2.0 ANALYSIS AND DESIGN 2.2 STRUCTURAL ELEMENT BEAM Develop by :- NOR AZAH BINTI AIZIZ KOLEJ...
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Transcript of 2.0 ANALYSIS AND DESIGN 2.2 STRUCTURAL ELEMENT BEAM Develop by :- NOR AZAH BINTI AIZIZ KOLEJ...
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2.0 ANALYSIS AND DESIGN
2.2 STRUCTURAL ELEMENT
BEAMDevelop by :-
NOR AZAH BINTI AIZIZKOLEJ MATRIKULASI TEKNIKAL KEDAH
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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).
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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
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Examples of beams:-
i)Beam Slab Bridge
Bridge Over Sg. Muda, Kuala Muda
Guthrie Corridor Expressway Eleanor
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Types of beam
Primary Beam - Beam that supporting by column at the end
Secondary Beam- Beam that supporting by another beam at the end
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Types of beam
1.Identify primary beam and secondary beam.
A
1
2
CB
1a
4m 4m
2m
2m
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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 sideLy /Lx > 2 = one-way slabLy / x ≤ 2 = two-way slab
• Loads supported by precast concrete slab systems are distributed to beams in one direction only.
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One-way slab
Two-way slab
Two types of load distribution to beams
L y
L x
L y
L x
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Let’s do it now!!!!
A
1
2
CB
1a
5.5 m2.0 m
2.5 m
2.5 m
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 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.
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ANSWER
A
1
2
CB
1a
5.5 m2.0 m
2.5 m
2.5 m
Identify one way slab @ two way slab
Panel A-B/1-2LY/LX = 5 / 2 = 2.5 >2 :- one way slab
Panel B-C/1-1aLY/LX = 5.5 / 2.5 = 2.2 >2 :- one way slab
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ANSWER
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 mm
Self weight slab = 24 x 0.15 = 3.6 kN/m²Total characteristic dead load = 3.6 + 1 = 4.6 kN/m²Design load on slab, w = 1.4 gk + 1.6 qk
= 1.4 ( 4.6 ) + 1.6 ( 2.5 ) = 10.44 kN/m²
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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
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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
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Cross Section Detail
b
h d
b - width
d – depth
h – high
F
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Fcc = 0.405fcuAcc0.9 x
z= (d-0.9x/2)
0.87fy
0.45fcu b
Fst = 0.87 fy 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
AS – area of reinforcement cross section
M – Ultimate Moment
Equation∑Ma = 0Fcc (d-0.9x/2) – M = 0Fcc = Fst
Fcc = 0.405fcu Acc @ Fcc = 0.45fcu Acc
= 0.405 x fcu x bx = 0.45 x fcux 0.9xb
Fst = 0.87 fy As
section stress forceM
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Concrete compression
Acc = (0.9x) (125)
F cc = 0.45fcu x ACC
= 0.45fcu x
(0.9x)(125)
F st = 0.87 As
F cc
Acc
125mm0.9x
Fcc
Fst
Fst
Steel tension
0.9x
d
0.45fcu
0.87fy
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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
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6 mm
Factored load,G k = 2kN/m
h = 300mm
b = 125mm
BEAM DESIGN
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STEP 1 : Calculation of Moment
Moment at centre (max)
=WL2/ 8
= 2 x 62 /8
= 9kNm6 mm
gk = 2kN/m
9kNm
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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
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STEP 3 : Force Diagram
Fst
Fcc
d = 259mm
b = 125mm
As
∑Ma = 0
Fcc x ( d - 0.9x / 2) – M = 0
0.45fcu x Acc x (d - 0.9x / 2) – M = 0
z=(d-0.9x/2)
a
F cc
Fst
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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
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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
b = 125mm
As = 90.70 mm2
D = 8mm
:. size rebar to be provided is 2 T 8
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COLUMN
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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)
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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
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Design Column
A short reinforced concretecolumn 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 theconcrete 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 Asc2069 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
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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
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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