concrete beam design flow charts

The concrete beam design flow charts address the following subjects:

• For a rectangular beam with given dimensions: Analyzing the beam section 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 ess than the ultimate (factored) moment.

• For T-section concrete beam: Analyzing the beam T -section to determine

its 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 flanged

portion of the beam, i.e, the neutral axis N.A. depth is less than the slab 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 the

flange thickness, i.e. the neutral axis is located below the bottom of the 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, in order to handle repetitive beam shear reinforcement selection. See shear check introduction page for further details.

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

Introduction to Concrete Beam Design Flow Charts

STRUNETCONCRETE DESIGN AIDS

Strunet.com: Concrete Beam Design V1.01 -

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. A’s = area of reinforcement at compression side, in 2. 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 -side

reinforcement. d’ = distance from extreme compression fiber to centroid of compression -

side reinforcement. Es = modulus of elasticity of reinforcement, psi. f’c = 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. β1 = factor as defined by ACI 10.2.7.3. εc = concrete strain at extreme compression fibers, set at 0.003. ε 's = strain in compression-side reinforcement. εy = yield strain of reinforcement. ρ = ratio of tension reinforcement. ρb = ratio of tension reinforcement at balanced condition. ρf = ratio of reinforcement equivalent to compression force in slab of T -

section beam. ρmax = maximum ratio of tension reinforcement permitted by ACI 10.3.3. ρmin = minimum ratio of tension reinforcement permitted by ACI10 .5.1. ρreq’d = required ratio of tension reinforcement. φ = strength reduction factor.

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 YESACI 10.2.7.3

ACI 10.3.3

ACI 10.2.7.3

ACI 10.3.3


0 003

29 000 000

c

yy

s

s

.fE

E , , psi

ε

ε

=

=

=

finding maxρ

4000cf psi′ ≤

cb

c y

c d εε ε

= +

140000 85 0 05 0 65

1000cf. . .β′ − = − ≥

1 0 85.β =

1β

10 85 87 000

87 000c

by y

. f ,f , f

ρ β ′

= +

0 75max b.ρ ρ=

1b ba cβ=

0 75max ba . a=

0 9.φ =

bal 0 852max

n c maxaM . f ba dφ φ ′= −

bal u nM Mφ<

STRUNETCONCRETE DESIGN AIDS Moment Strength of Rectangular Concrete Beam


NO

ACI 10.5.2

YESNO

' maxreq dρ ρ> YESNO

ACI 10.5.1

STOP. go to rectangularbeam with tension and

compression steel

selectreinforcement, As

proceed toshear design

d

T

C

c b

a

0.85fc'

b

As

rectangular beamwith tension steel

only

YES

2u

uMRbdφ

=

0 85 21 10 85

c ureq' d

y c

. f Rf . f

ρ ′

= − − ′ 3

max of200

c

ymin

y

ff

f

ρ

′

=

req' d minρ ρ≥

1 33 req' d.ρ ρ=

minρ ρ<

minρ ρ= 1 33 req' d.ρ ρ= req' duseρ ρ=

sA bdρ=

0 85s y

c

A fa

. f b=

′

0 852n caM . f ba dφ φ ′= −


Rectangular Concrete Beam withTension Reinforcement


d'=2.5"

steel attension side

steel atcomp. side

selectreinforcement

As& A's

find the new d

proceed to checkcompression steel

yields

T

Ccc a

0.85fc'

b

Asd

A's

rectangular beamwith tension and

compression steel

1

Cs

cε

sε ′ d'

u u nbal

M M Mφ′ = −

( )( )0 85u

y c

Mf . f d d bd

ρφ

′′ =

′ ′− −

maxρ ρ ρ′= +

sA bdρ=

sA bdρ′ ′=

( )0 85s s y s

c s

A A f Aa. f b A

′− ′= +

′

1

acβ

=

s cc dc

ε ε′− ′ =


Rectangular Beam with Tension & Compression Reinforcement


compressionsteel yields

compressionsteel does NOT

yieldcompression steel may beneglected, and thus momentstrength is calculated based onthe tension steel only.Alternatively:


YESNO s yε ε′ >

alternatively

alternatively

1

( ) ( ) ( ) ( )( )

21

1

87 87 4 0 85 872 0 85

s y s s y s c s

c

A f A A f A . f b A dc

. f bβ

β

′ ′ ′ ′ ′− ± − +=

′

0 85c cC . f ab′=

s s c yc df E fc

ε′− = <

( )0 852n c s saM . f ba d A f d dφ φ ′ ′ ′= − + −

( )0 85s s y cC A f . f′ ′= −

( )2n c s uaM C d C d d Mφ φ ′= − + − ≥

( )n s yM A f d dφ ′ ′ ′= −

n n n ubal

M M M Mφ φ φ ′= + ≥


Rectangular Beam with Tension & Compression Reinforcement (cont.)


bw

As

be

t

T-SectionBeam

finding balancedmoment strength

@ a=tGiven:bw ,be ,d , f'c , fy , Mu Vu

ACI 9.3.2.1

NOYES

ACI 10.2.7.3

ACI 10.3.3

ACI 10.2.7.3

YESNO

let a=t

ACI 8.4.3

use T-Sectioncase 2

use T-Sectioncase 1

d

finding maxρ

4000cf psi′ ≤

1 0 85.β =1

40000 85 0 05 0 651000

cf. . .β′ − = − ≥

1β

10 85 87 000

87 000c

by y

. f ,f , f

ρ β ′

= +

( )0 85 c e wf

y w

. f b b tf b d

ρ′ −

=

( )0 75 wmax b f

e

b.b

ρ ρ ρ= +

2n ctM C dφ φ = −

u nM Mφ<

0 9.φ =

0 85c c eC . f b t′=

Moment Strength of T-Section BeamSTRUNET

CONCRETE DESIGN AIDS


NO

ACI 10.5.2

YES

ACI 10.5.1


verify depth ofcompression

block

STOP. go toT-Section case 2

STOP. usecompression steel

at T-Section

NO YES

Alternatively:

selectreinforcement, As

check momentstrength

continuation ofprevious sheet

YES

YES

NO

NO

T-Sectioncase 1

1

1

2ue

MuRb dφ

=

0 85 21 10 85

c ureq' d

y c

. f Rf . f

ρ ′

= − − ′

req' d maxρ ρ<

21 10 85

u

c

Ra d. f

= − − ′

3

max of200

c

ymin

y

ff

f

ρ

′

=

a t>

0 85s y

c

A fa

. f b=

′

0 852n caM . f ba dφ φ ′= −

0 85 c es

y

. f abAf′

=

s eA b dρ=

req' duseρ ρ=1 33 req' d.ρ ρ=minρ ρ=

minρ ρ<

1 33 req' d.ρ ρ=

req' d minρ ρ≥


T-Section Beam Case - 1


T-Sectioncase 2

YESNObw

As

be

ad

selectreinforcement,

As

STOP. revise toinclude

compression steel


t

( ) ( )2 2 0 520 85

e wu

c w w

t b b d . tMa d d. f b bφ

− − = − − − ′

[ ]0 85 cs w e wreq' d y

. fA ab t(b b )f

′= + −

s max emaxA b dρ=

s sreq' d maxA A≥

( )0 85

s ye w

w w

A f ta b b. f b b

= − −′

1 0 85c c wC . f b a′=

( )2 0 85c c e wC . f t b b′= −

1 22 2n c c ua tM C d C d Mφ = − + − ≥


T-Section Beam Case - 2

Introduction to Concrete Beam Shear Design



Concrete Beam Shear Design

Introduction and discussion:

The approach of the beam shear check chart is to define the nominal shear strength of the concrete, then compare it with the ultimate shear force at the critical section, and subsequent sections. Shear reinforcement calculation is performed, where applicable.

The shear charts are presented into two parts. One is the Shear Basic Chart, which is outlining the main procedures of the shear design in accordance with ACI applicable code provisions. The second, Shear Detailed Chart, is outlining the steps required for repetitive shear check. The detailed charts provide as much variables and or scenarios as needed to facilitate the creation of automated shear check applications.

