Full Beam Design Example

Full Beam Design Example CEE 3150 Reinforced Concrete Design Fall 003 Design the flexural (including cutoffs) and shear reinforcement for a typical interior span of a six span continuous beam with center-to-center spacing of 0 ft. Assume the supports are 1 inches wide. Check that deflection (excessive deflections will cause problems) and crack-width serviceability requirements are met. Given: SDL.15 kip/ft LL.65 kip/ft f c 4 kip/in f y 60 kip/in γ c 150 lb/ft 3 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 l/10 0 ft/10 (1 in/foot) 4 in. (B) Assume two rows of steel both top and bottom: d pos d neg h 3.5 in 0.5 in. (C) Use a narrow beam for better efficiency. Assuming r b/d pos 0.4, b 0.4d pos 0.4(0.5 in) 8. in, use b 9 in. (D) Compute M u and M n 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 bhw c (9 in)(4 in)(150 lb/ft 3 )(1 foot /144 in )(1 kip/1000 lb) 0.5 kip/ft DL SDL + SW (.15 kip/ft + 0.5 kip/ft).375 kip/ft w u 1.DL + 1.6LL 1.(.375 kip/ft) + 1.6(.65 kip/ft) 7.09 kip/ft Using the ACI design coefficients {ACI 8.3.3}: M u,pos w u l n/16 (7.09 kip/ft)(19 ft) /16 160 kip ft M n,pos M u,pos /φ (160 kip ft)/0.9 178 kip ft M u,neg w u l n/11 (7.09 kip/ft)(19 ft) /11 33 kip ft M n,neg M u,neg /φ ( 33 kip ft)/0.9 59 kip ft (E) Compute R Mn bd. R pos M n,pos bd pos R neg M n,neg bd neg 178 kip ft (1 in/foot)(1000 lb/1 kip) 564 psi (1) (9 in)(0.5 in) 59 kip ft (1 in/foot)(1000 lb/1 kip) 80 psi () (9 in)(0.5 in) 1

5 #7s #7s #5s 6 #5s M φ n Support 51 kip ft φ Mn Midspan 16 kip ft Figure 1: Reinforcing layout at the support and at midspan. Reading from the chart gives: (F) Compute A s,pos and A s,neg. ρ pos 0.0103 (3) ρ neg 0.0160 (4) A s,pos ρ pos bd pos 0.0103(9 in)(0.5 in) 1.90 in (5) A s,neg ρ neg bd neg 0.0160(9 in)(0.5 in).95 in (6) (G) Choose bars, make sure they fit. Positive: Use 6 #5, A s,pos 1.86 in. Negative: Use 5 #7, A s,neg 3.00 in. We are using slightly less steel for the positive reinforcement because #5 bars are small, resulting in a larger d pos 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: n b min cover + d bs + d bf,i + (n 1)c s,min (7) i1 where d bs is the diameter of the stirrup, d bf,i is the diameter of the i th flexural reinforcing bar, and c s,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, b min, is controlled by the 3 #7 (d b 0.875 in) bars: c s,min max(d b, 1 in) max(0.875 in, 1 in) 1 in (8) b min (1.5 in) + (0.375 in) + 3(0.875 in) + (1 in) 8.38 in (9) The bars will fit. Calculate the actual values for d pos and d neg. Because the positive steel is in two equal rows, the centroid is located at the center of the two rows. d pos 4 in 1.5 in 0.375 in 0.65 in 0.5 in 1.0 in (10)

