Reinforced Concrete Design

Transcription

1 Reinored Conrete Design Notation: a = depth o the eetive ompression blok in a onrete beam A g = gross area, equal to the total area ignoring any reinorement A s = area o steel reinorement in onrete beam design A s = area o steel ompression reinorement in onrete beam design A st = area o steel reinorement in onrete olumn design A v = area o onrete shear stirrup reinorement ACI = Amerian Conrete Institute b = width, oten ross-setional b E = eetive width o the lange o a onrete T beam ross setion b = width o the lange b w = width o the stem (web) o a onrete T beam ross setion = distane rom the top o beam to the neutral axis (also see x) = shorthand or lear over C = name or entroid = name or a ompression ore C = ompressive ore in the ompression steel in a doubly reinored onrete beam C s = ompressive ore in the onrete o a doubly reinored onrete beam d = eetive depth rom the top o a reinored onrete beam to the entroid o the tensile steel d = eetive depth rom the top o a reinored onrete beam to the entroid o the ompression steel d b = bar diameter o a reinoring bar D = shorthand or dead load DL = shorthand or dead load e E E = eentriity = modulus o elastiity or Young s modulus = shorthand or earthquake load = modulus o elastiity o onrete 1 E s y F F y h H = modulus o elastiity o steel = symbol or stress = onrete design ompressive stress = ompressive stress in the ompression reinorement or onrete beam design = yield stress or strength = shorthand or luid load = yield strength = ross-setion depth = shorthand or lateral pressure load h = depth o a lange in a T setion I transormed = moment o inertia o a multimaterial setion transormed to one material l d s l dh = development length or reinoring steel = development length or hooks l n = lear span rom ae o support to ae o support in onrete design L = name or length or span length, as is l = shorthand or live load L r = shorthand or live roo load LL = shorthand or live load M = internal bending moment M n = nominal lexure strength with the steel reinorement at the yield stress and onrete at the onrete design strength or reinored onrete beam design M u = maximum moment rom atored loads or LRFD beam design n = modulus o elastiity transormation oeiient or steel to onrete n.a. = shorthand or neutral axis (N.A.) P o = maximum axial ore with no onurrent bending moment in a reinored onrete olumn P n = nominal olumn load apaity in onrete design P u = atored olumn load alulated rom load ators in onrete design R = shorthand or rain or ie load

2 = radius o urvature in beam deletion relationships (see ) R n = onrete beam design ratio = M u /bd s = spaing o stirrups in reinored onrete beams S = shorthand or snow load t = name or thikness (as is h) T = name or a tension ore = shorthand or thermal load U = atored design value V = shear ore apaity in onrete V n = nominal shear ore V s = shear ore apaity in steel shear stirrups V u = shear at a distane o d away rom the ae o support or reinored onrete beam design w = unit weight o onrete w DL = load per unit length on a beam rom dead load w LL = load per unit length on a beam rom live load w sel wt = name or distributed load rom sel weight o member w u = load per unit length on a beam rom load ators W x y 1 t y balaned = shorthand or wind load = distane rom the top to the neutral axis o a onrete beam = distane rom the top o beam to the neutral axis (also see ) = vertial distane = oeiient or determining stress blok height, a, based on onrete strength, = strain = strain in the steel = strain at the yield stress = resistane ator = resistane ator or ompression = density or unit weight = radius o urvature in beam deletion relationships (see R) = reinorement ratio in onrete beam design = A s /bd = balaned reinorement ratio in onrete beam design = engineering symbol or normal stress = shear strength in onrete design Reinored Conrete Design Strutural design standards or reinored onrete are established by the Building Code and Commentary (ACI ) published by the Amerian Conrete Institute International, and uses strength design (also known as limit state design). = onrete ompressive design strength at 8 days (units o psi when used in equations) Materials Conrete is a mixture o ement, oarse aggregate, ine aggregate, and water. The ement hydrates with the water to orm a binder. The result is a hardened mass with iller and pores. There are various types o ement or low heat, rapid set, and other properties. Other minerals or ementitious materials (like ly ash) may be added.

