4
ACI DESIG N HANDBOOK-S P-17M(09)
Appendix C
$=0.57+67e,
0.003
H
--r·-;
:
"'41,....
j ,,,./'
Reinforcement
closest to the
tension face
.,·
Ld
~
E1 = 0.005
~~
,, :
\._)
~ i j·
/,.
,, .:
i
;
Chapter 9
cj>=0.48+83E,
J.....................L............L.........
E, = 0.002
E1 = 0.005
- - i'
f-.........
!
,
t
inimum permitted
for beams
E1 = 0.004 min. strain permitted
for pure flexure
Fig. 1.2-Strength reduction (</>)factors for Grade 420 reinforcement.
Aids Flexure 1 through Flexure 4, included at the end of the
chapter, were developed using this condition. Accordingly,
T=C
=
Asfl'
0.85fd b
(1 -8)
( J-1)
Asfy = 0.85/~ P1cb
p
Pie =
o.85/c' P1c
df)'
M 11 -- bd2[1
(1 -2)
(1-3)
Pfv ] pf
-~·~
uJ:
)'
\.J
(1 -9)
M11 = bd2KII
(1 -10)
pf,. ] .f'
Kn = [ 1 - l.7fd P1 y
(1 -11)
where
where
As
(1 -4)
p = bd
The c/d ratio in Eq. (1 -3) can be written in terms of the
steel strain Es illustrated in Fig. 1. 1. For sections with single
layer tension reinforcemen t, d = d 1 and Es = E1. The c/d ratio
for this case becomes
Flexure 1 through 4 contain ~K11 values computed by
Eq. (1-11 ), where the ~-factor is obtained from Fig. 1.2 for
selected values of i::1 listed in the design aids.
Flexure Examples 1 through 4 illustrate the application of
Flexure 1 through 4.
1.2.2 Rectangular sections with compression reinforce0.003
- = - = -d
0.003 + e,
d,
C
p =
C
o.85fc' p1 0.003
0.003 + E 1
fv
(1 -5)
(1-6)
Equation (1-6) was used to generate the values for reinforcement ratio p (%) in Flexure 1 through 4 for sections with
single layer tension reinforcemen t. For other sections, where
the centroid of tension reinforcemen t does not coincide with
the centroid of extreme tension layer, multiply the p values
given in Flexure 1 through 4 by d/d.
Compute the nominal moment strength from the internal
force couple as shown as follows
M ,, = A s~,(d From Eq. (1 -2),
\.J
Pt)
(1-7)
\._,I
ment-Gener ally, flexural members are designed for only
tension reinforcement . Any additional moment strength
required in the section is usually provided by increasing the
section s ize or the amount of tens ion reinforcemen t.
However, the cross-sectional dimensions of some applications
can be limi ted by architectural or functional requirements,
and the extra moment strength may have to be provided by
additional tension and compression reinforcement . The extra
steel generates an internal force couple, adding to the
sectional moment strength without changing the section's
ductility. In such cases, the total moment s trength consists of
two components: i) moment due to the tension reinforcemen t
that balances the compression concrete, and ii) moment
generated by the internal steel force couple consisting of
compression reinforcement and an equal amount of additional
tension reinforcemen t, as illustrated in Fig. 1.3.
M11=M1 +M2
(1 -12)
M 1 =K11bd2
(1-13)
.____,/
ACI DESIGN HANDBOOK-SP-17M(09)
6
b
- - ___ ___, ---· +-- ·- -
d
-
=
A
.: I
~,._,-.---,-.--=
Cf
71
T=!~-r;=
:-~
--~ -
l J~?~--L
~~-=- -~-~~-~ l"
~
~
b
hr
l_
-
- -
:
·-
iI
: :
A
:L_·__sf1:
M
(
+
nr
bw
r-
----~
---- -,
, ~
n.a.
1
I)
.:
"_,.~
Cw
-+-
\,_/
~-~-~-~(- ' '_:_~
M nw
A
I
SW
___.. ,._ Tr
••
- - Tw
Fig. 1.4- T-section behavior.
Asw
P..-
(1-23)
bwd
Moment components M,if and Mmv can be obtained from
Flexure 1 through 4 when the tables are entered with Pf and
Pw values. For design, p1 needs to be found first and this can
be done from the equilibrium of internal forces for the
portion of total tension steel balancing the overhang
concrete. This is illustrated as follows.
T1 =Cr
(1-24)
A,J/2, = 0.85f: hj(b - bw)
(1-25)
0.85fc'
PJ
= -fy
'::J
d
(1-26)
Equation (1-26) was used to generate Flexure 7 and 8.
Flexure Examples 6, 7, and 8 illustrate the use of Flexure 7 and 8.
When T-section flanges are in tension, part of the flexural
tension reinforcement is required to be distributed over an
effective area, as illustrated in Flexure 6, or a width equal to
1/10 the span, whichever is smaller (Section 10.6.6). This
requirement is intended to control cracking that can result
from widely spaced reinforcement. When 1/10 of the span is
smaller than the effective width, additional reinforcement
should be provided in the outer portions of the flange to
minimize wide cracks in these regions.
1.3-Minimum flexural reinforcement
Reinforced concrete sections that are larger than required
for strength, for architectural and other functional reasons,
may need to be protected against a brittle failure immediately
after cracking by a minimum amount of tension reinforcement.
Reinforcement in a section is effecti ve only after the
cracking of concrete. When the reinforcement area is too
small to generate a sectional strength that is less than the
cracking moment, the section cannot sustain its strength upon
cracking. To safeguard against such brittle failures, ACl 318M
requires a minimum area of tension reinforcement in positive
and negative moment regions (Section 10.5.1).
As, min
0.25 Jf: bwd
4f,.
but not less than l.4bwdffy
(1 -27)
The aforementioned requirement is indicated in Flexure 1
through 4 by a horizontal line above which the reinforcement
ratio p is less than that for minimum reinforcement.
For statically determinate members, when the T-section
flange is in tension, the minimum reinforcement required to
have a sectional strength above the cracking moment is
approximately twice that required for rectangular sections.
Therefore , Eq. (1-27) is used with bw replaced by 2bw or the
flange width, whichever is smaller (refer to Section 10.5.2).
When the steel area provided in every section of a member is
high enough to provide at least 1/3 greater flexural strength
than required by analysis, then the minimum steel require ment need not apply (refer to Section 10.5.3). This exception
prevents the use of excessive reinforcement in very large
members that have sufficient reinforcement.
For structural slabs and footings, minimum reinforcement
in the direction of the span is the same as that used for
shrinkage and temperature control (refer to Section 10.5.4).
The minimum area of such reinforcement is 0.0018 times the
gross area of concrete for Grade 420 deformed bars (refer to
Section 7 .12.2.1). Where higher grade reinforcement is used,
with yield stress measured at 0.35 % strain, the minimum
reinforcement ratio is proportionately adjusted as (0.0018 x
420)//2,· The maximum spacing of this reinforcement is
limited to three times the slab or footing thickness, or 450 mm,
whichever is smaller (refer to Section 10.5.4).
'._/
\._.I
\ _,/
1.4-Placement of reinforcement in sections
Flexural reinforcement is placed in a section with due
considerations given to reinforcement spacing, crack
control, and concrete cover. lt is usually preferable to use a
sufficient number of small bars, as opposed to fewer large
bars, while respecting spacing requirements.
1.4.1 Minimum spacing of longitudinal reinforcementLongitudinal reinforcement should be placed with sufficient
spacing to allow proper placement of concrete. Flexure 9 shows
the minimum spacing requirement for beam reinforcement.
1.4.2 Concrete protection for reinforcement-Flexural
reinforcement should be placed to maximize the lever arm
between internal fo rces for increased moment strength. This
implies that the main longitudinal reinforcement should be
placed as close to the concrete surface as possible. The
reinforcement should be protected against corrosion and
aggressive environments by a sufficiently thick concrete
cover (refer to Section 7.7), as indicated in Flexure 9. The
concrete cover should satisfy the requirements for fire
protection (referto Section 7.7 .7) .
\,_/
8
ACI DESIGN HANDBOOK-SP-17M(09}
1.5-Flexure examples
Flexure Example 1: Calculation o,f tension reinforcement area for a rectangular tension-controlled cross section
\,_/'
For a rectangular section subjected to a factored bending moment M u, detennine the required tension reinforcement area for the
dimensions given. Assume interior construction not exposed to weather.
.GiY.m;
b
h
fJ
fy
Mu
=
=
=
=
=
250mm
510mm
28 MPa
420MPa
122 kN·m
ACI318M-05
section
Procedure
7.7.1
Estimate d by allowing for clear
cover, the radius of longitudinal
reinforcement, and stirrup diameter.
I•
b
•I
hT[lf
Calculation
Considering a minimum clear cover of 38 mm for
interior exposure, allow 65 mm to the cenu-oid of main
reinforcement.
Compute q>Kn = M J(bd2 ) .
Select p from Flexure 1.
d = 510 - 65 = 445 mm
q>K11 = 122 x 10°/(250 x (445)2] = 2.46 MPa
For q>Kn = 2.46 MPa, p = 0.70%
Compute required steel area:
As= pbd.
As= pbd = 0.0070 x 250 x 445 = 778 rnm 2
Design aid
Flexure 9
\._,I
Flexure 1
'.._,/
Use three No. 19; (A 5 )prov = (3)(284) =
7.6.1
3.3.2
Detennine the provided steel area.
10.3.4
9.3.2
For reinforcement placement, refer to
Flexure Example 9.
852 mm 2
Flexure 9
Note: Three No. 19 can be placed within a 250 mm
width.
(p)prov= (852)/((250)(445)] = 0.75%
Note: for (p)prov = 0. 75%
&1 ::: 0.0163
c; = 0.0163 > 0.005 (tension-controlled section) and
¢ = 0.9
Flexure 1
'-.J
'-.-,,I
10
ACI DESIGN HANDBOOK-SP-17M(09)
Flexure Example 3: Calculation of tension reinforcement area for a rectangular cross section in the transition zane
For a rectangular section subjected to a factored bending moment Mu , determine the required area of tension reinforcement for
the dimensions given. Assume interior construction not exposed to weather.
.Gi.Y.en;_
h
.r;
As
=
ACI318M-05
section
Procedure
7.7.1
Estimated by allowing for clear
cover, the radius of longitudinal
reinforcement, and stirrup diameter.
•I
I'
Calculation
Design aid
Considering a minimum clear cover of 40 mm for interior
Flexure 9
exposure, allow 65 mm to the centroid of main reinforcement
Compute ~K11 =MJ(bd ) .
Select p from Flexure 1.
d = 660 - 65 = 595 mm
~Kn= 660 x 106 /[360 X (595 )2] = 5.18 MPa
For ~K11 = 5.18 MPa, p = 1.59%
Compute As = pbd.
A s= pbd = 0.0 159 x 360 x 595 = 3406 mm2
Determine the steel area provided.
Try No. 25 bars; 3406/510 = 6.7.
2
7.6.1
3.3.2
b
- I•
25 mm maximum aggregate size
b
= 360mm
h
= 660mm
= 28MPa
= 420MPa
.fy
Mu = 660k.N·m
'\_,,
Need seven No. 25 bars in a single layer, but seven No. 25
bars cannot be placed in a single layer within a 360 mm
width without violating spacing limits. Try placing in
two layers.
\J
Flexure 1
'-..,..I
Flexure 9
Allow 90 mm from the extreme tension fiber to the
centroid of two layers of reinforcement.
Revise d = 660 - 90 = 570 mm
~K11
= 660 x 106/[360 x (570)2] = 5.64 MPa
\._,I
Flexure 1
For ~K,, = 5.64 MPa, p = 1.77%
As= pbd = 0.01 77 x 360 x 570 = 3632 mm 2
Try No. 25 bars; 3632/510 =7 .1
7 .6.1
3 .3.2
Select eight No. 25 bars in two layers (four No. 25 bars
in each layer). Note that four No. 25 bars can be placed
within a 360 mm width.
Flexure 9
(AJpro v = (8)(510) = 4080 mm2
(p)prov = 4080/((360)(570)]
10.3.4
9.3 .2
= 0.020
Note: For (PJprov = 0.020, ¢Kn = 5.76 MPa
ci = 0.0042
st= 0.0042 < 0.005 (transition zane)
¢ = 0.83 and ¢Mn > Mu
Flexure 1
Flexure 1
'J
ACI DESIGN HANDBOOK- SP-17M(09)
12
Flexure Example 5: Calculation of tension and compression reinforcement area for a rectangular beam section subjected to
positive bending
~
For a rectangular section subjected to a factored positive moment Mu, determine the required tension and compression
reinforcement area for the dimensions given as follows.
Given:
b
=
h
=
d'
=
1;
Jy
Mu
=
=
=
360mm
620mm
65mm
28 MPa
420MPa
786kN·m
ACI318M-05
Procedure
section
Estimate d by allowing for clear
7.7.1
cover, the radius of longitudinal
reinforcement, and stirrup diameter.
Compute q>K11 = Mul(bd2 ).
Select p from Flexure l.
I•
ctr:
b
'" I
i
::11ct·
Design aid
Calculation
Flexure 9
Considering a minimum clear cover of 40 mm for interior
exposure, allow 65 mm to the centroid of main reinforcement.
°--._I
d = 620 - 65 = 555 mm
q>K11 = 786 x 106/(360 x (555)1) = 7.09 MPa
q>K11 = 7.09 MPa is outside the range of Flexure 1. This
Flexure 1
indicates that the amount of steel needed exceeds the
maximum allowed when only tension steel is provided.
Therefore, compression steel is needed.
7.6. J
3.3.2
Compute (As -A;).
Select a reinforcement ratio close to
the maximum allowed to reach the
full strength of compression
concrete. Select p = 1.8% (slightly
below Pmax= 2.06% so that when the
bars are placed, Pmax is not exceeded).
Select p = 0.018 (f:1 =0.005)
2
As -A.: = pbd = 0.ol8 x 360 x 555 = 3596 mm
Try No. 25 bars; 3596/510 =7 .05.
Select eight No. 25 bars for (As - A;).
However, eight No. 25 bars cannot be placed in a single
layer. Try two layers.
Flexure 1
\._I
Flexure 9
Allow 90 mm from the extreme tension fiber to the
centroid of two layers of No. 25 bars.
10.3.4
9.3.2
Revised = 620 - 90 = 530 mm.
As - A; = pbd = 0.018 X 360 x 530 = 3434 mm 2
Try No. 25 bars; 3434/510 = 6.73
Select seven No. 25 bars for (A 5 -A;) to be placed in
two layers. (As - A;) = (7)(510) = 3570 mm 2
Corresponding p = 3570/((360)(530)]
OK
= 0 .0187 < Pmax = 0.0206
For p = 0.0187, <j>K11 = 5.75 MPa, f: 1 =0.0047, and qi= 0.87
Compute moment to be resisted
q>M11 = q>K11 bd2
by compression concrete and
corresponding tension steel (A, -A; ). q>M11 = 5.75 x 360(530)2/10 6 = 58 1 kN·m
Compute moment to be resisted by
the steel couple (with an equal tension
and compression steel area of A; ).
\.._I
Flexure 1
q>M,; = Mu - q>M11
q>M,; = 786 - 581 =205 kN-m
\..._,/
ACI DESIGN HANDBOOK- SP-17M(09)
14
a
Flexure Example 6: Calculation of tension reinforcement area for a T-secrion subjected to positive bending, behaving as
rectangular section
For a T-section subjected to a factored bending moment Mu, determine the required tension reinforcement area for the dimensions
given.
Given:
b
=
bw =
d
=
=
ht
=
Iv =
Mu =
t:
760mm
360 mm
480 mm
65mm
28MPa
420MPa
312 k.N·m
ACI318M-0S
section
d =480 mm
As
••
f------+j
b.... = 360 mm
Procedure
Assume tension-controlled section
(<I>= 0.9).
Determine if the section behaves as a
T- or rectangular section.
When M,, > <p[0.85/; b'3/(d - hf12)]
T-section, otherwise rectangular
section behavior.
Compute q>K,, = M,,l(bd1.)
Select p from Flexure 1.
Compute As = pbd.
Find provided area of steel.
10.3.4
9.3.2
Read
"-"
£ 1 and
<I> from Flexure 1.
Design aid
Calculation
q>MII = 0.9(0.85/; bly(d - h1l2)]
= 0.9[0.85(28)( 760)(65)(480 - 65/2))
= 474 x 106 N -mm = 474 kN·m > M 11 = 312 kN·m
Therefore, the neutral axis is within the flange and the
section behaves as a rectangular section with width b =
760 mm.
q>K11 =(312)(10t,)/ [(760)(480l) = 1.78 MPa
For q>K,, = 1.78 MPa, p = 0.50%
2
A 5 = pbd = 0.0050 x 760 x 480 = 1824 mm
,.._;
Flexure 1
·, .._;
2
Use five No. 22 bars with As= (5)(387) = 1935 mm
p = 1935/((760)(4 60)) = 0.0055
For p = 0.0055, £ 1 = 0.025 > 0.005 (tension-cont rolled
section), and <I>= 0.9
Flexure 1
·, .._;
\_/
ACI DESIGN HANDBOOK- SP-17M(09)
16
Flexure Example 8: Calculation of the area of tension reinforcement for an L-beam section, subjected to positive bending
behaving as an L-section in the transition zone
\._../
for the
For an L-section subjected to a factored bending moment Mu, determine the required area of tension reinforcement
exposure.
interior
dimensions given. The beam has
Given:
20 mm maximum aggregate size
= 900mm
b
bw = 550 mm
= 900mm
h
= 75 mm
ht
Jc' = 28 MPa
!y = 420MPa
Mu = 2400 kN-m
ACI318M-05
Procedure
section
Estimated by allowing for clear
7.7.1
cover, radius of longitudinal
reinforcement, and stirrup diameter.
Assume tension-contr olled section
and determine when the section
behaves as an L-section or a
rectangular section.
b = 900 mm
- - -h=
11..
T
ooom{
h1= 75 mm
As
,....___
_ . __
__J
~
b.v = 550 mm
Design aid
Calculation
9
Flexure
imerior
fur
mm
40
of
cover
clear
minimum
a
Considering
exposure, allow 65 mm to the centroid of main reinforcement.
d = 900 - 65 = 835 mm
q,M 11 = 0.9[0.85fd bhr(d - h/2)]
= 0.9[(0.85)(28 )(900)(75)(835 - 75/2)]
= 1,1 53,065 N-mm = 11 53 kN-m <Mu = 2400 kN·m
T herefore, the neutral axis is below the fl ange and the
section behaves as an L-section.
\_,I
When M 11 > q>[0.85/d bhfd - hJ'2)]
L-section, otherwise rectangular
section behavior.
Compute the amount of steel that
balances compression concrete in the
flange o verhang from Flexure 7 .
Find the moment amo unt resisted by
P.r from Flexure 1.
7.6. J
3.3.2
'-"
dlht = 835/75 = 1 1. 13
Flexure 7
Pt = 0.51 %
For Pf = 0.51 %
<J>Kn = 1. 83 MPa, and q> = 0 .90
<J>Mi = q,K,/b - b....,)d2
Flexure 1
= 1.83(900 - 550)(835)2 = 447 kN-m
q>Mw = M 11 - <J>M1= 2400 - 447 = 1953 kN-m
Determine the amount of steel
q,K,, = q>M,,J[(bw)(d)2]
req uired to resist the remaining
6
moment. This additional moment is q,K,, = 1953 x 10 /((550)(835 )2] = 5.09 MPa
Flexure 1
For <!>Kn= 5.09 MPa, Pw = 1.56%
to be resisted by the web, Pw·
rolled).
Note: ¢ = 0.90 (tension-cont
Ai= pfb - bw)d = 0.005 1(900 - 550)(835) = 1490 mmz
Compute the total area of tension
2
Aw= Pnhwd = 0 .0156(550)(83 5) = 7 164 mm
reinforcemen t.
2
As= At+ Aw = 1490 + 7 164 = 8654 mm
Flexure 9
needed.
are
bars
29
Select No. 29 bars; fo urteen No.
Fourteen No. 29 bars cannot be placed in a single layer.
Therefore , use two layers of reinforcement and revise
the design.
d = 900 - 90 = 8 10 mm
Recalculate the effective depth d and
revise design. Asswne cover of 90 mm Note: Reduced d will result in increased area of steel
and the beam will continue behaving as a T-section (no
to the centroid of two layers of
need to check again).
reinforcement.
\....-,I
"-./
ACI DESIGN HANDBOOK-S P-17M(09)
18
Flexure Example 9: Placement of reinforcemen t in the recrangular beam section designed in Flexure Example 1
cover
Select and place flexural beam reinforcemen t in the section provided below, with due considerations given to spacing and
weather.
to
exposed
not
requirements. Assume interior construction
Given:
20 mm maximum aggregate size
No. 10 stirrups
2
= 787 mm
As
b
= 250mm
h
= 500mm
Jy = 420MPa
ACI318M-05
section
7.7.1
7.6
b
I•
•I
{ r
As
=
Calculation
Procedure
Determine bar size and number of bars. Select No. 19 bars; No. of bars = 787 /284 = 2.8.
Use three No. 19 bars.
Considering minimum clear cover of 40 mm on each
Determine bar spacing.
side for interior exposure and allowing two srirrup bar
diameters, s = [250 - 2(40) - 2(10)- 3(20))/2 = 45 mm
Check against minimum spacing.
\.._.,I
Design aid
Flexure 9
\._,I
Flexure 9
(s),nin = {db;( 1 ~) amax; 25 mm}
(s)min = {20 mm; ( 1~) (20 mm); 25 mm}= 20 mm
OK
s = 50 mm > 20 mm
10.6.4
Check against maximum spacing as
governed by crack control.
(S)111ax = 380(280ifs) - 2.5cc ~ 300(280ifs)
fs = 2/3fv = 2/3(420 MPa) = 280 MPa
Cc= (40·+ 10) = 50 mm
(s)max = 380(1) - 2.5(50) = 255 mm
s = 50 mm < 300 mm OK
Eq.(1-28)
Final bar placement.
Provide three No. 19 as indicated below.
Flexure 9
40
mm
_,
\.._,I
\._,I
1 0mm
,.
