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Design of Singly Reinforced Beam
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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
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