3 ASTM designations are Type I: Ordinary portland ement (OPC) Type II: Moderate heat o hydration and sulate resistane Type III: High early strength (rapid hardening) Type IV: Low heat o hydration Type V: Sulate resistant The proper proportions, by volume, o the mix onstituents determine strength, whih is related to the water to ement ratio (w/). It also determines other properties, suh as workability o resh onrete. Admixtures, suh as retardants, aelerators, or superplastiizers, whih aid low without adding more water, may be added. Vibration may also be used to get the mix to low into orms and ill ompletely. Slump is the measurement o the height loss rom a ompated one o resh onrete. It an be an indiator o the workability. Proper mix design is neessary or durability. The ph o resh ement is enough to prevent reinoring steel rom oxidizing (rusting). I, however, raks allow orrosive elements in water to penetrate to the steel, a orrosion ell will be reated, the steel will rust, expand and ause urther raking. Adequate over o the steel by the onrete is important. Deormed reinoring bars ome in grades 40, 60 & 75 (or 40 ksi, 60 ksi and 75 ksi yield strengths). Sizes are given as # o 1/8 up to #8 bars. For #9 and larger, the number is a nominal size (while the atual size is larger). Reinored onrete is a omposite material, and the average density is onsidered to be 150 lb/t 3. It has the properties that it will reep (deormation with long term load) and shrink (a result o hydration) that must be onsidered. Constrution Beause resh onrete is a visous suspension, it is ast or plaed and not poured. Formwork must be able to withstand the hydrauli pressure. Vibration may be used to get the mix to low around reinoring bars or into tight loations, but exess vibration will ause segregation, honeyombing, and exessive bleed water whih will redue the water available or hydration and the strength, subsequently. Ater asting, the surae must be worked. Sreeding removes the exess rom the top o the orms and gets a rough level. Floating is the proess o working the aggregate under the surae and to loat some paste to the surae. Troweling takes plae when the mix has hydrated to the point o supporting weight and the surae is smoothed urther and onsolidated. Curing is allowing the hydration proess to proeed with adequate moisture. Blak tarps and uring ompounds are ommonly used. Finishing is the proess o adding a texture, ommonly by using a broom, ater the onrete has begun to set. 3

4 Behavior Plane setions o omposite materials an still be assumed to be plane (strain is linear), but the stress distribution is not the same in both materials beause the modulus o elastiity is dierent. (=E) Ey R 1 1 E1 where R (or ) is the radius o urvature In order to determine the stress, we an deine n as the ratio o the elasti moduli: E n E Ey E R 1 n is used to transorm the width o the seond material suh that it sees the equivalent element stress. Transormed Setion y and I In order to determine stresses in all types o material in the beam, we transorm the materials into a single material, and alulate the loation o the neutral axis and modulus o inertia or that material. ex: When material 1 above is onrete and material is steel to transorm steel into onrete n E E 1 E E steel onrete to ind the neutral axis o the equivalent onrete member we transorm the width o the steel by multiplying by n to ind the moment o inertia o the equivalent onrete member, I transormed, use the new geometry resulting rom transorming the width o the steel onrete stress: steel stress: onrete steel I I My transormed Myn transormed 4

5 Reinored Conrete Beam Members Ultimate Strength Design or Beams The ultimate strength design method is similar to LRFD. There is a nominal strength that is redued by a ator whih must exeed the atored design stress. For beams, the onrete only works in ompression over a retangular stress blok above the n.a. rom elasti alulation, and the steel is exposed and reahes the yield stress, F y For stress analysis in reinored onrete beams the steel is transormed to onrete any onrete in tension is assumed to be raked and to have no strength the steel an be in tension, and is plaed in the bottom o a beam that has positive bending moment 5

6 The neutral axis is where there is no stress and no strain. The onrete above the n.a. is in ompression. The onrete below the n.a. (shown as x, but also sometimes named ) is onsidered ineetive. The steel below the n.a. is in tension. Beause the n.a. is deined by the moment areas, we an solve or x knowing that d is the distane rom the top o the onrete setion to the entroid o the steel: x bx na s( d x ) 0 x an be solved or when the equation is rearranged into the generi ormat with a, b & in the binomial equation: ax bx 0 by b x b 4a a T-setions I the n.a. is above the bottom o a lange in a T setion, x is ound as or a retangular setion. h h I the n.a. is below the bottom o a lange in a T setion, x is ound by inluding the lange and the stem o the web (b w ) in the moment area alulation: h x h b h x x h ( ) 0 bw nas d x b w b w Load Combinations (Alternative values are allowed) 1.4D 1.D + 1.6L + 0.5(L r or S or R) 1.D + 1.6(L r or S or R) + (1.0L or 0.5W) 1.D + 1.0W +1.0L + 0.5(L r or S or R) 1.D + 1.0E + 1.0L + 0.S 0.9D + 1.0W 0.9D + 1.0E Internal Equilibrium. h A s b d C x a= 1 x n.a. T 0.85 a/ T C atual stress Whitney stress blok 6

