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2.0 ANALYSIS AND DESIGN 2.2 STRUCTURAL ELEMENT BEAM Develop by :- NOR AZAH BINTI AIZIZ KOLEJ MATRIKULASI TEKNIKAL KEDAH.

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Presentation on theme: "2.0 ANALYSIS AND DESIGN 2.2 STRUCTURAL ELEMENT BEAM Develop by :- NOR AZAH BINTI AIZIZ KOLEJ MATRIKULASI TEKNIKAL KEDAH."— Presentation transcript:

1 2.0 ANALYSIS AND DESIGN 2.2 STRUCTURAL ELEMENT BEAM Develop by :- NOR AZAH BINTI AIZIZ KOLEJ MATRIKULASI TEKNIKAL KEDAH

2 BEAM  A beam is a structural member subject to bending.(Flexural member)  Its function carrying gravity load in the direction normal to its axis, which results in bending moment and shear force.  Bending occurs in member when a component of load is applied perpendicular to member axis, and some distance from a support.  Most beams span between two or more fixed points (support).

3 Three types of beams:- i)A Simply Supported Beams - both ends are supported by one pin and one roller ii)Cantilever Beams - one end is unsupported, but the other must rigidly built-in top prevent rotation. iii)A continuous Beams - beams with extra supports

4 Examples of beams:- i)Beam Slab Bridge Bridge Over Sg. Muda, Kuala Muda Guthrie Corridor Expressway Eleanor

5 Types of beam  Primary Beam - Beam that supporting by column at the end  Secondary Beam - Beam that supporting by another beam at the end

6 Types of beam 1.Identify primary beam and secondary beam. A 1 2 C B 1a 4m 2m

7 BEAM DISTRIBUTION OF LOADS FROM SLAB TO BEAMS Loads from a slab are transferred to its surrounding beams in either one-way @ two-way depend on the ratio L y /L x L y = longer side, L x = shorter side L y /L x > 2 = one-way slab L y / x ≤ 2 = two-way slab Loads supported by precast concrete slab systems are distributed to beams in one direction only.

8 One-way slab Two-way slab Two types of load distribution to beams L y L x L y L x

9 Let’s do it now!!!! A 1 2 C B 1a 5.5 m 2.0 m 2.5 m Concrete density: 24 kN/m 3 Dead load characteristic: 1.0 kN/m² (excluding the slab self-weight) Live load characteristic: 2.5 kN/m² Floor thickness: 150 mm 1)sketch the floor tributary areas all for beams. 2)calculate the ultimate design load supported by beam A/1-2 in kN/m considering all floor loadings. Ignoring selfweight of beam. 3) Calculate the maximum shear force and maximum bending moment.

10 ANSWER A 1 2 C B 1a 5.5 m 2.0 m 2.5 m Identify one way slab @ two way slab Panel A-B/1-2 LY/LX = 5 / 2 = 2.5 >2 :- one way slab Panel B-C/1-1a LY/LX = 5.5 / 2.5 = 2.2 >2 :- one way slab

11 ANSWER Concrete density: 24 kN/m 3 Dead load characteristic: 1.0 kN/m² (excluding the slab self-weight) Live load characteristic: 2.5 kN/m² Floor thickness: 150 mm Self weight slab = 24 x 0.15 = 3.6 kN/m² Total characteristic dead load = 3.6 + 1 = 4.6 kN/m² Design load on slab, w = 1.4 g k + 1.6 q k = 1.4 ( 4.6 ) + 1.6 ( 2.5 ) = 10.44 kN/m²

12 ANSWER Design load on beam A/1-2 ( kN/m) = 0.5 x w x l x = 0.5 x 10.44 x 2 = 10.44 kN/m Design load on beam A/1-2 ( kN) = 10.44 kN/m x 5m = 52.2 kN

13 ANSWER Maximum shear force V = wL/2 = 10.44 x 5 /2 = 26.1 kN Maximum bending moment M = wL 2 / 8 = 10.44 (5) 2 / 8 = 32.63 kN/m

14 Cross Section Detail b h d b - width d – depth h – high F

15 F cc = 0.405f cu A cc 0.9 x z= (d-0.9x/2) 0.87fy 0.45fcu b F st = 0.87 f y A s AsAs x d a  Where: f cu - Characteristic of concrete strength (30N/mm2) f y - Characteristic of reinforcement strength (460N/mm2) A – area of beam cross section A S – area of reinforcement cross section M – Ultimate Moment Equation ∑M a = 0 F cc (d-0.9x/2) – M = 0 F cc = F st F cc = 0.405f cu A cc @ F cc = 0.45f cu A cc = 0.405 x f cu x b x = 0.45 x f cu x 0.9 x b F st = 0.87 f y A s section stress force M