The concept in selecting stirrups is based on an input of the bar diameter (db) of the stirrups to be used, usually #3, 4, or 5, as well as the number of legs and thus finding the spacing (s) required.

The shear chart intentionally did not include the following ACI provisions due to p ractical and economical justifications:

• Detailed method of ACI §11.3.2.1 for calculating nominal shear strength of concrete, vc . The reason is the value Vud/Mu is not constant along the beam span. Although the stirrups spacing resulting from the detailed method may be 1.5 larger than that of the direct method using ACI 11.3.1.1 at the critical section only, the use of the detailed method is not practically justified beyond this critical section, i.e. beyond distance d from the face of support.

• Shear reinforcement as inclined stirrups per §11.5.6.3, and bent up bars per §11.5.6.4 and §11.5.6.5. Only vertical stirrups per §11.5.6.2 are used, since other types of shear reinforcement are not economically justified.

bw = Width of beam (web) d = flexural depth of the beam, in. f’c = concrete compressive strength fct = average splitting tensile strength of lightweight concrete fy = reinforcement yield strength L = beam clear span, from support face to other support face. N = number of stirrups required within a given segment of the beam Nl = number of legs for each stirrup Vc = concrete nominal shear strength Vs = nominal shear strength provide by the shear reinforcement Vsb = nominal shear strength provided by shear reinforcement at the section where Vs is the

max permitted by ACI 11.12.1.1 . locating of this section is needed to define which

maximum s provisions applies, i.e. §11.5.4.1 or §11.5.4.3 Vs req’d = required nominal shear strength provided by shear reinforcement. Vu = factored shear force at the face of the beam support Vu d = factored shear force at distance d from the face of the support in

accordance with §11.3.1.1 this is the critical shear force provided that: • support is subjected to compressi ve force. • no concentrated load on the beam within the distance d.

Vu req’d = factored shear force at the mid-span of the beam, will not be zero if the beam is partially loaded with superimposed loads (i.e. live load on half the beam span)

φVn max = reduced shear strength of the beam section located along the beam span where minimum shear reinforcement is required in accordance with §15.5.5.1

s1 = spacing of stirrups within the critical section. sk = spacing of stirrups within any section subsequent t o the critical section. smax = maximum stirrups spacing permitted by §11.5.4.1 or §11.5.4.3 sreq’d = required stirrups spacing at the section considered xb = the distance along the beam at which Vsb occurs. for any beam section within the

distance xb, Vsb is based on §11.5.4.3, otherwise is based on §11.5.4.1 xmin = distance from the face of the support along the beam span after which minimum shear

reinforcement in accordance to §11.5.5.1 is no longer required. xmax = distance from the face of the support along the beam span after which stirrups shall be

placed with the maximum spacing per §11.5.4.1, and §11.5.4.3 s = incremental in stirrups spacing between the subsequent sections, suggested to be 1,

2, and or 3 inches

Notations for Concrete Beam Shear Design



∆

Normal or LightWt ConcreteLIGHT

NO

ACI 11.3

ACI 11.3.1.1

ACI 11.2.1

ACI 11.2.1.2YES

ACI 11.2.1.1

ACI 9.3.2.3

NORMAL

Finding Vc

STOP. increasef'c or d or bw

smax is the min. of: d/2 24"

h=<10"h<2.5th<0.5bw

STOP. no min.shear reinf. req'd


' 100cf psi≤

smax= min. of: d/2 24"

smax= min. of: d/4 12"

Smax s is the min. of:smaxsreq'd

loop for othervalues of Vu

loop for othervalues of Vu

ACI 11.1.1

ACI 11.1.1

ACI 11.12.1.1

ACI 11.12.1.1ACI 11.1.2

ACI 11.5.4.1

ACI 11.2.1.1ACI 11.5.4.3

ACI 11.5.6.8

ACI 11.5.6.2

ACI 11.1.1

ACI 11.5.6.2

ACI 11.5.5.1

ACI 11.5.5.3

ACI 11.5.5.1


ctis f given?