7 C = ompression in onrete = stress x area = 0.85 ba T = tension in steel = stress x area = A s y C = T and Mn = T(d-a/) where = onrete ompression strength a = height o stress blok 1 = ator based on (0.05) 0.65 x or = loation to the neutral axis 1000 b = width o stress blok y = steel yield strength A s = area o steel reinorement d = eetive depth o setion = depth to n.a. o reinorement A s y With C=T, Asy = 0.85 ba so a an be determined with a b Criteria or Beam Design For lexure design: Mu Mn = 0.9 or lexure (when the setion is tension ontrolled) so or design, Mu an be set to Mn =T(d-a/) = Asy (d-a/) Reinorement Ratio The amount o steel reinorement is limited. Too muh reinorement, or over-reinoring will not allow the steel to yield beore the onrete rushes and there is a sudden ailure. A beam with the proper amount o steel to allow it to yield at ailure is said to be under reinored. As The reinorement ratio is just a ration: ρ (or p). The amount o reinorement is bd limited to that whih results in a onrete strain o and a minimum tensile strain o When the strain in the reinorement is or greater, the setion is tension ontrolled. (For smaller strains the resistane ator redues to 0.65 beause the stress is less than the yield stress in the steel.) Previous odes limited the amount to 0.75 balaned where balaned was determined rom the amount o steel that would make the onrete start to rush at the exat same time that the steel would yield based on strain ( y ) o d y The strain in tension an be determined rom t (0.003). At yield, y. Es The resistane ator expressions or transition and ompression ontrolled setions are: ( t y) (0.005 ) ( t y) (0.005 ) y y or spiral members (not less than 0.75) or other members (not less than 0.65) 7

8 Flexure Design o Reinorement One method is to wisely estimate a height o the stress blok, a, and solve or A s, and alulate a new value or a using M u. 1. guess a (less than n.a.). ba. As 0 85 y 3. solve or a rom setting Mu = Asy (d-a/) : M u a d As y 4. repeat rom. until a ound rom step 3 mathes a used in step. Design Chart Method: M n 1. alulate Rn bd. ind urve or and y to get 3. alulate A s and a, where: A s bd and a As y b Any method an simpliy the size o d using h = 1.1d rom Reinored Conrete, 7th, Wang, Salmon, Pinheira, Wiley & Sons, 007 Maximum Reinorement Based on the limiting strain o in the steel, x(or ) = 0.375d so a 1 ( d ) to ind A s-max ( 1 is shown in the table above) Minimum Reinorement Minimum reinorement is provided even i the onrete an resist the tension. This is a means to ontrol raking. Minimum required: As 3 (bwd ) y 00 (tensile strain o 0.004) but not less than: As (bwd ) y where is in psi. This an be translated to 3 00 min but not less than y y 8

9 Cover or Reinorement Cover o onrete over/under the reinorement must be provided to protet the steel rom orrosion. For indoor exposure, 1.5 inh is typial or beams and olumns, 0.75 inh is typial or slabs, and or onrete ast against soil, 3 inh minimum is required. Bar Spaing Minimum bar spaings are speiied to allow proper onsolidation o onrete around the reinorement. The minimum spaing is the maximum o 1 in, a bar diameter, or 1.33 times the maximum aggregate size. T-setions (pan joists) Beams ast with slabs have an eetive width, b E, that sees ompression stress in a wide lange beam or joist in a slab system with positive bending. For interior T-setions, b E is the smallest o L/4, b w + 16t, or enter to enter o beams For exterior T-setions, b E is the smallest o b w + L/1, b w + 6t, or b w + ½(lear distane to next beam) When the web is in tension the minimum reinorement required is the same as or retangular setions with the web width (b w ) in plae o b. M n =C w (d-a/)+c (d-h /) (h is height o lange or t) When the lange is in tension (negative bending), the minimum reinorement required is the greater value o where is in psi, b w is the beam width, and b is the eetive lange width A s 6 3 ( bwd) or As ( b d) y y Compression Reinorement I a setion is doubly reinored, it means there is steel in the beam seeing ompression. The ore in the ompression steel that may not be yielding is C s = A s ( s ) The total ompression that balanes the tension is now: T = C + C s. And the moment taken about the entroid o the ompression stress is Mn = T(d-a/)+C s (a-d ) where A s is the area o ompression reinorement, and d is the eetive depth to the entroid o the ompression reinorement Beause the ompression steel may not be yielding, the neutral axis x must be ound rom the ore equilibrium relationships, and the stress an be ound based on strain to see i it has yielded. 9

10 Slabs One way slabs an be designed as one unit - wide beams. Beause they are thin, ontrol o deletions is important, and minimum depths are speiied, as is minimum reinorement or shrinkage and rak ontrol when not in lexure. Reinorement is ommonly small diameter bars and welded wire abri. Maximum spaing between bars is also speiied or shrinkage and rak ontrol as ive times the slab thikness not exeeding 18. For required lexure reinorement the spaing limit is three times the slab thikness not exeeding 18. Shrinkage and temperature reinorement (and minimum or lexure reinorement): Minimum or slabs with grade 40 or 50 bars: or As-min = 0.00bt bt Minimum or slabs with grade 60 bars: or As-min = bt bt A s A s Shear Behavior Horizontal shear stresses our along with bending stresses to ause tensile stresses where the onrete raks. Vertial reinorement is required to bridge the raks whih are alled shear stirrups (or stirrups). The maximum shear or design, V u is the value at a distane o d rom the ae o the support. Nominal Shear Strength The shear ore that an be resisted is the shear stress ross setion area: V b d w The shear stress or beams (one way) where so V b d One-way joists are allowed an inrease o 10% V i the joists are losely spaed. Av yd Stirrups are neessary or strength (as well as rak ontrol): Vs s 8 bwd (max) where A v = area o all vertial legs o stirrup s = spaing o stirrups d = eetive depth 10 b w = the beam width or the minimum width o the stem. = 0.75 or shear w

