x EL

Axial load on column:

for static cases,

Pu = ((W d + W L)/1.3) x A bay + 1.4 x (column weight + drop panel) + VL1

for seismic cases,Pu = ((W d + W L)/1.3)x A bay +(1.05 + 1.40 x v o)x(column weight + drop panel)+ VL1

Pu//0 = P u/0.7500

Flexural load on column:

Since columns form part of two or more intersecting lateral-force-resistingsystems, the effects of the two orthogonal direction moments (M 2b,x, and M 2b,y )wilbe combined using the square root of the sum of the squares (SRSS) method.Conservatively we will assume that the moments generated on the columns by thegravity loads and moment magnifiers are the same in both directions.

M2b,x = ADOSS (See Appendix)

M2b,min = 1.3 x Pu (0.6 + 0.03 x d column ) / 12.00 in/ft ACI 318-95, Eq. (10-15)

M2b = / (M2b,x2 + M 2b,y

2)

wher e M 2b, x = M 2b, y = maxi mum val ue f r om M 2b, x1 , M2b, x2 , and M 2b, mi n

Assuming both ends of the columns are constrained against rotation, the momentdue to the imposed deflection during seismic loading will be calculated usingload factors from ACI 318-95, Sect. (9.2): (1.1 x 1.7 x 0.75 = 1.4), and ACI350R-95, Sect. (2.6.5a):(1.30)

Md = 1.3 x 1.4 x 12 x E x I x delta x L column = 1.3 x 1.4 x 6 x E x I x delta 2 x L column

3 L column2

Mi = Moment caused by the inertia of the column in an earthquake assuming bothends of the column are constrained against rotation.

= 1.3 x 1.4 x column weight x L column x u o/12The force caused by the water accelerating against the column in an earthquakewill be calculated based on Reference 10.


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Mw = Moment caused by the water accelerating against the column in an earthquakeassuming both ends of the column are constrained against rotation (see

Reference 10).

= 1.3 x 1.4 x mass of displaced fluid x L column x u o/12

= M i x gamma fluid /gamma concrete= M i x 62.50/150.0

Moment Magnification:

Cm = 1.0 (conservative - assumes equal and opposite end moments)

Bd = ratio of maximum factored axial dead load to maximum total factored axialload

Since the critical load of a column will be reduced due to creep and sustained loads,the factor B d will be calculated based on gravity loads which effect the columnduring static load conditions. While calculation of a separate B d could be done forseismic cases, the resulting critical load of the column would be artificiallyinflated due to the reduced load factor for dead load in the seismic cases.Therefore, calculation of B d based on static loads only will provide an acceptablyconservative result for the critical load of the column. (W d x A bay )/1.3 + 1.4 x column weight + drop panelBd =

Pu (E x I gross ) ACI 318-95, EQ. (10-13)EI =

2.5 x (1 + B d)

Pc = ( B2 x EI)/(k x L column )

2 ACI 318-95, EQ. (10-11)

Cm ACI 318-95, EQ. (10-10)delta b =

1 - P u//0P cMc = delta b x (M 2b + M d + M i + M w)

Eccentricity:

e min = 0.6 + 0.03 x d columne = M c /P uif e < e min , e = e min


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Variables for STATIC condition


interior longest shortestcolumn end span end span

DL (psf) 250.0 250.0 250.0LL (psf) 150.0 150.0 150.0EL (psf) 0 0 0Wd (psf) 455.0 455.0 455.0WL (psf) 331.5 331.5 331.5Pu (k) 364.7 312.6 279.9Pu//0 (k) 486.2 416.8 373.2alpha ec 0.2972 0.2972 0.2972M2b,x (ADOSS)(ft-k) 74.50 68.30 57.00

M2b,min (ft-k) 52.15 44.70 40.03M2b 105.4 96.59 80.61E (psi) 3.605e+06 3.605e+06 3.605e+06I (in 4) 1.629e+04 1.629e+04 1.629e+04delta (in) 0 0 0Md (ft-k) 0 0 0Mi (ft-k) 0 0 0Mw (ft-k) 0 0 0Cm 1.000 1.000 1.000Bd 0.4977 0.4843 04733EI (lb-in 2) 1.568e+10 1.582e+10 1.594e+10Pc (kips) 1263. 1275. 1284.delta b 1.626 1.486 1.410Mc (ft-k) 171.3 143.5 113.6e (in) 5.636 5.509 4.871


