UNIT 22- Lecture 02

8/11/2019 UNIT 22- Lecture 02

1/28

UNIT 22: STRUCTURAL BEHAVIOUR AND DETAILING

Learning outcomesLO1- Ability to determine the properties of structural materials

LO2- Understanding of fundamental structural concepts

LO3- Analysis of statically determinate structures

LO4- Design of structural elements

LO5- Detailing of structural elements

LinksUnit 26 : Properties and performance of construction materialsUnit 34 : Structural analysis and design

10/10/2014

Prepared by: Eng. Chamil Duminda MahagamageB.Sc.Eng (Hons), C Eng, MIE(SL)

1International College of Business and Technology

M/601/1282 - Lecture Note: 02

8/11/2019 UNIT 22- Lecture 02

2/28

10/10/2014




Loads acting on structures

Dead Loads (DL)

The loads which constant in magnitude and fixed in location through out the lifetime of thestructure. Major part is the weight of structure itself and all the other permanent

construction including services of a permanent nature.

The characteristic dead loads can be estimated using schedule of weights of building

materials given in BS 648 (Table 2.1), ReynoldsHandbook or manufacturersliterature.

Symbols

gk -: Uniformly distributed characteristic dead loads

Gk-: Total characteristic dead loads

Gk

8/11/2019 UNIT 22- Lecture 02

3/28

10/10/2014




8/11/2019 UNIT 22- Lecture 02

4/28

10/10/2014




Live loads(LL), Imposed loads (IL)

The load assumed to be produced by the intended occupancy or use, including the weight of

movable partitions, distributed, concentrated, impact & inertia loads. Their magnitude and

distribution any given time are uncertain and even their maximum intensities throughoutthe lifetime of the structure are not known with precision.

BS 6399: Part 1: 1996 Code of Practice for Dead and Imposed Loads gives typical

characteristic imposed floor loads for different classes of structure.

Symbols

qk -: Uniformly distributed characteristic live loads

Qk-: Total characteristic live loads

Qk

8/11/2019 UNIT 22- Lecture 02

5/28

10/10/2014




8/11/2019 UNIT 22- Lecture 02

6/28

8/11/2019 UNIT 22- Lecture 02

7/28

10/10/2014




8/11/2019 UNIT 22- Lecture 02

8/28

10/10/2014




Environmental Loads

Consist mainly Snow load, Wind pressure and suction, Earthquake load (i.e inertia forces

caused by earthquake motions), Soil pressures on subsurface portions of structure, loads of

possible ponding of rain water on flat surfaces and forces caused by temperaturedifferentials. These loads are uncertain both magnitude & distribution.

Wind Loads (WL)

Wind pressure can either add to the other gravitational forces acting on the structure or

equally well, exert suction or negative pressures on the structure. Under particular

situations, the latter may well lead to critical conditions and must be considered in the

design.The characteristic wind loads acting on a structure can be assessed in accordance with the

recommendations given in CP3: Chapter V: Part 2: 1972 Wind Loads or BS 6399 : Part 2:

1997 Code of Practice for Wind Loads.

Symbols

wk -: Uniformly distributed characteristic live loads

Wk-: Total characteristic live loads

8/11/2019 UNIT 22- Lecture 02

9/28

10/10/2014




Estimation of Wind Loads

1. The basic wind speed Vdepends on the location in the country.

2. The design wind speed Vs= VS1S2S3

where,

S1 -: topography factor (normally taken as 1)

S2 -: factor depends on ground roughness, building size and height

above the ground

S3 -: statistical factor (normally taken as 1)

3. The dynamic pressure q = 0.613Vs2N/m2(SI units)is the pressure on a surface normal tothe wind and is modified by the dimensions of the building and by openings in the building

4. The wind force on a surface F = (Cpe-Cpi)qA

where,

Cpe-: external pressure coefficient

Cpi-: internal pressure coefficientA -: area of the surface

5. The wind load on a building as a whole F = CfqAe

where,

Cf-: force coefficient , Ae-: effective frontal area of the building

8/11/2019 UNIT 22- Lecture 02

10/28

10/10/2014




8/11/2019 UNIT 22- Lecture 02

11/28

10/10/2014




8/11/2019 UNIT 22- Lecture 02

12/28

10/10/2014




8/11/2019 UNIT 22- Lecture 02

13/28

10/10/2014




8/11/2019 UNIT 22- Lecture 02

14/28

10/10/2014




8/11/2019 UNIT 22- Lecture 02

15/28

10/10/2014




DESIGN METHODS

Relationship between stress and strength

1. Permissible stress design

In permissible stress design, sometimes referred to as modular ratio or elastic design, the

stresses in the structure at working loads are not allowed to exceed a certain proportion of

the yield stress of the construction material, i.e. the stress levels are limited to the elastic

range.

