Structural Design (Portal Frame - Plastic Design)

TABLE OF CONTENTS

1. INTRODUCTION 21. Introduction 31.1. Lecture Objectives. 41.2. Individual (student) Objectives. 4

2. PRODUCT DESIGN SPECIFICATION 52. Product Design Specification (PDS). 6

3. CONCEPT DESIGN 133.1. Concept 1. 143.2. Concept 2. 153.3. Concept 3. 163.4. Concept 4. 173.5. Concept 5. 18

4. Design Loads Acting On Structures 214.1. Dead Loads. 24

4.1.1. Dead loads calculations. 244.2. Imposed Loads. 30

4.2.1. Snow loads calculations. 314.3. Wind Loads. 33

4.3.1. Wind Load Coefficient from BS6399-2. 334.3.1.2. External Pressure Coefficients for walls. 364.3.1.3. External Pressure Coefficient for Roofs. 394.3.1.4. Internal Pressure Coefficient. 44

4.3.2. Wind Load Calculations. 45

5. STRUCTURE ANALYSIS 555.1. Verify Appropriate Section Properties. 58

5.1.1 Plastic Moment under Dead Load. 585.1.2. Plastic Moment under Dead and Imposed Load. 695.1.3. Plastic Moment under Dead, Imposed and Wind Load. 75

5.1.3.1. Suction wind pressure on windward roof. 755.1.3.2. Pressing wind pressure on windward roof. 82

5.2. Structure Stability. 905.2.1. Resistance to Lateral-Torsional Buckling. 90

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5.2.2. Segment Adjacent to Plastic Hinge. 975.2.3. Haunch and Stanchion Stability. 995.2.4. Sway Stability. 100

6. DRAWING 107

7. EVALUATION AND DISCUSSION 116

8. FUTURE DEVELOPMENT 119

9. REFERENCES 121

10. APPENDIX 126

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1. Introduction

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1. INTRODUCTION

This project is raised with the objective to design a single-storey warehouse according to

plastic design criteria and British Standards specifications where is appropriate. The designed

single-storey warehouse is erected by several steel members and is arranged in a regular

geometrical form, in where they can interact between them through out structural connections or

joint to support loads and maintain the structure under equilibrium or stability.

Basically all steel structure is composed by a number of interconnected elements such as

beams, columns, stanchion, rafter, eaves and foundations. As in group, they enable the force or

load that acting on the structure to be safely transmitted to the ground. Nevertheless, the structure

member must have the capacity to carry loads in a variety of ways, and may act in tension,

compression, shear, torsion or in combination of these. To deliver a steel structure without any

interruption during it service life. It is imperative that the structure is been interpreted and several

analyses should be carry out to precisely predict the response of the structure under natural

phenomena influence, such as wind, snow, heat, moisture, rain penetration, fire protection and

etc. To sustain numerous of influence, the structure basically must be strong, stiff and stable to

withstand the load or force that subjected to it.

This paper, consist of design document in three section; structure specification, structure

calculations, and drawings. It emphasis the structure calculations with the inclusion of related

material specifications and important structural details. British Standard will be the guide line of

this project where the appropriate terminology, notations, and citations will be use.

Objective:

Following are lecture objective and individual (student) objective of this project.

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1.1. Lecture Objectives.

Design a single-storey warehouse according to elastic design.

Warehouse is located at 10 km from the centre of Manchester.

The warehouse is to be 36m long and 18m wide, whereby the height should be sufficient

to allow access of forklift truck, van or small pick-up truck.

Steel framed building with cladding.

Warehouse design is base on British Standard specification.

1.2. Individual (student) Objectives.

Design a single-storey warehouse according to plastic design.

Gain understanding of the design process and the responsibilities of a designer.

Gain understanding of the structure under influence of loading.

Develop/gain project management skills.

Familiarisation of exploration of all different resources available for design a structure

e.g. Internet, Books, British Standards and others.

Gain experience of CAD software and understanding their role in the engineering field.

A traditional design approach has been taken to design the structure. The approach is;

Product Design Specification (PDS)

Concept Design

Structure Analysis

Detail Design

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2. Product Design Specification (PDS).

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2. PRODUCT DESIGN SPECIFICATION

This project aim to design a warehouse or steel structure which can be operated safe

under nature phenomena. This section focuses on the design specifications and the

considerations to achieve a successful design, which meet all lecture objective and as well as

student objective. These specifications are being very closely considered during the design

process.

Following are the specifications with technical and quantitative information wherever it

is available and necessary for the structure design;

Function.

Site and environment of the structure.

Type and size of buildings.

Type of joints.

Loads subjected to the structure.

Material selections.

Stability of the structure.

Fire resistance.

Stress concentration and Residual stress.

Corrosion and protection of the steel work.

Weldability of the steel structure.

Process and installation.

Aim and advantages of plastic design..

Function.

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This section describes the functional requirements of the structure in the form of brief

statement;

Used to store good, spare components such as electric motors, gear boxes, coupling and

etc.

Prevent rain penetration from the roof and wall that might spoil the component keep

inside the warehouse.

Warehouse is the winning key component for the supply chain business.

Distribution point of materials or good between manufacturer to customer and all points

in between.

Safe and reliability under service life.

Site and environment of the structure.

The designed warehouse or structure is located in 10 km from the centre of Manchester

and there is no obstruction or building within 50m from the warehouse. Manchester lies in a

bowl-shaped land area bordered to the north and east by the Pennine hills and to the south by the

Cheshire plain [3]. Manchester has a relatively damp climate and is a rainy city. Thus, the

structure might expose to a different situation under different environment conditions.

Environment condition around Manchester,

Condition: Sun, Rain, Wind and Snow.

Temperature: average lies between 19°C to 1°C.

Ground Level: between 35 and 42 meter above sea level.

Type and size of buildings.

Several type of warehouse or steel structure under consideration; single bay portal frame,

double-bay portal frame with valley beam, double bay portal frame, lattice girder structure, roof

truss and etc. Advantage and disadvantages for each type of structure will be considered in

chapter 3. The size or dimension of the structure must meet the lecture objective.

Type of joints.

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A structural connection may be defined as an assembly of components which are

arranged to transmit forces from one member to another member. Normally, connection of a

joint will subjected with axial force, shear force and moment, but for calculation purposes the

joint will generally reduce the forces in one plane. The contraction or expansion of the structure

may occur due to temperature change, shrinkage, creep and etc. The joint that used in this design

is assume to be watertight but don’t transmitted any forces, and the joint is sufficient stiff to

sustain the shear force, axial force or moment that subjected to the joint.

Load subjected to the structure.

Forces that induced into the members of structure from dead load, imposed load, wind

load, earthquake forces, thermal stress and etc. However, for this project only dead load,

imposed load and wind load will be considered. The forces that induced into a structure from the

gravitational self weight or dead load of the member are significant for large span beam. Dead

load should always be under consideration since it exists all the time. Another form of force that

created by gravitational as well is define as imposed load or live load which are applied to the

structure; human, furniture, vehicle, stored material, fluid, snow and etc. The amount of snow

load subjected to the structure, can obtain from BS6399-3. Wind load is another form of imposed

load but they aren’t produce by gravitational. The pressure values are relying on geographical

locality, slope of the roof and shape of the structure. The pressure can be strut pressure or

suction pressure.

Material selection.

For material selection, the design strength, young’s modulus, bending strength and other

properties of the steel, I-section, bolt and nut must according to the grade and product standard

specified in BS 5950-2. For plastic design, one of the requirements is to make sure that the

plastic hinge is occurring at the steel member, before the plastic hinge can occur. The steel

member must undergo an elastic behaviour. Thus, the ductility of the steel member will be

concern.

Stability of the structure.

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For a structure to stand, it must be stable under the action of loads. The resultant forces

that applied on the structure in the vertical direction or horizontal direction and pass through the

center of gravity (CG) of the structure, the structure is say to be stable. However, if the resultant

forces subjected on the structure pass through slightly eccentric from the center of gravity. A

moment which defines as overturning moment will occur, which cause instability of structure.

For an individual problem, a steel member, I-section is very efficiency and strong when load

through it shear center, but it is inherently weak when subjected to lateral bending (minor axis)

or torsion, particularly the universal beam section. Such phenomenon is known as lateral

torsional buckling. Emphasize against the possibility of failure caused by lateral torsional

buckling of a member across the cross-section must be taken care.

Fire resistance.

In design of steel structure, the ‘critical temperature’ is important and is defined as the

temperature in which the strength of the steel member is reduced to a certain level that the

structure will collapse. As the temperature increase, it may induce thermal stress on the steel

member. BS5950 part 8 is the standard that been using for designing a steel structure which can

resist fire.

Stress concentration and residual stress.

Stress concentration normally occurs at the holes for bolts which use to connect the

structural member and small cracks along the surface of the I-section which cause fatigue.

Meanwhile, residual stresses are induced into the steel member during rolling, welding of the

steel member to certain geometry, lifting and transportation, flame cut or drilling. During rolling

process, the whole steel is initially at a uniform temperature, however as the cooling process

undergo, some portion of the steel member which have a thinner cross section than others and

consequently the cooling effect will be faster than the thicker cross section. Thus, as for welded

joints, the part that cool 1st may have compression residual stress and the part that cool last have

the tensile residual stress. Since the cooling rate affected the yield strength of the materials, the

thinner section tends to have higher yield strength than the thicker section. It is believe that, for

plastic design, when a plastic hinge occur at the steel member, the residual stress are relieved by

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the large strains, thus it doesn’t affect the ultimate strength of the steel member unless lateral

torsional buckling occurs.

Corrosion and protection of the steel work.

Corrosion is a chemical reaction between the steel member, water and oxygen which

produce hydrated iron oxide called rust. The effects of corrosion on the steel structure are

reducing of cross-section of steel member and may create a small conical pit which may cause

stress concentration effect. For this project, drainage holes are included at appropriate location

along the steel member.

Weldability of the steel structure.

Welding is a joining process in which the joint production can be achieved with the use

of high temperature, high pressure or both. For this project, only the high temperature welding to

produce a joint will be used and the most common method of welding structural steel is MMA

(manual metal arc welding). Before the two steel members is weld, the welding edge between the

two steel members must undergo a fillet process or chamfer to compose the welding joint

stronger.

Process and Installation.

For plastic design, the steel member will undergo a process where the plastic hinge is

created. To induce a plastic hinge in the steel members, the steel member will simply supported

at both end and a pressure or load will apply along the steel member until the plastic hinge occur.

This ensure that the rafter or stanchion meet the requirement for plastic design.

Aim and advantages of plastic design.

The main objective using plastic design methods is to calculate the collapse load or

collapse moment of the structures. Thus, plastic theory concentrates on one criterion which is the

strength of the steel member. In the simple form of plastic design, the plastic design theory

makes no attempt to assess the deflection of the structure, since it is relative small compare with

the structure size or to enquire the stability of the structure. For the project, stability will be

concern, especially lateral torsional buckling along the rafter. Plastic design has several

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advantages; realistic the steel frame actual behaviour, simplifies the analysis of steel structures, a

uniform margin of safety of a structure and the utilization of the steel member [15].

Summary of PDS.

Following are brief statement of the structure requirement, which summarize all above

specifications.

Steel structure can sustain 3 type of loading; dead load, imposed load and wind load

which obtain from British Standard 6399-2 and British Standard 6399-3

Stability of the structure, avoid lateral torsional buckling and corrosion protection.

Analysis of steel structure meets the British Standard 5950-1.