The negative steel has two unequal rows so the centroid is found by: d neg,1 4 in 1.5 in 0.375 in (0.875 in)/ 1.7 in (11) d neg, d neg,1 0.875 in 1 in 19.8 in (1) d neg 3d neg,1 + d neg, 3(1.7 in) + (19.8 in) 0.9 in (13) 5 5 (H) Analyze the section, check φm n M u. 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: φm n,pos 16.4 kip ft > M u,pos 160 kip ft (14) φm n,neg 50.8 kip ft > M u,neg 33 kip ft (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 w u l n /. Thus, for the present case: V u,sup w ul n (7.09 kip/ft)(19 ft) 67.4 kip (16) Assume that the maximum shear at midspan is as derived for a simple beam: V u,mid 1.6(LL)l n 8 1.6(.65 kip/ft)(19 ft) 8 10.1 kip (17) The equation which defines the shear envelope is then given by: [ ] Vu,sup V u,mid V u (x) V u,sup x (18) l n / [ ] 67.4 kip 10.1 kip 67.4 kip x (19) 19 ft/ 67.4 kip (6.03 kip/ft)x (0) () Check that V u,max φ(v c + 8 f cbd). Recall that V u,max is the shear at the critical section distance d from the face of the support. Assume that d d pos 1.0 in(1 foot/1 in) 1.75 ft so that: V u,max V u (d pos ) V u (1.75 ft) (1) 67.4 kip (6.03 kip/ft)(1.75 ft) 56.8 kip () V c f cbd pos (3) 4, 000 lb/in (9 in)(1.0 in)(1 kip/1000 lb) (4) 3.9 kip (5) φ(v c + 8 f cbd pos ) φ(5v c ) 0.75(5)(3.9 kip) 89.7 kip. (6) Thus, V u,max 56.8 kip < φ(v c + 8 f cbd pos ) 89.7 kip indicating that the section is large enough. The shear envelope is plotted in Figure. 3

V u,sup 60 50 V u (x) (kip) 0 10 V u,max V u,mid 1 3 x (feet) 6 7 8 9 Figure : The shear envelope. (3) Pick bars and set spacing. Assume #3 stirrups will be used. The spacing near the support, dependent on V u,max, is (again, using d pos for d): s A vf y d pos V u,max φ 0. in (60 kip/in )(1.0 in) 56.8 kip V c 3.9 kip 0.75 5.35 in (7) Use s 5.0 in and compute the capacity of the stirrups: V s A vf y d pos s 0. in (60 kip/in )(1.0 in) 5.0 in 55.4 kip (8) To find the maximum spacing, compare V s to 4 f cbd pos : 4 f cbd pos 4 4, 000 lb/in (9 in)(1.0 in)(1 kip/1000 lb) 47.8 kip (9) Thus, because V s > 4 f cbd pos the maximum spacing is given by d pos /4 1.0 in/4 5.5 in. The chosen spacing, s 5.0 in < 5.5 in, meets the maximum spacing. The total capacity of the beam, φv n, for s 5.0 in is then: φv n φ(v c + V s ) 0.75(3.9 kip + 55.4 kip) 59.5 kip > V u,max 56.8 kip (30) As the shear drops, the maximum spacing will increase to d pos / (1.0 in)/ 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 V s A vf y d pos 0. in (60 kip/in )(1.0 in) 7.7 kip (31) s 10.0 in φ(v c + V s ) 0.75(3.9 kip + 7.7 kip) 38.7 kip (3) Solving Equation 0 for V u (x) 38.7 kip gives x 4.76 ft 4 ft 9 in. Note that at s 10.0 in, V s < 4 f cbd pos so that the maximum stirrup spacing is indeed d/. Check that 4

8 @ 5" 4 @ 7" 4 @ 10" " 4" 1 3 4 5 6 7 8 9 x (feet) Figure 3: Initial shear design. this meets the minimum steel requirements: ( b A v,min max 0.75 f c w s, 50b ) ws f y f y ( (9 in)10.0 in max 0.75 4, 000 lb/in 60, 000 lb/in, 50 ) lb/in (9 in)10.0 in 60, 000 lb/in (33) (34) max ( 0.071 in, 0.075 in ) 0.075 in (35) Using a #3 stirrup, A v (0.11 in ) 0. in > A v,min, so we meet the minimum area requirement of the code. Choose one more spacing interval, s 7 in, resulting in: V s A vf y d pos 0. in (60 kip/in )(1.0 in) 39.6 kip (36) s 7 in φ(v c + V s ) 0.75(3.9 kip + 39.6 kip) 47.6 kip (37) Because V s < 4 f cbd pos for s 7 in (compare Equation 36 to Equation 9), d/ spacing controls. Solving Equation 0 for V u (x) 47.6 kip gives x 3.8 ft 3 ft 3 in. Zone A starts when V u drops below φv c /: φv c / 0.75(3.9 kip)/ 8.96 kip < V u,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 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 51 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

