CEE 3150 â Reinforced Concrete Design â Fall 2003. Design the flexural (including cutoffs) and shear reinforcement for a typical interior span of ...

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SDL = 2.15 kip/ft LL = 2.65 kip/ft fc0 = 4 kip/in2 fy = 60 kip/in2 γc = 150 lb/ft3

FLEXURAL DESIGN (A) Choose the beam depth, h. Because we are concerned about deflection, use a fairly deep beam (architect said it was O.K.): h = `/10 = 20 ft/10 (12 in/foot) = 24 in. (B) Assume two rows of steel both top and bottom: dpos = dneg = h − 3.5 in = 20.5 in. (C) Use a narrow beam for better efficiency. Assuming r = b/dpos = 0.4, b = 0.4dpos = 0.4(20.5 in) = 8.2 in, use b = 9 in. (D) Compute Mu and Mn at the midspan (maximum positive moment) and support (maximum negative moment). With our estimated beam height and width, we can calculate the selfweight, SW , and determine the total dead load, DL. SW = = = DL = = wu =

bhwc (9 in)(24 in)(150 lb/ft3 )(1 foot2 /144 in2 )(1 kip/1000 lb) 0.225 kip/ft SDL + SW = (2.15 kip/ft + 0.225 kip/ft) 2.375 kip/ft 1.2DL + 1.6LL = 1.2(2.375 kip/ft) + 1.6(2.65 kip/ft) = 7.09 kip/ft

Using the ACI design coefficients {ACI 8.3.3}: Mu,pos Mn,pos Mu,neg Mn,neg

(E) Compute R =

= = = =

wu `2n /16 = (7.09 kip/ft)(19 ft)2 /16 = 160 kip · ft Mu,pos /φ = (160 kip · ft)/0.9 = 178 kip · ft wu `2n /11 = (7.09 kip/ft)(19 ft)2 /11 = −233 kip · ft Mu,neg /φ = (−233 kip · ft)/0.9 = −259 kip · ft

Mn . bd2

Mn,pos 178 kip · ft = (12 in/foot)(1000 lb/1 kip) = 564 psi 2 bdpos (9 in)(20.5 in)2 |Mn,neg | | − 259 kip · ft| = = (12 in/foot)(1000 lb/1 kip) = 820 psi 2 bdneg (9 in)(20.5 in)2

Rpos =

(1)

Rneg

(2)

1

5 #7s

2 #7s

2 #5s

6 #5s

Support

Midspan

φ M n = −251 kip−ft

φ M n = 162 kip−ft

Figure 1: Reinforcing layout at the support and at midspan. Reading from the chart gives: ρpos ≈ 0.0103 ρneg ≈ 0.0160

(3) (4)

As,pos = ρpos bdpos = 0.0103(9 in)(20.5 in) = 1.90 in2 As,neg = ρneg bdneg = 0.0160(9 in)(20.5 in) = 2.95 in2

(5) (6)

(F) Compute As,pos and As,neg .

(G) Choose bars, make sure they fit. Positive: Use 6 #5, As,pos = 1.86 in2 . Negative: Use 5 #7, As,neg = 3.00 in2 . We are using slightly less steel for the positive reinforcement because #5 bars are small, resulting in a larger dpos than originally estimated. The positive reinforcement will be two rows of three bars, the negative reinforcement will have a top row of three and a bottom row of two bars (see Figure 1) and it is assumed that two bars from both positive and negative reinforcing will be continuous. A generic formula for the minimum width of a beam is given by: bmin = 2 × cover + 2dbs +

n X

dbf,i + (n − 1)cs,min

(7)

i=1

where dbs is the diameter of the stirrup, dbf,i is the diameter of the ith flexural reinforcing bar, and cs,min is the minimum clear spacing between bars. We will use 1.5 inches clear cover and assume a #3 stirrup. The minimum beam width, bmin , is controlled by the 3 #7 (db = 0.875 in) bars: cs,min = max(db , 1 in) = max(0.875 in, 1 in) = 1 in bmin = 2(1.5 in) + 2(0.375 in) + 3(0.875 in) + 2(1 in) = 8.38 in

(8) (9)

The bars will fit. Calculate the actual values for dpos and dneg . Because the positive steel is in two equal rows, the centroid is located at the center of the two rows. dpos = 24 in − 1.5 in − 0.375 in − 0.625 in − 0.5 in = 21.0 in 2

(10)

The negative steel has two unequal rows so the centroid is found by: dneg,1 = 24 in − 1.5 in − 0.375 in − (0.875 in)/2 = 21.7 in dneg,2 = dneg,1 − 0.875 in − 1 in = 19.8 in 3dneg,1 + 2dneg,2 3(21.7 in) + 2(19.8 in) dneg = = = 20.9 in 5 5

(11) (12) (13)

(H) Analyze the section, check φMn ≥ Mu . Because some of the positive steel will continue into the negative moment region, and vice-versa, sections will be analyzed as doubly reinforced. It is assumed that two bars on top and bottom will continue throughout the beam. The spreadsheet doubly.xls in the course directory is used to analyze the sections and perform code checks. From the spreadsheet: φMn,pos = 162.4 kip · ft > Mu,pos = 160 kip · ft φMn,neg = 250.8 kip · ft > |Mu,neg | = 233 kip · ft

