BSF - DESIGN OF REINFORCEMENT, CANTILEVERED BEAM- BEAM

Transcription

1 MEMO 55 BSF - DESIGN OF REINFORCEMENT, CANTILEVERED BEAM-BEAM Dato: Siste rev.: Dok. nr.: K4-10/55E Sign.: Sign.: Control: sss sss ps DESIGN BSF - DESIGN OF REINFORCEMENT, CANTILEVERED BEAM- BEAM CONTENT BSF - DESIGN OF REINFORCEMENT, CANTILEVERED BEAM-BEAM... 1 PART 1 BASIC ASSUMPTIONS GENERAL STANDARDS QUALITIES DIMENSIONS AND CROSS-SECTION PARAMETERS LOADS TOLERANCES... 7 PART PRINCIPAL DESIGN OF REINFORCEMENT - BSF BEAM BOX BEAM BOX - EQUILIBRIUM BEAM BOX ANCORING REINFORCEMENT BEAM BOX- HORIZONTAL ANCHORING EVALUATION OF REINFORCEMENT IN THE END OF THE CONCRETE BEAM STRUT AND TIE MODEL REINFORCEMENT IN TOP OF BEAM BOND AND ANCHORING SHEAR STIRRUPS IN BEAM END SHEAR COMPRESSION IN BEAM END HORIZONTAL BARS IN BEAM END PART 3 BSF BEAM BOX ANCHORING REINFORCEMENT BEAM BOX HORIZONTAL ANCHORING EXAMPLE REINFORCEMENT IN BEAM END... 0 Page 1 of 41

2 3.3.1 REINFORCEMENT IN TOP OF BEAM BOND AND ANCHORING SHEAR STIRRUPS IN BEAM END SHEAR COMPRESSION IN BEAM END HORIZONTAL BARS IN BEAM END... PART 4 - BSF BEAM BOX ANCHORING REINFORCEMENT BEAM BOX HORIZONTAL ANCHORING EXAMPLE REINFORCEMENT IN BEAM END REINFORCEMENT IN TOP OF BEAM BOND AND ANCHORAGE SHEAR STIRRUPS IN BEAM END SHEAR COMPRESSION IN BEAM END HORIZONTAL BARS IN BEAM END... 8 PART 5 BSF BEAM BOX ANCHORING REINFORCEMENT BEAM BOX HORIZONTAL ANCHORING EXAMPLE REINFORCEMENT IN BEAM END REINFORCEMENT IN TOP OF BEAM BOND AND ANCHORAGE SHEAR STIRRUPS IN BEAM END SHEAR COMPRESSION IN BEAM END HORIZONTAL BARS IN BEAM END PART 6 - BSF BEAM BOX ANCHORING REINFORCEMENT BEAM BOX HORIZONTAL ANCHORING EXAMPLE REINFORCEMENT IN BEAM END REINFORCEMENT IN TOP OF BEAM BOND AND ANCHORAGE SHEAR STIRRUPS IN BEAM END SHEAR COMPRESSION IN BEAM END HORIZONTAL BARS IN BEAM END Page of 41

3 PART 1 BASIC ASSUMPTIONS 1.1 GENERAL This memo deals with BSF used as beam-beam connections for continuous beams. Standard BSF-units and beam boxes are used. Reinforcement in the beam with the BSF knife is found in Memo 51. Therefore, only reinforcement related to the BSF beam-box is discussed. As the cross sections of the two connected beams will vary, there may be issues with the local force transfer in the end of the beam that is not covered by the examples given in this Memo. Therefore, the following calculations of anchorage of the units and the resulting reinforcement must be considered as an example to illustrate the calculation model. The EC- shall always be applied as the governing design document for the beam reinforcement. 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 both the relevant standards, and the structural behaviour of concrete and steel structures. 1. STANDARDS The calculations are in accordance with: Eurocode : Design of concrete structures. Part 1-1: General rules and rules for buildings. Eurocode 3: Design of steel structures. Part 1-1: General rules and rules for buildings. Eurocode 3: Design of steel structures. Part 1-8: Design of joints. The selected values for the NDP s in the following calculations are: Parameter γ c γ s α cc α ct Value 1,5 1,15 0,85 0,85 Table 1: NDP-s in EC. Parameter γ M0 γ M1 γ M Value 1,1 1,1 1,5 Table : NDP-s in EC3. Page 3 of 41

4 1.3 QUALITIES Concrete C35/45: f ck 35,0 MPa EC, Table 3.1 f cd α cc f ck/γ c 0,85 35/1,5 19,8 MPa EC, Clause 3.15 f ctd α ct f ctk,0,05/γ c 0,85,/1,5 1,4 MPa EC, Clause 3.16 f bd,5 η1 η f ctd,5 1,0 1,0 1,4,79 MPa EC, Clause 8.4. Note: For simplicity, good bond conditions are assumed when calculating f bd. This assumption may not be correct in all situations and has to be evaluated in each case. EC indicates poor bond conditions for anchoring in top of the beam. Reinforcement 500C (EN , Annex C): f f yk/γ s 500/1, MPa EC, Clause 3.. Note: Reinforcement steel of different qualities may be chosen provided that the calculations take into account the actual yield strength (f y 500 MPa) and that the bendability is sufficient for fitting the vertical suspension reinforcement to the half round steel. Steel Sxxx (EN 1005-): Steel S355: Tension: f f y/ γ M0 355/1,1 3 MPa Compression: f f y/ γ M0 355/1,1 3 MPa Shear: f sd f y/(γ M0 3) 355/(1,1 3) 186 MPa Weld S355: f u f w, d 6MPa γ M 3 β w 1,5 3 0,9 Threaded bars/nut: 8.8 quality steel: f 0,9 f u/ γ M 0,9 800/1,5 576 MPa 1.4 DIMENSIONS AND CROSS-SECTION PARAMETERS UNIT HALF ROUND STEEL HORIZONTAL ANCHORING 1) INTERNAL OPENING BEAM BOX D [mm] L [mm] Steel grade (WIDTH HEIGHT DEPTH) BSF5 Ø S355 M1, 8.8+ nut, L650mm 30mm 15mm 80mm BEAM BOX & st.pl , S355 BSF300 Ø S355 M1, 8.8+ nut, L650mm 30mm 55mm 80mm BEAM BOX & st.pl , S355 BSF450 Ø S355 1 M0, 8.8+ nut, L750mm 40mm 70mm 9,5mm BEAM BOX & st.pl , S355 BSF700 BEAM BOX Ø S355 M0, 8.8+ nut, L750mm & st.pl , S355 50mm 310mm 105mm Table 3: Dimensions BSF beam box. 1) See also Table 4. Note: The steel plate anchoring both the M0 bars for the BSF700 is designed only for the actual design force of 10kN, not the tensile capacity of two M0 bars. Page 4 of 41

