The Basic Design Philosophies

Published Date: 02 Nov 2017

The aim of design is the achievement of an acceptable probability that structures being designed will perform satisfactorily during their intended life. With an appropriate degree of safety, they should sustainall the loads and deformations of normal construction and use and have adequate durability and resistance to the effects of misuse and fire.

Design method

The method recommended in this code is that of limit state design. Account should be taken of accepted theory, experiment and experience and the need to design for durability. Calculations alone do not produce safe, serviceable and durable structures. Suitable materials, quality control and good supervision are equally important.

Durability, workmanship and materials

It is assumed that the quality of the concrete, steel and other materials and of the workmanship, as verified by inspections, is adequate for safety, serviceability and durability.

Design process

Design, including design for durability, construction and use in service should be considered as a whole.The realization of design objectives requires conformity to clearly defined criteria for materials, production, workmanship and also maintenance and use of the structure in service.

Strucural design

Well-detailed and properly-erected structures designed by the limit state method will have acceptable probabilities that they will not reach a limit state, i.e.will not become unfit for their purpose by collapse, overturning, buckling (ultimate limit states), deformation, cracking, vibration, etc. (serviceability limitstates) and that the structure will not deteriorate unduly under the action of the environment over the design life, i.e.will be durable. The usual approach is to design on the most critical limit state and then to check that the remaining limit states will not be reached.

Ultimate limit state (ULS)

Structural stability

The structure should be so designed that adequate means exist to transmit the design ultimate dead, wind and imposed loads safely from the highest supported level to the foundations. The layout of the structure and the interaction between the structural members should be such as to ensure a robust and stable design. The engineer responsible for the overall stability of the structure should ensure the compatibility of thedesign and details of parts and components, even where some or all of the design and details of those part sand components are not made by this engineer. The design strengths of materials and the design loads should be , as appropriate for the ULS. The design should satisfy the requirement that no ULS is reached by rupture of any section, by overturning or by buckling under the worst combination of ultimate loads.

 Robustness

Structures should be planned and designed so that they are not unreasonably susceptible to the effects of accidents. In particular, situations should be avoided where damage to small areas of a structure or failure of single elements may lead to collapse of major parts of the structure.

Unreasonable susceptibility to the effects of accidents may generally be prevented if the following precautions are taken.

applied at each floor or roof level simultaneously(2 floor)

All buildings are provided with effective horizontal ties

1) around the periphery

2) internally

3) to columns and walls.

Serviceability limit states (SLS)

General

The design properties of materials and the design loads should be those given in Section 3 of BS8110-2:1985 as appropriate for SLS. Account should be taken of such effects as temperature, creep, shrinkage, sway, settlement and cyclic loading as appropriate.

 Deflection due to vertical loading 

The deformation of the structure or any part of it should not adversely affect its efficiency or appearance. Deflections should be compatible with the degree of movement acceptable by other elements including finishes, services, partitions, glazing and cladding; in some cases a degree of minor repair work or fixing adjustment to such elements may be acceptable. Where specific attention is required to limit deflections to particular values, reference should be made to

3.2 of BS8110-2:1985; otherwise it will generally be satisfactory to use the span/effective depth ratios given for reinforced concrete

(b).

Minimum cover is the design depth of concrete to all steel reinforcment ,including links.it is the dimension used to in design and indicated on the drawings.the actual cover to all reinforcement should never less than the nominal cover minus 5mm.the nominal cover should,

comply with the recommendation for bar size ,aggregate size and concrete cast against uneven surface

protect the steel from corrosion

protect the steel from fire

allow for surface treatments such as bush hammering

Eg -:

Nominal cover to all reinforcmnet (including links) to meet durability requirmnet

condition of exposure

Nominal cover

mild

25mm

20mm

20mm

20mm

20mm

moderate

-

35

30

25

20

severe

-

-

40

30

25

very severe

-

-

50t

40t

30

extreme

-

-

-

60t

50

maximum free water/cement ratio

0.65

0.60

0.55

0.50

0.45

minimum cement content

275

300

325

350

400

lowest grade of concrete

C30

C35

C40

C45

C50

(c)

A majority of structural cracks occur as a result of the following conditions:

Design deficiency

Construction deficiency

Settlement or heaving of soil

Reinforcement corrosion Sometimes structural cracks manifest themselves with some side effects. Doors and windows do not open and close easily. Floors feel uneven.

