Concrete Lecture 1
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stephaniejeancortez522Concrete Lecture 1
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stephaniejeancortez522DESIGN OF BEAMS AT ROOF
BEAM BR-9 ( SHORT ) EXTERIOR
Mu = 54 kN.m Vu = 30 kN
54000000 N.mm 30000 N
L= 4.2 m
SPAN 4200 mm 90 mm steel covering
b= 200 mm fc' = 21 mpa
d= 210 mm fy = 228 mpa
h= 300 mm Ø= 0.9 Ø= 0.85
H/B= 1.5 FALSE β= 0.85 H/B MIN >1.6<1.8, OK
USING 16 mm steel bar diameter. 1.6 1.8
N= 9 STEEL BARS <1.6, NOT OK
>1.8, NOT OK
Check if only tension bars are needed:
Mu(1x106) = Øfc'bd2w(1-0.59w)
0.59Øfc'bd2w2 - Øfc'bd²w + Mu(1x106) = 0
w= 0.436
p =wfc'/fy = 0.0402
Pb = (0.85fc'β600)/fy(600+fy)
Pb = 0.0482
Pmax = 0.75Pb
Pmax = 0.0362
P > Pmax (NEEDS COMPRESSION REINFORCEMENT), DOUBLY DESIGN
As = pbd
As = 1687 mm2
Using 16 mm dia. Steel bar
πØ2 N/4 = As
N= 8.39 SAY 9 STEEL BARS AS1 AS2
SAY 4 BARS 3 3
T=C AC1 AC2
Asfy = 0.85 fc' ab
a= 107.77
Mu = ØT(d-a/2)
T = Asfy
Mu = ØAsfy(d-a/2)
Mu = 54055972 >Mu given, SAFE
a = βc
c= 126.8
Es/(d-c) - 0.003/c
Єs = 0.00197
Єy =fy/Es
Єy = 0.001140
Єs > Єy (STEEL YIELDS) Assumption is correct
CHECK FOR SHEAR: Ø= 0.85 (FOR SHEAR)
Vu = 30000 N
Vc = 1/6 √fc' bd
Vc = 32078 N
32.08 N
ØVc/2 = 13633 N
Vu > ØVc/2 (NEED STIRRUPS)
Vs = Vu/Ø - Vc
Vs = 3216.087782
S = Avfy d/Vs
USING 10 mm Ø STIRRUPS
Av = (π/4) Ø2 (2)
Av = 157 mm2
S= 2337.4 SAY 2338.0 mm
S= 0.33bd*(fc')^1/2
Smax= 209 mm or 300mm
Max. S = d/2 = 105 mm OR 600MM
1/3√fc'bd = 64156 >VC, safe
DEVELOPMENT LENGTH
Ld = 0.02Abfy/√fc' = 200 mm
Min Ld = 0.06dbfy = 218.88 mm USE 300 mm DEVELOPMENT LENGTH
300 mm (minimum) 300 mm
CHECK BAR ARRANGEMENT
Ac=2cc+2ds+(n-1)*db+ndb Ac<B, ok
Ac= 444 Ac>B, not ok, rearrange bars. Ac>B, not ok, rearrange
bars. AS1 AS2
198 Ac<B, ok 3 3
Check for ductility C=600d/(fy+600)
As act=nb(ab) c= 152.173913
As act = 1809.5616 mm^2 a=0.85C
Asb= 0.85*fc'b*(a)/fy a= 129.3478261
Asb = 2025.314645 mm^2
Asmax=0.75Asb <Asact<Asmax, safe
Asmax = 1518.985984 mm^2 failed
Asmin=1.4bd/fy 1809.5616 1518.985984
Asmin = 257.8947368 mm^2 <Asact<Asmax, safe
CHECK MOMENT CAPACITY
COMPRESSION= TENSION
C=T
a=Asact*fy/0.85*fc'd
a= 110.065 > Muact, SAFE
<Muact, FAILED
Mu= Ø C(d-0.50a)/1000^2
Mu= 57.5428 kN.m > Muact, SAFE
check for crackings:
fs= 0.60fy
fs = 136.8 mpa
AC= 2dsb ds= cc+(2*ds)+0.50db
AC = 27200 mm^2 ds = 68 mm
A= AC/n dc= cc+(2*ds)+db+25+0.50db
A= 3022.222222 mm^2 dc = 109 mm
Z=fs(Adc)^1/3/1000 (Adc)6!/3= 69.06387855
Z= 9.45 < Z limits, SAFE. ZLIMITS 25
< Z limits, SAFE.
> Z limits, Failed
CHECK BY DEFLECTIONS:
DEFELCTION ALLOWED= L/360
L/360=11.67 mm
Ig= bh^3/12
Ig = 450000000 mm^4
Yt= h/2
Yt = 150.00 mm
Fr= 0.70(fc')^1/2
Fr = 3.207802986 mpa
Ec=4700(fc')^1/2
Ec = 21538.11 mPa
Mcr= FR*Ig/Yt (Mcr/Mu)^3 < 1 , use K factor.
Mcr = 9.62 kn.m NOT OK.
(Mcr/Mu)= 0.17
(Mcr/Mu)^3= 0.00468 (Mcr/Mu)^3 < 1 , use K factor.
k= -Pn +(Pn^2+2Pn)^1/2 Pn= -0.325496997
k= 1.5667 Pn^2+2Pn= 3.580466962
Sqrt= 1.892212187
y=kd 329.0
Ie=(Mcr/Mu)^3*Ig + [1-(mcr/Mu)^3]Icr
Ie = 2571260948 mm^4
Icr=by^3/3 +nAs(d-y)^2 d-y= -119.0 mm
Icr = 2581229787 mm^4
Sustainable deflection:
Sd= 5Mu*1000^2*L^2/48EcIg
Sd= 10.237669 mm^4
Pact= 0.043
time factor to sustain loads( 3months)
e= 1
x= (e/1+50P) <L/360, SAFE
x= 0.32 > l/360, FAILED
Long term deflection
Ld= Sd*x
Lt = 3.25 mm <L/360, SAFE
Summary: b h stirrups main bars LD Smax Nbars C- BARS
T- BARS
200 300 10 16 300 105 9 3 3
SAY 9