Concrete and Masonry Structures 1 Office hours

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1 Concrete and Masonry Structures 1 Petr Bílý, office B731 Courses in English Concrete and Masonry Structures 1 Office hours

2 Credit receiving requirements General knowledge of design of concrete structures (e.g. 133FSTD Fundamentals of Structural Design) Working out of the homework assigned during the semester. Homework delivered: the next week 3 points 1 week delay 2 points 2 weeks delay 1 point 3 or more weeks delay 0 points, but you still have to deliver!!! Tests at the lectures max. 3 points each Reach at least 45 points out of total of 72 points (12 tests, 12 pieces of homework)

3 structures

4 Homework 7 tasks Page layout A4 onesided Write by pencil (easy to correct) M Ed = 1/8*f*L 2 M Ed = 1/8*8*5 2 M Ed = 25 knm 5 cm Load Char. γ F Design 1,35 1,50. TOTAL

5 Rules for structural analysis elaboration Well-arranged, clear, controllable Page numbers (cross reference to previous calculations and results) All calculations, assumptions write down in the analysis State formula substitute calculate results, quote units Calculation of loads in tables Sketches, figures

6 1st task: FrameStructure

7 Individual Parameters see excel spreadsheet R, a [m] distance of axes in the plan (spans) h [m] floor height n number of floors concrete class Permanent load (except self weight) for typical floor (gg 0 ) floor,k [kn/m 2 ] Permanent load (except self weight) for roof (g-g 0 ) roof,k [kn/m 2 ] Variable load for typical floor q floor,k [kn/m 2 ] Variable load for roof q k = 0,75 kn/m 2 Exposure class related to environmental conditions Design working life

8 Our goal will be to: Design dimensions of all elements Do detailed calculation of 2D frame - calculation of bending moments, shear and normal forces using FEM software Design frame reinforcement Draw layout of reinforcement

9 Design of dimensions Depth of the slab Dimensions of the beam Column dimensions Sketch of the structure

10 Depth of the slabh S One-way slab Empirical estimation: Effective depth d: hs 1 1 = d = h c s 2 l Diameter of steel bars, 10 mm Cover depth

11 Cover depth c c = c= c + c min c dev = 10 mm (technology allowance) c min,b = 10 mm (cover depth necessary for good mechanical bond between steel and concrete, equal to diameter of steel bars) c min,dur see table (cover depth necessary for good resistance to unfavourable effects of the environment) dev ( c c ) max ; ;10 mm min min,b min,dur

13 Depth of the slabh S Span/depth ratio (deflection control): See table, for slabs consider the value for 0,5 % reinf. ratio l λ = λ κ κ κ λ lim = d Effect of shape 1.0 c1 c2 c3 d,tab Effect of span 1.0 Effect of reinforcement 1.2 If λ λ min, detailed calculation of deflections may be omitted However, usually the slab is uneconomical if the condition is satisfied

14 for outer span of the continuous beam/slab Concrete class for inner span of the continuous beam/slab Concrete class

15 Depth of the slabh S Usually the slab is uneconomical if the span/depth condition is satisfied => just adjust the empirical design with respect to span/depth ratio If λ > λ min, increase the depth of the slab by some mm, depending on the difference between empirical design and design according to span/depth ratio

16 Design of the beam Empirical estimation hb = lb bb = h B To reach sufficient stiffness of the beam: h B 2.5h S

17 Preliminary check of the beam To avoid troubles during detailed check Theoretical maximum values of internal forces in the beam: MEd,max = fblb VEd,max = fblb 8 8 Load per 1 m of the beam in kn/m Real internal forces will be lower

18 Preliminary check of the beam Preliminary check of bending M Ed,max table (see web) µ = ξ 2 bd B Bfcd Relative height of compressed part of the beam (x/d) Relative bending moment (a factor expressing to what extent the beam is utilized by applied bending moment) Effective height of the beam, estimated diameter of rebars mm If ξ < > design is correct If ξ < 0.15 you should decrease h B and/or b B If ξ > 0.40 you have to increase h B and/or b B

