Page 1 of Sven Alexander Last revised SBProduksjon STATICAL CALCULATIONS FOR BCC 250


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1 Page 1 of 18 CONTENT PART 1 BASIC ASSUMPTIONS PAGE 1.1 General 1. Standards 1.3 Loads 1. Qualities PART ANCHORAGE OF THE UNITS.1 Beam unit equilibrium 3. Beam unit anchorage in front..1 Check of capacity.. Anchorage.3 Beam unit anchorage of the horizontal force Check of capacity 6.3. Anchorage length 6. Column unit anchorage in the column 7..1 Check of capacity 7.. Anchorage of the bolts 7 PART 3  REINFORCEMENT 3.1 Beam reinforcement in the rear 7 3. Beam shear stirrups Beam check of shear compression 7 3. Column splitting stress Conclusion reinforcement in the beam 8 PART DESIGN OF STEEL UNITS.1 Equilibrium of knife 9. Design of knife.3 Beam unit transverse bending of top flange of rectangular tube 1. Beam unit transverse bending of bottom flange of rectangular tube 1.5 Column unit equilibrium 15.6 Column unit bending of plates Bottom plate Top plate 17.7 Conclusion steel units 18
2 Page of 18 PART 1 BASIC ASSUMPTIONS 1.1 GENERAL In these calculations certain assumptions has been made about dimensions and qualities in the precast concrete elements that may not always be the case. Therefore the following calculations of anchorage of the units and the resulting reinforcement must be considered as an example to illustrate the calculation model. In beams it must always be checked that the forces from the anchorage reinforcement can be transferred to the beam s main reinforcement. The recommended shear reinforcement (stirrups) includes all necessary stirrups in the beam end; i.e. the normal shear reinforcement in beam ends and an addition due to the cantilever action of the beam unit from the beam. The information found here and in the memos assumes that the design of the elements and the use of the units in structural elements are carried out under the supervision of a structural engineer with knowledge about the flow of forces. 1. STANDARDS The calculations are carried out according to: Eurocode : Design of concrete structures. Part 11. General rules and rules for buildings. Eurocode 3: Design of steel structures. Part 11: General rules and rules for buildings. Eurocode 3: Design of steel structures. Part 18: Design of joints. EN 080: Steel for the reinforcement of concrete. Weldable reinforcing steel. General. For all NPDs in the Eurocodes the recommended values are used. These NDPs are: In EC : γ c ; γ s ; α cc ; α ct ; k 1 ; k and Ø m,min. In EC 3: γ M0 ; γ M1 and γ M. 1.3 LOADS Vertical ultimate limit state load = F V = 50 kn Horizontal ultimate limit state load = F H = 0,3 F V = 75 kn The design is carried out with a horizontal force of 30 % of the vertical force. Recommended capacity for transfer of active forces (as wind loads) may be assumed to be 0 % of smallest vertical load present at the same time. 1. QUALITIES Concrete grade C5/55: f cd = α cc f ck /γ c = 1,0 5/1,5 = 30,0 MPa f ctd = α ct f ctk,0,05 /γ c = 1,0,/1,5 = 1,80 MPa f bd =,5 η 1 η f ctd =,5 1,0 1,0 1,80 =,05 MPa Reinforcement B500C: f yd = f yk /γ s = 500/1,15 = 35 MPa Tension in threaded bars: 8.8 quality steel: f yd = f y /γ M = 60/1,5 = 51 MPa Nominal diameter (mm) M M1 M16 M0 M M30 M33 M36 Equivalent diameter (mm) 8,6, 1,1 17,7 1, 6,7 9,7 3,6 Stress area (mm ) Structural steel S355: Tension and compression: f yd = f y / γ M0 = 355/1,00 = 355 MPa < f u /γ M = 5/1,5 = 08 MPa Shear: f sd = f y /(γ M0 3) = 355/(1,00 3) = 05 MPa f Welds: u f vw d = = 6 γ 3 β 1,5 3 0,9 MPa M w
