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Structures for Architects IIARC 460ARC 460
Lecture (11)Design Concrete Beams (2)
1
Depth of Beam for Preliminary Design
The ACI code prescribes minimum values of h, height ofbeam, for which deflection calculations are not required.
Minimum values of h to avoid deflection calculations
Type of beam
construction
simply supported
one end continuous
both ends continuous
cantilever
beams or joists
l /16 l /18.5 l /21 l /8
one way slabs
l /20 l /24 l /28 l /10
ACI Code Requirements for Strength DesignUltimate Strength
nu MM Mu = Moment due to factored loads (required
ultimate moment)ultimate moment)Mn = Nominal moment capacity of the cross-section
using nominal dimensions and specified material strengths.
= Strength reduction factor (Accounts for variability in dimensions, material strengths, approximations in strength equations.
ACI Code Requirements for Strength DesignUltimate Strength
nu MM 9.0
VV 85.0 nu VV 85.0
nu PP 70.0
Factored Load Combinations
U = 1.2 D +1.6 L Always check even if other load types are present.
U = 1.2(D + F + T) + 1.6(L + H) + 0.5 (Lr or S or R)U = 1.2D + 1.6 (Lr or S or R) + (L or 0.8W)U = 1.2D + 1.6 (Lr or S or R) + (L or 0.8W)U = 1.2D + 1.6 W + 1.0L + 0.5(Lr or S or R) U = 0.9 D + 1.6W +1.6HU = 0.9 D + 1.0E +1.6H
U = Required Strength to resist factored loadsD = Dead LoadsL = Live loadsW = Wind Loads E = Earthquake LoadsH = Pressure or Weight Loads due to soil , ground
6
H = Pressure or Weight Loads due to soil , ground water , etc.
F = Pressure or weight Loads due to fluids with well defined densities and controllable maximum heights.
T = Effect of temperature, creep, shrinkage, differential settlement, shrinkage compensating.
No Reinforcing Brittle failure
Reinforcing < balance Steel yields before concrete fails
ductile failure
bdAs
=
yf200
min =
Failure Modes
ductile failure
Reinforcing = balance Concrete fails just as steel yields
Reinforcing > balance Concrete fails before steel yields Sudden failure
+
=
yy
cbal ff
f87000
8700085.0 '1
bal 75.0max =
maxmin
11 is a factor to account for the
non-linear shape of the compression stress block.
f'c 1ca 1=
f'c 10 0.85
1000 0.852000 0.853000 0.854000 0.855000 0.86000 0.757000 0.78000 0.659000 0.65
10000 0.65
Useful Tables:
Example (1)Concrete rectangular Beams with d= 17.5 , b = 12 , As = 3 x #8 Fc = 4000 psiFy = 60000 psi
Required:
10
Required:
1- Check
Solution
11
As = 3 x #8 = 3x 0.79 = 2.37
d= 17.5 , b = 12As = 3 x #8 = 3x 0.79 = 2.37Fc = 4000 psiFy = 6000 psi
0113.05.1712
37.2===
xbdAs
f'c 10 0.85
1000 0.852000 0.853000 0.854000 0.855000 0.86000 0.757000 0.7
=cf 8700085.0 '1
12
7000 0.78000 0.659000 0.65
10000 0.65
+
=
yy
cbal ff
f87000
8700085.0 1
0285.06000087000
8700060000
400085.085.0=
+
=
xxbal
0033.060000
200200min ===
yf
0214.00285.075.075.0max === xbal0.0033
AssessmentConcrete rectangular Beams with d= 25 , b = 16 , As = 3 x #10 Fc = 4000 psiFy = 60000 psi
Required:
13
Required:
1- Check
actual ACI equivalentstress block stress block
Flexure Equations
bdAs
=
Data: Section dimensions b, h, d, (span) Steel area - As Material properties fc, fy
Required: Strength (of beam) Moment - Mn Required (by load) Moment Mu Load capacity
'' 85.085.0 cy
c
ys
fdf
orbf
fAa
=
Rectangular Beam Analysis
Load capacity
1. Find = As/bd(check min< < max)
2. Find a3. Find Mn4. Calculate5. Determine max. Mu (or span)
=
2adfAM ysn
8
2)6.12.1( lLLDLuM
+=
nu MM nu MM
Useful Tables:
Example (2)Concrete rectangular Beams with d= 17.5 , b = 12 , As = 3 x #8 Fc = 4000 psiFy = 60000 psiSpan of Beam = 25 FeetBeam Is simple support
17
Beam Is simple supportDL=200PsfLL= 400 PsfRequired:1- Find Mu2- Check If beam is Safe
Solution49.3
12400085.06000037.2
85.0 '===
xx
x
bffA
ac
ys
2241111249.35.176000037.2
2=
=
= xx
adfAM ysn
IbinxMM nu == _201700022411119.0
18
nu
KftMu _1681000/12/2017000U = 1.2 D +1.6 L = 1.2x200+1.6x400=880 Psf
75.6881000
258808
22
===
x
xWlM u Safe