5 NOMINAL DIAMETER M1 M16 M0 Equivalent diameter: Ø eq [mm] Stress area: A s [mm ] Tensile capacity (8.8): F cap f A s [kn] With across flats: NV [mm] Required dim. of square steel plate anchoring F cap: 1 b req [F cap/f cd+π Ø nom/4] 0.5 [mm] Select b b Net area for compression anchorage: A neta steel plate-π Ø nom/4 [mm ] Concrete stress: σ cf cap/a net [MPa] Required thickness of steel plate, S355: 1 a( 0.5 b-nv)/ -> t 1 a (σ c/f ) 0,5 [mm] cb/-nv/ -> t 3 0,5 c (σ c/f ) 0,5 [mm] t>[t 1,t ] Standard height of nut: (H) [mm] Required thread length in blind holes: S355 10,4 14,1 17, ,4 Select a5,9 c15,5 Table 4: Dimensions - threaded bars and anchoring steel plates. 69 Select Select ,1 19,1 18,1 t 16,5 t 6,7 a37,5 c3 t 19,1 t 9,7 a48,6 c30 t 111,5 t 1,3 Select t8mm Select t10mm Select t1mm 10,0 13,0 16,0 18mm 4mm 30mm 1 An illustration, and background for the formulas, can be found in the Memo BSF-Design of steel units. The listed dimensions are based on the concrete quality and parameters given in Section 1. and Section1.3. Note: The steel plate anchoring both the M0 bars for the BSF700 is designed only for the actual design force of 10kN, not the tensile capacity of two M0 bars. Page 5 of 41

6 1.5 LOADS Vertical ultimate limit state load: F V According to Table 5. Horizontal ultimate limit state load - in axial direction: F H0kN (see notes below) Horizontal ultimate limit state load - in transverse direction: F T0kN *Note on loads: The BSF beam box is a product designed to transfer primarily vertical load. Significant horizontal loading on the unit may also occur if imposed deformation (shrinkage, temperature differences etc.) in the pre-cast element is resisted. When the occurring horizontal force exceeds the potential friction force the knife will slide and the force will be partly relieved. The static friction factor steel-steel at support is assumed to be within the range (0,-0,5). The maximum friction force due to gradually increasing imposed deformations will however be associated with vertical service loads. The steel parts of the unit, and anchoring of these parts into the concrete are designed for the following unfavourable load combination: Vertical force 1,0F v + Horizontal force 0,3F v In some cases transfer of static global horizontal load via the unit may be requested. The magnitude of this force would be limited by the minimum friction factor at the support and vertical load present at the same time. This will imply uncertainty in resistance, and it s recommended to transfer the horizontal forces by proper reinforcement through the joint. In case of dynamic loads, the horizontal resistance should always be assumed to be zero. Horizontal anchoring of the steel parts assumes minimum concrete grade C35 in column and beam. UNIT VERTICAL ULTIMATE LIMIT STATE LOAD F v [kn] VERT. 1,0F v [kn] LOAD BEAM BOX HOR. 0,3F v [kn] BSF ,5 BSF BSF BSF Table 5: Design loads Page 6 of 41

7 1.6 TOLERANCES The design nominal gap between two beams is 0mm, with a tolerance of ±10mm.The tolerances are handled with the cantilevering of the knife from the beam. If the gap is 30mm, the knife is pushed out an extra 10mm and vice versa if the gap is only 10mm. Thus, the load point in the beam box will always be the same. The knife shall always be pushed out until it bottoms against the back of the beam box. The tolerance on location of the reinforcement for the beam box is ±mm. Figure 1: Tolerances. (cog center of gravity) Page 7 of 41

8 PART PRINCIPAL DESIGN OF REINFORCEMENT - BSF BEAM BOX.1 BEAM BOX - EQUILIBRIUM Figure : Equilibrium. The assumed flow of forces is: Vertical force: Suspension reinforcement designed for the load is to be placed at the load point R VF V. Horizontal force. Anchored with threaded bars. R HF H. The bending moment associated with the small vertical shift in the horizontal force is neglected. Page 8 of 41

9 . BEAM BOX ANCORING REINFORCEMENT 1) Required cross section for suspension reinforcement: FV A s f ) Mandrel diameter Bending of reinforcement - EC, clause 6.5.4: Figure 3: Bending of reinforcement. Minimum mandrel diameter: FV Øm f ck beff 0,6 (1 ) f cd 0,5 50 b eff effective beam width. If the compression strut crosses the unit the width of the unit shall be extracted. Normally this will not be the case. Ø m Mandrel diameter of reinforcement θ Concrete strut assumed in 45degrees, sinθ cosθ0,5, see also Memo 51, Part. Select appropriate mandrel diameter. The minimum mandrel diameter shall comply with the requirements of EN , 8.3. Page 9 of 41

