Concrete Beam Design Flow Charts

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

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Notations for Concrete Beam Design Flow Charts a ab amax As A’s b be bw c cb Cc Cs d d’ Es f’c fy Mn M n bal Mu Ru t

β1 εc ε 's εy ρ ρb ρf ρmax ρmin ρreq’d φ

= depth of equivalent rectangular stress block, in. = depth of equivalent rectangular stress block at balanced condition, in. = depth of equivalent rectangular stress block at maximum ratio of tension-reinforcement, in. = area of tension reinforcement, in2. = area of reinforcement at compression side, in 2 . = width of beam in rectangular beam section, in. = effective width of a flange in T-section beam, in. = width of web for T-section beam, in. = distance from extreme compression fiber to neutral axis, in. = distance from extreme compression fiber to neutral axis at balanced condition, in. = compression force in equivalent concrete block. = compression force in compression reinforcement. = distance from extreme compression fiber to centroid of tension -side reinforcement. = distance from extreme compression fiber to centroid of compression side reinforcement. = modulus of elasticity of reinforcement, psi. = specified compressive strength of concre te. = specified tensile strength of reinforcement. = nominal bending moment. = moment strength at balanced condition. = factored (ultimate) bending moment. = coefficient of resistance. = slab thickness in T-section beam, in. = factor as defined by ACI 10.2.7.3. = concrete strain at extreme compression fibers, set at 0.003. = strain in compression-side reinforcement. = yield strain of reinforcement. = ratio of tension reinforcement. = ratio of tension reinfo rcement at balanced condition. = ratio of reinforcement equivalent to compression force in slab of T section beam. = maximum ratio of tension reinforcement permitted by ACI 10.3.3. = minimum ratio of tension reinforcement permitted by ACI10 .5.1. = required ratio of tension reinforcement. = strength reduction factor.

Strunet.com: Concrete Beam Design V1.01 - Page 2

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Moment Strength of Rectangular Concrete Beam

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Rectangular Beam Given: b, d, fc', fy, Mu ,Vu

finding balanced moment strength

ACI 10.3.3

finding ρmax ACI 10.2.7.3

ε c = 0.003 εy =

fy

NO

Es

fc′ ≤ 4000psi

Es = 29,000,000 psi

 εc cb = d  ε +ε y  c

 fc′ − 4000   ≥ 0.65  1000 

β1 = 0.85 − 0.05 

  

β1 ab = β1cb

ACI 8.4.3

ACI 10.3.3

ρb =

amax = 0.75ab

ACI 9.3.2.1

φ = 0.9

0.85fc′  87,000 β1  fy  87,000 + fy

ACI 10.3.3





a



φMn bal = φ  0.85fc′bamax  d − max   2   

NO

try rectangular beam with tension and compression steel

Mu < φMn bal

YES

ACI 10.2.7.3

YES

use rectangular beam with tension steel only

Strunet.com: Concrete Beam Design V1.01 - Page 3

ρmax = 0.75ρb

  

β1 = 0.85

Rectangular Concrete Beam with Tension Reinforcement

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rectangular beam with tension steel only

Ru =

ACI 10.5.1

3 fc′

ρmin = max of

ρreq' d =

fy 200 fy

Mu φ bd 2

0.85fc′  2Ru  1− 1−  fy  0.85fc′ 

ρreq' d ≥ ρmin

NO

YES

ρ = 1.33ρreq' d NO

ρ = ρmin

ρ < ρmin

YES

ρreq 'd > ρmax

NO

ρ = 1.33ρreq' d

useρ = ρreq' d

YES

STOP. go to rectangular beam with tension and compression steel

ACI 10.5.2

As = ρbd 0.85fc '

Asfy 0.85fc′b

d

a=

C

cb

a

select reinforcement, As

As



 

a 

φMn = φ  0.85fc′ba  d −   2 

proceed to shear design Strunet.com: Concrete Beam Design V1.01 - Page 4



T

b

Rectangular Beam with Tension & Compression Reinforcement

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rectangular beam with tension and compression steel

Mu′ = Mu − φMn

bal

d'=2.5"

Mu′ φ ( fy − 0.85fc′ ) ( d − d ′) bd

εc

ρ = ρmax + ρ′

0.85fc'

ε s′

d'

ρ′ =

steel at comp. side

As′ = ρ′bd

Cc

c

As = ρbd

A's

d

steel at tension side

a

Cs

As T

select reinforcement As& A's

find the new d

proceed to check compression steel yields

a=

( As − As′ ) fy 0.85fc′b c=

+

As′ As

a

β1

 c − d′    c 

ε s′ = ε c 

1 Strunet.com: Concrete Beam Design V1.01 - Page 5

b

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Rectangular Beam with Tension & Compression Reinforcement (cont.)