20mm
40mmT
I
I
'
'
50mm
250mm
.1
\._,I
ACI DESIGN HANDBOOK-SP-17M(09)
20
1.6-Flexure design aids
~
r
Flexure 1: Flexural coefficients for rectangular beams with tension reinforcement;
fy = 420 MPa
2
P =A/bd
<j)M,,~Mu
<j)M,,=4>Knbd
fr =420 MPa
fc'. M Pa :
c,
0.2
0.15
21
:F=---c
'Wii:,'C~ ,
where M,, is in kN·m; K,, is in M Pa; band dare in mm.
o ,0l .003n
~
28
35
40
0.75
0.0039
131:
0.85
0.85
0.80
Pmin:
0.0033
0.0033
0.0035
hpp C
p,%
<j)K,,,MPa
p,%
<j)K,, , MPa
p,%
<j)K,, , MPa
p,%
<j)K,, , MPa
0.9
0.05
0.20
0.07
0.27
0.08
0.31
0.09
0.34
0.9
0.9
0.07
0.27
0.09
0.35
0.11
0.42
0.12
0.45
0.9
0.11
0.39
0. 14
0.52
0.17
0.62
0.18
0.66
qi
0.9
0.1
0.9
O.Q75
0.9
0.9
0.14
0.52
0.19
0.69
0.22
0.81
0.23
0.87
0.05
0 .9
0.9
0.20
0.75
0.27
1.01
0.32
l.19
0.34
l.27
0.04
0.9
0.9
0.25
0.92
0.34
1.23
0.40
1.45
0.42
1.56
0.035
0.9
0.9
0.29
1.04
0.38
1.39
0.45
1.64
0.48
1.76
0.52
0.55
0.03
0.9
0.9
0.33
l.19
O.Q25
0.9
0.9
0.39
1.40
0.44
0.52
1.59
1.86
0.61
J.88
2.20
0.65
2.02
2.36
2.24
0.74
2.65
0.79
2.85
0.02
0.9
0.9
0.47
1.68
0.63
0.019
0.9
0.9
0.49
1.75
0.66
2.34
0.77
2.76
0.83
2.97
1.83
0.69
2.44
0.81
2.89
0.87
3.10
0.01 8
0.9
0.9
0.52
0.Ql 7
0.9
0.9
0.54
1.92
0.72
2.56
0.85
3.02
0.91
3.25
0.016
O.QJ5
0.9
0.9
0.57
2.01
0.76
2.68
0.89
3.17
0.96
3.41
0 .9
0.9
0.60
2. 11
0.80
2.82
0.94
3.33
1.0 1
3.59
0.014
0.9
0.9
0.64
2.23
0.85
2.97
1.00
3.51
1.07
3.78
4.00
4 .12
0.013
0.9
0.9
0.68
2.36
0.90
3.14
1.06
3.72
l.14
O.QJ25
0.9
0.9
0.70
2.43
0.93
3.23
1.10
3.82
I. I 8
0.012
0.9
0.9
0.72
2.50
0.96
3.33
1.13
3.94
1.21
4.25
1.1 7
4.06
1.26
4.38
0.0115
0.9
0.9
0.75
2.58
1.00
3.44
0 .01 1
0.9
0.9
0.77
2.66
1.03
3.55
1.21
4.20
1.30
4.52
0.80
2.75
1.07
3.66
1.26
4.34
1.35
4.68
0.9
0.83
2.84
1.11
3.79
1.31
4.49
1.40
4.84
0.87
2.94
1.16
3.92
1.36
4.65
1.46
5.01
0.0105
O.Ql
0.9
0.9
0.9
0.0095
0.9
0.9
0.009
0.9
0.9
0.90
3.05
1.20
4.07
1.42
4.82
1.52
5.20
0.0087
0.9
0.9
0.93
3.12
1.24
4 .16
l.45
4.93
1.56
5.32
0.0084
0.9
0.9
0.95
3.19
1.27
4 .26
1.49
5.04
1.60
5.44
0.0081
0.9
0.9
0.98
3.27
1.30
4.36
1.53
5.16
1.64
5.57
0.0077
0.9
0.9
1.01
3.37
1.35
4.50
1.59
5.33
1.70
5.76
0 .0074
0.9
0.9
1.04
3.46
1.39
4.6 1
1.63
5.47
1.75
5.90
6.06
0.0071
0.0068
0.0065
0.0062
0.0059
0.9
0.9
0.9
0.9
0.9
0.9
0.9
0.9
1.07
1.11
1.14
3.54
3.64
3.73
4.73
1.68
5.61
l.80
4.85
1.73
5.75
1.86
6.22
1.52
4.98
1.79
5.91
1.92
6.39
l.85
6.07
1.98
6.57
1.43
l.47
0.9
1. 18
3.84
l.57
5. 11
0.9
].22
3.94
1.62
5.26
1.9 I
6.25
2.05
6.76
l.68
5.41
1.98
6.43
2.12
6.96
0.9
0.9
1.26
4.06
0.0053
0.9
0.9
1.31
4.18
1.74
5.57
2.05
6.62
2.19
7.17
0.005
0.9
0.9
1.35
4.30
1.81
5.74
2.13
6.83
2.28
7.40
0.0048
0.88
0.89
1.39
4.31
1.85
5.75
2.18
6.84
2.34
7.41
0.0046
0.87
0.87
1.43
4.32
1.90
5.76
2.24
6.85
2.40
7.43
0.0044
0.85
0.86
l.46
4.33
1.95
5.77
2.30
6.87
2.46
7.45
0.0043
0.84
0.85
1.48
4.33
5.78
5.78
2 .33
6.88
2.50
7.46
0.0042
0.83
0.85
I.SI
4.33
2.01
5.78
2.36
6.89
2.53
7.47
0.0041
0.82
0.84
1.53
4.34
2.04
5.79
2.39
6.90
2.57
7.48
0.004
0.82
0.83
1.55
4.34
2.06
5.79
2.43
6.90
2.60
7.49
0.0056
Notes: The values of p above th e rule are less than p,.;,,. <l>Kn values are based on q,-factors provided in Chapter 9. \\'hen Appendix C values of d, are used, q,K,, values in the transition
zone may be up to 2.4 % hi gher (more conservative).
\_/
'-,I
'-,i
\...,/
,.._,I
ACI DESIGN HANDBOOK-SP-17M(09)
22
I~ ,.f,,,p'..-c,
Flexure 3: Flexural coefficients for rectangular beams with tension reinforcement;
fy= 520 MPa
<\>M,.= $K,.bd2
<\>M,,2M"
where M,, is in kN·m; K11 is in MPa; band dare in mm.
fv =520MPa
0.15
0.1
0.075
0.05
0.04
40
35
28
fc' , MPa:
21
~l :
0.85
0.85
0.80
0.75
0.0027
0.0027
0.0028
0.0031
Pmr,,:
r.,
<I>
0.9
0.9
0.9
0.9
0.9
<l>AppC
0.9
0.9
0.9
0.9
0.9
p, %
0.06
0.08
0.11
0.17
0.20
<j>K,.,MPa
0.27
0.39
0.52
0.75
0.92
\_,/
'1w2J L
p=A/bd
<j>K11 ,MPa
p,%
(j)K11 ,MPa
0.09
0.42
0. 10
0.45
0.13
0.62
0.14
0.66
0.69
0.18
0.81
0.19
0.87
0.22
I.OJ
0.26
1.19
0.28
1.27
0.27
1.23
0.32
1.45
0.34
1.56
0.31
1.39
0.36
1.64
0.39
1.76
p,%
0.08
0.1 I
0.15
<j>K,.,MPa
0.35
0.52
p,%
0.035
0.9
0.9
0.23
1.04
0.03
0.9
0.9
0.27
1.19
0.35
J.59
0.42
1.88
0.45
2.02
0.31
1.40
0.42
1.86
0.49
2.20
0.53
2.36
0 .025
0.9
0.9
0.02
0 .9
0.9
0.38
2.24
0.64
0.9
0.40
0.53
2.34
0.60
0.62
2.65
0.9
1.68
1.75
0.51
0.019
2.76
0.67
2.85
2.97
0.018
0.9
0.9
0.42
1.83
0.56
2.44
0.65
2.89
0.70
3.10
0.017
0.9
0.9
0.44
1.92
0.58
2.56
0.69
3.02
0.74
3.25
3.41
3.59
0.016
0.9
0.9
0.46
2.01
0.6 1
2.68
0.72
3. 17
0.77
3.33
0.82
0.015
0.9
0.9
0.49
2. 1 I
0.65
2.82
0.76
0.014
0.9
0.9
0.51
2.23
0.69
2.97
0.81
3.51
0.87
3.78
0.86
3.72
0.92
4 .00
0.013
0.9
0.9
0.55
2.36
0.73
3.14
0.0125
0.9
0.56
2.43
0.75
3.23
0.89
3.82
0.95
4.12
0 .012
09
0.9
0.9
0.58
2.50
0.78
3.33
0 .92
3.94
0.98
4.25
0.0115
0.9
0.9
0.60
2.58
0.80
3.44
0.95
4.06
1.01
4.38
0.011
0.9
0.9
0.63
2.66
0.83
3.55
0.98
4.20
1.05
4.52
0.0105
0.9
0.9
0.65
2.75
0.86
3.66
J.02
4.34
1.09
4.68
0.01
0.9
0.9
0.67
2.84
0.90
3.79
1.06
4.49
1.13
4.84
0.0095
0.9
0.9
0.70
2.94
0.93
3.92
1.10
4.65
1.18
5.01
0.009
0.9
0.9
0.73
3.05
0.97
4.07
1.14
4 .82
1.23
5.20
1.26
5.32
0 .0087
0.9
0.9
0.75
3.12
1.00
4.16
1.17
4 .93
0.0084
0.9
0.9
0.77
3.19
1.02
4.26
1.20
5.04
1.29
5.44
1.05
4.36
1.24
5 .16
1.33
5.57
1.28
5.33
1.37
5.76
0.0081
0.9
0.9
0.79
3.27
0.0077
0.9
0.9
0.82
3.37
1.09
4.50
0.0074
0.9
0.9
0.84
3.46
1.1 2
4.61
1.32
5.47
1.41
5.90
0.9
0.87
3.54
I. I 6
4.73
1.36
5.6 1
1.46
6.06
3.64
1.19
4.85
1.40
5.75
I.SO
6.22
0.0071
0.9
0.0068
0.9
0.9
0.89
0.0065
0.9
0.9
0.92
3.73
1.23
4 .98
1.45
5.91
J.55
6.39
0.0062
0.9
0.9
0.95
3.84
1.27
5.11
1.49
6.07
J.60
6.57
0.0059
0.9
0.9
0.98
3.94
1.31
5.26
1.54
6.25
1.65
6 .76
0.0056
0.9
0.9
J.02
4.06
1.36
5.41
1.60
6.43
1.71
6.96
0 .0053
0.9
0.9
1.05
4.18
1.41
5.57
1.65
6.62
1.77
7.17
0.005
0.9
0.9
1.09
4.30
1.46
5.74
1.72
6.83
1.84
7.40
0 .0048
0.88
0.89
1.12
4.30
I.SO
5.73
1.76
6.82
1.89
7.39
0.0046
0.87
0.87
1.15
4.34
1.54
5.78
1.8 1
6.88
1.94
7.46
1.58
5.77
1.86
6.87
1.99
7.45
1.88
6.87
2.02
7.45
7.44
0.0044
0.85
0.86
1.1 8
4.33
0.0043
0.84
0.85
1.20
4.32
1.60
5.76
0.0042
0.83
0.85
1.22
4.32
J.62
5.76
1.91
6.86
2 .04
4.31
1.64
5.75
1.93
6.85
2.07
7.44
4.36
J.67
5.81
1.96
6.93
2.10
7.52
0.0041
0.004
0.82
0.82
0.84
0.83
1.23
1.25
Notes: The values of p above the rule are less than Pmin- tj,K,, values are based on <!>-factors provided in Chapter 9. When Appendix C values of 4> are used, tj,K,, values in the
Lransition w ne may be up to 2.4% higher (more conservative).
"-"
\_,/
\_,/
'-_,I
ACI DESIG N HANDBOOK-SP-17M(09)
24
Flexure 5: Reinforcement ratio p' for compression reinforce ment
p = (A,-A;)tbd
<j,M,, + <j,M,; ;,_ M11
cj,M,,=<j,K,,bd2
(from Flexure l through 4)
2
p'=A;tbd
<j,M,; = <j,K~ bd
420MPa
d'ld
I
0.02
0.06
0.1
0.28
As
-•-•
0.03
O.Q7
0.04
O.Q7
0.04
O.D7
A,'
'--"
T'
520MPa
0.14
0.18
0.22
0,02
0.06
0.1
I
0. 14
0.18
0.22
0.03
p',%
p',%
K,;,MPa
0 .14
d11~1·af:_A, •--'
where M,, is in kN-m; K,, is in MPa; band dare in mm.
fv
A'
_!.--c·
0.04
0.08
0.04
0.08
0.04
0.09
0,03
O.o3
0.03
0.03
0.05
0.06
0.06
0.06
0.07
O.o7
0.08
0.09
0.09
0.09
0.10
0.10
0.03
0.41
0.10
0.11
0.11
0.12
0.12
0. 13
0.55
0.14
0.14
0.15
0.16
0.16
0.17
0.11
0.1 1
0. 12
0.12
0.13
0. 14
0.21
0.14
0.14
0.15
0.16
0.16
0.17
0.70
0.17
0.18
0.19
0.19
0.20
0.83
0.20
0.21
0.22
0.23
0.24
0.26
0.16
0.17
0.18
0.19
0.20
0.21
0.26
0.27
0.28
0.30
0. 19
0.20
0.21
0.22
0.23
0.24
0.33
0.34
0.22
0.23
0.24
0.25
0.26
0.27
0.97
1.10
1.24
0.24
0.25
0.27
0.28
0.30
0.31
0.31
0.32
0.33
0.35
0.37
0.38
0.24
0.26
0.27
0.28
0.29
0.31
0.39
0.41
0.43
0 .27
0.28
0.30
0.31
0.33
0.34
1.40
0.34
0.35
0.37
1.52
0.37
0 .39
0.41
0.43
0.45
0.47
0.30
0.31
0.33
0.34
0.36
0 .38
1.65
0.41
0.43
0.44
0.47
0.49
0.51
0.33
0.34
0.36
0.37
0.39
0.41
1.80
0.44
0.46
0.48
0.50
0.53
0.56
0.35
0.37
0.39
0.40
0.42
0.44
1.93
0.48
0.50
0.52
0.54
0.57
0.60
0.38
0.40
0.4 1
0.43
0.46
0.48
0.43
0.44
0.47
0.49
0.51
2.10
0.51
0.53
0.56
0.58
0.61
0.64
0.41
0.68
0.44
0.45
0.47
0.50
0.52
0.55
0.58
2.20
0.54
0.57
0.59
0.62
0.65
2.34
0.58
0 .60
0.63
0.66
0.69
0.73
0.46
0.48
0.50
0.53
0.55
0.77
0.49
0.5 1
0.53
0.56
0.59
0.62
0.56
0.59
0.62
0.65
2.48
0.61
0.64
0.67
0 .70
0.73
0.74
0.77
0.8 1
0.52
0.54
2.62
0.65
0.67
0.70
2.80
0.68
0.71
0. 74
0.78
0.81
0.85
0.54
0.57
0.59
0.62
0.65
0.68
0.74
0.78
0.81
0.85
0.90
0.57
0.60
0.62
0.65
0.68
0.72
2.90
0.71
3.03
0.75
0.78
0.81
0.85
0.89
0.94
0.60
0.62
0.65
0.68
0.72
0.75
3.17
0.78
0.82
0.85
0.89
0.93
0.98
0.63
0.65
0.68
0.71
0.75
0.79
3.31
0.82
0.85
0.89
0.93
0.98
1.03
0.65
0.68
0.71
0.74
0.78
0.82
0.68
0.71
0.74
0.78
0.81
0.85
0.71
0.74
0.77
0.8 1
0.85
0.89
3.50
0.85
0.89
0.93
0.97
1.02
1.07
3.59
0.88
0.92
0.96
1.01
1.06
1.11
3.72
0.92
0.96
1.00
I.05
1.10
1.15
0.73
0.77
0.80
0.84
0.88
0.92
1.09
l.14
1.20
0.76
0.79
0.83
0.87
0.91
0.96
1.24
0.79
0.82
0.86
0.90
0.94
0.99
3.86
0.95
0.99
1.04
4.00
0.99
1.03
1.07
1.12
1.18
4. 14
1.02
1.06
1.11
1.16
1.22
1.28
0.82
0.85
0.89
0.93
0.98
1.03
1.26
1.32
0 .84
0.88
0.92
0.96
I.OJ
1.06
4.27
4.41
1.05
1.10
J. 15
1.20
1.09
1.13
1.19
1.24
1.30
1.37
0.87
0.91
0.95
0.99
1.04
1.09
1.17
1.22
1.28
1.34
1.41
0.90
0.94
0.98
1.02
1.07
1.13
4 .55
1.1 2
4.69
1.16
1.21
1.26
1.32
1.38
1.45
0.93
0.96
1.01
1.05
l.11
1.16
4.82
1.19
1.24
1.30
1.36
1.42
1.50
0.95
0.99
1.04
1.09
1.14
1.20
4.96
1.22
1.28
1.33
1.40
1.46
1.54
0.98
1.02
1.07
1.1 2
1. 17
1.23
l.1 0
1.15
1.20
1.26
1.24
1.30
1.33
5.10
1.26
1.31
1.37
1.43
1.50
1.58
1.01
1.05
5.24
1.29
1.35
1.41
1.47
1.54
1.62
1.03
1.08
1.13
1.18
5.38
1.33
1.38
1.44
1.51
1.59
1.67
1.06
J.1 1
l. 16
1.21
1.27
1.71
1.09
1.13
1.19
1.24
1.30
1.37
1.1 2
1.16
1.2 1
1.27
1.33
1.40
5.50
1.36
1.42
1.48
1.55
l.63
5.65
1.39
l.45
l.52
1.59
1.67
1.75
5.79
1.43
1.49
1.56
1.63
J.71
1.79
1.14
1.19
1.24
1.30
1.37
1.44
1.59
1.67
1.75
1.84
1.17
1.22
1.27
l.33
1.40
1.47
5.93
1.46
1.52
'--"
'-.-,/
'--,I
\_/
ACI DESIGN HANDBOOK-SP-17M(09)
26
Flexure 7: Reinforcement ratio p,(%) balancing concrete in overhang(s) in T- or
L-beams; fy = 420 MPa
<j,M,if+ <!>Mm,· 2'. M"
'' -._,I
b
Pr A~l[(b- b,.)d]
~
, hf
'----i
~
i----tJ----7
r--'
-----
j _ _ ~- - - r - - i
T
P,r = A,..J(b.,d)
d
I
I
:
\
+
:A:
As
As = A.ef+Asw
'
I
Sf l
I •
I
L - - -1
t
-----
A
•
SW
•
Use Flexure 1 or 2 with p1 and (b - b..,) to find <j>M,if Use Flexure I or 2 with p,.. and b,.. to fin d <j>M,,_..
f,,=420MPa
48
55
62
70
4.25
4.96
5.67
2.83
3.3 1
3.78
6.38
4.25
7.08
2.36
1.77
2.13
2.48
2.83
3. 19
3.54
2.27
2.55
2.83
21
28
35
40
2
2.13
2.83
3.54
3
1.42
1.89
4
I.06
1.42
fc', MPa:
Pf·%
dlh1
5
6
7
0 .85
0.71
0.61
1.13
0.94
0.8J
1.42
1.70
I.98
4.72
1.18
1.42
l.Ul
1.21
1.65
1.42
1.89
1.62
2.13
1.82
2.36
2.02
0.89
1.06
1.24
1.42
1.59
1.77
8
0.53
0.71
9
0.47
0.63
0.79
0.94
1.10
1.26
1.42
1.57
0.71
0.85
0.99
1.13
1.28
1.42
1.29
0.43
0.57
0.39
0.52
0.64
0.77
0.90
1.03
1.16
12
0.35
0.47
0.59
0.71
0.83
0.94
1.06
1.1 8
13
0.33
0.44
0.54
0.65
0 .76
0 .87
0.98
1.09
14
0.30
0.40
0.51
0.61
0.71
0.81
0.91
1.01
15
0.28
0.38
0.47
0.57
0.66
0.76
0.85
0.94
16
0.27
0.35
0.44
0.53
0.62
0.7 1
0.80
0.89
0.67
0.75
0.83
10
11
17
0.25
0.33
0.42
0.50
0.58
!8
0.24
0.31
0.39
0.47
0.55
0.63
0.71
0.79
0.67
0.75
19
0.22
0.30
0.37
0.52
20
0.21
0.28
0.35
0.45
0.43
0.60
0.50
0 .57
0.64
0.7!
0.40
0.47
0.54
0.61
0.67
21
0.20
0.27
0.34
22
0.19
0.26
0.32
0.39
0.58
0.64
23
0.18
0.31
0.37
0.45
0.43
0.52
0.25
0.49
0.55
0.62
24
0.18
0.24
0.30
0.35
0.41
0.47
0.53
0.59
25
0.17
0.23
0.28
0.34
0.40
0.45
0 .51
0.57
26
0.16
0.22
0.27
0.33
0.38
0.44
0.49
0.54
27
0.1 6
0.21
0.26
0.31
0.37
0.42
0.47
0.52
28
0.15
0.20
0.25
0 .30
0.35
0.40
0.46
0.5 1
29
0.15
0.20
0.24
0.29
0.34
0.39
0.44
0.49
30
0.14
0.19
0.24
0.28
0.33
0.38
0.43
0.47
31
0.14
0.18
0.23
0.27
0.32
0.37
0.41
0.46
0.40
0.44
0.39
0.31
0.35
0.30
0.34
33
0.13
0.17
0.21
0.27
0.26
34
0. 13
0.17
0.21
0.25
0.29
0.33
0.38
0.43
0.42
0.12
0.16
0.20
0.24
0.28
0.32
0.36
0.40
36
0.12
0.16
0 .20
0.24
0.28
0.31
0.35
0.39
37
0. 11
0.15
0.19
0.23
0.27
0.3 1
0.34
0.38
38
0.11
0.15
0.19
0.22
0.26
0.30
0.34
0.37
39
0.11
0.15
0.1 8
0.22
0.25
0.29
0.33
0.36
40
0.11
0. 14
0.18
0.21
0.25
0.28
0.32
0.35
32
35
0.13
0.18
0.22
\,_,/
\._I
\.J
\._I
ACI DESIGN HANDBOOK-S P-17M(09)
28
Flexure 9: Bar spacing and cover requirements
\.J
s' ~ 25 mm
I
I
I
lt---+I
s
db
s ~
{
a
1¼ a m ax
ouu
= Max. aggregate size
\._/
25mm
Minimum cover f<>r protection of reinforceme nt (Section 7.7.1)
Not exposed to weather or in contact with ground
Beams and columns
Slabs, walls. and joists with No. 36 and smaller bars
Slabs, walls, and joists with No. 43 and 57 bars
40 mm
20 mm
40mm
Exposed to earth or weather
Members with No. 16 and smaller bars
Members with No. 19 through 57 bars
\._,I
40mm
50mm
75 mm
Cast against and permanently exposed to earth
Notes:
i) The minimum cover is measured from the concrete surface to the outermost surface of stirrups, or to the outer·
most surface of main bars when more than one layer is used without sti rrups.
ii) In corrosive environments or othe r severe exposure conditions. the amount of cover shall be suitably increased
(Section 7.7.5).
iii) The minimum cover shall also satisfy the fire protection requirement (Section 7.7.7).