10 3) Anchoring of reinforcement - EC, clause and 8.4.4: Figure 4: Anchoring of reinforcement. l bd α 1 α α 3 α 4 α 5 l b,reqd l b,min l b,reqd Ø σ 4 f sd bd FV Stress in stirrup: σ sd As A s Total area of selected reinforcement bars. l b,min max(0,3 l b,reqd; 10 Ø; 100mm) Table 8.: Straight bar: α 1 1,0 Table 8.: Concrete cover: α 1 0,15 (c d 3 Ø)/Ø Neglecting any positive effect of concrete cover, selecting α 1,0 Table 8.: Confinement by reinforcement: α 3 1 K λ Neglecting any positive effect of transverse reinforcement, selecting α 3 1,0 Table 8.: Confinement by welded transverse reinforcement: α 4 1,0 Not relevant. Table 8.: Confinement by transverse pressure: α 5 1,0 Not relevant. α α 3 α 5 > 0,7 Page 10 of 41

11 4) Lap of stirrups - EC, clause 8.7.3: Figure 5: Lap of reinforcement. l 0 α 1 α α 3 α 5 α 6 l b,reqd l 0,min Required lap length: l b,reqd as calculated in clause. 3. l 0,min max(0,3 α 6 l b,reqd; 15 Ø; 00mm) Table 8.: α 1, α, α 3 and α 51,0 as calculated in clause 3. Table 8.3: α 61.5 (All reinforcement is lapped) l 0 1,0 1,0 1,0 1,0 1,5 l b,reqd 5) Anchoring of main reinforcement: It must be ensured the horizontal part of the reinforcement is continued until the main reinforcement is sufficient anchored to carry the load..3 BEAM BOX- HORIZONTAL ANCHORING The beam box is anchored for a total horizontal load of F H0,3F V. The knife will be in contact with the half round steel and the horizontal force is transferred by friction between the two steel parts. The half round steel is anchored with threaded bars. Page 11 of 41

12 The required dimension of threaded bar and machined thread lengths in the half round steel is found from Table 4..4 EVALUATION OF REINFORCEMENT IN THE END OF THE CONCRETE BEAM.4.1 STRUT AND TIE MODEL The beam box will be located in the upper part of the beam cross section. Compatibility in strains through the cross section implies that some of the force will bypass the half round steel and spread into the underlying concrete. This is illustrated with at strut and tie model in Figure 6. Horizontal force in compression strut: The horizontal force in the assumed compression strut must be anchored with horizontal reinforcement inwards from the beam end. For design purpose, the horizontal force may be thought of as smeared, giving horizontal force intensity towards the vertical end the beam: For the case of zb, the horizontal force per unit height of the beam becomes: 1/ F v/(z/) F v/z For the case of z3b, the horizontal force per unit height of the beam becomes: 1/3 F v/(z/3) F v/z The above evaluation illustrates that the force intensity towards the end of the beam always becomes F V/z. Thus, the intensity is depending on the beam height. Narrow stirrups (Just a bit wider than the half round steel) distributed just under the half round steel is recommended. It is important these stirrups are sufficient anchored inwards. Vertical force in compression strut: The vertical force in the compression strut will never exceed F v. When the ordinary beam shear reinforcement (designed for the shear force F v ) runs until the end of the beam, it will ensure integrity for the vertical force. Splitting force in transverse direction: Due to the shape of the half round steel, it is recommended always to include some reinforcement for splitting stress below the unit. This reinforcement may be designed according to EC clause Wide stirrups (as wide as the beam) distributed below the unit according to the recommendations may be applied. Page 1 of 41

13 Figure 6: Strut and tie model in beam end. (Should be printed in colour) Page 13 of 41

14 .4. REINFORCEMENT IN TOP OF BEAM BOND AND ANCHORING As illustrated in Figure 7, drawing no. 1, the tension force at the top of the truss is one ahead of the compression force at the bottom of the truss. Proper anchoring of the reinforcement in top of the beam may conservatively be ensured at a distance z form the support. (This corresponds to a shift in the bending moment diagram a distance z; see also EC, clause and clause (7). Figure 7: Illustration. (Should be printed in colour) Page 14 of 41

15 The tension in the reinforcement at distance z from the support equals the tension force at support. (Note: forces from other loads on the beam will come in addition): Fv ( l a) S z Estimate, required reinforcement: A s S/f sd Anchoring length for top reinforcement (fully anchored): π Ø / 4 435MPa n ln π Øn f bd Ø diameter of bar Ø n diameter of bar. Equivalent diameter when bundled nnumber of bars f bd bond stress Control 1: Anchoring at support: Equivalent amount of fully anchored reinforcement: A eqva s,selected (l-contrete cover)/l n A s,selected Total amount of top reinforcement l ncalculated anchoring length for fully anchored top reinforcement Control: A eqv>a s (Anchored suspension reinforcement may also be added to A eqv) Control : Anchoring at distance z from support: Equivalent amount of fully anchored reinforcement: A eqva s,selected (l-z- contrete cover)/l n A s,selected Total amount of top reinforcement l ncalculated anchoring length for fully anchored top reinforcement Control: A eqv>a s (Tension force to be anchored at distance z is equal to the tension force to be anchored at support. Anchored suspension reinforcement may also be added to A eqv) Control 3: Anchoring at the end of the suspension reinforcement: This is relevant if the suspension reinforcement bars ends outside the distance z from the support, and if the suspension reinforcement is included in A eqv in control 1 and. Control 3 is done in the same way as control 1&. The tension is calculated for a situation as illustrated in drawing 3 in Figure 6. Fv ( l a b) S z Control 4: Bond/transfer of force into reinforcement at top of beam: Page 15 of 41

16 Sufficient bond in order to transfer the increase in tension along the beam into the top reinforcement must be ensured. This is a relevant issue for large concentrated cantilevered loads. Increase in tension in the reinforcement per/mm: S(x)/dx ( M(x)/dx)/zV(x)/z Capacity for increase in force by bond per/mm: S bond/dx f bd Ø n π n Control: S bond(x)/dx > S(x) )/dx.4.3 SHEAR STIRRUPS IN BEAM END The shear at the end of the beam equals F V: A V s Rd, s FV s z f z f.4.4 SHEAR COMPRESSION IN BEAM END Shear compression: EC, clause V Rd,max α cw b w z υ 1 f cd/(cot θ + tan θ) b wb beam.4.5 HORIZONTAL BARS IN BEAM END Narrow stirrups for the horizontal force according to strut and tie model: As FV s z f A total cross section area equal to: A s/s H, shall be included. (Both legs on the stirrups is active) Wide stirrups for splitting stress, EC, clause : If b<h/: 1 b a 1 As Fv 4 b f If b>h/: 1 a 1 As (1 0,7 ) Fv 4 H / f Conservative simplification: 1 1 As Fv 4 f (Only one leg per stirrup is active in transverse direction) Page 16 of 41

