strunet - concrete beam design
Post on 24-Oct-2014
494 Views
Preview:
TRANSCRIPT
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 - Page 1
Notations for Concrete Beam Design Flow Charts
STRUNETCONCRETE DESIGN AIDS
Strunet.com: Concrete Beam Design V1.01 - Page 2
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
Strunet.com: Concrete Beam Design V1.01 - Page 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
Strunet.com: Concrete Beam Design V1.01 - Page 4
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 ′= −
STRUNETCONCRETE DESIGN AIDS
Rectangular Concrete Beam withTension Reinforcement
Strunet.com: Concrete Beam Design V1.01 - Page 5
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′− ′ =
STRUNETCONCRETE DESIGN AIDS
Rectangular Beam with Tension & Compression Reinforcement
Strunet.com: Concrete Beam Design V1.01 - Page 6
compressionsteel yields
compressionsteel does NOT
yieldcompression steel may beneglected, and thus momentstrength is calculated based onthe tension steel only.Alternatively:
proceed toshear design
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 ′= + ≥
STRUNETCONCRETE DESIGN AIDS
Rectangular Beam with Tension & Compression Reinforcement (cont.)
Strunet.com: Concrete Beam Design V1.01 - Page 7
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
Strunet.com: Concrete Beam Design V1.01 - Page 8
NO
ACI 10.5.2
YES
ACI 10.5.1
proceed toshear design
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≥
STRUNETCONCRETE DESIGN AIDS
T-Section Beam Case - 1
Strunet.com: Concrete Beam Design V1.01 - Page 9
T-Sectioncase 2
YESNObw
As
be
ad
selectreinforcement,
As
STOP. revise toinclude
compression steel
proceed toshear design
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 = − + − ≥
STRUNETCONCRETE DESIGN AIDS
T-Section Beam Case - 2
Introduction to Concrete Beam Shear Design
STRUNETCONCRETE DESIGN AIDS
Strunet.com: Concrete Beam Design V1.01 - Page 10
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
STRUNETCONCRETE DESIGN AIDS
Strunet.com: Concrete Beam Design V1.01 - Page 11
∆
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
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
Strunet.com: Concrete Beam Design V1.01 - Page 12
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=
n c sV V Vφ φ φ= +
2req' ds cV V>
ACI 11.5.4.1
STRUNETCONCRETE DESIGN AIDS
Beam Shear Basic Chart
Strunet.com: Concrete Beam Design V1.01 - Page 13
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=
STRUNETCONCRETE DESIGN AIDS
Beam Shear Detailed Chart
Strunet.com: Concrete Beam Design V1.01 - Page 14
STOP. increasef'c or d or b
smax = min. of: d/2 24"
h=<10"h<2.5th<0.5b
STOP. no min.shear reinf. req'd
STOP. no min.shear reinf. req'd
Smax
' 100cf psi≤
s = min. of:smaxsreq'd
use: N @ s
smax = min. of: d/2 24"
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=
n c sV V Vφ φ φ= +
minxNs
=
STRUNETCONCRETE DESIGN AIDS
Beam Shear Detailed Chart (cont. 1)
Strunet.com: Concrete Beam Design V1.01 - Page 15
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= +
STRUNETCONCRETE DESIGN AIDS
Beam Shear Detailed Chart (cont. 2)
Strunet.com: Concrete Beam Design V1.01 - Page 16
smax = min. of: d/4 12"
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
+ −=
STRUNETCONCRETE DESIGN AIDS
Beam Shear Detailed Chart (cont. 3)
top related