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4 ACI DESIG N HANDBOOK-S P-17M(09) 0.003 Appendix C $=0.57+67e, H --r·-- ; /,. ~ i j· \._) ~~ : "'41,.... ,, .: ,, : j ,,,./' i ; - - i' , Chapter 9 .,· f-......... ! cj>=0.48+83E, J.....................L............L......... Reinforcement closest to the ~ Ld inimum permitted f...

Description

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, 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 (= 0.9). Determine if the section behaves as a T- or rectangular section. When M,, > [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 % Kn = 1. 83 MPa, and q> = 0 .90

Mi = q,K,/b - b....,)d2

Flexure 1

= 1.83(900 - 550)(835)2 = 447 kN-m q>Mw = M 11 - 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