LRFD sla negative moment.mcd 7/1/3 1 of 3 Beam Data mp := 1 Beam length (ft) = length := 77. Composite sla strength (psi) = fc := 4 Concrete unit weitht (kcf) = γc :=.15 Initial strength of concrete (ksi) = fci := 5.8 Final Strength of concrete (ksi) = fcf := 7. Modulus of eam concrete ased on final (ksi) = Ec := 33 γc 1.5 fcf Ec = 57.41 Modulus of sla concrete (ksi) = Esl := 33 γc 1.5 fc Esl = 3834.54 Haunch thickness (in) = ha := Effective compression sla width (in) = ETFW := 11 Sla thickness (in) = ts := 8.5 Note: 8.5 was used in sla calculations, the value of 8.5 use here is aritrary. Live Load distriution factor for moment = LLDF :=.85 Non compostie DL (excluding eams)(k/ft) = DLnc := 1.3 Span Lengths (ft) = span := 77 77 Numer of spans = numer := ns :=.. numer 1 sp ns := 1 + ns 1 Strength of reinforcing steel (ksi) = fy := 6 Factor for slas = β1 :=.85 Size of ar to try = size := 9 As of one ar Asone := 1. Bar diameter (in) = d := 1.18 clear cover (in) = cover :=.5
LRFD sla negative moment.mcd 7/1/3 of 3 Beam type to use type := 3 1 = AASHTO TYPE I = AASHTO TYPE II 3 = AASHTO TYPE III 4 = AASHTO TYPE IV 5 = BT54 6 = BT63 7 = BT7 Beam area (in^) = Area := eam_data type, 1 Area = 56 Dist from ottom to cg (in) = y := eam_data type, y =.7 Section inertia (in^) = Inc := eam_data type, 3 Inc = 1539 Ic := Inc Beam weight (k/ft) = wt := eam_data type, 4 wt =.583 Maximum span (ft) = max_span := eam_data type, 5 max_span = 8 Total eam depth (in) = h := eam_data type, 6 h = 45 Width of top flange (in) = fwt := eam_data type, 7 fwt = 16 5 Section 4 3 eam xa, 1 1 1 3 eam xa,
LRFD sla negative moment.mcd 7/1/3 3 of 3 Composite moment of Inertia Haunch thickness (in) = ha = Effective compression sla width (in) = ETFW = 11 Modular ratio = η := fc η =.756 fcf Transformed sla width (in) = 1 := ETFW η 1 = 83.15 Sla thickness (in) = ts = 8.5 Composite distance from ottom to c.g. (in) = ts 1 ts h + ha + + Area y yc := yc = 36.439 1 ts + Area Composite N.A. to top eam (in) = yt := h yc yt = 8.561 Composite N.A. to top sla (in) = yts := h + ts yc yts = 17.61 1 ts 3 Composite moment of inertia (in^t) = Ic Inc + Area ( y yc) := + + 1 ts yts 1 ts Ic = 395.137 Composite Section Modulus Section modulus ottom of eam (in^3) = Section modulus top eam (in^3) = Section modulus top concrete (in^3) = Sc := Ic yc Sc = 1759.59 St := Ic yt St = 45795.9 Stc := Ic 1 yts η Stc = 3398.91 Non-Composite Section Modulus Section modulus ottom of eam (in^3) = Section modulus top eam (in^3) = S := Inc y S = 6185.989 St := Inc h y St = 57.36
LRFD sla negative moment.mcd 7/1/3 4 of 3 Non compostie Moments Total load including eam weight (k/ft) = DL := wt + DLnc DL = 1.586 Define range (ft) = x1 ns := j ns ceil Indicate which span = a1 ns := j ns ceil ns +.1 1 j1 ns sp ns j ns 1 j1 ns 1 ( ) span( jns ) ns +.1 1 span( jns ) 1 wt x1 ns Calculate moment at tenth points from eam = M ns := ( ) a1 ns x1 ns DLnc x1 ns Calculate moment at tenth points from other = Mo ns := ( ) a1 ns x1 ns
