1. 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

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 i)Beam Slab Bridge Examples of beams:-
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. Types of beam 1. Identify primary beam and secondary beam. B C A 1 1a2 C B 1a 4m 2m
1. Identify primary beam and secondary beam.

7. 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.

8. 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

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/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

10. ANSWER Identify one way slab @ two way slab Panel A-B/1-2C 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

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

12. 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

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

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

15. 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

16. 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

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

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

20. 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

21. 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

22. 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

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 = 300mm
b = 125mm
As = mm2
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. 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)

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 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 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.