(14) (15)

The initial flexural design is done. Next, we design the shear reinforcement. SHEAR DESIGN (1) Find the shear envelope. The ACI coefficients give shear at the face of interior supports as wu `n /2. Thus, for the present case: wu `n (7.09 kip/ft)(19 ft) = = 67.4 kip 2 2 Assume that the maximum shear at midspan is as derived for a simple beam: Vu,sup =

1.6(LL)`n 1.6(2.65 kip/ft)(19 ft) = = 10.1 kip 8 8 The equation which defines the shear envelope is then given by: Vu,mid =

"

(16)

(17)

#

Vu,sup − Vu,mid Vu (x) = Vu,sup − x `n /2 " # 67.4 kip − 10.1 kip = 67.4 kip − x 19 ft/2 = 67.4 kip − (6.03 kip/ft)x

(18) (19) (20)

q

(2) Check that Vu,max ≤ φ(Vc + 8 fc0 bd). Recall that Vu,max is the shear at the critical section— distance d from the face of the support. Assume that d = dpos = 21.0 in(1 foot/12 in) = 1.75 ft so that: Vu,max = Vu (dpos ) = Vu (1.75 ft) = 67.4 kip − (6.03 kip/ft)(1.75 ft) = 56.8 kip

(21) (22)

q

(23)

q

(24) (25)

Vc = 2 fc0 bdpos = 2 4, 000 lb/in2 (9 in)(21.0 in)(1 kip/1000 lb) = 23.9 kip q

φ(Vc + 8 fc0 bdpos ) = φ(5Vc ) = 0.75(5)(23.9 kip) = 89.7 kip. q

(26)

Thus, Vu,max = 56.8 kip < φ(Vc + 8 fc0 bdpos ) = 89.7 kip indicating that the section is large enough. The shear envelope is plotted in Figure 2. 3

Vu,sup 60

Vu (x) (kip)

50

20

10

Vu,max Vu,mid 1

2

3

6

x (feet)

7

8

9

Figure 2: The shear envelope. (3) Pick bars and set spacing. Assume #3 stirrups will be used. The spacing near the support, dependent on Vu,max , is (again, using dpos for d): 0.22 in2 (60 kip/in2 )(21.0 in) Av fy dpos = 5.35 in s ≤ Vu,max = 56.8 kip − 23.9 kip − Vc 0.75 φ

(27)

Use s = 5.0 in and compute the capacity of the stirrups: Vs =

Av fy dpos 0.22 in2 (60 kip/in2 )(21.0 in) = = 55.4 kip s 5.0 in

(28)

q

To find the maximum spacing, compare Vs to 4 fc0 bdpos : q

q

4 fc0 bdpos = 4 4, 000 lb/in2 (9 in)(21.0 in)(1 kip/1000 lb) = 47.8 kip

(29)

q

Thus, because Vs > 4 fc0 bdpos the maximum spacing is given by dpos /4 = 21.0 in/4 = 5.25 in. The chosen spacing, s = 5.0 in < 5.25 in, meets the maximum spacing. The total capacity of the beam, φVn , for s = 5.0 in is then: φVn = φ(Vc + Vs ) = 0.75(23.9 kip + 55.4 kip) = 59.5 kip > Vu,max = 56.8 kip

(30)

As the shear drops, the maximum spacing will increase to dpos /2 = (21.0 in)/2 = 10.5 in. So, we will use s = 10.0 in near the midspan. The shear capacity of the stirrups and total capacity at that spacing is Av fy dpos 0.22 in2 (60 kip/in2 )(21.0 in) Vs = = = 27.7 kip s 10.0 in φ(Vc + Vs ) = 0.75(23.9 kip + 27.7 kip) = 38.7 kip

(31) (32)

Solving Equation 20 q for Vu (x) = 38.7 kip gives x = 4.76 ft ≡ 4 ft 9 in. Note that at s = 10.0 in, Vs < 4 fc0 bdpos so that the maximum stirrup spacing is indeed d/2. Check that 4

8 @ 5"

4 @ 7"

4 @ 10"

2"

4"

1

2

3

4

5

6

7

8

9

x (feet)

Figure 3: Initial shear design. this meets the minimum steel requirements: Av,min

bw s 50bw s , = max 0.75 fc0 fy fy q

!

(9 in)10.0 in 50 lb/in2 (9 in)10.0 in , = max 0.75 4, 000 lb/in 60, 000 lb/in2 60, 000 lb/in2 q

2

= max 0.071 in2 , 0.075 in2 = 0.075 in2

(33) !