\._/
Flexure 1 O: Skin reinforcement
Tension face - Negative bending
s
s
h
2
s
s
h
2
s
s
s
-....__,;
Tension face - Positive bending
ACI DESIGN HANDBOOK-SP-17M(09)
30
s ~ (<j,AJy 1 d)/(Vu - <j,Vc)
(2-1)
The quantity (Vj<j, - Ve) represents Vs, the nominal shear
strength provided by reinforcement. ACI 318M, Section
11.5.6. 1 requires the placement of shear reinforcement in all
beams for which the required strength is more than half <j, VeThe full development of a critical shear crack between stirrups is prevented by Section 11.5.5, which sets the maximum
spacing of stirrups at d/2 when Vs < 0.33 Ji; b 1,,d. Because
Ve= 0.17
b.,.d, maximum spacings can be d/2 or< 300 mm.
bwd.
as long as VJ<j,, which equals (Ve+ Vs)::; 3Vc or 0.5
Maximum spacing is d/4 or 150 mm. when Vj<j, <'.. 3Ve. ACI
318M, Section 11.5.7.9 sets a maximum value on V5 as 4Vc
or 0.66
b.,,d. Concrete compression struts cannot sustain
more shear when the required amount of V5 exceeds 4 Ve =
0.66
bwd regardless of additional shear reinforcement.
Thus, a beam section must be made larger when Vul<j, >
0.83
bwd.
A graph in design aid Shear 1 displays limits of nominal
shear stress values of Vnl(q>bwd) for concrete :suength J;
from 21 MPa to 70 MPa. The graph shows stress ranges for
which design requirements change, and is not intended for
precise evaluation of member strength, as precise strength
values are provided in other design aids. No shear reinforcements (stirrups) are required when V,/(bwd) is less than
The strength Ve of concrete in sections reinforced
0.08
for shear is 0.17 jj; bwd. Stirrup strength can be added to the
concrete strength Ve to determine the total strength of a
section. Stirrups should be spaced no more than d/2 apart
Where Vn/Cbwd) > 0.5
where V,/(bwd) ::; 0.5
maximum stirrup spacing becomes d/4. The compressive
strut strength of concrete is reached when V,/(bvA =
Additional stirrups cannot increase section shear
0.83
strength, as the concrete strength is considered exhausted
when V,/(bwd) > 0.83
Design aid Shear 2 consists of three tables that can be used
to determine shear strength for rectangular sections of width
b or bw from 250 to 810 mm and thickness h from 250 to
1220 mm. It is assumed that depth dis 65 mm less than thickness for h < 7 50 mm, but that larger longitudinal bars would
make d"' h- 75 mm for deeper beams.
Table 2(a) gives values Krc= J(Jc' /28) to be used as modifiers of Kvc when members are made with concrete strength
different from f,' = 28 MPa. ln conjunction with required
stirrups, the nominal shear strength of concrete Ve= KteK vc·
Table 2(b) gives values Kvs for determining nominal
stirrup strength Vs= KvsCA)s).
Table 2(c) gives values Kvc in kN. Kvc is the shear strength
of concrete when required stirrups are used in members
made withf! = 28 MPa concrete.
The nominal strength of a rectangular section is the sum of
concrete strength Ve and reinforcement strength Vs to give V11
= VJ<j, = K_rcKvc + Kv/ A/s).
Shear 3 is a design aid for use when Grade 420 stirrups
larger than No. 16 are used and sections must be deep
enough for tension strength bar development of larger stirrups or closed ties. Required thickness of section values are
tabulated for concrete strengths from 21 to 70 MPa and for
JI:
Ji:
Ji:
Ji:
Ji:
JI: .
J1Z .
JI: .
Ji: .
Ji: ,
No. 19, No. 21, and No. 25 stirrups. ACI 318M-05, Section
11.5.2 limits the yield strength of reinforcement bar stirrups
to no more than 420 MPa.
Section 11.5.6.3 sets lower limits on the amount of shear
reinforcement used when such reinforcement is required for
strength. These limits can prevent stirrups from yielding
upon shear crack formation. The limit of Av must exceed
0.062
b 11.s/fyr > 0.35b....slfyr- The first quantity governs
whenfd is greater than 31 MPa.
Shear reinforcement design includes the selection of
stirrup size and stirrup spacing along the beam. Design aids
Shear 4.1 and Shear 4.2 give strength values Vs of No. 10
U-stirrups and No. 13 U-stirrups (two vertical legs) as
shear reinforcement tabulated for depth values d from 200 to
1000 mm and stirrup spacing s from 50 mm to maximum
spacings= d/2. Each table also lists the maximum section width
for which each stirrup size can be used without violating the
required minimum shear reinforcement. Shear 4.1 applies for
Grade 280 stirrups, and Shear4.2 applies for Grade 420 stirrups.
'-J
Ji:
\._,,I
2.4-Shear strength of two-way slabs
Loads applied to a relatively small slab area create shear
stress perpendicular to the edges of the load area. Columns
that support flat plate slabs and columns supported by footings
are the most common examples. Section 11.1 2.2.1 provides
expressions for determining shear strength in such conditions for which shear failure is assumed to occur near the
column face(s). Failure is assumed to occur on the prism
face(s) located at a distance of d/2 from each column face.
The prism perimeter b 0 multiplied by the slab depth d is
taken as the area of the failure surface.
Three expressions are given for computing a critical stress
on the failure surface. A coefficient o.5 is used to accommodate columns in different locations:
o.s = 40 for interior columns;
0.5 = 30 for edge columns; and
o.s = 20 for corner columns.
The critical (failure) stress may be taken as the least value
of either 0.33 $ , 0.1 7(1 + 2/P)Ji: , or 0.083(a5 dlb 0 +
2)
The quantity p is the ratio of long side to short side
of the column. The first expression governs for centrally
loaded footings and for interior columns unless the ratio ~
exceeds 2 or the quantity 40dlb 0 is Jess than 2. Shear strength
at edge columns and corner columns that support flat plates
must be adequate for the direct force at the column and fo r
additional shear forces associated with moment transfer at
such columns. Diagrams for the prism at slab sections for
columns are shown with Shear Examples 5, 7, and 8.
Design aid Shear 5.1 gives shear strength values of twoway slabs at columns as limited by potential failure around
the column perimeter.
Table 5 .1(a) gives values of Kl as a fu nction of slab d and
column size b and h.
Table 5.l(b) gives values of the shear stress factor K2 as a
function of the ratio Pc between the longer side and the
shorter side of rectangular column sections.
\_,I
\.._/
Ji: .
\..,J'
ACI DESIGN HANDBOOK-SP-17M(09)
32
strength by bending or by torsion and can be ignored. An upper
limit to the torque resistance of concrete functioning as
DETERMINATE TORSION
'--"
compression struts is taken from ACI 318M-05, Eq. (11-18) as
= ~0.083A.(A 0 1,)
Tm ax
2
Ji! fp1,
(2-3)
Torsion reinforcement requires closed ties and longitudinal bars located in the section periphery. With torsion
cracks assumed at an angle 0 from the member axis, torsion
strength from closed ties is computed as
Tn
= (2A£1i1.f;,1cot0)/s
(2-4)
The angle 0 must be greater than 30 degrees and less than
60 degrees. This chapter uses 0 = 45 degrees for design aids.
Solid concrete sections must be large enough to resist flexural shear Vu and torsion shear Tu within the upper limits
established for each. ACI 3 l 8M, Eq. (11-18) gives
JrV,,l(b.d) ]' + [T.p.f( 1.7A!.)]
2
s; ~[ Vrl(b •. d) + 0.66 Ji;]
(2-5)
In addition, ACI 318M, Eq. (1 1-22) requires that longitudinal bars with an areaAe be placed around the periphery of
sections.
Ae =At pis
(2-6)
Longitudinal spacing of transverse closed ties must be no
greater than p1,l200, or 300 mm. The spacing between longitudinal bars in the periphery of sections must be no greater than
300 mm. Where torsion reinforcement is required, the area
of two legs of a closed tie (Av + 2A 1) must be greater than
0.062(bwsffy1)
but not be less than 0.35bwslfyDesign aid Shear 6.1 displays critical torsion strength
values for rectangular sections made with concrete strength
f) = 28 MPa. When concrete stren gth/) is different from
28 MPa, multiply the torque values T11 from Table 6. l (a) and
Tcr from Table 6.1 (b) with the corresponding correction
factor Kfc from Table 2(a).
JI:
INDETERMINATE TORSION
\.......,I
Fig. 2.4~Detenninate torsion versus indeterminate torsion.
Table 6.1(a) gives values of K 1 , the maximum torque limTn
a section can resist as a function of section thickness h and
width b. Assume that the distance from section surface to the
center of closed ties is 45 mm.
Table 6.1 (b) gives values K,cr of torque Tcr that will cause
sections to crack as a function of section dimensions band h.
Design aid Shear 6.2 can be used to determine the torsion
strength of closed ties. Numbers K 15 for width b and thickness h in the charts are multiplied by the ratio between tie
area A 1 and tie spacing s to compute the nominal torque Ts
resisted by closed ties. The distance from section surface to
tie centerline is 45 mm.
Table 6.2(a) applies for Grade 280 ties. Table 6.2(b )
applies for Grade 420 ties.
\_I
\.......,/
-..._;
ACI DESIGN HANDBOOK-SP-17M(09)
34
11.5.6.3
Step 8-Detennine minimum required
shear reinforcement.
ls Av,prov > Av,min?
Av,min = 0.062Jf:b""s
Av.min == 0.062
!y
but not less than
11.5.6.1
0.35b
/2,
\,_,I
.S
"
Step 9-Determine position beyond
which no stirrups are required. No
stirrups required if Vu< 0.5~ Ve
With zero shear at midspan, the
distance z from midspan to V11 =0.5Vc
becomes z = 0.5~ V }wu.
Stirrups are required in the space
(3000- 762) = 2238 mm from face of
each support. Compute in mm.
Begin with a half space = 100 mm
and compute n = number of stirrup
spaces required.
)28(350 )(230)
2
2
8
== 63 mm <Av.prov== 142 mm
420
Av == 0.75 (0.35)(350)(230 ) == 50.3 mm2 < 142 mm2
420
OK
z = 0.5(0.75)132.5 kN/65.2 kN/m x 103 = 762 mm
n
= (2238 mm -
100)/9 = 237.5 mm
Use ten No. 10 U-stirrups spaced at 230 mm starting
100 mm from each support.
\_,I
\J
\_,I
\_,I
ACI DESIGN HANDBOOK- SP-17M(09)
36
Shear Example 3: Vertical U-stirrupsfor beam with triangular shear diagram
Detennine the size and spacing of stirrups for a beam.
Vn (k)
Given:
=
=
=
=
=
=
=
350mm
737mm
d
h
810mm
28 MPa
fj
420MPa
fvr
796kN
W 11
169.3 kN/m
Nom1alweight concrete
bw
vii
\,.,_/
lv
1061 kN
944.5 kN
894.7 kN
f--------"~
· ••••••• ••"· ••••·• ••.·,, ••.5. ;'.,~q9 .mm
725.3 kN
542.7 kN 1.• · • .• • • ••• •••• • ••. i • .• •• • • ••• •••• x ••• .5. ;'. :'.~~ .1:lm
Step 2, Ve = 326 kN
t--- -- - - + - -- ----lf.----:::::,._~
163 kN
Distance from
face of su pport (m)
#13 stirrup spacing
0
·7
0.3 0 .6
0.9 1.
10@125mm
= 1250 mm
1.5 1.s 2.1
I 2.4
5 @200 mm
=1000mm
-..._/
5@350mm
= 1750 mm
Total= 4000 mm
ACI318M-0S
section
11.1 .3.1
Calculation
Procedure
Vj<j> = 796/(0.75) = 1061 kN
Step I-Determine Vn = maxVj <j>
(<j> = 0.75).
wj<j> = 169.3 kN/m/(0.75) = 225.7 kN/m
I Compute wj<j>.
Vj<j> at d = 1061 - 225 .7(737) x 10- 3 = 894.7 kN
Compute V j<j> at d from face of
support.
At d from face, Vj<j> = Vj<j> - (wj<j>)d
V,, must exceed 894.7 kN at support.
11 .3.1.1
I Step 2-
9.3.2.3
11.5.6.1
11.5.7.2
11.5.5.3
Detennine Ve= KreKVC"
For f j = 28 MPa
For b = 500 mm and h = 810 rrun
Show this Ve line on graph above.
I Step 3- Compute distance evover
which stirrups are required.
(Vn- 0.5VJ
ev
(w,/<j>)
0.5(326.0 kN)
225.7 kN·m
say 4000 mm
I
Shear 2
Table 2(a)
Table 2(c)
Kfe = 1
Kve = 326.0 kN
Ve= (1 )(326.0 kN) = 326.0 kN
I e,. = 1061 kN -
Design aid
I
\,,_/
I
-..._/
= 3.98 m = 3980 mm,
Step 4-Select stirrup size for
maximum V5 .
I Shear 4.2
Compute maximum V5 = VJ<j>atd- Ve. I Maximum V5 = 894.7 - 326.0 = 568.7 kN
Read stirrup spacing ford= 737 mm
I With No. 10 stirrups, s must be < 75 mm
Table 4.2(a)
I and v11 = 568 .7 kN
With No. 13 stirrups, scan be 125 mm and Vs= 618.5 kN Table 4.2(b)
I
Select No. 13 stirrups, and use 125 mm
spacing from face of support and Vs=
618.27 kN
Note, if Vs> 2VC' s must be< d/4 =
184 mm
Since s = 125 mm is< 184 mm,
spacing is OK.
I Compute 2Vc = 2(326.5) = 653 kN
\.J
ACI DESIGN HANDBOOK- SP-17M(09)
38
Shear Example 4: Vertical U-srirrups for beam with trapezaidal and triangular shear diagram
\,_I
Determine the required spacing of vertical No. IO stirrups for the shear diagram shown.
r-
.iliY.rn;_
= 330 mm
= 500mm
1; = 28MPa
= 420MPa
!yr
Pui = 67kN
vu
= 258 kN
Wu
= 67 kN/m
X]
= 1.4 m
= 0.75 for shear
<I>
Nonualweight concrete
b
d
Vu= 258 kN
$Ve= 111.7 kN
-
d = 500 m
Face of support
ACI318M-05
section
Procedure
Step 1-Detennine, at d from face of
11.1.3.1
support, the value of maximum Vu =
Vu-wud/12.
11.3.1.1
Step 2-Detennine the value
<j> Ve= 0. l 7<j>( Ji: )bd
or use design aid for b = 330 mm and
9.3 .2.3
h = 572 mm; Kvc = 146 kN
Step 3- Detennine required Vu each
side of Pui·
Left of Pu1, vu= Vu- WuX1
Right of Pui, change in Vu= Pu1
11. J.l
Step 4-Determine spacing s 1
11.5.7.2
required for No. 10 U-stirrups at face
of support.
A v= 2(7 1) = 142 mm2
s 1 = <j>Avfyt d/(maximum Vu - q> Ve)
11.5.5.1
11.5.7.2
11.1.1
X
Step 5- Since maximum spacing
S 111ax = d/2 with d = 250 mm, determine value of <!>Vu = <!>Ve+ <j>Avfy1dls.
X1=1 .4m
tv1
Calculation
Maximum Vu
= 258 kN - 67 kN/m(500 mm) x 10- 3 = 224.5 kN
<J>Vc = 0.17(0.75)( J 28 MPa)330 mm(500 mm)
= 111.3 kN
<J>Ve = 0.75KfeKvc = 0.75(1.00) 149 = 111.7 kN
Design aid
\....._,/
Shear 2
Table 2(c)
Left Pui, Vu= 258 kN - 67 kN/m(l.4 m) = 164.2 kN
Right Pu 1, Vu= 164.2 - 67 = 97.2 kN
_ (0.75)142 mm\ 420 MPa)(500 mm) x 10- 3
(224.5 - 111.7)
= 198 mm
Maximum spacing s111ax = 500 mm/2 = 250 mm
\,_I
SI -
Shear 4.2
\....._,/
0
<j>V
"
= lll. 7 kN + 0.75(142 mm- )(420 MPa)(500 mm)
250 mm
= 201 kN
11.5.6.1
or use design aid for Vs when s
= 250mm
<J>Vu = <J>Vc + <J>Vs
Vs= 119 kN for No. 10 stirrups at 250 mm spacing
Conclude: uses = 175 mm until
<!>Vu < 201 kN and uses= 250 mm
until <J>Vu < 0 .5<J>Vc
From face of support, use 75 mm space
then fi ve spaces at 175 mm (890 mm) and fi ve spaces at
250 mm (1 250 mm); 2160 mm > 2032 mm
<!>Vu= 11 1.7kN + 0.75(1 19 kN) = 201 kN
Step 6- Detennine distance x from
face of support to point at which Vu =
201 kN
x = (change in shear)/wu
x = (258 kN - 201 kN)/67 kN/m = 0.85 m = 850 mm
Step 7-Determine distance 1\,i,
distance beyond x I at which no stirrups
are required.
t;,1 = (97.2 kN - 111.7 kN/2)/67 kN/m =0.61 7 m = 617 mm
Find l;,1 = (Vu - Vcf2)lwu
X] + f;v ] = 1400 + 617 = 2017 mm
Compute x 1 + evl
\...J
ACI DESIGN HANDBOOK-SP-17M(09)
40
Shear Example 6: Detennination of thickness required for perimeter shear strength ofa flat slab at an interior rectangular column
-"-../
Given:
= 400mm
= 600 mm
J; = 35 MPa
f yt
= 420MPa
Vu = 792 kN
Normalweight concrete
Refer to Shear Example 5 for diagram of shear perimeter and Code clauses
be
he
ACI318M-05
section
11.12.2.1
9.3.2.3
11.12.1.2
9.1.1
Procedure
Step 1-Set up expression for ~Ve·
~Ve= 0.33~( Jf: )b0 d
= 0.33q>(
)2(he + d + b C + d)d
Step 2-Equate Vu to qiVe and solve
ford.
JI:
Calculation
<j>Vc= 0.33(0.75)(,/35 MPa)2(600 mm+ d + 400 mm
+ d)dmm
792(1 O-' N) = 1.46 N/mnl" (1 OOOd + 2d°1.) mm1
271,233 mm 2 = (1000d + 2d2 ) mm2
135,616 + 62,500 = 62,500 + 250d + d 2
d = (J198,116 ) - 250 = 195 mm
h 5 = 195 mm + 20 mm + 16 mm= 231 mm, say, h 5 =
230mm
Step 3- Allow for 20 mm clear cover
of tension bars to make h5 = d + 20 +
bar diameter (estimated)
ALTERNATE METHOD using Design aid Shear 5.1
9.3.2.3
Step I-Compute minimum Vn = V j<j> v,, = 792 kN/0.75 = 1056 kN
(he + be)= (600 mm+ 400 mm)= 1000 mm
and compute (he + be)
11.12.1.2
1056 kN is between:
Step 2- Withfj = 35 MP a and
V11 = 1056 kN
Vn = 887 kN and V,, = 1 I 83 kN
Use (he + be)= 1000 mm, interpolate
K1K2 = 150 + 50(10 56 - 887 ) = 178.5 MPa
K1K2
(1183-887)
7 .7.1
11.12.2.l
7.7.1
Step 3- Compute J3e = he/be
Design aid
J3e = 600 mm/400 mm = 1.5 < 2, so K2 = 8.2
and Kl = KIK2 MPa/K2 = 21.77 MPa
\J
\_,I
Table 5. l (c)
Table 5.l(b)
Step 4- Table 5.1 (a) with (he+ be) =
Table 5.l (a)
1000 and KI = 21.77 MPa
Interpolate for d
d= 150mm+ 16mm = 166mm
Step 5- Allow for 20 mm clear cover h 5 = 189.88 mm + 20 mm + 16 mm= 225.88 mm slab,
of tension bars to make h 5 = d + 20 + say, h 5 = 230 mm
bar diameter (estimated)
\J
_;
FOREWORD
This edition of the AC! Design Handbook was prepared in a new format based on the feedback received
from handbook users over the years. The advances in programmable and nonprogrammable electronic
calculators and personal computers have enabled the elimination of numerous design aids that were
intended to carry out relatively simple design calculations. Instead, explanatory materials have been added
to each chapter, while maintaining essential design aids and illustrative examples. This is the first edition
of the handbook after the introduction of explicit strain limits for tension- and compression-controlled
sections in flexure, with variable strength reduction factors q> within the transition zone. This necessitated
the development of a new set of design aids for members subjected to flexure. The increased use of higher
strength concretes and higher grade reinforcement has been recognized and the design aids were developed
for concrete strengths of up to 70 MPa for beams and 85 MPa for columns; and steel Grades 420 and 520.
The column interaction diagrams of Chapter 3 were developed for nominal quantities and in a non-dimensional form to facilitate the use of interaction diagrams with any system of units and any strength reduction
factor. The use of nominal quantities also facilitates column investigations.
The chapters of the handbook were developed by individual authors, as indicated on the first page of each
chapter. The authors were long-standi ng members of the fonner ACT Committee 340, Design Aids for
Building Codes, which had the mandate to develop the previous editions of the handbook in accordance
with the ACI 318 Building Code. Committee 340 was first established in 1958 and was discharged in 2003.
The first handbook was developed on the basis of the 1963 edition of the ACI Building Code. The
committee subsequently engaged in revising the handbook in the years to come, and developed updated
editions of the handbook for the 1971, 1977, 1983 and 1995 ACI Codes. The current edition is based on
ACI 318M-05.
Although the handbook chapters were developed by individual authors, the first draft of the document
was reviewed by the former members of ACI Committee 340. Their input is gratefully acknowledged.
Furthermore, many individuals and former members of ACI Committee 340 contributed to the earlier
editions of the handbook which formed the basis for the current edition. Their contributions, as well as the
administrative and technical assistance received from ACI Staff, are gratefully acknowledg ed.
Murat Saatcioglu
Editor
\_,,I
...._;
\_,I
\...J
'-.J
42
ACI DESIGN HANDBOOK-SP-17M(09)
Shear Example 8: Determination of required thickness of a footing to satisfy perimeter shear strength at a rectangular column
be
= 400 mm
h,
= 400 mm
21MPa normalweight concrete
=
fd
�= 420MPa
Pu = 1165 kN
Interior column, a.s = 40
Normalweight concrete
ACl318M-05
section
1 1.12.2.1
9. 3.2. 3
Procedure
Step I-Determine net bearing
pressure under factored load P11
f br = Pj(footing area)
Step 2-Express V11
= fb,.(footing area- prism area)
= fbr[2100 X 2100 - (400 + d)]
Step 3-Express <We=
d)[0.33 (
)b0d]
= d)[0.33 ( jf; )4(400+d)d]
Step 4-Equate V,, = d)Ve and solve
for d.
Ji:
7.7.1
Step 5-AlJow 7 5 mm clear cover
below steel plus one bottom bar
diameter to make h "" d + 100.
ALTERNATEMETHOD using Design aid Shear 5 .1
Step I-Determine net bearing
pressure under factored load P 11
fbr = P j(footing area).
Step 2-Estimate that bearing area of
shear prism is 1 0% of footing area.
Compute V 11 = fbr(I - 0.1 0)Aftg ·
Compute Vn = V jd).
9. 3.2. 3
11.12.2.1
Step 3-Find KIK2 with V11 = 1397 kN
and/; = 21 MPa
Note that since h/be < 2, K2 = 0.0 85.
Thus,
11.12.2.1
Step 4-Compute he+ be
With Kl = 35 88 MPa and he+ be =
800 mm, find d.