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Variables for SEISMIC condition



DL (psf) 250.0 250.0 250.0LL (psf) 150.0 150.0 150.0EL (psf) 85.54 85.54 85.54Wd (psf) 341.3 341.3 341.3WL (psf) 405.3 405.3 405.3Pu (k) 350.9 301.6 270.5Pu//0 (k) 467.9 402.1 360.7alpha ec 0.2972 0.2972 0.2972M2b,x (ADOSS)(ft-k) 56.20 54.30 44.00M2b,min (ft-k) 50.19 43.13 38.69M2b 79.48 76.79 62.23E (psi) 3.605e+06 3.605e+06 3.605e+06I (in 4) 1.629e+04 1.629e+04 1.629e+04delta (in) 0.1250 0.1250 0.1250Md (ft-k) 54.52 54.52 54.52Mi (ft-k) 14.86 14.86 14.86Mw (ft-k) 6.191 6.191 6.191Cm 1.000 1.000 1.000Bd 0.4977 0.4843 0.4733EI (lb-in 2) 1.568e+10 1.582e+10 1.594e+10Pc (kips) 1263. 1275. 1284.delta b 1.588 1.461 1.390Mc (ft-k) 246.3 222.6 191.6e (in) 8.420 8.856 8.500


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After time, all of the soil will be considered dead load. The columns were analyzed forthis additional load case, where 200.0 psf dead load and 50.00 psf live load were used.

Variables for STATIC condition



DL (psf) 350.0 350.0 350.0LL (psf) 50.00 50.00 50.00EL (psf) 0 0 0Wd (psf) 637.0 637.0 637.0WL (psf) 110.5 110.5 110.5Pu (k) 351.5 302.0 270.9Pu//0 (k) 468.6 402.7 361.2alpha ec 0.2972 0.2972 0.2972M2b,x (ADOSS)(ft-k) 74.50 68.30 57.00M2b,min (ft-k) 50.26 43.19 38.74M2b 105.4 96.59 80.61E (psi) 3.605e+06 3.605e+06 3.605e+06I (in 4) 1.629e+04 1.629e+04 1.629e+04delta (in) 0 0 0Md (ft-k) 0 0 0Mi (ft-k) 0 0 0Mw (ft-k) 0 0 0Cm 1.000 1.000 1.000Bd 0.6921 0.6659 0.6445EI (lb-in 2) 1.388e+10 1.410e+10 1.428e+10Pc (kips) 1118. 1136. 1151.delta b 1.721 1.549 1.458Mc (ft-k) 181.4 149.6 117.5e (in) 6.193 5.946 5.205


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Variables for SEISMIC condition



DL (psf) 350.0 350.0 350.0LL (psf) 50.00 50.00 50.00EL (psf) 85.54 85.54 85.54Wd (psf) 477.8 477.8 477.8WL (psf) 238.9 238.9 238.9Pu (k) 340.9 293.4 263.6Pu//0 (k) 454.5 391.3 351.5alpha ec 0.2972 0.2972 0.2972M2b,x (ADOSS)(ft-k) 56.20 54.30 44.00M2b,min (ft-k) 48.73 41.95 37.69M2b 79.48 76.79 62.23E (psi) 3.605e+06 3.605e+06 3.605e+06I (in 4) 1.629e+04 1.629e+04 1.629e+04delta (in) 0.1250 0.1250 0.1250Md (ft-k) 54.52 54.52 54.52Mi (ft-k) 14.86 14.86 14.86Mw (ft-k) 6.191 6.191 6.191Cm 1.000 1.000 1.000Bd 0.6921 0.6659 0.6445EI (lb-in 2) 1.388e+10 1.410e+10 1.428e+10Pc (kips) 1118. 1136. 1151.delta b 1.721 1.549 1.458Mc (ft-k) 266.8 236.0 200.9e (in) 9.393 9.653 9.146