LRFD sla negative moment.mcd 7/1/3 5 of 3 Input the values straight from qcon here. This program was set up to use the loads generated y the program Qcon Bridge. The live load values input here shall not have had the LL dist factor applied, that will e done later. The Live loads for Qcon are in lanes per eam. The LRFD dist factor is on lanes. So the get the service I LL moments, simply multiply the Qcon values y the LL distriution factor. fullq fp1 := fn1 fp1 fn1 DC LOADS (non-comp) DW Loads LL + I FATIGUE TRUCk LOCATION self wt other (sla) (comp) M (+) M (-) M(+) M(-) 1........ 1.1 155.55 67.61 111. 69. -91. 39. -5. 1. 76.53 475.74 188. 1169. -181. 641. -1. 1.3 36.94 64.41 31. 1443. -71. 794. -149. 1.4 414.79 713.61 39. 1565. -36. 836. -199. 1.5 43.8 743.35 13. 1534. -453. 793. -48. 1.6 414.79 713.61 154. 1375. -454. 747. -98. 1.7 36.94 64.41 6. 16. -635. 591. -347. 1.8 76.53 475.74-69. 68. -75. 344. -397. 1.9 155.55 67.61-3. 37. -1. 149. -447.... -47.. -16.. -496..1 155.55 67.61-31. 37. -1. 149. -447.. 76.53 475.74-69. 68. -75. 344. -397..3 36.94 64.41 6. 16. -635. 591. -347..4 414.79 713.61 154. 1375. -454. 747. -98..5 43.8 743.35 13. 1534. -453. 793. -48..6 414.79 713.61 39. 1565. -36. 836. -199..7 36.94 64.41 31. 1443. -71. 794. -149..8 76.53 475.74 188. 1169. -181. 641. -1..9 155.55 67.61 111. 69. -91. 39. -5. 3........ ( M Mo)
LRFD sla negative moment.mcd 7/1/3 6 of 3 Service I loads (moment) Live load distriution factor has een applied to the Live Loads shown here. full SI1 SI SI3 SI4 SI5 SI6 SI7 := SI1 SI SI3 SI4 SI5 SI5 SI7. DC LOADS ( DW Loads LL + I TOTAL LOADS LOCATION self wt other (sla) (comp) M (+) M (-) M (+) M (-) 1........ 1.1 155.55 67.61 111. 69. -91. 16.15 443.15 1. 76.53 475.74 188. 1169. -181. 19.7 759.7 1.3 36.94 64.41 31. 1443. -71. 661.36 947.36 1.4 414.79 713.61 39. 1565. -36. 93.41 15.41 1.5 43.8 743.35 13. 1534. -453. 9.4 935.4 1.6 414.79 713.61 154. 1375. -454. 657.41 88.41 1.7 36.94 64.41 6. 16. -635. 19.36 41.36 1.8 76.53 475.74-69. 68. -75. 1311.7-41.73 1.9 155.55 67.61-3. 37. -1. 43.15-88.85... -47.. -16. -47. -9..1 155.55 67.61-31. 37. -1. 49.15-89.85. 76.53 475.74-69. 68. -75. 1311.7-41.73.3 36.94 64.41 6. 16. -635. 19.36 41.36.4 414.79 713.61 154. 1375. -454. 657.41 88.41.5 43.8 743.35 13. 1534. -453. 9.4 935.4.6 414.79 713.61 39. 1565. -36. 93.41 15.41.7 36.94 64.41 31. 1443. -71. 661.36 947.36.8 76.53 475.74 188. 1169. -181. 19.7 759.7.9 155.55 67.61 111. 69. -91. 16.15 443.15 3........ fullq