Calculation
Design aid
fbr = 1165 kN/(2100 mm x 2100 mm)
= 2.64 x 1o--4 kN/mm2
V 11 = 2.64 x 10 -4(4.41 x 10° mmL- (400 mm+d mm) 2]
= J 122.8 kN- (0.211 kN/mm)d- (2.64 X 10--4 kN/
mm 2)d2
<We= 0.75[0.33(J21 MPa)4(400+ d) mm(d) mm] x 10-5
= 4.537 x 10-3 kN/mm2(400+ d2) mm2
(1122.8- 0.21ld-2.64 X 10-4d.l) JcN = 4.537 X 10-j
(400d+d2) kN
= 4. 801 X 10-3d2 + 2.026d = 1122. 8
d2 + 422d +211 2 = 2 33 , 867. 94 + 44,521
(d + 211 ) 2 = 27 8,389
d+ 211 = 5 27.62
d= 316.64 mm
Use footing h = 316.6 + 100 = 416.6 mm
Make h = 420 mm
fbr = 1165 /(2100
X
2100) = 2.64 X } 0--4 kN/mm 2
v11 = 2.64 x 10--4 kN/mm2(0.90)(2100)(2100) = 1048.6 kN
vn = 1048.6/0.7 5 = 13 98 kN
K1K2 =
3 00+( 400- 300 )
= 305.0 MPa
3
Kl = o5.0 = 3588 MPa
1398 -l37 5
1 833 - 137 5
Table 5.1(c)
Table 5. l(b)
K2
400 mm+ 400 mm= 800 mm
8- 3360
= 31 3.4 mm
d = 300 + ( 350- 300) 358
4200-3360
As above, make footing h = 420 mm
he+ be
=
f
Note: In ALTERNATE METHOD, check assumed Step 2 proponion o shear prism area tofooting area.
%footing area= JO0[(d + bJ]2l(Aft0) = 100£(400 mm+ 320 nun/1(2100 x 2100) = 11.8% (estimare was 10%).
Vu should have been (1 - O. ll8)(1165) = l028 kN instead of estimated 1122 kN.
Table 5.l(a)
44
ACI DESIGN HANDBOOK-SP-17M(09)
Shear Example 10: Determination of thickness required for a flat slab based on required perimeter shear strength at an inte­
rior round column
Determine the thickness required for a two-way slab to resist an ultimate shear force of Vu= 676 kN, based on perimeter shear
strength at an interior circular column of 500 mm diameter when fc' = 28MPa for the normalweight slab concrete.
/0\
\,,! ,//
Gjven:
vn
O.s
ho
vu
he
f:
=
=
0.083(a.sdlb0 + 2)(§: )brfl �4(§: )b0d
40 for interior column,
30 for edge column, and
20 for corner column
= perimeter of shear prism= n(hc+ d)
= 676 kN
= 500 mm diameter
= 28MPa
/
a, = 40 for interior column,
30 for edge column,
20 for comer column
.,,,.----
..... '
'
I
Mechanism of shear prism
representing perimeter
shear strength surface
V, =0.083(a,d/b0 + 2)(,ff; )bod
S 0.33(�)b0d
/
I ·.
b0 = perimeter of shear prism
=x(hc + d)
Slab
h•
I
____
d/2
Diameter
he
.,,,,..
d/2
Plan at column
·•
J .·
I
··.\
I ·._ .
r-.
·
___.___,, 1
1
I.·
Averaged
.·1
.-· I
( I
.
·
•
Section at column
ACl318M-05
section
Procedure
Calculation
Design aid
11.12.1.2
Step 1-Set up equation
<I> V11 = 0.33(0.75)( Ji8 MPa)n(500 mm+ d mm)d
11.12.2.l
q> V11 = 0.33<)>( J1:) rt(hc + a) X d.
= 4.109 MPa(500d + d 2) mm2
9.3.2.3
11.12.2.1
Step 2-Equate Vu to <I> V11 and
676,000 N = 4.109 MPa(500d + d 1') mm'
solve for d.
1.0d2 + (500)d (mm)+ [(0.5)(500)) 2 = 164,517+ [(0.5)(500))2
(d + 250) 2= 227,017
d= ( J227,0l 7 mm2) - 250 mm= 476.5 mm - 250 mm=
226.5 mm
7.7.lc
Step 3-Make hs deep enough for hs = 226.5 mm+ 20 mm+ 22 mm= 268.5 mm
20 mm concrete cover plus diameter Use hs = 270 mm
or top bars. Estimate No. 22 bars.
ALTERNATEMETHOD with Design aid Shear 5.2
9.3.2.3
Table 5.2(b)
v11= 676//(0.75) = 901 kN
Step I-Compute V11 = Vu /<)>.
11.12.1.2
Withf; =28MPa and V11=901 kN, K3 = 160 x 10 3 + ( 200 x l0 3 - 160 x ]0 3 ) <901 - 847)
(1058-847)
obtain K3.
= 170. 2 x 10 3 mm 2
11.12.2. l
7.7.lc.2.1
Step 2-With h,= 500 mm and
K3= 170.2 x 103 mm2 and
assume No. 22 bars, with 20 mm
cover plus bottom bar diameter
3
3
Find d = 200 + (230 - 200) 170.2 x 1� - 145 x 1�
174 X JO� - 145 X JO
d= 226 mm
hs ::::: 226 + 20 + 22 = 268 mm
Use 270 mm
Table 5.2(a)
46
ACI DESIGN HANDBOOK-SP-17M(09)
Shear Example 12: Determination of closed ties required for the beam shown to resist flexural shear and determinate torque
Given:
J; = 35 MPa norrnalweight concrete
Grade 420 reinforcement
1111 = 271 kN
T11 = 72 kN·m determinate
Wu = 69 kN/m
= 420MPa
• • •
d =546m
600mm
(v
• •
I
40 mm clear cover
1-- (typical)
i-•--4-'-00 _m_m_-·�1
ACI318M-05
section
Procedure
11.0
Step I-Determine section properties
for torsion, allowing 6 mm as radius
of ties.
A cp=hwh
A oh = (bw - 90)(h - 90)
11.6.1
11.6.3.1
11.5.6.2
11.6.3.6
11.6.5.2
A cp =400 mm(600 mm) = 240,000 mm2
A 0h =(400 mm - 90 mm)(600 mm - 90 mm)
=158,100 mm2
A0 =0.85(158,100 mm 2)= 134,385 mm2
Pep = 2(400 mm+ 600 mm) = 2000 mm
Ph=2(400 mm- 90 mm+600 mm - 90 mm)= 1640 mm
A 0= 0.85A 0h
Pep 2(bw+ h)
Ph =2(bw - 90+ h - 90)
Step 2--Compute cracking torsion Tcr
Tc,.=0.33(0.75)(,/35 MPa)(240,000 mm 2)2/2000 mm
Tc r=0.33<\>(JJ: )A c/lPcp
=42,164, 817 N·mm
Compute threshold torsion=0.25Tcr · Tc ,-= 42,169,817 N-mm/106 =42.2 kN-m
Since T11=72 kN·m > 10.5 kN·m, ties Threshold torsion=0.25(42.2 kN-m) =10.5 kN·m
for torsion are required.
Step 3-ls section large enough?
fv =271 kNI(400 mm x 546 mm)= 1.24 Nlmm2
Compute!,, = Vj(b11.d)
f,,1=72 kN-m(l640 mm)l[l.7(158,000 mm2)2 ]
Computefvt =TuP,/(l.7A 0/)
=2.78 Nlmm2
Compute limit
Limit= 0.75(0.17 + 0.66](J3s MPa) =3.68 Nlmm2
=(j>[(0.17Jj; +U66Jj;)l1000
=
9.3.2.3
Calculation
ls
+fv�] < limit?
Therefore, section is large enough.
Step 4-Compute
(b wd)]l(<\>J;.d )
A v is= [Vu - 0.17<\>
J[f.;
JI:
Compute A,fs = T/[2<\>A 0J;,cot0]
Compute (A)s + 2A 1 ls)
J[I.24 + 2.78 ) =3.04 Nlm.m 2 < 3.68 Nlmm 2
2
2
3
- 0.17(0.75)(J35 MPa)(400)(510)
A 1• Is=271 ( 10 ) kN(0.75)(4
20 MPa)(510)
=0.73 mm 2lmm
72( 10 ) N-mm
=0.85 mm2 lmm
A 1Is= ( 2
)(0.75)(134,385)(4 20)cot45
AJs + 2A,fs=0.73+ 2(0.85) =2.43 mm
6
Use No. 13 ties for which (A v + 2A 1 )1
s =258 mm, and compute s=
s =2.5812.43=106.2 mm
285 mm!
s + 2A,fs).
Use 100 mm
ls 0.062( Jt
'Jc' )b,)fy, < (A 11 ls + 2A,ls)? 0.062(
)4001420 =0.349 mm YES
J3s
Design aid
48
ACI DESIGN HANDBOOK-SP-17M(09)
Shear Example 13: Determination of closed ties required for the beam of Example 12 to resistflexural shear and indetemunaie torque
�
Use the same data as that for Shear Example 12, except that the required torsion estimate of 72 kN-m is based on an indeterminate
analysis, not an equilibrium requirement.
= 35 MPa normalweight concrete
f/
bw = 400 mm
vu = 271 kN
= !yr =420MPa
= 600 mm
h
Tu = 72 kN-m (based on indeterminate analysis)
.t;,
ACl318M-05
section
Procedure
11.3.1.l
Step 1-Look up parameters for
11.5.7.2
Jc' 35 MPa, Grade 420 reinforcement,
bw=400 mm, h = 600 mm
l l.6.2.2a
11.6.3.1
11.6.3.6
Calculation
=
(1.118)193
kN
= 215.8 kN
KfcK vc
=
l l.6.2.2a
Kvs=224.7kN/m
K1�1=(1.118)(113.7kN-m)=127.1 kN-m
K_r�icr= (1.118)(50.3 kN·m) = 56.2 kN·m
K1s 56.44 k:N-m/mm
=
Step 2-If indeterminate T11 >
q>KfcK rc, (0.75)56.2 k:N·m=42.2 kN·m,
q>Kf�tcr , q>KfcKrcr can be used as Tu. Tu 72 k:N·m > 42.2 k:N-m
Use T11 =42.2 kN·m
0.25
,
42.2 k:N·m > 0.25(56.2 kN·m)=14.1 kN·m.
=
42.2
>
KfcKtcr
Step 3-If T,,
ties are required.
Therefore, ties are required.
Step 4-Section is large enough if
2 71 kN
4 2 . 2 kN·m
2
2
=
J£(V)(5�K1,K .,)] + [T.l(�K 1,K,)] < I
=
=
11.6. la
,
{ 5(0.75)(215.8)} + { 0.75 x 1 27.I kN-m
2
J{ 0.335} 2 + { 0.443}
11.6.3.7
11.6.5.3
Step 5-Compute A)s =
(V11 /q> - Kf�vc)/ Kvs
Compute 2A,Js= T,,f(q>K,s)
Compute (A vis+ 2A,Js)
No. 13 ties provide 258 mm2 /mm
Compute s< 258/(A/s+2A/s)
Step 6---Compute Ph 2(b + h - 7).
From Step 4, 2A,ls=1 .0 rnm 2
Compute At = (A,Js)(phcot 2 45)
Is At> At .min 0.42(Jj: )Ac/fy
-p0 1 ls?
=
=
r
=
= 0.555 < 1
Therefore, section is large enough.
A,Js = (271 kN/0.75 -215.8)/224.7 kN/m = 0.6S nun:;/mm
2A,Js=42.2 k:N·m/[0.75(56.44)] =1.0 nun2/mm
(A 11 /s + 2A,Js) = 1.65 mm2/mm
s< 258/1.65= 156.4 mm. Use 150 mm spacing.
Ph= 2(400 mm+ 600 mm -190 mm)= 1620 mm
Thus A,ls =(1/2)(1 .0)= 0.5 mm
A e = (0.5 mm)(1620 mm)(1.0) = 810 mm 2
A e .min = 0.42( ,.J35 MPa)240,000 mm 2 /420 MPa
2
2
- 1620(0.5) mm = 609.8 mm YES
11.6.6.1
Ae ,min =810 mm 2 governs. (No. 10
bars provide sufficient area, but
select No. 13 as a minimum.
Maximum spacing of longitudinal
bars=pJ8 or 300 mm
1 l .6.6.2
For 10 positions, use No. 13 bars. Place (See Shear Example 12)
three No. 13 in bottom and two No. 13
in each side face. Excess flexural
capacity in top at d from support can
serve in place of three No. 13 in top.
pJ8 = 1620/8 = 203 mm
Design aid
Shear 2
Table 2(a) &
2(c)
Shear 2(a) &
2(b)
Shear 6.l(a)
Shear 6.1(b)
Shear 6.2(b)
50
ACI DESIGN HANDBOOK-SP-17M(09)
Shear 2: Shear strength coefficients Kfc, Kvc• and Kvs
Reference: Sections 11.3. l. l and 11.5.7.2
Ve = 0.17(jJ:)bwd=0.9KrA,.d
Vs
Vn
=
=
Kfc =
Kvs =
K11, =
A.jy<i,ls = A,J(vJs
Ve+ Vs
J(fc' /28) (Table 2(a))
f,,.d (kN/mm) (Table 2(b))
(0.17
)b,.,d/1000 (kN) (Table 2(c))
J28
Section 11.2.1.1: When J,, is specified for lightweight concrete, substitute f,,10.56 for
Table 2(a)-Values Krc for various values of fc'
//. MPa
21
0.866
28
1.000
35
I.I 18
40
1.225
55
1.414
For Table 2(c):
d = h - 65 mm when h s 750 mm
d = h - 75 mm when h > 150 mm
Table 2(b)
Values K.,, kN/m
Beamh,mm
250
300
350
400
450
500
550
600
650
700
750
800
850
900
950
1000
1050
1100
1150
1200
f.v orJy,, MPa
280
51.8
65.8
79.8
93.8
107.8
121.8
135.8
149.8
163.8
177.8
189.0
203.0
217.0
231.0
245.0
259.0
273.0
287.0
301.0
315.0
420
77.7
98.7
I 19.7
140.7
161.7
182.7
203.7
224.7
245.7
266.7
283.5
340.5
325.5
346.5
367.5
388.5
409.5
430.5
451.5
472.5
70
1.581
Ji: , but keep fer 10.56 s Ji: .
ACI DESIGN HANDBOOK-SP-17M(09)
52
Shear 3: Minimum beam height to provide development length required for No. 19, No. 22, and No. 25 Grade 420 stirrups
Reference: Sections 12.13.2.1 and 12.13.2.2.
Section 11.5.2 states "Design yield strength of shear reinforcement (bars) shall not exceed 420 MPa, ...."
0.17 do(420)
--Ir;
h
Minimum beam height h = 2[0.17 db (420)/ Ji[+ 40] in millimeters.
Concrete J; , MPa
21
28
35
40
55
70
Minimum beam height h, mm
Stirrup size
No.19
No.22
660
582
528
490
434
396
757
666
605
559
493
450
No. 25
856
752
681
627
554
503
Nore: Values shown in the rable are for 40 mm clear cover over srirmps. For cover greater than 40 mm, add
2/cover-40 mm) to tabulated values.
Example: Determine whether a beam 600 mm high (h = 600 mm) with 35 MPa concrete will provide sufficient development
length for No. 19 Grade 420 vertical stirrups.
Solution: With/; = 35 MPa, for No. 19 stirrups, minimum beam height reads 528 mm for beams with 40 mm clear cover over
stirrups. Because h = 600 mm, the beam is deep enough.
54
ACI DESIGN HANDBOOK-SP-17M(09)
Shear 4.2: Shear strength Vs with Grade 420 U-stirrups
Reference: Sections 11.5.5.3 and 11.5.6.3
V5 = V11
-
Ve = A vfw(dls)
Ma.ximumbw = AJy,l(0.35s) whenfj :$ 31 MPa
Ma.ximumhw = Avfy/(0.062s Ji[) whenfj 2 31 MPa
Table 4.2(a)
Stirrup Beam
d.
size
mm
200
250
300
350
400
450
500
550
No.10
U600
stirrups 650
700
750
800
850
900
950
1000
Maximumbw
(mm) for
f/<;31 MPa
Values of V5 , kN
50
75
100
125
150
175
200
-
Spacings, mm
230
--
250
280
300
239
298
358
417
477
537
596
656
716
775
835
895
954
1014
1074
1133
1193
159
199
239
278
318
358
398
437
477
517
557
596
636
676
716
755
795
119
149
179
209
239
268
298
328
358
388
417
447
477
507
537
567
596
119
143
167
191
215
239
262
286
310
334
358
382
406
429
453
477
119
139
159
179
199
219
239
258
278
298
318
338
358
378
398
119
136
153
170
187
204
222
239
256
273
290
307
324
341
119
134
149
164
179
194
209
224
239
253
268
283
298
117
130
143
156
169
182
194
207
220
233
246
259
119
131
143
155
167
179
191
203
215
227
239
117
128
138
149
160
170
181
192
202
213
119
129
139
149
159
169
179
189
199
3408
2272
1704
1363
1136
974
852
741
682
609
568
280
300
·------ ---
-
--
--
-
--
-·---
--
350
400
- --
�--
-
-
-
450
-
Spacing > d/2
is not allowed
-
-119
128
136
145
153
162
170
500
,_
___
-
I
I
_]
I
I
I
=--cc....
--
-�--.�. -1
-·-
-- --
-
-----·-
I
_I
--
-II
119
127
134
142
149
119
126
133
I 19
487
426
379
341
350
400
450
500
--
Table 4.2(b)
Valu�s uf V,, kN
Sti_rrup
size
No. 13
U-
stirrups
Beam
d,
mm
200
250
300
350
400
450
500
550
600
650
700
750
800
850
900
950
JOO
Maxi1num b u .
(mm) for
f/<;31 MPa
50
75
100
125
150
175
200
Spacings. mm
230
250
---
433
542
650
759
867
975
1084
1192
1300
1409
1517
1625
1734
1842
1950
2059
2167
289
361
433
.506
578
650
722
795
867
939
1011
1084
1156
1228
1300
1373
1445
217
271
325
379
433
488
542
596
650
704
759
813
867
921
975
1029
1084
217
260
303
347
390
433
477
520
563
607
650
694
737
780
824
867
217
253
289
325
361
397
433
470
506
542
578
614
650
686
722
217
248
279
310
341
372
402
433
464
495
526
557
588
619
217
244
271
298
325
352
379
406
433
461
488
515
542
212
236
259
283
306
330
353
377
400
424
448
471
217
238
260
282
303
325
347
368
390
412
433
213
232
252
271
290
310
329
348
368
387
217
235
253
271
289
307
325
343
361
217
232
248
263
279
294
310
217
230
244
257
271
217
229
241
6192
4128
3096
2477
2064
1769
1548
1346
1238
1106
1032
885
774
688
..........,_
-
-
Spacing> d/2
is not allowed
I
I
-· �---
--
- -
- ----- �
I
I
I
���1
I
I
I
_I
- I
I ·- I
217
619
56
ACI DESIGN HANDBOOK-SP-17M(09)
Shear 5.2: Shear strength of slabs based on perimeter shear at interior round columns when no shear reinforcement is used
Reference: Sections 11.12.1.2 and 11.12.2.1
V11 =Ve= (K3) Jf: (kN)
he = column diameter (mm)
d = slab depth (mm)
K3 = 0.33nd(d + he) when he< 5.37d
K3 = 0.17nd(hc + 7.37d) when he> 5.37d for which Table 5.2(a) values are in bold type
Table 5.2(a)
Column
h,mm
200
250
300
350
400
450
500
550
600
650
700
750
800
850
900
950
l000
Values K3 (x 10 3 ). mm2
75
21
25
29
33
37
40
42
44
46
48
50
52
54
56
58
60
62
100
31
36
41
47
52
57
62
125
42
49
55
62
68
75
81
87
94
100
69
71
74
77
79
82
85
87
90
93
108
112
115
118
122
125
128
150
54
62
70
78
86
93
10)
109
117
124
)32
140
148
175
68
77
86
95
104
113
122
132
141
150
159
168
177
186
195
157
161
165
169
209
214
d,mm
200
83
93
104
114
124
135
145
156
166
176
187
197
207
218
228
238
249
230
l03
114
126
138
150
162
174
186
198
210
222
234
246
258
269
281
293
250
117
130
143
156
168
181
194
207
220
233
246
259
272
285
298
311
324
300
156
171
187
202
218
233
249
264
280
295
311
327
342
358
373
389
404
350
200
218
236
254
272
290
308
327
345
363
38]
399
417
435
454
472
490
400
249
270
290
311
332
352
373
394
415
435
456
477
498
518
539
560
581
450
303
327
350
373
397
420
443
467
490
5)3
537
560
583
606
630
653
676
Table 5.2(b)
Values Vn � V,, 1$, kN
Jc', MPa
K3 (x 10 3 ), rnm 2
21
40
80
120
160
200
250
300
400
500
600
800
900
1000
28
183
367
550
733
917
1146
1375
1833
2291
2750
3666
4124
4583
5041
5499
5957
6416
6874
7332
8249
35
212
423
635
847
1058
1323
1587
2ll7
2646
3175
4233
4762
5292
5821
6350
6879
7408
7937
8466
9525
237
473
710
947
1183
1479
1775
2366
2958
3550
4733
5324
5916
6508
7099
769)
8283
8874
9466
10,649
I JOO
1200
1300
1400
1500
1600
1800
40
253
506
759
1012
1265
1581
1897
2530
3162
3795
5060
5692
6325
6957
7589
8222
8854
9487
10,119
11,384
55
297
593
890
l )87
1483
1854
2225
2966
3708
4450
5933
6675
7416
8158
8899
9641
10,383
11,124
11,866
13,349
70
335
669
1004
1339
1673
2092
2510
3347
4183
5020
6693
7530
8367
9203
10,040
10,877
11.713
12,550
13,387
15,060
500
363
389
415
441
467
492
518
544
570
596
622
648
674
700
726
752
778
58
ACI DESIGN HANDBOOK-SP-17M(09)
Shear 6.2: Shear and torsion coefficients K1 s
Reference: Section 11.6.3.6
T,, = (2A 0 Arfyfs)cot0 = 2K1s(A/s) kN-m
A 0 = O.85(h - 3.5)(b - 3.5)
0 = 45 degrees
Table 6.2(a)
Values K1s (kN-mlmm) with Grade 280 ties
Beam/z,
mm
250
300
350
400
450
500
550
600
650
700
750
800
850
900
950
1000
250
6.09
8.00
9.90
11.80
13.71
15.61
17.52
19.42
21.32
23.23
25.13
27.04
28.94
30.84
32.75
34.65
300
8.00
10.50
12.99
15.49
17.99
20.49
22.99
25.49
27.99
30.49
32.99
35.49
37.98
40.48
42.98
45.48
350
9.90
12.99
16.09
19.18
22.28
25.37
28.46
31.56
34.65
37.75
40.84
43.93
47.03
50.12
53.22
56.31
400
11.80
15.49
19.18
22.87
26.56
30.25
33.94
37.63
41.32
45.01
48.69
52.38
56.07
59.76
63.45
67.14
450
13.71
17.99
22.28
26.56
30.84
35.13
39.41
43.70
47.98
52.26
56.55
60.83
65.12
69.40
73.68
77.97
Beam b, mm
500
15.61
20.49
25.37
30.25
35.13
40.01
44.89
49.77
54.64
59.52
64.40
69.28
74.16
79.04
83.92
88.80
550
17.52
22.99
28.46
33.94
39.41
44.89
50.36
55.83
61.31
66.78
72.26
77.73
83.20
88.68
94.15
99.63
600
19.42
25.49
31.56
37.63
43.70
49.77
55.83
61.90
67.97
74.04
80.11
86.18
92.25
98.32
104.39
110.46
650
21.32
27.99
34.65
41.32
47.98
54.64
61.31
67.97
74.64
81.30
87.96
94.63
101.29
107.96
114.62
121.28
700
23.23
30.49
37.75
45.01
52.26
59.52
66.78
74.04
81.30
88.56
95.82
103.08
110.34
117.60
124.85
132.11
750
25.13
32.99
40.84
48.69
56.55
64.40
72.26
80.11
87.96
95.82
103.67
111.53
119.38
127.23
135.09
142.94
600
29.13
38.23
47.34
56.44
65.55
74.65
83.75
92.86
101.96
111.06
120.17
129.27
138.37
147.48
156.58
165.68
650
31.99
41.98
51.98
61.98
71.97
81.97
91.96
101.96
111.96
121.95
l 31.95
141.94
151.94
161.94
171.93
181.93
700
34.84
45.73
56.62
67.51
78.40
89.29
100.17
111.06
121.95
132.84
143.73
154.62
165.51
176.39
187.28
198.17
750
37.70
Table 6.2(b)
Values K1s (kN-m/mm) with Grade 420 ties
Beam h.
mm
250
300
350
400
450
500
550
600
650
700
750
800
850
900
950
1000
250
9.14
12.00
14.85
17.71
20.56
23.42
26.28
29.13
31.99
34.84
37.70
40.56
43.41
46.27
49.12
51.98
300
12.00
15.74
19.49
23.24
26.99
30.74
34.49
38.23
41.98
45.73
49.48
53.23
56.98
60.73
64.47
68.22
350
14.85
19.49
24.13
28.77
33.42
38.06
42.70
47.34
51.98
56.62
61.26
65.90
70.54
75.18
79.83
84.47
400
17.71
23.24
28.77
34.31
39.84
45.37
50.91
56.44
61.98
67.51
73.04
78.58
84.11
89.64
95.18
100.71
450
20.56
26.99
33.42
39.84
46.27
52.69
59.12
65.55
71.97
78.40
84.82
91.25
97.68
104.10
110.53
116.95
Beam b, mm
500
23.42
30.74
38.06
45.37
52.69
60.01
67.33
74.65
81.97
89.29
96.60
103.92
111.24
118.56
125.88
133.20
550
26.28
34.49
42.70
50.91
59.12
67.33
75.54
83.75
91.96
100.17
108.39
116.60
124.81
133.02
141.23
149.44
49.48
61.26
73.04
84.82
96.60
108.39
120.17
l 31.95
143.73
155.51
167.29
179.07
190.85
202.63
214.41
60
ACI DESIGN HANDBOOK-SP-17M(09)
-�
h
i::=i;
f
Cross-Section
t .,y
i:,=0.005
Strain Distribution
Stress Distribution
Fig. 3.2-Column section analysis.
arranged in a circle have been published in The AC! Struc­
tural Journal (Everard 1997). The interaction diagrams
contained in SP-7 were subsequently published in SP-17A
(ACI Committee 340 1970).