V (x) and φ u V n (x) (kip) 60 50 40 30 0 10 φv n(x) V u(x) 1 3 4 5 6 7 8 9 x (feet) Figure 4: The shear envelope. Support Cutoff #1 Cutoff # φ Mn 51 kip ft φ Mn 163 kip ft φ Mn 111 kip ft Figure 5: Negative reinforcing layouts and section moment resistance. 6

c v c c 1 h Figure 6: Dimensions c 1, c h, and c v. 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: l d 3 f y αβγλ 40 f c+k tr d b (39) c d b Note that α 1.3 because there are more than 1 inches of concrete below the reinforcing. Recall that c min(c 1, c ) where c 1 is the distance from the center of the bar to the nearest concrete surface and c is one-half the center-to-center spacing of the bars being developed in the horizontal or vertical direction. Designating c h as the horizontal dimension and c v as the vertical dimension (see Figure 6): c 1 1.5 in + 0.375 in + (0.875 in)/.31 in (40) c h (b/ c 1 ) (9 in/.31 in).19 in (41) c v (d b + 1 in) / (0.875 in + 1 in) / 0.938 in (4) c min(c 1, c h, c v ) 0.938 in (43) Conservatively estimating that K tr 0, gives: c + K tr d b 0.938 in + 0 0.875 in 1.07 (44) However, because we have stirrups and more than d b clear spacing in both the horizontal and vertical direction, the expression from the table in {ACI 1..} can be used for a #7 bar: l d f yαβλ 0 d b 60, 000 lb/in (1.3)(1.0)(1.0) f c 0 0.875 in 4, 000 lb/in (45) 54.0 in 4.50 ft 0.37l n (46) Cutoff # 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 (c c h ). Again, we will use the general 7

Midspan Cutoff #1 Cutoff # φ Mn 16 kip ft φ Mn 138 kip ft φ Mn 88 kip ft Figure 7: Positive reinforcing layouts and section moment resistance. formula and compute c first: c 1 1.5 in + 0.375 in + (0.875 in)/.31 in (47) c (b/ c 1 )/ ((9 in)/.31 in)/ 1.10 in (48) c min(c 1, c ) 1.10 in (49) c + K tr 1.10 in + 0 1.5 (50) d b 0.875 in The expression c+ktr d b is less than 1.5, so it is more advantageous to use the equations in {ACI 1..} which gives the same development length as before: l d 0.37l n. At the cutoff locations, it is required that a buffer of length max(d, 1d b ) be provided (Rule #1). max(d, 1d b ) max(d neg, 1d b ) max(0.9 in, 1(0.875 in)) 0.9 in (51) Using a buffer of d neg 0.9 in 1.74 ft 0.09l 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. c 1 1.5 in + 0.375 in + (0.65 in)/.19 in (5) c h (b/ c 1 )/ ((9 in)/.19 in)/ 1.16 in (53) c v (d b + 1 in)/ (0.65 in + 1 in)/ 0.81 in (54) c min(c 1, c h, c v ) min(.19 in, 1.16 in, 0.81 in) 0.81 in (55) c + K tr 0.81 in + 0 1.3 (56) d b 0.65 in 8

