RC Design Spreadsheets- working sheet.xlsx
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Transcript of RC Design Spreadsheets- working sheet.xlsx
Leh Lev S1 S2 a bin in in in in in
2 2 2 2 2 2
SN Column section Base plate
d tw bf tf Fy Fu Width (N)in in in in ksi ksi in
39.37
Design for Small Moment
Design Check
cin
2
Base plate Bolt
Height (B) Thickness (t) Fy Fu diameter Length Fy Fuin in ksi ksi in in ksi ksi
7.87 1.00 36.00 58.00 1.00 15.70
Weld Pedestal Forces Calculations
Strength Thickness (D/16) Width Height fc' Axial Momentksi in in in ksi Kips Kip-in A1 (in2) A2 (in2)
70.00 40.00 10.00 3.40 -133.00 235.00 309.84 400.00
B.Plate Area
Pedestal Area
Sqrt(A2/A1) fp (max) qmax ecrit Design typein ksi kips/in in
1.77 1.14 2.134377 16.79755 15.72609 Design for Small Moment
eccentricity (e)
Design of Spread Footing
Design of a Square Spread Footing Kiran AcharyaRef: PCI notes section 15, (pg 1131) Date: Jan 10 2015
1) InputsDead load= 674.4 Kips 3000 kNLive load= 179.84 Kips 800 KNsurcharge= 0.104 ksf 5 KN/m2Assume average weight of soil and
concrete above footing base= 100 pcfAllowable soil pressure at bottom of footing= 7.70 ksf 370 KN/m2
b, mm= 1985 mmColumn Dimensions= 78 13 in d, mm= 375 mm
fc'= 6000.0 psi 0.0018 fc'= 20.684 MpaSqrt(fc')= 77.5 rebar dia 1 inAssume Side cover 3 infooting thickness= 47.24409 in 1200 mmCover= 5 in
0.75αs= 40fy= 60000 psi
2) Determination of Base Area
Base area is determined using service (unfactored) loads with net permissible soil pressureNet allowable soil pressure= 6.946 ksf (find out why 0.75 has been subtracted)Required base area of footing= 123.0 ft2
use 11 X 11 ft square footingArea of footing, Af= 121 ft2
3) Factored loads and soil reaction:Pu= 1097.024 KipsLoad intensity, qs= 9.1 ksf
ρmin=
, Shear=
Design of Spread Footing
4) Depth of FootingFooting depth is determined based in shear strength without shear reinforcement.Both wide-beam action and two-way action for strength computation need to be investigated
Footing effective thickness, d= 3.5 fta) Wide Beam Action
bw for beam action= 11 ft2Tributary area= 26.7 ft2
242.3 Kips647.9 Kips
DC Ratio= 0.37 OK
b) Two-way Action
Tributary area for two way action= 74.6 ft2Perimeter of Critical Section, bo= 351.68404 in
676.7 Kipsβ= 5.9
= 1150785.3
= 2.7
= 6.8
2309 KipsDC Ratio= 0.29 OK
Shear force, Vu=Shear Capacity, Vn=
Shear Force, Vu=
Vn=
Design of Spread Footing
5) Design for Footing Reinforcement
Rebar fy= 60000 psiqs= 9.1 ksfFooting length= 11 ft
a) Critical section for moments is at face of columnCritical section location= 4.9498239 ftMu= 1222 kip-ft= 0.9
b)Required Rn= 69 Psi
0.0012d= 42.244094 in
0.0010
c) Check Minimum As required for footings of uniform thickness; for Grade 60
0.0018 (from 7.12.2.1)
1.73 Not OK
d) Required area of Steel, AsAs= 6.5 in2One Rebar dia= 1 inOne Rebar area= 0.79 in2No. of rebar required = 9 nosTotal Area= 7.1 in2
Compute Required As assuming tension-controlled section (=0.9)
ρ=
ρ (gross area)=
ρmin=
ρ (gross area) > ρmin
Design of Spread Footing
e) Check Maximum rebar spacing
Concrete side cover= 3 inActual spacing= 15.6Maximum limit of rebar spacing= 18 in Ref: 15.10.4
Maximum spacing limit is OKNote: lesser amount of reinforcement is required in the perpendicular direction due to lesser Mu, but for ease uniform reinforcement will be provided.
f) Check Net Tensile Strain
Depth of compression block, a= 0.63Neutral Axis Depth, c= 0.74Footing depth,d= 42.244094 in
0.1681Minimum tensile strain (10.3.5)= 0.004
OKSince the actual tensile strain is more than the minimum requiredthe section is tension-controlled and initial assumption is valid.
