Reinforced Concrete Design

Transcription

1 Reinfored Conrete Design Notation: a = depth of the effetive ompression blok in a onrete beam A = name for area A g = gross area, equal to the total area ignoring any reinforement A s = area of steel reinforement in onrete beam design A s = area of steel ompression reinforement in onrete beam design A st = area of steel reinforement in onrete olumn design A v = area of onrete shear stirrup reinforement ACI = Amerian Conrete Institute b = width, often ross-setional b E = effetive width of the flange of a onrete T beam ross setion b f = width of the flange b w = width of the stem (web) of a onrete T beam ross setion = shorthand for lear over C = name for entroid = name for a ompression fore C = ompressive fore in the ompression steel in a doubly reinfored onrete beam C s = ompressive fore in the onrete of a doubly reinfored onrete beam d = effetive depth from the top of a reinfored onrete beam to the entroid of the tensile steel d = effetive depth from the top of a reinfored onrete beam to the entroid of the ompression steel d b = bar diameter of a reinforing bar D = shorthand for dead load DL E E E s f = shorthand for dead load = modulus of elastiity or Young s modulus = shorthand for earthquake load = modulus of elastiity of onrete = modulus of elastiity of steel = symbol for stress 1 f = ompressive stress f = onrete design ompressive stress f pu = tensile strength of the prestressing reinforement f s = stress in the steel reinforement for onrete design f s = ompressive stress in the ompression reinforement for onrete beam design f y = yield stress or strength F = shorthand for fluid load F y = yield strength G = relative stiffness of olumns to beams in a rigid onnetion, as is h = ross-setion depth H = shorthand for lateral pressure load h f = depth of a flange in a T setion I transformed = moment of inertia of a multimaterial setion transformed to one material k = effetive length fator for olumns = length of beam in rigid joint b l d l dh = length of olumn in rigid joint = development length for reinforing steel = development length for hooks l n = lear span from fae of support to fae of support in onrete design L = name for length or span length, as is l = shorthand for live load L r = shorthand for live roof load LL = shorthand for live load M n = nominal flexure strength with the steel reinforement at the yield stress and onrete at the onrete design strength for reinfored onrete beam design M u = maximum moment from fatored loads for LRFD beam design n = modulus of elastiity transformation oeffiient for steel to onrete n.a. = shorthand for neutral axis (N.A.)

2 ph = hemial alkalinity P = name for load or axial fore vetor P o = maximum axial fore with no onurrent bending moment in a reinfored onrete olumn P n = nominal olumn load apaity in onrete design P u = fatored olumn load alulated from load fators in onrete design R = shorthand for rain or ie load R n = onrete beam design ratio = M u /bd s = spaing of stirrups in reinfored onrete beams S = shorthand for snow load t = name for thikness T = name for a tension fore = shorthand for thermal load U = fatored design value V = shear fore apaity in onrete V s = shear fore apaity in steel shear stirrups V u = shear at a distane of d away from the fae of support for reinfored onrete beam design w = unit weight of onrete w DL = load per unit length on a beam from dead load w LL = load per unit length on a beam from live load w self wt = name for distributed load from self weight of member w u = load per unit length on a beam from load fators W x y 1 balaned = shorthand for wind load = horizontal distane = distane from the top to the neutral axis of a onrete beam = vertial distane = oeffiient for determining stress blok height, a, based on onrete strength, f = elasti beam defletion = strain = resistane fator = resistane fator for ompression = density or unit weight = radius of urvature in beam defletion relationships = reinforement ratio in onrete beam design = A s /bd = balaned reinforement ratio in onrete beam design = shear strength in onrete design Reinfored Conrete Design Strutural design standards for reinfored onrete are established by the Building Code and Commentary (ACI ) published by the Amerian Conrete Institute International, and uses ultimate strength design (also known as limit state design). f = onrete ompressive design strength at 8 days (units of psi when used in equations) Materials Conrete is a mixture of ement, oarse aggregate, fine aggregate, and water. The ement hydrates with the water to form a binder. The result is a hardened mass with filler and pores. There are various types of ement for low heat, rapid set, and other properties. Other minerals or ementitious materials (like fly ash) may be added.