(34) (35)

Using a #3 stirrup, Av = 2(0.11 in2 ) = 0.22 in2 > Av,min , so we meet the minimum area requirement of the code. Choose one more spacing interval, s = 7 in, resulting in: Av fy dpos 0.22 in2 (60 kip/in2 )(21.0 in) = = 39.6 kip s 7 in φ(Vc + Vs ) = 0.75(23.9 kip + 39.6 kip) = 47.6 kip Vs =

(36) (37)

q

Because Vs < 4 fc0 bdpos for s = 7 in (compare Equation 36 to Equation 29), d/2 spacing controls. Solving Equation 20 for Vu (x) = 47.6 kip gives x = 3.28 ft ≡ 3 ft 3 in. Zone A starts when Vu drops below φVc /2: φVc /2 = 0.75(23.9 kip)/2 = 8.96 kip < Vu,mid = 10.1 kip

(38)

Thus, there is no Zone A—stirrups must be used throughout the beam. Using the calculations above, the stirrups are designed as shown in Figure 3 with the first stirrup at 2 inches from the face of the support, then 8 additional stirrups at 5 inch spacing, 4 at 7 inch spacing, and 4 at 10 inch spacing. This results in a shear capacity as shown in Figure 4. FLEXURAL STEEL CUTOFFS Negative Reinforcement Cutoffs At the support there are five #7 bars as tension reinforcement (at the top of the beam since the moment is negative). It is assumed that two #5 compressive bars continue from the positive moment reinforcement. The nominal moment capacity for this arrangement is 251 kip · ft as calculated above. The steel configuration at the support is shown in Figure 5 along with configurations used after cutting off steel as explained below. 5

Vu (x) and φVn (x) (kip)

60

50

40

30

20

10

φVn(x) Vu(x) 1

2

3

4

5

6

7

8

9

x (feet)

Figure 4: The shear envelope.

Support φ M n = −251 kip−ft

Cutoff #1 φ M n = −163 kip−ft

Cutoff #2 φ M n = −111 kip−ft

Figure 5: Negative reinforcing layouts and section moment resistance.

6

c2v

c1

c2h

Figure 6: Dimensions c1 , c2h , and c2v . Cutoff #1 We first wish to cut off the two #7s in the bottom row of the tension steel. This results in a moment capacity of 163 kip · ft. Determine the development length using the general formula: ld =

3 fy αβγλ q db tr 40 fc0 c+K db

(39)

Note that α = 1.3 because there are more than 12 inches of concrete below the reinforcing. Recall that c = min(c1 , c2 ) where c1 is the distance from the center of the bar to the nearest concrete surface and c2 is one-half the center-to-center spacing of the bars being developed in the horizontal or vertical direction. Designating c2h as the horizontal dimension and c2v as the vertical dimension (see Figure 6): c1 c2h c2v c

= = = =

1.5 in + 0.375 in + (0.875 in)/2 = 2.31 in (b/2 − c1 ) = (9 in/2 − 2.31 in) = 2.19 in (db + 1 in) /2 = (0.875 in + 1 in) /2 = 0.938 in min(c1 , c2h , c2v ) = 0.938 in

(40) (41) (42) (43)

Conservatively estimating that Ktr = 0, gives: c + Ktr 0.938 in + 0 = = 1.07 db 0.875 in

(44)

However, because we have stirrups and more than db clear spacing in both the horizontal and vertical direction, the expression from the table in {ACI 12.2.2} can be used for a #7 bar:

`d =

f αβλ y q

20 fc0

db =

60, 000 lb/in2 (1.3)(1.0)(1.0) √ 0.875 in 20 4, 000 lb/in2

= 54.0 in ≡ 4.50 ft = 0.237`n

(45) (46)

Cutoff #2 Next, we will cut the center tension bar of the remaining three #7s. As there is only one row left, we do not need to worry about vertical spacing (c2 = c2h ). Again, we will use the general

7

Midspan

Cutoff #1

φ M n = 162 kip−ft

φ M n = 138 kip−ft

Cutoff #2 φ M n = 88 kip−ft

Figure 7: Positive reinforcing layouts and section moment resistance. formula and compute c first: c1 c2 c c + Ktr db

= 1.5 in + 0.375 in + (0.875 in)/2 = 2.31 in = (b/2 − c1 )/2 = ((9 in)/2 − 2.31 in)/2 = 1.10 in = min(c1 , c2 ) = 1.10 in 1.10 in + 0 = = 1.25 0.875 in

(47) (48) (49) (50)

tr The expression c+K is less than 1.5, so it is more advantageous to use the equations in {ACI 12.2.2} db which gives the same development length as before: `d = 0.237`n . At the cutoff locations, it is required that a buffer of length max(d, 12db ) be provided (Rule #1).

max(d, 12db ) = max(dneg , 12db ) = max(20.9 in, 12(0.875 in)) = 20.9 in

(51)

Using a buffer of dneg = 20.9 in ≡ 1.74 ft = 0.09`n , the cutoff locations are determined as sketched on the attached plot. Positive Reinforcement Cutoffs At the midspan there are six #5 bars as tension reinforcement. It is assumed that two #7 compressive bars continue from the negative moment reinforcement. The steel configuration is shown in Figure 7 along with configurations used after cutting off steel as explained below. Cutoff #1 We first wish to cut off the middle #5 in the top row of the tension steel. This results in a moment capacity of 138 kip · ft. c1 c2h c2v c c + Ktr db

