Direct-Design-Method1.pdf
- 1. Design Of Reinforced Concrete Structures ii Two-Way Slabs 1 1. Inroduction When the ratio (L/S) is less than 2.0, slab is called two-way slab, as shown in the fig. below. Bending will take place in the two directions in a dish-like form. Accordingly, main reinforcement is required in the two directions.
- 2. Design Of Reinforced Concrete Structures ii Two-Way Slabs 2 2.Types Of Two Way Slabs 3. Design Methods Two-way slabs Slabs without beams Flat plates Flat slabs Slabs with beams Two-way edge- supported slab Two-way ribbed slab Waffle slabs Two-way Edge- supported ribbed slabs Design methods Simplified Design Methods Grashoff method. Marcus method Egyptian Code method Direct Design Method"DDM" Equivalent Frame Method " EFM "
- 3. Design Of Reinforced Concrete Structures ii Two-Way Slabs 3 4. Direct Design Method "D.D.M" Before Discussion Of this Method, we have to study some concepts: 1. Limitations: 1. Three or more spans in each direction. 2. Variation in successive spans 33% ( . 3. LL 2 DL 4. Column offset 10% in each direction. 5. L/B 2. 6. For slabs on beams, for one panel . 2. Determination of Two way slab thickness: Case 1 : interior and edge beams are exist. = = Where: : is the largest clear distance in the longest direction of panels. : is the clear distance in the short direction in the panel. = = Example for finding : for fig. shown: For panel 1 … For panel 5 … So h to be used should be : hmin< h < hmax
- 4. Design Of Reinforced Concrete Structures ii Two-Way Slabs 4 Case 2: interior beams are not existing, thickness can be found according to table 8.8, page 339. 3. Estimating dimensions of interior and exterior beams sections: Dimensions can be estimated from the following figures: Where: b = beam width, h = slab thickness, a =beam thickness.
- 5. Design Of Reinforced Concrete Structures ii Two-Way Slabs 5 ► Design Procedures Discussion will be done to one representative strip in the horizontal and vertical directions; the same procedure can be used for the other strips. a- Determination of total factored Static Moment : = Strip width /8 : total factored load in t/m2 . = clear distance in the direction of strip, and not less than 0.65 .
- 6. Design Of Reinforced Concrete Structures ii Two-Way Slabs 6 b- Distribution of the total factored static moment to negative and positive moments: I. For interior Spans: According to the code, the moments can be distributed according to factores shown in the figure: II. For Edge Spans : Static Mom. Mo can be distributed, according to factors given in the table 8.9, page 341.
- 7. Design Of Reinforced Concrete Structures ii Two-Way Slabs 7 c- Distribution of the positive and negative factored moments to the Column and middle strips: Note: width of column strip is equal to 0.25l1 or 0.25l2 which is smaller. l1: length in the direction of strip, center to center between columns. l2: length in the direction perpendicular to l1. I. Determination of factored moments on column and middle strips: Finding α and βt: α = α : is ratio of flexural stiffness . Ib : Moment of inertia of the beam in the direction of strip… can be found from fig.8.14 and fig.8.15, pages 310 and 311. Is : Moment of inertia of slab = , where is slab thickness. βt = , βt: Ratio of torsional stiffness and are the modulus of elastisity of concrete for beam and slab. Note: βt is given only for edge beams perpendicular to the strip Note: α is given only for the beams in the direction of the strip
- 8. Design Of Reinforced Concrete Structures ii Two-Way Slabs 8 C: Cross sectional constant defines torsional properties C = X: smallest dimension in the section of edge beam. Y: Largest dimension in the section of edge beam. Note: the C relation is applicable directly for rectangular section only, but when used for L-Shape beams, we should divide it to two rectangular sections and find C. C "A" = C1 + C2 for A and C "B" = C1 + C2 for B. C to be used = Max (C "A" , C "B" ). When α and βt are found, factors for moment can be found from table 8.10 page 343 for the column strip. Notes: α l2/l1 = 0.0 , when there is no interior beams in the direction of strip under consideration. βt = 0.0 , when there is no extirior “edge” beams perpendicular to the strip under consideration.