16 Concrete compression A cc = (0.9 x) ( 125) F cc = 0.45f cu x A CC = 0.45f cu x (0.9 x )(125) F st = 0.87 A s F cc A cc 125mm 0.9 x F cc F st Steel tension 0.9x d 0.45f cu 0.87f y

17 Example: The beam 6m long shown in Figure with ultimate load of 2kN/m has characteristic material strengths of f cu = 30N/mm 2 for the concrete and f y = 460 N/mm 2 for the steel. Calculate steel area (A s ) and size of rebar to be provided for the beam. 2kN/m 6m

18 6 mm Factored load,G k = 2kN/m h = 300mm b = 125mm BEAM DESIGN

19 STEP 1 : Calculation of Moment Moment at centre (max) =WL 2 / 8 = 2 x 6 2 /8 = 9kNm 6 mm gk = 2kN/m 9kNm

20 h = 300mm b = 125mm STEP 2 : Calculation of d d = h - cover – Φ link – Φ rebar = 300 – 25 – 10 – 12/2 = 259 mm d = mm

21 STEP 3 : Force Diagram F st F cc d = 259mm b = 125mm AsAs ∑M a = 0 F cc x ( d - 0.9x / 2) – M = 0 0.45f cu x A cc x (d - 0.9x / 2) – M = 0 z=(d-0.9x/2) a F cc F st 

22 STEP 3 : Force Diagram 0.405 x 30 x 125 x x ( 259 – 0.9x / 2 ) – 9x10 6 = 0 1518.8x x (259 - 0.45x) – 9 x 10 6 = 0 393369.2x - 683.46x 2 - 9 x10 6 = 0 683.46x 2 – 393369.2x + 9x10 6 = 0 x = -b + b 2 -4ac x = 551.7mm @ 23.9mm F cc = 0.405 x 30 x 23.9 x 125 = 36298N = 36.3kN 2a

23 F cc = F st 36298N = 0.87f y x A s A s = 36298 / 0.87(460) = 90.70 mm 2 So size rebar A = Ωj 2 = Ω D 2 / 4 = 90.70mm 2 A = 90.70 /2 = 45.35mm D = 45.35 x 4 / Ω D = 7.6 mm for 2 bar So size rebar for the beam is 8mm. h = 300mm b = 125mm A s = 90.70 mm2 D = 8mm :. size rebar to be provided is 2 T 8

24 COLUMN

25  Vertical elements which are normally loaded in compression.(compression member)  2 types :- i) Strut – small member in a framed structure ii) Column – larger member as a main support for a beam in a building  Axial loaded compression members can fail in two principal ways: i) short fat member fail by crushing or splitting of the material. ( strength criterion) ii) long thin members fail by sideways buckling. (stiffness criterion)

26 DESIGN COLUMN Ultimate compressive load capacity, N = sum of the strengths of both the concrete and steel components. N= 0.4 f cu A c + 0.75 f y A sc f cu = characteristic concrete cube crushing strength f cu = area of concrete f y = characteristic yield stress of steel A sc = area of steel Table 1 Diameters and areas of reinforcing bars Bar dia.(mm)6 8 10 12 16 20 25 32 40 C/s area (mm 2 )28 50 79 113 201 314 491 804 1256

27 Design Column A short reinforced concrete column is to support the following axial loads : characteristic dead load : 758 kN characteristic live load : 630 kN If the column is to measure 325 mm x 325 mm and the concrete characteristic strength is 30 N/mm2, determine the required size of high yield reinforcing bars. Design load = 1.4 Gk + 1.6 Qk = 1.4 (758) + 1.6 (630) = 2069 KN N= 0.4 fcu Ac + 0.75fy Asc 2069 x 10 3 = 0.4 ( 30 ) 325 2 + 0.75 ( 460) Asc 801500 = 0.75 x 460 x Asc Asc = 2323 mm 2 Consider 4 bars are used: Asc = 2323 mm 2 4 = 581 mm2 From Table 1 ; area 32 mm dia. Bar = 804 mm 2 Size of rebar required = 4T32

28  The foundation of a building is that part of walls, piers and columns in direct contact with, and transmitting loads to, the ground.  The building foundation is sometimes referred to as the artificial foundation, and the ground on which it bears as the natural foundation. FOUNDATION DESIGN

29  The primary functional requirement of a foundation is strength and stability. Strength and stability  The combined, dead, imposed and wind loads on a building must be transmitted to the ground safely, without causing deflection or deformation of the building or movement of the ground that would impair the stability of the building and/or neighboring structures.  Foundations should also be designed and constructed to resist any movements of the subsoil. FOUNDATION DESIGN


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