3 ASTM designations are Type I: Ordinary portland ement (OPC) Type II: Low temperature Type III: High early strength Type IV: Low-heat of hydration Type V: Sulfate resistant The proper proportions, by volume, of the mix onstituents determine strength, whih is related to the water to ement ratio (w/). It also determines other properties, suh as workability of fresh onrete. Admixtures, suh as retardants, aelerators, or superplastiizers, whih aid flow without adding more water, may be added. Vibration may also be used to get the mix to flow into forms and fill ompletely. Slump is the measurement of the height loss from a ompated one of fresh onrete. It an be an indiator of the workability. Proper mix design is neessary for durability. The ph of fresh ement is enough to prevent reinforing steel from oxidizing (rusting). If, 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 further raking. Adequate over of the steel by the onrete is important. Deformed reinforing bars ome in grades 40, 60 & 75 (for 40 ksi, 60 ksi and 75 ksi yield strengths). Sizes are given as # of 1/8 up to #8 bars. For #9 and larger, the number is a nominal size (while the atual size is larger). Reinfored onrete is a omposite material, and the average density is onsidered to be 150 lb/ft 3. It has the properties that it will reep (deformation with long term load) and shrink (a result of hydration) that must be onsidered. Constrution Beause fresh 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 flow around reinforing bars or into tight loations, but exess vibration will ause segregation, honeyombing, and exessive bleed water whih will redue the water available for hydration and the strength, subsequently. After asting, the surfae must be worked. Sreeding removes the exess from the top of the forms and gets a rough level. Floating is the proess of working the aggregate under the surfae and to float some paste to the surfae. Troweling takes plae when the mix has hydrated to the point of supporting weight and the surfae is smoothed further 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 of adding a texture, ommonly by using a broom, after the onrete has begun to set. 3

4 Behavior Plane setions of 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 of elastiity is different. (f=e) E y ρ 1 f1 E1ε E y ρ f Eε In order to determine the stress, we an define n as the ratio of the elasti moduli: E n E n is used to transform the width of the seond material suh that it sees the equivalent element stress. 1 Transformed Setion y and I In order to determine stresses in all types of material in the beam, we transform the materials into a single material, and alulate the loation of the neutral axis and modulus of inertia for that material. ex: When material 1 above is onrete and material is steel to transform steel into onrete n E E 1 E E steel onrete to find the neutral axis of the equivalent onrete member we transform the width of the steel by multiplying by n to find the moment of inertia of the equivalent onrete member, I transformed, use the new geometry resulting from transforming the width of the steel onrete stress: steel stress: f onrete f steel I I My transformed Myn transformed 4

5 Reinfored Conrete Beam Members Ultimate Strength Design for Beams The ultimate strength design method is similar to LRFD. There is a nominal strength that is redued by a fator whih must exeed the fatored design stress. For beams, the onrete only works in ompression over a retangular stress blok above the n.a. from elasti alulation, and the steel is exposed and reahes the yield stress, F y For stress analysis in reinfored onrete beams the steel is transformed 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 of 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. is onsidered ineffetive. The steel below the n.a. is in tension. Beause the n.a. is defined by the moment areas, we an solve for x knowing that d is the distane from the top of the onrete setion to the entroid of the steel: x bx na s( d x ) 0 x an be solved for when the equation is rearranged into the generi format with a, b & in the binomial equation: ax bx 0 by b x b 4a a T-setions If the n.a. is above the bottom of a flange in a T setion, x is found as for a retangular setion. If the n.a. is below the bottom of a flange in a T setion, x is found by inluding the flange and the stem of the web (b w ) in the moment area alulation: h b h x f f f x h f b w x h f na (d x ) 0 s f b w h f f b w h f 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 6

7 Internal Equilibrium h A s b d C x a= 1 x n.a. T 0.85f a/ T C atual stress Whitney stress blok C = ompression in onrete = stress x area = 0.85 f ba T = tension in steel = stress x area = A s f y C = T and Mn = T(d-a/) where f = onrete ompression strength a = height of stress blok 1 = fator based on f x = loation to the neutral axis b = width of stress blok f y = steel yield strength A s = area of steel reinforement d = effetive depth of setion = depth to n.a. of reinforement With C=T, Asfy = 0.85 f ba so a an be determined with a As f y f b Criteria for Beam Design For flexure design: Mu Mn = 0.9 for flexure (when the setion is tension ontrolled) so for design, Mu an be set to Mn =T(d-a/) = Asfy (d-a/) Reinforement Ratio The amount of steel reinforement is limited. Too muh reinforement, or over-reinforing will not allow the steel to yield before the onrete rushes and there is a sudden failure. A beam with the proper amount of steel to allow it to yield at failure is said to be under reinfored. As The reinforement ratio is just a fration: ρ (or p) and must be less than a value bd determined with a onrete strain of and tensile strain of (minimum). When the strain in the reinforement is or greater, the setion is tension ontrolled. (For smaller strains the resistane fator redues to 0.65 see tied olumns - 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 from the amount of steel that would make the onrete start to rush at the exat same time that the steel would yield based on strain. 7