External sulfate attack

This is the more common type and typically occurs where water containing dissolved sulfate penetrates the concrete . A fairly well-defined reaction front can often be seen in polished sections; ahead of the front the concrete is normal,or near normal. Behind the reaction front, the composition and microstructure of the concrete will have changed. The effect of these changes is an overall loss of concrete strength

These changes may vary in type or severity but commonly include:

Extensive cracking

Expansion

Loss of bond between the cement paste and aggregate

Alteration of paste composition

Internal sulfate attack

Internal sulfate attack

occurs when a source of sulfate is incorporated into the concrete when mixed. Examples include the use of sulfate-rich aggregate ,excess of added gypsum in the cement or contamination. Proper screening and testing procedures should generally avoid internal sulfate attack

vertical crack

A loading crack is a result of the loading to which the structure is subjected .A properly designed structure would not exhibit these cracks, but an improperlydesigned structure is very susceptible to this damage. Vertical cracking at theend of a structure is typically due to a concentrated force being applied at thetop of a structure which exceeds the shear capacity within the end section of the structure. This type crack typically maintains a tight appearance at the topand at the bottom but may show a wider gap at approximately mid-height of thestructure. This would tend to indicate a bulging effect of the end segment of thestructure away from the remainder of the structure. This crack is illustrated below

Plastic shrinkage crack

 When water evaporates from the surface of freshly placed concrete faster  thanit is replaced by bleed water , the surface concrete shrinks. Due to the restraint provided by the concrete below the drying surface layer, tensile stresses develop in the weak, stiffening plastic concrete, resulting in shallow cracks of varying depth. These cracks are often fairly wide at the surface.

Alkali-aggregate reaction crack

 Alkali-aggregate reactivity is a type of concrete deterioration that occurs when the active mineral constituents of some aggregates react with the alkalihydroxides in the concrete.

.

Thermal crack

 

Temperature rise (especially significant in mass concrete) results from the heat of hydration of cementitious materials. As the interior concrete increases in temperature and expands, the surface concrete may be cooling and contracting . This causes tensile stresses that may result in thermal cracks at the surface if the temperature differential between the surface and center is too great.

ssettlement crack

Loss of support beneath concrete structures, usually caused by settling or washout of soils and sub base materials, can cause a variety of problems in concrete structures, from cracking and performance problems to structural failure. Loss of support can also occur during construction due to inadequate formwork

support or premature removal of forms.

Corrosion Crack

Steel reinforcement , in the alkaline environment provided by concrete, is

in a stable condition because a protective oxide layer  forms on the steel surface, which stops corrosion. There are, however, two situations where this passivating environment at the reinforcement can be disrupted . The first is known as carbonation . This is when atmospheric carbon dioxide dissolves in water to form carbonic acid, which neutralises the concrete alkalinity. The carbonation proceeds through the concrete cover and eventually reaches there in for cement, at which point the passive layer is no longer sustained and corrosion occurs. The second disruptive effect is chloride attack

Preventing methods for cracks

Curing method

Rapid drying of the slab will significantly increase the possibility of cracking. Thechemical reaction which causes concrete to go from the liquid or plastic state to asolid state requires water. This chemical reaction, or hydration, continues tooccur for days and weeks after concrete was poured. Engineers must make surethat the

necessary water is available for this reaction by adequately curingthe slab . The use of liquid curing compounds, covering the slab with plastic,wet burlap , and other methods can be used to cure concrete

Solid ground

The ground upon which the concrete will be placed must be compacted . Never pour concrete on frozen ground as once the ice melt it will cause void in the soil.

Proper usage of material

Many people had wondered why ancient structures are so strong and stillstanding till now. Engineers had found that these buildings were over-designedor in other words, maximum usage of construction material. Let’s take anexample of a driveway concrete slab. A 5-inch thick slab is definitely better insustaining heavy vehicles than a 4-inch thick slab which is more likely to crackunder loading. Some contractors might suggest that 4-inch is just enough whencost comes into consideration but a 5-inch thick is even safer in reality.

Thicker concrete is a good idea for better load bearing structure (for this case, it wasslab). Cracking can be minimized by following other guidelines as well. Installing proper strength concrete for intended use is always a good practice asconcrete is available in many different strengths.

Cover for reinforcement

The deterioration of reinforced concrete is mainly due to reinforcement corrosion. The mechanism of this deterioration is to be reminded. Reinforcements corrode when they are in contact with a high amount of aggressive agents. This is the reason why, the prevention of reinforcement corrosion, in structures to be built, is obtained mainly by controlling the thickness and the quality of the concretecover  . The concrete cover thickness around reinforcement also depends on the environment aggressiveness.

Coating on concrete

In some cases, concrete reinforcement corrodes in a slow pace. Therefore, its cover seems physically satisfactory and has neither crack nor delamination. But, it is then convenient to slow down this corrosion rate, even to stop it. The methods which can be proposed are either  concrete impregnation with water-proof products (sealants) or inhibitors. Paintings and coatings of various thicknesses can also be applied on concrete to improve its resistance to liquid penetration.

