2.0 ANALYSIS AND DESIGN

2.2 STRUCTURAL ELEMENT

BEAMDevelop by :-

NOR AZAH BINTI AIZIZKOLEJ 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

1.Identify primary beam and secondary beam.

A

1

2

CB

1a

4m 4m

2m

2m

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.

One-way slab

Two-way slab

Two types of load distribution to beams

L y

L x

L y

L x

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.

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

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

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

b

h d

b - width

d – depth

h – high

F

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

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

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

6 mm

Factored load,G k = 2kN/m

h = 300mm

b = 125mm

BEAM DESIGN

STEP 1 : Calculation of Moment

Moment at centre (max)

=WL2/ 8

= 2 x 62 /8

= 9kNm6 mm

gk = 2kN/m

9kNm

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

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

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

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

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 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

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.

FOUNDATION DESIGN