LRFD sla negative moment.mcd 7/1/3 7 of 3 Service III loads (moment) SIII1 SIII SIII3 SIII4 SIII5 SIII6 SIII7 := SIII1 SIII SIII3 SIII4 SIII5 SIII5 SIII7 DC LOADS (non-comp) DW Loads LL + I TOTAL LOADS LOCATION self wt other (sla) (comp) M (+) M (-) M (+) M (-) 1........ 1.1 155.55 67.61 111. 553.6-7.8 187.75 461.35 1. 76.53 475.74 188. 935. -144.8 1875.47 795.47 1.3 36.94 64.41 31. 1154.4-16.8 37.76 11.56 1.4 414.79 713.61 39. 15. -89.6 619.41 177.81 1.5 43.8 743.35 13. 17. -36.4 615.6 16. 1.6 414.79 713.61 154. 11. -363. 38.41 919.1 1.7 36.94 64.41 6. 849.6-58. 1896.96 539.36 1.8 76.53 475.74-69. 5.4-58. 1185.67 13.7 1.9 155.55 67.61-3. 189.6-81.6 38.75-68.45... -47.. -181.6-47. -178.6.1 155.55 67.61-31. 189.6-81.6 381.75-69.45. 76.53 475.74-69. 5.4-58. 1185.67 13.7.3 36.94 64.41 6. 849.6-58. 1896.96 539.36.4 414.79 713.61 154. 11. -363. 38.41 919.1.5 43.8 743.35 13. 17. -36.4 615.6 16..6 414.79 713.61 39. 15. -89.6 619.41 177.81.7 36.94 64.41 31. 1154.4-16.8 37.76 11.56.8 76.53 475.74 188. 935. -144.8 1875.47 795.47.9 155.55 67.61 111. 553.6-7.8 187.75 461.35 3........ full
LRFD sla negative moment.mcd 7/1/3 8 of 3 Strength I loads (moment) Maximum 1.5*DW + 1.5*DW + 1.75*(LL + IM) Minimum.9*DC +.65*DW + 1.75*(LL + IM) The loads shown in the DL columns reflect the values from Service I. The appropriate load comination (max or min) is shown in the total loads columns. The minimum load factors for dead load are used when dead load and future wearing survace stresses are of opposite sign to that of the live load. STI1 STI STI3 STI4 STI5 STI6 STI7 := STI1 STI STI3 STI4 STI5 STI6 STI7 DC LOADS (non-comp) DW Loads LL + I TOTAL LOADS LOCATION self wt other (sla) (comp) M (+) M (-) M (+) M (-) 1........ 1.1 155.55 67.61 111. 69. -91. 196.44 93.74 1. 76.53 475.74 188. 1169. -181. 368.9 48.49 1.3 36.94 64.41 31. 1443. -71. 415.95 564.5 1.4 414.79 713.61 39. 1565. -36. 457.76 537.4 1.5 43.8 743.35 13. 1534. -453. 4473.8 43.58 1.6 414.79 713.61 154. 1375. -454. 447.76 31.17 1.7 36.94 64.41 6. 16. -635. 318.7-183.63 1.8 76.53 475.74-69. 68. -75. 1994.49-695.1 1.9 155.55 67.61-3. 37. -1. 794.19-1717.66... -47.. -16.. -3444..1 155.55 67.61-31. 37. -1. 793.54-1719.16. 76.53 475.74-69. 68. -75. 1994.49-695.1.3 36.94 64.41 6. 16. -635. 318.7-183.63.4 414.79 713.61 154. 1375. -454. 447.76 31.17.5 43.8 743.35 13. 1534. -453. 4473.8 43.58.6 414.79 713.61 39. 1565. -36. 457.76 537.4.7 36.94 64.41 31. 1443. -71. 415.95 564.5.8 76.53 475.74 188. 1169. -181. 368.9 48.49.9 155.55 67.61 111. 69. -91. 196.44 93.74 3........ full