11 For shear design: V U V V = 0.75 or shear C S Spaing Requirements Stirrups are required when V u is greater than V Eonomial spaing o stirrups is onsidered to be greater than d/4. Common spaings o d/4, d/3 and d/ are used to determine the values o V s at whih the spaings an be inreased. Av yd Vs s This igure shows the size o V n provided by V + V s (long dashes) exeeds V u / in a step-wise untion, while the spaing provided (short dashes) is at or less than the required s (limited by the maximum allowed). (Note that the maximum shear permitted rom the stirrups is 8 b d w The minimum reommended spaing or the irst stirrup is inhes rom the ae o the support. 11

12 Torsional Shear Reinorement On oasion beam members will see twist along the axis aused by an eentri shape supporting a load, like on an L-shaped spandrel (edge) beam. The torsion results in shearing stresses, and losed stirrups may be needed to resist the stress that the onrete annot resist. Development Length or Reinorement Beause the design is based on the reinorement attaining the yield stress, the reinorement needs to be properly bonded to the onrete or a inite length so it won t slip. This is reerred to as the development length. Providing suiient length to anhor bars that need to reah the yield stress near the end o onnetions are also speiied with hook lengths. Detailing reinorement is a tedious job.. Splies are also neessary to extend the length o reinorement that ome in standard lengths. The equations are not provided here. Development Length in Tension With the proper bar to bar spaing and over, the ommon development length equations are: d bfy #6 bars and smaller: ld or 1 in. minimum 5 dbfy #7 bars and larger: ld or 1 in. minimum 0 Development Length in Compression 0. 0dbFy ld d bfy Hook Bends and Extensions The minimum hook length is l dh d 100 b 1

13 Modulus o Elastiity & Deletion E or deletion alulations an be used with the transormed setion modulus in the elasti range. Ater that, the raked setion modulus is alulated and E is adjusted. Code values: E 57, 000 (normal weight) E w , w = 90 lb/t lb/t 3 Deletions o beams and one-way slabs need not be omputed i the overall member thikness meets the minimum speiied by the ode, and are shown in Table 9.5(a) (see Slabs). Criteria or Flat Slab & Plate System Design Systems with slabs and supporting beams, joists or olumns typially have multiple bays. The horizontal elements an at as one-way or two-way systems. Most oten the lexure resisting elements are ontinuous, having positive and negative bending moments. These moment and shear values an be ound using beam tables, or rom ode speiied approximate design ators. Flat slab two-way systems have drop panels (or shear), while lat plates do not. Criteria or Column Design (Amerian Conrete Institute) ACI Code and Commentary: P u P n where P u is a atored load is a resistane ator P n is the nominal load apaity (strength) Load ombinations, ex: 1.4D (D is dead load) 1.D + 1.6L (L is live load) For ompression, = 0.75 and P n = 0.85P o or spirally reinored, = 0.65 and P n = 0.8P o or tied olumns where P ( A A ) A and P o is the name o the o g st y st maximum axial ore with no onurrent bending moment. Columns whih have reinorement ratios, A st ρg, in the Ag range o 1% to % will usually be the most eonomial, with 1% as a minimum and 8% as a maximum by ode. Bars are symmetrially plaed, typially. Spiral ties are harder to onstrut. 13

14 Columns with Bending (Beam-Columns) Conrete olumns rarely see only axial ore and must be designed or the ombined eets o axial load and bending moment. The interation diagram shows the redution in axial load a olumn an arry with a bending moment. Design aids ommonly present the interation diagrams in the orm o load vs. equivalent eentriity or standard olumn sizes and bars used. Rigid Frames Monolithially ast rames with beams and olumn elements will have members with shear, bending and axial loads. Beause the joints an rotate, the eetive length must be determined rom methods like that presented in the handout on Rigid Frames. The harts or evaluating k or non-sway and sway rames an be ound in the ACI ode. 14