= = = =

1.5 in + 0.375 in + (0.625 in)/2 = 2.19 in (b/2 − c1 )/2 = ((9 in)/2 − 2.19 in)/2 = 1.16 in (db + 1 in)/2 = (0.625 in + 1 in)/2 = 0.81 in min(c1 , c2h , c2v ) = min(2.19 in, 1.16 in, 0.81 in) = 0.81 in 0.81 in + 0 = = 1.3 0.625 in

8

(52) (53) (54) (55) (56)

We have more than db clear spacing both horizontally and vertically and we are using stirrups, so again we will use the table in {ACI 12.2.2} for #5 reinforcing steel:

`d =

f αβλ y q

25 fc0

db =

60, 000 lb/in2 (1.0)(1.0)(1.0) √ 0.625 in 25 4, 000 lb/in2

= 23.7 in ≡ 1.98 ft = 0.104`n

(57) (58)

Cutoff #2 Next, we will cut the remaining #5s in the top row. The remaining three bars in the bottom row will continue into the supports. When cutting these two #5s, the vertical spacing is the same as before and will control the development length so that `d = 0.104`n again. In this case our buffer is dpos = 21.0 in ≡ 1.75 ft = 0.092`n , and the cutoff locations are determined as sketched on the attached plot. Checking the Rules Negative Cutoff #1 Rule #1 Bars must extend the longer of d or 12db past flexural cutoff points except at supports of simple spans or ends of cantilevers. {ACI 12.10.3} – The buffer of max(d, 12db ) is shown on the attached moment plot and the cutoff was determined so that this rule was satisfied. Rule #2 Bars must extend at least `d from the point of maximum bar stresses or from the point at which adjacent bars which are cut or bent are no longer required to resist flexure. {ACI 12.10.2, 12.10.4, 12.12.2} – Cutting these bars `d from the support results in a cutoff location 4.5 ft from face of the support. Rule #3 Bars cannot be cutoff in tension regions unless conditions in {ACI 12.10.5} are met. – The #7s being cut are in tension so we must satisfy {ACI 12.10.5} which stipulates we must meet one of the conditions given in {ACI 12.10.5.1, 12.10.5.2, or 12.10.5.3}. {ACI 12.10.5.1}: To satisfy this condition, we need Vu ≤ 2φVn /3 or Vu /φVn ≤ 2/3. Looking at the shear design at x = 4.50 ft (the location of cutoff #1) we are using s = 7 in as spacing so that φVn = 47.6 kip, and Vu = Vu (4.50 ft) = 67.4 kip − (6.03 kip/ft)(4.50 ft) = 40.3 kip. 40.3 kip 2 Vu = = 0.85 > φVn 47.6 kip 3

(59)

Thus, we do not satisfy this section. {ACI 12.10.5.2}: To satisfy this, we may need to add extra stirrups over the last 0.75dneg of the bar being cut. The required spacing of the additional bars, s+ is: Av fy dneg s+ ≤ min , 2 (60 lb/in )b 8βb

!

(60)

where βb is the ratio of the area of bars being cut to the total area. In this case, βb = 2/5 since two #7s out of five are being cut. s+ ⇒

0.75dneg s+

0.22 in2 (60, 000 lb/in2 ) 20.9 in ≤ min , 60 lb/in2 (9 in) 8(2/5) 0.75(20.9 in) = = 2.40 stirrups 6.53 in/stirrup

!

= 6.53 in

(61) (62)

At the point of the cut we are using s = 7 in spacing, so 0.75dneg 0.75(20.9 in) 15.7 in = = = 2.24 s 7 in 7 in 9

(63)