LRFD sla negative moment.mcd 7/1/3 9 of 3 LRFD 5.7.3.3. minimum reinforcement Unless specified, at any section of a flexural component, the amount of prestressed and nonprestressed tensile reinforcement shall e adequate to develop a factored flexural resistance, Mr, at least equal to the lesser of 1.* the cracking strength 1.33 times the factored moment required y Strength I What I shall do is calculate required steel ased on the cracking moment. Then I shall calculate the required steel ased on strength. If any of the required are less than 1.*Mcr I shall increase the required y 4/3. LRFD 5.4..6 the modulus of rupture (ksi) = fr :=.4 fc fr =.48 Cracking moment (k*ft) = Mcr := fr Stc 1 Mcr = 115.956 Effective section depth (in) = d := h + ts d = 49.5 assume steel in center of sla Acr := As := Ar := Art := Required steel ased on 1.*Mcr (in^) = Acr := root 1 Mcr.9 Acr fy d Acr fy 1.7 fc, Acr Acr = 5.756 CALCULATIONS BASED ON RECTANGULAR SECTION Required steel (pure strength)(in^) = Ar ns := j ns root 1 STI7 ns ( ) if STI7 ns.9 Ar fy d Ar fy 1.7 fc, Ar Ar mp = 18.53 j ns Required Steed (in^) = As1 ns := if STI7 ns As1 mp = 18.53 4 3 Ar ns if Ar ns < Acr Ar ns Distance etween the N.A. and the compression flange (in) = c1 ns := As1 ns fy.85 fc β1 c1 mp = 17.5
LRFD sla negative moment.mcd 7/1/3 1 of 3 CALCULATIONS BASED ON "T" SECTION hf := tf Required steel for "T" section Art ns := root 1 STI7 ns.9 Art fy d Art fy β1 fc ( w) hf 1.7 fc.85 fc ( w) β1 hf Art fy β1 fc ( w) hf 1.7 fc hf, Art Apply the requirements of 1.*Mcr to "T" section ehavior Required Steed (in^) = As ns := if STI7 ns 4 3 Art ns if Art ns < Acr Art ns As mp = 16.955 Check rectangular section ehavior Thickness of ottom flange (compression flange) (in) = tf = 7 Check ck1 ns := "R" if tf > c1 ns "T" Actual steel to use (in^) = As ns := As1 ns if ck1 ns = "R" As ns disp ns, := sp ns disp ns, 1 := x1 ns disp ns, := STI7 ns disp ns, 3 := Ar ns disp ns, 4 := As1 ns disp ns, 5 := c1 ns disp ns, 6 := ck1 ns disp ns, 7 := Art ns disp ns, 8 := As ns disp ns, 9 := As ns
LRFD sla negative moment.mcd 7/1/3 11 of 3 column = span point column 1 = range column = Strength I Negative moment column 3 = required steel for negative moment ased on rectangular section column 4 = Required steel using 1.*Mcr column 5 = "c" value ased on rectangular section column 6 = check rectangular section "R" denotes rectangular "T" denotes "T" section column 7 = required steel ased on "T" section non considering 1.*Mcr column 8 = required steel ased on "T" section and using 1.*Mcr column 9 = actual steel to use disp = 1 3 4 5 6 7 8 9 1 11 1 13 14 15 16 17 18 19 1 3 4 5 6 7 8 9 1 "R".617 1.1 7.7 93.737 "R" 1.851 1. 15.4 48.494 "R".656 1.3 3.1 564.51 "R" 3.9 1.4 3.8 537.417 "R".89 1.5 38.5 43.58 "R".318 1.6 46. 31.167 "R" 1.967 1.7 53.9-183.69.834 1.11 1.5 "R" 1.385 1.847 1.11 1.8 61.6-695.6 3.1 4.95 4.53 "R" 3.576 4.768 4.95 1.9 69.3-1717.663 8.313 8.313 7.845 "T" 8.197 8.197 8.197-3444 18.53 18.53 17.5 "T" 16.955 16.955 16.955.1 7.7-1719.163 8.31 8.31 7.85 "T" 8.4 8.4 8.4. 15.4-695.6 3.1 4.95 4.53 "R" 3.576 4.768 4.95.3 3.1-183.69.834 1.11 1.5 "R" 1.385 1.847 1.11.4 3.8 31.167 "R" 1.967.5 38.5 43.58 "R".318.6 46. 537.417 "R".89.7 53.9 564.51 "R" 3.9.8 61.6 48.494 "R".656.9 69.3 93.737 "R" 1.851 3 "R".617 Required steel Ar ns 15 As ns Acr 1 5 1 1. 1.4 1.6 1.8..4.6.8 sp ns
