Slab Shear Capacity

Page 1

This program calculates Shear Capacityof a beam without shear reo.

1

1

1b 1000d 400Ast 1000mm2 This program calculates Shear Capacityf'c 20MPa in accordance with Equation 19.9Vuc 147.4kN Vuc 304.1kN0.7Vuc 103.2kN 0.7Vuc 212.9kN

b1

b2

b3

Short Columns

Page 2

This program calculates the axial load capacity of a short column

b 350D 720Ag 252000mm2Ast 8000mm2 okfsy 400MPaf'c 32MPa

Nuo 10054.4kNPhi 0.6Phi Nuo 6032.6kN Bearing Stress 24MPa

Symbol Symbola a A Ab b B Bc c C Cd d D De e E Ef f F Fg g G Gh h H HI I I Ij j J Jk k K Kl l L L

m m M Mn n N No o O Op p P Pq q Q Qr r R Rs s S St t T Tu u U Uv v V Vw w W Wx x X Xy y Y Yz z Z Z

JOB No: 3312656

Level 8, 68 Grenfell Street

ADELAIDE SA 5000 DATE: 4/19/2023

TELEPHONE (08) 8235 6600

FACSIMILE (08) 8235 6694 DESIGN: GAB

SHEET:

Beam Deflections of reinforced concrete beams

Short term load

G 45.0kN/m0.7

Q 15.0kN/m

55.5kN/m

Span Lef 15.0m

Mo 1560.9kNm

234.0kNm

1171.0kNm

858.4kNm

EMAIL adlmail@ghd.com.au

ys

G+ysQ

ML

MR

MM

This spreadsheet calculates the moment capacity of a singly reinforced concrete beam to AS3600-2001 amendments 1 & 2

The cracking moment Mcr and Ieff, based on the Branson Formula, are calculated in accordance with AS3600

Enter the following data

2198mm2 M*= 290.0kNm 200.0kNm

500MPa 5.41MPa #ADDIN? Section Cracked

40MPa 3.79MPa #ADDIN?

46.00MPa #ADDIN?

430mm 0.70MPa #ADDIN?

1200mm 34290MPa 0.0006

200000MPa

30mm n = 5.83 #ADDIN?

12mm 7.95E+09mm4 #ADDIN?

20mm 114.4kNm #ADDIN?

378mm ###

g = 0.766 Z = 3.70E+07mm3 #ADDIN?

0.0930 ### ku < 0.4 ie section under-reinforced OK

35.2mm ###

0.0208 ### pmax when ku=0.4

0.0048 pmin slabs supported by columns 9.1.1(a) 0.0025

pmin slabs supported by beams/walls 9.1.1(b) 0.002

pmin beams 8.1.4.1 0.0022 #ADDIN?

320.5kNm ### > M* OK

400.6kNm 168.4kNm #ADDIN? OK

a=0.5dn 600 Calculates dn by equating first moments of the compressive

b=nAst 12820.16012997 and tensile areas about the neutral axis

c=-nAstd -4846020.52913 0.5bdn2=nAst(d-dn) or 0.5bdn^2+nAstd-nAstd=0

79.8mm solve quadratic for dn

1.34E+09mm4 ###

2.58E+09mm4 ###

4.77E+09mm4 ### Note Ief>Ie,max Use Ie,max

Ast = Ms =

fsy = fb =

f'c = f'cf =

f'cm =

D = fcs =

b = Ec = ecs =

Es =

Cover =

Lig dia = Ig =

bd = Mcr =

d =

ku =

dn =

pmax =

p actual =

f Muo =

Muo = Muo,min =

dn =

Icr =

Ief =

Ie,max =

= Ms · 1000000 / Z [Eqn. 1]

= 0.6(f 'c) [Eqn. 2]

= 1.15f 'c [Eqn. 3]

= ([1.5p actual] / [1 + 50p actual]) · 200000ecs [Eqn. 4]

= Es / Ec [Eqn. 5]

= bD3 / 12 [Eqn. 6]

= (f 'cf - fcs)Ig / ([0.5D] · 1000000) [Eqn. 7]

= D - (Cover + Lig dia + 0.5bd) [Eqn. 8]

= b(D2) / 6 [Eqn. 9]

= fsyAst / (0.85gf 'cbd) [Eqn. 10]

= kud [Eqn. 11]

= 0.85g · 0.4f 'c / fsy [Eqn. 12]

= 0.22(D / d)2f 'cf / fsy [Eqn. 13]

= 0.8 · 0.85f 'cgku(1 - 0.5gku)bd2 · 0.000001 [Eqn. 14]

= 1.2Zf 'cf / 1000000 [Eqn. 15]

= ([1 / 3]bdn3) + (nAst[d - dn]

2) [Eqn. 16]

= Icr + (Ig - Icr)([Mcr / Ms]3) [Eqn. 17]

= if(p actual<0.005 , 0.6Ig , Ig) [Eqn. 18]

b

dd

Ast

Dd

Ief, design 2.58E+09mm4

= if(p actual<0.005 , 0.6Ig , Ig) [Eqn. 18]

b

dd

Ast

Dd