3 Page 3 of 18 PART ANCHORAGE OF THE UNITS.1 BEAM UNIT  EQUILIBRIUM R CO a = l co/ Rectangular tube 160x80x5, L = 5 d d 1 l cu/ Threaded bar M16 F H F H F V a = Figure 1. Designations and assumed dimensions. R CU Neglecting the moment created by the small vertical shift in the transfer of F H. R CU = F V d 1 /d R CO = F V + R CU l CU = R CU /(f cd b) (b is the width of the beam unit) l CO / is decided by the location of the front reinforcement. Assume Ø1: l CO / = +( )/ = 78 mm d 1 = a 1 a +l CO / d = a l CO / l CU / Use a spread sheet. Assume d 1 /d, calculate resulting d 1 /d. Change assumed value until they are equal. (Spread sheet EquilibriumBCC50beamunit.exc.) Equilibrium of BCC 50 beam unit: Concrete grade = C 5 /55 Material coefficient for concrete = 1,5 Geometry: a 1 = 51 3 mm f cd = 30,0 MPa a = 5 5 mm Width of the beam unit = 80 mm F V = 50 kn l CO / = 78 mm Assumed d 1 / d = 0, R CU = 95 kn d 1 = 136 mm 3 R CO = 5 kn d = 357 mm l CU = 0 mm Calculated d 1 /d = 0,
4 Page of 18. BEAM UNIT ANCHORAGE IN FRONT..1 Check of capacity A s,reqd = R CO /f yd,reinf = 35/0,35 = 793 mm Assumed Ø1 = 113 = 90 mm ok.. Anchorage In order to achieve equilibrium in the joint where the reinforcement bringing R CO down is anchored, the same amount of reinforcement must be supplied horizontally in order to carry a 5º compression diagonal (strut and tie model). Select Ø1 Ushaped stirrups. As illustrated in the left section in figure it will be congested if the beam has minimum dimensions. This must be taken into account when the anchorage length of the horizontal U shaped bars is calculated. (The main reinforcement of the beam is not shown in the side view.) Select to check the anchorage for a situation as illustrated in the right hand section in figure. According to EC clause 8..: l bd = α 1 α α 3 α α 5 l b,reqd l b,min 00 (minimum) 00 (minimum) 35 Ø σ sd 1 0,90 l b,reqd = = = 83 mm f,05 bd l b,min = max(0,3 l b.reqd ; Ø; 0 mm) = mm Figure. Anchorage of front reinforcement. 95 Anchorage in the node: Lower bars: c = = 65 mm Upper bars: c = = 1 mm Ushaped anchorage: c d = c = 65 mm > 3 Ø = 36 mm ; i.e. α 1 = 0,7 Concrete cover: α = 10,15 (c d 3 Ø)/Ø = 10,15 (503 1)/1 = 0,85 Confinement by reinforcement: Assume transverse reinforcement in the anchorage zone to be Ø1 bars (see figure 5): 113 0,5 113 α 3 = 1 K λ = 1 0,05 = 0,813 Confinement by welded transverse reinforcement: α = 1,0 Confinement by transverse pressure: α 5 = 1,0 113 α α 3 α 5 = 0,85 0,813 1,0 = 0,671 < 0,7 i.e. use 0,7. l bd = 0,7 0,7 1,0 83 = 139 mm No contribution to the anchorage is required from the main reinforcement in the beam.
