Presentation on theme: "2.2 STRUCTURAL ELEMENT BEAM"— Presentation transcript:
1 2.2 STRUCTURAL ELEMENT BEAM 2.0 ANALYSIS AND DESIGN2.2 STRUCTURAL ELEMENTBEAMDevelop by :-NOR AZAH BINTI AIZIZKOLEJ MATRIKULASI TEKNIKAL KEDAH
2 BEAMA beam is a structural member subject to bending.(Flexural member)Its function carrying gravity load in the directionnormal to its axis, which results in bending momentand shear force.Bending occurs in member when a componentof load is applied perpendicular to member axis, and some distance from a support.Most beams span between two or more fixedpoints (support).
3 BEAM Three types of beams:- i) A Simply Supported Beams - both ends are supported by onepin and one rollerii) Cantilever Beams- one end is unsupported, but theother must rigidly built-in topprevent rotation.iii) A continuous Beams- beams with extra supports
4 BEAM i)Beam Slab Bridge Examples of beams:- Bridge Over Sg. Muda, Kuala MudaGuthrie 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. B C A 1 1a 2CB1a4m2m1. Identify primary beam and secondary beam.
7 DISTRIBUTION OF LOADS FROM SLAB TO BEAMS Loads from a slab are transferred to itssurrounding beams in either one-way@ two-way depend on the ratio Ly/LxL y= longer side , Lx= shorter sideLy /Lx > 2 = one-way slabLy / x ≤ = two-way slabLoads supported by precast concrete slab systemsare distributed to beams in one direction only.
8 BEAM Two types of load distribution to beams One-way slab Two-way slab L yL xOne-way slabTwo-way slabTwo types of load distribution to beams
9 Let’s do it now!!!! Concrete density : 24 kN/m3 Dead load characteristic: 1.0 kN/m² (excluding the slab self-weight)Live load characteristic: 2.5 kN/m²Floor thickness : 150 mmsketch the floor tributary areas all for beams.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.A12CB1a5.5 m2.0 m2.5 m
10 ANSWER Identify one way slab @ two way slab Panel A-B/1-2 CB1a5.5 m2.0 m2.5 mIdentify one way two way slabPanel A-B/1-2LY/LX = 5 / 2 = 2.5 >2:- one way slabPanel B-C/1-1aLY/LX = 5.5 / 2.5 = 2.2 >2
11 ANSWERConcrete density : 24 kN/m3 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 = = 4.6 kN/m² Design load on slab, w = 1.4 gk qk = 1.4 ( 4.6 ) ( 2.5 ) = kN/m²
12 ANSWERDesign load on beam A/1-2 ( kN/m) = 0.5 x w x lx = 0.5 x x 2 = kN/m Design load on beam A/1-2 ( kN) = kN/m x 5m = 52.2 kN
13 ANSWERMaximum shear force V = wL/2 = x 5 /2 = 26.1 kN Maximum bending moment M = wL2/ 8 = (5) 2 / 8 = kN/m
15 BEAM DESIGN x d z= (d-0.9x/2) M section stress force Fcc = 0.405fcuAcc0.9 xz= (d-0.9x/2)0.87fy0.45fcubFst = 0.87 fy AsAsxdaMsectionstressforceWhere:f cu - Characteristic of concrete strength (30N/mm2)f y - Characteristic of reinforcement strength(460N/mm2)A – area of beam cross sectionAS – area of reinforcement cross sectionM – Ultimate MomentEquation∑Ma = 0Fcc (d-0.9x/2) – M = 0Fcc = FstFcc = 0.405fcu Acc @ Fcc = 0.45fcu Acc= x fcu x bx = 0.45 x fcux 0.9xbFst = 0.87 fy As
16 Concrete compression Fst Steel tension 0.9x d 0.45fcu 0.87fy Acc Fcc 125mm0.9xFccFstF ccAcc = (0.9x) (125)F cc = 0.45fcu x ACC= 0.45fcu x (0.9x)(125)F st = 0.87 As
17 Example: The beam 6m long shown in Figure with ultimate load of 2kN/m has characteristic material strengths offcu = 30N/mm2 for the concrete and fy = 460 N/mm2for the steel.Calculate steel area (As) and size of rebar to be provided for the beam.2kN/m6m
19 STEP 1 : Calculation of Moment Moment at centre (max) =WL2/ 8 = 9kNm6 mmgk = 2kN/m9kNm
20 d = h - cover – Φ link – Φ rebar = 300 – 25 – 10 – 12/2 = 259 mm STEP 2 : Calculation of dd = h - cover – Φ link – Φ rebar= 300 – 25 – 10 – 12/2= 259 mmd = mmh = 300mmb = 125mm
21 STEP 3 : Force Diagram ∑Ma = 0 Fcc x ( d - 0.9x / 2) – M = 0 FstFccd = 259mmb = 125mmAsz=(d-0.9x/2)aF ccFst∑Ma = 0Fcc x ( d - 0.9x / 2) – M = 00.45fcu x Acc x (d - 0.9x / 2) – M = 0
22 STEP 3 : Force Diagram0.405 x 30 x 125 x x ( 259 – 0.9x / 2 ) – 9x106 = 01518.8x x ( x) – 9 x 106 = 0x x2 - 9 x106 = 0683.46x2 – x + 9x106 = 0x = -b + b2-4acx = 23.9mmFcc = x 30 x 23.9 x 125= 36298N= 36.3kN2a
23 :. size rebar to be provided is 2 T 8 Fcc= Fst 36298N = 0.87fy x As As= / 0.87(460) = mm2 So size rebar A = Ωj2= Ω D2 / 4 = 90.70mm2 A = /2 = 45.35mm D = x 4 / Ω D = 7.6 mm for 2 bar So size rebar for the beam is 8mm.h = 300mmb = 125mmAs = mm2D = 8mm:. size rebar to be provided is 2 T 8
25 COLUMN Vertical elements which are normally loaded in compression.(compression member)2 types :-i) Strut – small member in a framed structureii) Column – larger member as a main support for a beam in a buildingAxial 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 fcu = characteristic concrete cube crushing strength DESIGN COLUMNUltimate compressive load capacity,N = sum of the strengths of both the concrete and steel components.N= 0.4 fcu Ac fy Ascfcu = characteristic concrete cube crushing strengthfcu = area of concretefy = characteristic yield stress of steelAsc = area of steelTable 1Diameters and areas of reinforcing barsBar dia.(mm)C/s area (mm2)
27 Design Column A short reinforced concrete column is to support the following axial loads :characteristic dead load : 758 kNcharacteristic live load : 630 kNIf the column is to measure325 mm x 325 mm and theconcrete characteristic strength is30 N/mm2, determine therequired size of high yieldreinforcing bars.Design load = 1.4 Gk Qk = 1.4 (758) (630) = 2069 KN N = 0.4 fcu Ac fy Asc 2069 x 103 = 0.4 ( 30 ) ( 460) Asc = 0.75 x 460 x Asc Asc = 2323 mm2 Consider 4 bars are used: Asc = 2323 mm2 4 = 581 mm2 From Table 1 ; area 32 mm dia. Bar = 804 mm2 Size of rebar required = 4T32
28 FOUNDATION DESIGNThe 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.
29 FOUNDATION DESIGNThe primary functional requirement of a foundation is strength and stability.Strength and stabilityThe combined, dead, imposed and wind loads on a buildingmust be transmitted to the ground safely,without causing deflection or deformation of the buildingor movement of the ground that wouldimpair the stability of the building and/or neighboring structures.Foundations should also be designed and constructedto resist any movements of the subsoil.