8 Flexure Design of Reinforement One method is to wisely estimate a height of the stress blok, a, and solve for A s, and alulate a new value for a using M u. 1. guess a (less than n.a.). A s. f ba 0 85 f 3. solve for a from y setting Mu = Asfy (d-a/): M u a d As f y 4. repeat from. until a found from step 3 mathes a used in step. Design Chart Method: M n 1. alulate Rn bd. find urve for f and f y to get 3. alulate A s and a, where: A s bd and a As f y f b Any method an simplify the size of d using h = 1.1d for whih is permitted to be 0.9 from Reinfored Conrete, 7th, Wang, Salmon, Pinheira, Wiley & Sons, 007 Maximum Reinforement Based on the limiting strain of in the steel, x(or ) = 0.375d so a 1 ( d ) to find A s-max ( 1 is shown in the table above) Minimum Reinforement Minimum reinforement is provided even if the onrete an resist the tension. This is a means to ontrol raking. f Minimum required: As 3 ( bwd f y ) 00 but not less than: As ( b d ) (tensile strain of 0.004) w f y f where f is in psi. This an be translated to 3 min but not less than f y 8 00 f y

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

10 Slabs One way slabs an be designed as one unit - wide beams. Beause they are thin, ontrol of defletions is important, and minimum depths are speified, as is minimum reinforement for shrinkage and rak ontrol when not in flexure. Reinforement is ommonly small diameter bars and welded wire fabri. Maximum spaing between bars is also speified for shrinkage and rak ontrol as five times the slab thikness not exeeding 18. For required flexure reinforement the spaing limit is three times the slab thikness not exeeding 18. Shrinkage and temperature reinforement (and minimum for flexure reinforement): Minimum for slabs with grade 40 or 50 bars: or As-min = 0.00bt bt Minimum for 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 reinforement is required to bridge the raks whih are alled shear stirrups (or stirrups). The maximum shear for design, V u is the value at a distane of d from the fae of the support. Nominal Shear Strength The shear fore that an be resisted is the shear stress ross setion area: V b d w The shear stress for beams (one way) where f so V f b d One-way joists are allowed an inrease of 10% V if the joists are losely spaed. Av f yd Stirrups are neessary for strength (as well as rak ontrol): Vs s 8 f bwd (max) where A v = area of all vertial legs of stirrup s = spaing of stirrups d = effetive depth 10 b w = the beam width or the minimum width of the stem. = 0.75 for shear w

11 For shear design: V U V V = 0.75 for shear C S Spaing Requirements Stirrups are required when V u is greater than V Eonomial spaing of stirrups is onsidered to be greater than d/4. Common spaings of d/4, d/3 and d/ are used to determine the values of V s at whih the spaings an be inreased. Av f yd Vs s This figure shows the size of V n provided by V + V s (long dashes) exeeds V u / in a step-wise funtion, 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 from the stirrups is 8 f b d ) w The minimum reommended spaing for the first stirrup is inhes from the fae of the support. 11

12 Torsional Shear Reinforement 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 for Reinforement Beause the design is based on the reinforement attaining the yield stress, the reinforement needs to be properly bonded to the onrete for a finite length (both sides) so it won t slip. This is referred to as the development length, l d. Providing suffiient length to anhor bars that need to reah the yield stress near the end of onnetions are also speified by hook lengths. Detailing reinforement is a tedious job. Splies are also neessary to extend the length of reinforement 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 f dbfy #7 bars and larger: ld or 1 in. minimum 0 f Development Length in Compression 0. 0dbFy ld d bfy f Hook Bends and Extensions The minimum hook length is l dh d 100 f b 1