LRFD sla negative moment.mcd 7/1/3 1 of 3 LRFD 5.7.3.4 CONTROL OF CRACKING BY DISTRIBUTION OF REINFORCEMENT The provisions specified herein shall apply to the reinforcement of all concrete components, except that of deck slas designed in accordance with Article 9.7., in which tension in the cross-section exceeds 8 percent of the modulus of rupture, specified in article 5.4..6, at applicale service limit state load comination specified in Tale 3.4.1-1. Since this is the main negative moment reinforcement parallel to traffic I shall apply the provisions of this article. Components shall e so proportioned that the tensile stress in the mild steel reinforcement at the service limit state, fsa does not exceed: Z fsa =.6 fy 5.7.3.4-1 1 ( dc A) 3 dc = depth of concrete measured from extreme tension fier to center of ar or wire located closest thereto; for calculation purposes, the thickness of clear coner used to compute dc shall not e taken to e greater than in. A = area fo concrete having the same centroid as the principal tensile reinforcement and ounded y the surfaces of hte cross-section and a straight line parallel to the neutral axis, divided y the numer of ars or wires; for calculation purposes the thickness of clear concrete cover used to compute A shall not e taken to e greater than. in. Z = crack width parameter - use 17 k/in for moderate exposure. Max Required numer of ars using input ar size = (formula will chose an even numer of ars) ( ) max As nmax := j ceil nmax = 18 Asone j if ceil j j = j + 1 Maximim steel (in^) = Asmax := nmax Asone Asmax = 18 Minimum numer of ars (after cutoff) = nmax nmin := nmin = 9 Minimum steel (in^) = Asmin := nmin Asone Asmin = 9
LRFD sla negative moment.mcd 7/1/3 13 of 3 SL moment to e used (use if the moment is positive) M1 ns := SI7 ns if SI7 ns < M1 mp = 9 F := Stress in steel for rectangular section SL Stress in max steel (psi) = fsmax1 ns := root 1 M1 ns Asmax F d Asmax F 1.7 fc, F fsmax1 mp = 9.67 SL Stress in min steel (psi) = fsmin1 ns := root 1 M1 ns Asmin F d Asmin F 1.7 fc, F fsmin1 mp = 59.14 Stress in steel for "T" section fsmax ns := root 1 M1 ns.9 Asmax F d Asmax F β1 fc ( w) hf 1.7 fc.85 fc ( w) β1 hf Asmax F β1 fc ( w) hf 1.7 fc hf, F fsmax mp = 3.59 fsmin ns := root 1 M1 ns.9 Asmin F d Asmin F β1 fc ( w) hf 1.7 fc.85 fc ( w) β1 hf Asmin F β1 fc ( w) hf 1.7 fc hf, F fsmin mp = 64.518 Chose values of stress to use (rectangular or "T" stress) Max stress (ksi) = fsmax ns := if M1 ns = fsmax1 ns if ck1 ns = "R" fsmax mp = 3.59 fsmax ns Minimum stress (ksi) = fsmin ns := if M1 ns = fsmin ns if ck1 ns = "R" fsmin mp = 64.518 fsmin ns
