elementos de concreto armado

7/30/2019 ELEMENTOS DE CONCRETO ARMADO

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The concrete beam design flow charts address the following subjects:

For a rectangular beam with given dimensions: Analyzing the beamsection to determine its moment strength and thus defining the beam

section to be at one of the following cases:

Case 1: Rectangular beam with tension reinforcement only. This

case exists if the moment strength is larger that the ultimate(factored) moment.

Case 2: Rectangular Beam with tension and compression

reinforcement. This case may exist if the moment strength is l essthan the ultimate (factored) moment.

For T-section concrete beam: Analyzing the beam T -section to determineits moment strength and thus defining the beam section to be one of the

following cases:

Case 1: The depth of the compression block is within the flangedportion of the beam, i.e, the neutral axis N.A. depth is less than theslab thinness, measured from the top of the slab. This case exists if

moment strength is larger than ultimate moment.

Case 2: The depth of the compression block is deeper t han theflange thickness, i.e. the neutral axis is located below the bottom ofthe slab. This case exists if the moment strength of T -section beam

is less that the ultimate (factored) moment.

Beam Section Shear Strength: two separate charts outline in det ails Shear

check. One is a basic shear check, and two is detailed shear check, inorder to handle repetitive beam shear reinforcement selection. See shearcheck introduction page for further details.

In any of the cases mentioned above, detailed procedure s and equations areshown within the charts cover all design aspects of the element underinvestigations, with ACI respective provisions.

Introduction to Concrete Beam Design Flow Charts

STRUNETCONCRETE DESIGN AIDS

Strunet.com: Concrete Beam Design V1.01 - Page 1


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Notations for Concrete Beam Design Flow Charts



a = depth of equivalent rectangular stress block, in.

ab = depth of equivalent rectangular stress block at balanced condition, in.amax = depth of equivalent rectangular stress block at maximum ratio of

tension-reinforcement, in.

As = area of tension reinforcement, in2.

As = area of reinforcement at compression side, in2.

b = width of beam in rectangular beam section, in.

be = effective width of a flange in T-section beam, in.

bw = width of web for T-section beam, in.

c = distance from extreme compression fiber to neutral axis, in.

cb = distance from extreme compression fiber to neutral axis at balanced

condition, in.Cc = compression force in equivalent concrete block.

Cs = compression force in compression reinforcement.

d = distance from extreme compression fiber to centroid of tension -sidereinforcement.

d = distance from extreme compression fiber to centroid of compression -

side reinforcement.

Es = modulus of elasticity of reinforcement, psi.

fc = specified compressive strength of concre te.

fy = specified tensile strength of reinforcement.

Mn

= nominal bending moment.

Mn bal = moment strength at balanced condition.

Mu = factored (ultimate) bending moment.

Ru = coefficient of resistance.

t = slab thickness in T-section beam, in.

1 = factor as defined by ACI 10.2.7.3.

c = concrete strain at extreme compression fibers, set at 0.003.

's = strain in compression-side reinforcement.

y = yield strain of reinforcement.

= ratio of tension reinforcement.

b = ratio of tension reinforcement at balanced condition.f = ratio of reinforcement equivalent to compression force in slab of T -

section beam.

max = maximum ratio of tension reinforcement permitted by ACI 10.3.3.

min = minimum ratio of tension reinforcement permitted by ACI10 .5.1.

reqd = required ratio of tension reinforcement.

= strength reduction factor.


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ACI 9.3.2.1

ACI 8.4.3

ACI 10.3.3

YESNO

try rectangularbeam with tensionand compression

steel

use rectangularbeam with tension

steel only

RectangularBeam

finding balancedmoment strength

Given:

b, d, fc', fy, Mu,Vu

NO YES

ACI 10.2.7.3

ACI 10.3.3

ACI 10.2.7.3

ACI 10.3.3


0 003

29 000 000

c

y

y

s

s

.

f

E

E , , psi

=

=

=

finding max

4000cf psi

cb

c y

c d

= +

1

4000

0 85 0 05 0 651000cf

. . .

= 1

0 85. =

1

1

0 85 87 000

87 000

c

b y y

. f ,

f , f

=

+

0 75max b. =

1b ba c=

0 75max ba . a=

0 9. =

bal 0 852max

n c max

aM . f ba d

=

bal u nM M

mid minu nV V>0 5 1 minnmin

u

Vx . L

V

=

0 5 1 min mid

mid

n u

min

u u

V Vx . L

V V

=

minx

v l bA N A=


Beam Shear Detailed Chart


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STOP. increasef'cordorb

smax= min. of:d/224"

h=

( )0 50 5

mid

d mid

u u

u u

V VV . L d V

. L

= +

( )0 5

0 5du

u

VV . L d

. L

=

duV

du cV V>

s u cV V V = 2c

u

VV

>

req'd

ss

VV

=

4req' ds c

V V>

2req' ds c

V V>

2bs c

let V V =

bn c sV V V = +

0 0midu

V .>

0 5 1 bn

b

u

Vx . L

V

=

0 5 1 b mid

mid

n u

b

u u

V Vx . L

V V

=

bx

maxs

50

v y

req' d

A fs

b=5000

50

v y

eq' d

c

A f

b f

=

v y

s

A f dV

s=

n c sV V V = +

minxNs

=


Beam Shear Detailed Chart (cont. 1)


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use: N @ smax

STOP. go tostirrups number

Loop as long:xmin>xk , and

sk

0 0midu

V .>minxNs

=

0 5 1 maxn

max

u

Vx . L

V

=

0 5 1 max mid

mid

n u

max

u u

V Vx . L

V V

=

maxx

1

req' d

v y

s

A f ds

V=

1s

1klet s s s= +

k max s s0 0009 24 12s yc bl . f d " =d r dbl k l

=

0 020 0003

b yc

db b yc

cc

. d fl . d f

f

< 3000cf psi

= 1 33s sl . l=s sl l

sl

>d aval l

= 6aval h

( )2 max s dbl l ,l =


Column Dowels Development

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