2 'c c wV f b d=2

6 7

6 7

ctc w

'ctc

fV b d.

f f.

=

≤

( )( )

0 75 2

0 85 2

'c c w

'c c w

All Light wt : V . f b d

Sand Light Wt : V . f b d

− =

=

0 85.φ =cVφuV

u cV Vφ>

s u cV V Vφ φ= −2c

uVV φ

>

req' d

ss

VV φφ

=

4req' ds cV V>

n c sV V Vφ φ φ= +

v ys

A f dV

s=

500050

v yreq' d

w c

A fs

b f

= ′ 50v y

req' dw

A fs

b=

req' d

v y

s

A f ds

V=


2req' ds cV V>

ACI 11.5.4.1


Beam Shear Basic Chart


Shear BeamCheck

Normal or LightWt Concrete

?ctis f given

LIGHT

NO

ACI 11.3

ACI 11.3.1.1

ACI 11.2.1

ACI 11.2.1.2

YES

ACI 11.2.1.1

ACI 9.3.2.3

NORMAL

Finding Vc

STOP. xmindoes not exist

1

2 'c c wV f b d=2

6 7

6 7

ctc w

'ctc

fV b d.

f f.

=

≤

( )( )

0 75 2

0 85 2

'c c w

'c c w

All Light wt : V . f b d

Sand Light Wt : V . f b d

− =

=

0 85.φ =cV

cVφ

2min

cn

VV φφ =

0 0miduV .>

mid minu nV Vφ>0 5 1 minnmin

u

Vx . L

Vφ

= −

0 5 1 min mid

mid

n umin

u u

V Vx . L

V Vφ −

= − −

minx

v l bA NA=


Beam Shear Detailed Chart


STOP. increasef'c or d or b

smax = min. of: d/2 24"

h=<10"h<2.5th<0.5b



Smax

' 100cf psi≤

s = min. of:smaxsreq'd

use: N @ s


1

2

0 0miduV .>

( )0 50 5

mid

d mid

u uu u

V VV . L d V

. L−

= − +

( )0 50 5d

uu

VV . L d. L

= −

duV

du cV Vφ>

s u cV V Vφ φ= −2c

uVV φ

>

req' d

ss

VV φφ

=

4req' ds cV V>

2req' ds cV V>

2bs clet V V=

bn c sV V Vφ φ φ= +

0 0miduV .>

0 5 1 bnb

u

Vx . L

Vφ

= −

0 5 1 b mid

mid

n ub

u u

V Vx . L

V Vφ −

= − −

bx

maxs

50v y

req' d

A fs

b=5000

50v y

req' dc

A fs

b f

= ′

v ys

A f dV

s=


minxNs

=


Beam Shear Detailed Chart (cont. 1)


use: N @ smax

STOP. go tostirrups number

Loop as long:xmin>xk , and

sk<smax

2

3

4

max

v ys

max

A f dV

s=

max maxn c sV V Vφ φ φ= +

d maxu sV Vφ>

0 0miduV .>minxN

s=

0 5 1 maxnmax

u

Vx . L

Vφ

= −

0 5 1 max mid

mid

n umax

u u

V Vx . L

V Vφ −

= − −

maxx

1req' d

v y

s

A f ds

V=

1s

1 klet s s s= +∆

k maxs s<

k

v ys

k

A f dV

s=

k kn c sV V Vφ φ φ= +





STOP. go tostirrups number

stirrupsnumber

Loop until:xk+1=xmin , andxk=xmax

revise Smax

3

4 0 0miduV .>

0 5 1 k mid

mid

n uk

u u

V Vx . L

V Vφ −

= − − 0 5 1 kn

ku

Vx . L

Vφ

= −

kx

k bx x>

k minx x>

1 k klet s s s+ = +∆

1k ks s +=

11

kxNs

=

1k kk

k

x xNs

+ −=



concrete beam design flow charts

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