That is, we have 2.24 stirrups from the initial design and we need to add 2.40 additional stirrups for a total of 4.64 stirrups over 15.7 inches at the end of the cut bars. This works out to a spacing of 15.7 inches/4.64 stirrups=3.4 inch/stirrup. So, we could reduce our stirrup spacing to 3 inches near the end of the cut bars. This is less than the ideal minimum of 4 inch spacing, but we are only doing it for a short distance because of concerns about the cutoff. {ACI 12.10.5.3}: In order to meet this criterion, we must have Vu /φVn ≤ 3/4. Considering Equation 59, this is not met. Thus, to meet rule #3 we must add stirrups near the end of the bars being cut since we are cutting in tension. However, the point of inflection (P.I.) occurs at 4 ft 7 in from the face of the support so that by extending the two #7s by an inch we would no longer be cutting in tension and rule #3 would not apply. This is what we will do. Rule #6 Negative moment reinforcement must be anchored into or through supporting columns of members. {ACI 12.12.1} – We plan to continue the negative reinforcement into the next span, through the support—this rule is satisfied. Rule #7 We are considering an interior beam, so the applicable rule is: At least one-third of the negative moment reinforcement must be extended by the maximum of d, 12db , or l/16 past the negative moment P.I. {ACI 12.12.3} – The two #7s that continue throughout the beam represent 2/5 of the negative moment reinforcement which is greater than 1/3. This rule is satisfied. We have satisfied the cutoff rules for negative cutoff #1. The only change we needed to make was to add an extra inch to the cutoff location so that we are not cutting these bars in tension. Negative Cutoff #2 Rule #1 Bars must extend the longer of d or 12db past flexural cutoff points except at supports of simple spans or ends of cantilevers. {ACI 12.10.3} – The buffer is plotted on the chart and by inspection, this rule is met. Rule #2 Bars must extend at least `d from the point of maximum bar stresses or from the point at which adjacent bars which are cut or bent are no longer required to resist flexure. {ACI 12.10.2, 12.10.4, 12.12.2} – The adjacent bars were cut at cutoff #1. Consulting the attached plot, it is seen that the continuing bars continue more than `d from the point at which the bars from cutoff #1 are no longer needed. Rule #3 Bars cannot be cutoff in tension regions unless conditions in {ACI 12.10.5} are met. – The #2 cutoff occurs in a region of positive moment, thus the bars are not being cut in tension. We do not need to consult {ACI 12.10.5}. Rule #6 Negative moment reinforcement must be anchored into or through supporting columns of members. {ACI 12.12.1} – As above, we plan to continue the negative reinforcement into the next span, through the support, satisfying this rule. Rule #7 – The same reasoning from cutoff #1 applies and this rule is satisfied. The cutoff rules are satisfied for negative cutoff #2. Rule checking was made considerably easier because we are not cutting this bar in tension. Positive Cutoff #1 Rule #1 The buffer is plotted—this rule is met. Rule #2 Bars must extend at least `d from the point of maximum bar stresses or from the point at which adjacent bars which are cut or bent are no longer required to resist flexure. {ACI 12.10.2, 12.10.4, 12.12.2} – The bar at cutoff #1 develops its full strength before maximum moment at midspan. This rule is satisfied. Rule #3 Bars cannot be cutoff in tension regions unless conditions in {ACI 12.10.5} are met. – The #5 being cut is in tension so we must satisfy one of the conditions given in {ACI 12.10.5.1, 12.10.5.2, or 12.10.5.3}. {ACI 12.10.5.1}: Looking at the shear design at x = 5.00 ft (the location of positive cutoff #1)

10

we are using s = 7 in spacing so that φVn = 47.6 kip, and Vu = Vu (5.00 ft) = 67.4 kip − (6.03 kip/ft)(5.00 ft) = 37.3 kip. 37.3 kip 2 Vu = = 0.78 > φVn 47.6 kip 3

(64)

We do not satisfy this section. {ACI 12.10.5.2}: The required spacing over the last 0.75d of the bar being cut of the additional stirrups, s+ is: !

s+

⇒

0.75dpos s+

dpos Av fy ≤ min , 2 (60 lb/in )b 8βb ! 0.22 in2 (60, 000 lb/in2 ) 21.0 in = min , = 15.75 in 60 lb/in2 (9 in) 8(1/6) 0.75(21.0 in) = = 1.00 stirrup 15.75 in/stirrup

(65) (66) (67)

where βb = 1/6 since one #5 out of six are being cut. At the point of the cut we are using s = 7 in spacing, so 0.75dpos 0.75(21.0 in) 15.75 in = = = 2.25 stirrups s 7 in 7 in/stirrup

(68)

That is, we have 2.25 stirrups from the initial design and we need to add an additional 1.00 stirrup for a total of 3.25 stirrups over 15.75 in at the end of the cut bars. This works out to a spacing of 15.75 in/3.25 stirrups= 4.85 inch/stirrup. So, we will reduce our spacing to 4 inches near the end of the cut bar. {ACI 12.10.5.3}: In order to meet this criterion, we must meet three conditions: 1. “No. 11 bars or smaller.” We are cutting a #5, so we meet this. 2. “Continuing reinforcement provides double the area required for flexure at the cutoff point.” That is, φMn /Mu ≥ 2 at the cutoff point, where φMn is the moment capacity of the continuing bars. After positive cutoff #1 the moment capacity is φMn = 138 kip − ft, and reading from the graph, the moment demand, Mu , is 96 kip-ft. φMn 138 kip − ft = = 1.53 < 2 Mu 96 kip − ft

(69)

Thus, this criterion is not met. 3. “Factored shear does not exceed three fourths the design shear strength, φVn .” That is, Vu /φVn ≤ 3/4. Based on the results from Equation 64, this criterion is not met either. Overall, the stipulations of {ACI 12.10.5.3} are not met based on the moment capacity. The end result is that we will be adding extra stirrups near the end of the first cut bar as required by {ACI 12.10.5.2}. Rules #4 and #5 are based on conditions near the supports and will be checked for Positive Cutoff #2.