5 Page 5 of 18 Anchorage for lap splicing with the main reinforcement in the beam: Straight bar: α 1 = 1,0 Concrete cover: c d = min(a/; c 1 ; c) = min(8/; ; 5+15) = mm α = 1 0,15 (c d Ø)/Ø = 1 0,15 (1)/1 = 0,85 Confinement by reinforcement: Assume transverse reinforcement Ø1c/c 80 mm (see figure 5): 113 0,5 113 α 3 = 1 K λ = 1 0,05 80 = 0, Confinement by welded transverse reinforcement: α = 1,0 Confinement by transverse pressure: α 5 = 1,0 α α 3 α 5 = 0,85 0,881 1,0 = 0,75 > 0,7 l bd = 1,0 0,75 1,0 83 = 1 mm Minimum total horizontal length of the Ushaped bars = = 351 mm It must always be checked that the beam s main reinforcement has sufficient anchorage at the end of the horizontal part of the front anchorage. This will probably lead to much greater lengths for the horizontal part of the front anchorage than calculated here. It is recommended to make sure the anchorage lengths always are ample in order to avoid anchorage failures, as anchorage failures normally will be brittle failures. Compressive strut: According to EC, clause 6.5. and 6.5..c): = 15/50 = 0,8 σ Rd,max = k 3 f cd = 0,75 0,8 30 = 18,5 MPa F cd = R CO = 35 = 88 kn Required concrete area = 88 3 /18,5 = 6373 mm Height of compressive strut is at least equal to the anchorage length = 139 = 197 mm Required beam width = (6373/197)+80 = 1 mm width of BCC unit
6 Page 6 of 18 a End of horizontal part of front anchorage. z x x 1 Center og gravity of the beam s main reinforcement Main reinforcement in the beam Figure 3. Anchorage of the beam s main reinfocement. a FV x1 The force is approximately [ FV FH z + + ]. Additional force in the beam s top z x reinforcement is about [ F H ], but this force leads to a reduction in the compressive z stresses and may be neglected. If there is a large distance between the horizontal part of the front anchorage and the beam s main reinforcement, the length of the horizontal part of the front anchorage must be increased with this distance. (This distance is approximately zero in the figure.).3 BEAM UNIT ANCHORAGE OF HORIZONTAL FORCES.3.1 Check of capacity F H = 75 kn (see clause 1.3) A s,reqd = 75/0,51 = 16 mm (see clause 1.) M16 threaded bar (157 mm ) is assumed  ok.3. Anchorage length l bd = α 1 α α 3 α α 5 l b,reqd l b,min No information exists on the bond characteristics of threaded bars. Assume η 1 to be 0,7. 75 Ø σ sd 16 0,157 f bd =,5 0,7 1,0 1,80 =,8 MPa; l b,reqd = = = 673 mm fbd,8 l b,min = max(0,3 l b.reqd ; Ø; 0 mm) = 0 mm Straight bar: α 1 = 1,0 Concrete cover: c d = min(a/; c 1 ; c) = min(50/; 0; 0) = 5 mm α = 1 0,15 (c d Ø)/Ø = 1 0,15 (516)/16 = 0,916 Confinement by reinforcement: Assume transverse reinforcement Ø1c/c150 mm: ,5 113 α 3 = 1 K λ = 1 0, = 0, Confinement by welded transverse reinforcement: α = 1,0 Confinement by transverse pressure: α 5 = 1,0 l bd = 1,0 0,916 0,91 1,0 1,0 673 = 560 mm Recommended length 666 mm (18 pieces from a 1 m bar).
7 Page 7 of 18. COLUMN UNIT ANCHORAGE IN THE COLUMN..1 Check of capacity M = F H (15+/)+F V [(/)35] = = 50 (0,3 6+0) = 1950 knmm S = 1950/5 = 7,96 kn A s = 7,96/0,51 = 16 mm M at the top is ok (see table in clause 1.) F H S = 0, = 67 kn A s = 67/0,51 = 131 mm M16 in the bottom is ok.. Anchorage of the bolts See separate calculations: Design of anchorages used in the BCC units. See also Memo 7. PART 3 REINFORCEMENT 3.1 BEAM REINFORCEMENT IN THE REAR A s,reqd = R CU /f sd = 95/0,35 = 18 mm, corresponding to 18/( 113) = 0,96; i.e. 1 Ø 1 stirrup. 3. BEAM SHEAR STIRRUPS Use a strutandtie model with compressive diagonals at 5º. The shear force within the central part of the beam unit is assumed to be R CO = 35 kn. Assume z = 300 mm and stirrup diameter Ø1. s = A sw z f ywd cot θ/v s = cot 5º / 35 3 = 85 mm Select Ø1 c/c 80 mm. This reinforcement should be brought approximately 00 mm past the end of the beam unit in order to absorb any splitting effects from the anchorage of the threaded bar. 3.3 BEAM CHECK OF SHEAR COMPRESSION Assume beam width of 00 mm and beam depth of 50 mm. (See also clause..) This means that z min = 300 mm with one layer of reinforcement in the top of the beam and three layers in the bottom. The shear reinforcement is calculated based on 5º compressive field. The check of shear compression will be carried out under the same assumption: V Rd,max = α cw b w z υ 1 f cd /(cot θ + tan θ) V Rd,max = {1,0 (00 80) 300 0,6 [1 (5/50)] 30,0/(cot 5º+tan 5º)} 3 V ccd = 531 MPa > 35 MPa  OK 3. COLUMN SPLITTING STRESS Splitting stress under the column unit according to EN , clause 6.5.3: T = FV = 0,19 50 = 8 kn 300 A S = 8/0,35 = 1 mm Ø stirrups ( 78,5 mm ) are sufficient. 5 M threaded 80 insert M16 threaded insert Figure. Assumed dimensions. Equilibrium of the column unit.
