DESIGN OF ARCH SLAB :
Unit no. (2)coveringmanual design
done by
stud. / mahmoud abdalsamad abo zied stud. /bahaa eldeen elsayed
khalil stud. /medhat asem elmorsy aboahf
pesented toprof. / ahmed ata
DESIGN OF ARCH SLABArea properties :
Span = 13 ms. --Length =60 ms. -Designing an arch slab covering
with spacing of 6 ms between columns & span of 13.00 ms
.1-Concrete dimensions :
slab thickness :-at crown>>>>>>>> ts =
100 mm
at quarter>>>>>>>> ts = 120 mm
at foot >>>>>>>> ts= 140 mm
-flight = L/(6-8) =13/(6-8) = 2.167-1.625 m = 1.8 m
-vertical beam :
thickness = span/18 =13000/18 = 750 mm
width = 250 mm
-horizontal beam :thickness = t vertical beam + (50-100) =
750+100 =850 mm
-tie :
assume b X t = 300X300 mm
-hanger :
assume b X t = 200X200 mm with spacing =3.25 m
three hangers
-post :
assume b X t = 300X300 mm
that
T=(H/10 = 2000/10 = 200 mm)
Or
T=300 mm >>>>> minimum
______________________________________________
2-Loads :
D.L = (hanger o.w + tie o.w + 1.09 c ts + F.C )=
(.2*.2*25*4.5+.3*.3*25+1.09*25*.12+1.5)
= 11.52 t/m'
L.L = 1 t/m'
Wu = 1.5(1+11.52) = 18.78 t/m'
Cases of loading :
Case 1 : case of total load :
Yu = W*L/2 = 18.76*13/2 = 122.1 t
Hu = W*L2/8f = 18.78*132/8*1.8 = 220.4 t
3-Designing of each component:*(Slab):1-design of crown point :M
= 0.0 >>>> hingeN = H = 220.4 t
As = (.8/100)*1000*100=800 mm2 = 416/m'Check :
Pu=.35 fcu Ac + .67 Fy Asc =
.35*25*100*1000+.67*360*.01*100*1000 = 1339440 t
>>>>>>>>>> Nu .. ok
2-Design of foot point :
Nmax = Hu2+Yu2 = 122.12+220.42 = 251.96 t
M= 0.0
As = .8/100 * 1000*140 = 1120 mm2 = 616/m'
Check :
Pu=.35 fcu Ac + .67 Fy Asc =
.35*25*120*1000+.67*360*.01*120*1000 = 1339440 t
>>>>>>>>>> Nu .. ok
3-Design of quarter point :
-16.13*13*6.5-1.6*6.5*9.75+Ya*13 = 0.0
Ya = 112.65 t
Hu = gL2/8f + PL2/16f = 144.6 t
N =Hu(1+(2f/L))= 144.6(1+(2*1.8/13)2) = 150tM=Pu*L2/8f =
1.6*132/8*1.8 = 18.8 t.me=M/N =18.8/150= .125 > .05 consider
moment
e/t = .125/.12 = 1.04K = N/fcu*b*t =150*103/30*1000*120 = .042
>.04
Using interaction diagram :
K=.042K*e/t =.06*1.04=.0624Fy=360 t/m2 = 1.00
= .8
from interaction diagram :
take = 1.50
= 1.5*30*10-4 = 4.5*10-3As = *b*t = 4.5*10-3 * 1000*120 = 540
mm2Asmin = The bigger of >.225fcu/fy *b*d /fy = 410mm2 >
1.1*b*d/fy = 366.7
== 410 mm2 The small of > 410 mm2 &1.3As = 468 mm2==410
mm2The bigger of .15*b*d/100 = 180 mm2 & 410
mm2o.k.......................................
So As = 540 mm2 = 316/m'
*(Design of vertical beam) :
1- loads :
Wu = 112.65+1.4(25*.25*.75)+1.4(25*.35*.85)
= 129.625 t
2- straining action :
Mmax = W*L2/10 = 129.625*62/10 = 466.65 t.m
Qu = .6*6*129.625 = 466.65 t
700=c1(466.65*106)/(30*250)
C1 = 2.8
J = .721
3- RFT :
As = (466.65*106)/(360*.721*700) = 2568.4 mm2 = 722 /m'As' = .2
As = 543.7 = 316 Check for shear :
qcu = .24(fcu/c) = .24(30/1.5) = 1.07 N/mm2qumax = .7(fcu/c) =
.7(30/1.5)
= 3.13 N/mm2 40 >>>. bXT = 300*300 = 90000 mm2 40 =
40*380.13= 15205.2 mm2O. KPost design:
P = Ya + post weight = 112.65 + .3*.3*2*25 = 117.15 t
So the reaction which will be reflected on the reinforced frame
= 117.15 t......