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C. Column Reinforcing

The column reinforcing will be assumed as follows:

vertical reinforcing = 10 - #8 bars

A bar = 0.7900 sqin

dbar = 1.000 in

A s = 10 bars x 0.7900 sqin/bar = 7.900 sqin

cover over spiral reinforcing = 2.000 in

cover over vertical reinforcing = 2.000 + 0.5000 in = 2.500 in

spiral reinforcing = #4 bars

spiral pitch = 3.000 in

Depth to reinforcing from compression edge of column:

d1 = d column /2 + (d column /2 - cover - d bar /2) x sin(72)

= (24.00in)/2 + ((24.00in/2) - 2.500in - (1.000in/2)) x sin(72) = 20.56 ind2 = d column /2 + (d column /2 - cover - d bar /2) x sin(36)

= (24.00in)/2 + ((24.00in/2) - 2.500in - (1.000in/2)) x sin(36) = 17.29 in

d3 = d column /2 = (24.00 in)/2 = 12.00 in

d4 = d column - d2 = 24.00 in - 17.29 in = 6.710 in

d5 = d column - d1 = 24.00 in - 20.56 in = 3.440 in

1. Check Minimum and Maximum Reinforcing Ratios

A g = B x (d column /2)2 = B x (24.00 in/2) 2 = 452.4 sqin

rho min = 0.01 ACI 318-95, Sect. (10.9.1) As min = 0.01 x A g = 0.01 x 452.4 sqin = 4.524 sqin

rho max = 0.08 ACI 318-95, Sect. (10.9.1)

As max = 0.08 x A g = 0.08 x 452.4 sqin = 36.19 sqin

As min < A s < As max2. Calculation of / 0P n(max)

/0P n(max) = / 0 x 0.85 x [0.85 x f'c x (A g - A s ) + fy x A s ]

/0 = 0.7500

/0P n(max) = 0.7500 x 0.85 x [0.85 x 4000 psi x (452.4 - [10 bars x 0.7900 sqin/bar])+ 60000 psi x 10 bars x 0.7900 sqin/bar] x 0.001 k/lb

= 1266. k

ADOPT: / 0P n(max) = 1266. k


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3. Calculation of e 1

In the following calculations, the strain will be set equal to the yield strain forany bar which has been strained beyond the yield point.

Assuming the neutral axis (NA 1) is 24.08 in from the compression face of the columnand the crushing strain in the concrete is 0.003000:

Sy = 0.002069

Ss1 = (d1 - NA 1)/NA 1 x 0.003000

= (20.56 in - 24.08 in)/24.08 in x 0.003000 = -0.0004385

Ss2 = (d2 - NA 1)/NA 1 x 0.003000

= (17.29 in - 24.08 in)/24.08 in x 0.003000 = -0.0008459

Ss3 = (d3 - NA 1)/NA 1 x 0.003000

= (12.00 in - 24.08 in)/24.08 in x 0.003000 = -0.001505

Ss4 = (d4 - NA 1)/NA 1 x 0.003000

= (6.710 in - 24.08 in)/24.08 in x 0.003000 < -S ytherefore, S s4 = -S y

Ss5 = (d5 - NA 1)/NA 1 x 0.003000


Forces in the vertical reinforcing:

Fs1 = 2 x A bar x S s1 x E s= 2 x 0.7900 in 2 x -0.0004385 x 2.900e+04 ksi = -20.09 k



Fs4 = 2 x A bar x S s4 x Es

= 2 x 0.7900 in 2 x -0.002069 x 2.900e+04 ksi = -94.80 k

Fs5 = 2 x A bar x S s5 x Es

= 2 x 0.7900 in 2 x -0.002069 x 2.900e+04 ksi = -94.80 k

Compression force in the concrete: A c,1 = A g - d column

2 x (- 0 - sin- 0) - 10 x A bar 8

-0 = 2 x cos -1 ( a - d column /2 ) ____________ d column /2

a = Beta 1 x NA 1 = 0.8500 x 24.08 in = 20.47 in

-0 = 2 x cos -1 (20.47 - 24.00/2) = 1.574 rad 24.00/2

A c,1 = 452.4 sqin - 24.002 x [1.574 - sin(1.574)] - 10 x 0.7900 = 403.2 sqin

8


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Cc,1 = 0.85 x f'c x A c= 0.85 x 4000 psi x 403.2 sqin x 0.001 k/lb = 1371. k