11

Positive Cutoff #2 Rule #1 Satisfied by inspection. Rule #2 Bars must extend at least `d from the point of maximum bar stresses or from the point at which adjacent bars which are cut or bent are no longer required to resist flexure. {ACI 12.10.2, 12.10.4, 12.12.2} – The bar at cutoff #1 was adjacent to the bars being checked now. The bars from cutoff #2 reach their full strength before bars from cutoff #1 are needed, so this rule is satisfied. Rule #3 Bars cannot be cutoff in tension regions unless conditions in {ACI 12.10.5} are met. – The edge of the buffer occurs at x = 3.14 ft. If we cut here, it will be in a region of positive moment, thus we must meet one of the conditions of {ACI 12.10.5}. At the cutoff location, x = 3.14 ft, we are using s = 5 in spacing so that φVn = 56.8 kip, and Vu = Vu (3.14 ft) = 67.4 kip− (6.03 kip/ft)(3.14 ft) = 48.5 kip. Vu 48.5 kip = = 0.85 φVn 56.8 kip

(70)

The ratio in Equation 70 is greater than 2/3 and 3/4, so we cannot meet the provisions of {ACI 12.10.5.1} or {ACI 12.10.5.3}. We could add extra stirrups near the cutoff location as stipulated in {ACI 12.10.5.2}. However, we are cutting the bars near the positive P.I. so that by extending the cutoff a few inches (4.33 inches to be exact) towards the support we will not be cutting in tension. We elect to extend the bars and the rule is satisfied since we will no longer be cutting in tension. Rule #4 (a) Simple supports—At least one-third of the positive moment reinforcement must extend six inches into the support. {ACI 12.11.1} (b) Continuous interior beams with closed stirrups—At least one-fourth of the positive moment reinforcement must extend six inches into the support. {ACI 12.11.1, 7.13.2.3} (c) Continuous interior beams without closed stirrups—At least one-fourth of the positive moment reinforcement must be continuous. {ACI 12.11.1, 7.13.2.4} (d) Continuous perimeter beams—At least one-fourth of the positive moment reinforcement must be continuous around the perimeter of the building and enclosed within closed stirrups or stirrups with 135-degree hooks. {ACI 7.13.2.2} We are considering a continuous interior beam. One-half of the positive reinforcing (three of six bars) will be continuing through the support. Thus, whether the stirrups are closed or not, we meet this rule. Rule #5 At the positive moment point of inflection (P.I.), and at simple supports, the positive moment reinforcement must satisfy: qMn `d ≤ + `a (71) Vu where Vu is the factored shear at the P.I. or the support, for a P.I. `a is the larger of d or 12db and is less than the actual embedment past the P.I, for a simple support `a is the end anchorage past the center of the support, and q is 1.3 if the ends of reinforcement are confined by compressive reaction (generally true for simple supports) and 1.0 otherwise. {ACI 12.11.3} – The positive moment P.I. occurs at 0.146`n = 0.146(19 ft) = 2.77 ft. Calculating the needed quantities at the P.I.: 88 kip·ft 88 kip·ft = = 97.8 kip·ft φ 0.9 = 67.4 kip − (6.03 kip/ft)(2.77 ft) = 50.7 kip = max(dpos , 12db ) = max(21.0 in, 12(0.625 in)) = 21.0 in

φMn = 88 kip·ft ⇒ Mn = Vu `a

12

(72) (73) (74)

Because this is a P.I. (and not a support), q = 1.0. The development length from cutoff #1 is `d = 23.7 in so that qMn (1.0)97.8 kip·ft + `a = (12 in/foot) + 21.0 in = 44.1 in > `d = 23.7 in Vu 50.7 kip

(75)

Rule #5 is satisfied and we are done with the flexural steel cutoff design. FINAL DESIGN Stirrup at Center

As required by ACI 12.10.5.2:

8 @ 5"

2 @ 7"

5 @ 4"

3 @ 10"

8" 2"

Positive Cutoff #2 @ 2ft 9in

1

2

Negative Cutoff #1 @ 4ft 7in

3

4

Positive Cutoff #1 @ 5ft 0in

5 x (feet)

Figure 8: The final design.

13

6

Negative Cutoff #2 @ 6ft 0in

7

8

CL 9

SERVICEABILITY—DEFLECTION (1) Would large deflections cause a problem? Yes they would (given). (2) Calculate Ig , Mcr , and Icr . 9 in(24 in)3 bh3 = = 10, 368 in4 12 12 1 1 q Mcr = fr bh2 = 7.5 4, 000 lb/in2 (9 in)(24 in)2 6 6 = 409, 831 lb · in ≡ 34.2 kip · ft Es 29, 000, 000 lb/in2 √ n = = = 8.04 Ec 57, 000 4, 000 lb/in2 Ig =

(76) (77)

(78)

Note that Equation 77 is valid because this is a rectangular section.1 Calculate Icr at the midspan (Icr,mid ) and support (Icr,sup ). At midspan, there are 6 #5 for the tensile steel (As = 1.86 in2 ) and 2 #7 for the compressive steel (A0s = 1.20 in2 ). The distance from the top to the neutral axis, c, is found by solving the quadratic formula: bc2 2

