User Guide for FB-Deep F.C. Townsend
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2 User Guide or FB-Deep F.C. Townsend Intro: FB-Deep (ormerly SHAFT-SPT) is a combination o SHAFT 98, which uses the FHWA O Neill & Reese method or drilled shats and intermediate geomaterials, and SPT97, which is the FDOT version or driven piles. The program is designed such that the SPT boring log only is entered once, and either shats or piles can be analyzed. SPT97 Example SPT 97 Uniorm Sand (N=15) Example Depth N ST Sand ST= t N = 15 blows t x t psc Opening screen Units = English, Type = Pile Unit wt. = 150 pc, Section = Square (psc) Click Insert Pile Single, L=0t, W= Click boring log icon in Toolbar
3 Boring Log Screen Click Insert Layer Depths = 0 to 50 t Soil Type = 3 (sand) SPT N = 15 blows Click OK Note: Last entry doesn t = 0 As or old SPT97 Main Screen Click Cap. Report Results
4 SPT 97 Uniorm Sand (N=15) Example Depth N ST Sand ST= t N = 15 blows t x t F. Bearing Capacity 8 B above = 0 8() = t 3.5 B below = () = 7t Elev N q t (ts) or sand (ST = 3) q t = 3.N/3 = 3.(15)/3=16ts Ave[AQPTA] = (16)/ = 16ts Ave = (16 +16)/=16ts Ave [AQPTB] =(16)/ = 16ts Check depth o embedment to see i correction required: D c = 9B = 9()=18<D a =0; NO corection required. Maximum BC = 16ts (x) = 6 T Corrected CMAXB( no correction) = 6 T E. Skin Friction Cpacity A. Skin riction above bearing layer = 0 (air) B. Untimate skin riction in bearing layer (note: elva = 0 N=15, uniorm distribution.) 0t s = 0.019N = 0.019(15) = 0.85 ts 0t C. Ult. SFBL = 0.85(x)(0t) = 91. T Since uniorm no corections required. G. Pile Capacity Estimated Davisson (USF + CMAXB) = 91. T +6 T = 155. T (310 k ) Allowable = Davisson = 155. / = 77.6 T Ultimate (or driving) USF + 3(CMAXB) = 91. T + 3(6 T ) = 83.6 T Davisson Oset (x) = B/10 = /10 = 0.35 k PL 310. (0x1) = = 0.058" AE (15ksi) δ = = 310 k
5 To design a range o pile Lengths and widths, Click Insert Range Range 0 to 0 t long in increments o 5t. Click Graph cap. Graphics screen SHAFT98 Example CLAYS: Example File: Clay1.dat 1. Multi Layer Clay with Casing Casing 0 t 6 t Clay N =8, γ = 100 pc, q c = 16 ts 0 t 3 t Clay N =1, γ = 100 pc, q c = 30 ts
6 Opening Screen:Drilled Shat Units = English Water Table = deep R% = 0.0 Insert Shat Case = 6t, L = 0t, Diam =36, Bell L = 0.0t Bell Diam = 36 Click Boring Log Icon In upper toolbar to get Boring log. Boring Log Screen Ground Surace = 0.0t C u Calcs = CPT (Lower Let) Click Insert Layer Depth ST γ CPT 0 1(clay) 100pc 16ts Rock Friction Not included Click OK to return to main menu
7 Click Cap. Report For results Results Settlement For 0.3, R % = Click Cap. Report R Results show Cap = 0.3
8 APPENDIX A - Examples CLAYS: Example File: Clay1.dat. Multi Layer Clay with Casing 3. Multi Layer Clay with Casing B > 75 0 t 6 t Clay Casing N =8, = 100 pc, q c = 16 ts 0 t Clay N =1, = 100 pc, q c = 30 ts 3 t q σ c = c * *100 Clay Layer # 1 : c = =, ps ( ts ) * *100 Clay Layer # : c = = 3,800 ps (1.90 ts ) 15 30* 000 0*100 Clay Layer # Tip c = = 3,733.3 ps (1.867 ts ) Multi Layer Clay with Casing: Full Capacity (0 t Shat) a) Skin Friction: π * 3.0 * (0 6 )(0.55*1.033) + (0 3 )(0.55*1.9) = 9.8*[ ] =. Tons Q s = [ ] b) End Bearing: Q b = q b π b.,
9 0 q b = N ccu, N c = 6.0 * = > 9 ( use 9) 3 π 3 Q b = (9*1.867ts ). = Tons c) Total Capacity = Skin Friction + End Bearing = = Tons (ultimate) d) Calculation o Skin Friction: *0.55 = ts *0.55 =1.05 ts 35 0 e) Settlement: S = (i) 0.3 and (ii) S = 3.0 (i) Q s =. T, Q b = T, Q T = T S *100 B = 0.3* = > 0.7, q st = ( ) *. = 3.70 T q Q br b = 1.183E 0*(0.833) = 0.935* = T Q = = T E 03(0.833) +.9E 0(0.833) (0.833) (
10 For range o shat lengths Click Insert Range Min L = 0t Max L =0t Increment = t. Click Cap. Graph Graphical Results Example: Sand overlying Rock (IGM) Socket 5' 0' Diam = 36-in Sand N=15 ST=3 γ = 100pc Rock q u = 10ts q t = 1.0ts q b = 0.5q u E m = 115q u Depth N ST q u q t ' ' ' ' ' ' ' ' '+ 10ts 1.0ts 5 10ts 1.0ts 50 10ts 1.0ts 55' 10ts 1.0ts 60' 10ts 1.0ts 65 10ts 1.0ts