17 Figure 8: Illustration horizontal stirrups in beam end. PART 3 BSF BEAM BOX ANCHORING REINFORCEMENT (Note: In the example calculations, «good» bond conditions are assumed when calculating f bd. This may not be the case at the top of the beam, see EC, clause 8.4. ()) 1) Required cross section for reinforcement: FV 5kN As 517mm f 435MPa Ø16 stirrups 01mm x4804mm Capasity of selected reinforcement: 804mm 435MPa349kN ) Mandrel diameter Bending of reinforcement - EC, clause 6.5.4: FV 5000 Ømf min f ck 35 beff 0,6 (1 ) f cd 0, ,6 (1 ) 19,8MPa 0, b eff effective width of beam. Assume: b effb beam 300mm Ø mf Mandrel diameter of reinforcement. Concrete strut assumed in 45degrees, se Part., 147 mm Page 17 of 41

18 Select: Ø00mm 3) Anchoring of reinforcement, EC clause and 8.4.4: Figure 9: Anchoring of reinforcement. l bd α 1 α α 3 α 4 α 5 l b,reqd l b,min Ø σ sd l b,reqd 4 f bd Stress in reinforcement: σ 5kN sd 80MPa 804mm l b,reqd 401mm 4, 79 l b,min max(0,3 l b,reqd; 10 Ø; 100mm)160mm Table 8.: Straight bar: α 1 1,0 Table 8.: Concrete cover: α 1 0,15 (c d 3 Ø)/Ø Neglecting any positive effect of concrete cover, selecting α 1,0 Table 8.: Confinement by reinforcement: α 3 1 K λ Neglecting any positive effect of transverse reinforcement, selecting α 3 1,0 Table 8.: Confinement by welded transverse reinforcement: α 4 1,0 Not relevant. Table 8.: Confinement by transverse pressure: α 5 1,0 Not relevant. Page 18 of 41

19 α α 3 α 51,0 1,0 1,01,0 > 0,7 OK l bd 1,0 1,0 1,0 1,0 1,0 401mm401mm 4) Lap of stirrups, EC clause 8.7.3: Figure 10: Lap of reinforcement. l 0 α 1 α α 3 α 5 α 6 l b,reqd l 0,min Required lap length: l b,reqd 401mm, see evaluation in clause 3. l 0,min max(0,3 α 6 l b,reqd; 15 Ø; 00mm) Table 8.: α 1, α, α 3 and α 51,0 as calculated in clause 3. Table 8.3: α 61.5 (All reinforcement is lapped) l 0 1,0 1,0 1,0 1,0 1,5 401mm60mm Select: l 0600mm 3. BEAM BOX HORIZONTAL ANCHORING Horizontal anchoring of half round steel: R H0,3xF V67,5kN: Select: M1 threaded bars, 8.8 with nut & steel plate 48kN 96kN Page 19 of 41

20 3.3 EXAMPLE REINFORCEMENT IN BEAM END Assume: Columns with five meters spacing. Beam-beam connection at 1m cantilevering from column. Cross section as illustrated in Figure 11. z0,9 d0,9 476mm48mm Horizontal part of the suspension reinforcement is 600mm ( equals the minimum calculated lap length). I.e. the bars end at x mm. (The final required length is found from the calculations) Neglecting self-weight. Assumed dead and live loads 0kN/m Figure 11: Example Beam with BSF5 beam box. (Note, the illustrated reinforcement does not represent the conclusion from the evaluations, follow the calculations below.) REINFORCEMENT IN TOP OF BEAM BOND AND ANCHORING The tensile force in the reinforcement at top of the beam at distance z from the support: 5kN ( ) mm S 496kN 48mm Estimate, required reinforcement: A s 496kN / 435MPa 1141mm Assume main reinforcement at top of beam: 4Ø5 bundled + (1963mm ) Equivalent diameter of Ø5 bundled: Page 0 of 41

21 Ø n Ø 5 35mm Anchoring length of a bundle: π 1,5 435MPa π 1,5 435MPa 47kN L mm n π Ø f π 35,79MPa 0,3067kN / mm 139 n bd Control 1: Anchoring at support (x1000mm): Equivalent fully anchored reinforcement: A eqv 1963mm /139mmx( )mm1367mm A eqv >1141mm OK. Control : Anchoring at distance z from support (x mm). Equivalent fully anchored reinforcement: A eqv 1963mm x(57-30)mm/139mm 764mm A eqv <1141mm NOT OK. Selected solution in this case: Use of the suspension reinforcement. These bars will have sufficient anchoring towards the end of the beam, but anchoring towards the support must be evaluated: Force anchored in Ø5: S Ø5854kN/139mmx(57-30)mm33kN Not anchored: S496kN-33kN164kN Required anchoring length 4Ø16: N N L mm n π Ø fbd 4 π 16,79MPa 4 9 Transfer of force to the main reinforcement with lap of bars. Select l 01,5 l n1,5 9mm438mm Available length: L Ø1603mm, see Figure 14. Solution: Horizontal part of suspension reinforcement is elongated 300mm. Control 3: Anchoring at the end of the suspension reinforcement (x775mm): In the example, this point is within a distance z from the support. Thus, the tension and the required reinforcement will be as calculated in control 1: Equivalent fully anchored reinforcement: A eqv1963mm x(775-30)mm/139mm 1045mm A eqv<1141mm NOT OK. Selected solution in this case: Use of the suspension reinforcement. These bars will have sufficient anchoring towards the end of the beam, but anchoring towards the support must be evaluated: Force anchored in Ø5: S Ø5 A eqv x435mpa1045mm x435mpa 455kN Not anchored: S496kN-455kN41kN Required anchoring length 4Ø16: Page 1 of 41