(9 in)c2 2

+ [nAs + (n − 1)A0s ] c − nAs dpos − (n − 1)A0s d0pos (79) + 8.04(1.86 in2 ) + 7.04(1.20 in2 ) c − 8.04(1.86 in2 )(21.0 in) − 7.04(1.20 in2 )(2.3 in) (80) 4.5c2 + 23.4c − 333.5 (81)

Solving gives c = 6.39 in. We can then solve for the cracked moment of inertia, Icr,mid . Icr,mid

bc3 = + (n − 1)A0s (c − d0pos )2 + nAs (dpos − c)2 3 (9 in)(6.39 in)3 = + 7.04(1.20 in2 )(6.39 in − 2.3 in)2 3 +8.04(1.86 in2 )(21.0 in − 6.39 in)2 = 4116 in4

(82) (83) (84)

At the support, there are five #7 bars for tensile steel at the top since moments are negative (As = 3.0 in2 ), the bottom has 3 #5 (A0s = 0.93 in2 ) and is in compression so that cneg is the distance from the bottom to the neutral axis. The distances to the steel, d0neg and dneg are also measured from the bottom of the section. bc2neg 2

(9 in)c2neg 2

+ [nAs + (n − 1)A0s ] cneg − nAs dneg − (n − 1)A0s d0neg (85) + 8.04(3.0 in2 ) + 7.04(0.93 in2 ) cneg − 8.04(3.0 in2 )(20.9 in) − 7.04(0.93 in2 )(2.2 in) (86) 4.5c2neg + 30.7cneg − 518.5 (87)

Solving gives cneg = 7.85 in which leads to bc3neg + (n − 1)A0s (cneg − d0neg )2 + nAs (dneg − cneg )2 3 (9 in)(7.85 in)3 = + 7.04(0.93 in2 )(7.85 in − 2.2 in)2 3 +8.04(3.0 in2 )(20.9 in − 7.85 in)2 = 5768 in4

Icr,sup =

(3) Find ∆I,D , ∆I,D+L , ∆I,L , ∆I,SU ST . 1

For other sections use the definition: Mcr =

fr I g yt .

14

(88) (89) (90)

Deflection under Service Dead Loads Using the coefficients from {ACI 8.3.3} with service loads (wD = DL): MD,mid = wD `2n /16 = (2.375 kip/ft)(19 ft)2 /16 = 53.6 kip · ft MD,sup = wD `2n /11 = (2.375 kip/ft)(19 ft)2 /11 = 77.9 kip · ft

(91) (92)

Both exceed the uncracked moment of inertia, Mcr = 34.2 kip · ft, so that Ie,mid

Mcr = Icr,mid + (Ig − Icr,mid ) MD,mid

!3

(93)

34.2 kip · ft = 4116 in + (10, 368 in − 4116 in ) 53.6 kip · ft 4

Ie,sup

4

Mcr = Icr,sup + (Ig − Icr,sup ) MD,sup

⇒ Ie

4

= 5740 in4

(94)

!3

(95)

34.2 kip · ft = 5768 in + (10, 368 in − 5768 in ) 77.9 kip · ft = 0.70Ie,mid + 0.30Ie,sup = 0.70(5740 in4 ) + 0.30(6157 in4 ) = 5865 in4 4

!3

4

!3

4

= 6157 in4

(96) (97)

An approximate formula for the deflection of a continuous beam is: ∆=

5w`4n MA,sup `2n MB,sup `2n − − 384Ec Ie 16Ec Ie 16Ec Ie

(98)

where MA,sup and MB,sup are the moments at the supports at side A and B respectively, and Ec is the modulus of elasticity for concrete: q

q

Ec = 57, 000 fc0 = 57, 000 4, 000 lb/in2 (1 kip/1000 lb) = 3605 kip/in2

(99)

Thus, with MA,sup = MB,sup = MD,sup the immediate deflections from dead loads are: 5wD `4n MD,sup `2n − 384Ec Ie 8Ec Ie 5(2.375 kip/ft)(19 ft)4 (12 in/foot)3 = 384(3605 kip/in2 )(5865 in4 ) 77.9 kip · ft(19 ft)2 (12 in/foot)3 − 8(3605 kip/in2 )(5865 in4 ) = 0.042 in

∆I,D =

∆I,D

(100) (101) (102) (103)

Deflection under Service Dead + Live Loads The service distributed load is wD+L = DL+LL = 2.375 kip/ft+2.65 kip/ft = 5.025 kip/ft. MD+L,mid = wD+L `2n /16 = (5.025 kip/ft)(19 ft)2 /16 = 113 kip · ft MD+L,sup = wD+L `2n /11 = (5.025 kip/ft)(19 ft)2 /11 = 165 kip · ft 15

(104) (105)

!3

Mcr

Ie,mid = Icr,mid + (Ig − Icr,mid )

(106)

MD+L,mid

34.2 kip · ft = 4116 in + (10, 368 in − 4116 in ) 113 kip · ft 4

4

Mcr

Ie,sup = Icr,sup + (Ig − Icr,sup )

= 4289 in4 (107)

!3

(108)

