Design MEMO 54a Reinforcement design for RVK 41

Transcription

1 Page of 5 CONTENTS PART BASIC ASSUMTIONS... 2 GENERAL... 2 STANDARDS... 2 QUALITIES... 3 DIMENSIONS... 3 LOADS... 3 PART 2 REINFORCEMENT... 4 EQUILIBRIUM... 4

2 Page 2 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 2: Design of concrete structures. Part : General rules and rules for buildings. Eurocode 3: Design of steel structures. Part : General rules and rules for buildings. Eurocode 3: Design of steel structures. Part 8: Design 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 k /3 f ck /2 0.5 Parameter M0 M M2 Recoended value Table 2: NDP s in EC 3.

3 Page 3 of 5 QUALITIES Concrete grade C35/45: f ck = 35,0 MPa EC2, Table 3. f cd = α cc f ck /γ c = 35/,5 = 23,3 MPa EC2, Pt.3.5 f ctd = α ct f ctk,0,05 /γ c = 2,20/,5 =,46 MPa EC2, Pt.3.6 f bd = 2,25 η η 2 f ctd = 2,25 0,7,0,46 = 2,3 MPa EC2, Pt 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) = 205 MPa EC2, Pt Inner tube: HUP 70x40x4, Cold formed, S355 Outer tube: HUP 80x50x4, Cold formed, S355 LOADS Vertical ultimate limit state load = F V = 40kN.

4 Page 4 of 5 PART 2 REINFORCEMENT EQUILIBRIUM Figure : Forces acting on the unit. F V = External force on the inner tube R i, R 2i = Internal forces between the inner and the outer tubes. R, R 2, R 3 = Support reaction forces of the outer tube. g= distance to the middle plane of the anchoring stirrups in front of the unit.

5 Page 5 of 5 I) Equilibrium inner tube: Figure 2: Forces acting on the inner tube. Equilibrium equations of the inner tube: ): M=0: F v (L b e) R i (L b a g e)=0 () 2): F y =0: F v R i +R 2i =0 (2) Assuming nominal values: L =275, a=75, b=35, g=35, e=0 Solving R i from eq. : Fv ( L b e) R i (3) ( L b a g e) Solving R 2i from eq. 2: R 2i = R i F v (4) Results: 40kN ( ) R i 76. 7kN ( ) R i 76.7kN 40kN 36. 7kN 2

6 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 2. R 3 is assumed 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 3 g d) (R 2i R 3 ) (L 3 g c d)=0 (5) 2): F y =0: R 2 +R 3 +R i R 2i R =0 (6) Assuming nominal values: L=298, c=20, g=35, e=0, d=0; (c=l b a g e= =20) Solving R from eq. 5: ( R R ) ( L 3 g d) ( R (76.7 R ) ( ) (36.7 0) ( ) R R i kN 250 2i R ) ( L 3 g c d) 0 3

7 Page 7 of 5 Solving R 2 from eq. 6: R R R i R i kN 2 2 2) 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 2 becomes zero. The forces follow directly from the assumption: R = R i R 3 = R 2i and R 2 =0 R R R kN 0kN 36.7kN 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 2 and R 3 : Figure 4: Forces. Reinforcement necessary to anchor the unit to the concrete: Reinforcement R : A s = R /f sd 76.7kN/435Mpa =76 2 Select 2 Ø8 = =200 2 Capacity selected reinforcement: R= MPa=87kN Reinforcement R 3 : A s3 = R 3 /f sd = 36.7kN/435MPa =84 2 Select Ø8 = 2 50 = 00 2 Capacity selected reinforcement: R= MPa=43.5kN Reinforcement R 2 : A s2 = R 2 /f sd = 7.6kN/435MPa =4 2 Select Ø8 = 2 50 = 00 2 Capacity selected reinforcement: R= MPa=43.5kN

8 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 =275, a=75, b=35, g=35+5=40, e=0 Gives: 40kN ( ) R 80. 0kN ( ) R 80.0kN 40kN 40. 0kN 2 Alt 2) Assume: L =275, a=75, b=35, g=35+5=40, e=0+5, Gives: 40kN ( ) R 8. 8kN ( ) R 8.8kN 40kN 4. 8kN 2 Conclusion tolerances: Alt 2 represents the most unfavorable position of reinforcement allowed without exceeding the reinforcement capacity. Thus, the assembling tolerances for P, P2 and P4 should be ±5. For recoended reinforcement pattern, see Memo 55a. Transverse reinforcement: One transverse bar with the same diameter as the anchorage bar to be placed in the bend of every anchoring bar.

9 Page 9 of 5 Figure 5: Anchoring reinforcement.