22 41000N 41000N L mm n π Ø f 4 π 16,79MPa 4 73 bd This is ok if the horizontal part of the suspension reinforcement is elongated 300mm as stated in control. Control 4: Bond/transfer of force into reinforcement at top of beam: Increase in force per/mm: S(x)/dx ( M(x)/dx)/zV(x)/z5kN/48mm55N/mm Capacity for increase in force by bond per/mm: S bond(x)/dx f bd Ø n π,79 35 π 613N/mm S bond(x)/dx > S(x)/dx OK. The bond to the main reinforcement is sufficient to take the increase in force SHEAR STIRRUPS IN BEAM END Use a strut-and-tie model with compression diagonal at 45. The shear at the end of the beam is F V5kN. Beam as illustrated in Figure A V s Rd, s 5 10 N 109mm / m s z f 0,48m 435MPa Assume stirrup diameter Ø10 Select Ø10c/c100 (1570mm /m) SHEAR COMPRESSION IN BEAM END Shear compression: EC, clause Beam as illustrated in Figure 11. V Rd,max α cw b w z υ 1 f cd/(cot θ + tan θ) b wb beam300mm V Rd,max {1, ,6 [1 (35/50)] 19,8/(1+1)} 10 3 V Rd,max 655 kn (>V Rd OK) HORIZONTAL BARS IN BEAM END Example: Beam as illustrated in Figure 11: Narrow stirrups for horizontal force: Page of 41

23 Assume z0,9 d As FV h s z f 5000N h 17mm 91mm 0,9 48mm 435MPa Select two narrow u-bars: Ø1π 6 445mm. Placed just below the unit. Simplified: Horizontal length of bar: L(z-H)+40Ø(476-17)mm+40 1mm 800mm Wide stirrups for splitting force: 1 Fv N As 130mm 4 f 4 435MPa Select two u-bars: Ø1π 6 6mm. Distributed below the unit. Simplified: Horizontal length of bar: L40Ø40 1mm 500mm. PART 4 - BSF BEAM BOX ANCHORING REINFORCEMENT (Note: In the example calculations, «good» bond conditions are assumed when calculating f bd. This may not be the case at the top of the beam, see EC, clause 8.4. ()) 1) Required cross section for reinforcement: A F 300kN 435MPa V s f 689mm Ø16 stirrups 01mm x4804mm Capasity of selected reinforcement: 804mm 435MPa349kN ) Mandrel diameter Bending of reinforcement - EC, clause 6.5.4: FV Ømf min f ck 35 beff 0,6 (1 ) f cd 0, ,6 (1 ) 19,8MPa 0, b eff effective width of beam. Assume: b effb beam300mm Ø mf Mandrel diameter of reinforcement. Concrete strut assumed in 45degrees, se Part., Select: Ø00mm 195 mm Page 3 of 41

24 3) Anchoring of reinforcement, EC clause and 8.4.4: Figure 1: Anchoring of reinforcement. l bd α 1 α α 3 α 4 α 5 l b,reqd l b,min Ø σ sd l b,reqd 4 f bd Stress in reinforcement: σ 300kN sd 373MPa 804mm l b,reqd 535mm 4, 79 l b,min max(0,3 l b,reqd; 10 Ø; 100mm)160mm Table 8.: Straight bar: α 1 1,0 Table 8.: Concrete cover: α 1 0,15 (c d 3 Ø)/Ø Neglecting any positive effect of concrete cover, selecting α 1,0 Table 8.: Confinement by reinforcement: α 3 1 K λ Neglecting any positive effect of transverse reinforcement, selecting α 3 1,0 Table 8.: Confinement by welded transverse reinforcement: α 4 1,0 Not relevant. Table 8.: Confinement by transverse pressure: α 5 1,0 Not relevant. α α 3 α 51,0 1,0 1,01,0 > 0,7 OK Page 4 of 41

25 l bd 1,0 1,0 1,0 1,0 1,0 535mm535mm 4) Lap of stirrups, EC clause 8.7.3: Figure 13: Lap of reinforcement. l 0 α 1 α α 3 α 5 α 6 l b,reqd l 0,min Required lap length: l b,reqd 535mm, see evaluation in clause 3. l 0,min max(0,3 α 6 l b,reqd; 15 Ø; 00mm) Table 8.: α 1, α, α 3 and α 51,0 as calculated in clause 3. Table 8.3: α 61.5 (All reinforcement is lapped) l 0 1,0 1,0 1,0 1,0 1,5 535mm80mm Select: l 0800mm 4. BEAM BOX HORIZONTAL ANCHORING Horizontal anchoring of half round steel: R H0,3xF V90kN: Select: M1 threaded bars, 8.8 with nut & steel plate 48kN 96kN Page 5 of 41

26 4.3 EXAMPLE REINFORCEMENT IN BEAM END Assume: Columns with 4,8m spacing. Beam-beam connection at 1m cantilevering from column. Cross section as illustrated in Figure 14. z0,9 d0,9 518mm466mm Horizontal part of the suspension reinforcement is 800mm ( equals the minimum calculated lap length). I.e. the bars end at: x mm. (The final required length is found from the calculations) Neglecting self-weight. Assumed dead and live loads 0kN/m Figure 14: Example Beam with BSF300 beam box. (Note, the illustrated reinforcement does not represent the conclusion from the evaluations, follow the calculations below.) REINFORCEMENT IN TOP OF BEAM BOND AND ANCHORAGE The tensile force in the reinforcement at top of the beam at distance z from the support: 300kN ( ) mm S 608kN 466mm Estimate, required reinforcement: A s 608kN / 435MPa 1398mm Assume main reinforcement at top of beam: 4Ø3 bundled + (316mm ) Equivalent diameter of Ø3 bundled: Page 6 of 41