Xc,1 = distance from column centerline to centroid of A c,1 = (d column x sin(- 0/2))

3/12 - X s,1 x A bar A c,1

where, X s,1 = sum of moment arms for reinforcing in compression

= 0.0000 in

Xc,1 = (24.00 in x sin(1.574/2))3/12 - 0.0000 in x 0.7900 sqin = 1.015 in

403.2 sqin

Nominal moment capacity of the column:

Mn,1 = C c,1 x X c,1 + F s1 x (d1 - d column /2) + F s2 x (d2 - d column /2) + F s3 x

(d3 - d column /2) + F s4 x (d4 - d column /2) + F s5 x (d5 - d column /2)

= 1371. k x 1.015 in + -20.09 k x (20.56 in - 24.00 in/2) + -38.76 k x(17.29 in - 24.00 in/2) + -94.80 k x (6.710 in - 24.00 in/2) + -94.80 kx (3.440 in - 24.00 in/2)

= 2327. in-k

/0Mn,1 = 0.7500 x 2327. in-k x 1 ft/12 in = 145.4 ft-k

Check axial capacity:

from statics, / 0P n(max) = / 0P nwhere,

/0P n = / 0 x (C c,1 - F s1 - F s2 - F s3 - F s4 - F s5 )

= 0.7500 x (1371. k - -20.09 k - -38.76 k - -68.96 k - -94.80 k - -94.80 k)

= 1266. k

1266. k.

1266. k therefore, O.K.e 1 = / 0Mn,1 //0P n(max) = 0.7500 x 2327. in-k/1266. k = 1.379 in

ADOPT: e 1 = 1.379 in


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4. Calculation of e b


Neutral Axis location for balanced condition (NA b):

NAb = 0.003000 x d1 = 0.003000 x 20.56 in = 12.17 in 0.003000 + S y 0.003000 + 0.002069

Strain in reinforcing bars:

Sy = 0.002069

Ss1 = (d1 - NA b)/NA b x 0.003000

= (20.56 in - 12.17 in)/12.17 in x 0.003000 < S ytherefore, S s1 = S y

Ss2 = (d2 - NA b)/NA b x 0.003000

= (17.29 in - 12.17 in)/12.17 in x 0.003000 = 0.001262

Ss3 = (d3 - NA b)/NA b x 0.003000= (12.00 in - 12.17 in)/12.17 in x 0.003000 = -0.00004191

Ss4 = (d4 - NA b)/NA b x 0.003000

= (6.710 in - 12.17 in)/12.17 in x 0.003000 = -0.001346

Ss5 = (d5 - NA b)/NA b x 0.003000



Fs1 = 2 x A bar x S s1 x E s= 2 x 0.7900 in 2 x 0.002069 x 2.900e+04 ksi = 94.80 k




Fs5 = 2 x A bar x S s5 x E s

= 2 x 0.7900 in2

x -0.002069 x 2.900e+04 ksi = -94.80 k


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Compression force in the concrete:

A c,b = d column2 x (- 0 - sin- 0) - 6 x A bar

8

-0 = 2 x cos -1 ( d column /2 - a ) ____________

d column /2a = Beta 1 x NA b = 0.8500 x 12.17 in = 10.34 in

-0 = 2 x cos -1 (24.00/2 - 10.34) = 2.864 rad 24.00/2

A c,b = 24.002 x [2.864 - sin(2.864)] - 6 x 0.7900 = 181.7 sqin

8

Cc,b = 0.85 x f'c x A c,b = 0.85 x 4000 psi x 181.7 sqin x 0.001 k/lb = 617.8 k

Xc,b = distance from column centerline to centroid of A c = (d column x sin(- 0/2))

3/12 - X s,b x A bar A c,b

where, X s,b = sum of moment arms for reinforcing in compression

= 6 x d column /2 - 2 x d3 - 2 x d4 - 2 x d5

= 6 x 24.00in/2 - 2 x 12.00in - 2 x 6.710in - 2 x 3.440in = 27.70 in

Xc,b = (24.00 in x sin(2.864/2))3/12 - 27.70 in x 0.7900 sqin = 6.039 in

181.7 sqin

Nominal moment capacity of column:

Mn,b = C c,b x X c,b +Fs1 x (d1-d column /2)+F s2 x (d2-d column /2)+F s3 x (d3-d column /2)

+ F s4 x (d4 - d column /2) + F s5 x (d5 - d column /2)

= 617.8 k x 6.039 in + 94.80 k x (20.56 in - 24.00 in/2) + 57.83 k x(17.29 in - 24.00 in/2) + [-61.67 k x (6.710 in - 24.00 in/2)] +

[-94.80 k x (3.440 in - 24.00 in/2)]

= 5986. in-k

/0Mn,b = 0.7500 x 5986. in-k x 1 ft/12 in = 374.1 ft-k

Nominal axial capacity of column:

/0P n,b = / 0 x (C c,b - F s1 - F s2 - F s3 - F s4 - F s5 )

= 0.7500 x (617.8 k - 94.80 k - 57.83 k - -1.920 k - -61.67 k - -94.80 k)

= 467.7 k

e b = / 0Mn,b //0P n,b = (0.7500 x 5986. in-k)/467.7 k = 9.599 in

ADOPT: e b = 9.599 in


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5. Calculation of / 0M o


Assume the neutral axis (NA o) is 6.089 in from the compression face of the column andthe crushing strain in the concrete is 0.003000.

Sy = 0.002069

Ss1 = (d1 - NA o)/NA o x 0.003000

= (20.56 in - 6.089 in)/6.089 in x 0.003000 > S ytherefore, S s1 = S y

Ss2 = (d2 - NA o)/NA o x 0.003000


Ss3 = (d3 - NA o)/NA o x 0.003000


Ss4 = (d4 - NA o)/NA o x 0.003000

= (6.710 in - 6.089 in)/6.089 in x 0.003000 = 0.0003060

Ss5 = (d5 - NA o)/NA o x 0.003000

= (3.440 in - 6.089 in)/6.089 in x 0.003000 = -0.001305








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Compression force in the concrete:

A c,o = d column2 x (- 0 - sin- 0) - 2 x A bar

8

-0 = 2 x cos -1 ( d column /2 - a ) ____________

d column /2a = Beta 1 x NA o = 0.8500 x 6.089 in = 5.176 in

-0 = 2 x cos -1 (24.00/2 - 5.176) = 1.932 rad 24.00/2

A c,o = 24.002 x [1.932 - sin(1.932)] - 2 x 0.7900 = 70.17 sqin

8

Cc,o = 0.85 x f'c x A c,o = 0.85 x 4000 psi x 70.17 sqin x 0.001 k/lb = 238.6 k

Xc,o = distance from column centerline to centroid of A c,o = (d column x sin(- 0/2))

3/12 - X s,o x A bar A c,o

where, X s,o = sum of moment arms for reinforcing in compression

= 2 x d column /2 - 2 x d 5 = 2 x 24.00/2 - 2 x 3.440 = 17.12 in

Xc,o = (24.00 in x sin(1.932/2))3/12 - 17.12 in x 0.7900 sqin = 8.946 in

70.17 sqin

Nominal moment capacity of the column:

Mo = C c,o x X c,o + F s1 x (d1 - d column /2) + F s2 x (d2 - d column /2) +

Fs3 x (d3 - d column /2) + F s4 x (d4 - d column /2) + F s5 x (d5 - d column /2)

= 238.6 k x 8.946 in + 94.80 k x (20.56 in - 24.00 in/2) + 94.80 k x(17.29 in - 24.00 in/2) + 14.02 k x (6.710 in - 24.00 in/2) + -59.80 k x(3.440 in - 24.00 in/2)

= 3885. in-k

/0Mo = 0.7500 x 3885. in-k x 1 ft/12 in = 242.8 ft-k

Check axial capacity:

from statics, P n = 0

where,

Pn = C c,o - F s1 - F s2 - F s3 - F s4 - F s5 = 238.6 k - 94.80 k - 94.80 k - 94.80 k - 14.02 k - -59.80 k

= -0.02000 k

-0.02000 k . 0 therefore, O.K.

