Date: 04.0.0 sss Page of 5 CONTENTS PART BASIC ASSUMTIONS... GENERAL... STANDARDS... QUALITIES... 3 DIMENSIONS... 3 LOADS... 3 PART REINFORCEMENT... 4 EQUILIBRIUM... 4
Date: 04.0.0 sss Page of 5 PART BASIC ASSUMTIONS GENERAL The following calculations of anchorage of the units and the corresponding reinforcement must be considered as an example illustrating the design model. It must always be checked that the forces from the anchorage reinforcement can be transferred to the main reinforcement of the concrete components. The recoended reinforcement includes only the reinforcement necessary to anchor the unit to the concrete. In the vicinity of the unit the element must be designed for the force R. STANDARDS The calculations are carried out in accordance with: Eurocode : of concrete structures. Part : General rules and rules for buildings. Eurocode 3: of steel structures. Part : General rules and rules for buildings. Eurocode 3: of steel structures. Part 8: of joints. EN 0080: Steel for the reinforcement of concrete. Weldable reinforcing steel. General. For all NDPs (Nationally Determined Parameter) in the Eurocodes the recoended values are used. NDP s are as follows: Parameter c s α cc α ct C Rd,c v min k Recoended value Table : NDP s in EC..5.5.0.0 0. 0.035k /3 f ck / 0.5 Parameter M0 M M Recoended.0.0.5 value Table : NDP s in EC 3.
Date: 04.0.0 sss Page 3 of 5 QUALITIES Concrete grade C35/45: f ck = 35,0 MPa EC, Table 3. f cd = α cc f ck /γ c = 35/,5 = 3,3 MPa EC, Pt.3.5 f ctd = α ct f ctk,0,05 /γ c =,0/,5 =,46 MPa EC, Pt.3.6 f bd =,5 η η f ctd =,5 0,7,0,46 =,3 MPa EC, Pt.8.4. Reinforcement B500C: Structural steel S355: DIMENSIONS f yd = f yk /γ s = 500/,5 = 435 MPa Tension: f yd = f y / γ M0 = 355/,0 = 355 MPa Compression: f yd = f y / γ M0 = 355/,0 = 355 MPa Shear: f sd = f y /( M0 3) = 355/(,0 3) = 05 MPa EC, Pt.3..7 Inner tube: HUP 00x50x6, Cold formed, S355 Outer tube: HUP 0x80x5, Cold formed, S355 LOADS Vertical ultimate limit state load = F V =.
Date: 04.0.0 sss Page 4 of 5 PART REINFORCEMENT EQUILIBRIUM Figure : Forces acting on the unit. F V = External force on the inner tube R i, = Internal forces between the inner and outer tubes. R, R, R 3 = Support reaction forces the outer tube.
Date: 04.0.0 sss Page 5 of 5 I) Equilibrium inner tube: Figure : Forces acting on the inner tube. Equilibrium equations of the inner tube: ): M=0: F v (L b e) R i (L b a 5 g e)=0 () ): F y =0: F v R i + =0 () Assuming nominal values: L =347, a=75, b=35, g=35, e=0 Solving R i from eq. : Fv ( L b e) R i (3) ( L b a 5 g e) Solving from eq. : = R i F v (4) Results: (347 35 0) 6. 5kN (347 35 75 5 35 0) 6.5kN 6. 5kN
Date: 04.0.0 sss Page 6 of 5 II) Equilibrium outer tube: Figure 3: Forces acting on the outer tube. Exact distribution of forces depends highly on the behavior of the outer tube. Both longitudinal bending stiffness and local transverse bending stiffness in the contact points between the inner and the outer tubes affects the equilibrium. Two situations are considered: ) Rigid outer tube. Outer tube rotates as a stiff body. This assumption gives minimum reaction force at R, and maximum reaction force at R. R 3 becomes zero. (The force R 3 will actually be negative, but since no reinforcement to take the negative forces is included at this position, it is assumed to be zero.) Equilibrium equations of the outer tube: ): M=0: (R i R ) (L 5 g d) ( R 3 ) (L 5 g c d)=0 (5) ): F y =0: R +R 3 +R i R =0 (6) Assuming nominal values: L=397, c=87, g=35, e=0, d=0 ; (c=l b a 5 g e=347 35 75 5 35 0=87, see Figure ) Solving R from eq. 5: ( R R ) ( L 5 g d) ( R (6.5 R ) (397 5 35 0) (6.5 0) (397 5 35 87 0) 0 56040 347R R i 9840 0 4600 33.kN 347 Solving R from eq. 6: i R ) ( L 5 g c d) 0 3
Date: 04.0.0 sss Page 7 of 5 R R Ri R i R3 33. 6.5 6.5 33. kn ) Outer tube without bending stiffness. No forces transferred to outer tube at the back of inner tube. This assumption gives maximum reaction forces R and R 3. R becomes zero. The forces follow directly from the assumption: R = R i R 3 = and R =0 R R R 3 6.5kN 0kN 6.5kN The magnitude of the forces will be somewhere in between the two limits, and the prescribed reinforcement ensures integrity for both situations. Reinforcement is to be located at the assumed attack point for support reactions. Reinforcement for R, R and R 3 : Figure 4: Forces. Reinforcement necessary to anchor the unit to the concrete: Reinforcement R : A s = R /f sd = 6.5kN/435Mpa =37 Select Ø = 3 =45 Capacity selected reinforcement: R=45 435MPa=96.6kN Reinforcement R 3 : A s3 = R 3 /f sd = 6.5kN/435MPa =4 Select Ø = 3 = 6 Capacity selected reinforcement: R=6 435MPa=98.3kN Reinforcement R : A s = R /f sd = 33.kN/435MPa =76 Select Ø = 3 = 6 Capacity selected reinforcement: R=6 435MPa=98.3kN
Date: 04.0.0 sss Page 8 of 5 Tolerances on the positioning of the reinforcement: Due to the small internal distances, the magnitude of the forces will change when changing the position of the reinforcement. Thus, strict tolerances are required. Alt ) Assume: L =347, a=75, b=35, g=35+5=40, e=0 (347 35 0) 66kN (347 35 75 5 40 0) 66kN 66kN Alt ) Assume: L =347, a=75, b=35, g=35 5=30, e=0 (347 35 0) 57. 3kN (347 35 75 5 30 0) 57.3kN 57. 3kN Alt 3) Assume: L =347, a=75, b=35, g=35, e=0 5=5 (347 35 5) 59. 9kN (347 35 75 5 35 5) 59.9kN 59. 9kN Alt 4) Assume: L =347, a=75, b=35, g=35, e=0+5=5 (347 35 5) 63. kn (347 35 75 5 35 5) 63.kN 63. kn
Date: 04.0.0 sss Page 9 of 5 Alt 5) Assume: L =347, a=75, b=35, g=35+5=40, e=0+5=5 (347 35 5) 67. 8kN (347 35 75 5 40 5) 67.8kN 67. 8kN Conclusion tolerances: The assembling tolerances for P, P and P4 should be ±5. For recoended reinforcement pattern, see Memo 57. Transverse reinforcement: One transverse bar to be placed in the bend of every anchoring bar, see Figure 5. One transverse bar with the same diameter as the anchorage bar to be placed in the bend of every anchoring bar. Figure 5: Anchoring reinforcement.