Q2

5KN/m =UDL

20 KN 25 KN

4m 4m 4m

12m

This can represent as,

20KN 25KN 2) 5KN/m

+

Finding the support reaction of (1),

20 25

From equilibrium,

A

= 23.33

so,

= 21.66

Different segment of loading

20 25

i) iii)

ii)

21.66 23.33

21.66 V

M ; M – 21.66 x = 0

M = 21.66 x

x = 0 ,M = 0

x = 4, M = 86.64

20

21.66 v

;

20 25

(x – 4)

21.66 v

;

M = -23.34 + 280

x = 8,M = 93.28

x = 12, M = 0

Finding support reaction (2),

12

5KN/m

Considering equilibrium,

it is symetric so,

30 V

;

Maximum bending,

when,

When

Shear force

51.66

31.66

20

11.66

0 4 8 12

-8.34

25

-33.34

-53.34

Bending Moment

86.64+80 89.96+90

(166.64) (179.96 93.28+80

(173.28)

0 4 6 8 12

Q3.

Design a rectangular beam to take an ultimate load 150kN/m.

as a singly reinforced beam

beam whose overall depth is limited to 500 mm.

Refference

Calculation

Resalts

cl 3.4.4.4

BS 8110

Table 3.27

singly R/F section

lets assume d/b =2.0

depth for single R/F section we should assume that x/d = 0.5

k = k1 = 0.156

k =

0.156 = = 425.29

choose ,

d = 475mm

= d + 60 = (475 + 60) = 535mm

b = = 237.5 mm

choose,

225mm

Now chek k,

k =

= 0.118 0.156

so, single R/F is OK

Z = 401 0.95d

401 0.9 475

401 451

so, OK

Area of steel

=

hence,

from the table ,

2T 25

)

=

= 0.815 0.13 OK

overall depth should be 500mm

if overall depth restrict to 500mm, h = 500mm

h – d = 6omm so, d = 440mm

assume b = 225mm

then,

k = k1 = 0.156

= 0.2170.156

=

= 443.9mm from the table

[ we select 402m]

= 0.357 0.2 its OK

lever arm,

Z =

= 440[0.5 +[0.25 -

= 222m 0.95d

= 222m 0.95

=222m 418mm OK

As =

=

= 2304m

from the table,

1963mm2 – 4T 25

2410mm2 – 3T 32

we select - 3T 32

OK

(a) 225mm (b) 225mm

535mm 500mm

= 475mm

b = 225mm

b & d OK

2T 25

compression R/f is required

16 2T (402m

32 3T (2410m

compression steel

16 2T

tension steel

32 3T

Q4.

b

x h

Steel area

Bendy moment=’M’

Assume cover is x

As define area of tension reinforcement,

So,

Step – finding k

Step – determining lever-arm from

Step – finding area steel

Step – checking the area of steel accuracy within provided limit

Q5

The current code of practice called "Structural Use of Steelwork in Building" is BS 5950 Part 1. This code gives specific guidance on the strength and stiffness of steel structures for buildings to allow numerical calculations to be made.

Beam capacity is checked by comparing the ultimate strength with factored loadings.

It checks the strength of a structure by ensuring ultimate strength is not less than working load x load factor

The load factor varies with the type of load.

Different load factors are applied to different types of load and load combinations. This reflects the varying degree of confidence in the values of dead, imposed and wind loads used. Values of the load factors to be used are given in Table 2 of BS 5950.

The material strength is specified in relation to steel grade.

The ultimate strength is dependent on yield stress. Stresses are given for three grades of steel called S275, S (These were formerly referred to as Grades 43, 50 and 55). Grade S275 (formerly Grade 43) is commonly used, although S355 is popular on larger projects where it can offer significant economies. Higher grades are rarely used, except for bridges, and special applications

The yield stresses py corresponding to the different grades are given in BS 5950.

Specific guidance is given for determining the maximum moment and shear capacity of a beam cross-section.

The maximum moment capacity is the lesser of:

Mc = py.Sx & Mc = 1.2 py.Zx

where,

Sx is the plastic section modulus and Zx is the elastic section modulus.

The maximum shear capacity is

Pv = 0.6 py Av

where Av is the shear area.

Cross-sections are classified according to their proportions.

Open sections are classified as plastic, compact, semi-compact or slender. The classification depends on the proportions of the webs and flanges. (Note that rolled sections are seldom classified as slender).

Shear area for beam sections

The moment capacity of some sections may be reduced if shear forces are high.