We have more than d b clear spacing both horizontally and vertically and we are using stirrups, so again we will use the table in {ACI 1..} for #5 reinforcing steel: l d f yαβλ 5 d b 60, 000 lb/in (1.0)(1.0)(1.0) f c 5 0.65 in 4, 000 lb/in (57) 3.7 in 1.98 ft 0.104l n (58) Cutoff # 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 l d 0.104l n again. In this case our buffer is d pos 1.0 in 1.75 ft 0.09l 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 1d b past flexural cutoff points except at supports of simple spans or ends of cantilevers. {ACI 1.10.3} The buffer of max(d, 1d b ) is shown on the attached moment plot and the cutoff was determined so that this rule was satisfied. Rule # Bars must extend at least l 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 1.10., 1.10.4, 1.1.} Cutting these bars l 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 1.10.5} are met. The #7s being cut are in tension so we must satisfy {ACI 1.10.5} which stipulates we must meet one of the conditions given in {ACI 1.10.5.1, 1.10.5., or 1.10.5.3}. {ACI 1.10.5.1}: To satisfy this condition, we need V u φv n /3 or V u /φv n /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 φv n 47.6 kip, and V u V u (4.50 ft) 67.4 kip (6.03 kip/ft)(4.50 ft) 40.3 kip. V u φv n 40.3 kip 47.6 kip 0.85 > 3 Thus, we do not satisfy this section. {ACI 1.10.5.}: To satisfy this, we may need to add extra stirrups over the last 0.75d neg of the bar being cut. The required spacing of the additional bars, s + is: ( A v f y s + min (60 lb/in )b, d ) neg 8β b where β b is the ratio of the area of bars being cut to the total area. In this case, β b /5 since two #7s out of five are being cut. 0.75d neg s + (59) (60) ( 0. in (60, 000 lb/in ) ) 0.9 in s + min, 6.53 in (61) 60 lb/in (9 in) 8(/5) 0.75(0.9 in) 6.53 in/stirrup At the point of the cut we are using s 7 in spacing, so.40 stirrups (6) 0.75d neg s 0.75(0.9 in) 7 in 15.7 in 7 in.4 (63) 9

That is, we have.4 stirrups from the initial design and we need to add.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 stirrups3.4 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 1.10.5.3}: In order to meet this criterion, we must have V u /φv n 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 1.1.1} 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, 1d b, or l/16 past the negative moment P.I. {ACI 1.1.3} The two #7s that continue throughout the beam represent /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 # Rule #1 Bars must extend the longer of d or 1d b past flexural cutoff points except at supports of simple spans or ends of cantilevers. {ACI 1.10.3} The buffer is plotted on the chart and by inspection, this rule is met. Rule # Bars must extend at least l 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 1.10., 1.10.4, 1.1.} The adjacent bars were cut at cutoff #1. Consulting the attached plot, it is seen that the continuing bars continue more than l 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 1.10.5} are met. The # cutoff occurs in a region of positive moment, thus the bars are not being cut in tension. We do not need to consult {ACI 1.10.5}. Rule #6 Negative moment reinforcement must be anchored into or through supporting columns of members. {ACI 1.1.1} 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 #. 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 # Bars must extend at least l 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 1.10., 1.10.4, 1.1.} 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 1.10.5} are met. The #5 being cut is in tension so we must satisfy one of the conditions given in {ACI 1.10.5.1, 1.10.5., or 1.10.5.3}. {ACI 1.10.5.1}: 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 φv n 47.6 kip, and V u V u (5.00 ft) 67.4 kip (6.03 kip/ft)(5.00 ft) 37.3 kip. V u φv n 37.3 kip 47.6 kip 0.78 > 3 (64) We do not satisfy this section. {ACI 1.10.5.}: The required spacing over the last 0.75d of the bar being cut of the additional stirrups, s + is: 0.75d pos s + ( A v f y s + min (60 lb/in )b, d ) pos (65) 8β b ( 0. in (60, 000 lb/in ) ) 1.0 in min, 15.75 in (66) 60 lb/in (9 in) 8(1/6) 0.75(1.0 in) 15.75 in/stirrup 1.00 stirrup (67) where β b 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 0.75d pos s 0.75(1.0 in) 7 in 15.75 in 7 in/stirrup.5 stirrups (68) That is, we have.5 stirrups from the initial design and we need to add an additional 1.00 stirrup for a total of 3.5 stirrups over 15.75 in at the end of the cut bars. This works out to a spacing of 15.75 in/3.5 stirrups 4.85 inch/stirrup. So, we will reduce our spacing to 4 inches near the end of the cut bar. {ACI 1.10.5.3}: 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.. Continuing reinforcement provides double the area required for flexure at the cutoff point. That is, φm n /M u at the cutoff point, where φm n is the moment capacity of the continuing bars. After positive cutoff #1 the moment capacity is φm n 138 kip ft, and reading from the graph, the moment demand, M u, is 96 kip-ft. φm n M u 138 kip ft 96 kip ft 1.53 < (69) Thus, this criterion is not met. 3. Factored shear does not exceed three fourths the design shear strength, φv n. That is, V u /φv n 3/4. Based on the results from Equation 64, this criterion is not met either. Overall, the stipulations of {ACI 1.10.5.3} 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 1.10.5.}. Rules #4 and #5 are based on conditions near the supports and will be checked for Positive Cutoff #. 11