MD+L,sup

34.2 kip · ft = 5768 in + (10, 368 in − 5768 in ) 165 kip · ft = 0.70Ie,mid + 0.30Ie,sup = 0.70(4289 in4 ) + 0.30(5809 in4 ) = 4745 in4 4

⇒ Ie

!3

4

4

!3

4

= 5809 in4 (109) (110)

We can then calculate the deflections due to dead + live and live only: 5wD+L `4n MD+L,sup `2n − 384Ec Ie 8Ec Ie 5(5.025 kip/ft)(19 ft)4 (12 in/foot)3 = 384(3605 kip/in2 )(4745 in4 ) 165 kip · ft(19 ft)2 (12 in/foot)3 − 8(3605 kip/in2 )(4745 in4 ) = 0.109 in = ∆I,D+L − ∆I,D = 0.109 in − 0.042 in = 0.067 in

∆I,D+L =

∆I,L

(111) (112) (113) (114) (115)

Deflection under Sustained Loads Assume that 50% of the live load is sustained: ML,mid MSU ST ,mid ML,sup MSU ST ,sup

= = = =

MD+L,mid − MD,mid = 113 kip · ft − 53.6 kip · ft = 59.4 kip · ft MD,mid + 0.5ML,mid = 53.6 kip · ft + 0.5(59.4 kip · ft) = 83.3 kip · ft MD+L,sup − MD,sup = 165 kip · ft − 77.9 kip · ft = 87.1 MD,sup + 0.5ML,sup = 77.9 kip · ft + 0.5(87.1 kip · ft) = 121 kip · ft

(116) (117) (118) (119)

Solving for the effective moments of inertia: !3

Mcr

Ie,mid = Icr,mid + (Ig − Icr,mid )

34.2 kip · ft = 4116 in + (10, 368 in − 4116 in ) 83.3 kip · ft 4

4

⇒ Ie

!3

4

= 4549 in4 (121)

!3

Mcr

Ie,sup = Icr,sup + (Ig − Icr,sup )

(122)

MSU ST ,sup

34.2 kip · ft = 5768 in + (10, 368 in − 5768 in ) 121 kip · ft = 0.70Ie,mid + 0.30Ie,sup = 0.70(4549 in4 ) + 0.30(5872 in4 ) = 4946 in4 4

(120)

MSU ST ,mid

4

4

!3

= 5872 in4 (123) (124)

Solving for the deflection: wSU ST

= DL + 0.5LL = 2.375 kip/ft + 0.5(2.65 kip/ft) = 3.70 kip/ft

16

(125)

∆I,SU ST

∆I,SU ST

5wSU ST `4n MSU ST ,sup `2n − 384Ec Ie 8Ec Ie 5(3.70 kip/ft)(19 ft)4 (12 in/foot)3 = 384(3605 kip/in2 )(4946 in4 ) 121 kip · ft(19 ft)2 (12 in/foot)3 − 8(3605 kip/in2 )(4946 in4 ) = 0.079 in =

(126) (127) (128) (129)

(4) Find long-term creep and shrinkage deflection, ∆CS . As defined in the code, ∆CS = λ∆I,SU ST =

ξ ∆I,SU ST 1 + 50ρ0

(130)

Using a 5-year timespan (ξ = 2.0) and ρ0 at the midspan: 2(0.60 in2 ) A0s = = 0.00635 bdpos (9 in)(21.0 in)

(131)

2.0 (0.079 in) = 0.120 in 1 + 50(0.00635)

(132)

ρ0 = Substituting all values: ∆CS =

(5) Check code requirements. The serviceability deflection, ∆S , is given by ∆S = ∆CS + ∆I,L = 0.120 in + 0.067 in = 0.187 in

(133)

Because large deflections could cause problems: `n 19 ft = (12 in/foot) = 0.475 in 480 480 We meet the deflection requirement, the beam is adequately designed. ∆S = 0.187 in ≤

(134)

SERVICEABILITY—CRACK WIDTH To meet the requirements for crack width, the center-to-center spacing of the tension steel, sc−c , must satisfy 540 kip/in sc−c ≤ − 2.5cc (135) fs At the midspan, we have 3 #5 bars and a #3 stirrup so that sc−c is: 9 in − 2(1.5 in) − 2(0.375 in) − 0.625 in = 2.31 in (136) 2 With 1.5 inch clear cover, cc = 1.5 in + 0.375 in = 1.875 in, and use fs = 0.6fy = 0.6(60 ksi) = 36 ksi so that the equations to check are sc−c =

540 kip/in − 2.5(1.875 in) = 10.3 in 36 kip/in2 ! ! 36 ksi 36 ksi = 2.31 in ≤ 12 in = 12 in = 12 in fs 36 ksi

sc−c = 2.31 in ≤

(137)

sc−c

(138)

This requirement is satisfied at the midspan. At supports there are three #7s in the tension zone so that 9 in − 2(1.5 in) − 2(0.375 in) − 0.875 in = 2.19 in (139) 2 The quantities cc and fs are the same as at the midspan, and since sc−c is actually smaller at the support, we know that the section at the support meets crack-width requirements. sc−c =

THE END 17