13 Modulus of Elastiity & Defletion E for defletion alulations an be used with the transformed setion modulus in the elasti range. After that, the raked setion modulus is alulated and E is adjusted. Code values: E 57, 000 f (normal weight) E w f, w = 90 lb/ft lb/ft 3 Defletions of beams and one-way slabs need not be omputed if the overall member thikness meets the minimum speified by the ode, and are shown in Table 9.5(a) (see Slabs). Criteria for 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 often the flexure resisting elements are ontinuous, having positive and negative bending moments. These moment and shear values an be found using beam tables, or from ode speified approximate design fators. Flat slab two-way systems have drop panels (for shear), while flat plates do not. Criteria for Column Design (Amerian Conrete Institute) ACI Code and Commentary: P u P n where P u is a fatored load is a resistane fator 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 for spirally reinfored, = 0.65 and P n = 0.8P o for tied olumns where P f ( A A ) f A and P o is the name of the o g st y st maximum axial fore with no onurrent bending moment. Columns whih have reinforement ratios, A st ρg, in the Ag range of 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 fore and must be designed for the ombined effets of 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 form of load vs. equivalent eentriity for standard olumn sizes and bars used. Rigid Frames Monolithially ast frames with beams and olumn elements will have members with shear, bending and axial loads. Beause the joints an rotate, the effetive length must be determined from methods like that presented in the handout on Rigid Frames. The harts for evaluating k for non-sway and sway frames an be found in the ACI ode. Frame Columns Beause joints an rotate in frames, the effetive length of the olumn in a frame is harder to determine. The stiffness (EI/L) of eah member in a joint determines how rigid or flexible it is. To find k, the relative stiffness, G or, must be found for both ends, plotted on the alignment harts, and onneted by a line for braed and unbraed fames. where EI l G EI l b E = modulus of elastiity for a member I = moment of inertia of for a member l = length of the olumn from enter to enter l b = length of the beam from enter to enter For pinned onnetions we typially use a value of 10 for. For fixed onnetions we typially use a value of 1 for. 14

15 Braed non-sway frame Unbraed sway frame Example 1 15

16 Example h M n 3 f =0.80 in bd, F y M u M u lb-in M n lb-in mm M n Example 3 16

17 Example 3 (ontinued) 17

18 Example 4 A simply supported beam 0 ft long arries a servie dead load of 300 lb/ft and a live load of 500 lb/ft. Design an appropriate beam (for flexure only). Use grade 40 steel and onrete strength of 5000 psi. SOLUTION: Find the design moment, Mu, from the fatored load ombination of 1.D + 1.6L. It is good pratie to guess a beam size to inlude self weight in the dead load, beause servie means dead load of everything exept the beam itself. Guess a size of 10 in x 1 in. Self weight for normal weight onrete is the density of 150 lb/ft 3 multiplied by the ross setion 1ft area: self weight = 150 lb 3 (10in)(1in) ( ) = 15 lb/ft ft 1in wu = 1.(300 lb/ft + 15 lb/ft) + 1.6(500 lb/ft) = 1310 lb/ft wl The maximum moment for a simply supported beam is 8 : Mu = w u l lb ft 8 (0ft) 65,500 lb-ft Mn required = Mu/ = 65, lb ft = 7,778 lb-ft To use the design hart aid, find 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: If there are stirrups, you must also subtrat the diameter of the stirrup bar.) lbft 7,778 Rn = (1 in ft) = 831 psi (10in)(10. 5in) orresponds to approximately 0.03, so the estimated area required, As, an be found: As = bd = (0.03)(10in)(10.5in) =.36 in The number of bars for this area an be found from handy harts. (Whether the number of bars atually fit for the width with over and spae between bars must also be onsidered. If you are at max do not hoose an area bigger than the maximum!) Try As =.37 in from 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 reinfored beam!!) Find the moment apaity of the beam as designed, Mn a = Asfy/0.85f b =.37 in (40 ksi)/[0.85(5 ksi)10 in] =.3 in.3in 1 Mn = Asfy(d-a/) = 0.9(.37in )(40ksi)(10in ) ( ) 63. k-ft 65.5 k-ft needed (not OK) 1 in So, we an inrease d to 13 in, and Mn = 70.3 k-ft (OK). Or inrease As to # 10 s (.54 in ), for a =.39 in and Mn of 67.1 k-ft (OK). Don t exeed max ft 18