8 Page 8 of CONCLUSION REINFORCEMENT IN THE BEAM For show the beam end reinforcement more clearly, the beam s main reinforcement is not show in the side view. Between the shown stirrups in each end of the beam a normal calculation of the shear reinforcement must be carried out. The main reinforcement in the beam must of course also be calculated. 1Ø1P. as support 7Ø1P.c/c 80 for top reinforcement Ø1P.1 1Ø1P. stirrup extra by the end of the beam unit Ø1P.3 Theraded bar M16 Ø1P.1 9Ø1P. Ø1P.3 At the end of the horizontal part of the front anchorage (P.3) it must be checked that the beam s main reinforcement has sufficient anchorage. See clause... Figure 5. Reinforcement in the beam end.
9 Page 9 of 18 PART DESIGN OF STEEL UNITS.1 EQUILIBRIUM OF KNIFE R VO 55 R l / HO o Rectangular tube 160x80x5, L = 5 d d 1 l u / h 1 = hxt = 160x5 = 150 R HU Threaded bar M16 F H R VU F V Figure 6. Equilibrium of knife. F V d 1 + F H h 1 = R VU d + R HU h 1, dvs. F V d 1 +F H 150=R VU d +R HU 150 F V + R VU = R VO ; F H = R HO + R HU For R VU is f yd reduced with about 5% to 66 MPa due to the rounding of the tip of the knife. l o = R VO / (f yd t k ) = R VO / (0,355 5) = 0,117 R VO l u = R VU / (f yd t k ) = R VU / (0,355 0,75 5) = 0,150 R VU d 1 = (5 ) l o / = 58 + l o / d = (95+75) l o / l u / = 3 l o / l u / R HU = 0,3 R VU R HO = F H R HU F V d 1 +0,3 F V 150=R VU d +0,3 R VU 150 F V (d 1 +5)=R VU (d +5) R VU = F V (d 1 +5) / (d +5) R VO = F V + F V (d 1 +5)/(d +5) = F V [1 + (d 1 +5)/(d +5)] Use a spread sheet. Assume (d 1 +5)/(d +5), calculate resulting (d 1 +5)/(d +5). Change assumed value until they are the same. (Spread sheet EquilibriumBCC50knife.exc.) F V = 50 kn F H = 75 kn Assumed (d 1 +5)/(d +5) = 0,3900 R VU = 78 kn R HU = 3 kn R VO = 38 kn R HO = 5 kn l o = 37 mm d 1 = 76 mm l u = 1 mm d = 36 mm Calculated (d 1 +5)/(d +5) = 0,3900
10 Page of 18. DESIGN OF KNIFE kn /37 = 8,86 kn/mm 117 t = /5 = 5,56 kn/mm 1 78/1 = 6,50 kn/mm 1 75 kn x Figure 7. Forces on the knife. 50 kn 0 x 5: 5 x : h = x/ = 30+1,67x h = 30+1,67x1 = 18+1,67x V = 5,56x; N = 75x/5 = 1,667x V = 50; N = 75 x ,67x M = 5,56 + x 5 M = 50(x )+75[1+(18+1,67x)/] M =,78x + 0,83x(30 + 1,67x) M = 31x 395 x 80: 80 x 117: h = 17 h = 17 V = 50; N = 75 V = 50 8,86(x 80); N = 75 51(x 80)/37 M = 50(x )+75 17/ M = 50x+13 8,86(x 80) /+(51/37)(x 80)(17/) M = 50x+13 M = 50x+13,3(x 80) +1(x 80) 117 x 3: h = 17 V = = 78 kn; N = = M = 50x+13 38(x 98,5)+51 17/ = x W = 5 h /6 =,17 h ; A = 5 h Design plastic shear resistance = V pl,rd : V pl,rd = A v (f y / 3)/γ M0 =h 5 (0,355/ 3)/1,00 = 5,1 h; For V/V pl,rd 0,5 the effect of shear on the moment resistance may be neglected. σ M = N/A + M/W
11 Page 11 of 18 x h V N M V pl,rd A V/V pl,rd W σ M (mm) (mm) (kn) (kn) (knmm) (kn) (mm ) (mm 3 ) (MPa) , , , , , , , , , , , , , , , , , , < 0,5 < ok  ok