15 Example 1 h M n 3 =0.80 in bd, F y M u M u M n M n Example (pg 43) 15

16 Example (ontinued) 16

17 Example 3 A simply supported beam 0 t long arries a servie dead load o 300 lb/t and a live load o 500 lb/t. Design an appropriate beam (or lexure only). Use grade 40 steel and onrete strength o 5000 psi. SOLUTION: Find the design moment, Mu, rom the atored load ombination o 1.D + 1.6L. It is good pratie to guess a beam size to inlude sel weight in the dead load, beause servie means dead load o everything exept the beam itsel. Guess a size o 10 in x 1 in. Sel weight or normal weight onrete is the density o 150 lb/t 3 multiplied by the ross setion 1t area: sel weight = 150 lb t 3 (10in)(1in) ( ) = 15 lb/t 1in wu = 1.(300 lb/t + 15 lb/t) + 1.6(500 lb/t) = 1310 lb/t wl The maximum moment or a simply supported beam is 8 : Mu = w u l lb t 8 (0t) 65,500 lb-t Mn required = Mu/ = 65, lb t = 7,778 lb-t To use the design hart aid, ind Rn = M n, estimating that d is about 1.75 inhes less than h: bd d = 1in 1.75 in (0.375) = 10.5 in (NOTE: I there are stirrups, you must also subtrat the diameter o the stirrup bar.) lbt 7,778 Rn = (1 in t) = 831 psi (10in)(10. 5in) orresponds to approximately 0.03 (whih is less than that or strain o ), so the estimated area required, As, an be ound: As = bd = (0.03)(10in)(10.5in) =.36 in The number o bars or this area an be ound rom handy harts. (Whether the number o bars atually it or the width with over and spae between bars must also be onsidered. I you are at max do not hoose an area bigger than the maximum!) Try As =.37 in rom 3#8 bars d = 1 in 1.5 in (over) ½ (8/8in diameter bar) = 10 in Chek =.37 in /(10 in)(10 in) = whih is less than max = OK (We annot have an over reinored beam!!) Find the moment apaity o the beam as designed, Mn a = Asy/0.85 b =.37 in (40 ksi)/[0.85(5 ksi)10 in] =.3 in.3in 1 Mn = Asy(d-a/) = 0.9(.37in )(40ksi)(1 0in ) ( ) 63. k-t 65.5 k-t needed (not OK) 1in So, we an inrease d to 13 in, and Mn = 70.3 k-t (OK). Or inrease As to # 10 s (.54 in ), or a =.39 in and Mn o 67.1 k-t (OK). Don t exeed max or max i you want to use =0.9 t 17

18 Example 4 A simply supported beam 0 t long arries a servie dead load o 45 lb/t (inluding sel weight) and a live load o 500 lb/t. Design an appropriate beam (or lexure only). Use grade 40 steel and onrete strength o 5000 psi. SOLUTION: Find the design moment, Mu, rom the atored load ombination o 1.D + 1.6L. I sel weight is not inluded in the servie loads, you need to guess a beam size to inlude sel weight in the dead load, beause servie means dead load o everything exept the beam itsel. wu = 1.(45 lb/t) + 1.6(500 lb/t) = 1310 lb/t wl The maximum moment or a simply supported beam is : Mu = 8 Mn required = Mu/ = 65, lb t = 7,778 lb-t M n 1310 lb w l ( 0 t u t 65,500 lb-t 8 8 ) To use the design hart aid, we an ind Rn =, and estimate that h is roughly 1.5- times the size o b, and h = 1.1d (rule o bd thumb): d = h/1.1 = (b)/1.1, so d 1.8b or b 0.55d. We an ind Rn at the maximum reinorement ratio or our materials, keeping in mind max at a strain = is o o the hart at about 1070 psi, with max = Let s substitute b or a untion o d: lb t Rn = 1070 psi = 7,778 (1 in ) t (0.55d)( d) Rearranging and solving or d = 11.4 inhes That would make b a little over 6 inhes, whih is impratial. 10 in is ommonly the smallest width. So i h is ommonly 1.5 to times the width, b, h ranges rom 14 to 0 inhes. (10x1.5=15 and 10x = 0) Choosing a depth o 14 inhes, d (lear over) - ½(1 diameter bar guess) -3/8 in (stirrup diameter) = in. Now alulating an updated Rn = 7,778 lb t (10in)(11. 65in) (1 in ) t 646.psi now is 0.00 (under the limit at strain o ), so the estimated area required, As, an be ound: As = bd = (0.00)(10in)(11.65in) = 1.98 in The number o bars or this area an be ound rom handy harts. (Whether the number o bars atually it or the width with over and spae between bars must also be onsidered. I you are at max do not hoose an area bigger than the maximum!) Try As =.37 in rom 3#8 bars. (or.0 in rom #9 bars. 4#7 bars don t it...) d(atually) = 14 in. 1.5 in (over) ½ (8/8 in bar diameter) 3/8 in. (stirrup diameter) = in. Chek =.37 in /(10 in)(11.65 in) = whih is less than max = OK (We annot have an over reinored beam!!) Find the moment apaity o the beam as designed, Mn a = Asy/0.85 b =.37 in (40 ksi)/[0.85(5 ksi)10 in] =.3 in Mn = Asy(d-a/) =.3in 1 0.9(.37in )(40ksi)(11.65in ) ( ) 74.7 k-t > 65.5 k-t needed 1in OK! Note: I the setion doesn t work, you need to inrease d or As as long as you don t exeed max t