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Design Flow Chart

Section Description:

Loading: Determine the load subjected to the structure.Analysis: Analyse the structure according to British Standards.Stability: Analyse the stability of structure according to British Standards.

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Preliminary Design

Loading

Analysis

Stability

Drawings

END

Yes

Yes

No

No

3. Concept Design.

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3. CONCEPT DESIGN

3.1. Concept 1.

Advantages.

Easy to erect.

Light structure.

Size – medium.

Suitable for warehouse.

Large space inside the building.

Wide and low.

Easy to design.

Cheap cost.

Disadvantages.

Complex maintenance.

Strong but not stiff.

Manufacturing difficulties.

3.2. Concept 2.

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Truss Frame System

Advantages.

Light structure.

Size – medium.



Wide and low.

Easy to erect.

Strong and stiff.

Disadvantages.


Difficult to design.


Expensive cost.

3.3. Concept 3.

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Arched Roof System

Advantages.

Light structure.

Size – medium.



Wide and low.

Strong and stiff.

Easy to design.

Easy and fast to erect.

Disadvantages.



Moderate cost.

3.4. Concept 4.

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Portal Frame System

Advantages.

Size – large



Wide building.

Strong and stiff.

Design for minimum direct sun light

Disadvantages.

Difficulties in design.



Snow load, dust, leaves stuck between the roof pitch.

High building.

Expensive cost.

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Saw-tooth Roof System

3.5. Concept 5.

Advantages.

Size – large


Very large space inside the building.

Wide building.

Strong and stiff.

Disadvantages.

Difficulties in design.



Snow load, dust, leaves stuck at the center of the roof.

Expensive cost.

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Portal Frame with Alternative Valley Columns Ommited

This section is produced with several of concept sketches and each of the sketches is listed

with the advantages and disadvantages which will be utilize in the evaluation matrix. The

evaluation matrix enables the concept design to be chosen under certain important criteria which

satisfied lecture objectives and individual objectives. The concept designs that score the highest

score on the evaluation matrix are the most desirable option and the design is forwarded to the

detail design stage.

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4. Design Loads Acting On Structures.

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4. Design Loads Acting On Structures

All structure is composed of a number of interconnected elements such as roof deck,

purlins, slabs, beams, columns and foundations. Collectively, it enables the internal or external

loads that acting on the structure safely transmitted to the foundations then to the ground. The

design loads are used to calculate the bending moments, shear force, stress and deflections at

critical points along the steel member and finally suitable dimension, section or properties for the

element can be determined.

Loads acting on the structure can be classified as static and dynamic. Static loads are

applied on the structure slowly and gradually and these are steady-state which mean doesn’t vary

with time. Dynamic loads are defined as sudden application on the structure and variation of the

magnitude respect with time. The design static loads acting on the structures are divided into four

basic types: dead load, imposed or live load, wind load and force due to thermal effect in which

partly can be obtain from the British Standard (BS). In static, loads are force that acting on

structural element that can represent as surface loads, line load and concentrated loads.

A force that applied on the large size of area is considered as a surface load, and this type

of load has units of force per unit area which is classified as pressure (N.m -2). In general, wind

load and snow load is considered as a surface pressure which will further discuss in chapter 4.3.

A force that applied over a long, narrow area may considered a line load and assumed to be

distributed along the long dimension, line loads have units of force per unit length (N.m -1) and

often vary with position along the line. Self-weight of the beam, purlin and roof deck can treated

as a line load and the force that distributed along the length of the member is classified as

Uniformly Distributed Load (UDL). A force that applied over a small area can considered as a

concentrated load or point load, and this type of load has a unit of Newton (N). Consider a purlin

“rested” itself on the haunch frame or structure. A reaction will occur at the both end of the

purlin, in which the reaction is acting on the haunch frame. The reaction produce by the purlin is

treated as a point or concentrated load.

The effect of Static Loads consist of Dead loads, Snow load and Wind load when

subjected on the haunch portal frame will be concern. The design loads are obtain by multiplying

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the characteristics loads by the partial factor of loads γf, which can be obtain from [1:p12: table

12]. Several load combinations with their design loads are used to analysis the haunch portal

frame as shown in table 1. Analysis on the bending moment, shear forces, deflections and sway

stability will be carry on in chapter 4.

4.1. Dead Loads.

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Table 1

Dead load or classified as gravity load, is the weight that act permanently and it remains

constant in term of magnitude, location and direction. Dead loads always act vertically

downwards and such loads normally consist of the self-weight of the structure as well as the

weight of the building elements, which is non-moving partition can be concern as dead load. By

referring to the schedule of weights of buildings material given in [British Standard 648], the

characteristics dead loads can be estimated or from the manufacture’s literature or BRITISH

STANDARD 6399-1: 1996. In general the symbol Gk and gk is used to denote the total load or

UDL of the dead loads. According to the preliminary design, the deal load acting on the portal

frame will be calculated.

4.1.1. Dead loads calculations.Refer to appendix,

Unit weight of the roof deck is given, 100.5 N.m-2

Unit mass of the purlin for (Sr.No. 9) is given, 6.25 kg.m-1

Unit weight of the I-Beam (W410 x 0.73) is given, 0.73 x 103 N.m-1

By referring to figure 1, it shows the concentrated load or point load provided by the purlin is

acting on the haunch portal frame. In the preliminary design, the spacing between each purlin is

1.5 m and the height of the roof (hr) as indicated in figure 1 is 3 m, meanwhile the height of the

column from the foundation or ground is 3 m. The width or W as indicate in figure 1 is 18 m and

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hr

hc

A E

D1B1

B B1B2

B3B4

B5 D5 D4 D3 D2 D1 D

C

W

Figure 1

18.43°

the slope of the roof as indicate in figure 1 is 18.43°. The distance between two haunch frames is

11.83m. The roof deck “rested” itself on the purlin and the purlin “rested” itself on the portal

frame. To calculate the point load or the reaction of the purlin acting on the haunch frame, the

haunch portal frame is ‘cut’ at the apex as shown in figure 2.

Concentrated load calculation.

The concentrated load acting on point C,

Given unit weight of roof deck = 100.5 N.m-2

Assume the weight of the roof deck within the

half of the distance between the purlin is acting

on point C. Thus,

The UDL of the roof deck acting on the purlin =

= 71.61 N.m-1

The unit mass of the purlin = 6.25 kg.m-1

Unit Weight of the purlin =

= 61.31 N.m-1

Total line load along purlin, Gk = Line load purlin + line load roof

= 61.31 + 71.61

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A

B1

B B1B2

B3B4

B5C

‘cut’

Figure 2

= 132.94 N.m-1

Refer to table 1,

Given partial factor for dead load, γf = 1.4

Ultimate Design Load = 1.4 Gk

= 1.4 x 134.94 N.m-1

= 186.09 N.m-1

The UDL acting on the purlin and the reaction (R1,R2) acting on the haunch frame is illustrated

in figure 3.

Ultimate UDL acting purlin C = 186.09 N.m-1

Given the length of the purlin = 11.83 m

Reaction on both end of purlin (R1, R2) =

=

= 1.1 k N

Unit weight of the I-Beam is given, = 0.73 x 103 N.m-1

Assume the weight of the beam within the


on point C. Hence, concentrated load for beam

acting on point C, Gk =

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Figure 3

R1 R2

UDL, ω

L

= 520.13 N.

Ultimate Design Load = 1.4Gk

= 1.4 x 520.13

= 728.18 N

Total Ultimate Design Dead Load = 1.1 k + 728.18

= 1.83 k N

The concentrated load acting on point B,

By referring to figure 2, at point B the concentrated load that it sustains is the reaction from the

purlin where the reaction is the half of the combination of roof deck self-weight and purlin self-

weight. The self-weight of the beam is excluded at point B. Thus, the concentrated load on point

B is 1.1 k N.

The concentrated load acting on point B1, B2, B3, B4 and B5,

The concentrated load acting on point Bx,

Where x can be 1, 2, 3, 4 and 5.

Given unit weight of roof deck = 100.5 N.m-2

Assume the weight of the roof deck within the

distance between the purlin is acting

on point Bx. Thus,


= 143.21 N.m-1

Total line load along purlin, Gk = Line load purlin + line load roof

= 61.31 + 143.21

= 204.52 N.m-1

Refer to table 1,

Given partial factor for dead load, γf = 1.4

Ultimate Design Load = 1.4 Gk

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= 1.4 x 204.52 N.m-1

= 286.33 N.m-1

Ultimate UDL acting purlin Bx = 286.33N.m-1



(Refer to figure 3) =

= 1.69 k N

Unit weight of the I-Beam is given, = 0.73 x 103 N.m-1

Assume the weight of the beam within the

distance between the purlin is acting

on point Bx. Hence, concentrated load for beam

acting on point Bx, Gk =

= 1.04 k N

Ultimate Design Load = 1.4Gk

= 1.4 x 1.04 k

= 1.46 k N

Total Ultimate Design Dead Load = 1.69 k + 1.46 k

= 3.15 k N

Base on the Preliminary design, the whole structure consist of 4 portal frames as shown in figure

4.

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Reg3

Reg1 = Region 1

Reg2 = Region 2

Reg3 = Region 3

1 = Portal Frame 1

2 = Portal Frame 2

3 = Portal Frame 3

Assuming the structure is under influence of dead, imposed, or wind load and the influence of

the load is separated into 3 regions as indicated in figure 4. Portal frame 1 and portal frame 2,

each of them sustain half of the loading at Region 1. At region 2, portal frame 2 and portal frame

3, each of them sustain half of the load at region 2. In the other way around, two portal frames

lies between regions sustain half of the load from the region. Thus, portal frame 2 and 3 sustain

twice the load at portal frame 1 and portal frame 4.

Table 2 show the total ultimate concentrated load for dead load on the haunch portal frame at

different location along the rafter of the haunch portal frame under partial factor, γ f of 1.4. From

table 2 it shown that, the ultimate concentrated load for portal frame 2 and portal frame 3 sustain

the higher value of point load compare with portal frame 1 and portal frame 4. Thus, it will

provide larger bending moment, in which will be use to estimate the section of properties of the

portal frame which will be carrying on in chapter 5. The ultimate concentrated load in Table 2

will be concern with load combination (dead + imposed).

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Figure 4

4

3

2

1

Reg1

Table 2

Table 3 show the total ultimate concentrated load value for dead load on the haunch portal frame

at different point along the rafter under partial fraction, γf of 1.2, the ultimate concentrated load

for table 3 will be concern under load combination (dead + imposed + wind). The method to

obtain the value in table 2 and table 3 is the same.

4.2. Imposed Loads.

Imposed load or refer as live loads is represents the load due weight of the occupants, or

user, furniture, moveable partitions, distributed, impact, inertia and roof loads including snow

but the excluding wind load. When live loads are applied slowly and gradually on the structure,

they are static load. When they are applied suddenly and their magnitude varies rapid with time,

they are dynamic load. The snow loads are often significant, especially for the roof of the

building. Snow is moveable load, because the snow load will not cover the whole entire of the

roof top, and some member supporting the roof may receive the maximum stress or deflection,

with the snow cover only a portion of the roof. Snow loads on the roof vary widely and depend

on such factors as elevation, latitude, wind frequency, duration of snow fall, roof size, geometry,

snow density and slope of the roof. Generally the symbol Qk or qk is used to donate the amount

of imposed load. The snow load acting on the roof for the preliminary design structure can be

obtained from BRITISH STANDARDS 6399-3: 1988.

Base on the preliminary design, where the roof slope is 12°. The minimum design snow load

acting on the roof;

Refer to [7:p2:4.3.1], stated:

- A uniformly distributed load of 0.6 k N.m-2 measured on plan for roof slope of 30° or less.