27 Ø n Ø 3 45mm Anchoring length of a bundle: π MPa π MPa 700kN L mm n π Ø f π 45,79MPa 0,3944kN / mm 1774 n bd Control 1: Anchoring at support (x1000mm): Equivalent fully anchored reinforcement: A eqv316mm x( )mm/1774mm 1758mm A eqv>1398mm OK. Control : Anchoring at distance z from support (x mm). Equivalent fully anchored reinforcement: A eqv316mm x(534-30)mm/1774mm 914mm A eqv<1398mm NOT OK. Selected solution in this case: Use of the suspension reinforcement. These bars will have sufficient anchoring towards the end of the beam, but anchoring towards the support must be evaluated: Force anchored in Ø3: S Ø3 A eqv x435mpa914mm x435mpa 398kN Not anchored: S608kN-398kN10kN Required anchoring length 4Ø16: 10000N 10000N L mm n π Ø fbd 4 π 16,79MPa Transfer of force to the main reinforcement with lap of bars. Select l 01,5 l n1,5 374mm561mm Available length: L Ø16441mm, see Figure 14. Solution: Horizontal part of suspension reinforcement is elongated 00mm. Control 3: Anchoring at the end of the suspension reinforcement (x975mm): In the example, this point is close to the support. Thus, the tension and the required reinforcement will be as calculated in control 1: Equivalent fully anchored reinforcement: A eqv316mm x(975-30)mm/1774mm 1713mm A eqv>1398mm OK. Page 7 of 41

28 Control 4: Bond/transfer of force into reinforcement at top of beam: Increase in force per/mm: S(x)/dx ( M(x)/dx)/zV(x)/z300kN/466mm644N/mm Capacity for increase in force by bond per/mm: S bond(x)/dx f bd Ø n π,79 45 π 788N/mm S bond(x)/dx > S(x)/dx OK. The bond to the main reinforcement is sufficient to take the increase in force SHEAR STIRRUPS IN BEAM END Use a strut-and-tie model with compression diagonal at 45. The shear at the end of the beam is F V300kN. Beam as illustrated in Figure A V s Rd, s N 1480mm / m s z f 0,466m 435MPa Assume stirrup diameter Ø1 Select Ø1c/c100 (61mm /m) SHEAR COMPRESSION IN BEAM END Shear compression: EC, clause Beam as illustrated in Figure 14. V Rd,max α cw b w z υ 1 f cd/(cot θ + tan θ) b wb beam350mm V Rd,max {1, ,6 [1 (35/50)] 19,8/(1+1)} 10 3 V Rd,max 833 kn (>V Rd OK) HORIZONTAL BARS IN BEAM END Example: Beam as illustrated in Figure 14: Narrow stirrups for horizontal force: Assume z0,9 d As FV N h h 7mm 336mm s z f 0,9 518mm 435MPa Select two narrow u-bars: Ø1π 6 445mm. Placed just below the unit. Simplified: Horizontal length of bar: L(z-H)+40Ø(518-7)mm+40 1mm 800mm Page 8 of 41

29 Wide stirrups for splitting force: 1 Fv N A 4 f 4 435MPa s 17mm Select two u-bars: Ø1π 6 6mm. Distributed below the unit. Simplified: Horizontal length of bar: L40Ø40 1mm 500mm. PART 5 BSF BEAM BOX ANCHORING REINFORCEMENT (Note: In the example calculations, «good» bond conditions are assumed when calculating f bd. This may not be the case at the top of the beam, see EC, clause 8.4. ()) 1) Required cross section for reinforcement: A F 450kN 435MPa V s f 1035mm 3Ø16 stirrups 01mm x6106mm Capasity of selected reinforcement: 106mm 435MPa54kN ) Mandrel diameter Bending of reinforcement - EC, clause 6.5.4: FV Ømf min f ck 35 beff 0,6 (1 ) f cd 0, ,6 (1 ) 19,8MPa 0, b eff effective width of beam. Assume: b effb beam 350mm Ø mf Mandrel diameter of reinforcement. Concrete strut assumed in 45degrees, se Part., Select: Ø30mm 51 mm Page 9 of 41

30 3) Anchoring of reinforcement, EC clause and 8.4.4: Figure 15: Anchoring of reinforcement. l bd α 1 α α 3 α 4 α 5 l b,reqd l b,min Ø σ sd l b,reqd 4 f bd Stress in reinforcement: σ 450kN sd 373MPa 106mm l b,reqd 535mm 4, 79 l b,min max(0,3 l b,reqd; 10 Ø; 100mm)160mm Table 8.: Straight bar: α 1 1,0 Table 8.: Concrete cover: α 1 0,15 (c d 3 Ø)/Ø Neglecting any positive effect of concrete cover, selecting α 1,0 Table 8.: Confinement by reinforcement: α 3 1 K λ Neglecting any positive effect of transverse reinforcement, selecting α 3 1,0 Table 8.: Confinement by welded transverse reinforcement: α 4 1,0 Not relevant. Table 8.: Confinement by transverse pressure: α 5 1,0 Not relevant. α α 3 α 51,0 1,0 1,01,0 > 0,7 OK Page 30 of 41

31 l bd 1,0 1,0 1,0 1,0 1,0 535mm535mm 4) Lap of stirrups, EC clause 8.7.3: Figure 16: Lap of reinforcement. l 0 α 1 α α 3 α 5 α 6 l b,reqd l 0,min Required lap length: l b,reqd 535mm, see evaluation in clause 3. l 0,min max(0,3 α 6 l b,reqd; 15 Ø; 00mm) Table 8.: α 1, α, α 3 and α 51,0 as calculated in clause 3. Table 8.3: α 61.5 (All reinforcement is lapped) l 0 1,0 1,0 1,0 1,0 1,5 535mm80mm Select: l 0800mm 5. BEAM BOX HORIZONTAL ANCHORING Horizontal anchoring of half round steel: R H0,3xF V135kN: Select: 1 M0 threaded bars, 8.8 with nut & steel plate 141kN Page 31 of 41