The related equations were derived considering the
following:
(a) For rectangular and square columns having steel bars
placed on the end faces only, the reinforcement was
assumed to consist of two equal thin strips parallel to the
compression face of the section;
(b) For rectangular and square columns having steel bars
equally distributed along all four section faces, the
reinforcement was considered to consist of a thin
rectangular or square tube; and
(c) For square and circular sections having steel bars
arranged in a circle, the reinforcement was considered to
consist of a thin circular tube.
The interaction diagrams were developed using the rectan­
gular stress block, provided in Section 10.2.7. In all cases,
for reinforcement within the compressed portion of the depth
perpendicular to the compression face of the concrete (a =
fk), the compression stress was reduced by 0.85/; to account
for the concrete area displaced by the reinforcement bars
within the compression stress block.
The interaction diagrams were plotted in nondimensional
form. The vertical coordinate [Kil = P11 /(f;Ag)] represents
the nondimensional form of the nominal axial load strength
of the section. The horizontal coordinate [R11 = M,,l(fd Ai)J
represents the nondimensional nominal bending moment
strength of the section. The nondirnensional forms were used
so that the interaction diagrams could be used with any
system of units (SI or in.-lb units). Because ACI 3 l 8M-05
contains different <!>-factors in Chapter 9, Chapter 20, and
Appendix C, the strength reduction factor <I> was considered
as 1.0 so the nominal values in the interaction diagrams
could be used with any set of <!>-factors.
It is important to note that the <!>-factors provided in
Chapter 9 are based on the strain values in the tension reinforce­
ment farthest from the compression face of a member, or at
the centroid of the tension reinforcement. Code Section 9.3.2
references Sections 10.3.3 and 10.3.4 where the strain values for
tension conrrol and compression conrrol are defined.
It should also be noted that the eccentricity ratios (elh =
MIP), sometimes included as diagonal lines, are not included
in the interaction diagrams. Using the eccentricity ratio as a
coordinate with Kil or R,, can lead to inaccuracies because the
e/h lines converge rapidly at the lower ends of the diagrams.
Straight lines for tension steel stress ratios !,.If;, have been
plotted to assist in designing splices for the reinforcement.
Further, the ratio fsf/2, = 1.0 represents steel strain E:y = f;JE,
which is the boundary point for the compression control <!>­
factor, and the beginning of the transition zone for linear
increase of the <!>-factor to that for tension control.
To provide interpolation for the <!>-factor, other strain lines
were plotted. The strain line for f;1 = 0.005, the beginning of
the tension control zone, has been plotted on all diagrams. The
intermediate strain line for E:, = 0.035 has been plotted for steel
yield strength 420 MPa. The intermediate strain line for E:1 =
0.038 has been plotted for steel yield strength 520 MPa. Note
that all strains refer to the reinforcement bar or bars farthest
from the compression face of the section. Discussions and
tables related to the strength reduction factors a.re contained in
two articles in Concrete International (Everard 2002a,b).
To illustrate designs prohjbited by Section 10.3.5, strain
lines for f;1 = 0.004 have also been plotted. Designs between
the lines for i::, = 0.004 and K11 < 0.10 are not permitted by
ACI 318M-05. This includes tension axial loads, with K,,
negative. Tension axial loads are not included in the interaction
diagrams. However, the interaction diagram lines for tension
axial loads are nearly linear from Kil = 0.0 to R,, = 0.0 with
[Kil= A51f..lifd Ag)]. This is discussed in the next section.
Straight lines for Kmax are also provided on each interac­
tion diagram. Here, Kmax refers to the maximum permissible
nominal axial load on a column laterally reinforced with ties
conforming to Section 7.10.5. Defining Ko as the theoretical
axial compression strength of a member with R11 = 0.0, K111ax
= 0.80K0 or considering Eq. (10-2) without the <!>-factor,
Pn,max = 0.8[0.85/d (Ag -As1 ) + f,.A51]
(3-5)
Then,
(3-6)
For columns with spirals conforming to Section 7.10.4,
multiply K1110x values from the interaction diagrams by 0.85/
0.80 ratio.
The number of longitudinal reinforcement bars that can be
contained is not limited to the number shown on the interaction
62
ACI DESIGN HANDBOOK-SP-17M(09)
pll
•
;:-�ilure
�rfac:e
Fig. 3.7-Faillfre surface S3, which is reciprocal ofsurface S 1.
Fig. 3.5-Failure surface S 1.
Fig. 3.8-Graphical representation of reciprocal load
method.
Fig. 3.6-Failure surface S 2 .
pni
1
P,u
1
1
pn_v
Po
= -+---
(3-10)
where
approximation of nominal axial load strength at
eccentricities ex and ey
nominal axial load strength for eccentricity ey
,
P v:
along the y-axis only (x-axis is axis of bending)
P11Y = nominal axial load strength for eccentricity ex
along the x-axis only (y-axis is axis of bending)
Po = nominal axial load strength for zero eccentricity
For design purposes, when <j) is constant, 1/P11; in Eq. (3-10)
can be used. The variable K11 = P,/(f; Ag) can be used
directly in the reciprocal equation, as follows
P11 ;
1
1
1
-+---
K nx
K,,Y
Ko
(3-11)
where the K values refer to the corresponding P11 values as
defined above.
Once a preliminary cross section with an estimated steel
ratio Pg is selected, calculate R,v: and R11v using the actual
bending moments about the cross section x- and y-axes.
Obtain the corresponding values of K,,x and K11 l' from the
interaction diagrams presented in this chapter as the intersec­
tion of R,, value and the assumed steel ratio curve for P g ·
Then, obtain the theoretical compression axial load strength
Ko at the intersection of the steel ratio curve and the vertical
axis for zero R,,.
3.3.2 Load contour method-The load contour method
uses the failure surface S2 (Fig. 3.6) and works with a load
contour defined by a plane at a constant value of P11 (Fig. 3.9).
The load contour defining the relationship between M11x and
M,,y for a constant P,, can be expressed nondimensionally as
(3-12)
ACI DESIGN HANDBOOK-SP-17M(09)
64
3.4-Columns examples
Columns Example 1: Determination of required steel area for a rectangular tied column with bars on four faces with slender­
ness ratio below critical value
For a rectangular tied column with bars equally distributed along four faces, find steel area.
Given:
Loading
Pu = 2491 kN, and Mu = 443 kN·m
Assume <I> =0.70
Nominal axial load P11 =249110.70 =3559 kN
Nominal moment M11 =443/0.70 =633 kN-m
Materials
Compressive strength of concrete Jc' =28 MPa
Yield strength of reinforcementfy =420 MPa
Normalized maximum size of aggregate is 25 mm
Design conditions
Short column braced against sidesway
ACI318M-05
section
p
+
f A
A) Compute K,, =
c
8
M
B) Compute R,, = --"-
f,.' A 8 h
.
h-5
C) Estnnate y :::: --
K,, =
R,, =
3559
(28)(2
X
X
10
3
10
5
)
633 X 10
(28)(2
X
5
=0.64
6
J0 )(500)
=0.23
500- 125 = 0.75
500
D) Determine the appropriate inter- For a rectangular tied column with bars along four faces,
action diagrams.
fd =28 MPa,fv =420 MPa, and an estimated y of 0.75,
enter diagram R28-420.7 and R28-420.8 with K11 =0.64
and R11 = 0.23, respectively.
E) Read Pg for K11 and R,, values from Read Pg = 0.041 for y =0.7 and Pg= 0.039 for y =0.8. Columns
3.2.2
Interpolating pR =0.040 for y = 0.75
appropriate interaction diagrams.
(R28-420.7)
F) Compute required A51 from A 51 = Required A 51 = 0.040 x 2 x 10::, mm2 = 8000 mm
and 3.2.3
PgAg ·
(R28-420.8)
h
10.2
10.3
Design aid
Procedure
Calculation
Determine column section size.
Given: h = 500 mm and b =400 mm
Determine reinforcement ratio Pg
P,, = 3559 kN
using known values of variables on M,, =633 kN·m
appropriate interaction diagrams and h= 500 mm
compute required cross section area b = 400 mm
A 51 of longitudinal reinforcement.
A £ =b x h = 500 x 400 =2 x 1o 5 mm2
y::::
L
66
ACI DESIGN HANDBOOK-SP-17M(09)
Columns Example 3: Selection of reinforcement for a square spiral column with slenderness ratio below critical value
For the square spiral column section shown, select reinforcement.
h=450mm
�
Loading
P,, =2936kN andM,, =298kN·m
Assume <I>= 0.70
Nominal axial load Pn= 2936/0.70 =4194kN
Nominal moment Mn= 298/0.70 =426 kN·m
E
E
Materials
Compressive strength of concrete Jc'= 28 MPa
Yield strength of reinforcement.t;, = 420 MPa
Normalized maximum size of aggregate is 25 mm
..c:
Design conditions
Column section size h = b =450 mm
Slenderness effects may be neglected because kljh isknown to be below critical value
ACl318M-05
section
Procedure
Determine reinforcement ration p8
usingknown values of variables on
appropriate interaction diagrams and
compute required cross section area
Ast of longitudinal reinforcement.
Kn =
Mn
B) Compute R n =-Jc' A 8 h
Rn =
E) Read Pg for Kn and R,, values.
Design aid
Mn =426kN-m
h=450 mm
b=450 mm
A�= bx h= 450 x 450 = 2.03 x 105 mm2
A) Compute Kn = pn
Jc' A s
.
- 125
C) Estimate y "" h
- -h
D) Determine the appropriate
interaction diagrams.
10.2
10.3
Calculation
P11 = 4194kN
4194 X 10
=0.74
5
X 10 )
28)(2.03
(
3
6
426
X
]0
(28)(2.03
X
J0 )(450)
5
=0.17
450- 125 =0.72
450
For a square spiral column,/) =28 MPa,JY = 420 MPa,
and an estimated y of 0.72, use interaction diagram
S28-420.7 and S28-420.8.
For K11 = 0.73 and Rn= 0.16, and for
Pg= 0.035
y = 0.70:
Pg = 0.031
y =0.80:
Pg = 0.034
y =0.72:
5
As1 = 0.034 x 2.03 x 10 mm2 = 6902 mm2
Y ""
Columns
3.20.2
(S28-420.7)
and 3.20.3
(S28-420.8)
68
10.2
10.3
ACI DESIGN HANDBOOK-SP-17M(09)
.
125
--C) Est1mate y "" hh
D) Determine the appropriate
interaction diagrams.
125
r� 380380
= 0.67
For a rectangular tied column withfc' = 35 MPa,.t;, =
420 MPa, and y = 0.67, use interaction diagrams
R35-420.6 and R35-420.7.
E) Read Pg for K11 and R11 values from For K11 = 0.264, R,,= 0.188, and for
appropriate interaction diagrams.
y = 0.60:
Pg= 0.043
y = 0.70
Pg = 0.034
y = 0.67:
P .e = 0.037
5
F) Compute required A 51 from Ast = Required A51 = 0.037 x 1.44 x 10 mm L = 5343 mmL
PgA8 and add approx.imately 15% for Use A st � 6100 mm 2
skew bending.
Columns
3.3.1
(R35-420.6)
and 3.3.2
(R35-420.7)
ACI DESIGN HANDBOOK-SP-17M(09)
70
p
A) Compute Kn = _n_
Jc'Ac
K,. =
M
B) Compute R,. = --11fc'Ai
Rn =
.
h- 125
C) Estunate y "' -h
D) Determine the appropriate interaction diagrams.
E) Read p8 for K11 and R 11 values from
appropriate interaction diagrams.
F) Compute required As, from As, =
p i'Ai'.
5973 X ]0
3
(35)(1.45 X 10 5 )
77
X
10
= 1.16
6
5
( 35)(1.45 X 10 )(432)
= 0.0352
430 - 125 =0.71
430
For a circular column with/; =35 MPa andfy =
420 MPa. Use interaction diagram C35-420.7.
For K11 =1.18, R11=0.0356, and
y=0.71:
pi' =0.040
Required As,=0.040 x 1.45 x 1o:i mmz
=5808 mm 2
y"'
Columns
3.15.2
(C35-420.7)
72
ACI DESIGN HANDBOOK-SP-17M(09)
COLUMNS 3.1.2 - Nominal load-moment strength interaction diagram, R21-420.7
2.4
r-T""T"T'""T""T',-,-"'T""T--r-T"""T"""1rir------------�r--------,
INTERACTION DIAGRAM R21-420.7
= 21MPa
2.2 --------------l',-------+----1 fc
J;,= 420MPa
2.0
r = o.1
�--f"'-.-.----l�__b,,=====,,..,..,,..==,,...,,,....�=""'r"""...........j
�----------,��,-----+--�
1 .6 �......
O>
--
_
'
0
1 .2
•••
•
I
I
1.8
<!.
h
••
•
------f . I
p
t-'-.------1---,--f,- �c--+-�----'---/------'x----,----->'<-----+-�---+------+--�
0.0 L..I....L...1.....L....J....i....&....a.::..i....L...1.....L...1.........::.L...1-.L....L...�....;J;:...........J.L&....,_.....L...1L..Q;;;;;....,_.i...J...a......i....i....i.....1.ZC�...&....I..L.L...L...1-i.....l
0.00
0.05
0.10
0.15
0.20
0.25
0.30
0.35
0.40
0.45
0.50
74
ACI DESIGN HANDBOOK-SP-17M(09)
COLUMNS 3.1.4 -Nominal load-moment strength interaction diagram, R21-420.9
2.4
2.2
2.0
1.8
1.6
1.4
II
1.2
... ..,
1 •I"' • ·1
INTERACTION DIAGRAM R21-420.9
fc = 21 MPa
!y= 420MPa
b
]'fl
••
r = o.9
•
•
I
I
�---�����I
I
-7
1.0
0.8
0.6
0.4
0.2
0.0 L.L.J....&..L.J..J..l...i...lr:::J..J...u...J....L.1.�L...L..L...&..L..i.a...u..L.J.J...L.L.J...J..I..U...UI.-I....L.l.....,_w..&....&..L..i...:lca..L.U...,L..J,.Li....L.1.....,_L.1
0.00 0.05 0.10 0.15 0.20 0.25 0.30 0.35 0.40 0.45 0.50 0.55 0.60 0.65
76
ACI DESIGN HANDBOOK-SP-17M(09)
COLUMNS 3.2.2 - Nominal load-moment strength interaction diagram, R28-420.7
IE
2.0 ........,....,_.,......,..___..........-.-_______________--.--------,
INTERACTIONDIAGRAM R28-420.7
fc = 28 MPa
1.8 .......,.,�----------1--------, J;, = 420 MPa
y = 0.7
h
I""'
I
rh "" I ,..
.--:----+--:..........,
e
e
• • ••
•
I
I.
e
Pn
i:::i
1.2
-
a>
<!.
--
C)
'+-
II
........_--J�'t -�
1.0
0.8
0.6
0.0 L-L....i......1....1--1.....i..:::::i......i......a......1.....i..:::.i......1......i....J�...L-.L...-1.-L...i..:::-.i.�...L....J._.J,..J._.L...-l,-=-i.......i......L..4:...L....J._.J,...J....1
0.00
0.05
0.10
0.15
0.20
0.25
0.30
0.35
0.40
78
ACI DESIGN HANDBOOK-SP-17M(09)
COLUMNS 3.2.4 - Nominal load-moment strength interaction diagram, R28-420.9
I,. I,.
2.0 ......._...........__.._,........_...,...,r--_____ _______...------------.
INTERACTION DIAGRAM R28-420.9
fc = 28 MPa
.....,.,�--+----...-+--�
8
1.
fy = 420 MPa
r = o.9
� ., I ..I
----------,
e
e
•
•
• • •
h
1.2
-....
1.0
II
0.8
--
<)
0.6
0.2
��.,__._�----&....l,,j.........----'....._..........__,__._........_.�.................--.._,,_........._._.
0.0 ..............----'....._.......___,_........
0.00
0.05
0.10
0.15
0.20
0.25
0.30
0.35
0.40
0.45
0.50
80
ACI DESIGN HANDBOOK-SP-17M(09)
COLUMNS 3.3.2 -Nominal load-moment strength interaction diagram, R35-420.7
IE
1.8 .....................-.....--..--.---,---.-.....--,---------------..-----------,,
I
h
INTERACTION DIAGRAM R35-420.7
J-= � ""I ..
fc = 35 MPa
..-------------.
e
e
1.6 t---"',.-----------T""l')---<=-ttt,M---1 f'y = 420 MPa
I
r = o.1
1.2 -'-<------......-��----+--I
--
-
(.)
C:
C.
II
.
C:
1.0
•
•
• • •
I
I
e·
______
P
I
0.8
0.6
0.0 L--L--'-......-'--�--'---'-�.L.......l__,_�....L...-'--I....Ll.__._....._._;;:-..,.""""""'.....,_..,__"'--1,--'-.=6..,__.L.....L�............
0.35
0.30
0.25
0.20
0.15
0.10
0.05
0.00
82
ACI DESIGN HANDBOOK-SP-17M(09)
COLUMNS 3.3.4 - Nominal load-moment strength interaction diagram, R35-420.9
1.8 ....,.._,...,...........,,.....,......--,,_..,......-,--____________________----,
_e .
I
I
1 .2
-....
--
�----r--1�,�,--��--t---------------------------
P
I
1.0
'-'
C
0...
II
0.8
C
0.6
0.0 .........................._........................._........................_�..............._................................._._............................................_.__................_._..........
0.00
0.05
0.10
0.15
0.20
0.25
0.30
0.35
0.40
0.45
84
ACI DESIGN HANDBOOK-SP-17M(09)
COLUMNS 3.4.2 - Nominal load-moment strength interaction diagram, R40-420.7
I,. I
1.6 .--.--...--,--,,.......,,.......,---,---,--.,---------------------h
INTERACTION DIAGRAM R40-420.7
________
1.4
_""'"
__,
fc = 40 MPa
J;, =
420 MPa
E
]'ii
•
r = o.1
I
I
I--.
•
•
..
..--.----.
.___• __•_•__.
---II
c
�
0.6 --------+------......,,-�---
0.0
"--L......L......-----.l.....:;.i---1---1----o11:;.......i......i.---i.'---I-.....L.---'-'�--'----i,:;---'----'-�-'--'-.....,__,_....K......................................
0.00
0.05
0.10
0.15
0.20
0.25
0.30
84
ACI DESIGN HANDBOOK-SP-17M(09)
COLUMNS 3.4.2 - Nominal load-moment strength interaction diagram, R40-420.7
I,. I
1.6 .--.--...--,--,,.......,,.......,---,---,--.,---------------------h
INTERACTION DIAGRAM R40-420.7
________
1.4
_""'"
__,
fc = 40 MPa
J;, =
420 MPa
E
]'ii
•
r = o.1
I
I
I--.
•
•
..
..--.----.
.___• __•_•__.
---II
c
�
0.6 --------+------......,,-�---
0.0
"--L......L......-----.l.....:;.i---1---1----o11:;.......i......i.---i.'---I-.....L.---'-'�--'----i,:;---'----'-�-'--'-.....,__,_....K......................................
0.00
0.05
0.10
0.15
0.20
0.25
0.30
82
ACI DESIGN HANDBOOK-SP-17M(09)
COLUMNS 3.3.4 - Nominal load-moment strength interaction diagram, R35-420.9
1.8 ....,.._,...,...........,,.....,......--,,_..,......-,--____________________----,
_e .
I
I
1 .2
-....
--
�----r--1�,�,--��--t---------------------------
P
I
1.0
'-'
C
0...
II
0.8
C
0.6
0.0 .........................._........................._........................_�..............._................................._._............................................_.__................_._..........