Positive Cutoff # Rule #1 Satisfied by inspection. Rule # Bars must extend at least l 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 1.10., 1.10.4, 1.1.} The bar at cutoff #1 was adjacent to the bars being checked now. The bars from cutoff # 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 1.10.5} 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 1.10.5}. At the cutoff location, x 3.14 ft, we are using s 5 in spacing so that φv n 56.8 kip, and V u V u (3.14 ft) 67.4 kip (6.03 kip/ft)(3.14 ft) 48.5 kip. V u φv n 48.5 kip 56.8 kip 0.85 (70) The ratio in Equation 70 is greater than /3 and 3/4, so we cannot meet the provisions of {ACI 1.10.5.1} or {ACI 1.10.5.3}. We could add extra stirrups near the cutoff location as stipulated in {ACI 1.10.5.}. 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 1.11.1} (b) Continuous interior beams with closed stirrups At least one-fourth of the positive moment reinforcement must extend six inches into the support. {ACI 1.11.1, 7.13..3} (c) Continuous interior beams without closed stirrups At least one-fourth of the positive moment reinforcement must be continuous. {ACI 1.11.1, 7.13..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..} 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: l d qm n V u + l a (71) where V u is the factored shear at the P.I. or the support, for a P.I. l a is the larger of d or 1d b and is less than the actual embedment past the P.I, for a simple support l 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 1.11.3} The positive moment P.I. occurs at 0.146l n 0.146(19 ft).77 ft. Calculating the needed quantities at the P.I.: φm n 88 kip ft 88 kip ft 88 kip ft M n 97.8 kip ft φ 0.9 (7) V u 67.4 kip (6.03 kip/ft)(.77 ft) 50.7 kip (73) l a max(d pos, 1d b ) max(1.0 in, 1(0.65 in)) 1.0 in (74) 1

Because this is a P.I. (and not a support), q 1.0. The development length from cutoff #1 is l d 3.7 in so that qm n V u + l a (1.0)97.8 kip ft (1 in/foot) + 1.0 in 44.1 in > l d 3.7 in (75) 50.7 kip Rule #5 is satisfied and we are done with the flexural steel cutoff design. FINAL DESIGN As required by ACI 1.10.5.: 8 @ 5" @ 7" 5 @ 4" 3 @ 10" Stirrup at Center 8" " Positive Cutoff # @ ft 9in Negative Cutoff #1 @ 4ft 7in Positive Cutoff #1 @ 5ft 0in Negative Cutoff # @ 6ft 0in CL 1 3 4 5 6 7 8 9 x (feet) Figure 8: The final design. 13

SERVICEABILITY DEFLECTION (1) Would large deflections cause a problem? Yes they would (given). () Calculate I g, M cr, and I cr. I g bh3 1 9 in(4 in)3 1 10, 368 in 4 (76) M cr 1 6 f rbh 1 6 7.5 4, 000 lb/in (9 in)(4 in) (77) 409, 831 lb in 34. kip ft n E s 9, 000, 000 lb/in E c 57, 000 8.04 (78) 4, 000 lb/in Note that Equation 77 is valid because this is a rectangular section. 1 Calculate I cr at the midspan (I cr,mid ) and support (I cr,sup ). At midspan, there are 6 #5 for the tensile steel (A s 1.86 in ) and #7 for the compressive steel (A s 1.0 in ). The distance from the top to the neutral axis, c, is found by solving the quadratic formula: bc + [na s + (n 1)A s] c na s d pos (n 1)A sd pos (79) (9 in)c + [ 8.04(1.86 in ) + 7.04(1.0 in ) ] c 8.04(1.86 in )(1.0 in) 7.04(1.0 in )(.3 in) (80) 4.5c + 3.4c 333.5 (81) Solving gives c 6.39 in. We can then solve for the cracked moment of inertia, I cr,mid. I cr,mid bc3 3 + (n 1)A s(c d pos) + na s (d pos c) (8) (9 in)(6.39 in)3 + 7.04(1.0 in )(6.39 in.3 in) 3 (83) +8.04(1.86 in )(1.0 in 6.39 in) (84) 4116 in 4 At the support, there are five #7 bars for tensile steel at the top since moments are negative (A s 3.0 in ), the bottom has 3 #5 (A s 0.93 in ) and is in compression so that c neg is the distance from the bottom to the neutral axis. The distances to the steel, d neg and d neg are also measured from the bottom of the section. bc neg + [na s + (n 1)A s] c neg na s d neg (n 1)A sd neg (85) (9 in)c neg + [ 8.04(3.0 in ) + 7.04(0.93 in ) ] c neg 8.04(3.0 in )(0.9 in) 7.04(0.93 in )(. in) (86) 4.5c neg + 30.7c neg 518.5 (87) Solving gives c neg 7.85 in which leads to I cr,sup bc3 neg 3 + (n 1)A s(c neg d neg) + na s (d neg c neg ) (88) (9 in)(7.85 in)3 + 7.04(0.93 in )(7.85 in. in) 3 (89) +8.04(3.0 in )(0.9 in 7.85 in) (90) 5768 in 4 (3) Find I,D, I,D+L, I,L, I,SUST. 1 For other sections use the definition: M cr frig y t. 14