32 5.3 EXAMPLE REINFORCEMENT IN BEAM END Assume: Columns with six meters spacing. Beam-beam connection at 1,m cantilevering from column. Cross section as illustrated in Figure 17. z0,9 d0,9 665mm599mm Horizontal part of the suspension reinforcement is 800mm ( equals the minimum calculated lap length). I.e. the bars end at x mm. (The final required length is found from the calculations) Neglecting self-weight. Assumed dead and live loads 0kN/m Figure 17: Example Beam with BSF450 beam box. (Note, the illustrated reinforcement does not represent the conclusion from the evaluations, follow the calculations below.) REINFORCEMENT IN TOP OF BEAM BOND AND ANCHORAGE The tensile force in the reinforcement at top of the beam at distance z from the support: 450kN (100 6,5) mm S 855kN 599mm Estimate, required reinforcement: A s 855kN / 435MPa 1966mm Assume main reinforcement at top of beam: 4Ø3 bundled + (316mm ) Page 3 of 41

33 Equivalent diameter of Ø3 bundled: Ø n Ø 3 45mm Anchoring length of a bundle: π MPa π MPa 700kN L mm n π Ø f π 45,79MPa 0,3944kN / mm 1774 n bd Control 1: Anchoring at support: (x100mm): Equivalent fully anchored reinforcement: A eqv316mm x(100-30)mm/1774mm 11mm A eqv>1966mm OK. Control : Anchoring at distance z from support (x mm). Equivalent fully anchored reinforcement: A eqv316mm x(601-30)mm/1774mm 1035mm A eqv<1966mm NOT OK Selected solution in this case: Use of the suspension reinforcement. These bars will have sufficient anchoring towards the end of the beam, but anchoring towards the support must be evaluated: Force anchored in Ø3: S Ø3 A eqv x435mpa1035mm x435mpa 450kN Not anchored: S855kN-450kN405kN Required anchoring length 6Ø16: N N L mm n π Ø fbd 6 π 16,79MPa Transfer of force to the main reinforcement with lap of bars. Select l 01,5 l n1,5 481mm71mm Available length: L Ø16451mm, see Figure 17. Solution: Horizontal part of suspension reinforcement is elongated 300mm. Control 3: Anchoring at the end of the suspension reinforcement (x105). In the example, this point is within a distance z from the support. Thus, the tension and the required reinforcement will be as calculated in control 1: Equivalent fully anchored reinforcement: A eqv316mm x(105-30)mm/1774mm 185mm A eqv<1966mm NOT OK. Selected solution in this case: Use of the suspension reinforcement. These bars will have sufficient anchoring towards the end of the beam, but anchoring towards the support must be evaluated: Force anchored in Ø3: S Ø3 A eqv x435mpa185mm x435mpa805kn Page 33 of 41

34 Not anchored: S855kN-805kN50kN Required anchoring length 6Ø16: 50000N 50000N L mm n π Ø f 6 π 16,79MPa 6 60 bd This is ok if the horizontal part of the suspension reinforcement is elongated 300mm as stated in control. Control 4: Bond/transfer of force into reinforcement at top of beam: Increase in force per/mm: S(x)/dx ( M(x)/dx)/zV(x)/z450kN/599mm75N/mm Capacity for increase in force by bond per/mm: S bond(x)/dx f bd Ø n π,79 45 π 788N/mm S bond(x)/dx > S(x)/dx OK. The bond to the main reinforcement is sufficient to take the increase in force SHEAR STIRRUPS IN BEAM END Use a strut-and-tie model with compression diagonal at 45. The shear at the end of the beam is F V450kN. Beam as illustrated in Figure A V s Rd, s N 177mm / m s z f 0,599m 435MPa Assume stirrup diameter Ø1. Select Ø1c/c100 (61mm /m) SHEAR COMPRESSION IN BEAM END Shear compression: EC, clause Beam as illustrated in Figure 17. V Rd,max α cw b w z υ 1 f cd/(cot θ + tan θ) b wb beam 400mm V Rd,max {1, ,599 0,6 [1 (35/50)] 19,8/(1+1)} 10 3 V Rd,max 13 kn (>V Rd OK) Page 34 of 41

35 5.3.4 HORIZONTAL BARS IN BEAM END Example: Beam as illustrated in Figure 17: Narrow stirrups for horizontal force: Assume z0,9 d As FV N h h 36mm 66mm s z f 0,9 665mm 435MPa Select three narrow u-bars: Ø1π mm. Placed just below the unit. Simplified: Horizontal length of bar: L(z-H)+40Ø(665-36)mm+40 1mm 800mm Wide stirrups for splitting force: 1 Fv N A 4 f 4 435MPa s 59mm Select three u-bars: Ø1π mm. Distributed below the unit. Simplified: Horizontal length of bar: L40Ø40 1mm 500mm. PART 6 - BSF BEAM BOX ANCHORING REINFORCEMENT (Note: In the example calculations, «good» bond conditions are assumed when calculating f bd. This may not be the case at the top of the beam, see EC, clause 8.4. ()) 1) Required cross section for reinforcement: FV 700kN As 1609mm f 435MPa Ø5 stirrups 490mm x41960mm Capasity of selected reinforcement: 1960mm 435MPa85kN ) Mandrel diameter Bending of reinforcement - EC, clause 6.5.4: FV Ømf min fck 35 beff 0,6 (1 ) fcd 0, ,6 (1 ) 19,8MPa 0, b eff effective width of beam. Assume: b effb beam 550mm, 49 mm Page 35 of 41

36 Ø mf Mandrel diameter of reinforcement. Concrete strut assumed in 45degrees, se Part. Select: Ø30mm 3) Anchoring of reinforcement, EC clause and 8.4.4: Figure 18: Anchoring of reinforcement. l bd α 1 α α 3 α 4 α 5 l b,reqd l b,min Ø σ sd l b,reqd 4 f bd Stress in reinforcement: σ 700kN sd 357MPa 1960mm l b,reqd 800mm 4, 79 l b,min max(0,3 l b,reqd; 10 Ø; 100mm)50mm Table 8.: Straight bar: α 1 1,0 Table 8.: Concrete cover: α 1 0,15 (c d 3 Ø)/Ø Neglecting any positive effect of concrete cover, selecting α 1,0 Table 8.: Confinement by reinforcement: α 3 1 K λ Neglecting any positive effect of transverse reinforcement, selecting α 3 1,0 Table 8.: Confinement by welded transverse reinforcement: α 4 1,0 Not relevant. Page 36 of 41