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Table 3

Snow load acting on the roof, (Qk) = 0.6 k N.m-2

Snow load is considered as a surface load. To transfer surface load to concentrated load, similar

step can be carry out as finding the concentrated load for the roof deck.

4.2.1. Snow loads calculations.The concentrated load acting on point C, refer

figure 2.

Given unit weight of snow load = 0.6 k N.m-2

Assume the weight of the snow load within the


on point C. Thus,


Line load along purlin for snow, Qk = 427.5 N.m-1

Refer to table 1,

Combinations load of Dead + Imposed

Given partial factor for imposed load, γf = 1.6

Ultimate Design Load = 1.6 Qk

= 1.6 x 427.5 N.m-1

Ultimate UDL for snow load acting purlin C = 684 N.m-1



=

= 4.05 k N

Reaction at point C = 4.05 k N

Reaction at point B = 4.05 k N

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Reaction at point B1, B2, B3, B4 and B5 = 4.05 x 2 = 8.1 k N

Table 4 show the value of the total ultimate concentrated load for imposed load at different point

of the haunch portal frame under partial factor, γ f of 1.6. The value shown in table 4 will be

concern under load combination (dead + imposed)

Table 5 show the total ultimate concentrated load for imposed load on the haunch portal frame at

different point of the haunch portal frame under partial factor, γ f of 1.2. The ultimate

concentrated load value shown in table 5 will be concern when under load combination (dead +

imposed + wind). The method to obtain the value in table 5 is the same as method using in table

4.

4.3. Wind Loads.

When a wind is subjected to a structure, the wind will deflect or stopped by a structure. In

which the kinetic energy of the wind come from the free stream velocity is stopped by the wall or

structure, thus transformed into potential energy of pressure or suction (Bernoulli’s equation), in

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Table 4

Table 5

which the magnitude of the pressure is depend on the free stream velocity, the surface roughness

on which the wind acts, the density of the air, the structure geometrical shape, dimension and

orientation of the structure. The wind load is assume to be static effect, doesn’t vary with time.

Generally the symbol Wk or wk is used to donate the amount of wind load subjected to a structure

or building and can obtain from BRITISH STANDARD 6399-2: 1997.

4.3.1. Wind Load Coefficient from BS6399-2.The wind load acting on the haunch portal frame can be obtained from BS 6399-2, section 2,

standard methods. Base on the preliminary design, haunch portal frame which is consider as

equal-duo pitch structure. The wind load subjected to the structure is two orthogonal load cases

where the wind direction normal to the faces of the building.

To acquire the pressure subjected to the haunch portal frame. Dynamic Pressure (qs) must obtain

in which the value of the dynamic pressure is influence by several factor which will discuss in

this chapter.

4.3.1.1. Dynamic pressure

Refer to [6:p13: 2.1.2], Dynamic pressure is defined as;

qs – Dynamic Pressure (N.m-2)

Ve – Effective Wind Speed (m.s-1)

Effective wind speed

Refer to [6:p27: 2.2.3], Effective wind speed is defined as;

Ve = Vs x Sb

Vs – Site of the wind speed

Sb – Terrain and building factor

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Terrain and building factor

Refer to [6:p28: 2.2.3.3], Terrain and building factor, Sb

The Sb value can be obtain from [6:p28; table 4]

But, the effective height is needed to indicate the Sb value.

Where the effective height can obtain from [6:p10: 1.7.3.2] and [6:p8: figure 2]

The effective height He may be conservatively taken as the reference height Hr.Hr = 6 mSince the location of the structure is located at 10km within Manchester and the nearest distance from Manchester to seaside > 100km (refer to google earth). The value of the terrain and building factor Sb is obtain from column (site in town, extending ≥ 2km upwind from the site) is chosen and the closest distance to sea upwind ≥ 100km. Interpolation may needed, where the He

is between 5 and 10.

Interpolation,He - 56 – 5 = 1

= 0.044

1 x 0.044 = 0.0441.36 + 0.044 = 1.40

Sb = 1.40

Site wind speedRefer to [6:p18: 2.2.2], Site wind speed is define as;

Vs = Vb x Sa x Sd x Ss x Sp

Vb – is the basic wind speed.Sa – altitude factor.Sd – direction factor.Ss – Seasonal factor.Sp – Probability factor.

Basic Wind SpeedRefer to [6:p19: figure 6]At location 10 km around Manchester Vb = 22 m.s-1

Altitude factorRefer to [6:p20: 2.2.2.2], altitude factor is define as;

Sa = 1+ 0.001Δs

Kah Hin Tan - 35 -

Δs – Site altitude (in meters above mean sea level).

Refer to [22]The altitude for Manchester is between 35 and 42 meter above sea level.

Sa = 1+ 0.001(42)= 1 + 0.042= 1.042

Direction factorRefer to [6:p27: 2.2.2.3], Direction factor stated;If the orientation of the building is unknown or ignored, the value of the direction factor should be taken as Sd = 1.00 for all directions.

Seasonal factorRefer to [6:p27: 2.2.2.4], Seasonal factor stated;The seasonal factor Ss is used for buildings which are expected to be exposed to the wind for specific subannual period. Assume the Ss = 1 for worst case might happen.

Probability factorRefer to [6:p27: 2.2.2.5] and [6:p103: Annex D] Probability factor stated;Sp = 0.749 for Q = 0.632Note 1 stated, the annual mode, corresponding to the most likely annual maximum value.

Given,Vb = 22 m.s-1

Sa = 1.042Sd = 1Ss = 1Sp = 0.749

Site wind, Vs = 22 x 1.042 x 1 x 1 x 0.749= 17.17m.s-1

Effective wind speed,Ve = Vs x Sb

= 17.17 x 1.40= 24.04 m.s-1

Dynamic Pressure,

= 0.613 (24.04)2

= 354.27 N.m -2

Since the dynamic pressure proportional with the effective velocity, the effective velocity proportional with terrain and building factor with assuming the site wind speed remain constant.

Kah Hin Tan - 36 -

While the terrain and building factor proportional with the height of the building. Thus, the dynamic pressure is proportional with the height of the building, but it is assume to remain constant and doesn’t change with the height of the building and the maximum value of the dynamic pressure will be use.

4.3.1.2. External Pressure Coefficients for walls.

Base on the preliminary design, the wind load subjected to the structure is two orthogonal load

cases where the wind direction normal to the faces of the building. Wind load can separated into

2 cases, where it is side wind and gable wind. Figure 5 indicate the side wind direction and figure

6 will indicate the gable wind direction.

Given the dimension for the preliminary design is,

Length = 35.49 m

Width = 18 m

Height of the wall = 3 m

External pressure coefficient for side wind.

Refer [6:p30: 2.4.1.2],

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Figure 5Side Wind Direction

Figure 6Gable Wind Direction

Wind directionWind direction

The value for the pressure coefficient for windward and leeward can be obtained in [6:p31: table

5] by using a ratio of the building dimension, D/H. Where D is the depth of the building in which

varies with the wind direction. H is defined as the height of the wall.

D = 18 m

H = 3 m

Windward (front),

Cpe (front) = +0.6

Leeward (rear),

Cpe (rear) = -0.5

Refer [6:p30: 2.4.1.3] stated;

Scaling length (b), where

b = B or,

b = 2H

B – Crosswind breadth of the building

H – Height of the wall

Whichever is smaller.

b = Length of the building.

= 35.49 m

b = 2H

= 2 (6.02)

= 12.04 m

b = 12.04 m

Refer [6:p31: figure 12 b]

D>b, where D = 18m and b = 12.04m

Thus, 3 side wall faces subjected to pressure.

Refer [6:p30: 2.4.1.4]

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As stated in the PDS, there is no other building located near the warehouse within 50m. Isolated

value should be used;

Side pressure coefficient,

Zone A, -1.3

Zone B, -0.8

Zone C, -0.5

External pressure coefficient for gable wind.

Refer [6:p30: 2.4.1.2],

The methods to obtain the external pressure coefficient for gable wind are similar with the

external pressure coefficient side wind.

D = 35.49 m

H = 3 m

Cpe (front) = 0.6

Cpe (rear) = -0.5

Refer [6:p30: 2.4.1.3] stated;

Scaling length (b), where

b = B or,

b = 2H

Whichever is smaller.

b = Width of the building

= 18 m

b = 2H

= 2 (6.02)

= 12.04 m

b = 12.04 m

Refer [6:p31: figure 12 b]

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D>b, where D = 18m and b = 12.04m

Thus, 3 side wall faces subjected to pressure.

Refer [6:p30: 2.4.1.4]

As stated in the PDS, there is no other building located near the warehouse within 50m. Isolated

value should be used;

Side pressure coefficient,

Zone A, -1.3

Zone B, -0.8

Zone C, -0.5

External pressure coefficient.

4.3.1.3. External Pressure Coefficient for Roofs.

Refer [6:p40: 2.5.2.4.1]

External pressure coefficient for duopitch roof can be obtain from [6:p43: table 10] using the key

in [6:p42: figure 20].

According to the preliminary design of the roof, the pitch angle is 12°.

Two set of pressure coefficients for roofs is given in table 10, whereby zone for wind direction θ

= 0° (side wind) and zone for wind direction θ = 90° (gable wind). Interpolation for the Cpe value

for both wind directions is required.

Pressure Coefficient for wind direction θ = 0° (side wind),

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Table 6

Interpolation for Cpe value A,

Cpe value for A,

Cpe min = -0.98

Cpe max = +0.32

Interpolation for Cpe value B,

Cpe value for B,

Cpe min = -0.74

Cpe max = +0.26

Interpolation for Cpe value C,

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at θ = +15° at θ= +30°Cpe = -1.1 Cpe = -0.5

Min Value for θ = 18°-1.1 - (-0.5) = -0.6

= -0.04

-0.04 x 3 = -0.12-1.1 - (-0.12) = -0.98

at θ = +15° at θ= +30°Cpe = +0.2 Cpe = +0.8

Max Value for θ = 18°0.2 - (0.8) = -0.6

= -0.04

-0.04 x 3 = -0.120.2 - (-0.12) = +0.32

at θ = +15° at θ= +30°Cpe = -0.8 Cpe = -0.5

Min Value for θ = 18°-0.8 - (-0.5) = -0.3

= -0.02

-0.02 x 3 = -0.06-0.8 - (-0.06) = -0.74

at θ = +15° at θ= +30°Cpe = +0.2 Cpe = +0.5

Max Value for θ = 18°0.2 - (0.5) = -0.3

= -0.02

-0.02 x 3 = -0.060.2 - (-0.06) = +0.26

Cpe value for C,

Cpe min = -0.36

Cpe max = +0.24

Interpolation for Cpe value E,

Cpe value for E,

Cpe = -1.22

Interpolation for Cpe value F,

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at θ = +15° at θ= +30°Cpe = -0.4 Cpe = -0.2

Min Value for θ = 18°-0.4 - (-0.2) = -0.2

= -0.013

-0.013 x 3 = -0.039-0.4 - (-0.039) = -0.36

at θ = +15° at θ= +30°Cpe = -1.3 Cpe = -0.9

Min Value for θ = 18°-1.3 - (-0.9) = -0.4

= -0.027

-0.027 x 3 = -0.081-1.3 - (-0.081) = -1.22

at θ = +15° at θ= +30°Cpe = +0.2 Cpe = +0.4

Min Value for θ = 18°0.2 - (0.4) = -0.2

= -0.013

-0.013 x 3 = -0.0390.2 - (-0.039) = +0.24

Cpe value for F,

Cpe = -0.82

Cpe value for G,

Cpe = -0.5

Table 7 shown the value for the external pressure coefficient Cpe for wind acting at θ = 0°.