19 Preliminary check of the beam Preliminary check of reinforcement ratio M A s,rqd ρ s,rqd = = Ac Required reinforcement ratio Ed,max ζd f bd B yd B B table (see web) Relative value of lever arm of internal forces (z/d) If ρ s,rqd >0.04 you have to increase h B and/or b B

20 Preliminary check of the beam Preliminary check of load-bearing capacity in shear ( compressed diagonals ) cotθ V = ν f b ζ d V 1+ cot θ Rd,max cd B B 2 Ed,max Load-bearing capacity of compressed diagonals in shear Coefficient expressing effect of shear cracks and transversal deformations ν= f ck 0, Cotangent of slope of shear cracks, cotθ = 1,5 If the condition is not checked, you have to increase h B and/or b B

21 Preliminary check of the beam Span/depth ratio (deflection control) same procedure as for slabs Select a row in the table for λ d,tab (outer span) according to value of ρ s,rqd calculated If the condition is not checked, you have to increase h B (unlike slabs, it is usually a good idea to meet the condition for beams)

22 Dimensions of the column Calculate design load in the foot of the column (N Ed ) N Ed N Rd 0,02 A c 400 MPa N = 0,8A f + Aσ N Rd c cd s s Ed A c cd N Ed 0,8f + 0, 02σ s => dimensions of rectangular column

23 Adjustment of dimensions Round dimensions to 50 mm Round slab dimensions to 10 mm Round beam dimensions to 50 mm If the difference between column width and beam widthis less than 100 mm, use the bigger dimension for both elements Reason: dimensions of formwork systems

24 Sketch of the structure

25 For the next week We will focus on detailed calculation of internal forces Are you able to use any Finite Element Analysis software? If not, check easy-to-use free software Idea Statica on and tryto getfamiliarwithit

26 Example Two-way slabs supported on four sides concrete class C30/37, cover depth 25 mm, 6 mm steel bars, 4 floors

27 Slab depth design lx + ly hs = = = 173 mm d = hs c = = 145mm 2 2 Deflection control: l /d = 6000 / 145 = 41 λ lim = 1,0*1,0*1,2*30,8 = 37 => h s has to be increased Slab height h S = 190 mm

28 Calculation of loads Slab load charakteristic γ F design kn/m 2 kn/m 2 Permanent other permanent load 0,50 self weight 0,19m. 25kN/m 3 4,75 Total g k = 5,25 1,35 g d = 7,09 Variable (kategorie C1) q k = 3,00 1,5 q d = 4,50 Total (g+q) k = 8,25 (g+q) d = 11,59 Roof load charakteristic γ F design kn/m 2 kn/m 2 Permanent other permanent load 2,00 self weight 0,19m. 25kN/m 3 4,75 Total g k = 6,75 1,35 g d = 9,11 Variable (kategorie C1) q k = 0,75 1,5 q d = 1,125 Total (g+q) k = 7,5 (g+q) d = 10,24

29 Beam hb = l= 7 0,5 m hb 2. 5h S ( ) b= 0,33 0,5 h= 0,25m

30 Column tributing area A = 6,5 x 6 = 39m 2

31 load from the slab 3x typical floor 3 x 39 m2 x 11,59 kn/m2 = 3x 452 = 1356 kn 1x roof 1 x 39 m2 x 10,24 kn/m2 = 339,4 kn 1755,4 kn load from the beam (0,5-0,19)m x 0,25m x 25 kn/m3 = 0,08 * 25 = 2 kn/m (6,5 + 6)m * 2 kn/m= 25 kn x 4 floors = 100kN estimate self weight of the column 25kN N Ed = = 1880 N Ed = 0, 8 A c f cd + A s σ ,8A c ,02A = C min. area: A c =0,078m column 300 x 300 mm s 2

Bending and Shear in Beams

Bending and Shear in Beams Lecture 3 5 th October 017 Contents Lecture 3 What reinforcement is needed to resist M Ed? Bending/ Flexure Section analysis, singly and doubly reinforced Tension reinforcement,