19 Example 5 A simply supported beam 0 ft long arries a servie dead load of 45 lb/ft (inluding self weight) and a live load of 500 lb/ft. Design an appropriate beam (for flexure only). Use grade 40 steel and onrete strength of 5000 psi. SOLUTION: Find the design moment, Mu, from the fatored load ombination of 1.D + 1.6L. If self weight is not inluded in the servie loads, you need to guess a beam size to inlude self weight in the dead load, beause servie means dead load of everything exept the beam itself. wu = 1.(45 lb/ft) + 1.6(500 lb/ft) = 1310 lb/ft wl The maximum moment for a simply supported beam is : Mu = 8 Mn required = Mu/ = 65, lb ft = 7,778 lb-ft M n 1310 lb w l ( 0 ft u ft 65,500 lb-ft 8 8 ) To use the design hart aid, we an find Rn =, and estimate that h is roughly 1.5- times the size of b, and h = 1.1d (rule of bd thumb): d = h/1.1 = (b)/1.1, so d 1.8b or b 0.55d. We an find Rn at the maximum reinforement ratio for our materials off of the hart at about 100 psi, with max = (Pratial max at a strain = is ). Let s substitute b for a funtion of d: lb ft Rn = 100 psi = 7,778 (1 in ) ft (0.55d)( d) Rearranging and solving for d = 11.0 inhes That would make b roughly 6, whih is impratial. 10 in is ommonly the smallest width. So if h is ommonly 1.5 to times the width, b, h ranges from 14 to 0 inhes. (10x1.5=15 and 10x = 0) Choosing a depth of 14 inhes, d (lear over) - ½(1 diameter bar guess) -3/8 in (stirrup diameter) = in. Now alulating an updated Rn = 7,778 lb ft (10in)(11. 65in) (1 in now is 0.00, so the estimated area required, As, an be found: As = bd = (0.00)(10in)(11.65in) = 1.98 in ) ft 646.psi The number of bars for this area an be found from handy harts. (Whether the number of bars atually fit for the width with over and spae between bars must also be onsidered. If you are at max do not hoose an area bigger than the maximum!) Try As =.37 in from 3#8 bars. (or.0 in from #9 bars. 4#7 bars don t fit...) 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 reinfored beam!!) Find the moment apaity of the beam as designed, Mn a = Asfy/0.85f b =.37 in (40 ksi)/[0.85(5 ksi)10 in] =.3 in Mn = Asfy(d-a/) =.3in 1 0.9(.37in )(40ksi)(11.65in ) ( ) 74.7 k-ft > 65.5 k-ft needed 1in OK! Note: If the setion doesn t work, you need to inrease d or As as long as you don t exeed max 19 ft