37 Table 8.: Confinement by transverse pressure: α 5 1,0 Not relevant. α α 3 α 51,0 1,0 1,01,0 > 0,7 OK l bd 1,0 1,0 1,0 1,0 1,0 800mm800mm 4) Lap of stirrups, EC clause 8.7.3: Figure 19: Lap of reinforcement. l 0 α 1 α α 3 α 5 α 6 l b,reqd l 0,min Required lap length: l b,reqd 800mm, see evaluation in clause 3 l 0,min maks(0,3 α 6 l b,reqd; 15 Ø; 00mm) Table 8.: α 1, α, α 3 and α 51,0 as calculated in clause 3. Table 8.3: α 61.5 (All reinforcement is lapped) l 0 1,0 1,0 1,0 1,0 1,5 800mm100mm Select: l 0100mm 6. BEAM BOX HORIZONTAL ANCHORING Horizontal anchoring of half round steel: R H0,3xF V10kN: Select: M0 threaded bars, 8.8 with nut & steel plate 8kN Page 37 of 41

38 6.3 EXAMPLE REINFORCEMENT IN BEAM END Assume: Columns with 7,m spacing. Beam-beam connection at 1,4m cantilevering from column. Cross section as illustrated in Figure 0. z0,9 d0,9 735mm66mm Horizontal part of the suspension reinforcement is 100mm ( equals the minimum calculated lap length). I.e. the bars end at x mm. (The final required length is found from the calculations) Neglecting self-weight. Assumed dead and live loads 0kN/m. Figure 0: Example Beam with BSF700 beam box. (Note, the illustrated reinforcement does not represent the conclusion from the evaluations, follow the calculations below.) REINFORCEMENT IN TOP OF BEAM BOND AND ANCHORAGE The tensile force in the reinforcement at top of the beam at distance z from the support: 700kN ( ) mm S 1401kN 66mm Estimate, required reinforcement: A s 1401kN / 435MPa 30mm Assume main reinforcement at top of beam: 6Ø3 bundled ++ (485mm ) Equivalent diameter of Ø3 bundled: Page 38 of 41

39 Ø n Ø 3 45mm Anchoring length of a bundle: π MPa π MPa 700kN L mm n π Ø f π 45,79MPa 0,3944kN / mm 1774 n bd Control 1: Anchoring at support (x1400mm): Equivalent fully anchored reinforcement: A eqv485mm x( )mm/1774mm 376mm A eqv>30mm OK. Control : Anchoring at distance z from support (x mm). Equivalent fully anchored reinforcement: A eqv485mm x(738-30)mm/1774mm 195mm A eqv<30mm NOT OK. Selected solution in this case: Use of the suspension reinforcement. These bars will have sufficient anchoring towards the end of the beam, but anchoring towards the support must be evaluated: Force anchored in Ø3: S Ø3 A eqv x435mpa195mm x435mpa 837kN Not anchored: S1401kN-837kN564kN Required anchoring length 4Ø5: N N L mm n π Ø fbd 4 π 5,79MPa Transfer of force to the main reinforcement with lap of bars. Select l 01,5 l n1,5 609mm914mm Available length: L Ø579mm, se Figure 14. Solution: Horizontal part of suspension reinforcement is elongated 00mm. Control 3: Anchoring at the end of the suspension reinforcement (x1467). In the example, this point is on the inside of the support The anchoring at x1467mm is sufficient, see control 1. (The anchoring is further improved when the reinforcement is elongated 00mm as stated in control. Control 4: Bond/transfer of force into reinforcement at top of beam: Increase in force per/mm: S(x)/dx ( M(x)/dx)/zV(x)/z700kN/66mm1058N/mm Capacity for increase in force by bond per/mm: S bond(x)/dx f bd Ø n π 3,79 45 π 31183N/mm S bond(x)/dx > S(x)/dx OK. The bond to the main reinforcement is sufficient to take the increase in force. Page 39 of 41

40 6.3. SHEAR STIRRUPS IN BEAM END Use a strut-and-tie model with compression diagonal at 45. The shear at the end of the beam is F V700kN. Beam as illustrated in Figure 0. 3 A V s Rd, s N 430mm / m s z f 0,66m 435MPa Assume stirrup diameter Ø1. Select Ø1c/c80 (87mm /m) SHEAR COMPRESSION IN BEAM END Shear compression: EC, clause Beam as illustrated in Figure 0. V Rd,max α cw b w z υ 1 f cd/(cot θ + tan θ) b wb beam550mm V Rd,max {1, ,6 [1 (35/50)] 19,8/(1+1)} 10 3 V Rd,max 1860 kn (>V Rd OK) HORIZONTAL BARS IN BEAM END Example: Beam as illustrated in Figure 0. Narrow stirrups for horizontal force: Assume z0,9 d As FV N h h 45mm 596mm s z f 0,9 735mm 435MPa Select three narrow u-bars: Ø1π mm. Placed just below the unit. Simplified: Horizontal length of bar: L(z-H)+40Ø(665-36)mm+40 1mm 800mm Wide stirrups for splitting force: 1 Fv N A 4 f 4 435MPa s 40mm Select two u-bars: Ø16π 8 40mm. Distributed below the unit. Simplified: Horizontal length of bar: L40Ø40 16mm 700mm Page 40 of 41

41 REVISION Date: Description: First edition Changed the half round steel on the BSF700 unit. Corrected Table Changed position of the M0 threaded bars in the half round steel BSF 700 unit. Changed steel plate anchoring M0 threaded bars BSF 700 unit Updated Table 4. Required thread length in blind holes Included a nut on the front side of the steel plate anchoring the threaded bars. (To ensure correct position of the plate when casting the concrete) New template Page 41 of 41