19 Example 5 A simply supported beam 5 t long arries a servie dead load o k/t, an estimated sel weight o 500 lb/t and a live load o 3 k/t. Design an appropriate beam (or lexure only). Use grade 60 steel and onrete strength o 3000 psi. SOLUTION: Find the design moment, Mu, rom the atored load ombination o 1.D + 1.6L. I sel weight is estimated, and the seleted size has a larger sel weight, the design moment must be adjusted or the extra load. wu = 1.( k/t k/t) + 1.6(3 k/t) = 7.8 k/t So, Mu = Mn required = Mu/ = k t = k-t M n wul k t( 5 t k-t 8 ) To use the design hart aid, we an ind Rn =, and estimate that h is roughly 1.5- times the size o b, and h = 1.1d (rule o bd thumb): d = h/1.1 = (b)/1.1, so d 1.8b or b 0.55d. We an ind Rn at the maximum reinorement ratio or our materials o o the hart at about 700 psi with max = Let s substitute b or a untion o d: k t lb / k ( 1000 ) Rn = 700 psi = ( 1 in t ) ( 0. 55d )( d ) Rearranging and solving or d = 7.6 inhes That would make b 15. in. (rom 0.55d). Let s try 15. So, h d (lear over) +½(1 diameter bar guess) +3/8 in (stirrup diameter) = = in. Choosing a depth o 30 inhes, d (lear over) - ½(1 diameter bar guess) -3/8 in (stirrup diameter) = 7.65 in. Now alulating an updated Rn = 677,100 lb t (15in)(7. 65in) (1in ) t 710psi This is larger than Rn or the strain limit! We an t just use max The way to redue Rn is to inrease b or d or both. Let s try inreasing h to 31 in., then Rn = 661 psi with d = 8.65 in.. That puts us under max We d have to remember to keep UNDER the area o steel alulated, whih is hard to do. From the hart, 0.013, less than the max o , so the estimated area required, As, an be ound: As = bd = (0.013)(15in)(9.65in) = 5.8 in The number o bars or this area an be ound rom handy harts. Our harts say there an be 3 6 bars that it when ¾ aggregate is used. We ll assume 1 inh spaing between bars. The atual limit is the maximum o 1 in, the bar diameter or 1.33 times the maximum aggregate size. Try As = 6.0 in rom 6#9 bars. Chek the width: 15 3 (1.5 in over eah side) 0.75 (two #3 stirrup legs) 6*1.18 5*1.18 in. = in NOT OK. Try As = 5.08 in rom 4#10 bars. Chek the width: 15 3 (1.5 in over eah side) 0.75 (two #3 stirrup legs) 4*1.7 3*1.7 in. =.36 OK. d(atually) = 31 in. 1.5 in (over) ½ (1.7 in bar diameter) 3/8 in. (stirrup diameter) = 8.49 in. Find the moment apaity o the beam as designed, Mn a = Asy/0.85 b = 5.08 in (60 ksi)/[0.85(3 ksi)15 in] = 8.0 in 8.0in 1 Mn = Asy(d-a/) = 0.9(5.08in )(60ksi)( 8.49in ) ( ) k-t < 609 k-t needed!! (NO GOOD) 1 in t More steel isn t likely to inrease the apaity muh unless we are lose. It looks like we need more steel and lever arm. Try h = 3 in. AND b = 16 in., then Mu * (with the added sel weight o 33.3 lb/t) = 680. k-t, 0.01, As = 0.01(16in)(9.4in)=5.66 in. 6#9 s won t it, but 4#11 s will: = 0.013, a = 9.18 in, and Mn = 697. k-t whih is inally larger than 680. k-t OK 19

20 Example 6 (pg 437) 4. y 0

21 Example 7 Design a T-beam or a loor with a 4 in slab supported by -t-span-length beams ast monolithially with the slab. The beams are 8 t on enter and have a web width o 1 in. and a total depth o in.; = 3000 psi and y = 60 ksi. Servie loads are 15 ps and 00 ps dead load whih does not inlude the weight o the loor system SOLUTION: 0.004(66)(19) = 3.01 in. Use 3#9 (A s = 3.00 in. ) 7.15 in in. (O.K.) 1.( ) + 1.6(1.00) = 4.7 kip/t 4.7() 58 t-kips 3.00 in = (66)(19) = in. > 3.00 in. (O.K) 1. Veriy the moment apaity: (Is M M ) u n a = (3.00)(60)/[0.85(3)(66)] = 1.07 in M n 0.9(3.00)(60)( ) 1 1 = t-kips (Not O.K) Choose more steel, A s = 3.16 in rom 4-#8 s d = 19.6 in, a = 1.13 in M n = 71.0 t-kips, whih is OK 13. Sketh the design 58 R n = ksi a R n o ksi