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1

NO

s y

− 87As′ ) ±

YES

compression steel does NOT yield

compression steel may be neglected, and thus moment strength is calculated based on the tension steel only. Alternatively:

(A f c=

ε s′ > ε y

compression steel yields

(A f

Cc = 0.85fc′ab

− 87As′ ) + 4 (0.85fc′β1b )(87 As′d ′ ) 2

s y

2 ( 0.85fc′β1b )

 c − d′  fs = Esεc   < fy  c 

 a   φMn = φ  0.85fc′ba  d −  + As′fs (d − d ′)   2   

Cs = As′ ( fy − 0.85fc′ )



 

a







φMn = φ Cc  d −  + Cs ( d − d ′) ≥ Mu 2 

alternatively

alternatively

φMn′ = As′fy ( d − d ′)

φMn = φMn + φMn′ ≥ Mu bal

proceed to shear design

Strunet.com: Concrete Beam Design V1.01 - Page 6

Moment Strength of T-Section Beam

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ACI 10.3.3

finding ρmax

T-Section Beam

ACI 10.2.7.3

finding balanced moment strength @ a=t

NO

ACI 10.2.7.3

fc′ ≤ 4000psi

 fc′ − 4000   ≥ 0.65  1000 

β1 = 0.85 − 0.05 

Given: bw ,be ,d , f'c , fy , Mu Vu let a=t

β1 ACI 8.4.3

ρb =

Cc = 0.85fc′bet ACI 9.3.2.1

0.85fc′  87,000 β1  fy  87,000 + fy

φ = 0.9  

t 

φMn = φCc  d −  2

ρf =

0.85fc′ ( be − bw ) t fy bw d

ρmax = 0.75

YES

be

use T-Section case 1

d

use T-Section case 2

Mu < φMn

bw ( ρb + ρf ) be

t

NO

  

As bw

Strunet.com: Concrete Beam Design V1.01 - Page 7

YES

β1 = 0.85

T-Section Beam Case - 1

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T-Section case 1

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Ru =

ρreq' d =

Mu φ bed 2

0.85fc′  2Ru  1− 1−  fy  0.85fc′ 

ρreq' d < ρmax

NO

YES verify depth of compression block

STOP. use compression steel at T-Section

 2Ru  a = d 1− 1−  0.85fc′   ACI 10.5.1

3 fc′

ρmin = max of

NO

a>t

YES

fy continuation of previous sheet

200 fy

ACI 10.5.2

ρreq' d ≥ ρmin

NO

STOP. go to T-Section case 2 1 YES

ρ = 1.33ρreq' d NO

ρ < ρmin

ρ = ρmin

select reinforcement, As

YES

ρ = 1.33ρreq' d

check moment strength

useρ = ρreq' d a=

As = ρbed Alternatively:



0.85fc′abe As = fy

Strunet.com: Concrete Beam Design V1.01 - Page 8

Asfy 0.85fc′b  

a 

φMn = φ  0.85fc′ba  d −   2 

1

proceed to shear design



T-Section Beam Case - 2

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T-Section case 2

 2Mu 2t ( be − bw )( d − 0.5t )  a = d − d2 −  −  bw  0.85fc′φ bw 

0.85fc′ [abw + t(be − bw )] fy

be

a

req' d

=

t

As

max

NO

As

= ρmax bed

req' d

d

As

As

≥ As

YES

max

bw

STOP. revise to include compression steel

select reinforcement, As

a=

Asfy t − ( be − bw ) 0.85f ′bw bw

Cc1 = 0.85fc′bwa

Cc 2 = 0.85fc′t ( be − bw )  

a

 

t 

φMn = Cc1  d −  + Cc 2  d −  ≥ Mu 2 2 

proceed to shear design

Strunet.com: Concrete Beam Design V1.01 - Page 9

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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 ultimat e shear force at the critical section, and subs equent sections. Shear reinforc ement calculation is performed, where applic able. 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 accordanc e 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 automat ed 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 economic al justifications: •

•

Detailed method of ACI §11.3.2.1 for calculating nominal shear strength of concrete, vc . The reason is the value V ud/Mu is not constant along the beam span. Although the stirrups spacing resulting from the detailed met hod 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 det ailed 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.

Strunet.com: Concrete Beam Design V1.01 - Page 10

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Notations for Concrete Beam Shear Design

bw d f’c f ct fy L N Nl Vc Vs Vsb

Vs req’d Vu Vu d

= = = = = = = = = = =

Width of beam (web) flexural depth of the beam, in. concrete compressive strength average splitting tensile strength of lightweight concrete reinforcement yield strengt h beam clear span, from support face to other support face. number of stirrups required wit hin a given segment of the beam number of legs for each stirrup concrete nominal shear strengt h nominal shear strength provide by the shear reinforcement 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 = required nominal shear strength provided by shear reinforcement. = factored shear force at the face of the beam support = factored shear force at distance d from the fac e of the support in accordance with §11.3. 1.1 this is the critical shear force provided that:

Vu req’d

=

φVn max

=

s1 sk s max s req’d xb

= = = = =

x min

=

x max

=

∆s

=

• support is subjected to compressi ve force. • no concentrated load on the beam within the distance 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) reduced shear strength of the beam section locat ed along the beam span where minimum shear reinforcement is required in accordance with §15.5.5.1 spacing of stirrups within the critical section. spacing of stirrups within any section subsequent t o the critical section. maximum stirrups spacing permitted by §11.5.4.1 or §11.5.4.3 required stirrups spacing at the section considered the distance along the beam at which Vsb occurs. for any beam section within the distance xb, V sb is based on §11.5. 4.3, otherwise is based on §11.5.4.1 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. 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 incremental in stirrups spacing between the subsequent sections, suggested to be 1, 2, and or 3 inches

Strunet.com: Concrete Beam Design V1.01 - Page 11

ACI 11.3

Beam Shear Basic Chart

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Finding Vc

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ACI 11.2.1 Normal or Light Wt Concrete

LIGHT

is fct given?