Pressure Coefficient for wind direction θ = 90° (gable wind),

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at θ = +15° at θ= +30°Cpe = -0.9 Cpe = -0.5

Min Value for θ = 18°-0.9 - (-0.5) = -0.4

= -0.027

-0.027 x 3 = -0.081-0.9 - (-0.081) = -0.82

Table 7

Interpolation for Cpe value A,

Cpe value for A,

Cpe = -1.52

Interpolation for Cpe value B,

Cpe value for B,

Cpe = -1.42

Cpe value for C,

Cpe = -0.6

Interpolation for Cpe value D,

Kah Hin Tan - 44 -

at θ = +15° at θ= +30°Cpe = -1.6 Cpe = -1.2

Min Value for θ = 18°-1.6 - (-1.2) = -0.4

= -0.027

-0.027 x 3 = -0.08-1.6 - (-0.08) = -1.52

at θ = +15° at θ= +30°Cpe = -1.5 Cpe = -1.1

Min Value for θ = 18°-1.5 - (-1.1) = -0.4

= -0.027

-0.027 x 3 = -0.08-1.5 - (-0.08) = -1.42

Cpe value for E,

Cpe = -0.42

4.3.1.4. Internal Pressure Coefficient.

Based on the preliminary design, the haunch portal frame structure is considered as an

enclosed building. The portal frame structure it is cover up with roof deck on the roof and each

side of the walls. It is assume that the dominant opening doesn’t occur and as stated in [6:p53:

2.6.1.1], the internal pressure coefficient for an enclosed building may be obtain from [6:p54:

table 16]. The internal pressure coefficient is -0.3.

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at θ = +15° at θ= +30°Cpe = -0.4 Cpe = -0.5

Min Value for θ = 18°-0.4 - (-0.5) = 0.1

= 0.007

0.007 x 3 = 0.02-0.4 - (0.02) = -0.42

Table 8

4.3.2. Wind Load Calculations.Wind load is considered as a lateral loading. Based on the preliminary design, the haunch

portal frame structure is low, wide and long structure. Gable wind loading doesn’t cause any

instability of the structure, since the wind direction is inwind depth with the length of the

structure. However, the side wind loading will be checked base on the value that obtains from

chapter 4.3.1.

The pressure coefficients that obtain from 4.3.1 on table 6 and table 7, it can be represent by

figure 7 and figure 8.

Figure 7 and figure 8 are treated as separated case.

Kah Hin Tan - 46 -

Figure 7

0.6-0.3

-0.5

-0.98 -1.22

0.6-0.3

-0.5

0.32 -1.22

Figure 8

Winddirection

Winddirection

Concentrated load calculation base on figure 7.

The wind load is consider as a surface load, and the surface load acting on the roof and wall

surface will be calculated to transfer the surface load to point load acting on the haunch portal

frame.

Net pressure coefficient for figure 7 is shown in figure 9.

Point load acting on the haunch portal frame is shown in figure 10.

The distance between A and A1 is 1m, A1 and A2 is 1m. Meanwhile the other dimension for the

haunch portal frame remains the same as shown in figure 1.

Kah Hin Tan - 47 -

0.9 -0.2

-0.68 -0.92

Figure 9

B B1B2

B3B4

B5

CB , CD

D5 D4 D3 D2 D1 D

E2

E1

A2

A1

EA

Figure 10

Given Dynamic Pressure from chapter 4.3.1.1,

qs = 354.27 N.m-2

Assume the side effect factor Ca = 1

Refer to [6:p13: 2.1.3.3] stated;

Net pressure for enclosed buildings,

P = Pe - Pi

Where,

Pe – External Pressure

Pi – Internal Pressure

And given Pe and Pi from [6:p13: 2.1.3.1] and [6:p13: 2.1.3.2];

Pe = qs.Cpe.Ca

Pi = qs.Cpi.Ca

Net Pressure = External Pressure – Internal Pressure.

P = qs (Cpe – Cpi) Ca

Ca = 1

P = qs (Cpe – Cpi)

Where (Cpe – Cpi) value can be obtain from figure 9 for each side of the wall and roof.

Concentrated load calculation on windward wall.

Assume purlin A1 sustain half of the windward pressure and the other half of the windward

pressure is sustain by purlin A2.

Given dynamic pressure, qs = 354.27 N.m-2

Given net Pressure Coefficient, Cp = 0.9 (refer figure 9)

Pressure, P = 354.27 x 0.9

= 318.84 N.m-2

UDL acting on purlin A1 and purlin A2, = Pressure x (half of the column height)

= 318.84 x 1.5

= 478.26 N.m-1

Kah Hin Tan - 48 -

Ultimate UDL, = 1.2 Wk

= 1.2 (478.26)

= 573.91 N.m-1

Given length of the purlin = 11.83

Reaction of the purlin acting on the haunch

portal frame. [Refer figure 3] =

=

= 3.39 k N

Reaction of the purlin at point A1 and A2 is 3.39 k N.

Concentrated load calculation on windward roof.

The concentrated load acting on point CB ,

[refer figure 10].


Given net Pressure Coefficient, Cp = -0.68 (refer figure 9)

Pressure, P = 354.27 x -0.68

= -240.9 N.m-2

UDL acting on purlin C, =

= -171.64 N.m-1


= 1.2 (-171.64)

= -205.97 N.m-1



portal frame. [refer figure 3] =

Kah Hin Tan - 49 -

=

= -1.22 k N

Concentrated load on the haunch portal frame,

At point CB = -1.22 k N

At point B = -1.22 k N

At point B1, B2, B3, B4 and B5 = -1.22 k x 2 = -2.44 k N

Concentrated load calculation on leeward roof.

The concentrated load acting on point CD ,

[refer figure 10].


Given net Pressure Coefficient, Cp = -0.92 (refer figure 9)

Pressure, P = 354.27 x -0.92

= -325.93 N.m-2


= -232.23 N.m-1


= 1.2 (-232.23)

= -278.68 N.m-1



portal frame. [refer figure 3] =

=

= -1.65 k N

Kah Hin Tan - 50 -


At point CD = -1.65 k N

At point D = -1.65 k N

At point D1, D2, D3, D4 and D5 = -1.65 k x 2 = -3.3 k N

Concentrated load calculation on leeward wall.

Assume purlin E1 sustain half of the windward pressure and the other half of the windward

pressure is sustain by purlin E2.


Given net pressure Coefficient, Cp = -0.2 (refer figure 9)

Pressure, P = 354.27 x -0.2

= -70.85 N.m-2

UDL acting on purlin A1 and purlin A2, = Pressure x (half of the column height)

= -70.85 x 1.5

= -106.28 N.m-1


= 1.2 (-106.28)

= -127.54 N.m-1




=

= -0.754 k N

Reaction of the purlin at point E1 and E2 is -0.754 k N.

Kah Hin Tan - 51 -

Table 9 shows the ultimate concentrated load acting on different portal frame under side wind

direction with wind up or suction pressure at the wind ward roof.

Kah Hin Tan - 52 -

Table 9

Concentrated load calculation base on figure 8.

The wind load is consider as a surface load, and the surface load acting on the roof and wall

surface will be calculated to transfer the surface load to point load acting on the haunch portal

frame.

Net pressure coefficient for figure 8 is shown in figure 11.

Compare figure 11 and figure 9, the different for the net pressure coefficient on each surface

happen at the windward roof surface. So, only concentrated load under windward roof need to

recalculate.

Concentrated load calculation on windward roof.

The concentrated load acting on point CB ,

[refer figure 10].


Given net Pressure Coefficient, Cp = 0.62 (refer figure

11)

Pressure, P = 354.27 x 0.62

= 219.65 N.m-2


= 156.5 N.m-1

Kah Hin Tan - 53 -

0.98 -0.2

0.62 -0.88

Figure 11


= 1.2 (156.5)

= 187.8 N.m-1




=

= 1.11 k N


At point CB = 1.11 k N

At point B = 1.11 k N

At point B1, B2, B3, B4 and B5 = 1.11 k x 2 = 2.22 k N

Kah Hin Tan - 54 -

Table 10 shows the ultimate concentrated load acting on different portal frame under side wind

direction with wind down or pressing pressure at the wind ward roof.

In this section the value of the concentrated load acting on the purlin, will be carry to

chapter 5 to calculate the free bending moment, reactant moment, plastic moment, sway stability

and lateral torsional buckling.

Kah Hin Tan - 55 -

Table 10

5. Structure Analysis.

Kah Hin Tan - 56 -

5. STRUCTURE ANALYSIS

Base on the preliminary design, haunch portal frame is classified as a continuous

construction. For steel member such as beam, rafter or stanchion are combining together using

several kind of joint. These joint are capable to convey moment, shear and thrust over the entire

structure when is subjected with load, and due to the advantages provided by the continuous

structure, in where the greater participation of steel member of the structure to resist the applied

load. Thus, the structure becomes more tough, rigid and stable.

Initially, there are two types of design where can be use to designed the haunch portal

frame; elastic design and plastic design. The disadvantages of elastic design is mainly about the

utilisation of the steel member is within the elastic region of the stress strain curve, meanwhile

the plastic design limits the steel member at the ultimate tensile stress, thus it bring an

advantages for plastic design to utilise the steel member is higher order. Plastic design often

simplifies the design, analysis and calculation. The main criteria for plastic design is to

concentrate on the strength of the steel member and in the simple form of plastic design, the

plastic design theory make no attempt on the deflections or the stability of the structure, but the

analysis on the stability and lateral torsional buckling of the structure will be taken count in this

chapter, and emphasize on the lateral torsional buckling effect. Since the plastic design consist of

several advantages compare with elastic design, plastic will be use to design and analyses the

haunch portal frame in this chapter. [8:p152]

To use plastic design, the steel member must sustain loading until the excess of yield, so

that the plasticity (plastic hinge) can develops in the certain region of the steel members. Thus, a

ductile behaviour is required for the steel member to yield over the entire cross-sections. The

steel member should be fabricated at certain of high standard where it can be obtained from

[1:p117: 5.2.3.3]. For fixed-base haunch portal frame, collapse under vertical load will normally

occur at the joints, and some point between the rafters. Indeed the plastic moment or collapse

moment for the structure when subjected with load need to be determined and the skill required

is; transform or ‘cut’ the portal frame into half at the apex of the portal frame and three reactions

which is horizontal and vertical shear force and bending moment. Since the vertical load is

transfer from the roof deck to purlin and from purlin to the main frame at the purlin points, the

plastic hinge will from at whichever of the purlin point load corresponds to the maximum

Kah Hin Tan - 57 -

sagging moment in the rafter. Failure to locate the exact point of the plastic hinge, will lead to a

violation of the fundamental yield condition of plastic theory. Base on BRITISH STANDARD

5950-1 section 5, it will be the guide line for the plastic analysis on the haunch portal frame.

Collapse plastic moment analysis for the haunch portal frame under 4 type of load

combination as shown in table 1, bending moment diagram at collapse moment for the frame will

be determine, Overall sway stability of the frame will be check, rafter stability, buckling

resistance moment, deflection of the rafter, shear force at the bolt, stanchion stability and finally

haunch stability.

In this chapter, assumptions had to be made before analysis can carry on, the list below

stated several assumption;

The steel is ideal elasto plastic material, which means the behaviour of the material

idealized the stress-strain curve with sufficient accuracy.