12 Page 1 of 18.3 BEAM UNIT TRANSVERSE BENDING OF TOP FLANGE OF RECTANGULAR TUBE Room for P. stirrup in front 50 mm 0 mm (115 kn) (30 kn) 7 (39, kn) 150 (88,8 kn) a) Forces acting on the front of the beam unit p p p b) Forces on the front plate 115 = 1,533 kn/mm 75 Contact area 5x37 mm Assumed fracture lines (bending up) =31,5 kn,5,5,5,5 Assumed fracture lines (bending down) = 7,1 kn/mm 5 c) Possible fracture lines =3,067 kn/mm d) Forces on the front part of the reinforcing plate =115 kn e) Forces on the rear part of the reinforcing plate Figure 8. Reinforcing of top flange. The knife will come to rest partly against the front plate and partly against the rectangular tube (figure 8a). The front plate with a thickness of mm can absorb (38/37) = 88,6 kn of R VO. The remaining 3888,6 = 39, kn will be carried by 37 = 7 mm of the rectangular steel tube (based on l o in spread sheet in clause.1). Some of this will be transferred back to the front plate through the weld between the front plate and the reinforcing plate. Considering the upper 55 mm of the front plate as a beam, the forces on the front plate will be as shown in figure 8b. Based on the stress capacity of the smallest section of the front plate: p 1 = 0,355 = 3,55 kn/mm Contact area between the knife and the front plate: p = 5 0,355 = 88,8 kn /35 =,536 kn/mm
13 Page 13 of 18 Remaining capacity of the front plate: p 3 =, ,55 5 = 88,8 kn / = 1,68 kn/mm Moment at the center line of the beam : M =, ,55 1,68 = 1831kNmm W req d = 1831/0,355 = 5157 mm 3 W el = 55 /6 = 50 mm 3 W pl = 55 / = 7563 mm 3 OK Required weld size between front plate and reinforcing plate = 1,68/0,6 =,8 mm Select 5 mm throat value. The forces must be transferred from the front plate to the tube through welds on the vertical sides of the tube: Required throat value = 88,8 3 /[6 (160)] =,3 mm Select mm throat value. The slight moment created in the front plate by the distance between the welds and the center of gravity of the beam s support reactions are taken by the front plate in diaphragm action. Conservatively assume that only 150 kn is carried by the front plate, instead of p 1 +p 3 = 3, ,8 = 177,5 kn, as calculated here. The remaining load, R VO 150 = = 178 kn, must be carried by the front part of the reinforcing plate. As a simplified model it is assumed that this is carried by 50 mm of the reinforcing plate as a oneway slab supported on the tube walls. Possible fracture lines are shown in figure 8c. This force is counteracted by the uniformly distributed reaction from the concrete on the reinforcing plate. Since only 1/3 of the reinforcing plate is considered, the force must be 1/3 of R CO = 35/3 = 115 kn. Forces are shown in figure 8d. Maximum moment: M = 31,5 7,1 + 1,533 = 13kNmm W req d = 13/0,355 = 797 mm 3 W el = 50 (0 +5 )/6 = 35 mm 3 W pl = 50 (0 +5 )/ = 5313 mm 3 OK Weld between reinforcing plate and tube = 31,5 3 /(6 50) =, mm Select mm throat value.