LRFD sla negative moment.mcd 7/1/3 14 of 3 Allowale Stress Cracking factor = z := 17 Total Section Height (in) = height := h + ts height = 53.5 cover d Distance from extreme tension fier to CL. reinforcing dc := min +.75 + dc = 3.314 Effective tension flange area dc ETFW Amax := Amax = 4.54 nmax dc ETFW Amin := Amin = 81.9 nmin 1 Allowale Stress fsamax ns := min 3 fsamax ( dc Amax) mp = 33. z.6 fy fsamin ns := min z ( dc Amin).6 fy 1 3 fsamin mp = 6.35 Check max steel for cracking moment ckmax ns := "fail max" if fsamax ns < fsmax ns ckmax mp = "pass" "pass" Check min steel for cracking moment ckmin ns := "fail min" if fsamin ns < fsmin ns ckmax mp = "pass" "pass" Prepare output display disp := disp ns, := sp ns disp ns, 1 := fsmax ns disp ns, 4 := fsmin ns disp ns, := fsamax ns disp ns, 5 := fsamin ns disp ns, 3 := ckmax ns disp ns, 6 := ckmin ns
LRFD sla negative moment.mcd 7/1/3 15 of 3 disp = 1 3 4 5 6 7 8 9 1 11 1 13 14 15 16 17 18 19 1 1 3 4 5 6 1 33. "pass" 6.35 "pass" 1.1 33. "pass" 6.35 "pass" 1. 33. "pass" 6.35 "pass" 1.3 33. "pass" 6.35 "pass" 1.4 33. "pass" 6.35 "pass" 1.5 33. "pass" 6.35 "pass" 1.6 33. "pass" 6.35 "pass" 1.7 33. "pass" 6.35 "pass" 1.8.566 33. "pass" 5.7 6.35 "pass" 1.9 13.575 33. "pass" 7.151 6.35 "fail min" 3.59 33. "pass" 64.518 6.35 "fail min".1 13.59 33. "pass" 7.18 6.35 "fail min"..566 33. "pass" 5.7 6.35 "pass".3 33. "pass" 6.35 "pass".4 33. "pass" 6.35 "pass".5 33. "pass" 6.35 "pass".6 33. "pass" 6.35 "pass".7 33. "pass" 6.35 "pass".8 33. "pass" 6.35 "pass".9 33. "pass" 6.35 "pass" 3 33. "pass" 6.35 "pass" column = span point column 1 = stress on max column = allowale on max column 3 = pass/fail for max column 4 = stress on min column 5 = allowale on min column 6 = pass/fail for min column = span point column 1 = stress on max column = allowale on max column 3 = pass/fail for max column 4 = stress on min column 5 = allowale on min column 6 = pass/fail for min
LRFD sla negative moment.mcd 7/1/3 16 of 3 LRFD 5.5.3. Fatigue The stress range in straight reinforcement resulting from the fatigue load comination, specified in Tale 3.4.1-1 shall not exceed: ff = 1.33 fmin + 8 r h ff = stress range (ksi) fmin = the minimum live load stress resulting from the fatigue load comination specified in Tale 3.4.1-1 r/h = ratio f ase radius to height of rolled-on trasnverse deformations; if the actual value is not known,.3 may e used Apply LLDF to the fatigue truck = fp ns := fp1 ns LLDF fp mp = Using maximum steel fn ns := fn1 ns LLDF fn mp = 4.59 Determine if section is rectangular or "T" section for max and min amounts of steel used. The previouse determination was done ased on the required steel of a rectangular section. This calculation will e ased on the actual steel used (max and min). "c" value for "T" section ct := Asmax fy.85 β1 fc ( w) hf.85 fc β1 w ct = 38.386 "c" value for rectangular cr := Asmax fy.85 fc β1 cr = 16.986 determine rectantular or "T" ck := "R" if cr tf "T" ck = "T" jd for rectangular section jdrmax := d Asmax fy.85 fc jdrmax = 34.811 jd for "T" section jdtmax := d β1 ct jdtmax = 3.936 Pick value of jd for rectangular or "T" section jdmax := jdrmax if ck = "R" jdtmax jdmax = 3.936