The steel materials behave identically in tension and compression.

The structural connections are rigid to extent that they are capable to transferring the

redistributed effects.

No construction technique or structural details obstructing the development of plastic

hinge.

The structural remain at the same location even subjected with load.

The steel structure is ductile enough to undergo the plastic region.

5.1. Verify Appropriate Section Properties.

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Suitable section properties for the rafter will be determine in this chapter, in where plastic

collapse moment will be determine 1st, then by using the maximum plastic moment under 4 type

of load combination stated in table 1, suitable section of properties will be determine.

5.1.1 Plastic Moment under Dead Load.

Dead load combination.

Refer to table 2 to obtain the concentrated load acting on the haunch portal frame at the purlins

point shown in figure 12 and calculate free bending moment diagram.

Take moments,

MB5 = 2.93 k x 1.5

= 4.4 k N.m

MB4 = (2.93 k x 3) + (4.84k x 1.5)

= 16.05 k N.m

MB3 = [2.93 k x 4.5] + [4.84 k x (3 + 1.5)]

= 34.97 k N.m

MB2 = [2.93 k x 6] + [4.84 k x (4.5 + 3 + 1.5)]

= 61.14 k N.m

Kah Hin Tan - 59 -

A

B1

B B1B2

B3B4

B5C

‘cut’

Figure 12

MB1 = [2.93 k x 7.5] + [4.84 k x (6 + 4.5 + 3 + 1.5)]

= 94.58 k N.m

MB = [2.93 k x 9] + [4.84 k x (7.5 + 6 + 4.5 + 3 + 1.5)]

= 135.27 k N.m

MB = MA

Since the concentrated load value for the other side of the haunch portal frame is the same. The

free bending moment is the same. Figure 13 show the free bending moment diagram for haunch

portal frame under dead load with partial fraction, γf 1.4.

When the haunch portal frame is ‘cut’ at the apex point, there are 3 equal and opposite reaction;

moment ‘M’, horizontal shear force ‘H’ and vertical shear force ‘V’ as showing in figure 14. By

Kah Hin Tan - 60 -

Figure 13

using the equilibrium equation, maximum plastic moment occur (plastic hinge) on the rafter of

the structure will be calculate.

Refer [8:p161], figure 15 and figure 16 shows two possible collapse mechanisms for a haunch

portal frame. To determine the plastic moment, reactant moment must determine 1st. Thus,

reactant moment equation are required at A, E, B1, B1 and B5.

Kah Hin Tan - 61 -

hr

hc

hh

A E

W

B1B5

H H

V

V

M MD5 D1

Figure 14

B1 D1

Figure 15Collapse mechanism ‘A’

A E

B5 D5

B1 D1

Where,hr = 3 mhh = 0.5 mhc = 2.5 mW = 18 m

Taking moment about possible plastic hinge may occur on haunch portal frame (refer to figure

14, figure 15 and figure 16):

(Assume anti-clockwise moment positive for left hand side and clockwise positive for right hand

side of ‘cut’ at apex)

At, A; reactant moment = M + 6H + 9V

At, A1; reactant moment = M + 5H + 9V

At, A2; reactant moment = M + 4H + 9V

At, B1; reactant moment = M + 3.5H + 9V

At, B; reactant moment = M + 3H + 9V

At, B1; reactant moment = M + 2.5H + 7.5V

At, B2; reactant moment = M + 2H + 6V


At, B4; reactant moment = M + H + 3V


At, D5; reactant moment = M + 0.5H - 1.5V

At, D4; reactant moment = M + H - 3V

At, D3; reactant moment = M + 1.5H - 4.5V

At, D2; reactant moment = M + 2H - 6V

At, D1; reactant moment = M + 2.5H – 7.5V

At, D; reactant moment = M + 3H - 9V

At, D1; reactant moment = M + 3.5H - 9V

At, E2; reactant moment = M + 4H - 9V

At, E1; reactant moment = M + 5H - 9V

At, E; reactant moment = M + 6H - 9V

Investigation of the collapse mode ‘A’ under dead load by using equilibrium equation.

Kah Hin Tan - 62 -

Figure 16Collapse mechanism ‘B’

A E

B5 D5

B1 D1

Table 11Reactant Moment at purlin point

Refer to figure 17, figure 15 and table 11.set up the equilibrium equation.

At A; 135.27 = reactant moment = M + 6H + 9V

135.37 - M - 6H - 9V = 0 equation 1

At B1; 135.27 = reactant moment + Mp = M + 3.5H + 9V + Mp

135.27 - M - 3.5H - 9V = +Mp equation 2

At B5; 4.4 + Mp = reactant moment = M + 0.5H + 1.5V

4.4 - M - 0.5H - 1.5V = -Mp equation 3

At E; 135.27 = reactant moment = M + 6H - 9V

135.27 - M - 6H + 9V = 0 equation 4

Solve the simultaneous equation;

Consider equation 4 and equation 1.

Equation 1;

M = 135.37 – 6H – 9V equation 5

Equation 4;

M = 135.27 - 6H + 9V equation 6

Kah Hin Tan - 63 -

Figure 17Free bending moment and reactant moment under collapse mechanism ‘A’

Solve equation 6 and equation 5,

135.37 – 6H – 9V = 135.27 - 6H + 9V

V = 0

Substitute equation 5 into equation 2,

135.27 – (135.37 – 6H – 9V) - 3.5H - 9V = +Mp

135.27 – 135.27 + 6H + 9V -3.5H -9V = +Mp

2.5H = +Mp equation 7

Substitute equation 5 and equation 7 into equation 3,

4.4 – (135.37 – 6H – 9V) - 0.5H - 1.5V = -2.5H

8H = 130.97

H = 16.37 k N

Substitute H into equation 5 and equation 7, to obtain M and Mp.

M = 135.37 – 6(16.37)

= 37.15 k N.m

Mp = 2.5 (16.37)

= 40.93 k N.m

Mp > M (obey fundamental yield condition of plastic theory).

Location for the plastic hinge to occur on the portal frame is at B1, B5, D1 and D5. To check

whether the plastic hinge is located at the correct position on the rafter, calculation of the

reactant moment on all the purlin point is needed to check the maximum plastic moment can

occur along the rafter. Table 12 show the amount of plastic moment along the rafter on each

point of purlin under influence of dead load.

Kah Hin Tan - 64 -

From table 12, the maximum plastic moment is at point B and D, but the plastic moment occurs

at point B and point D is considered as moment in the haunch. It is assumed the plastic hinge that

occurs at the eaves of the haunch is shifted to the column top immediately below the lower end

of the haunch, point B1 and D1. Thus, the plastic hinge occurs at point D1, D5, B5 and B1. Figure

18 show the free bending moment line, reactant moment line and plastic moment base on the

values that obtain from table 12.

Kah Hin Tan - 65 -

Table 12

Investigation of the collapse mode ‘B’ under dead load will be carry on by using equilibrium

equation.

Kah Hin Tan - 66 -

Figure 19Free bending moment and reactant moment under collapse mechanism ‘B’

Figure 18



135.37 - M - 6H - 9V = 0 equation 8

At B1; 94.58 = reactant moment + Mp = M + 2.5H + 7.5V + Mp

94.58 - M - 2.5H – 7.5V = +Mp equation 9


4.4 - M - 0.5H - 1.5V = -Mp equation 10


135.27 - M - 6H + 9V = 0 equation 11



Equation 1;

M = 135.37 – 6H – 9V equation 12

Equation 4;

M = 135.27 - 6H + 9V equation 13


135.37 – 6H – 9V = 135.27 - 6H + 9V

V = 0


94.58 – (135.37 – 6H – 9V) - 2.5H – 7.5V = +Mp

94.58 – 135.27 + 6H + 9V -2.5H -9V = +Mp

3.5H - 40.69 = +Mp equation 14


4.4 – (135.37 – 6H – 9V) - 0.5H - 1.5V = -3.5H + 40.69

9H = 171.66

H = 19.07 k N

Kah Hin Tan - 67 -


M = 135.37 – 6(19.07)

= 20.95 k N.m

Mp = 3.5 (19.07) – 40.69

= 26.06 k N.m


Location for the plastic hinge to occur on the portal frame is at B1, B5, D1 and D5. To check

whether the location of the plastic hinge is correct or wrong, calculation of the moment at critical

point can be done,

Let take B1,

At B1; Moment in envelope = free bending moment - reactant moment

= 135.27 - M - 3.5H - 9V

= 135.27 – 20.95 – 3.5(19.07)

= 47.57 k N.m (> Mp)

Yield violated the fundamental condition of plastic theory; mechanism ‘B’ is invalid.

Mechanism ‘A’ critical = 40.93 k N.m

Suitable cross-section of properties for the rafter to sustain the plastic moment occurs on collapse

mechanism ‘A’.

Refer [1:p32: table 11], [1:p29: figure 5] and [1:p30: 3.5.2];

Refer to appendix, given section properties for the rolled I-Section,

The classification of this rolled I-section is determined by referring [1:p32: table11] and [1:p29:

figure 5]

b = 180/2 = 90

T = 16

d = 413 – (2 x 16) = 381

t = 9.65

Kah Hin Tan - 68 -

py = design strength = 275 N.mm-2 refer [1:table 9]

= = 1

Outstand element of compression flange for rolled section;

5.625 < 9ε : Class 1 Plastic

Web of an I-Section, neutral axis at mid depth;

39.48 < 80ε : Class 1 Plastic

Class 1 Plastic – Cross-sections with plastic hinge rotation capacity.

Since the section is class 1 plastic, it may use for plastic design.

Given formula where,

Sxx required =

Where Mp is maximum plastic moment and py is design strength.

From previous calculation,

Mp = 40.93 k N.m (under dead load)

py = 275 M N.m-2

= 0.149 x 10-3 m3

Zxx required = 0.149 x 10-3 m3

Zxx provided by section = 1.51 x 10-3 m3.

The rafter section can sustain the plastic moment.

Kah Hin Tan - 69 -

5.1.2. Plastic Moment under Dead and Imposed Load.Dead load + Imposed load combination.

Refer table 2 and table 4 to obtain the concentrated load acting on the haunch portal frame at the

purlin points shown in figure 12 and calculate the free bending moment under imposed load and

dead load, where the imposed load is refer to snow load.

Table 12 indicate the total concentrated load acting on each of the purlin point of the haunch

portal frame.

Take moments,

MB5 = 11.03 k x 1.5

= 16.55 k N.m

MB4 = (11.03 k x 3) + (21.04k x 1.5)

= 64.65 k N.m

MB3 = [11.03 k x 4.5] + [21.04 k x (3 + 1.5)]

= 144.32 k N.m

MB2 = [11.03 k x 6] + [21.04 k x (4.5 + 3 + 1.5)]

= 255.54 k N.m

MB1 = [11.03 k x 7.5] + [21.04 k x (6 + 4.5 + 3 + 1.5)]

= 398.33 k N.m

MB = [11.03 k x 9] + [21.04 k x (7.5 + 6 + 4.5 + 3 + 1.5)]

= 572.67 k N.m

MB = MA

Kah Hin Tan - 70 -

Table 13

Figure 20 is the free bending moment diagram for the haunch portal frame under dead

load and imposed load with partial fraction, γf = 1.4 for dead load and γf = 1.6 for imposed load.

Plastic moment need to be calculated under load combination (dead load + imposed load) to

ensure the I-Section that been choose can sustain the plastic moment produce by the load.

Possible collapse mechanism ‘A’ will be check under (dead + imposed) load.

