2.0 ANALYSIS AND DESIGN

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

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

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 Ly/Lx L y= longer side , Lx= shorter side Ly /Lx > 2 = one-way slab Ly / x ≤ 2 = two-way slab • Loads supported by precast concrete slab systems are distributed to beams in one direction only.

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

9. A C B 1 2 1a 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/m² Floor thickness : 150 mm sketch 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. 2.5 m 2.5 m 2.0 m 5.5 m

10. A C B 1 2 1a ANSWER 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 2.5 m 2.5 m 2.0 m 5.5 m

11. ANSWER Concrete 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 = 3.6 + 1 = 4.6 kN/m² Design load on slab, w = 1.4 gk + 1.6 qk = 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 lx = 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 = wL2/ 8 = 10.44 (5) 2 / 8 = 32.63 kN/m

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

15. BEAM DESIGN b 0.45fcu Fcc = 0.405fcuAcc 0.9 x x d z= (d-0.9x/2) Where: f cu - Characteristic of concrete strength (30N/mm2) f y - Characteristic of reinforcement strength (460N/mm2) A – area of beam cross section AS – area of reinforcement cross section M – Ultimate Moment Equation ∑Ma = 0 Fcc (d-0.9x/2) – M = 0 Fcc = Fst Fcc = 0.405fcu Acc @ Fcc = 0.45fcu Acc = 0.405 x fcu x bx = 0.45 x fcux 0.9xb Fst = 0.87 fyAs M stress Fst = 0.87 fy As section force As  a 0.87fy

16. Concrete compression 0.45fcu 0.9x F cc 125mm 0.9x d Acc Fst Fcc Fst 0.87fy Steel tension Acc = (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 of fcu = 30N/mm2 for the concrete and fy = 460 N/mm2 for the steel. Calculate steel area (As) and size of rebar to be provided for the beam. 2kN/m 6m

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

19. STEP 1 : Calculation of Moment Moment at centre (max) =WL2/ 8 = 2 x 62 /8 = 9kNm gk = 2kN/m 6 mm 9kNm

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

21. STEP 3 : Force Diagram F cc z=(d-0.9x/2) Fst  a Fcc ∑Ma = 0 Fcc x ( d - 0.9x / 2) – M = 0 0.45fcu x Acc x(d - 0.9x / 2) – M = 0 d = 259mm As Fst b = 125mm

22. STEP 3 : Force Diagram 0.405 x 30 x 125 x x ( 259 – 0.9x / 2 ) – 9x106 = 0 1518.8x x (259 - 0.45x) – 9 x 106= 0 393369.2x - 683.46x2 - 9 x106= 0 683.46x2 – 393369.2x + 9x106 = 0 x = -b + b2-4ac x = 551.7mm @ 23.9mm Fcc = 0.405 x 30 x 23.9 x 125 = 36298N = 36.3kN 2a

23. Fcc= Fst 36298N = 0.87fy x As As= 36298 / 0.87(460) = 90.70 mm2 So size rebar A = Ωj2= Ω D2 / 4 = 90.70mm2 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 As = 90.70 mm2 b = 125mm D = 8mm :. size rebar to be provided is 2 T 8

24. COLUMN

25. COLUMN • 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 fcu Ac + 0.75 fy Asc fcu = characteristic concrete cube crushing strength fcu = area of concrete fy = characteristic yield stress of steel Asc = 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 (mm2) 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 103 = 0.4 ( 30 ) 3252 + 0.75 ( 460) Asc 801500 = 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 DESIGN • 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.

29. FOUNDATION DESIGN • 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.