0.00
0.05
0.10
0.15
0.20
0.25
0.30
0.35
0.40
0.45
80
ACI DESIGN HANDBOOK-SP-17M(09)
COLUMNS 3.3.2 -Nominal load-moment strength interaction diagram, R35-420.7
IE
1.8 .....................-.....--..--.---,---.-.....--,---------------..-----------,,
I
h
INTERACTION DIAGRAM R35-420.7
J-= � ""I ..
fc = 35 MPa
..-------------.
e
e
1.6 t---"',.-----------T""l')---<=-ttt,M---1 f'y = 420 MPa
I
r = o.1
1.2 -'-<------......-��----+--I
--
-
(.)
C:
C.
II
.
C:
1.0
•
•
• • •
I
I
e·
______
P
I
0.8
0.6
0.0 L--L--'-......-'--�--'---'-�.L.......l__,_�....L...-'--I....Ll.__._....._._;;:-..,.""""""'.....,_..,__"'--1,--'-.=6..,__.L.....L�............
0.35
0.30
0.25
0.20
0.15
0.10
0.05
0.00
78
ACI DESIGN HANDBOOK-SP-17M(09)
COLUMNS 3.2.4 - Nominal load-moment strength interaction diagram, R28-420.9
I,. I,.
2.0 ......._...........__.._,........_...,...,r--_____ _______...------------.
INTERACTION DIAGRAM R28-420.9
fc = 28 MPa
.....,.,�--+----...-+--�
8
1.
fy = 420 MPa
r = o.9
� ., I ..I
----------,
e
e
•
•
• • •
h
1.2
-....
1.0
II
0.8
--
<)
0.6
0.2
��.,__._�----&....l,,j.........----'....._..........__,__._........_.�.................--.._,,_........._._.
0.0 ..............----'....._.......___,_........
0.00
0.05
0.10
0.15
0.20
0.25
0.30
0.35
0.40
0.45
0.50
76
ACI DESIGN HANDBOOK-SP-17M(09)
COLUMNS 3.2.2 - Nominal load-moment strength interaction diagram, R28-420.7
IE
2.0 ........,....,_.,......,..___..........-.-_______________--.--------,
INTERACTIONDIAGRAM R28-420.7
fc = 28 MPa
1.8 .......,.,�----------1--------, J;, = 420 MPa
y = 0.7
h
I""'
I
rh "" I ,..
.--:----+--:..........,
e
e
• • ••
•
I
I.
e
Pn
i:::i
1.2
-
a>
<!.
--
C)
'+-
II
........_--J�'t -�
1.0
0.8
0.6
0.0 L-L....i......1....1--1.....i..:::::i......i......a......1.....i..:::.i......1......i....J�...L-.L...-1.-L...i..:::-.i.�...L....J._.J,..J._.L...-l,-=-i.......i......L..4:...L....J._.J,...J....1
0.00
0.05
0.10
0.15
0.20
0.25
0.30
0.35
0.40
74
ACI DESIGN HANDBOOK-SP-17M(09)
COLUMNS 3.1.4 -Nominal load-moment strength interaction diagram, R21-420.9
2.4
2.2
2.0
1.8
1.6
1.4
II
1.2
... ..,
1 •I"' • ·1
INTERACTION DIAGRAM R21-420.9
fc = 21 MPa
!y= 420MPa
b
]'fl
••
r = o.9
•
•
I
I
�---�����I
I
-7
1.0
0.8
0.6
0.4
0.2
0.0 L.L.J....&..L.J..J..l...i...lr:::J..J...u...J....L.1.�L...L..L...&..L..i.a...u..L.J.J...L.L.J...J..I..U...UI.-I....L.l.....,_w..&....&..L..i...:lca..L.U...,L..J,.Li....L.1.....,_L.1
0.00 0.05 0.10 0.15 0.20 0.25 0.30 0.35 0.40 0.45 0.50 0.55 0.60 0.65
72
ACI DESIGN HANDBOOK-SP-17M(09)
COLUMNS 3.1.2 - Nominal load-moment strength interaction diagram, R21-420.7
2.4
r-T""T"T'""T""T',-,-"'T""T--r-T"""T"""1rir------------�r--------,
INTERACTION DIAGRAM R21-420.7
= 21MPa
2.2 --------------l',-------+----1 fc
J;,= 420MPa
2.0
r = o.1
�--f"'-.-.----l�__b,,=====,,..,..,,..==,,...,,,....�=""'r"""...........j
�----------,��,-----+--�
1 .6 �......
O>
--
_
'
0
1 .2
•••
•
I
I
1.8
<!.
h
••
•
------f . I
p
t-'-.------1---,--f,- �c--+-�----'---/------'x----,----->'<-----+-�---+------+--�
0.0 L..I....L...1.....L....J....i....&....a.::..i....L...1.....L...1.........::.L...1-.L....L...�....;J;:...........J.L&....,_.....L...1L..Q;;;;;....,_.i...J...a......i....i....i.....1.ZC�...&....I..L.L...L...1-i.....l
0.00
0.05
0.10
0.15
0.20
0.25
0.30
0.35
0.40
0.45
0.50
ACI DESIGN HANDBOOK-SP-17M(09)
70
p
A) Compute Kn = _n_
Jc'Ac
K,. =
M
B) Compute R,. = --11fc'Ai
Rn =
.
h- 125
C) Estunate y "' -h
D) Determine the appropriate interaction diagrams.
E) Read p8 for K11 and R 11 values from
appropriate interaction diagrams.
F) Compute required As, from As, =
p i'Ai'.
5973 X ]0
3
(35)(1.45 X 10 5 )
77
X
10
= 1.16
6
5
( 35)(1.45 X 10 )(432)
= 0.0352
430 - 125 =0.71
430
For a circular column with/; =35 MPa andfy =
420 MPa. Use interaction diagram C35-420.7.
For K11 =1.18, R11=0.0356, and
y=0.71:
pi' =0.040
Required As,=0.040 x 1.45 x 1o:i mmz
=5808 mm 2
y"'
Columns
3.15.2
(C35-420.7)
68
10.2
10.3
ACI DESIGN HANDBOOK-SP-17M(09)
.
125
--C) Est1mate y "" hh
D) Determine the appropriate
interaction diagrams.
125
r� 380380
= 0.67
For a rectangular tied column withfc' = 35 MPa,.t;, =
420 MPa, and y = 0.67, use interaction diagrams
R35-420.6 and R35-420.7.
E) Read Pg for K11 and R11 values from For K11 = 0.264, R,,= 0.188, and for
appropriate interaction diagrams.
y = 0.60:
Pg= 0.043
y = 0.70
Pg = 0.034
y = 0.67:
P .e = 0.037
5
F) Compute required A 51 from Ast = Required A51 = 0.037 x 1.44 x 10 mm L = 5343 mmL
PgA8 and add approx.imately 15% for Use A st � 6100 mm 2
skew bending.
Columns
3.3.1
(R35-420.6)
and 3.3.2
(R35-420.7)
66
ACI DESIGN HANDBOOK-SP-17M(09)
Columns Example 3: Selection of reinforcement for a square spiral column with slenderness ratio below critical value
For the square spiral column section shown, select reinforcement.
h=450mm
�
Loading
P,, =2936kN andM,, =298kN·m
Assume <I>= 0.70
Nominal axial load Pn= 2936/0.70 =4194kN
Nominal moment Mn= 298/0.70 =426 kN·m
E
E
Materials
Compressive strength of concrete Jc'= 28 MPa
Yield strength of reinforcement.t;, = 420 MPa
Normalized maximum size of aggregate is 25 mm
..c:
Design conditions
Column section size h = b =450 mm
Slenderness effects may be neglected because kljh isknown to be below critical value
ACl318M-05
section
Procedure
Determine reinforcement ration p8
usingknown values of variables on
appropriate interaction diagrams and
compute required cross section area
Ast of longitudinal reinforcement.
Kn =
Mn
B) Compute R n =-Jc' A 8 h
Rn =
E) Read Pg for Kn and R,, values.
Design aid
Mn =426kN-m
h=450 mm
b=450 mm
A�= bx h= 450 x 450 = 2.03 x 105 mm2
A) Compute Kn = pn
Jc' A s
.
- 125
C) Estimate y "" h
- -h
D) Determine the appropriate
interaction diagrams.
10.2
10.3
Calculation
P11 = 4194kN
4194 X 10
=0.74
5
X 10 )
28)(2.03
(
3
6
426
X
]0
(28)(2.03
X
J0 )(450)
5
=0.17
450- 125 =0.72
450
For a square spiral column,/) =28 MPa,JY = 420 MPa,
and an estimated y of 0.72, use interaction diagram
S28-420.7 and S28-420.8.
For K11 = 0.73 and Rn= 0.16, and for
Pg= 0.035
y = 0.70:
Pg = 0.031
y =0.80:
Pg = 0.034
y =0.72:
5
As1 = 0.034 x 2.03 x 10 mm2 = 6902 mm2
Y ""
Columns
3.20.2
(S28-420.7)
and 3.20.3
(S28-420.8)
ACI DESIGN HANDBOOK-SP-17M(09)
64
3.4-Columns examples
Columns Example 1: Determination of required steel area for a rectangular tied column with bars on four faces with slender­
ness ratio below critical value
For a rectangular tied column with bars equally distributed along four faces, find steel area.
Given:
Loading
Pu = 2491 kN, and Mu = 443 kN·m
Assume <I> =0.70
Nominal axial load P11 =249110.70 =3559 kN
Nominal moment M11 =443/0.70 =633 kN-m
Materials
Compressive strength of concrete Jc' =28 MPa
Yield strength of reinforcementfy =420 MPa
Normalized maximum size of aggregate is 25 mm
Design conditions
Short column braced against sidesway
ACI318M-05
section
p
+
f A
A) Compute K,, =
c
8
M
B) Compute R,, = --"-
f,.' A 8 h
.
h-5
C) Estnnate y :::: --
K,, =
R,, =
3559
(28)(2
X
X
10
3
10
5
)
633 X 10
(28)(2
X
5
=0.64
6
J0 )(500)
=0.23
500- 125 = 0.75
500
D) Determine the appropriate inter- For a rectangular tied column with bars along four faces,
action diagrams.
fd =28 MPa,fv =420 MPa, and an estimated y of 0.75,
enter diagram R28-420.7 and R28-420.8 with K11 =0.64
and R11 = 0.23, respectively.
E) Read Pg for K11 and R,, values from Read Pg = 0.041 for y =0.7 and Pg= 0.039 for y =0.8. Columns
3.2.2
Interpolating pR =0.040 for y = 0.75
appropriate interaction diagrams.
(R28-420.7)
F) Compute required A51 from A 51 = Required A 51 = 0.040 x 2 x 10::, mm2 = 8000 mm
and 3.2.3
PgAg ·
(R28-420.8)
h
10.2
10.3
Design aid
Procedure
Calculation
Determine column section size.
Given: h = 500 mm and b =400 mm
Determine reinforcement ratio Pg
P,, = 3559 kN
using known values of variables on M,, =633 kN·m
appropriate interaction diagrams and h= 500 mm
compute required cross section area b = 400 mm
A 51 of longitudinal reinforcement.
A £ =b x h = 500 x 400 =2 x 1o 5 mm2
y::::
L
62
ACI DESIGN HANDBOOK-SP-17M(09)
pll
•
;:-�ilure
�rfac:e
Fig. 3.7-Faillfre surface S3, which is reciprocal ofsurface S 1.
Fig. 3.5-Failure surface S 1.
Fig. 3.8-Graphical representation of reciprocal load
method.
Fig. 3.6-Failure surface S 2 .
pni
1
P,u
1
1
pn_v
Po
= -+---
(3-10)
where
approximation of nominal axial load strength at
eccentricities ex and ey
nominal axial load strength for eccentricity ey
,
P v:
along the y-axis only (x-axis is axis of bending)
P11Y = nominal axial load strength for eccentricity ex
along the x-axis only (y-axis is axis of bending)
Po = nominal axial load strength for zero eccentricity
For design purposes, when <j) is constant, 1/P11; in Eq. (3-10)
can be used. The variable K11 = P,/(f; Ag) can be used
directly in the reciprocal equation, as follows
P11 ;
1
1
1
-+---
K nx
K,,Y
Ko
(3-11)
where the K values refer to the corresponding P11 values as
defined above.
Once a preliminary cross section with an estimated steel
ratio Pg is selected, calculate R,v: and R11v using the actual
bending moments about the cross section x- and y-axes.
Obtain the corresponding values of K,,x and K11 l' from the
interaction diagrams presented in this chapter as the intersec­
tion of R,, value and the assumed steel ratio curve for P g ·
Then, obtain the theoretical compression axial load strength
Ko at the intersection of the steel ratio curve and the vertical
axis for zero R,,.
3.3.2 Load contour method-The load contour method
uses the failure surface S2 (Fig. 3.6) and works with a load
contour defined by a plane at a constant value of P11 (Fig. 3.9).
The load contour defining the relationship between M11x and
M,,y for a constant P,, can be expressed nondimensionally as
(3-12)
60
ACI DESIGN HANDBOOK-SP-17M(09)
-�
h
i::=i;
f
Cross-Section
t .,y
i:,=0.005
Strain Distribution
Stress Distribution
Fig. 3.2-Column section analysis.
arranged in a circle have been published in The AC! Struc­
tural Journal (Everard 1997). The interaction diagrams
contained in SP-7 were subsequently published in SP-17A
(ACI Committee 340 1970).
The related equations were derived considering the
following:
(a) For rectangular and square columns having steel bars
placed on the end faces only, the reinforcement was
assumed to consist of two equal thin strips parallel to the
compression face of the section;
(b) For rectangular and square columns having steel bars
equally distributed along all four section faces, the
reinforcement was considered to consist of a thin
rectangular or square tube; and
(c) For square and circular sections having steel bars
arranged in a circle, the reinforcement was considered to
consist of a thin circular tube.
The interaction diagrams were developed using the rectan­
gular stress block, provided in Section 10.2.7. In all cases,
for reinforcement within the compressed portion of the depth
perpendicular to the compression face of the concrete (a =
fk), the compression stress was reduced by 0.85/; to account
for the concrete area displaced by the reinforcement bars
within the compression stress block.
The interaction diagrams were plotted in nondimensional
form. The vertical coordinate [Kil = P11 /(f;Ag)] represents
the nondimensional form of the nominal axial load strength
of the section. The horizontal coordinate [R11 = M,,l(fd Ai)J
represents the nondimensional nominal bending moment
strength of the section. The nondirnensional forms were used
so that the interaction diagrams could be used with any
system of units (SI or in.-lb units). Because ACI 3 l 8M-05
contains different <!>-factors in Chapter 9, Chapter 20, and
Appendix C, the strength reduction factor <I> was considered
as 1.0 so the nominal values in the interaction diagrams
could be used with any set of <!>-factors.
It is important to note that the <!>-factors provided in
Chapter 9 are based on the strain values in the tension reinforce­
ment farthest from the compression face of a member, or at
the centroid of the tension reinforcement. Code Section 9.3.2
references Sections 10.3.3 and 10.3.4 where the strain values for
tension conrrol and compression conrrol are defined.
It should also be noted that the eccentricity ratios (elh =
MIP), sometimes included as diagonal lines, are not included
in the interaction diagrams. Using the eccentricity ratio as a
coordinate with Kil or R,, can lead to inaccuracies because the
e/h lines converge rapidly at the lower ends of the diagrams.
Straight lines for tension steel stress ratios !,.If;, have been
plotted to assist in designing splices for the reinforcement.
Further, the ratio fsf/2, = 1.0 represents steel strain E:y = f;JE,
which is the boundary point for the compression control <!>­
factor, and the beginning of the transition zone for linear
increase of the <!>-factor to that for tension control.
To provide interpolation for the <!>-factor, other strain lines
were plotted. The strain line for f;1 = 0.005, the beginning of
the tension control zone, has been plotted on all diagrams. The
intermediate strain line for E:, = 0.035 has been plotted for steel
yield strength 420 MPa. The intermediate strain line for E:1 =
0.038 has been plotted for steel yield strength 520 MPa. Note
that all strains refer to the reinforcement bar or bars farthest
from the compression face of the section. Discussions and
tables related to the strength reduction factors a.re contained in
two articles in Concrete International (Everard 2002a,b).
To illustrate designs prohjbited by Section 10.3.5, strain
lines for f;1 = 0.004 have also been plotted. Designs between
the lines for i::, = 0.004 and K11 < 0.10 are not permitted by
ACI 318M-05. This includes tension axial loads, with K,,
negative. Tension axial loads are not included in the interaction
diagrams. However, the interaction diagram lines for tension
axial loads are nearly linear from Kil = 0.0 to R,, = 0.0 with
[Kil= A51f..lifd Ag)]. This is discussed in the next section.
Straight lines for Kmax are also provided on each interac­
tion diagram. Here, Kmax refers to the maximum permissible
nominal axial load on a column laterally reinforced with ties
conforming to Section 7.10.5. Defining Ko as the theoretical
axial compression strength of a member with R11 = 0.0, K111ax
= 0.80K0 or considering Eq. (10-2) without the <!>-factor,
Pn,max = 0.8[0.85/d (Ag -As1 ) + f,.A51]
(3-5)
Then,
(3-6)
For columns with spirals conforming to Section 7.10.4,
multiply K1110x values from the interaction diagrams by 0.85/
0.80 ratio.
The number of longitudinal reinforcement bars that can be
contained is not limited to the number shown on the interaction
58
ACI DESIGN HANDBOOK-SP-17M(09)
Shear 6.2: Shear and torsion coefficients K1 s
Reference: Section 11.6.3.6
T,, = (2A 0 Arfyfs)cot0 = 2K1s(A/s) kN-m
A 0 = O.85(h - 3.5)(b - 3.5)
0 = 45 degrees
Table 6.2(a)
Values K1s (kN-mlmm) with Grade 280 ties
Beam/z,
mm
250
300
350
400
450
500
550
600
650
700
750
800
850
900
950
1000
250
6.09
8.00
9.90
11.80
13.71
15.61
17.52
19.42
21.32
23.23
25.13
27.04
28.94
30.84
32.75
34.65
300
8.00
10.50
12.99
15.49
17.99
20.49
22.99
25.49
27.99
30.49
32.99
35.49
37.98
40.48
42.98
45.48
350
9.90
12.99
16.09
19.18
22.28
25.37
28.46
31.56
34.65
37.75
40.84
43.93
47.03
50.12
53.22
56.31
400
11.80
15.49
19.18
22.87
26.56
30.25
33.94
37.63
41.32
45.01
48.69
52.38
56.07
59.76
63.45
67.14
450
13.71
17.99
22.28
26.56
30.84
35.13
39.41
43.70
47.98
52.26
56.55
60.83
65.12
69.40
73.68
77.97
Beam b, mm
500
15.61
20.49
25.37
30.25
35.13
40.01
44.89
49.77
54.64
59.52
64.40
69.28
74.16
79.04
83.92
88.80
550
17.52
22.99
28.46
33.94
39.41
44.89
50.36
55.83
61.31
66.78
72.26
77.73
83.20
88.68
94.15
99.63
600
19.42
25.49
31.56
37.63
43.70
49.77
55.83
61.90
67.97
74.04
80.11
86.18
92.25
98.32
104.39
110.46
650
21.32
27.99
34.65
41.32
47.98
54.64
61.31
67.97
74.64
81.30
87.96
94.63
101.29
107.96
114.62
121.28
700
23.23
30.49
37.75
45.01
52.26
59.52
66.78
74.04
81.30
88.56
95.82
103.08
110.34
117.60
124.85
132.11
750
25.13
32.99
40.84
48.69
56.55
64.40
72.26
80.11
87.96
95.82
103.67
111.53
119.38
127.23
135.09
142.94
600
29.13
38.23
47.34
56.44
65.55
74.65
83.75
92.86
101.96
111.06
120.17
129.27
138.37
147.48
156.58
165.68
650
31.99
41.98
51.98
61.98
71.97
81.97
91.96
101.96
111.96
121.95
l 31.95
141.94
151.94
161.94
171.93
181.93
700
34.84
45.73
56.62
67.51
78.40
89.29
100.17
111.06
121.95
132.84
143.73
154.62
165.51
176.39
187.28
198.17
750
37.70
Table 6.2(b)
Values K1s (kN-m/mm) with Grade 420 ties
Beam h.
mm
250
300
350
400
450
500
550
600
650
700
750
800
850
900
950
1000
250
9.14
12.00
14.85
17.71
20.56
23.42
26.28
29.13
31.99
34.84
37.70
40.56
43.41
46.27
49.12
51.98
300
12.00
15.74
19.49
23.24
26.99
30.74
34.49
38.23
41.98
45.73
49.48
53.23
56.98
60.73
64.47
68.22
350
14.85
19.49
24.13
28.77
33.42
38.06
42.70
47.34
51.98
56.62
61.26
65.90
70.54
75.18
79.83
84.47
400
17.71
23.24
28.77
34.31
39.84
45.37
50.91
56.44
61.98
67.51
73.04
78.58
84.11
89.64
95.18
100.71
450
20.56
26.99
33.42
39.84
46.27
52.69
59.12
65.55
71.97
78.40
84.82
91.25
97.68
104.10
110.53
116.95
Beam b, mm
500
23.42
30.74
38.06
45.37
52.69
60.01
67.33
74.65
81.97
89.29
96.60
103.92
111.24
118.56
125.88
133.20
550
26.28
34.49
42.70
50.91
59.12
67.33
75.54
83.75
91.96
100.17
108.39
116.60
124.81
133.02
141.23
149.44
49.48
61.26
73.04
84.82
96.60
108.39
120.17
l 31.95
143.73
155.51
167.29
179.07
190.85
202.63
214.41
56
ACI DESIGN HANDBOOK-SP-17M(09)
Shear 5.2: Shear strength of slabs based on perimeter shear at interior round columns when no shear reinforcement is used
Reference: Sections 11.12.1.2 and 11.12.2.1
V11 =Ve= (K3) Jf: (kN)
he = column diameter (mm)
d = slab depth (mm)
K3 = 0.33nd(d + he) when he< 5.37d
K3 = 0.17nd(hc + 7.37d) when he> 5.37d for which Table 5.2(a) values are in bold type
Table 5.2(a)
Column
h,mm
200
250
300
350
400
450
500
550
600
650
700
750
800
850
900
950
l000
Values K3 (x 10 3 ). mm2
75
21
25
29
33
37
40
42
44
46
48
50
52
54
56
58
60
62
100
31
36
41
47
52
57
62
125
42
49
55
62
68
75
81
87
94
100
69
71
74
77
79
82
85
87
90
93
108
112
115
118
122
125
128
150
54
62
70
78
86
93
10)
109
117
124
)32
140
148
175
68
77
86
95
104
113
122
132
141
150
159
168
177
186
195
157
161
165
169
209
214
d,mm
200
83
93
104
114
124
135
145
156
166
176
187
197
207
218
228
238
249
230
l03
114
126
138
150
162
174
186
198
210
222
234
246
258
269
281
293
250
117
130
143
156
168
181
194
207
220
233
246
259
272
285
298
311
324
300
156
171
187
202
218
233
249
264
280
295
311
327
342
358
373
389
404
350
200
218
236
254
272
290
308
327
345
363
38]
399
417
435
454
472
490
400
249
270
290
311
332
352
373
394
415
435
456
477
498
518
539
560
581
450
303
327
350
373
397
420
443
467
490
5)3
537
560
583
606
630
653
676
Table 5.2(b)
Values Vn � V,, 1$, kN
Jc', MPa
K3 (x 10 3 ), rnm 2
21
40
80
120
160
200
250
300
400
500
600
800
900
1000
28
183
367
550
733
917
1146
1375
1833
2291
2750
3666
4124
4583
5041
5499
5957
6416
6874
7332
8249
35
212
423
635
847
1058
1323
1587
2ll7
2646
3175
4233
4762
5292
5821
6350
6879
7408
7937
8466
9525
237
473
710
947
1183
1479
1775
2366
2958
3550
4733
5324
5916
6508
7099
769)
8283
8874
9466
10,649
I JOO
1200
1300
1400
1500
1600
1800
40
253
506
759
1012
1265
1581
1897
2530
3162
3795
5060
5692
6325
6957
7589
8222
8854
9487
10,119
11,384
55
297
593
890
l )87
1483
1854
2225
2966
3708
4450
5933
6675
7416
8158
8899
9641
10,383
11,124
11,866
13,349
70
335
669
1004
1339
1673
2092
2510
3347
4183
5020
6693
7530
8367
9203
10,040
10,877
11.713
12,550
13,387
15,060
500
363
389
415
441
467
492
518
544
570
596
622
648
674
700
726
752
778
54
ACI DESIGN HANDBOOK-SP-17M(09)
Shear 4.2: Shear strength Vs with Grade 420 U-stirrups
Reference: Sections 11.5.5.3 and 11.5.6.3
V5 = V11
-
Ve = A vfw(dls)
Ma.ximumbw = AJy,l(0.35s) whenfj :$ 31 MPa
Ma.ximumhw = Avfy/(0.062s Ji[) whenfj 2 31 MPa
Table 4.2(a)
Stirrup Beam
d.
size
mm
200
250
300
350
400
450
500
550
No.10
U600
stirrups 650
700
750
800
850
900
950
1000
Maximumbw
(mm) for
f/<;31 MPa
Values of V5 , kN
50
75
100
125
150
175
200
-
Spacings, mm
230
--
250
280
300
239
298
358
417
477
537
596
656
716
775
835
895
954
1014
1074
1133
1193
159
199
239
278
318
358
398
437
477
517
557
596
636
676
716
755
795
119
149
179
209
239
268
298
328
358
388
417
447
477
507
537
567
596
119
143
167
191
215
239
262
286
310
334
358
382
406
429
453
477
119
139
159
179
199
219
239
258
278
298
318
338
358
378
398
119
136
153
170
187
204
222
239
256
273
290
307
324
341
119
134
149
164
179
194
209
224
239
253
268
283
298
117
130
143
156
169
182
194
207
220
233
246
259
119
131
143
155
167
179
191
203
215
227
239
117
128
138
149
160
170
181
192
202
213
119
129
139
149
159
169
179
189
199
3408
2272
1704
1363
1136
974
852
741
682
609
568
280
300
·------ ---
-
--
--
-
--
-·---
--
350
400
- --
�--
-
-
-
450
-
Spacing > d/2
is not allowed
-
-119
128
136
145
153
162
170
500
,_
___
-
I
I
_]
I
I
I
=--cc....