Kah Hin Tan - 71 -

Figure 20



572.67 - M - 6H - 9V = 0 equation 15


572.67 - M - 3.5H - 9V = +Mp equation 16


16.55 - M - 0.5H - 1.5V = -Mp equation 17


572.67 - M - 6H + 9V = 0 equation 18



Equation 15;

Kah Hin Tan - 72 -

Figure 21Free bending moment and reactant moment under collapse mechanism ‘A’

M = 572.67 – 6H – 9V equation 19

Equation 18;

M = 572.67 - 6H + 9V equation 20


572.67 – 6H – 9V = 572.67 - 6H + 9V

V = 0


572.67 – (572.67 – 6H – 9V) - 3.5H - 9V = +Mp

572.67 – 572.67 + 6H + 9V -3.5H -9V = +Mp

2.5H = +Mp equation 21


16.55 – (572.67 – 6H – 9V) - 0.5H - 1.5V = -2.5H

8H = 556.12

H = 69.52 k N


M = 572.67 – 6(69.52)

= 155.55 k N.m

Mp = 2.5 (69.52)

= 173.8 k N.m


Reactant moment along the rafter need to be discovered, to ensure the plastic moment occurs on

B1, B5, D5 and D1 is the maximum value. Table 14 show the reactant moment value along the

column and rafter of the frame.

H 69.52V 0M 155.55

All unit in (k N.m)

Kah Hin Tan - 73 -

Plastic Moment under Dead Load + Imposed Load

Point Reactant Moment F.B.M Reactant MomentPlastic

MomentA = M + 6H + 9V 572.67 572.67 0

B^1 = M + 3.5H + 9V 572.67 398.87 173.8B = M + 3H + 9V 572.67 364.11 208.56B1 = M + 2.5H + 7.5V 398.33 329.35 68.98B2 = M + 2H + 6V 255.54 294.59 -39.05B3 = M + 1.5H + 4.5V 144.32 259.83 -115.51B4 = M + H + 3V 64.65 225.07 -160.42B5 = M + 0.5H + 1.5V 16.55 190.31 -173.76C = M 0 155.55 -155.55

D5 = M + 0.5H - 1.5V 16.55 190.31 -173.76D4 = M + H - 3V 64.65 225.07 -160.42D3 = M + 1.5H - 4.5V 144.32 259.83 -115.51D2 = M + 2H - 6V 255.54 294.59 -39.05D1 = M + 2.5H - 7.5V 398.33 329.35 68.98D = M + 3H - 9V 572.67 364.11 208.56

D^1 = M + 3.5H - 9V 572.67 398.87 173.8E = M + 6H - 9V 572.67 572.67 0

From table 14, it shows that at point B1, B5, D5 and D1 is the maximum point for the plastic

moment to occur. Figure 22 shows the free bending moment, reactant moment and plastic

moment of the portal frame under dead load and imposed load.

Kah Hin Tan - 74 -

Table 14

Plastic moment provided by the I-section for the rafter need to be check to ensure the plastic

moment provided is more than the plastic moment occurs on the haunch portal frame under dead

load and imposed load.

Plastic Moment provided by the section;

Mp = Zxx x py

=1.51 x 10-3 x 275 x 106

= 415.25 k N.m

Plastic Moment required;

Mp = 173.76 k N.m

Plastic Moment required for the collapse mode ‘A’ < Plastic Moment provided by the section.

The haunch portal frame is said to be satisfactory for load combination (1.4Gk + 1.6Qk).

5.1.3. Plastic Moment under Dead, Imposed and Wind Load.To investigate the plastic moment of the haunch portal frame under dead, imposed and

wind load, it is more complex compare with portal frame which under combination of dead and

Kah Hin Tan - 75 -

Figure 22

imposed load only. This is due to the wind load is subjected on the wall of the structure, and it

will either increase or decrease the free bending moment value on the base of the portal frame.

Thus, the location of the highest plastic moment or plastic hinge might occur along the haunch

portal frame, need to be analysing again. The wind load acting on the roof purlin and wall purlin

already been calculated on section 4. Two separate case need to be analyses, where the windward

roof is under suction pressure and pressing pressure.

5.1.3.1. Suction wind pressure on windward roof.

The reaction of each purlins on the haunch portal frame when subjected with dead load,

imposed load and wind load, need to add it up together to obtain the total concentrated load

acting on each purlins. Table 15 show the total load for each purlin acting on the haunch portal

frame. The value of ultimate concentrated load of the purlin for dead, imposed and wind load,

can be obtain from table 3, table 5, and table 9.

Kah Hin Tan - 76 -

Table 15

BB1

B2B3

B4B5

CB , CDD5

D4D3

D2 D1D

Refer table 15 and figure 23, calculate free bending moment diagram.

Take moments,

MB5 = 6.12 k x 1.5

= 9.18 k N.m

MB4 = (6.12 k x 3) + (11.39k x 1.5)

= 35.45 k N.m

MB3 = [6.12 k x 4.5] + [11.39 k x (3 + 1.5)]

= 78.8 k N.m

MB2 = [6.12 k x 6] + [11.39 k x (4.5 + 3 + 1.5)]

= 139.23 k N.m

MB1 = [6.12 k x 7.5] + [11.39 k x (6 + 4.5 + 3 + 1.5)]

= 216.75 k N.m

MB = [6.12 k x 9] + [11.39 k x (7.5 + 6 + 4.5 + 3 + 1.5)]

= 311.36 k N.m

MA2 = 311.36 k N.m

MA1 = 311.36 + [6.78 x 1]

= 318.14 k N.m

MA = 311.36 + [6.78 x (2+1)]

= 331.7 k N.m

MD5 = 5.26 k x 1.5

= 7.89 k N.m

MD4 = (5.26 k x 3) + (9.67k x 1.5)

= 30.29 k N.m

MD3 = [5.26 k x 4.5] + [9.67 k x (3 + 1.5)]

= 67.19 k N.m

MD2 = [5.26 k x 6] + [9.67 k x (4.5 + 3 + 1.5)]

Kah Hin Tan - 77 -

E2

E1

A2

A1

EA

Figure 23

= 118.59 k N.m

MD1 = [5.26 k x 7.5] + [9.67 k x (6 + 4.5 + 3 + 1.5)]

= 184.5 k N.m

MD = [5.26 k x 9] + [9.67 k x (7.5 + 6 + 4.5 + 3 + 1.5)]

= 264.92 k N.m

ME2 = 264.92 k N.m

ME1 = 264.92 + [-1.508 x 1]

= 263.41 k N.m

ME = 264.92 + [-1.508 x (2+1)]

= 260.4 k N.m

Figure 24 show the free bending moment diagram under combination dead + imposed + wind

(up) load with partial fraction, γf = 1.2. Base on the reaction moment from table 11 for each

purlin points, plastic moment will be calculated. By referring to figure 24, the free bending

moment at the windward side is higher than the leeward side. It is predicted that the plastic

moment will occur at the eaves of the haunch on the leeward side of the haunch portal frame, yet

the plastic moment on the windward side will be check. [9]

Kah Hin Tan - 78 -

Figure 24

Refer to figure 25, figure 23 and table 11 to set up the equilibrium equation.


331.7 - M - 6H - 9V = 0 equation 22


311.36 - M - 3.5H - 9V = +Mp equation 23


9.18 - M - 0.5H - 1.5V = -Mp equation 24


260.4 - M - 6H + 9V = 0 equation 25

Solve the simultaneous equation

Recall equation 22 and equation 25,

Equation 22;

M = 331.7 – 6H – 9V equation 26

Equation 25;

M = 260.4 – 6H + 9V equation 27

Solve equation 26 and equation 27;

331.7 – 6H – 9V = 260.4 – 6H + 9V

18V = 71.3

Kah Hin Tan - 79 -

Figure 25

V = 3.96 k N

Substitute equation 26 and V into equation 23;

311.36 – (331.7 – 6H – 9V) - 3.5H - 9V = +Mp

311.36 – 331.7 + 6H – 3.5H = +Mp

-20.34 + 2.5H = +Mp equation 28

Substitute equation 28, equation 26 and V into equation 24;

9.18 – (331.7 – 6H – 9V) - 0.5H - 1.5V = 20.34 – 2.5H

8H = 313.16

H = 39.15 k N

Substitute H and V into equation 28 and equation 26.

+Mp = -20.34 + 2.5H

= -20.34 + 2.5(39.15)

= 77.54 k N.m

M = 331.7 – 6H – 9V

= 331.7 – 6(39.15) – 9(3.96)

= 61.16 k N.m

Plastic moment is obtain at point B1 and B5, other point along the frame need to be check. To

ensure the highest plastic moment value doesn’t excess the plastic moment provided by the

section. Table 16 show the value of plastic moment, reactant moment and free bending moment

under dead + imposed + wind load. The negative and positive sign for the plastic moment,

indicate the I-section either under sagging or hogging.

H 39.15 V 3.96 M 61.16 All unit in (k N.m)

Kah Hin Tan - 80 -

Plastic Moment under Dead + Imposed + Wind(up) Load

Point Reactant Moment F.B.MReactant Moment

Plastic Moment

A = M + 6H + 9V 331.7 331.7 0A1 = M + 5H + 9V 318.14 292.55 25.59A2 = M + 4H + 9V 311.36 253.4 57.96B^1 = M + 3.5H + 9V 311.36 233.825 77.535B1 = M + 2.5H + 7.5V 216.75 188.735 28.015B2 = M + 2H + 6V 139.23 163.22 -23.99B3 = M + 1.5H + 4.5V 78.8 137.705 -58.905

B4 = M + H + 3V 35.45 112.19 -76.74B5 = M + 0.5H + 1.5V 9.18 86.675 -77.495D5 = M + 0.5H - 1.5V 7.89 74.795 -66.905D4 = M + H - 3V 30.29 88.43 -58.14

D3 = M + 1.5H - 4.5V 67.19 102.065 -34.875D2 = M + 2H - 6V 118.59 115.7 2.89D1 = M + 2.5H - 7.5V 184.5 129.335 55.165D^1 = M + 3.5H - 9V 264.92 162.545 102.375E2 = M + 4H - 9V 264.92 182.12 82.8E1 = M + 5H - 9V 263.41 221.27 42.14E = M + 6H - 9V 260.4 260.42 -0.02

Figure 26 show the graph of the free bending moment, reactant moment and plastic moment

travel along the frame.

Kah Hin Tan - 81 -

Table 16

Refer to table 16; the maximum plastic moment is at point D1, 102.375 k N.m. The plastic

moment provided by the cross section (W410 x 0.73) is 415.25 k N.m. Plastic Moment required

under dead, imposed and wind load combination with partial fraction, γf = 1.2 < Plastic Moment

provided by the section. The haunch portal frame is said to be satisfactory for load combination

1.2(Gk + Qk + Wk).

5.1.3.2. Pressing wind pressure on windward roof.

The method to check the I-section of the haunch portal frame is the same that carried out on

chapter 5.1.3.1. The total load acting on each purling point need to be determine, then the

Kah Hin Tan - 82 -

Figure 26

bending moment and the reactant moment on the frame, and finally, the maximum plastic

moment that occur mustn’t exist the plastic moment provided by the section.

The total ultimate concentrated load acting on each purlin points can be determine by combining

table 3, table5 and table 10. Table 17 shows the load combination of dead, imposed and wind

load.

Compare the total ultimate concentrated load of table 17 and table 15, the different between two

tables is the ultimate concentrated load value at the windward roof. Thus, only the windward side

moment need to recalculate and the leeward side moment can be obtain from chapter 6.1.3.1.