ADOPT: / 0M o = 242.8 ft-k


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6. Summary of Vertical Reinforcing

By plotting the four points calculated above and connecting them, a conservativeinteraction diagram for the column can be created. All of the load cases can thenbe plotted on the interaction diagram, and all fall within the acceptable range forthe column.

ADOPT: 10 - #8 vertical bars with 2.000" cover to 4 spirals .7. Splice and Development Lengths for Vertical Reinforcing

For splices in the vertical reinforcing (see Design Drawings for locations), theminimum length as required by ACI 318-95 is as follows:

Tension:

L d = 3 x d b x fy x " x $ x 8 x m ACI 318-95, Sect (12.2.3)

40 x %(f'c) x (c + K tr /d b)

= 3 x 1.000 in x 60000 psi = 28.46 in 40 x %(4000) psi x 2.5

where, " = reinforcement location factor ACI 318-95, Sect. (12.2.4)

= 1.000

$ = coating factor ACI 318-95, Sect. (12.2.4

= 1.000

8 = lightweight aggregate concrete factor ACI 318-95, Sect. (12.2.4)

= 1.000

m = reinforcement size factor ACI 318-95, Sect. (12.2.4)

= 1.000

(c + K tr )/d b) = 2.5 (conservative)

Class B Splice Length = L splice = 1.3 x 28.46 in = 37.00 in ACI 318-95, Sect. (12.15)

Compression:

Lsplice = 0.0005 x d b x fy ACI 318-95,Sect.(12.16.1)

= 0.0005 x 1.000 in x 60000 psi = 30.00 in

Since the splice length for tension is greater than that for compression,

ADOPT: L splice = 42.00 in


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The development length of the vertical reinforcing in the footing can be calculatedas follows:

Tension (standard hook):

Ldh = 0.7000* x 0.8000 ** x L hb ACI 318-95, Sect. (12.5.1)

where, L hb = 1200 x d b ACI 318-95, Sect. (12.5.2) _________ %(f'c)

= 1200 x 1.000 in = 20.28 in %(3500 psi)

* ACI 318-95, Sect. (12.5.3.2)** ACI 318-95, Sect. (12.5.3.3)

L dh = 0.7000 x 0.8000 x 20.28 in = 11.36 in

Compression (deformed bar):

Ldh = 0.7500*** x L db ACI 318-95, Sect. (12.3.1)

where, L db = 0.02 x d b x f y ACI 318-95, Sect. (12.3.2) ______________ %(f'c)

= 0.02 x 1.000 in x 60000 psi = 20.28 in %(3500 psi)*** ACI 318-95, Sect. (12.3.3.2)

Ldh = 0.7500 x 20.28 in = 15.21 in

ADOPT: L development = 15.21 in


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8. Spiral Reinforcing Design

rho s = 0.45 x (A g/A c - 1) x f'c/fy ACI 318-95, EQ. (10-5)

where, A g = gross area of column

= B x (d column /2)2 = B x (24.00/2 in) 2

= 452.4 sqin

A c = area of core of column to O.D. of spirals

= B x (d column /2 - spiral cover)2 = B x (24.00 in/2 - 2.000 in) 2

= 314.2 sqin

rho s = 0.45 x (452.4 sqin/314.2 sqin - 1) x 3500 psi/60000 psi = 0.01320

where, rho s = ratio of spiral volume to core volume

where,

spiral volume = V s = 0.20 sqin x B x (d column - 2 x spiral cover - d spiral )/pitch

= 0.2 x B x(24.00 - 2 x 2.00 - 0.5000)/pitch = 12.25/pitch in3

/incore volume = 314.2 in 3/in

rho s = 12.25 = 0.01320314.2 x pitch

pitch = 12.25 = 2.955 in 314.2 x 0.01320

The minimum and maximum clear spacing of spiral reinforcing are given in the ACI 318-95,Sect. (7.10.4.3). These spacings correspond to pitches of 1.500 inches and 3.500 inchesrespectively.

The minimum lap splice in spiral reinforcing is 48d spiral (ACI 318-95, Sect. (7.10.4.5)).Conservatively, one full revolution will be provided for all laps.

ADOPT: 0.5000"/ 0 spirals at 3.000" pitch with 1 full revolution lap length


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D. Column Moment Transfer

In accordanc

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