22 Example 8 Design a T-beam or the loor system shown or whih b w and d are given. M D = 00 t-k, M L = 45 t-k, = 3000 psi and y = 60 ksi, and simple span = 18 t. SOLUTION First assume a h (whih is very oten the ase. Then the design would proeed like that o a retangular beam with a width equal to the eetive width o the T beam lange. The beam ats like a T beam, not a retangular beam, and i the values or and a above are not orret. I the value o a had been h, the value o A s would have been simply bd = 0.007(54)(4) = 9.33 in. Now break the beam up into two parts (Figure 5.7) and design it as a T beam. Assuming = 0.90 Chek minimum reinoring: A s min = 3 bd w y y (15)(4) = 60, 000 but not less than A s min = 00 b d 00(15)(4) w = = 1.in 60, 000 Only rows it, so try 8-#10 bars, A s = in or equilibrium: T = C w + C T = A s y = (10.16)(60) = k = in C = 0.85 (b-b w )h and C w = 0.85 ab w C w = T-C = k (0.85)(3)(54-15)3 = k a = 311.5/(0.85*3*15) = 8.14 in Chek strain ( t ) and : = a/ 1 = 8.14 in/0.85 = 9.58 d (0.003) (0.003) ! t 9.58 We ould try 10-#9 bars at 10 in, T =600 k, C w = k, a = 7.89, t = ; = 0.9! Designing a retangular beam with b w = 15 in. and d = 4 in. to resist 417 k-t Finally hek the apaity: a h M C ( ) ( ) n w d C d = [301.65(4-7.89/) (4-3/)]1t/1in = k-t So: M n = 0.9(1063.5) = 957. k-t 90 k-t (OK)

23 Example 9 (pg 448)

24 Example 10 1.w DL + 1.6w LL 1.(0.075) + 1.6(0.400) kip/t 0.73(10) 9.15 t-kips 11. Veriy the moment apaity: (Is M M ) u n R n : R n = 9.15(1) ksi R n = 0.457, the required = > (0.50)(60) a 0. 74in 0.85(4)(1) M 60)( n 0.9(0.50)(. ) 1 1 = 10.6 t-kips OK) 1. A design sketh is drawn: (1)(4.88)=0.45 in. /t 4

25 Example 11 (pg 461) 8. 5

26 Example 1 For the simply supported onrete beam shown in Figure 5-61, determine the stirrup spaing (i required) using No. 3 U stirrups o Grade 60 ( y = 60 ksi). Assume = 3000 psi. with legs, then (0.75) 3.0 V + V s s req d V s = V u - V = = 18.0 kips (< in. ( )( 0. in )( 60ksi )( 3. 5in) 18. 0k s req d V i V >V u >, but 16 (d/)would be the maximum as well. Use #3 16 max spaing 6

27 Example 13 Design the shear reinorement or the simply supported reinored onrete beam shown with a dead load o 1.5 k/t and a live load o.0 k/t. Use 5000 psi onrete and Grade 60 steel. Assume that the point o reation is at the end o the beam in 111 in SOLUTION: Shear diagram: Find sel weight = 1 t x (7/1 t) x 150 lb/t 3 = 338 lb/t = k/t wu = 1. (1.5 k/t k/t) ( k/t) = 5.41 k/t (= k/in) Vu (max) is at the ends = wul/ = 5.41 k/t (4 t)/ = 64.9 k Vu (support) = Vu (max) wu(distane) = 64.9 k 5.4 1k/t (6/1 t) = 6. k Vu or design is d away rom the support = Vu (support) wu(d) = 6. k 5.41 k/t (3.5/1 t) = 51.6 k Conrete apaity: We need to see i the onrete needs stirrups or strength or by requirement beause Vu V + Vs (design requirement) V = bwd = 0.75 () Stirrup design and spaing We need stirrups: Av = Vss/yd Vs Vu - V = 51.6 k 9.9 k = 1.7 k 5000 psi (1 in) (3.5 in) = lb = 9.9 kips (< 51.6 k!) Spaing requirements are in Table 3-8 and depend on V/ = 15.0 k and V = 59.8 k legs or a #3 is 0. in, so sreq d Avyd/Vs = 0.75(0. in )(60 ksi)(3.5 in)/1.7 k = 10.7 in Use s = 10 our maximum alls into the d/ or 4, so d/ governs with in Our 10 is ok. This spaing is valid until Vu = V and that happens at (64.9 k 9.9 k)/0.451 k/in = 78 in We an put the irst stirrup at a minimum o in rom the support ae, so we need 10 spaes or (78-6 in)/10 in = 7 even (8 stirrups altogether ending at 78 in) Ater 78 we an hange the spaing to the required (but not more than the maximum o d/ = in 4in); s = Avy / 50bw = 0. in (60,000 psi)/50 (1 in) = in We need to ontinue to 111 in, so ( in)/ 11 in = 3 even 7 in 8 - #3 U stirrups at 10 in Loating end points: 9.9 k = 64.9k k/in x (a) a = 78 in 15 k = 64.9k k/in x (b) b = 111 in. 3 - #3 U stirrups at 11 in