20 Example 6 A simply supported beam 5 ft long arries a servie dead load of k/ft, an estimated self weight of 500 lb/ft and a live load of 3 k/ft. Design an appropriate beam (for flexure only). Use grade 60 steel and onrete strength of 3000 psi. SOLUTION: Find the design moment, Mu, from the fatored load ombination of 1.D + 1.6L. If self weight is estimated, and the seleted size has a larger self weight, the design moment must be adjusted for the extra load. wu = 1.( k/ft k/ft) + 1.6(3 k/ft) = 7.8 k/ft So, Mu = Mn required = Mu/ = k ft = k-ft M n wul k ft( 5 ft k-ft 8 ) To use the design hart aid, we an find Rn =, and estimate that h is roughly 1.5- times the size of b, and h = 1.1d (rule of bd thumb): d = h/1.1 = (b)/1.1, so d 1.8b or b 0.55d. We an find Rn at the maximum reinforement ratio for our materials off of the hart at about 770 psi, with max = (Pratial max at a strain = is ). Let s substitute b for a funtion of d: k ft lb / k Rn = 770 psi = ( 1000 ) ( 1 in ft ) ( 0. 55d )( d ) Rearranging and solving for d = 6.6 inhes That would make b 13.3 in. (from 0.55d). Let s try 14. So, h d (lear over) +½(1 diameter bar guess) +3/8 in (stirrup diameter) = = in. Choosing a depth of 9 inhes, d (lear over) - ½(1 diameter bar guess) -3/8 in (stirrup diameter) = 6.65 in. Now alulating an updated Rn = 677,100 lb ft (14in)(6.65in) (1in ) ft 819psi OOPS! This is larger than the hart 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 30 in., then Rn = 760 psi with d = 7.65 in.. That puts us at max. We d have to remember to keep UNDER the area of steel alulated, whih is hard to do. Let s inrease h again to 31 in., then Rn = psi with d = 8.65 in. From the hart, 0.013, so the estimated area required, As, an be found: As = bd = (0.013)(14in)(8.65in) = 5. in The number of bars for this area an be found from handy harts. Our harts say there an be 3 6 bars that fit when ¾ aggregate is used. We ll assume 1 inh spaing between bars. The atual limit is the maximum of 1 in, the bar diameter or 1.33 times the maximum aggregate size. Try As 5.0 = in from 5#9 bars. Chek the width: 14 3 (1.5 in over eah side) 0.75 (two #3 stirrup legs) 5*1.18 4*1 in. = OK. d(atually) = 31 in. 1.5 in (over) ½ (1.7 in bar diameter) 3/8 in. (stirrup diameter) = 8.5 in. Find the moment apaity of the beam as designed, Mn a = Asfy/0.85f b = 5 in (60 ksi)/[0.85(3 ksi)14 in] = 8.4 in 8.4in 1 Mn = Asfy(d-a/) = 0.9(5in )(60ksi)(8.49in ) ( ) k-ft < 609 k-ft needed!! (NO GOOD) 1 in ft 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. for d = 9.49 in. AND b = 15 in., then As = 0.013(15in)(9.49in)=5.75in. 6#9 s won t fit, so inrease b to 16 in. and Mn = k-ft (!!!) Chek = 6 in /(16 in)(9.49 in) = whih is less than max = GOOD (We annot have an over reinfored beam!!) Chek self weight: (16in)(3in)/(1 in/ft ) *150 lb/ft 3 =533 lb/ft. The new design moment is Mu = k-ft < Mn OK 0

21 Example 7 1

22 Example 8 Design a T-beam for a floor with a 4 in slab supported by -ft-span-length beams ast monolithially with the slab. The beams are 8 ft on enter and have a web width of 1 in. and a total depth of in.; f = 3000 psi and f y = 60 ksi. Servie loads are 15 psf and 00 psf dead load whih does not inlude the weight of the floor 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/ft 4.7() 58 ft-kips 3.00 in = (66)(19) = in. > 3.00 in. (O.K) 1. Verify the moment apaity: (Is ) a = (3.00)(60)/[0.85(3)(66)] = 1.07 in. = ft-kips (Not O.K) Choose more steel, A s = 3.16 in from 4-#8 s d = 19.6 in, a = 1.13 in M n = 71.0 ft-kips, whih is OK 13. Sketh the design 58 R n = ksi R n of ksi required =0.004

23 Example 9 Design a T-beam for the floor system shown for whih b w and d are given. M D = 00 ft-k, M L = 45 ft-k, f = 3000 psi and f y = 60 ksi, and simple span = 18 ft. SOLUTION retangular orret. If the. Now 3

24 Example 10 4

25 Example 11 1.w DL + 1.6w LL 1.(0.075) + 1.6(0.400) kip/ft 0.73(10) 9.15 ft-kips 11. Verify the moment apaity: (Is ) R n = R n : 9.15(1) ksi R n = 0.457, the required = = 10.6 ft-kips OK) 1. A design sketh is drawn: > (1)(4.88)=0.45 in. /ft 5

26 Example 1 6

27 Example 13 For the simply supported onrete beam shown in Figure 5-61, determine the stirrup spaing (if required) using No. 3 U stirrups of Grade 60 (f y = 60 ksi). Assume f = 3000 psi. with legs, then (0.75) 3.0 V + V s V s = V u - V = = 18.0 kips (< 64.1 s req d A v F y d ( )( 0. in )( 60ksi )( 3. 5in) 18. 0k in. s req d V when V >V u >, but 16 (d/) would be the maximum as well. Use #3 16 max spaing 7