--
-�--.�. -1
-·-
-- --
-
-----·-
I
_I
--
-II
119
127
134
142
149
119
126
133
I 19
487
426
379
341
350
400
450
500
--
Table 4.2(b)
Valu�s uf V,, kN
Sti_rrup
size
No. 13
U-
stirrups
Beam
d,
mm
200
250
300
350
400
450
500
550
600
650
700
750
800
850
900
950
JOO
Maxi1num b u .
(mm) for
f/<;31 MPa
50
75
100
125
150
175
200
Spacings. mm
230
250
---
433
542
650
759
867
975
1084
1192
1300
1409
1517
1625
1734
1842
1950
2059
2167
289
361
433
.506
578
650
722
795
867
939
1011
1084
1156
1228
1300
1373
1445
217
271
325
379
433
488
542
596
650
704
759
813
867
921
975
1029
1084
217
260
303
347
390
433
477
520
563
607
650
694
737
780
824
867
217
253
289
325
361
397
433
470
506
542
578
614
650
686
722
217
248
279
310
341
372
402
433
464
495
526
557
588
619
217
244
271
298
325
352
379
406
433
461
488
515
542
212
236
259
283
306
330
353
377
400
424
448
471
217
238
260
282
303
325
347
368
390
412
433
213
232
252
271
290
310
329
348
368
387
217
235
253
271
289
307
325
343
361
217
232
248
263
279
294
310
217
230
244
257
271
217
229
241
6192
4128
3096
2477
2064
1769
1548
1346
1238
1106
1032
885
774
688
..........,_
-
-
Spacing> d/2
is not allowed
I
I
-· �---
--
- -
- ----- �
I
I
I
���1
I
I
I
_I
- I
I ·- I
217
619
ACI DESIGN HANDBOOK-SP-17M(09)
52
Shear 3: Minimum beam height to provide development length required for No. 19, No. 22, and No. 25 Grade 420 stirrups
Reference: Sections 12.13.2.1 and 12.13.2.2.
Section 11.5.2 states "Design yield strength of shear reinforcement (bars) shall not exceed 420 MPa, ...."
0.17 do(420)
--Ir;
h
Minimum beam height h = 2[0.17 db (420)/ Ji[+ 40] in millimeters.
Concrete J; , MPa
21
28
35
40
55
70
Minimum beam height h, mm
Stirrup size
No.19
No.22
660
582
528
490
434
396
757
666
605
559
493
450
No. 25
856
752
681
627
554
503
Nore: Values shown in the rable are for 40 mm clear cover over srirmps. For cover greater than 40 mm, add
2/cover-40 mm) to tabulated values.
Example: Determine whether a beam 600 mm high (h = 600 mm) with 35 MPa concrete will provide sufficient development
length for No. 19 Grade 420 vertical stirrups.
Solution: With/; = 35 MPa, for No. 19 stirrups, minimum beam height reads 528 mm for beams with 40 mm clear cover over
stirrups. Because h = 600 mm, the beam is deep enough.
50
ACI DESIGN HANDBOOK-SP-17M(09)
Shear 2: Shear strength coefficients Kfc, Kvc• and Kvs
Reference: Sections 11.3. l. l and 11.5.7.2
Ve = 0.17(jJ:)bwd=0.9KrA,.d
Vs
Vn
=
=
Kfc =
Kvs =
K11, =
A.jy<i,ls = A,J(vJs
Ve+ Vs
J(fc' /28) (Table 2(a))
f,,.d (kN/mm) (Table 2(b))
(0.17
)b,.,d/1000 (kN) (Table 2(c))
J28
Section 11.2.1.1: When J,, is specified for lightweight concrete, substitute f,,10.56 for
Table 2(a)-Values Krc for various values of fc'
//. MPa
21
0.866
28
1.000
35
I.I 18
40
1.225
55
1.414
For Table 2(c):
d = h - 65 mm when h s 750 mm
d = h - 75 mm when h > 150 mm
Table 2(b)
Values K.,, kN/m
Beamh,mm
250
300
350
400
450
500
550
600
650
700
750
800
850
900
950
1000
1050
1100
1150
1200
f.v orJy,, MPa
280
51.8
65.8
79.8
93.8
107.8
121.8
135.8
149.8
163.8
177.8
189.0
203.0
217.0
231.0
245.0
259.0
273.0
287.0
301.0
315.0
420
77.7
98.7
I 19.7
140.7
161.7
182.7
203.7
224.7
245.7
266.7
283.5
340.5
325.5
346.5
367.5
388.5
409.5
430.5
451.5
472.5
70
1.581
Ji: , but keep fer 10.56 s Ji: .
48
ACI DESIGN HANDBOOK-SP-17M(09)
Shear Example 13: Determination of closed ties required for the beam of Example 12 to resistflexural shear and indetemunaie torque
�
Use the same data as that for Shear Example 12, except that the required torsion estimate of 72 kN-m is based on an indeterminate
analysis, not an equilibrium requirement.
= 35 MPa normalweight concrete
f/
bw = 400 mm
vu = 271 kN
= !yr =420MPa
= 600 mm
h
Tu = 72 kN-m (based on indeterminate analysis)
.t;,
ACl318M-05
section
Procedure
11.3.1.l
Step 1-Look up parameters for
11.5.7.2
Jc' 35 MPa, Grade 420 reinforcement,
bw=400 mm, h = 600 mm
l l.6.2.2a
11.6.3.1
11.6.3.6
Calculation
=
(1.118)193
kN
= 215.8 kN
KfcK vc
=
l l.6.2.2a
Kvs=224.7kN/m
K1�1=(1.118)(113.7kN-m)=127.1 kN-m
K_r�icr= (1.118)(50.3 kN·m) = 56.2 kN·m
K1s 56.44 k:N-m/mm
=
Step 2-If indeterminate T11 >
q>KfcK rc, (0.75)56.2 k:N·m=42.2 kN·m,
q>Kf�tcr , q>KfcKrcr can be used as Tu. Tu 72 k:N·m > 42.2 k:N-m
Use T11 =42.2 kN·m
0.25
,
42.2 k:N·m > 0.25(56.2 kN·m)=14.1 kN·m.
=
42.2
>
KfcKtcr
Step 3-If T,,
ties are required.
Therefore, ties are required.
Step 4-Section is large enough if
2 71 kN
4 2 . 2 kN·m
2
2
=
J£(V)(5�K1,K .,)] + [T.l(�K 1,K,)] < I
=
=
11.6. la
,
{ 5(0.75)(215.8)} + { 0.75 x 1 27.I kN-m
2
J{ 0.335} 2 + { 0.443}
11.6.3.7
11.6.5.3
Step 5-Compute A)s =
(V11 /q> - Kf�vc)/ Kvs
Compute 2A,Js= T,,f(q>K,s)
Compute (A vis+ 2A,Js)
No. 13 ties provide 258 mm2 /mm
Compute s< 258/(A/s+2A/s)
Step 6---Compute Ph 2(b + h - 7).
From Step 4, 2A,ls=1 .0 rnm 2
Compute At = (A,Js)(phcot 2 45)
Is At> At .min 0.42(Jj: )Ac/fy
-p0 1 ls?
=
=
r
=
= 0.555 < 1
Therefore, section is large enough.
A,Js = (271 kN/0.75 -215.8)/224.7 kN/m = 0.6S nun:;/mm
2A,Js=42.2 k:N·m/[0.75(56.44)] =1.0 nun2/mm
(A 11 /s + 2A,Js) = 1.65 mm2/mm
s< 258/1.65= 156.4 mm. Use 150 mm spacing.
Ph= 2(400 mm+ 600 mm -190 mm)= 1620 mm
Thus A,ls =(1/2)(1 .0)= 0.5 mm
A e = (0.5 mm)(1620 mm)(1.0) = 810 mm 2
A e .min = 0.42( ,.J35 MPa)240,000 mm 2 /420 MPa
2
2
- 1620(0.5) mm = 609.8 mm YES
11.6.6.1
Ae ,min =810 mm 2 governs. (No. 10
bars provide sufficient area, but
select No. 13 as a minimum.
Maximum spacing of longitudinal
bars=pJ8 or 300 mm
1 l .6.6.2
For 10 positions, use No. 13 bars. Place (See Shear Example 12)
three No. 13 in bottom and two No. 13
in each side face. Excess flexural
capacity in top at d from support can
serve in place of three No. 13 in top.
pJ8 = 1620/8 = 203 mm
Design aid
Shear 2
Table 2(a) &
2(c)
Shear 2(a) &
2(b)
Shear 6.l(a)
Shear 6.1(b)
Shear 6.2(b)
46
ACI DESIGN HANDBOOK-SP-17M(09)
Shear Example 12: Determination of closed ties required for the beam shown to resist flexural shear and determinate torque
Given:
J; = 35 MPa norrnalweight concrete
Grade 420 reinforcement
1111 = 271 kN
T11 = 72 kN·m determinate
Wu = 69 kN/m
= 420MPa
• • •
d =546m
600mm
(v
• •
I
40 mm clear cover
1-- (typical)
i-•--4-'-00 _m_m_-·�1
ACI318M-05
section
Procedure
11.0
Step I-Determine section properties
for torsion, allowing 6 mm as radius
of ties.
A cp=hwh
A oh = (bw - 90)(h - 90)
11.6.1
11.6.3.1
11.5.6.2
11.6.3.6
11.6.5.2
A cp =400 mm(600 mm) = 240,000 mm2
A 0h =(400 mm - 90 mm)(600 mm - 90 mm)
=158,100 mm2
A0 =0.85(158,100 mm 2)= 134,385 mm2
Pep = 2(400 mm+ 600 mm) = 2000 mm
Ph=2(400 mm- 90 mm+600 mm - 90 mm)= 1640 mm
A 0= 0.85A 0h
Pep 2(bw+ h)
Ph =2(bw - 90+ h - 90)
Step 2--Compute cracking torsion Tcr
Tc,.=0.33(0.75)(,/35 MPa)(240,000 mm 2)2/2000 mm
Tc r=0.33<\>(JJ: )A c/lPcp
=42,164, 817 N·mm
Compute threshold torsion=0.25Tcr · Tc ,-= 42,169,817 N-mm/106 =42.2 kN-m
Since T11=72 kN·m > 10.5 kN·m, ties Threshold torsion=0.25(42.2 kN-m) =10.5 kN·m
for torsion are required.
Step 3-ls section large enough?
fv =271 kNI(400 mm x 546 mm)= 1.24 Nlmm2
Compute!,, = Vj(b11.d)
f,,1=72 kN-m(l640 mm)l[l.7(158,000 mm2)2 ]
Computefvt =TuP,/(l.7A 0/)
=2.78 Nlmm2
Compute limit
Limit= 0.75(0.17 + 0.66](J3s MPa) =3.68 Nlmm2
=(j>[(0.17Jj; +U66Jj;)l1000
=
9.3.2.3
Calculation
ls
+fv�] < limit?
Therefore, section is large enough.
Step 4-Compute
(b wd)]l(<\>J;.d )
A v is= [Vu - 0.17<\>
J[f.;
JI:
Compute A,fs = T/[2<\>A 0J;,cot0]
Compute (A)s + 2A 1 ls)
J[I.24 + 2.78 ) =3.04 Nlm.m 2 < 3.68 Nlmm 2
2
2
3
- 0.17(0.75)(J35 MPa)(400)(510)
A 1• Is=271 ( 10 ) kN(0.75)(4
20 MPa)(510)
=0.73 mm 2lmm
72( 10 ) N-mm
=0.85 mm2 lmm
A 1Is= ( 2
)(0.75)(134,385)(4 20)cot45
AJs + 2A,fs=0.73+ 2(0.85) =2.43 mm
6
Use No. 13 ties for which (A v + 2A 1 )1
s =258 mm, and compute s=
s =2.5812.43=106.2 mm
285 mm!
s + 2A,fs).
Use 100 mm
ls 0.062( Jt
'Jc' )b,)fy, < (A 11 ls + 2A,ls)? 0.062(
)4001420 =0.349 mm YES
J3s
Design aid
44
ACI DESIGN HANDBOOK-SP-17M(09)
Shear Example 10: Determination of thickness required for a flat slab based on required perimeter shear strength at an inte­
rior round column
Determine the thickness required for a two-way slab to resist an ultimate shear force of Vu= 676 kN, based on perimeter shear
strength at an interior circular column of 500 mm diameter when fc' = 28MPa for the normalweight slab concrete.
/0\
\,,! ,//
Gjven:
vn
O.s
ho
vu
he
f:
=
=
0.083(a.sdlb0 + 2)(§: )brfl �4(§: )b0d
40 for interior column,
30 for edge column, and
20 for corner column
= perimeter of shear prism= n(hc+ d)
= 676 kN
= 500 mm diameter
= 28MPa
/
a, = 40 for interior column,
30 for edge column,
20 for comer column
.,,,.----
..... '
'
I
Mechanism of shear prism
representing perimeter
shear strength surface
V, =0.083(a,d/b0 + 2)(,ff; )bod
S 0.33(�)b0d
/
I ·.
b0 = perimeter of shear prism
=x(hc + d)
Slab
h•
I
____
d/2
Diameter
he
.,,,,..
d/2
Plan at column
·•
J .·
I
··.\
I ·._ .
r-.
·
___.___,, 1
1
I.·
Averaged
.·1
.-· I
( I
.
·
•
Section at column
ACl318M-05
section
Procedure
Calculation
Design aid
11.12.1.2
Step 1-Set up equation
<I> V11 = 0.33(0.75)( Ji8 MPa)n(500 mm+ d mm)d
11.12.2.l
q> V11 = 0.33<)>( J1:) rt(hc + a) X d.
= 4.109 MPa(500d + d 2) mm2
9.3.2.3
11.12.2.1
Step 2-Equate Vu to <I> V11 and
676,000 N = 4.109 MPa(500d + d 1') mm'
solve for d.
1.0d2 + (500)d (mm)+ [(0.5)(500)) 2 = 164,517+ [(0.5)(500))2
(d + 250) 2= 227,017
d= ( J227,0l 7 mm2) - 250 mm= 476.5 mm - 250 mm=
226.5 mm
7.7.lc
Step 3-Make hs deep enough for hs = 226.5 mm+ 20 mm+ 22 mm= 268.5 mm
20 mm concrete cover plus diameter Use hs = 270 mm
or top bars. Estimate No. 22 bars.
ALTERNATEMETHOD with Design aid Shear 5.2
9.3.2.3
Table 5.2(b)
v11= 676//(0.75) = 901 kN
Step I-Compute V11 = Vu /<)>.
11.12.1.2
Withf; =28MPa and V11=901 kN, K3 = 160 x 10 3 + ( 200 x l0 3 - 160 x ]0 3 ) <901 - 847)
(1058-847)
obtain K3.
= 170. 2 x 10 3 mm 2
11.12.2. l
7.7.lc.2.1
Step 2-With h,= 500 mm and
K3= 170.2 x 103 mm2 and
assume No. 22 bars, with 20 mm
cover plus bottom bar diameter
3
3
Find d = 200 + (230 - 200) 170.2 x 1� - 145 x 1�
174 X JO� - 145 X JO
d= 226 mm
hs ::::: 226 + 20 + 22 = 268 mm
Use 270 mm
Table 5.2(a)
42
ACI DESIGN HANDBOOK-SP-17M(09)
Shear Example 8: Determination of required thickness of a footing to satisfy perimeter shear strength at a rectangular column
be
= 400 mm
h,
= 400 mm
21MPa normalweight concrete
=
fd
�= 420MPa
Pu = 1165 kN
Interior column, a.s = 40
Normalweight concrete
ACl318M-05
section
1 1.12.2.1
9. 3.2. 3
Procedure
Step I-Determine net bearing
pressure under factored load P11
f br = Pj(footing area)
Step 2-Express V11
= fb,.(footing area- prism area)
= fbr[2100 X 2100 - (400 + d)]
Step 3-Express <We=
d)[0.33 (
)b0d]
= d)[0.33 ( jf; )4(400+d)d]
Step 4-Equate V,, = d)Ve and solve
for d.
Ji:
7.7.1
Step 5-AlJow 7 5 mm clear cover
below steel plus one bottom bar
diameter to make h "" d + 100.
ALTERNATEMETHOD using Design aid Shear 5 .1
Step I-Determine net bearing
pressure under factored load P 11
fbr = P j(footing area).
Step 2-Estimate that bearing area of
shear prism is 1 0% of footing area.
Compute V 11 = fbr(I - 0.1 0)Aftg ·
Compute Vn = V jd).
9. 3.2. 3
11.12.2.1
Step 3-Find KIK2 with V11 = 1397 kN
and/; = 21 MPa
Note that since h/be < 2, K2 = 0.0 85.
Thus,
11.12.2.1
Step 4-Compute he+ be
With Kl = 35 88 MPa and he+ be =
800 mm, find d.
Calculation
Design aid
fbr = 1165 kN/(2100 mm x 2100 mm)
= 2.64 x 1o--4 kN/mm2
V 11 = 2.64 x 10 -4(4.41 x 10° mmL- (400 mm+d mm) 2]
= J 122.8 kN- (0.211 kN/mm)d- (2.64 X 10--4 kN/
mm 2)d2
<We= 0.75[0.33(J21 MPa)4(400+ d) mm(d) mm] x 10-5
= 4.537 x 10-3 kN/mm2(400+ d2) mm2
(1122.8- 0.21ld-2.64 X 10-4d.l) JcN = 4.537 X 10-j
(400d+d2) kN
= 4. 801 X 10-3d2 + 2.026d = 1122. 8
d2 + 422d +211 2 = 2 33 , 867. 94 + 44,521
(d + 211 ) 2 = 27 8,389
d+ 211 = 5 27.62
d= 316.64 mm
Use footing h = 316.6 + 100 = 416.6 mm
Make h = 420 mm
fbr = 1165 /(2100
X
2100) = 2.64 X } 0--4 kN/mm 2
v11 = 2.64 x 10--4 kN/mm2(0.90)(2100)(2100) = 1048.6 kN
vn = 1048.6/0.7 5 = 13 98 kN
K1K2 =
3 00+( 400- 300 )
= 305.0 MPa
3
Kl = o5.0 = 3588 MPa
1398 -l37 5
1 833 - 137 5
Table 5.1(c)
Table 5. l(b)
K2
400 mm+ 400 mm= 800 mm
8- 3360
= 31 3.4 mm
d = 300 + ( 350- 300) 358
4200-3360
As above, make footing h = 420 mm
he+ be
=
f
Note: In ALTERNATE METHOD, check assumed Step 2 proponion o shear prism area tofooting area.
%footing area= JO0[(d + bJ]2l(Aft0) = 100£(400 mm+ 320 nun/1(2100 x 2100) = 11.8% (estimare was 10%).
Vu should have been (1 - O. ll8)(1165) = l028 kN instead of estimated 1122 kN.
Table 5.l(a)
DESIGN OF CONCRETE ELEMENTS IN ACCORDANCE WITH ACI 318M-05
85
COLUMNS 3.4.3 - Nominal load-moment strength interaction diagram, R40-420.8
1.6 -------------...........---------------..---------,
I
I.
e
pn
-�
I
1.0
�--J-/ _____._
°'
--
<r..u
-II
:::s::::
0.8
0.6
0.0 L.....L--1...-&.....i.....:::i.....i---1........1..-':J.-.L-..1---Jc:;....J.._....i-�.....&..-'-"""-........i--M...._......_..c.......__.,__._'""'-...__...__.,__._�
0.35
0.30
0.25
0.20
0.15
0.00
0.10
0.05
86
ACI DESIGN HANDBOOK-SP-17M(09)
COLUMNS 3.4.4- Nominal load-moment strength interaction diagram, R40-420.9
1.6 -----...------- --------------.--------,
1.4
-,.._
--
CJ
II
�
---------+---------'"'----l
INTERACTION DIAGRAM R40-420.9 ..
h
,
I"' Jb '" I .. I
.(c = 40 MPa
i---'----+--�
iy= 420MPa
•
e
r= o.9
•• • ••
0.8
0.6
0.0 l...J..--'-..l......li..-L....I--J.-L_..t:;...L-.L.....,L.....io:;...L....JL-,L....L..J.-L.....L...a-.L......L-1.L..L....JL.-L...A-..i.....L-"'...i.....L......L......,._........___._
0.35
0.40
0.30
0.25
0.20
0.15
0.05
0.00
0.10
87
DESIGN OF CONCRETE ELEMENTS IN ACCORDANCE WITH ACI 318M-05
COLUMNS 3.5.1 - Nominal load-moment strength interaction diagram, R65-520.6
1.6
1.4
1.2
INTERACTION DIAGRAM R65-520.6
f'c= 65MPa
J;,= 520MPa
y=0.6
-
-
I�•,..