Take moments,

MB5 = 10.78 k x 1.5

= 16.17 k N.m

Kah Hin Tan - 83 -

Table 17

MB4 = (10.78 k x 3) + (20.71 k x 1.5)

= 63.41 k N.m

MB3 = [10.78 k x 4.5] + [20.71 k x (3 + 1.5)]

= 141.71 k N.m

MB2 = [10.78 k x 6] + [20.71 k x (4.5 + 3 + 1.5)]

= 251.07 k N.m

MB1 = [10.78 k x 7.5] + [20.71 k x (6 + 4.5 + 3 + 1.5)]

= 391.5 k N.m

MB = [10.78 k x 9] + [20.71 k x (7.5 + 6 + 4.5 + 3 + 1.5)]

= 563 k N.m

MA2 = 563 k N.m

MA1 = 563 + [6.78 x 1]

= 569.78 k N.m

MA = 563 + [6.78 x (2+1)]

= 583.34 k N.m

Free bending moment under load combination [(dead + imposed + wind (down)] is shown in

figure 27.

Since the shape of the free bending moment shown figure 27 is quite similar with the free

bending moment shown in figure 24. It is expected that the maximum plastic moment or plastic

hinge will occur at the eaves of the haunch on the leeward side of the frame.

Kah Hin Tan - 84 -

Figure 27

Determine the plastic moment,


583.34 - M - 6H - 9V = 0 equation 29

At B1; 563 = reactant moment + Mp = M + 3.5H + 9V + Mp

563 - M - 3.5H - 9V = +Mp equation 30


16.17 - M - 0.5H - 1.5V = -Mp equation 31


260.4 - M - 6H + 9V = 0 equation 32

Solve the simultaneous equation

Use equation 29 and equation 32,

Equation 29;

M = 583.34 – 6H – 9V equation 33

Equation 32;

M = 260.4 – 6H + 9V equation 34

Solve equation 33 and equation 34;

583.34 – 6H – 9V = 260.4 – 6H + 9V

18V = 322.94

V = 17.94 k N

Substitute equation 33 and V into equation 30;

563 – (583.34 – 6H – 9V) - 3.5H - 9V = +Mp

311.36 – 331.7 + 6H – 3.5H = +Mp

-20.34 + 2.5H = +Mp equation 35

Substitute equation 33, equation 35 and V into equation 31;

16.17 – (583.34 – 6H – 9V) - 0.5H - 1.5V = 20.34 – 2.5H

8H = 452.96

Kah Hin Tan - 85 -

H = 56.62 k N

Substitute H and V into equation 33 and equation 35.

+Mp = -20.34 + 2.5H

= -20.34 + 2.5(56.62)

= 121.21 k N.m

M = 583.34 – 6H – 9V

= 583.34 – 6(56.62) – 9(17.94)

= 82.16 k N.m

Kah Hin Tan - 86 -

Table 18

Table 18 and figure 28 show the collapse plastic moment along the haunch portal frame under

[dead + imposed + wind (down)] load. From table 18, the maximum plastic moment is 146.05 k

N.m and it occurs at point D1. The plastic moment provided by the I-section is 415.25 k N.m.

Hence, the frame is say to be satisfactory under dead load, imposed load and wind (down) load

with partial fraction γf = 1.2.

Plastic moment on each load combination for haunch portal frame.

Figure 29 show the plastic moment distributed around the haunch frame under (1.4Gk).

Kah Hin Tan - 87 -

Figure 28

All units for the moment are in (k N.m).

Figure 30 show the plastic moment distributed around the haunch frame under (1.4Gk + 1.6Qk).

All units for the moment are in (k N.m).

Figure 31 show the plastic moment distributed around the haunch frame under 1.2(Gk + Qk +

Wk), in where the wind pressure is a suction pressure on the windward roof. All units for the

moment are in (k N.m).

Kah Hin Tan - 88 -

49.01

49.01

16.51

8.7526.76 37.47

40.94

37.15

40.9437.47 26.76

8.75

16.51

49.01

49.01

0 0

Figure 29

208.56

208.56

68.96

39.05115.51 160.42

173.76

155.55

0 0

Figure 30

208.56

208.56

68.96

39.05115.51160.42

173.76

Figure 32 show the plastic moment distributed around the haunch frame under 1.2(Gk + Qk +

Wk), in where the wind pressure is a pressing pressure on the windward roof. All units for the

moment are in (k N.m).

Through out the analysis and calculation that have carry out in this section, the haunch portal

frame under 4 type of load combination is satisfactory, where the maximum plastic moment

under 4 type of loading doesn’t excess the plastic moment provided by the I-section (W410 x

Kah Hin Tan - 89 -

97.11

97.11

28.02

23.99

58.9176.74

77.5

61.16

0 0

121.95

121.95

55.16

2.89

34.8858.14

66.91

Figure 31

77.54

57.96

25.59

102.38

82.8

42.14

92.9

149.52

33.24

51.97

106.11129.19

121.21

82.16

0 0

174.36

174.36

95.34

30.83

19.17

54.67

75.67

Figure 32

149.52

121.21

43.06

146.05

117.74

59.61

0.73). Nevertheless, there is several factor need to be consider for the haunch portal frame; the

resistance of the frame under lateral-torsional buckling, distance for the segment adjacent to a

plastic hinge, segment or member with one flanged restrained (haunch and stanchion), in-plane

stability under gravity load and out of plane stability. BS5950-1 will be use as a guide line to

carry out an investigation on the haunch portal frame in section 6.2.

5.2. Structure Stability.

Kah Hin Tan - 90 -

Structures consist of several members and when it is subjected to different kind of loads;

dead load, imposed load and wind load. Base on the theory, for a perfectly straight beam is

loaded in the plane of the web, the elastic critical load the beam will fail suddenly by deflecting

sideways and twisting about its longitudinal axis [8]. In the other way around, the theory is

referring to buckling. If such members buckle when under service condition, it will create

instability of the structure, consequently the building or structure will collapse.

5.2.1. Resistance to Lateral-Torsional Buckling.Base on BS5950-1, the standard method to check the buckling resistance of the structure

is under section [1:p49: 4.3.6.2].

BS5950-1, section 4.3.6.2 stated;

Where,

Mx - maximum major axis moment in the segment.

Mb - buckling resistance moment.

mLT- equivalent uniform moment factor for lateral-torsional buckling.

Thus, Mb and mLT need to be obtain,

Mb = pbSx [1:p50:4.3.6.4]

pb - bending strength [1:p51: table 16]

Sx - plastic modulus about the major axis

To find the bending strength pb from table 16, equivalent slenderness λLT must be obtain, and λLT

can be obtain from BS5950-1,section 4.3.6.7.

For I-section member λLT =

In which,

Kah Hin Tan - 91 -

where,

LE - effective length for lateral-torsional buckling.

ry - radius of gyration about the minor axis.

u - buckling parameter = 0.9 (for rolled I-section with equal flange)

Βw = 1 (for class 1 plastic)

The Effective length under lateral torsional buckling is the length of the rafter;

Given horizontal length of the rafter = 9m

Effective length of the rafter =

= 9.47 m

Radius of gyration = 0.17 m refer to appendix

= 55.71

for I-section with equal flange ν can be determined;

where,

x - torsional index, and is defined as;

D = 413 mm, depth of the section

T = 16 mm, flange thickness

x = 25.81

Kah Hin Tan - 92 -

= 0.95

Find the equivalent slenderness λLT;

λLT =

=

= 47.63

Determine pb bending strength from table 16;

pb = 238 (N/mm-2)

Sx = 1.51 x 10-3 m3 [refer to appendix]

Thus,

Mb =

= 359.38 k N.m.

In the previous analysis section 5.1, where the plastic moment is determine by using equilibrium

equation under four type of load combination as stated in table 1. The results that obtain from the

analysis are showing in figure 29, figure 30, figure 31 and figure 32. Hence, there are 6 different

form and value of plastic moments acting on the rafter. Therefore, 6 lateral torsional buckling

analyses will carry out, to ensure the rafter can sustain the lateral-torsional buckling effect.

Find the mLT value for six form of plastic moment,

Refer [1:p54: table 18]

Given mLT for beam;

But mLT ≥ 0.44

Where,

M2 and M4 - are the value at the quarter point.

M3 - are the value at mid-length.

1st case,

Refer to figure 29, the plastic moment on the windward and leeward of the rafter is the same,

thus it is treated as same case.

Kah Hin Tan - 93 -

2nd case,

Refer to figure 30, the plastic moment on the windward and leeward of the rafter is the same,

hence, it should treat as same case.

3rd case and 4th case,

refer to figure 31, the plastic moment on the windward and leeward of the rafter give different

form and value of plastic moment, therefore, it should treated as separated case.

3rd case: Windward mLT,

4th case: leeward mLT,

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5th case and 4th case,

refer to figure 31, the plastic moment on the windward and leeward of the rafter give different

form and value of plastic moment, therefore, it should treated as separated case.

5th case: windward mLT,

6th case: leeward mLT,

The leeward mLT is smaller than 0.44. Therefore, the mLT value for this case is treated as 0.44.

From BS5950-1, section 4.3.6.2. Given,

1st case,

mLT = 0.75

Mx max = 135.27 k N.m

Mb = 359.38 k N.m.

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= 479.17 k N.m

Mx < 479.17 k N.m

Lateral-torsional buckling is satisfactory for 1st case.

2nd case,

mLT = 0.58

Mx max = 572.67 k N.m

Mb = 359.38 k N.m.

= 619.62 k N.m

Mx < 619.62 k N.m

Lateral-torsional buckling is satisfactory for 2nd case.

3rd case,

mLT = 0.67

Mx max = 311.36 k N.m

Mb = 359.38 k N.m.

= 536.39 k N.m

Mx < 536.39 k N.m

Lateral-torsional buckling is satisfactory for 3rd case.

4th case,

mLT = 0.49

Mx max = 264.92 k N.m

Mb = 359.38 k N.m.

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= 733.43 k N.m

Mx < 733.43 k N.m

Lateral-torsional buckling is satisfactory for 4th case.

5th case,

mLT = 0.74

Mx max = 563 k N.m

Mb = 359.38 k N.m.

= 485.65 k N.m

Mx > 485.65 k N.m

Lateral-torsional buckling is not satisfactory for 5th case.

6th case,

mLT = 0.44

Mx max = 264.92 k N.m

Mb = 359.38 k N.m.

= 816.77 k N.m

Mx > 816.77 k N.m

Lateral-torsional buckling is satisfactory for 6th case.

Calculations prove that, the shape and dimension of the structure under influence of

loading or moment can resist 5 cases out of 6 for lateral-torsional buckling. To prevent the

structure collapse under influence of moment which cause lateral torsional buckling. The rafter

may be stiffening by fix in a stiffener at the rafter. It may reduce the failure cause by the lateral

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torsional buckling. The purlin which connected from one frame to the other frames may help to

prevent the lateral torsional buckling.

5.2.2. Segment Adjacent to Plastic Hinge.BS5950-1 section 5.3.3 stated that, the length of a segment adjacent to a plastic hinge

location, between the points at which compression flange if laterally restrained, the distance

shouldn’t exceed Lm. From the statement, it indicates the distance between each purlin shouldn’t

exceed the distance Lm which can be obtain from;

where,

fc - compressive stress (in N.mm-2) due to axial force.

py - design strength (in N.mm-2).

ry - radius of gyration about the minor axis.

x - torsional index = 25.81

By referring to previous calculation in chapter 6.1, when the haunch portal frame is ‘cut’ into

half at the apex position, it will induce 3 opposite and equal reactions which is Moment, shear

force at horizontal axis (H) and shear force at vertical axis (V). The shear at the horizontal axis

(H) is to calculate the compressive stress fc.

where,

A - is the cross-section of area of the I-section = 9480 mm2 [refer to appendix]

The maximum horizontal shear force (H) that obtains from previous calculation in chapter 6.1 is

69.52 k N under influence of dead load and imposed load.