28 Example 14 Design the shear reinforement for the simply supported reinfored onrete beam shown with a dead load of 1.5 k/ft and a live load of.0 k/ft. Use 5000 psi onrete and Grade 60 steel. Assume that the point of reation is at the end of the beam. SOLUTION: in 111 in Shear diagram: Find self weight = 1 ft x (7/1 ft) x 150 lb/ft 3 = 338 lb/ft = k/ft wu = 1. (1.5 k/ft k/ft) ( k/ft) = 5.41 k/ft (= k/in) Vu (max) is at the ends = wul/ = 5.41 k/ft (4 ft)/ = 64.9 k Vu (support) = Vu (max) wu(distane) = 64.9 k 5.4 1k/ft (6/1 ft) = 6. k Vu for design is d away from the support = Vu (support) wu(d) = 6. k 5.41 k/ft (3.5/1 ft) = 51.6 k Conrete apaity: We need to see if the onrete needs stirrups for strength or by requirement beause Vu V + Vs (design requirement) V = f bwd = 0.75 () Stirrup design and spaing We need stirrups: Av = Vss/fyd 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 for a #3 is 0. in, so sreq d Avfyd/Vs = 0.75(0. in )(60 ksi)(3.5 in)/1.7 k = 10.7 in Use s = 10 our maximum falls 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 first stirrup at a minimum of in from the support fae, so we need 10 spaes for (78-6 in)/10 in = 7 even (8 stirrups altogether ending at 78 in) After 78 we an hange the spaing to the required (but not more than the maximum of d/ = in 4in); s = Avfy / 50bw = 0. in (60,000 psi)/50 (1 in) = in We need to ontinue to 111 in, so ( in)/ 11 in = 3 even 8 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

29 Example 15 9

30 Example 15 (ontinued) A s-min = 0.1 in /ft No. 3 at 11 temperature reinforement No. 3 at 8 No. 3 at 8 No. 3 at 8 Example 16 No. 3 at 9 No. 3 at 11 B = 1.(93.8) + 1.6(50) = = 516. psf (design load) Beause we are designing a slab segment that is 1 in. wide, the foregoing loading is the same as 51.6 lb/ft or kip/ft. 30

31 Example 16 (ontinued) (0.513)(11) = 4.43 ft-kips (end span) (0.513)(11) = 3.88 ft-kips (interior span) (0.513)(11) = 6.0 ft-kips (end span - first interior support) (0.513)(11) = 5.64 ft-kips (interior span both supports) (0.513)(11) =.58 ft-kips (end span exterior support) Similarly, the shears are determined using the ACI shear equations. In the end span at the fae of the first interior support, 1.15(0.513) 3.4 kips (end span first interior support) =(0.513).8 kips 4. Design the slab. Assume #4 bars for main steel with ¾ in. over: d = ½(0.5) = 4.5 in. 5. Design the steel. (All moments must be onsidered.) For example, the negative moment in the end span at the first interior support: R M 6. 0( 1 )( 1000 ) 0. 9( 1 )( 4. 5 ) u ftkips n 340 bd so A s = bd = 0.006(1)(4.5) = 0.35 in per ft. width of slab Use #4 at 7 in. (16.5 in. max. spaing) The minimum reinforement required for flexure is the same as the shrinkage and temperature steel. (Verify the moment apaity is ahieved: a 0.67 in. and M n = 6.38 ft-kips > 6.0 ft-kips) For grade 60 the minimum for shrinkage and temperature steel is: A s-min = bt = (1)(5.5) = 0.1 in per ft. width of slab Use #3 at 11 in. (18 in. max spaing) 6. Chek the shear strength. V f bd 0. 75( ) 3000( 1 )( 4. 5) lb = 4.44 kips 6 V u V Therefore the thikness is O.K. 7. Development length for the flexure reinforement is required. (Hooks are required at the spandrel beam.) For example, #6 bars: d bfy ld or 1 in. minimum #3 at 11 o.. temperature reinforement 5 f #3 at 11 o.. #4 at 7 o.. #4 at 8 o.. With grade 40 steel and 3000 psi onrete: 6 8 in(40,000 psi) l d 1. 9in psi #4 at 1 o.. #4 at 15 o.. (whih is larger than 1 in.) 8. Sketh: 31

32 Example 17 A building is supported on a grid of olumns that is spaed at 30 ft on enter in both the north-south and east-west diretions. Hollow ore planks with a in. topping span 30 ft in the east-west diretion and are supported on preast L and inverted T beams. Size the hollow ore planks assuming a live load of 100 lb/ft. Choose the shallowest plank with the least reinforement that will span the 30 ft while supporting the live load. SOLUTION: The shallowest that works is an 8 in. deep hollow ore plank. The one with the least reinforing has a strand pattern of 68-S, whih ontains 6 strands of diameter 8/16 in. = ½ in. The S indiates that the strands are straight. The plank supports a superimposed servie load of 14 lb/ft at a span of 30 ft with an estimated amber at eretion of 0.8 in. and an estimated long-time amber of 0. in. The weight of the plank is 81 lb/ft. 3

