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Plain & Reinforced Concrete-1 CE3601 Lecture # 13, 14 &15 13 th to 20 th April 2012 Flexural Analysis and Design of Beams (Ultimate Strength Design of.

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Presentation on theme: "Plain & Reinforced Concrete-1 CE3601 Lecture # 13, 14 &15 13 th to 20 th April 2012 Flexural Analysis and Design of Beams (Ultimate Strength Design of."— Presentation transcript:

1 Plain & Reinforced Concrete-1 CE3601 Lecture # 13, 14 &15 13 th to 20 th April 2012 Flexural Analysis and Design of Beams (Ultimate Strength Design of Beams)

2 Plain & Reinforced Concrete-1 Load Carried by the Beam Beam Supporting One-way Slab lxlx lxlx Exterior Beam Interior Beam Width of slab supported by interior beam = l x Width of slab supported by exterior beam = l x/2 + Cantilever width lyly ( l y/ l x > 2)

3 Plain & Reinforced Concrete-1 Load Carrie by the Beam Beam Supporting Two-way Slab ( l y/ l x≤ 2) lxlx lxlx lyly lyly Exterior Long Beam Interior Long Beam Exterior Short Beam Interior Short Beam 45 o

4 Plain & Reinforced Concrete-1 Load Carrie by the Beam Beam Supporting Two-way Slab ( l y/ l x≤ 2) contd… 45 o l x/2 Area of Square Shorter Beams For simplification this triangular load on both the sides is to be replaced by equivalent UDL, which gives same M max as for the actual triangular load.

5 Plain & Reinforced Concrete-1 Load Carried by the Beam Beam Supporting Two-way Slab ( l y/ l x≤ 2) contd… 45 o Equivalent Rectangular Area Factor of 4/3 convert this VDL into UDL. Equivalent width supported by interior short beam lxlx Equivalent width supported by exterior short beam Cantilever

6 Plain & Reinforced Concrete-1 Load Carried by the Beam Beam Supporting Two-way Slab ( l y/ l x≤ 2) contd… lxlx lxlx lyly Exterior Long Beam Supported Area l x/2 l y - l x

7 Plain & Reinforced Concrete-1 Load Carried by the Beam Beam Supporting Two-way Slab ( l y/ l x≤ 2) contd… Exterior Long Beam where Factor F converts trapezoidal load into equivalent UDL for maximum B.M. at center of simply supported beam. For Square panel R = 1 and F = 4/3

8 Plain & Reinforced Concrete-1 Load Carried by the Beam Beam Supporting Two-way Slab ( l y/ l x≤ 2) contd… Exterior Long Beam Equivalent width l x/2 l y - l x Equivalent width lyly + Cantilever (if present)

9 Plain & Reinforced Concrete-1 Load Carried by the Beam Beam Supporting Two-way Slab ( l y/ l x≤ 2) contd… Interior Long Beam Equivalent width l x/2 l y - l x Equivalent width lyly

10 Plain & Reinforced Concrete-1 Wall Load (if present) on Beam tw (mm) H (m) UDL on beam (kN/m)

11 Plain & Reinforced Concrete-1 Wall Load on the Lintel Equivalent UDL on lintel if height of slab above lintel is greater than 0.866L 0.866L 60 o L kN/m tw = wall thickness in “mm” L = Opening size in “m” If the height of slab above lintel is less than 0.866L Total Wall Load + Load from slab in case of load bearing wall UDL = (Equivalent width of slab supported) x (Slab load per unit area) = m x kN/m 2 = kN/m

12 Plain & Reinforced Concrete-1 Slab Load per Unit Area Top Roof Slab Thickness = 125 mm Earth Filling = 100 mm Brick Tiles = 38 mm Dead Load Self wt. of R.C. slab Earth Filling Brick Tiles 554 kg/m 2 Total Dead Load, W d =

13 Plain & Reinforced Concrete-1 Slab Load per Unit Area (contd…) Top Roof Live Load W L = 200 kg/m 2 Total Factored Load, W u

14 Plain & Reinforced Concrete-1 Slab Load per Unit Area (contd…) Intermediate Floor Slab Thickness = 150 mm Screed (brick ballast + 25% sand) = 75 mm P.C.C. = 40 mm Terrazzo Floor = 20 mm Dead Load Self wt. of R.C. slab Screed Terrazzo + P.C.C 633 kg/m 2 Total Dead Load, W d =

15 Plain & Reinforced Concrete-1 Slab Load per Unit Area (contd…) Intermediate Floor Live Load Occupancy Live Load = 250 kg/m 2 Moveable Partition Load = 150 kg/m 2 W L = = 400 kg/m 2 Total Factored Load, W u

16 Plain & Reinforced Concrete-1 Slab Load per Unit Area (contd…) Self Weight of Beam Service Self Wight of Beam = b x h x 1m x 2400 Kg/m Factored Self Wight of Beam kN/m Self weight of beam is required to be calculated in at the stage of analysis, when the beam sizes are not yet decided, so approximate self weight is computed using above formula.