14 Page 1 of 18 The rear of the reinforcing plate must carry the compressive stress caused by R CO. If the grouting of the tube is neglected (see figure 8e): Maximum moment: M = 115 3,067 = 156kNmm W req d = 156/0,355 = 6077 mm 3 W el = 0 (0 +5 )/6 = 8 mm 3 W pl = 0 (0 +5 )/ = 66 mm 3 OK The welds are executed with reserve capacity to compensate for possible deviating force distribution.. BEAM UNIT TRANSVERSE BENDING OF BOTTOM FLANGE OF RECTANGULAR TUBE 5 78 = = 39 kn = 3,1 kn/mm 5 78 = = 39 kn M = 39 75/ 3,1 (5/) / M = 119 knmm Req d. W = 119/0,355 = 333 mm 3 Try 0 mm width of plate (in the longitudinal direction of the beam unit): 0 (5 +t u )/6 = 333 t u = 13, mm Select t u = 15 mm t U Figure 9. Reinforcing of bottom flange. The weld: Req d. a = 39 3 /(6 0) = 1,5 mm Select weld with throat measure a = mm.
15 Page 15 of 18.5 COLUMN UNIT  EQUILIBRIUM 5 FH F V 35 5 F H Assume bottom plate to be mm. A c0 = = 900 mm Partially loaded area: Assume smallest relevant column dimension to be mm. A c1 = = mm f ' cd = 30,0 900 Assume top plate to be 80 mm. Assume concrete compressive stress to be the same on bottom plate and cantilevering part of top plate: Area of bottom plate = = 900 mm Area of top plate = = 350 mm = 750 mm Concrete compressive stress = 50/7,5 = 3,5 MPa Figure. Assumed geometry of column unit A c0 A c1 3x= A c0 A c1 Figure 11. Load distribution in column. = 51,9 < 3,0 30,0 = 90 MPa 5 65 Center of gravity y (65 + ) y = = 38 mm Bottom plate carries 900 3,5 = 169 kn Top plate carries 350 3,5 = 81 kn Figure 1. Cantilevering part of top plate.
16 Page 16 of 18.6 COLUMN UNIT BENDING OF PLATES.6.1 Bottom plate 6 63 cg Center of gravity s a cg = 63 /( 63+6) cg = 3 mm F V 8 cg=3 P 81 kn F W cg = 0 F V = 50 kn P s,max y = 38 Figure 13. Equilibrium of bottom plate. / = kn 35 Equilibrium: 81 (38 35) = P {[8 (3/3)]+[35 (0/3)]} P = 5,8 kn 0 σ s,max (1/) = 5,8 3 σ s,max = 36 MPa << 355 MPa OK F w,ed,max = 36 = 15 N/mm a = 15/6 = 0,6 mm; Select a = mm 8 1,5 5,5 Design bottom plate for transverse bending for the average compressive stress (see figure 13). Consider a section 1 mm wide: σ s = σ s,max / = 36/ = 18 MPa Force in the weld in the section = 18 = 7 N 7 N x t 7 N 50/(5x5)= N/mm 16 0 x 8: M = M 1 = 3,5 x / 8 x,5: M = M 1 +7 (x 8) = M 1 +M,5 x 35: M = M 1 +M (x,5)/ = M 1 +M M 3 W = M/355 = t /6; dvs. t = (M/59,17) 3,5 N/mm Figure 1. Forces on the bottom plate. Use a spread sheet: BCC50colunitbotpl.exc. x (mm) M 1 (Nmm) M (Nmm) M 3 (Nmm) M (Nmm) t (mm) 0, , , , , , , , Select t = 15 mm
17 Page 17 of Top plate M = 3,5 13 / = 915 Nmm W reqd = 915/355 = 8,1 mm 3 t = (8,1 6) = 7,0 mm Select t = mm The weld: F w = 3,5 15 = 518 N/mm t 3,5 N/mm 3 67 a = 518/6 =,0 mm Select a = mm t 3,5 N/mm Figure 15. Forces on the top plate.
18 Page 18 of 18.7 CONCLUSION STEEL UNITS t = 5 50 KNIFE Plate 150xx0 0 Rectangular tube 160x80x5, L = 5 18 Plate 0x0x Threaded insert M16 xx60 Plate 0xx BEAM UNIT Bolt M Threaded insert M 18x18x50 80 The length of the bolts depends upon column dimensions. Bolt M Ø Threaded insert M16 xx60 Figure 16. Steel units. COLUMN UNIT
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