••
h
1h
•
�• 1
�I
••
I
I
:e
1.0
--
p
·
I
Cl
-<{
(.)
II
r'\
�
0.8
0.6
1.0
0.0 L-J.--I-.J-..1--L.......i.....:s:�.J....L.....1....i.::;J,....l--'-...L.....,,c;...i.--L-.J-..1i,::;,,..,i.;::.1.-�;........1,;;;:a,i.J;:....,_........__,_..__&...,1.-1-..i.....1
0.200
0.175
0.150
0.125
0.100
0.075
0.050
0.000
0.025
88
ACI DESIGN HANDBOOK-SP-17M(09)
COLUMNS 3.5.2 - Nominal load-moment strength interaction diagram, R65-520.7
..
I
1.6 ....,..T"""T'"..,...,...,.............,__,...,_.....,_..-------------r------�
h
INTERACTION DIAGRAM R65-520.7
Jb
e
.fc = 65 MPa
I
!y= 520MPa
e
1 .4 t--�,--i----------1---�
e
y=0.7
I
--
-
(.)
II
�
•
•
••
•
0.8
0.6
0.2
0.0 I....I....L....L..-'--L..J....1...i:;;...i--&....i....i.....i...4....L..L..J....l.....i::;.J--L..L....L...'-1....L..L.�....i.;;;;;a--.L..,lli;...i,_.L....L..�.....i.�....i.........i...i...........
0.000 0.025 0.050 0.075 0.100 0.125 0.150 0.175 0.200 0.225 0.250
,....--..__
89
DESIGN OF CONCRETE ELEMENTS IN ACCORDANCE WITH ACI 318M-05
COLUMNS 3.5.3 - Nominal load-moment strength interaction diagram, R65-520.8
I..,
1.6 _,........,,........,__,_____,.______________�--------,
1.4 ------------
--
°'
4:..
-
(J
II
C:
�
INTERACTION DIAGRAM R65-520.8
I""
fc = 65 MPa
!y= 520MPa
e
h
I
...--------eI---,.,
•
]b
..
• • ••
r= o.s
0.8
0.6
0.0
&.....l.......&.--'-....i:;;.--'-__.__..a.:;;...,_�'-',,.--'---'--'--..i.....l;.i._.i.......1"'--1,...........i,,........--....i.-.i..-.i.......i__,_........�
0.00
0.05
0.10
0.15
0.20
0.25
0.30
90
ACI DESIGN HANDBOOK-SP-17M(09)
COLUMNS 3.5.4 - Nominal load-moment strength interaction diagram, R65-520.9
1.6 ___,,_,.......,,.......,,.......,�---------------r---------,,
h
INTERACTION DIAGRAM R65-520.9 ..,
, IE 1ft '"'I
.fc = 65MP a
I
;,._.;....._-4_...:.,..
iy= 520MPa
e
e
1.4 --�---�--__, r= o.9
•
•
;o,,
_,
• • •
I
I.
e
i::j
PD
---II
� 0.6
1--------+-----l--".--7'.'------+---\.-�.-----:�+-�---'\----'-----;
0.0 L..-&.....-L..,....,iL--,1i,:;.....Ji.-.,li.-.,l�---l---l---"�___.,___.,i;,_i.__.i.,__,,,�_................__._____.____._.....L._._................_,
0.25
0.30
0.15
0.20
0.00
0.05
0.10
DESIGN OF CONCRETE ELEMENTS IN ACCORDANCE WITH ACI 318M-05
91
COLUMNS 3.6.1 - Nominal load-moment strength interaction diagram, RSS-520.6
1.4 ,.....,___,,.-,-......,,,_,......,_.....,....________________,,._______
1.2
INTERACTION DIAGRAM RSS-520.6
fc= 85MPa
iy= 520 MPa
��=---�c--------1
I
1.0
r=
o.6
------,.-----.--------1
.,_
h
••
•
•
••
•
I
I
s:'·1
0.8
--
--
<)
C:
0..
II
0.6
1.0
0.0 L--L--'-...J......L....J...-i.:::;;.J.--L....L.......L..o:::1-JL--1......J.,Q.-...L.-J....,ol�......L..�.J--.a...::,,i�......A...;:a,,,_�__._......_._...__.
0.175
0.150
0.125
0.100
0.075
0.000
0.050
0.025
92
ACI DESIGN HANDBOOK-SP-17M(09)
COLUMNS 3.6.2 -Nominal load-moment strength interaction diagram, RSS-520.7
1.4
1.2
INTERACTION DIAGRAM RSS-520.7
fc = 85MPa
f. =
520MPa
y=0.7
1.0
1� I E
•••
h
J¾l
•
I
I
""I --1
••
•
P.
I-�
I,.
V
0.8
---
-u
a.. 0.6
C
II
0.4
0.2 �-------_,.__,.
1.0
0.0 L--'-...&.......L......L......J.....li:::;.,_...L--L-1...,i:i:;....a......1.-J....,t:::...,i_L......L.,....c;...J......1-..&...o::1.-'-.L.-l<:;..i_.&.......L.,�L......L...A-1......I-...L--L-'--'--'
0.000
0.025
0.050
0.075
0.100
0.125
0.150
0.175
0.200
DESIGN OF CONCRETE ELEMENTS IN ACCORDANCE WITH ACI 318M-05
93
COLUMNS 3.6.3 - Nominal load-moment strength interaction diagram, R85-520.8
1.. 4
1.2
1.0
0.8
INTERACTION DIAGRAM RSS-520.8
fc = 85MPa
f.y = 520MPa
r = o.s
i-•IE
••
h
,-h
•
I
I
I
•
..I ..
••
eI
pn
-....<!..
°'
--
'-'
a.. 0.6
C:
II
0.0 ._,__....,,-'-.L..ll:�1...L....L....11�.........J...oli;...a...,.....�.a...i.....&.-1..::i...-'--1,�i-..,..&.-,1"""--'--1,-"-L..J-........._,_-...-'-'
0.000 0.025 0.050 0.075 0.100 0.125 0.150 0.175 0.200 0.225
94
ACI DESIGN HANDBOOK-SP-17M(09)
COLUMNS 3.6.4 - Nominal load-moment strength interaction diagram, RSS-520.9
1.4 ........................._...,..._......,..__r--------------..----------.
INTERACTION DIAGRAM RSS-520.9 ___h__
= 85 MPa
fy= 520MPa
1 .2 i----:::,,,,__,_---+-=------i-------1 r = o.9
fc
'-,----........-----------1
I" •
l'-E-
••
J-b�
-i
•
0.8
<!..Cl
-....
--
0
a. 0.6
C:
II
0.0 L..L....L....L...L...J__ir:::::L..J_..L...1.....i..o:i::..a.....Ji....L...J..ll'::;..i._i_...L..1...,,c;..L...L..J......L..:L.L....L....IL....w:::.i.....i.....L...L.A...J.--I-L...lo:::�.J....L...L..li-L....L....J....J
0.000 0.025 0.050 0.075 0.100 0.125 0.150 0.175 0.200 0.225 0.250
DESIGN OF CONCRETE ELEMENTS IN ACCORDANCE WITH ACI 318M-05
95
COLUMNS 3.7.1- Nominal load-moment strength interaction diagram, L21-420.6
2.4
2.2
2.0
.
1
--
<.>
II
�
•
I
I
. e
P.
j:7
1.6
-<t..-
i-j
..
--
I
1.8
C>
I
•
h
INTERACTION DIAGRAM L21-420.6
�
.fc = 21 MPa
�I
.-----!
e
fy= 420MPa
I
e
y=O�
.....,__--J--.....
1.4
1.2
1.0
0.8
0.6
"""-L........,;;�..............._...........................................
0.0 L............
0.00 0.05 0.10 0.15 0.20 0.25 0.30 0.35 0.40 0.45 0.50 0.55 0.60
I
Rn = Pn e / f c A9 h
-'-1,,..&.-'-.,.._........................................,_'--'-"'........
-'-'.........��........
96
ACI DESIGN HANDBOOK-SP-17M(09)
COLUMNS 3.7.2-Nominal load-moment strength interaction diagram,L21-420.7
2.4
INTERACTION DIAGRAM L21-420.7
fc = 21MPa
2.2
!y= 420MPa
y=0-7
2.0
!
I
I
I
"'I
I
••
•
-,
P•
I.
-----J�--'-
1.4
II
••
•
. e
1.6
u
1�
I
I
1.8
-
b
�
1.2
1.0
0.8
0.6
0.4
0.2
0.0
0.0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
97
DESIGN OF CONCRETE ELEMENTS IN ACCORDANCE WITH ACI 318M-05
COLUMNS 3.7.3 - Nominal load-moment strength interaction diagram, L21-420.8
2.4
INTERACTION DIAGRAM L21-420.8
2 MPa
2.2 --�....____---1---1 fc = 1
J;, = 420MPa
y=0.8
2.0
•
•
e
}fl
I
I
I
as
••
P
D
.....___v---�
1.4
II
""'
i::i
1.6
0
I --1 ----•I •1
h
I
I
1.8
-
""
i------
1.2
1.0
0.8
0.6
0.4
1.0
0.2
0.0
0.0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
98
ACI DESIGN HANDBOOK-SP-17M(09)
COLUMNS 3.7.4 -Nominal load-moment strength interaction diagram, L21-420.9
2.4
INTERACTION DIAGRAM
.(
= 21 MPa
C
2.2 --------'-----I fy=
420MPa
= 0.9
r
b
••
•
2.0
� '"I ..,
I
I
I
I
I
I·
1.8
••
•
I
·7
e
Pn
I
1.6
1.4
--
1.2
II
1.0
--
<.)
0.8
0.6
0.4
0.2
0.0
0.0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
99
DESIGN OF CONCRETE ELEMENTS IN ACCORDANCE WITH ACI 318M-05
COLUMNS 3.8.1-Nominal load-moment strength interaction diagram, L28-420.6
..
1 �•I ·1
;___,;.---�
2.0 ,-,-..,......-,,-_.,.........-,,-_.,..........-------------,--------.
INTERACTION DIAGRAM L28-420.6
.fc = 28 MPa
1.8 -,--..P..-----+----l fy = 420 MPa
y=Q6
1. 6
I
�--�'V----------,--------,----t-----1
b
Jh
--1-=
---i
e
I
••
I
I
.__•_____.
•
;
I
-,
. e
P•
I.
--J-�
°'
<!..
-
- 0
1.0
l--- -_.c:-,-WN.L----/..O�-�-----�---,--,X,---__.,....."______-,,..______1-----�
II
0.8
1------�'---+--�---.,,..-:--�-----"<---c,,�----+"<------>---------'------t
....
c:
'::s::..
0.2----i......L.-11:>,,11..-....&....1..i...-'--l.............i._.__,
0.0 &...I....L-i.....&...J....1"-..1...1,-'-"-1...............'""-i.....&......t�..&....l=..i....i.......
.
0.45
0.40
0.30
0.35
0.25
0.20
0.15
0.05
0 00
0.10
.i..,g.
......
100
ACI DESIGN HANDBOOK-SP-17M(09)
COLUMNS 3.8.2 - Nominal load-moment strength interaction diagram, L28-420.7
2.0
�r-r-,r-r-i-r-r-r-r""T""T""T""T"'T7'""____________________
INTERACTION DIAGRAM L28-420.7 ,.,
,
fc = 21 MPa
1. 8 �,-----+--�-+-----H fy = 420 MPa
y = 0. 7
h
J-h >
I .,
1---'-.--+-•
I
-
"' --1
•
I
•
I
••
I
! e
1.4
°'
<t:..
--
- 0
..._
pn
I.�
_......--J'-----
1.0
�--"WJ.U.__-�---l'.----�-,,L----"-ri----",c-L----"--!-----+---+-----------1
0.8
t------+----"oe---+--"'
c:
c.
II c:
':::C.
0.6
0.0 .......................-..................................................................................................._..........._.._.,__................__.........__........____
0.00 0.05 0.10 0.15 0.20 0.25 0.30 0.35 0.40 0.45 0.50 0.55
DESIGN OF CONCRETE ELEMENTS IN ACCORDANCE WITH ACI 318M-05
101
COLUMNS 3.8.3 - Nominal load-moment strength interaction diagram, L28-420.8
2.0 l'"'T'"'l�...,..,..�...,...,""T"T...,..,..-r-1'"°"-------------,-------.
h
INTERACTION DIAGRAM L28-420.8
I
,h
fc = 28MPa
I"' I ,
.
11----,----i--�----'---------+---I J;,
= 420 MPa
18
!
I
r = o.s
I
••
•
.. l
I
•••
I
! e
·7
P.
I
---J--...i..
C>
<!.
....--
_ " 1.0
II c:
0 .8
t------P-,.------+,'-"',.---"�------".--,--_.,,,---+-_
t-----+---"'-------.-----+--�
0.0 .........................
0.00 0.05 0.10 0.15 0.20 0.25 0.30 0.35 0.40 0.45 0.50 0.55 0.60
.a....i..
.........
.i....a..,:i................_,__........................................................................................"""'-'-...................................................................,
102
ACI DESIGN HANDBOOK-SP-17M(09}
COLUMNS 3.8.4 - Nominal load-moment strength interaction diagram, L28-420.9
2.0 .....-.........------....-----......,..---------------,,----------,
h
INTERACTION DIAGRAM L28-420.9 ..
,.
,
-.,
-�.fc = 28 MPa
.. I I
1
I
!
1.8 ----'----'--� fy = 420 MPa
I
y=0.9
I
•••
1 .4
1.2
-----+---'-----",.---+-----"'7------+--+----+--�--+---l
•••
I
I
I·
p
.......__J-
e
pn
I
°'
-....
1.0
II
0.8
-<.
--
(.)
0.6
0.2
0.0 &...I..J....I....L.l�.........-'-,1,...J�.L...L..J.-'-..cl..........&..L,.l.-....................:;.....................
0.00 0.05 0.10 0.15 0.20 0.25 0.30 0.35 0.40 0.45 0.50 0.55 0.60 0.65 0.70
L..&,..I,,..........................."--"-'..............................................
DESIGN OF CONCRETE ELEMENTS IN ACCORDANCE WITH ACI 318M--05
103
COLUMNS 3.9.1- Nominal load-moment strength interaction diagram, L35-420.6
1.8 ......._.........__.......,._______________-r--------,
INTERACTION DIAGRAM L35-420.6
a
.(c = 35 MP
a
-ux---1-..---1 J;, = 420 MP
r = o.6
-.....
1·•I..,
i----h---i,
••
�' ,. I ;,,
�
I
I
I
••
•
1.0
C)
<(
t.)
II
0.8
'
c::
�
0.6
0.2
0.0 i.........i..........i-..::;..i....1-..1,�.,._&....1,.,4-.....,i......i,c;..i......,i-""-...i..;;;-..............._�....i........i.i:..i.....i-.i.-i..........._................
0.35
0.40
0.30
0.00
0.25
0.20
0.15
0.05
0.10
DESIGN OF CONCRETE ELEMENTS IN ACCORDANCE WITH ACI 318M-05
105
COLUMNS 3.9.3 - Nominal load-moment strength interaction diagram, L35-420.8
..
1.8 ..........................__.................._________________--,--------,
1.6
.....,..,_,-------+,,_,_____
--1
INTERACTION DIAGRAM L35-420.8
.( = 35 MPa
f,, = 420 MPa
r = o.s
1.4
••
•••
'
I
l
I
I
I
I
.
·=i
\le
po
I
1.2
--"
1 •I..
h
rh
1.0
--
8
a.. 0.
C
II
0.6
1.0
0.0 L...I...J....J....L....J�.L....1......L.....l�.L...L....t.....t:...L..-L....L.....L...L...L...1....l.....i...a....L...1.....L....�.L....I.....L...l'-'-.L...L..,J...,,it....L....L....L.........i.....L...1....1.....&.-.L-1
0.00 0.05 0.10 0.15 0.20 0.25 0.30 0.35 0.40 0.45 0.50
Rn = pn e / f /C Ag h
DESIGN OF CONCRETE ELEMENTS IN ACCORDANCE WITH ACI 318M-05
107
COLUMNS 3.10.1 - Nominal load-moment strength interaction diagram, L40-420.6
1 ·6 -----------
,,•
- - _ _ _ _ _ _ _ _ _ __ _ _ _- _ .....- -----=-------­
h
E CTIO ND IA GRAM L40420. 6
INT RA
Jh
/c = 40MPa
,
!y= 420MPa
1.4 --------�-- r = o.6
..
••
•
••
I
I
-,
v. e
P•
I.
1.0
-.....
--
<.>
0.8
II
0.6
0.2
0.0 L.....&.---I......L-�L......J.---I.....A......L...-.i........a.--L....i......__.i.......i_--i.......i.....a;;:,,..i__,_...a........__J...J.1__,_....L........__.................--'-�
0.00
0.05
0.15
0.25
0.10
0.35
0.20
0.30
109
DESIGN OF CONCRETE ELEMENTS IN ACCORDANCE WITH ACI 318M-05
COLUMNS 3.10.3 - Nominal load-moment strength interaction diagram, L40-420.8
1.6 r-,-...,.....,,.....,..........,___,,_,___,.-,----------------.------------.
h
INTERACTION DIAGRAM L40-420.8
_,
]'it
fc = 40MPa
,... I ;,, I
!y= 420MPa
I
1.4 -�...__�--+------1 r= o.s
I
••
•
I
••
•
I
I
. e Pn
I-�
-
""
<t.
- 0
0.8
------�--------------<---_.,,..�___,_._--------+------
0 .6
-----+--------.......-+-"'-"7""'�r.--�-----'I;
.._
II
::t::
0.0 ---------................--......._.__........_...,._.__................_._........._._........__........_._.__........___,___
0.45
0.40
0.35
0.00
0.25
0.30
0.05
0.20
0.15
0.10
DESIGN OF CONCRETE ELEMENTS IN ACCORDANCE WITH ACI 31 SM-05
111
COLUMNS 3.11.1 - Nominal load-moment strength interaction diagram, L65-520.6
--1
,.
1.6 ...................--........�.-----..--------------.....--------.
I..•
INTERACTION DIAGRAM L65-520.6 ___
h
�
.fc = 65 MPa
I""
••
!y= 520MPa
1
.4 --�,---,.-�--,----f r= o.6
1.2
--
<t:
-
II
C>
(.)
�
1
I
I
!
••I ,.
•
I
I
I
I
I·
e
I-�
pn
0.8
0.6
1.0
0.0 L.i.....1...1...i..J._,_L.oc:C::::i....J....i....i...io::::::i:..J....&......W::S:::.1.-l...i-..L.oc:C:::J...i............-=i:::J......i.....i:;�......,,::J:;ar:;;J::;;:1::a,1..""""":.J...&...,_i....&..J
0.000 0.025 0.050 0.075 0.100 0.125 0.150 0.175 0.200 0.225 0.250
113
DESIGN OF CONCRETE ELEMENTS IN ACCORDANCE WITH ACI 318M-05
COLUMNS 3.11.3 -Nominal load-moment strength interaction diagram, L65-520.8
1.6 .---.�.....,........--.--.---.�.....,....-,-----------------.- -------,
1.4 -----------
INTERACTION DIAGRAM L65-520.8
.( = 65 MPa
!y= 520MPa
r= o.s
----------...,.....------1
I
I
1.2 -.::----�-..,,....,.,.,,...-=,-....---""""--------+---------1
E
I I"'
••
•
h
J-ii
I
I
I
I
I
••
•
I
I
. e
P.
I-�
--
--
(.)
0.8
C
II
C
�
0.6
1.0
0.0 .......,,_,1,,,_,_""'-...___.......,,'-'-_,_........�.......,,�_,_.......�.__._--'-"'--'--.&....,.,o:--"'--'--"'--'--""'---'--'---'-_._�
0.35
0.30
0.25
0.20
0.15
0.10.
0.00
0.05
115
DESIGN OF CONCRETE ELEMENTS IN ACCORDANCE WITH ACI 318M-05
COLUMNS 3.12.1 - Nominal load-moment strength interaction diagram, L85-520.6
�,
1.4 ................-.-.........-.--.....,............-.-.---------------r-------,
1.2
I"
h _
INTERACTION DIAGRAM L85-520.6 ___
Jb
I.,__.., __,
fc = 85 MPa
I
!y= 520 MPa
e
�---�---+---- r
= o.6
•
I
I
..•
••
I
I
· e
po
i:j
.....__--J---.......
0.8
--
--
<.)
a. 0.6
C:
II
0.2 ----------
0.0 L..L.....L...J....l-..1.,,,J,:::;,,r._.L.....L.....i.,,c....1-..L.....L..::1.....L.....L.....i..c....1-�'""'-l-l-..lc:..l,-'-..i.......,c;....l....i1-1-a.....l.-'-............l....i'-'-.....................
0.000 0.025 0.050 0.075 0.100 0.125 0.150 0.175 0.200 0.225
DESIGN OF CONCRETE ELEMENTS IN ACCORDANCE WITH ACI 318M-05
117
COLUMNS 3.12.3 - Nominal load-moment strength interaction diagram, LSS-520.8
1.4
r--1--.---.--.....--....--...---.....,....---------------.------.......
INTERACTION DIAGRAM LSS-520.8
fc = 85 MPa
IE
,
..
J,= 520MPa
r = o.s
-------------!
1.2 �......-�,,_,___--+---------t
1. 0
••
•
•••
'
I
I
I
I
I
I.
PD
·7
e
I
J0.8
---
C>
<{
Cl.
C
II
0.6
0.0 L-&.....i-..i�.......i......:;.i--i.__.,�_...-""-__,_---";.........L.�-.L..---'--'-.........�......................................................................
0.30
0.25
0.20
0.15
0.05
0.10
0.00
DESIGN OF CONCRETE ELEMENTS IN ACCORDANCE WITH ACI 318M-05
119
COLUMNS 3.13.1- Nominal load-moment strength interaction diagram, C21-420.6
2.4
--------.
r-T'-,-�,---r-,--,--T""""T'-,-INTERACTION
---�
-----DIAGRAM C21-420.6
i-a---- --i
2.2
1-------"...-+--------l
.fc = 21 MPa
!y= 420 MPa
r = 0.6
2.0 �-------"-------::"'-'---',,=-="'=F===-======-""1
•
I
•••
I
. e
1.6 ----�---------4---�...-�---------+-------f
"'
<t..
_ '-'
1.2
II
1.0
--
•
P•
.-=,
I
--J-�
� ---�--l------''k---------"-.----1---,Ac-,-�-j--�,-------,-----+------t
0.8
0.6
0.4
0.2
0.0
0.00
1.0
0.05
0.1 0
0.15
0.20
0.25
0.30
0.35