17 Plain & Reinforced Concrete-1 Bar Bending Schedule Serial # Bar Designation Number of Bars Length of one Bar Dia of bar Weight of Steel Required Shape of Bar #10#13#15#19#25 1M-1#25 2S-1#10 3H-1#15 Σ Total weight of steel = 1.05Σ, 5% increase, for wastage during cutting and bending

18 Plain & Reinforced Concrete-1 Bar Bending Schedule (contd…) Bent-up Bar h 45 o h Additional Length = h Total Length = L h L

19 Plain & Reinforced Concrete-1 Bar Bending Schedule (contd…) 90 o -Standard Hooks (ACI) dbdb R = 4d b for bar up-to #25 R = 5d b for bar #29 & #36 R = 12d b L Total Length = L + 18d b For R 4d b

20 Plain & Reinforced Concrete-1 Bar Bending Schedule (contd…) 180 o -Standard Hooks (ACI) dbdb L Total Length = L + 20d b 4d b Same as 90 o hook

21 Plain & Reinforced Concrete-1 Bar Bending Schedule (contd…) Example: Prepare bar bending schedule for the given beam. Clear cover = 40 mm 2-#20 +1-# #20 2-# c/c Longitudinal Section 1-# 15

22 Plain & Reinforced Concrete-1 Bar Bending Schedule (contd…) Example: Prepare bar bending schedule for the given beam. Clear cover = 40 mm 180 c/c #10 1-# 15 2-#20 (M-2) (M-1) (S-1) (H-1) Cross Section

23 Plain & Reinforced Concrete-1 Bar Bending Schedule (contd…) Example: M-1 M-1 = x 228 – 2 x x (18 x 20) = 5096 M-2 h = 375 – 2 x 40 – 2 x 10 – 2 (15/2) = 260 h M-2 = x 228 – 2 x 40 + (0.414 x 260) x 2 = 5091 H-1 H-1 = x 228 – 2 x 40 = 4376

24 Plain & Reinforced Concrete-1 Bar Bending Schedule (contd…) Example: Prepare bar bending schedule for the given beam. Clear cover = 40 mm Number of Bars = 4000 / = 24 Round-up Shear stirrups a a = 375 – 2 x 40 – 10 = 285 mm b b = 228– 2 x 40 – 10 =138 mm Total length of S-1 = 2 ( x 10) = 1206 mm

25 Plain & Reinforced Concrete-1 Bar Bending Schedule Serial # Bar Designation Number of Bars Length of one Bar (m) Dia of bar Weight of Steel Bars Shape of Bar #10#15#20 1M M H S Σ Total Weight = 64 kg

26 Plain & Reinforced Concrete-1 Design of Singly Reinforced Beam by Strength Method (for flexure only) Data: Load, Span (SFD, BMD) fc’, fy, E s Architectural depth, if any Required: Dimensions, b & h Area of steel Detailing (bar bending schedule)

27 Plain & Reinforced Concrete-1 Design of Singly Reinforced Beam by Strength Method (contd…) Procedure: 1. Select reasonable steel ratio between ρmin and ρ max. Then find b, h and A s. 2. Select reasonable values of b, h and then calculate ρ and A s.

28 Plain & Reinforced Concrete-1 Design of Singly Reinforced Beam by Strength Method (contd…) 2. Using Trial Dimensions I. Calculate loads acting on the beam. II. Calculate total factored loads and plot SFD and BMD. Determine Vu max and Mu max. III. Select suitable value of beam width ‘b’. Usually between L/20 to L/15. preferably a multiple of 75mm or 114 mm. IV. Calculate d min. h min = d min + 60 mm for single layer of steel h min = d min + 75mm for double layer of steel Round to upper 75 mm

29 Plain & Reinforced Concrete-1 Design of Singly Reinforced Beam by Strength Method (contd…) V. Decide the final depth. For strength For deflection Architectural depth Preferably “h” should be multiple of 75mm. Recalculate “d” for the new value of “h”

30 Plain & Reinforced Concrete-1 Design of Singly Reinforced Beam by Strength Method (contd…) VI. Calculate “ρ” and “A s ”. Four methods Design Table Design curves Using trial Method a) b) c) d)

31 Plain & Reinforced Concrete-1 Design of Singly Reinforced Beam by Strength Method (contd…) VII. Check A s ≥ A s min. A s min = ρ min bd (ρ min = 1.4/fy to fc’ ≤ 30 MPa) VIII. Carry out detailing IX. Prepare detailed sketches/drawings. X. Prepare bar bending schedule.

32 Plain & Reinforced Concrete-1 Design of Singly Reinforced Beam by Strength Method (contd…) 1. Using Steel Ratio I. Step I and II are same as in previous method. III. Calculate ρ max and ρ min & select some suitable “ρ”. IV. Calculate bd 2 from the formula of moment V. Select such values of “b” and “d” that “bd 2 ” value is satisfied. VI. Calculate A s. VII. Remaining steps are same as of previous method.

33 Concluded


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