- 9. Design Of Reinforced Concrete Structures ii Two-Way Slabs 9 After finding the moments on the column strip, Moments on the middle strip is the remain. II. For the moment on the beam “ if exist ” : If: α l2/l1 ≥ 1 … The beam moment is 85% of the moment of the column strip. α l2/l1 = 0 … there is no beam .. mom. = 0 0 < α l2/l1 < 1 … Interpolation have to be done between 0 and 85% to find percentage of moment on the beam from that of the column strip. ** The Mom. on the remain part of column strip = Tot. Mom. on the column strip – Mom. on the beam. Summary: 1- Find Mo : 2- Distribute M0 into +ve and –ve Mom. 3- Distribute Mom. Into column strip and Middle Strip. Column strip Middle Strip 4- Distribute Mom. In column strip into Mom. On beam and remained slab. On beam On remained Slab After calculating Moments, we can find the ρ, then Ast required
- 10. Design Of Reinforced Concrete Structures ii Two-Way Slabs 10 Example 1: For the given data, design strip 1-2-3-4 of the two way slab for flexure. Data: Columns are 30cm X 30cm, Equivalent partitions load=250 Kg/m2 , Live Load = 400Kg/m2 , = 280 kg/cm2 = 4200 Kg/cm2 , slab thickness = 16cm
- 11. Design Of Reinforced Concrete Structures ii Two-Way Slabs 11 Solution: Thickness is given 16cm, no need to be checked. 1- Calculate total factored load Wu "t/m2 ": Wu = 1.4 (0.16 2.5 + 0.25) + 1.7 (0.4) = 1.59 t/m2 . 2- Determine The Total Factored Static Moment (Mo) : Mo = = = 13.83t.m
- 12. Design Of Reinforced Concrete Structures ii Two-Way Slabs 12 3- Distribute Mo into +ve and –ve moments : The total factored static moment was distributed according to Table "8.9" in your text book as shown in the following Figure.
- 13. Design Of Reinforced Concrete Structures ii Two-Way Slabs 13 4- Moments on the column Strip : Evaluate the constant and Evaluation of : α = . "For beam in direction of strip" For a/h=50/16=3.125 and b/h=30/16=1.875, f=1.4 (Fig. 8.14) α = = 4.07 Evaluation of : βt = . " For edge beam perpendicular to direction of strip" CA = = 208,909.55 CB = = 128,532.48 C = Max (CA or CB) = 207393.8 βt = = 0.97 After α and βt are calculated, factors for the moment of column strip can be found from Table 8.10, page 343 α = 4.07 =4.88 1 , βt between 0 and 2.5 , = 1.2 – Ve exterior moment Factor : =1 =1.2 =2 βt = 0 100 100 100 βt = 0.97 0.903 0.8797 0.7866 βt = 2.5 75 69 45
- 14. Design Of Reinforced Concrete Structures ii Two-Way Slabs 14 +Ve interior moment Factor : =1 =1.2 =2 75 69 45 -Ve interior moment Factor : =1 =1.2 =2 75 69 45
- 15. Design Of Reinforced Concrete Structures ii Two-Way Slabs 15 5- Moments on the Middle Strip: "The remain moment": 6- Moment On Beam : As α >1 …. Beam willl resist 85% of the column strip moment. 7- Moment On Remained Slab :
- 16. Design Of Reinforced Concrete Structures ii Two-Way Slabs 16 By: Eng. AbdUlla Taisir Al-Madhoun Eng. Nour Nagi Al-Hindi . Notes: For each value of moment, can be calculated, then . Widths to used for design and ρ calculations are : For the remained slab of column strip: b = 1.25-0.3=0.95m For half middle strip: b= 3.15-1.25=1.9m Beam = 0.3m Beam should be designed for shear, according to specifications of code ACI 318"13.6.8", and reported in page 344 of your text book.