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).
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
Examples of beams:-
i)Beam Slab Bridge
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
C
B
A
1
2m
1a
2m
2
4m
4m
1. Identify primary beam and secondary beam.
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.
Ly
Lx
Ly
Lx
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
B
A
C
1
2.5 m
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.
1a
2.5 m
2
2.0 m
5.5 m
ANSWER
B
A
C
Identify one way slab @
two way slab
1
2.5 m
1a
Panel A-B/1-2
LY/LX = 5 / 2 = 2.5 >2
:- one way slab
2.5 m
Panel B-C/1-1a
LY/LX = 5.5 / 2.5 = 2.2 >2
:- one way slab
2
2.0 m
5.5 m
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
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
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
Cross Section Detail
F
b
b - width
h
d
d – depth
h – high
b
0.45fcu
x
Fcc = 0.405fcuAcc
0.9 x
d
z= (d-0.9x/2)
Fst = 0.87 fy As
As
0.87fy
section

M
force
a
stress
Where:
Equation
∑Ma = 0
Fcc (d-0.9x/2) – M = 0
Fcc = Fst
f
cu
- Characteristic of concrete strength (30N/mm2)
f
y
- Characteristic of reinforcement strength
(460N/mm2)
Fcc = 0.405fcu Acc
= 0.405 x fcu x bx
Fst = 0.87 fy As
@ Fcc = 0.45fcu Acc
A – area of beam cross section
= 0.45 x fcux 0.9xb
AS – area of reinforcement cross section
M – Ultimate Moment
Concrete compression
0.45fcu
0.9x
F cc
125mm
0.9x
d
Fst
Acc
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
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
Factored load,G k = 2kN/m
6 mm
h = 300mm
b = 125mm
STEP 1 : Calculation of Moment
Moment at centre (max)
=WL2/ 8
gk = 2kN/m
= 2 x 62 /8
= 9kNm
6 mm
9kNm
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
F cc
Fcc
d = 259mm
Fst
z=(d-0.9x/2)
As
Fst
b = 125mm
∑Ma = 0
Fcc x ( d - 0.9x / 2) – M = 0
0.45fcu x Acc x (d - 0.9x / 2) – M = 0
a 
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
2a
x = 551.7mm @ 23.9mm
Fcc = 0.405 x 30 x 23.9 x 125
= 36298N
= 36.3kN
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.
:. size rebar to be provided is 2 T 8
h = 300mm
As = 90.70 mm2
b = 125mm
D = 8mm
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)
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
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:
= 2323 mm2
4
= 581 mm2
From Table 1 ; area 32 mm dia. Bar = 804 mm2
Asc
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.