Concrete Beam Design Flow Charts

STRUNET CONCRETE DESIGN AIDS

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 2: Rectangular Rectangular Beam with tension and compression reinforc reinforcement. ement. This This case may exist if the moment moment strength is l ess than the ultimate (factored) moment.

For T-section concrete concrete beam: Analyzing Analyzing the beam T -section -section to determine determine its moment strength and thus defining the beam section to be 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 1: The The depth dept h of the compression 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 The depth dept h of the compression block block is i s deeper t han th t he flange thickness, i.e. the neutral axis is located below the bottom of the the slab. slab. This This case ex e xists if the t he moment moment strength of T -sectio -section n beam beam is less that the ultimate (factored) moment.

Beam Section S hear hear Strength: Strength: two t wo separat separate e charts charts outline outline in i n det ails Shear check. check. One is a basic shear shear ch c heck, an a nd two is detaile de tailed d shear check, check, i n order to handle repetitive beam shear reinforcement selection. See shear check introduction page for further details.

In any of the cases mentio mention ned above, detailed detailed procedure procedure s and and equations equations are shown within the charts cover all design aspects of the element under investigations, with ACI respective provisions. Strunet.com: Concrete Concrete Beam Design Design V1.01 - Page 1


Notations for Concrete Beam Design Flow Charts

a ab amax

= = =

A s A’ s b be bw c c b

= = = = = = =

C c c C s d

= = =

d’

=

E s f’ c c f y y M n M n ba b al M u u R u u t β 1 ε c c 's ε '

= = = = = = = = = = = = = = =

ε y y ρ ρ b ρ f f ρ max max ρ min min ρ req’d req’d φ

= = = =

depth of equivalent eq uivalent rectangular stress b lock, in. depth of equivalent eq uivalent rectangular stress b lock a t balanced conditio co ndition, n, i n. depth of equivalent eq uivalent rectangular stress b lock at ma ma ximum ratio of tension-reinforcement, in. area of tension reinforcement, re inforcement, i n 2. area of reinforcement at compression co mpression side, in 2 . width of beam i n rectangular rectangular beam beam section, in. effective effective width width of a flange i n T -sectio -section n beam, i n. width of web for T -section -section beam, in. distance from extreme compressio com pressio n fiber to to ne ne utral a xis, i n. distance from from extreme compression compressio n fiber to neutral a xis at balanced ba lanced condition, in. compression force i n equiv eq uiva a lent concrete b lock. compression force i n compression compressi on reinforcement. distance distance from from extreme extreme compression compression fiber to to centroid of tension tension -side reinforcement. distance distance from from extrem extreme e compression compression fiber to to centroid of compr compression ession side reinforcement. modulus of elasticity elastici ty of reinforcement, psi. specified compressive compressive strength of concre te. specified specifi ed tensile tensile strength stre ngth of reinforcement. nomi nomi nal bending moment. moment strength stre ngth at balanced ba lanced condition. co ndition. factored (ultimate) bending be nding moment. coefficient of resistance. slab thickness in T-sectio T-section n beam, i n. factor factor as defi defi ned by ACI AC I 10.2.7.3. 10.2.7.3. concret concrete e strain at extreme extreme compression compression fibers, fibers, set at 0.003. strain in compression -side reinforcement. yield s train of reinforcement. ratio of tension reinforcement. ratio of tension reinfo rcement at balanced condition. co ndition. ratio ratio of reinforcement reinforcement equi equi valent to to compression compression force force i n slab of T section beam. maxim maximum um ratio o f tensio tension n reinforcement permitted b y AC I 10.3.3. minimum minimum ratio ratio of tension tension reinforcement reinforcement permitted permitted by ACI10 ACI10 .5.1. .5.1. required ratio of tension reinforcement. strength reducti reduction on factor.