33 Example 18 Also, design for e = 6 in. 33

34 Example 19 Determine the apaity of 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 for tied olumns and P f ( A o g A st ) f Steel area (found from reinforing bar table for the bar size): Ast = 8 bars (1.7 in ) = in Conrete area (gross): Ag = 16 in 16 in = 56 in Grade 40 reinforement has fy = 40,000 psi and y A st f = 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 0 16 x 16 preast reinfored olumns support inverted T girders on orbels as shown. The unfatored loads on the orbel are 81 k dead, and 7 k live. The unfatored loads on the olumn are 170 k dead and 150 k live. Determine the reinforement required using the interation diagram provided. Assume that half the moment is resisted by the olumn above the orbel and the other half is resisted by the olumn below. Use grade 50 steel and 5000 psi onrete. orbel 34

35 Example 1 35

36 Example ACI 7.7: Conrete exposed to earth or weather: No. 6 through No. 18 bars... in. minimum (0.75)(4)(45) (0.75)(4)(45)(4) (0.0)(45) = 9.04 in. #8, A st = 9.48 in. 17 bars of #8 an be arranged in ACI 10.1: In nonsway frames it shall be permitted to ignore slenderness effets for kl ompression members that satisfy: u M r M 36

37 Fatored Moment Resistane of Conrete Beams, M n (k-ft) with f = 4 ksi, f y = 60 ksi a Approximate Values for a/d Approximate Values for b x d (in) x 14 #6 #8 3 # x 18 3 #5 #9 3 # x #7 3 #8 (3 #10) x 16 #7 3 #8 4 # x 0 #8 3 #9 4 # x 4 #8 3 #9 (4 #10) x 0 3 #7 4 #8 5 # x 5 3 #8 4 #9 4 # x 30 3 #8 5 #9 (5 #11) x 4 3 #8 5 #9 6 # x 30 3 #9 6 #9 (6 #11) x 36 3 #10 6 #10 (7 #11) x 30 3 # 10 7 # 9 6 # x 35 4 #9 5 #11 (7 #11) x 40 6 #8 6 #11 (9 #11) x 3 6 #8 7 #10 (8 #11) x 40 6 #9 7 #11 (10 #11) x 48 5 #10 (8 #11) (13 #11) a Table yields values of fatored moment resistane in kip-ft with reinforement indiated. Reinforement hoies shown in parentheses require greater width of beam or use of two stak layers of bars. (Adapted and orreted from Simplified Engineering for Arhitets and Builders, 11 th ed, Ambrose and Tripeny,

38 Column Interation Diagrams 38

39 Column Interation Diagrams 39

40 Beam / One-Way Slab Design Flow Chart Collet data: L,,, llimits, hmin (or tmin); find beam harts for load ases and atual equations (estimate wself weight = x A) Collet data: load fators, fy, f' Find Vu & Mu from onstruting diagrams or using beam hart formulas with the fatored loads (Vu-max is at d away from fae of support) Assume b & d (based on hmin or tmin for slabs) Determine Mn required by Mu/, hoose method Chart (Rn vs ) Selet min max Find Rn off hart with fy, f and selet min max Choose b & d ombination based on Rn and hmin (tmin slabs), estimate h with 1 bars (#8) Calulate As = bd Selet bar size and spaing to fit width or 1 in strip of slab and not exeed limits for rak ontrol Inrease h, find d* Find new d / adjust h; Is min max? NO YES or provide As min Calulate a, Mn Inrease h, find d Is Mu Mn? NO Yes (on to shear reinforement for beams) 40

41 Beam / One-Way Slab Design Flow Chart - ontinued Beam, Adequate for Flexure Determine shear apaity of plain onrete based on f, b & d Is Vu (at d for beams) V? NO Beam? NO YES YES Slab? Determine Vs = (Vu - V ) / Inrease h and re-evaluate flexure (As and Mn of previous page)* NO Is Vu < ½ V? YES Is Vs? YES Determine s & As NO Find where V = V and provide minimum As and hange s Find where V = ½ V and provide stirrups just past that point Yes (DONE) 41