The actual shear force (multiplied by the factor) is referred to by the symbol Fv. If Fv exceeds 60% of the shear capacity of the cross-section, Pv, then this is defined as a "high" shear load and the moment capacity for plastic and compact sections is reduced.

For beams which are not fully restrained, bending strength may be reduced due to lateral-torsional buckling; this is related to the slenderness ratio and cross-section details of the beam.

If the compression side of a beam is not fully restrained against lateral torsional buckling then the design stress py is reduced. This reduction depends on two factors called the slenderness ratio and the D/T ratio.

The slenderness ratio l is given by:

l = LE/ry

LE is the effective length (Table 9 of BS5950).

ry is the radius of gyration = the square root of (I/A). Standard tables give values for ry.

The slenderness ratio may be reduced using the slenderness correction factor n from table 20, to allow for the shape of actual bending moment.

The beam must be stiff enough to carry the working load without exceeding the deflection limits specified.

Limiting deflections are given in Table 8 of BS 5950 as a proportion of the beam span. These limiting values are compared with calculated deflections taking account of the load, length, support conditions, and cross-section of the beam. In calculating deflections it is normal to ignore any permanent loads, and to consider conditions in service (ie without any load factors)

Bending ULS

When a simply supported beam bends, the extreme fibres above the neutral axis are placed in compression. If the beam is a steel beam this means that the top flange of the section is in compression and correspondingly

the bottom flange is in tension. Owing to the combined effect of the resultant compressive loading and the vertical loading, the top flange could tend to deform sideways and twist about its longitudinal axis. This is termed lateral torsional buckling, and could lead to premature failure of the beam before it reaches its vertical moment capacity.

Bending ULS of laterally restrained beams

bending, the elastic moment of resistance (MR) of a steel beam is given by,

MR = Æ’Z

where f is the permissible bending stress value for the steel and Z is the elastic modulus of the section. This assumes that the elastic stress distribution over the depth of the section will be a maximum at the extreme fibres and zero at the neutral axis.

Bending ULS of laterally unrestrained beams

Laterally unrestrained beams are susceptible to lateral torsional buckling failure, and must therefore be designed for a lower moment capacity known as the buckling resistance moment M b. It is perhaps worth reiterating

that torsional buckling is not the same as local buckling, which also needs to be taken into account by reference to the section classification of plastic, compact, semi-compact or slender.

Shear ULS

The shear resistance of a beam is checked by ensuring that the ultimate shear force Fv does not exceed the shear capacity Pv of the section at the point under consideration:

Fv £ Pv

Where,

Fv - ultimate shear force at point under consideration

Pv - shear capacity of section: Pv = 0.6pyAv

Py - design strength of steel,

Av - area of section resisting shear: Av = tD for rolled sections, as shown

in Figure

t - total web thickness, from section tables

D - overall depth of section, from section tables

t

D

It is recommended in BS 5950 that the combination of maximum moment and coexistent shear, and the combination of maximum shear and coexistent moment, should be checked. The moment capacity of plastic and compact beam sections is reduced when high shear loads occur. A high

shear load is said to exist when the ultimate shear force exceeds 0.6 times the shear capacity of the section, that is when Fv > 0.6 Pv. however this is not usually a problem except for heavily

loaded short span beams.

Web buckling Resistance

When a concentrated load, such as the reaction, is transmitted through the flange of a beam to the web, it may cause the web to buckle. In resisting such buckling the web of the beam behaves as a strut. The length of web concerned is determined on the assumption that the load is dispersed at 45° from the edge of stiff bearing to the neutral axis of the beam, The buckling resistance Pw of the un stiffened web is calculated from the following expression:

Pw = (b1+ n 1) t pc

Where

b1 - stiff bearing length

n1 - length obtained by dispersion through half the depth of the section

t - web thickness, from section tables

pc- compressive strength of the steel

The compressive strength pc of the steel should be obtained from BS 5950 Table 27c in relation to the ultimate design strength of the steel and the web slenderness..

Web Bearing Resistance

The web bearing resistance of a beam is the ability of its web to resist crushing induced by concentrated loads such as the reactions. These are considered to be dispersed through the flange at a slope of 1:2.5 to the point where the beam flange joins the web, that being the critical position for web bearing.

The ultimate web bearing capacity Pcrip of a beam is given by the

following expression:

Pcrip = (b1 + n2) t pyw

Where

b1 - stiff bearing length

n2 - length obtained by dispersion at a slope of 1:2.5 through the flange to the flange to web connection: n2 = 2.5(r + T)

r - root radius of the beam, from section tables

t - beam flange thickness, from section tables

Pyw - design strength of the web: pyw = py