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=


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=


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


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=


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 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=


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 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=


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


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


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 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

rc design spreadsheets- working sheet.xlsx

Documents