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

STRUNET

Moment Strength of Rectangular Concrete Beam

CONCRETE DESIGN AIDS

Rectangular Beam Given: b, d, f c ' , f y , M u ,V u

finding balanced moment strength

ACI ACI 10.3 .3

finding ρ max ACI ACI 10. 2.7.3

ε c = 0.003 ε y =

f y

NO

E s

fc ′ ≤ 4000psi

ACI ACI 10.2 .7.3

Es = 29, 000,000 psi



 f c ′ − 4000  β 1 = 0.85  ≥ 0.65  1000 

β 1 = 0.85 − 0.05 

  ε ε + c y  

cb = d  

ε c

β 1

ab = β 1c b

ACI 8.4.3

ACI ACI 10.3 .3

ρb =

amax = 0.75ab

ACI ACI 9.3 .2.1

φ = 0.9

0.85fc ′ fy

 87,000   87,000 + f y   

β 1 

ACI ACI 10 .3.3

 amax  max   φMn bal = φ  0.85fc′bamax  d −  2   

NO

try rectangular beam with tension and compre compressi ssion on steel

YES

Mu < φ M n bal bal

YES

use rectangular beam with tension steel only


ρmax = 0.75ρ b

Rectangular Concrete Beam with Tension Reinforcement


rectangular beam with tension steel only

R u =

ACI ACI 10 .5.1

3 f c ′ ρ min = max of

ρ req' req' d =

M u φ bd 2

0.85fc′  2R u  1 1 − −   fy  0.85f c ′ 

f y 200 f y

ρreq' d ≥ ρ min

NO

YES

ρ = 1.33ρ req' req' d

NO

ρ = ρ min

ρ < ρ min min

YES

ρreq'd > ρ max

NO


use use ρ = ρ req' req' d

YES

STOP. go to rectangular beam with tension and compression steel

ACI ACI 10.5 .2

A s = ρ bd

'

0.85f c

select reinforcement, As a

a=

b

c

As f y 0.85fc ′b

C

d

As

a    φMn = φ  0.85fc ′ba  d −   2    proceed to shear design Strunet.com: Concrete Concrete Beam Design Design V1.01 - Page 4

T b

Rectangular Beam with Tension & Compression Reinforcement


rectangular beam with tension and compression steel

Mu′ = Mu − φ M n bal bal

d ' =2.5"

ρ ′ =

M u ′ φ ( fy − 0.85fc ′ ) (d − d ′) bd

ε

c

ρ= ρmax + ρ′

0.85f c'

ε ′

' d

s

C s

steel at tension side

A= ρ bd

c

s

a

C c

A' s

d

steel at comp. side

′A= ρ ′ bd

A s

s

T b

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 β 1

 c − d ′    c 

ε s′ = ε c 

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

STRUNET

Rectangular Beam with Tension & Compression Reinforcement Reinforcement (cont.) (cont. )


1

NO

ε s′ > ε y

YES

compression steel does NOT yield

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

compression steel yields

Cc = 0.85fc ′ab c =

( A f − 87 A′ ) ± ( A f − 87 A′ ) s y

s

s y

s

2

+ 4 (0 8 . 5 fc′ β 1 b) (87 As′ d′ )

2 ( 0.85 85fc ′ β 1b )

 c − d ′   < f y c  

fs = Esε 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







φ Mn′ = As′fy ( d − d ′)

φMn = φMn + φ Mn′ ≥ Mu bal

proceed to shear design


Moment Strength of T-Section Beam


ACI ACI 10.3.3 10 .3.3

finding ρ max

T-Section Beam

ACI 10.2.7.3

NO

finding balanced moment strength @ a=t

fc ′ ≤ 4000psi

ACI 10.2.7.3

 f c ′ − 4000   ≥ 0.65  1000 

β 1 = 0.85 − 0.05 

Given: bw ,b e ,d , f' c , f y , M u V u let a=t

β 1 ACI ACI 8.4 .3

ρb =

Cc = 0.85fc′bet

0.85fc ′ fy

 87,000  β 1   87,000 + f y   

ACI 9.3.2.1

φ = 0.9 t   φ Mn = φ Cc  d −  2 

ρ f =

0.85fc′ ( be − bw ) t fy bw d

ρax = 0.75 m

NO

use T-Sect T-Section ion case 2

Mu < φ M n

YES YES

bw ( bρ+ fρ) be

YES YES

be

use T-Sect T-Section ion case 1

t

d

A s

bw

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

β 1 = 0.85

T-Section Beam Case - 1

STRUNET

T-Section case 1


R u =

ρ req' req' d =

NO

Mu φ bed 2

0.85fc′  fy

 1− 1 −  

ρreq' d < ρ max

2R u 



0.85f c ′ 

YES verify depth of compression block

STOP. use compression steel at T-Sectio T-Section n

  

a = d 1− 1 −

2R u   0.85f c ′ 

ACI 10.5.1

3 f c ′ ρ min = max of

NO

a > t

YES

f y continuation of previous sheet

200 f y

STOP. go to T-Section case 2 1

ACI ACI 10.5 .2

NO

ρreq' d ≥ ρ min

YES


NO

ρ < ρ min min

ρ = ρ min min

select reinforcement, As

YES


check moment strength

use use ρ = ρ req' req' d a=

A bd s = ρ e

Alternatively:

0.85fc ′b

a    φMn = φ  0.85fc ′ba  d −   2   

0.85fc′abe As = f y


Asf y

1


T-Section Beam Case - 2


T-Section case 2



2t ( be − bw ) (d − 0.5t )  2M u −  ′ φ 0 . 85 f b b c w w  

a = d − d 2 − 

A s req' req' d

=

0.85f c ′ [ abw + t( be − bw ]) f y

be

t

a

= ρ max be d

As

d

max max

NO

A s

≥ As

As req ' d

max

STOP. revise to include compression steel

YES bw

select reinforcement, As

a=

Asf y 0.85f ′bw

−

t bw

( be − bw )

Cc1 = 0.85fc′bw a

Cc2 = 0.85fc′t ( be − bw )

a t    φ Mn = Cc1  d −  + Cc 2  d −  ≥ Mu 2 2  




Introduction to Concrete Beam Shear Design

Concrete Bea m Shear Design Design Introduction Introduction a nd discussio discussio n: The approach of the beam shear check chart is to define the nominal shear strength of the concrete, concrete, the n compare it with the ultimate shear force at t he critical section, and subsequent sections. Shear reinforcement reinforcement calculation calculation is performed, performed, where applic applic able. 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 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 automat automat ed shear check applications. applications. The concept in selecting stirrups is based on an input of the bar diameter ( d b) 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 intentionally did not include the following 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, v c . The reason is the value V ud /M u 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 stirrups p er §11.5.6.3, §11.5.6.3, a nd bent up bars pe r §11.5.6.4 and §11.5.6.5. §11.5.6. 5. Only vertical vertical stirrups per §11.5.6. 2 are used, used, since other types of shear she ar reinforcement reinforcement ar e not econ omically omically justified. justified.



Notations for Concrete Beam Shear Design

b w d

= Width of beam (web) = flexural depth of the beam, in.

f’ c f ct

= concrete concrete comp ressive strength = average splitting tensile strength of lightweight concrete

f y y L N

= reinforcement reinforcement y ield strengt strengt h = beam beam cle ar span, from support support face t o other support f ace. = number of stirrups required within a given segment of the beam

N l l V c

= number of legs for each stirrup = concrete concrete n ominal shear strength

V s V sb sb

= nominal shear strength provide provide by the shear reinforcement = nominal shear shear st rength provide provide d by shear reinforcement reinforcement at t he section where Vs is the max permitted by ACI 11.12.1.1 . locating of this section is needed to define which

V s req’d V u V u d

maximum s provisions applies, i.e. §11.5.4.1 or §11.5.4.3 = required nominal nominal shear strength provided by shear reinforcement. = factored shear force at the face of the beam support = factored shear force at distance d from d from the face of the support in accordance with §11.3.1.1 this is the critical shear force provided that: •

support is s ubjected ubjected to c ompressi ve force.

V u req’d

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

V n max φ V

partially loaded with superimposed loads (i.e. live load on half the beam span) = reduced shear strength of the beam section locat locat ed along the beam span where where

s1 s k

minimum shear reinforcement is required in accordance with §15.5.5.1 = spacing of stirrups within the critical section. = spacing of stirrups within within any section subsequent t o the critical section.

s max s req’d

= maximum stirrups spacing permitted by §11.5.4.1 or §11.5.4.3 = required stirrups stirrups spacing at the sect ion considered

x b x min

= the distance along the beam at which V sb occurs. for any beam section within the distance x distance x b, 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

x max

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

•

∆s

placed placed with the maximum maximum spacing per §11.5.4. §11.5.4. 1, and §11.5.4. §11.5.4. 3 = incremental incremental in stirrups s pacing pacing between th e subsequent s ections, suggested to be 1, 2, and or 3 inches


ACI 11.3

Beam Shear Basic Chart

STRUNET

Finding V c


ACI ACI 11.2 .1 Normal or Light Wt Concrete

LIGHT

is fct given?

NO

ACI ACI 11.2.1 .2

YES ACI ACI 11. 2.1.1

)

(

ACI ACI 11.3 .1.1

 f ct   bw d 6 . 7  

All − Light wt : Vc = 0.75 75 2 fc' bw d

Vc = 2 

)

(

NORMAL

Sand Li Light Wt Wt : Vc = 0.85 85 2 fc' bw d

Vc = 2 fc' bw d

 f ct  '   ≤ f c  6.7  ACI 9.3.2.3

V u

φ = 0.85

φ V c

ACI ACI 11.1 .1

Vu > φ V c

ACI ACI 11. 5.5.1

V u >

φ V c

ACI ACI 11.5 .4.1 smax is the the min. min. of: d/2 24"

h=<10" h<2.5t h<0.5b w

V sreq' = req' d

ACI ACI 11.2 .1.1

sreq' req' d =

psi ACI ACI 11.5 .5.3

Av f y  5000    50bw  f c ′ 

sreq' req' d =

Av f y 50bw

ACI ACI 11.5 .4.1

ACI ACI 11. 5.6.8 ACI ACI 11.5.4 .3

smax = min. min. of: of: d/4 12"