2. Load factor design

Load factor or plastic design was used to take account of the behaviour of the structure

once the yield point of the construction material had been reached. This approach involved

calculating the collapse load of the structure. The working load was derived by dividing the

collapse load by a load factor.

8/11/2019 UNIT 22- Lecture 02

16/28

10/10/2014




3. Limit State Design

The limit state design can be seen as a compromise between the permissible and load factor

methods. It is in fact a more comprehensive approach which takes into account both

methods in appropriate ways. Most modern structural codes of practice are now based onthe limit state approach such as,

BS 8110 for concrete BS 5950 for structural steelwork

BS 5400 for bridges BS 5628 for masonry

Code of practice for design in timber, BS5268 and old structural steelwork code, BS 449 are

based on permissible stress designs.

Ultimate limit state

The whole structure or its elements should not collapse, overturn or buckle when subjected

to the design loads. Considerations are,

Strength

The structure must be designed to carry the most severe combination of loads to which it is

subjected. The sections of the elements must be capable of resisting axial loads, shears andmoments derived from the analysis. The design is made for ultimate loads and design

strengths of materials with partial safety factors applied to loads and material strengths.

8/11/2019 UNIT 22- Lecture 02

17/28

10/10/2014




Stability

The layout should be such as to give a stable and robust structure. Overall stability should

ensure compatibility of design and details of parts and components. The structure should be

such as to transmit all loads, dead, imposed and wind, safely to the foundations.

Robustness

Damage to a small area or failure of a single element should not cause collapse of a major

part of a structure. This means that the design should be resistant to progressive collapse.

Serviceability limit state

The structure should not become unfit for use due to excessive deflection, cracking or

vibration. Considerations are,

Deflection

The deformation of the structure should not adversely affect its efficiency or appearance.

Cracking

Cracking should be kept within reasonable limits by correct detailing

8/11/2019 UNIT 22- Lecture 02

18/28

10/10/2014




Load factors and load combinations

8/11/2019 UNIT 22- Lecture 02

19/28

10/10/2014




Load Transfer

Fig: Sequence of load transfer between element of a structure

Support Conditions

Pinned Connection

In this connection joint allows attached member to rotate freely but does not allow

translation in any direction. Consequently, the joint cannot provide moment resistance but

can provide force resistance in any direction.

8/11/2019 UNIT 22- Lecture 02

20/28

10/10/2014




Roller Connection

In addition to rotation, this connection also allows attached member to translate freely

parallel to the surface of the support, i.e. does not provide any force resistance parallel to

the surface of the support. However, the joint resists translations in the direction

perpendicular to the surface of the support.

Fixed Connection

This connection completely restrains rotations and translations of the attached members in

any direction. Consequently, the joint can provide moment and force resistances in any

direction.

Important-: For an object to be stable in equilibrium, the supports must be capable of

providing specific minimum number of force restraints. Ex: for a simple beam subjected to

the vertical and horizontal forces, the support must provide three force restraints for its

equilibrium corresponding to three conditions of equilibrium namely,

Fx= 0,Fy= 0,Mz= 0

8/11/2019 UNIT 22- Lecture 02

21/28

10/10/2014




8/11/2019 UNIT 22- Lecture 02

22/28

10/10/2014




Problems

1. Calculate the self weight of a reinforced concrete beam of breadth 300mm, depth

600mm and length 6m.2. A composite floor consisting of a 150mm thick reinforced concrete slab supported on

steel beams spanning 5m and spaced at 3m centers is to be designed to carry an

imposed load of 3.5 kN/m2. Assuming that the unit mass of the steel beams is 50kg/m

run, calculate the design loads on a typical internal beam.

3. The floor shown below with an overall depth of 225mm is to be designed to carry an

imposed load of 3 kN/m2 plus floor finishes and ceiling loads of 1kN/m2. Calculate thedesign loads acting on beams B1-C1, B2-C2 and B1-B3 and columns B1 and C1. Assume

that all the column heights are 3m and that the beam and column dimensions are

225mmx350mm and 250mmx250mm respectively.

4. The cross section of a reinforced concrete building is shown in figure 4.1 below. The

frames are at 4.5m centers, the length of the building is 36m. Assuming the building is

to be constructed in center of the Colombo city, calculate the wind pressure acting on anend frame and a mid frame.