Deflection under Service Dead Loads Using the coefficients from {ACI 8.3.3} with service loads (w D DL): M D,mid w D l n/16 (.375 kip/ft)(19 ft) /16 53.6 kip ft (91) M D,sup w D l n/11 (.375 kip/ft)(19 ft) /11 77.9 kip ft (9) Both exceed the uncracked moment of inertia, M cr 34. kip ft, so that I e,mid I cr,mid + (I g I cr,mid ) ( Mcr M D,mid ) 3 (93) ( ) 3 34. kip ft 4116 in 4 + (10, 368 in 4 4116 in 4 ) 5740 in 4 (94) 53.6 kip ft ( ) 3 Mcr I e,sup I cr,sup + (I g I cr,sup ) (95) M D,sup ( ) 3 34. kip ft 5768 in 4 + (10, 368 in 4 5768 in 4 ) 6157 in 4 77.9 kip ft (96) I e 0.70I e,mid + 0.30I e,sup (97) 0.70(5740 in 4 ) + 0.30(6157 in 4 ) 5865 in 4 An approximate formula for the deflection of a continuous beam is: 5wl4 n M A,supl n M B,supl n (98) 384E c I e 16E c I e 16E c I e where M A,sup and M B,sup are the moments at the supports at side A and B respectively, and E c is the modulus of elasticity for concrete: E c 57, 000 f c 57, 000 4, 000 lb/in (1 kip/1000 lb) 3605 kip/in (99) Thus, with M A,sup M B,sup M D,sup the immediate deflections from dead loads are: I,D 5w Dl 4 n 384E c I e M D,supl n 8E c I e (100) 5(.375 kip/ft)(19 ft)4 (1 in/foot) 3 384(3605 kip/in )(5865 in 4 ) (101) 77.9 kip ft(19 ft) (1 in/foot) 3 8(3605 kip/in )(5865 in 4 ) (10) I,D 0.04 in (103) Deflection under Service Dead + Live Loads The service distributed load is w D+L DL+LL.375 kip/ft+.65 kip/ft 5.05 kip/ft. M D+L,mid w D+L l n/16 (5.05 kip/ft)(19 ft) /16 113 kip ft (104) M D+L,sup w D+L l n/11 (5.05 kip/ft)(19 ft) /11 165 kip ft (105) 15