LRFD sla negative moment.mcd 7/1/3 17 of 3 Stress in steel for fatigue limit state fmaxp ns := fp ns 1 Asmax jdmax if fp ns < fmaxp mp = fp ns [ yts ( height d) ] η 1 Ic fmaxn ns := fn ns 1 Asmax jdmax if fn ns < fmaxn mp = 8.554 fn ns [ yts ( height d) ] η 1 Ic Stress range for max steel Min stress for Max steel ffmax ns := fmaxp ns fmaxn ns fm1 ns := min fmaxp ns fmaxn ns ffmax mp = 8.554 fm1 mp = Allowale stress range fmax ns := 1.33 fm1 ns + 8.3 fmax mp = 3.4 Check stress range frmax ns := "fail" if ffmax ns > fmax ns frmax mp = "pass" "pass" disp := disp ns, := sp ns disp ns, 1 := ck disp ns, := jdmax disp ns, 3 := fmaxp ns disp ns, 4 := fmaxn ns disp ns, 5 := ffmax ns disp ns, 6 := fmax ns disp ns, 7 := frmax ns
LRFD sla negative moment.mcd 7/1/3 18 of 3 1 3 4 5 6 7 8 9 disp = 1 11 1 13 14 15 16 17 18 19 1 1 3 4 5 6 7 1 "T" 3.936 3.4 "pass" column = span point 1.1 "T" 3.936 -.198.86 1.6 3.465 "pass" column 1 = section type 1. 1.3 1.4 1.5 1.6 "T" "T" "T" "T" "T" 3.936 3.936 3.936 3.936 3.936 -.34 -.41 -.4 -.41 -.377 1.75.57 3.43 4.77 5.139.48.971 3.854 4.677 5.516 3.57 3.53 3.539 3.53 3.55 "pass" column = "jd" value "pass" column 3 = stress for positive moment column 4 = stress for negative moment "pass" column 5 = stress range "pass" column 6 = allowale stress range "pass" column 7 = pass/fail 1.7 "T" 3.936 -.99 5.984 6.83 3.499 "pass" 1.8 "T" 3.936 -.174 6.847 7. 3.457 "pass" 1.9 "T" 3.936 -.75 7.79 7.784 3.45 "pass" "T" 3.936 8.554 8.554 3.4 "pass".1 "T" 3.936 -.75 7.79 7.784 3.45 "pass". "T" 3.936 -.174 6.847 7. 3.457 "pass".3 "T" 3.936 -.99 5.984 6.83 3.499 "pass".4 "T" 3.936 -.377 5.139 5.516 3.55 "pass".5 "T" 3.936 -.41 4.77 4.677 3.53 "pass".6 "T" 3.936 -.4 3.43 3.854 3.539 "pass".7 "T" 3.936 -.41.57.971 3.53 "pass".8 "T" 3.936 -.34 1.75.48 3.57 "pass".9 "T" 3.936 -.198.86 1.6 3.465 "pass" 3 "T" 3.936 3.4 "pass"
LRFD sla negative moment.mcd 7/1/3 19 of 3 Using minimum steel Determine if section is rectangular or "T" section for max and min amounts of steel used. The previouse determination was done ased on the required steel of a rectangular section. This calculation will e ased on the actual steel used (max and min). "c" value for "T" section ct := Asmin fy.85 β1 fc ( w) hf.85 fc β1 w ct = 11.693 "c" value for rectangular cr := Asmin fy.85 fc β1 cr = 8.493 determine rectantular or "T" ck := "R" if cr tf "T" ck = "T" jd for rectangular section jdrmin := d Asmax fy.85 fc jdrmin = 34.811 jd for "T" section jdtmin := d β1 ct jdtmin = 44.8 Pick value of jd for rectangular or "T" section jdmin := jdrmin if ck = "R" jdtmin jdmin = 44.8