= 7.33 N.mm-2

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Torsional index, x = 25.81

Design strength, py = 275 N.mm-2

Radius of gyration, ry = 40.4 mm

Lu =

=

= 1225.22 mm

Actual hinge spacing or purlin spacing;

Spacing =

= 1578.95 mm.

The actual length between the plastic hinge is >Lu. Thus, the number of purlin at the hinge region

will be double up to reduce the concentrated load of the purlin hence reduce the shear force

acting on the horizontal direction (H).

5.2.3. Haunch and Stanchion Stability.BS5950-1 section 5.3.4 stated, the spacing Ly between restraints to the compression

flange shouldn’t exceed the limiting spacing Ls given as; for steel grade S275;

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where,

K1 = 1 + 0.25(Dh/Ds)2/3

Torsional index, x = 25.81

Radius of gyration, ry = 40.4 mm [refer to appendix]

Where Dh and Ds is the dimension of depth of the rafter and haunch. [Refer to BS5950:1: figure

17].

Ds = I-section depth / cos 18.43

=

= 434.74 mm

Dh =

= 282.63 mm

= 1.19

Limiting spacing provided, Ls;

= 2789.31 mm

Actual haunch length;

Length =

= 1578.95 mm.

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Given Ls > length of the haunch. The haunch is said to be stable.

It is permissible to check the stanchion stability using BS5950-1, section 5.3.4. Given

limiting spacing;

refer [1:p120]

K1 = 1 for an un-haunch segment;

= 3318.02 mm

Stanchion height = 2500 mm

Limiting spacing > stanchion height. Thus, the stanchion is stable.

5.2.4. Sway Stability.The overall stability of the structure is check by using BS5950-1 section 5.5.4.2.2.

The portal frame is stable if it satisfactory the equation stated;

in which,

for single bay frame

Ω is the arching ratio;

D - is the cross section depth of the rafter; refer to appendix

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Dh - is the the additional depth of the haunch; refer [1:p123: figure 17]

Ds - is the the depth of the rafter; refer [1:p123: figure 17]

h - is the mean column height;

Ic - is the in-plane second moment of area of the column; refer to appendix

Ir - is the in-plane second moment of area of the rafter; refer to appendix

L - is the span of the bay;

Lb - is the effective span of the bay;

Lh - is the length of a haunch; refer [1:p123: figure 17]

Lr - is the total developed length of the rafter; refer [1:p123: figure 18a]

pyr - is the design strength of the rafter in (N.mm-2)

Wo - is the value of Wr for plastic failure of the rafters; refer [1:p123: figure 18b]

Wr - is the total factored vertical load on the rafters; refer [1:p123: figure 18b]

Obtain effective span of the bay, Lb.

L = 18m

Lh = 1.5m

Dh = 0.28 m

Ds = 0.43 m

Find,

Lb =

= 16.82 m

= 40.73

Given Ic and Ir = 274 x 10-6 m4

Find,

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ρ = 12

Consider a fixed end beam with a length of span (L) supporting a uniform distributed load (ω),

the plastic moment for the hinge to occur is given as;

given,

py - design strength = 275 M N.m-2

Sxx - Section modulus [refer to appendix]

Thus,

and

Wo = ωL

So,

Wo =

Wo =

= 325.11 k N

Lr =

= 18.95 m

find,

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= and,

For the frame sway stability analysis, four cases will be considered under load combination as

stated in table 1.

1st case,

Dead load [refer to table2]

Total vertical load acting on the rafter,

Wr = k N

= 54.26 k N.

Thus,

= 0.17

Given,

and,

=

= 1118.11

40.73 ≤ 1118.11

2nd case,

Dead load + Imposed Load [refer to table13]


Wr = k N

= 232.46 k N.

Thus,

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Portal frame under dead load is satisfactory under sway stability check.

= 0.72

Given,

and,

=

= 264

40.73 ≤ 264

Portal frame under dead load + imposed load is satisfactory under sway stability check.

3rd case,

Dead load + Imposed Load + Wind Load (UP) [refer to table15]


Wr = k N

= 116.68 k N.

Thus,

= 0.36

Given,

and,

=

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= 528

40.73 ≤ 528

Portal frame under dead load + imposed + wind load (UP) is satisfactory under sway stability

check.

4th case,

Dead load + Imposed Load + Wind Load (DOWN) [refer to table17]


Wr = k N

= 167.94 k N.

Thus,

= 0.52

Given,

and,

=

= 365.54

40.73 ≤ 365.5

Portal frame under dead load + imposed + wind load (DOWN) is satisfactory under sway

stability check.

Through out the analysis that have been carry out in this section, the structure is rigid and

stiff enough under influence of 4 type of loading. But only one analysis in this section that prove

unsatisfactory which is; lateral torsional buckling for case number 5 (pushing windward pressure

acting on the windward roof). To preventing the lateral torsional buckling to occur, a stiffener

must fix in the I-section or rafter at the position where the maximum plastic moment occur. It is

assume, the stiffener that fix in the I-section, can resist the lateral torsional buckling.

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Kah Hin Tan - 107 -6. DRAWING

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7. Evaluation and Discussion.

The final design is development of the concept 3. This design overcomes most of the

negatives effect discusses on the PDS. In this project, only dead load, snow load and wind load is

considered. The analysis that had carried out in chapter 5 proves that the final design is

satisfactory under all effect of loading combination, except the lateral torsional buckling effect

when the structure is subjected with side wind. The solution for the lateral torsional buckling

effect is by fixing the stiffener on the rafter I-section at the point where the maximum plastic

moment occur. Figure 33 show the location where the stiffener is installed at the rafter by MMA

welding process. For dimension inquire (refer to drawing num 0003)

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7. EVALUATION AND DISCUSSION

The influence of wind pressure on the steel portal frame structure clearly been observed

in chapter 5, where the wind load subjected to the structure increase to a limit that the lateral

torsional buckling unsatisfactory meet the standard given in BS5950. Thus, the wind effect can

bring destructive to steel structure and the structure might collapse. In British Standard 5950-1

2.4.2.3, the resistance of the structure to the horizontal force is to provided the triangulated

bracing, moment resisting joints, cantilever column, shear walls and etc. the x-bracing had been

used in this design as shown in drawing number 2. The x-bracing act as a compression member

which tied up the portal frame together, thus prevent the structure sway and buckling effect.

For corrosion resistance, it can be preventing by made some changes on the design.

Corrosion will occur if two elements exist which is water or moisture and air. It is impossible to

seal a steel structure with plastic or any element that provided air tight. Thus, the water element

must be preventing to trap at the steel for long time, drainage with holes or opening can reduce

the corrosion to occur. Figure 34 show the final design for the stanchion to prevent corrosion.

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Figure 33

Basically, the final design had meet lecture objective and individual objective and the structure is

stiff and strong enough to sustain dead load, imposed load and wind load.

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Figure 33

Suggested Future Work:

The structure can attach with the overhead crane that useful for the warehouse for

warehouse purposes.

Elongated the length of the warehouse without change the concept design.

Expand the site of the warehouse examples; change single bay portal frame to multi-bay

portal frame.

Design a structure which can sustain the thermal effect such as thermal stress.

Design a structure base on BS5950-8 (fire resistance)

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8. FUTURE DEVELOPMENT

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References:

Following are the sources of information, which has been utilised during the conduct of the

report.

Standards:

1. British Standard 5950-1:2000

- Code of practice for design – Rolled and welded section.


- Specification for materials, fabrication and erection – rolled and welded section.


- Code of practice for fire resistant design.

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9. REFERENCES


- Specification for hot rolled section.


- Code of practice for dead and imposed load.


- Code of practice for wind loads.


- Code of practice for imposed roof loads.

Books:

8. Title: Limit States Design of Structural SteelworkAuthor: David NethercotPublisher: Van Nostrand Reinhold (UK) Co. LtdDate of Publication: 1986ISBN number: 0-442-31752-2

9. Title: Structural Engineering Design in Practice.Author: Roger WestbrookPublisher: Longman Group UK LimitedDate of Publication: 1988ISBN number: 0-582-0 1735-1

10. Title: Elementary Structural Design of Steelwork to BS449Author: P.C.L Croxton, L.H. Martin, J.A. PurkissPublisher: Edward Arnold LtdDate of Publication: 1984ISBN number: 0 7131 3531

11. Title: Design of Structural Steelwork (2nd edition)

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Author: Peter KnowlesPublisher: Blackie & Son LtdDate of Publication: 1987ISBN number: 0-903384-59-0

12. Title: Plastic Design of FramesAuthor: Sir John Baker & Jacques HeymanPublisher: Syndics of the Cambridge UniversityDate of Publication: 1969ISBN number: 69-19370

13. Title: Plastic Design of Steel StructuresAuthor: A.mrazik, M.Skaloud, M.TochacekPublisher: Ellis Horwood LimitedDate of Publication: 1987ISBN number: 0-85312-381-0

14. Title: Analysis of StructuresAuthor: T.S. ThandavamoorthyPublisher: Oxford University PressDate of Publication: 2005ISBN number: 0-19-567003-5

15. Title: Plastic Design of Steel StructuresAuthor: A.mrazik, M.Skaloud, M.TochacekPublisher: Ellis Horwood LimitedDate of Publication: 1987ISBN number: 0-85312-381-0

16. Title: Design of Structural ElementsAuthor: Chanakya AryaPublisher: Chapman & HallDate of Publication: 1994ISBN number: 0 419 17620 9

17. Title: Structural Detailing (2nd edition)Author: Peter H.NewtonPublisher: Macmillan Education LtdDate of Publication: 1991ISBN number: 0-333-55471-X

18. Title: Mechanics of Materials (4th edition)Author: J.M. Gere and S.P. TimoshenkoPublisher: Stanley ThornesDate of Publication: 1997

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ISBN number: 0-7487-3998-X

19. Title: Designing Steel Structures Methods and CasesAuthor: Sol E.Copper with Andrew C.ChenPublisher: Elsevier Applied ScienceDate of Publication: 1989ISBN number: 0-85334-362-4

20. Title: Handbook of Structural DesignAuthor: I.E. MorrisPublisher: Reinhold Publishing Corporation.Date of Publication: 1987ISBN number: 62-70728

21. Title: Structural DesignAuthor: B.Currie and R.A. SharpePublisher: Stanley Thornes LtdDate of Publication: 1996ISBN number: 0-7487-0417-5

Internet:22. http://en.wikipedia.org/wiki/Manchester23. http://en.wikipedia.org/wiki/Tensile_strength#Explanation24. http://www.weatherbase.com/weather/weather.php3?s=43330&refer=&units=metric25. http://www.thakkargroup.com/sectional_z_purlin.htm

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http://www.thakkargroup.com/sectional_z_purlin.htm

http://www.weatherbase.com/weather/weather.php3?s=43330&refer=&units=metric

http://en.wikipedia.org/wiki/Tensile_strength#Explanation

http://en.wikipedia.org/wiki/Manchester

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10. APPENDIX

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Structural Design (Portal Frame - Plastic Design)

Documents

Transcript of Structural Design (Portal Frame - Plastic Design)