11 Opening Screen: Units = English Type = Drilled Shat γ c = 130pc, slump=6.89 E c = 57,000(5000psi) 1/ =030 Click Insert Shat L = 5t, D=36 Click Boring Log Icon In top toolbar Enter Boring Log Data Depth ST N γ q u q t Check Use Deault values Depth q b E m RQD Socket 0 5 deault deault q b is not deault, but ½ q u E m is 115 q u x /1 RQD = 1.0, no reduction Socket = 0 = smooth sides McVay s Friction 1 = q u qt Click OK
12 Set R% = 0.1 Click Cap. Report Results Case 5) Shat in rock under sand with smooth socket (Example ile: 5shatsandsmoothrocksocket.spc) 5' 0' End Bearing User deined q b = ½q u = ½ (10ts) = 5ts 36 diameter γ c = 130pc E c = 030ksi Depth N ST q u q t ' ' 15 3 Sand N=15 ST=3 γ = 100pc Rock q u = 10ts q t = 1.0ts q b = 0.5q u E m = 115q u 15' ' ' ' ' ' '+ 10ts 1.0ts 5 10ts 1.0ts 50 10ts 1.0ts 55' 10ts 1.0ts 60' 10ts 1.0ts 65 10ts 1.0ts
13 Total End Bearing 3 t QT = qt * AB = 5.0ts * π * = tons Skin Friction For soil type 3, sand, skin riction is: z = 0 to 5 t 5 Qs = π 3 (1.)(100) zdz Qs = π 0 ()( 3 10) 5 0 z Qs = lbs = tons z = 5 to 0 t 0 Qs = π 3 ( z )(100) zdz Qs z z = π dz ( 3) 150 z 13.5 Qs = π z 5 5 Qs = lbs = lbs = tons Skin riction in sand layer = 7.069tons tons = tons For soil type skin riction based on McVay et al. (199): 1 1 SU = qu qt = 10ts 1.0ts = ts
14 QS = SU * π * D * L = 1.581ts * π *3 t *5 t = tons Total skin riction = tons tons =38.18tons Total Shat Capacity Shat Capacity 35.33tons tons = 17.91tons Settlement FHWA IGM Calculations: (Note: Must enter values or E c, slump, and E m ) E m = 115 q u = 115 (10.0ts) = 1150ts Ω = 1.1 L D 1/ Ω = Γ = L D 1/ 1 E log( E c m ) ts 1150ts ( ) 1/ 1/ 0.05( ) log( ) 0. = L D 1/ Γ = L D 1/ 1 E log( E c m ) ts 1150ts ( ) 1/ 1/ 0.15( ) log( ) = θ EmΩ 1 = ; su = w π L Γ su q u q t θ ts * = = t w π *5 t *0.508*(1.581ts ) Λ = E m L ( ) 00 D L ( D + 1) L L [ Ω] [ 1 + ] D π L Γ D 0.67 Λ = (1150ts ).667 [ ] [ ] 00 + π *5 t * = 71.35ts t -0.67
15 qu 10.0ts Determine n or deormation criteria (igure) = = σ 1.07ts Em ; σ n For a slump = 7in, M ( table) = 0.65 ; σ = 0.65*130 pc *.5 t = 3591 ps = 1.796ts n Em σ n σ = M γ Z n c c Since 1150ts = = n ts Select values o w or calculating Z c p 5 = 0 + =.5 t Q Q t t = π D = π D L θ L k su su πd + πd + q q b b or or θ < n θ > n ; q b = Λ w 0.67 Let w = 0.036in = t ; θ / w = t -1, θ = t -1 * 0.003t = 0.75 < n = Q t π * = π *3 t *5 t *0.75 *(1.581ts ) + = 0.51tons tons ( 3 t) *71.35ts * t 0.67 * ( t) 0.67 = tons Sand Layer Above Load Corresponding to R = 0.1% carried in skin riction: For R S S max S 3 =.16R + 6.3R 7.36R +. 15R S max =.16 3 ( 0.1) + 6.3( 0.1) 7.36( 0.1) +.15( 0.1) = Q S = (0.375) tons = tons
16 Total load at R= tons tons = tons Let w = 0.16 in = t. ; θ / w = t -1, θ = t -1 * 0.018t = 1.65 > n = ( θ n)(1 n) k = n + ( θ n + 1) Q t = 6.091tons tons ( )( ) = = (1.65 (0.655) + 1) π * = π *3 t *5 t *0.833 *(1.581ts ) + ( 3 t) *71.35ts * t 0.67 * ( t) 0.67 = 96.7 tons Sand Layer Above Load Corresponding to R = 0.6% carried in skin riction: For R S S max =.16 S 3 =.16R + 6.3R 7.36R +. 15R S max 3 ( 0.6) + 6.3( 0.6) 7.36( 0.6) +.15( 0.6) = Q S = (0.999) tons = 86.1tons Total load at R= tons tons = tons Let w = in = t. ; θ / w = t -1, θ = t -1 * 0.030t =.75 > n = ( θ n)(1 n) ( )( ) k = n + = = ( θ n + 1) (.75 (0.655) + 1) Q t = 66.96tons tons ( 3 ) π * = π *3 t *5 t *0.8988*(1.581ts ) + t *71.35ts * t 0.67 *0.030 t 0.67 = tons
17 Sand Layer Above Load Corresponding to R = 1.0% carried in skin riction: For R S S max = S 3 =.16R + 6.3R 7.36R +. 15R S max 3 ( ) + 6.3( 1.0) 7.36( 1.0) +.15( 1.0) = Q S = (0.9781) tons = tons Total load at R= tons tons = tons
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