LRFD sla negative moment.mcd 7/1/3 of 3 Stress in steel for fatigue limit state fminp ns := fp ns 1 Amin jdmin if fp ns < fminp mp = fp ns [ yts ( height d) ] η 1 Ic fminn ns := fn ns 1 Asmin jdmin if fn ns < fminn mp = 1.75 fn ns [ yts ( height d) ] η 1 Ic Stress range for min steel ffmin ns := fminp ns fminn ns ffmin mp = 1.75 Min stress for Min steel fm ns := min fminp ns fminn ns fm mp = Allowale stress range fmin ns := 1.33 fm1 ns + 8.3 fmin mp = 3.4 Check stress range frmin ns := "fail" if ffmin ns > fmin ns frmin mp = "pass" "pass" disp := disp ns, := sp ns disp ns, 1 := ck disp ns, := jdmin disp ns, 3 := fminp ns disp ns, 4 := fminn ns disp ns, 5 := ffmin ns disp ns, 6 := fmin ns disp ns, 7 := frmin ns
LRFD sla negative moment.mcd 7/1/3 1 of 3 1 3 4 5 6 7 8 9 disp = 1 11 1 13 14 15 16 17 18 19 1 1 3 4 5 6 7 1 "T" 44.8 3.4 "pass" column = span point 1.1 "T" 44.8 -.198 1.83 1.481 3.465 "pass" column 1 = section type 1. 1.3 1.4 1.5 1.6 "T" "T" "T" "T" "T" 44.8 44.8 44.8 44.8 44.8 -.34 -.41 -.4 -.41 -.377.565 3.83 5.15 6.36 7.645.889 4.4 5.58 6.763 8. 3.57 3.53 3.539 3.53 3.55 "pass" column = "jd" value "pass" column 3 = stress for positive moment column 4 = stress for negative moment "pass" column 5 = stress range "pass" column 6 = allowale stress range "pass" column 7 = pass/fail 1.7 "T" 44.8 -.99 8.9 9.1 3.499 "pass" 1.8 "T" 44.8 -.174 1.185 1.359 3.457 "pass" 1.9 "T" 44.8 -.75 11.468 11.543 3.45 "pass" "T" 44.8 1.75 1.75 3.4 "pass".1 "T" 44.8 -.75 11.468 11.543 3.45 "pass". "T" 44.8 -.174 1.185 1.359 3.457 "pass".3 "T" 44.8 -.99 8.9 9.1 3.499 "pass".4 "T" 44.8 -.377 7.645 8. 3.55 "pass".5 "T" 44.8 -.41 6.36 6.763 3.53 "pass".6 "T" 44.8 -.4 5.15 5.58 3.539 "pass".7 "T" 44.8 -.41 3.83 4.4 3.53 "pass".8 "T" 44.8 -.34.565.889 3.57 "pass".9 "T" 44.8 -.198 1.83 1.481 3.465 "pass" 3 "T" 44.8 3.4 "pass"
LRFD sla negative moment.mcd 7/1/3 of 3 Final steel selection ( ) ( ckmin ns "pass" ) ( ) Asf ns := Asmin if Asmin > As ns = frmin ns = "pass" ( ) ( ckmax ns "pass" ) ( ) Asmax if Asmax > As ns = frmax ns = "pass" "fail" 15 Asf ns As ns 1 5 1 1. 1.4 1.6 1.8..4.6.8 sp ns 15 d d Emedment length (in) emed := max emed = 77 span 1 1
LRFD sla negative moment.mcd 7/1/3 3 of 3 Final ar output data Required steel 15 As ns Acr Asf ns 1 5 1 1. 1.4 1.6 1.8..4.6.8 Size of ar used = size = 9 sp ns 1 3 4 5 6 7 8 9 disp = 1 11 1 13 14 15 16 17 18 19 1 1 1 9 1.1 9 1. 9 1.3 9 1.4 9 1.5 9 1.6 9 1.7 1.11 9 1.8 4.95 9 1.9 8.197 18 16.955 18.1 8.4 18. 4.95 9.3 1.11 9.4 9.5 9.6 9.7 9.8 9.9 9 3 9 column = span point column 1 = Total area of steel required column = Numer of ars per eam