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