NO

ACI 11.2.1.2

(

All − Light wt : Vc = 0.75 2 f bwd ' c

(

)

Sand Light Wt : Vc = 0.85 2 fc' bwd

NORMAL

YES ACI 11.2.1.1

ACI 11.3.1.1

 f  Vc = 2  ct  bw d  6 .7   fct  '   ≤ fc 6 7 .  

)

Vc = 2 fc' bw d

ACI 9.3.2.3

φ = 0.85

φVc

Vu ACI 11.1.1

Vu > φVc

ACI 11.5.5.1

Vu >

φVc

ACI 11.5.4.1 smax is the min. of: d/2 24"

h= 4Vc

smax= min. of: d/2 24"

s is the min. of: smax sreq'd

Av fy d Vsreq ' d

ACI 11.1.1 loop for other values of V u

Strunet.com: Concrete Beam Design V1.01 - Page 12

φVs φ ACI 11.12.1.1

ACI 11.12.1.1

fc' ≤ 100 psi sreq' d

φVs = Vu − φVc

ACI 11.5.5.1

2

STOP. no min. shear reinf. req'd

loop for other values of Vu

ACI 11.1.1

φVn = φVc + φVs

STOP. increase f'c or d or bw

Beam Shear Detailed Chart

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Shear Beam Check

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ACI 11.3 Finding Vc ACI 11.2.1

NO ACI 11.2.1.2

(

LIGHT

Normal or Light Wt Concrete

is f ct given ?

YES

All − Light wt : Vc = 0.75 2 fc' bwd

(

)

Sand Light Wt : Vc = 0.85 2 fc' bwd

NORMAL

ACI 11.3.1.1

ACI 11.2.1.1

 f  Vc = 2  ct  bw d  6 .7   fct  '   ≤ fc  6 .7 

)

Vc = 2 fc' bw d

ACI 9.3.2.3

Vc

φ = 0.85

φVc φVn = min

φVc 2

Vumid > 0.0  φVnmin  xmin = 0.5L 1 −  Vu  

Vumid > φVnmin

xmin

 φVnmin − Vumid = 0.5L 1 −  Vu − Vumid 

xmin Av = Nl Ab 1 Strunet.com: Concrete Beam Design V1.01 - Page 13

  

STOP. xmin does not exist

Beam Shear Detailed Chart (cont. 1)

1

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Vumid > 0.0  Vu − Vumid  Vud = ( 0.5L − d )   + Vumid  0.5L 

 V  Vud = ( 0.5L − d )  u   0.5L 

Vud

Vu >

Vud > φVc

φVc

φVs = Vu − φVc

2

STOP. no min. shear reinf. req'd

h= 4Vc

Smax

Vsreq' d > 2Vc

fc' ≤ 100 psi sreq' d

A f  5000  = vy  50b  fc′ 

φVs φ

sreq' d =

Av fy

smax = min. of: d/2 24"

STOP. increase f'c or d or b

let Vsb = 2Vc

50b

φVn = φVc + φVs

b

s = min. of: smax sreq'd

Vs =

Av fy d s

Vumid > 0.0  φVnb  xb = 0.5L 1 −  Vu  

 φVnb − Vumid xb = 0.5L 1−  Vu − Vumid 

φVn = φVc + φVs N=

xb

xmin s

smax use: N @ s Strunet.com: Concrete Beam Design V1.01 - Page 14

2

  

Beam Shear Detailed Chart (cont. 2)

2

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Vsmax =

φVn

max

Av fy d smax

= φVc + φVsmax

Vud > φVsmax N=

xmin s

use: N @ smax

Vumid > 0.0  φVnmax  xmax = 0.5L 1 −  Vu  

 φVnmax − Vumid xmax = 0.5L 1 −  Vu − Vumid 

xmax s1 =

Av fy d Vsreq ' d

s1 let sk = s1 + ∆s

sk < smax Loop as long: xmin>xk , and sk 0.0  φVnk  xk = 0.5L 1−  Vu  

 φVnk − Vumid xk = 0.5L 1−  Vu − Vumid 

  

xk xk > xb smax = min. of: d/4 12"

revise Smax

xk > xmin let sk +1 = sk + ∆s sk = sk +1 stirrups number

N1 =

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

Strunet.com: Concrete Beam Design V1.01 - Page 16

Nk =

xk s1

xk +1 − xk sk

STOP. go to stirrups number