# 2.2 STRUCTURAL ELEMENT BEAM

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

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).

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

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

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

Types of beam 1. Identify primary beam and secondary beam. B C A 1 1a
2 C B 1a 4m 2m 1. Identify primary beam and secondary 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 ≤ = two-way slab Loads supported by precast concrete slab systems are distributed to beams in one direction only.

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

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

ANSWER Identify one way slab @ two way slab Panel A-B/1-2
C B 1a 5.5 m 2.0 m 2.5 m Identify one way 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

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 = = 4.6 kN/m² Design load on slab, w = 1.4 gk qk = 1.4 ( 4.6 ) ( 2.5 ) = kN/m²

ANSWER Design 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

ANSWER Maximum shear force V = wL/2 = x 5 /2 = 26.1 kN Maximum bending moment M = wL2/ 8 = (5) 2 / 8 = kN/m

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

BEAM DESIGN x d z= (d-0.9x/2)  M section stress force
Fcc = 0.405fcuAcc 0.9 x z= (d-0.9x/2) 0.87fy 0.45fcu b Fst = 0.87 fy As As x d a M section stress force 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 = x fcu x bx = 0.45 x fcux 0.9xb Fst = 0.87 fy As

Concrete compression Fst Steel tension 0.9x d 0.45fcu 0.87fy Acc Fcc
125mm 0.9x Fcc Fst F cc Acc = (0.9x) (125) F cc = 0.45fcu x ACC = 0.45fcu x (0.9x)(125) F st = 0.87 As

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

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

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

d = h - cover – Φ link – Φ rebar = 300 – 25 – 10 – 12/2 = 259 mm
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

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

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

:. 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 = 300mm b = 125mm As = mm2 D = 8mm :. size rebar to be provided is 2 T 8

COLUMN

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)

fcu = characteristic concrete cube crushing strength
DESIGN COLUMN Ultimate compressive load capacity, N = sum of the strengths of both the concrete and steel components. N= 0.4 fcu Ac 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) C/s area (mm2)

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 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

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.

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.
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