ACI ACI 11. 5.6.2

s=

Avf y d s

φVn = φVc + φ V s

> 2V c Vsreq' req' d

S max

ACI ACI 11.5.6 .2

V s =

> 4V c Vsreq' req' d

smax = min. min. of: of: d/2 24"

s is the min. min. of: of: smax sreq'd

Avf y d V sreq'd req'd

ACI ACI 11.1 .1 loop for other values of V of V u


φ ACI ACI 11.12.1 .1

ACI ACI 11.1 2.1.1

≤f 100 100

φ V s

STOP. STOP. no min. min. shear shear reinf reinf.. req'd req'd

ACI 11.1.2 ' c

φVs = Vu − φ V c

ACI ACI 11.5.5 .1

2


loop for other values of V of V u

ACI ACI 11.1 .1

φVn = φVc + φ V s

STOP. increase f' c or d or d or or b b w

Beam Shear Detailed Chart

STRUNET

Shear Beam Check


ACI 11.3 Finding V c ACI 11.2.1

NO

LIGHT

Normal or Light Wt Concrete

is f ct giv given ?

YES

ACI ACI 11.2 .1.2

NORMAL

ACI ACI 11. 2.1.1

)

(

 f ct   bw d  6.7 

All − Light wt : Vc = 0 .75 75 2 fc' bw d

Vc = 2 

)

(

ACI ACI 11. 3.1.1

Sand Light Wt : Vc = 0.85 85 2 fc' bw d

Vc = 2 fc' bw d

 f ct  '   ≤ f c  6.7  ACI ACI 9.3 .2.3

V c

φ = 0.85

φ V c φ V nmin =

φ V c

2

Vu mid > 0.0



xmin = 0 .5 L1 −



φ V nmin 

Vumid > φ V nmin

 V u  xmin

 φ Vn −V u = 0 .5 L1 −  Vu − V u  min

mid

mid

x min Av = Nl Ab 1


  

STOP. x min does not exist

Beam Shear Detailed Chart (cont. 1)

1


Vu mid d > 0.0

 Vu − V u   + Vu 0 . 5 L  

 V  Vu d = ( 0.5L − d )  u   0.5L 

Vud = ( 0.5L − d ) 

mid

mid

V u d Vud > φ V c V u >

φ V c

φVs = Vu − φ V c

2


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

V sreq'd =

min. of: smax = min. d/2 24"


φ V s φ

Vsreq' > 4V c req' d S max

'

c

sreq' req' d =

≤ f 100 100

Av f y  5000 

> 2V c Vsreq' req' d psi sreq' req' d =

  50b  f c ′ 

Av f y

min. of: smax = min. d/2 24"

let Vsb = 2Vc

STOP. increase f' c or d or b

50b

φVn = φVc + φ V sb min. of: s = min. smax sreq'd

V s =

Vu mid > 0.0

Avf y d s



φ V nb 



V u 

xb = 0.5L 1 −

  

xb = 0.5L 1−



φVn = φVc + φ V s N =

x b

x min s

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

2

φ Vnb − V u mid 



Vu − V u mid 


2


V smax =

Av fy d smax

φVnmax = φVc + φ V smax

Vud > φ V smax N =

x min s

Vu mid > 0.0



use: N @ smax

xmax max = 0 .5 L1 −



φ V nmax 

  

 V u 

xmax = 0 .5 L1 −

φ Vnmax − V u mid 

 Vu − V u mid 

x max

s1 =

Av fy d V sreq' d

s1 let sk = s1 + ∆s

sk < smax Loop as long: x min>x k , and sk
STOP. go to stirrups number

4

V sk =

Av fy d sk

φVnk = φVc + φ V sk 3

Strunet.com: Concrete Beam Design V1.01 - Page 15



3

4

Vu

mid mid



xk = 0.5L 1 −

φ V n 



k

V u

> 0.0   

 

xk = 0.5L 1 −

φ Vn − V u  k

Vu

mid

− V u

mid

 

x k

xk smax =

> x b

min. min. of:

d/4

xk

12"

revise

S max

let sk +1 = sk + ∆s

= sk +1

sk

stirrups number

N 1

Loop until: x k+1 =x min , and x k =x max

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

N k

=

=

x k s1

xk +1 − x k sk

> x min STOP. go to stirrups number

Concrete Beam Design Flow Charts

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