28 Example 14 (pg 483) 1 1 A s-min = 0.1 in /t No. 3 at 11 temperature reinorement No. 3 at 8 No. 3 at 8 No. 3 at 8 8 No. 3 at 9 No. 3 at 11

29 Example The bars are seleted in the same manner as or beams with minimum areas based on shrinkage and rak reinorement. Moment and shear apaities should be satisied. 5. Development length or the lexure reinorement is required (93.8) + 1.6(50) ps 51.6 lb/t or kip/t. For example, #6 bars: d bfy ld or 1 in. minimum 5 With grade 40 steel and 3000 psi onrete: 6 8 in(40,000 psi) l d 1. 9in psi (whih is larger than 1 in.) (0.513)(11) = 4.43 t-kips (0.513)(11) = 3.88 t-kips (0.513)(11) = 6.0 t-kips 1.15(0.513) 3.4 kips (0.513)(11) = 5.64 t-kips (0.513)(11) =.58 t-kips =(0.513).8 kips 9

30 Example 16 A building is supported on a grid o olumns that is spaed at 30 t on enter in both the north-south and east-west diretions. Hollow ore planks with a in. topping span 30 t in the east-west diretion and are supported on preast L and inverted T beams. Size the hollow ore planks assuming a live load o 100 lb/t. Choose the shallowest plank with the least reinorement that will span the 30 t while supporting the live load. SOLUTION: The shallowest that works is an 8 in. deep hollow ore plank. The one with the least reinoring has a strand pattern o 68-S, whih ontains 6 strands o diameter 8/16 in. = ½ in. The S indiates that the strands are straight. The plank supports a superimposed servie load o 14 lb/t at a span o 30 t with an estimated amber at eretion o 0.8 in. and an estimated long-time amber o 0. in. The weight o the plank is 81 lb/t. 30

31 Example 17 (pg 510) Also, design or e = 6 in. 31

32 Example 18 Determine the apaity o a 16 x 16 olumn with 8- #10 bars, tied. Grade 40 steel and 4000 psi onrete. SOLUTION: Find Pn, with =0.65 and Pn = 0.80Po or tied olumns and P ( A o g A st ) Steel area (ound rom reinoring bar table or the bar size): Ast = 8 bars (1.7 in ) = in Conrete area (gross): Ag = 16 in 16 in = 56 in Grade 40 reinorement has y = 40,000 psi and y A st = 4000psi Pn = (0.65)(0.80)[0.85(4000 psi )(56 in in ) + (40,000 psi)(10.16 in )] = 646,06 lb = 646 kips Example x 16 preast reinored olumns support inverted T girders on orbels as shown. The unatored loads on the orbel are 81 k dead, and 7 k live. The unatored loads on the olumn are 170 k dead and 150 k live. Determine the reinorement required using the interation diagram provided. Assume that hal the moment is resisted by the olumn above the orbel and the other hal is resisted by the olumn below. Use grade 60 steel and 5000 psi onrete. orbel 3

33 Example 0 33

34 Beam / One-Way Slab Design Flow Chart Collet data: L,,, llimits, hmin; ind beam harts or load ases and atual equations (sel weight = area x density) Collet data: load ators, y, Find Vu & Mu rom onstruting diagrams or using beam hart ormulas with the atored loads (Vu-max is at d away rom ae o support) Assume b & d (based on hmin or slabs) Determine Mn required, hoose method Chart (Rn vs ) Selet min max Find Rn o hart with y, and selet min max Choose b & d ombination based on Rn and hmin (slabs), estimate h with 1 bars (#8) Calulate As = bd Selet bar size and spaing to it width or 1 in strip o slab and not exeed limits or rak ontrol Inrease h, ind d* Find new d / adjust h; Is min max? NO YES Calulate a, Mn Inrease h, ind d Is Mu Mn? NO Yes (on to shear reinorement or beams) 34

35 Beam / One-Way Slab Design Flow Chart - ontinued Beam, Adequate or Flexure Determine shear apaity o plain onrete based on, b & d, V Is Vu (at d or beams) V? NO Beam? NO YES YES Is Vu < ½ V? Determine Vs = (Vu - V ) Inrease h and re-evaluate lexure (As and Mn o previous page)* YES YES Slab? Is V s 8 bwd (4V ) YES Determine s & Av NO Find where V = V and provide minimum Av and hange s Find where V = ½ V and provide stirrups just past that point Yes (DONE) 35