g) Check the development reinforcementThe critical section for development is the same as that for moment
tensile strain, εt=
Design of Spread Footing
DL calculation
Tributary area 72.25 m2No. of stories 40Avg. Dead load 15 KN/m2Total dead load 43350 KNRequired column area
Summary
Factored Load= 1097 Kips = 4914.7 KNallowable soil pressure= 370 KN/m2fc' 20.684 Mpa
Required footing area= 11.42512 m2Effective footing depth= 1073 mm
Max Shear DC ratio= 0.37Required rebar area= 4174.809 mm2rebar dia 25.4 mmrequired no= 8.2
Design of Spread Footing
Design of a One-Way Solid Slab Kiran AcharyaRef: PCI notes Example 7.2, (pg 703) Date: Jan 11 2015
1) Inputs
Clear Span, ln= 18 ft Assume slab thickness= 6 infc'= 4000 psi Dead load= 73.2 psffy= 60000 psi SDL= 1.8 psfSqrt(fc')= 63.24555 Live load= 50 psfCover= 3 in surcharge= 0.1 ksf
0.9
0.01806 Table 6.1rebar dia 0.625 in #5
s
2) Compute Moment Strengths (approximate moment analysis permitted by 8.8.3)
Factored load, qu= 170 psfPositive moment at discontinuous end integral with support (+Mu):
3.93 kip-ft/ftNegative Moment at exterior face of first interior support (-Mu):
5.51 kip-ft/ft
3) Determine required slab thicknessChoose a reinforcement percentage ρ equal to about 0.5ρt, or one hafl the maximum permitted for tension-controlled sections, to have reasonable deflection control. From table 6-1
From table 6.1, ρt= 0.01806Set ρ= 0.00903Rn= 499 psiRequired depth, d= 3.5 inAssumed rebar dia= 0.625 in #5Required ha= 4.4 in 0.75 is the min concrete cover
, tension=
ρtmax=
Design of Spread Footing
The above calculations indicate a slab thickness of 4.40947826734268 in is adequateHowever, Table 9-5(a) indicates a minimum thickness of l/24 in., unless deflections are computed.
4) Compute required negative moment reinforcement
Provided ha= 4.5 in (input based on above calculated ha)d= 3.59 inRn, required= 473.8 psiρ, required= 0.0085808Negative As (required)= 0.37 in2/ft
4) Compute required positive moment reinforcementFor Positive reinforcement, use Table 7-1
= 0.0846121
0.09ρ= 0.006Positive As (required)= 0.26 in2/ft
From Table 7.1, ω=
Design of Spread Footing
Design of Spread Footing
Design of Spread Footing
Design of a Rectangular Beam Kiran AcharyaRef: PCI notes Example 7.3, (pg 705) Date: Jan 10 2015
1) Inputs
Clear Span, ln= 18 ft Dead load= 45.0 psfWidth,b= 12 in SDL= 1.8 psfDepth,dt= 30 in Live load= 50 psfd' 2.5 in surcharge= 0.1 ksffc'= 4000 psi DL Moment, Md 430 Kips-ftfy= 60000 psi LL Moemnt, ML= 175 Kips-ft
Cover= 1.5 in 0.01806 Table 6.10.9 Sqrt(fc')= 63.20.9 stirrup dia= 0.5 in
Tension rebar dia 1.27 in #10Compression rebar dia 0.75 in #6
2) Determine required reinforcementa) Determine if compression reinforcement is needed
Mu= 796 Kips-ft
Mn= 884.4 Kips-ft
Rn= 982.7
Rn, Max 911 (Input from Table 6-1 corresponding to fc and fy)Check if Rn is greater than the Rn,Max from Table 6-1 for tension-controlled sections without compression reinforcementRn<Rn,Max Not OKCompression Reinforcement is Required
It seems two layers of tension reinforcement is necessary, so assumeNo of tension rebar layer= 2d= 28.8 in
ρtmax=, tension=, bending=
Design of Spread Footing
b) Find nominal strength moment resisted by the concrete section without compression reinforcement
ρt= 0.0181ρ= 0.0188
0.282
0.2351 Input the value from table 7-1 corresponding to the value of ω, as calculated above.
Mnt= 780.0 Kips-ftRequired moment strength to be resisted by the compression reinforcement:Mn'= 104.4 Kips-ft
c) Determine Compression steel stress, fs'Check yielding of compression reinforcement. Since the section is designed at the tension-controlled, net tensile
εt= 0.005ca1/dt= 0.375ca1= 11.25 ind'/ca1= 0.22
ω=
strain limit εt=0.005, ca1/dt=0.375
Design of Spread Footing
Compression reinforcement yields at the nominal strength (fs'=fy) (find out more why?)
d) Determine the total required reinforcement:Compression steel, As'= 0.79 in2Total reinforcement, As= 7.30 in2
e) Check Moment CapacityWhen the compression reinforcement yields:
a= 9.56 in (find why divided by 12, not shown in formula)797 Kips-ft
Check Manually
3) Select reinforcement to satisfy control of flexural cracking criteria of 10.6a) Compression reinforcement:
Required Compression reinforcement area, A's=0.79 in2
Required no. of comp. bar= 2.00
b) Tension reinforcement:Required no. of Tension reinforcement area, A=
7.30 in2Required no. of tension bar= 6.00
c) Maximum spacing allowed
cc= 2 infs= 40000 psiMax s, calculated= 10 in
Mn=
If Mn=Mu, OK
Design of Spread Footing
s max,limit= 12 ins,max= 10 in
No. of layer of tension rebar= 2rebar in 1 layer= 3Spacing provided= 3.365 inProvided Spacing of tension bar is OK
3) Select reinforcement to satisfy control of flexural cracking criteria of 10.6Stirrups or ties are required throughout distance where compression reinforement is required for strength.
Max. Spacing is the minimum of the following= 7.10.5.216*comp. longitudinal bar dia = 12 in48* tie bar dia= 24 inleast dimension of the member= 12 in
Therefore, s,max= 12 in
Design of Spread Footing
Design of Spread Footing
Design of Spread Footing
Design of a One-Way Solid Slab Kiran AcharyaRef: PCI notes section 15, (pg 1131) Date: Jan 10 2015
1) Inputs
SN From To reverseForce Kips KN 4.448 0.2248
Pressure ksf KN/m2 48.1 0.0208psi Mpa 0.006894 145.03
Area Sq ft. m2 0.0929 10.76