I e,mid I cr,mid + (I g I cr,mid ) ( Mcr M D+L,mid ) 3 (106) ( ) 3 34. kip ft 4116 in 4 + (10, 368 in 4 4116 in 4 ) 489 in 4 (107) 113 kip ft ( ) 3 Mcr I e,sup I cr,sup + (I g I cr,sup ) (108) M D+L,sup ( ) 3 34. kip ft 5768 in 4 + (10, 368 in 4 5768 in 4 ) 5809 in 4 165 kip ft (109) I e 0.70I e,mid + 0.30I e,sup (110) 0.70(489 in 4 ) + 0.30(5809 in 4 ) 4745 in 4 We can then calculate the deflections due to dead + live and live only: I,D+L 5w D+Ll 4 n 384E c I e M D+L,supl n 8E c I e (111) 5(5.05 kip/ft)(19 ft)4 (1 in/foot) 3 384(3605 kip/in )(4745 in 4 ) (11) 165 kip ft(19 ft) (1 in/foot) 3 8(3605 kip/in )(4745 in 4 ) (113) 0.109 in (114) I,L I,D+L I,D 0.109 in 0.04 in 0.067 in (115) Deflection under Sustained Loads Assume that 50% of the live load is sustained: M L,mid M D+L,mid M D,mid 113 kip ft 53.6 kip ft 59.4 kip ft (116) M SUST,mid M D,mid + 0.5M L,mid 53.6 kip ft + 0.5(59.4 kip ft) 83.3 kip ft (117) M L,sup M D+L,sup M D,sup 165 kip ft 77.9 kip ft 87.1 (118) M SUST,sup M D,sup + 0.5M L,sup 77.9 kip ft + 0.5(87.1 kip ft) 11 kip ft (119) Solving for the effective moments of inertia: I e,mid I cr,mid + (I g I cr,mid ) ( M cr M SUST,mid ) 3 (10) ( ) 3 34. kip ft 4116 in 4 + (10, 368 in 4 4116 in 4 ) 4549 in 4 (11) 83.3 kip ft ( ) 3 M cr I e,sup I cr,sup + (I g I cr,sup ) (1) M SUST,sup ( ) 3 34. kip ft 5768 in 4 + (10, 368 in 4 5768 in 4 ) 587 in 4 11 kip ft (13) I e 0.70I e,mid + 0.30I e,sup (14) Solving for the deflection: 0.70(4549 in 4 ) + 0.30(587 in 4 ) 4946 in 4 w SUST DL + 0.5LL.375 kip/ft + 0.5(.65 kip/ft) 3.70 kip/ft (15) 16

I,SUST 5w SUST l 4 n M SUST,supl n 384E c I e 8E c I e (16) 5(3.70 kip/ft)(19 ft)4 (1 in/foot) 3 384(3605 kip/in )(4946 in 4 ) (17) 11 kip ft(19 ft) (1 in/foot) 3 8(3605 kip/in )(4946 in 4 ) (18) I,SUST 0.079 in (19) (4) Find long-term creep and shrinkage deflection, CS. As defined in the code, CS λ I,SUST Using a 5-year timespan (ξ.0) and ρ at the midspan: Substituting all values: CS ρ A s (0.60 in ) bd pos (9 in)(1.0 in) ξ 1 + 50ρ I,SUST (130) 0.00635 (131).0 (0.079 in) 0.10 in (13) 1 + 50(0.00635) (5) Check code requirements. The serviceability deflection, S, is given by S CS + I,L 0.10 in + 0.067 in 0.187 in (133) Because large deflections could cause problems: S 0.187 in l n 19 ft (1 in/foot) 0.475 in (134) 480 480 We meet the deflection requirement, the beam is adequately designed. SERVICEABILITY CRACK WIDTH To meet the requirements for crack width, the center-to-center spacing of the tension steel, s c c, must satisfy 540 kip/in s c c.5c c (135) f s At the midspan, we have 3 #5 bars and a #3 stirrup so that s c c is: 9 in (1.5 in) (0.375 in) 0.65 in s c c.31 in (136) With 1.5 inch clear cover, c c 1.5 in + 0.375 in 1.875 in, and use f s 0.6f y 0.6(60 ksi) 36 ksi so that the equations to check are 540 kip/in s c c.31 in.5(1.875 in) 10.3 in (137) 36 kip/in ( ) ( ) 36 ksi 36 ksi s c c.31 in 1 in 1 in 1 in (138) 36 ksi This requirement is satisfied at the midspan. At supports there are three #7s in the tension zone so that f s 9 in (1.5 in) (0.375 in) 0.875 in s c c.19 in (139) The quantities c c and f s are the same as at the midspan, and since s c c is actually smaller at the support, we know that the section at the support meets crack-width requirements. THE END 17