design notes connection.ppt

8/20/2019 design notes connection.ppt

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BEAM


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DIFFERENT METHODS OF DESIGNOF CONCRETE

1. Working Stress Method2. Limit State Method3. Ultimate Load Method4. Probabilistic M ethod of Design


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LIMIT STATE METHOD OF DESIGN

The object of the design based on the limit stateconcept is to achieve an acceptable probability, thata structure will not become unsuitable in it’s lifetimefor the use for which it is intended,

i.e. It will not reach a limit state

A structure with appropriate degree of reliabilityshould be able to withstand safely.

All loads, that are reliable to act on it throughout it’slife and it should also satisfy the subs abilityrequirements, such as limitations on deflection andcracking.


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It should also be able to maintain the requiredstructural integrity, during and after accident, suchas fires, e plosion ! local failure.

i.e. limit sate must be consider in design to ensurean adequate degree of safety and serviceabilityThe most important of these limit states, whichmust be e amine in design are as follows

"imit state of collapse # $le ure # %ompression

# &hear # Torsion This state corresponds to the ma imum load

carrying capacity.


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TYPES OF REINFORCED CONCRETEBEAMS

a) Singly reinforced beamb) Doubly reinforced beamc) Singly or Doubly r einforced anged

beams


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SINGLY REINFORCED BEAM

In singly reinforced simply supported beams

or slabs reinforcing steel bars are placednear the bottom of the beam or slabs wherethey are most effective in resisting thetensile stresses .


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Reinforcement in simply supported beam

COMPRESSION b

STEEL REINFORCEMENT D d

TENSION

SUPPORT SECTION A - A

CLEAR SPAN


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Reinforcement in a cantilever beam

A TENSION

D

d

COMPRESSION SECTION A - A

A

CLEAR COVER


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STRESS – STRAIN CURVE FOR CONCRETE

f ck

STRESS

.20 % .35%

STRAIN


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STRESS STARIN CURVE FOR STEEL―


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STRESS BLOCK PARAMETERS

!

0.0035 0.446 f ck

X 2 X 2 a

X 1 X 1


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' (epth of )eutral a isb ' breadth of sectiond ' effective depth of section

The depth of neutral a is can be obtained byconsidering the equilibrium of the normal forces ,

that is,*esultant force of compression ' average stress + area ' .- f ck b

*esultant force of tension ' ./0 f y A t$orce of compression should be equal to force of tension,

.- f ck b ' ./0 f y A t

=

Where A t = area of tension steel


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1oment of resistance with respect to concrete ' compressive force lever arm

' .- f ck b 2

1oment of resistance with respect to steel ' tensile force lever arm

' ./0 f y A t 2


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MAXIMUM DEPTH OF NEUTRAL AXIS

A compression failure is brittle failure.

The ma imum depth of neutral a is is limited to ensure that tensilesteel will reach its yield stress before concrete fails in compression,thus a brittle failure is avoided.

The limiting values of the depth of neutral a is m for different gradesof steel from strain diagram.


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MAXIMUM DEPTH OF NEUTRAL AXIS

f y N/mm 2 x m

250 0.53 d

415 0.48 d 500 0.46 d


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LIMITING VALUE OF TENSIONSTEEL AND MOMENT OF

RESISTANCE&ince the ma imum depth of neutral a is is limited,the ma imum value of moment of resistance is alsolimited .

M lim with respect to concrete = 0.36 f ck b x z = 0.36 f ck b x m (d – 0.42 x m)

M lim with respect to steel = 0.87 f ck A t (d – 0.42 x m)


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LIMITING MOMENT OF RESISTANCE VALUES, N MM

Grade of

concrete

Grade of steel

Fe 250 steel Fe 450 steel Fe 500 steel

General 0.148 f ck bd 0.138 f ck bd 0.133 f ck bd

M20 2.96 bd 2.76 bd 2.66 bd

M25 3.70 bd 3.45 bd 3.33 bd

M30 4.44 bd 4.14 bd 3.99 bd


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TYPES OF PROBLEM

a) Analysis of a sec

b) Design of a section


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a3 $or under reinforced section, the value of 4d is less than m4d value.

The moment of resistance is calculated by following equation5

1 u ' ./0 f y At d 6

a3 $or balanced section, the moment of resistance is calculated by thefollowing equation5

1 u ' ./0 f y At 7 d 6 .89 m3

a3 $or over reinforced section, the value of 4( is limited to m4d and themoment of resistance is computed based on concrete5

1u ' .- f ck b m 7 d 6 .89 m 3


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Analysis of section


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Determine the moment of resistance for the section shown in figure. (i) f ck = 20 N/mm , f y = 415 N/mm

Solution: (i) f ck = 20 N/mm , f y = 415 N/mm

breadth (b) = 250 mm

effective depth (d) = 310 mm

effective cover = 40 mm

Force of compression = 0.36 f ck b x

= 0.36 X 20 X 250x

= 1800x N


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Area of tensi t = 3 X 113 mm

Force of Tension = 0.87 f y A t

= 0.87 X 415 X 3 X 113

= 122400 N

Force of Tension = Force of compression

122400 = 1800x

x = 68 mm

x m = 0.48d

= 0.48 X 310

= 148.8 mm

148.8 mm > 68 mm

Therefore,

Depth of neutral axis = 68 mm

f y x m

415 0.48d

500 0.46d


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Lever arm z = d – 0.42x

= 310 – 0.42 X 68

= 281 mm

As x < x m ( It is under reinforced )o o Since t his i s a n under r einforced section, moment of resistance i s

governed by steel.

o Moment of resistance w.r.t steel = tensile force X zo M u = 0.87 f y A t zo = 0.87 X 415 X 3 X 113 X 281o Mu = 34.40kNm


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Design of a section


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Question : Design a rectangu ar !eam to resist a !en"ing moment e#ua to 45 $Nm using(i) %15 mi& an" mi " stee

Solution :

The beam will be desi gned so that under the ap plied m oment both materialsreach their maximum stresses.

Assume ratio of overall dep

Breadth of the beam = b

Overall depth of beam = D

therefore , D/b = 2

For a b alanced design,

Factored BM = moment of resistance with respect to concrete = moment of resistance with respect to steel

= load factor X B.M

= 1.5 X 45

= 67.5 kNm


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For ba lanced section,

Moment of resistance M u = 0.36 f ck b x m (d - 0.42 x m)

Grade for mild steel is F e250

For Fe250 st eel,

x m = 0.53d

M u = 0.36 fck b (0.53 d) (1 – 0.42 X 0.53) d

= 2.22bd

Since D/b =2 or, d/b = 2 or, b= d/2

Mu = 1.11 d

Mu = 67.5 X 10 Nmm

d=394 mm and b= 200mm

f y x m

250 0.53d

415 0.48d


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Adopt D = 450 mm

b = 250 mm

d = 415mm

Area of tensile steel At =

=

= 962 mm

= 9.62 cm

Minimum area of steel A o= 0.85


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=

= 353 mm

353 mm < 962 mm

In beams t he d iameter of main reinforced bars is u sually sel ectedbetween 12 mm and 25 mm.

Provide 2-20mm and 1-22mm bars giving total area

= 6.28 + 3.80

= 10.08 cm > 9.62 cm


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!


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SLIDES BY :HARSIMRAN SINGH TIWANAROLL NO :5059

UNIVERSITY : 514010017

GROUP MEMBERS :DILRAJ SINGH D3/CIVIL/5051

HARSIMRAN SINGH D3/CIVIL/5059 AMNINDER SINGH D3/CIVIL/5060

design notes connection.ppt

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