8/11/2019 UNIT 22- Lecture 02

23/28

10/10/2014




5. Determine the unknown support reactions RAand RB in the body due to applied load as

shown in Figure 5.1.

4m

4m

5.5m

8m6m

Figure 4.1

2a a2a

5w

2w

Figure 5.1

Lecture Note 02 Contd

8/11/2019 UNIT 22- Lecture 02

24/28

10/10/2014




5. Determine the unknown support reactions RAand RB in the body due to applied load as

shown in Figure 5.1.

4m

4m

5.5m

8m6m

Figure 4.1

2a a2a

5w

2w

Figure 5.1

Lecture Note 02 Contd

8/11/2019 UNIT 22- Lecture 02

25/28

10/10/2014




STRUCTURAL THEORY

Classification based on the nature of internal forces

1. Axial force member

A member that is subjected either to axial tensile or axial compressive force is referred to

as an axial force member. Ex: a cable and truss member

2. Bending and shear resisting member

Members are subjected predominantly to bending or flexural actions. Ex: a member

subjected to loads transverse to its length.3. Members subjected to torsion

Members are subjected predominantly to torsion or twisting actions. Ex : a shaft

transmitting motion from one shaft to other.

4. Members subjected to a combined action

Members are subjected to any combination of axial force, bending moment, shear force

and torsion

8/11/2019 UNIT 22- Lecture 02

26/28

10/10/2014




Basic rigid elements

1. Beams and Columns

Beams are generally horizontal, which carry loads applied transversely to their lengths

and transfer them to the supporting vertical columns or other supports. The beams bendunder transverse loads and are said to carry loads by bending. The elements carrying axial

compressive forces termed struts, when vertical they are termed columns.

2. Frames

A framed object or structure is made by assembling beam and column elements with rigid

joints.

3. Trusses

The truss is composed of short and straight discrete elements arranged in to triangulated

patterns. The truss is non-rigid, but it maintains its shape as a result of the exact way the

individual line elements are positioned relative to one another.

4. Arches

An arch is a curved line-forming structural member spanning between two points and

carry the loads to the supports while being subjected predominantly to axial compression.

5. Walls and Plates

These are rigid surface elements. A load-bearing wall can typically carry both vertically

and laterally acting loads along its length.

8/11/2019 UNIT 22- Lecture 02

27/28

10/10/2014




Internal Forces

1. Tension force

Tension force tend to pull an element apart. The strength of a tension member is

generally independent of its length and tension stresses are uniformly distributed across thecross section of the member.

2. Compression forces

Compression forces tend to crush or buckle the element. Short members tend to crush and

have higher strength compared to a tension member. The load carrying capacity of a long

member, however, decreases with the increase in the length. The long compression

members may become unstable and may suddenly snap out from beneath the load atcertain critical load levels. This phenomenon is called buckling. Because of this buckling

phenomenon, long compression members are not capable of carrying vey high loads.

3. Bending force

Bending force is a force state associate with bending of a member. The bending action

causes fibres on one face of the member to elongate, and hence are in tension, and fibres on

the opposite face to compress.4. Shearing force

Shearing force is a force state associated with the action of opposing forces that tend to

cause one part of the member to slide with respect to the adjacent part.

8/11/2019 UNIT 22- Lecture 02

28/28

10/10/2014

Prepared by: Eng. Chamil Duminda MahagamageB Sc Eng (Hons) C Eng MIE(SL)



5. Torsion

Torsion is a twisting action. Both tension and compression stresses are normally

developed in the member subjected to torsion.

6. Bearing stressesBearing stresses exist at the interface between two members when forces are transferred

from one member to another. They act perpendicular to contact surfaces. The bearing

stresses are also developed at the ends of beams where they rest on the walls.

Idealization of Structures for Analysis

The primary aim of the analysis is to determine the reactions, internal forces and

deformation at any point of the given body caused by applied loads and forces. To achieve

this objective, it becomes necessary to idealize a body in a simplified form emendable to

analysis procedure. The members are normally represented by their centroidal axis.

Conditions of equilibrium

A structure in general is subjected to a set of forces which include external or applied forces,

internal forces or reactions that are developed within the body at connection points and

gravity forces caused by the mass of the elements. The structure must be in the state of

static equilibrium with respect to these forces.

i) Translational equilibrium Fx=0, Fy=0, Fz=0

ii) Rotational equilibrium Mx=0, My=0, Mz=0

Documents

UNIT 22- Lecture 02