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vikram.jeet
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PostPosted: Sat May 15, 2021 6:03 am    Post subject: Reply with quote

Laps



Sir UH Varyani sahib's suggestion in his Book "STRUCTURAL DESIGN OF MULTISTOREYED BUILDINGS

If length of bars is short, Mix at lapping can be increased.
I think it is very good suggestion from this book.

By increasing the mix , the   lap required can be reduced from say 57D( for M20/Fe500)  to :-(laps for bending tension)
49D for increased mix M25
45D for increased mix M30
40D for increased mix  M 35
36D for increased mix M 40

But as a personal opinion, at least 50% bars still be welded, if laps fall short of original mix mentioned in drgs.



vikram.jeet wrote:
Vertical extension of building
Column reinforcements lapping

A building where rcc columns & their foundations are designed for future vertical extension( including rcc beams for seismic) , column bars of extension storey are lapped with existing col bars.

(I) If Existing column bars are of varying lengths,  ( say 50% bars of length Ld and rest 50% of 2*Ld - - -Only lapping will suffice

(ii) If Existing column bars  are of length Ld - - Lapping only will not suffice and Welding of at least  50% bars be needed to satisfy the codal provision of  'Not more than 50% bars be lapped at one section'
(iii) If existing column bars are less than Ld, All such bars will require Welding with new bars

Vikramjeet
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vikram.jeet
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PostPosted: Sat May 15, 2021 8:51 am    Post subject: Reply with quote

Importance of Design Criterion

Any Structural design, whether for a small structure  or a mega structure, at the ouset , the design criteria needs to given providing scope, Loadings,Load combinations, Materials to be used and permitted stresses, Method of Analysis/ software informstion , Method of design (LSM/WSM) , Any assumption s, etc.

If design is to be proof checked , it is better to get it approved from Vetter. Otherwise from senior official in heirarchy. Now a days it is digital transmission and approval.

This removes many problems to be faced by designer later/ Changes needed are much less.

Design criteria is the preface of any design.
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PostPosted: Sat May 15, 2021 12:18 pm    Post subject: Reply with quote

Laps
In addition to discussion earlier , In case where more than 50% bars spliced at a section, or where splices are made at points of maximum stress, PRECAUTIONS
Use closely spaced stirrups around the length of splice
Use spirals
Increase length of Lap

Reference 25 .2.5 IS456
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PostPosted: Sat May 15, 2021 12:27 pm    Post subject: Reply with quote

Laps spacing - STAGGERING


This shall be 1.3*Ld.
For column verticals - It implies that 50% bars have length Ld whereas rest 50% shall have 2.3*Ld to have lapping requirement fulfilled that is Not more than 50% bars lapped at one sections +rest 50% satisfies 1.3*Ld spacing between laps .
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PostPosted: Sat May 15, 2021 1:07 pm    Post subject: Reply with quote

Laps

When discussion on laps is there , it will be injustice to ignore the anchorage offered by bends (and Hooks- now not in practice due to high strength bars).
Anchorage value of 90°bend is 8D
For column bars in footing bent at 90°, the lap length in tension for flexural tension is (Ld - 8D).

Anchorage value of each 45° bend = 4D ( but maximum for 180° i.e.16D)

Anchorage value of Hook   = 16D

For working out lap length for compression, only projected length of hook or length of bend is applicable , and No anchorage value is applicable.
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PostPosted: Sat May 15, 2021 1:23 pm    Post subject: Reply with quote

Laps
Last but not the least : as we all know

LAP LENGTH FOR COMPRESSION BARS =  0.80*Ld

( Since bond stress in compression is 1.25 times the tension.)

For columns bars subject to direct compression AND bending :

Due to bending moments , there is flexural tension in bars on ten sion side and there fore Ld needs to based on tension. Due to moments being reversible in EQ loadings , all column bars are subject to flexural tension /compression in EQ event.
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PostPosted: Sat May 15, 2021 2:28 pm    Post subject: Reply with quote

Still something on LAPS

As already stated in earlier post, the lap length for bars under DIRECT TENSION  shall be 2*Ld.

For members with direct tension and bending , The lap length needs to be on weighted basis.
For example :
Reinforcement for bending tension  = 80%
Reinforcement for direct tensiin  = 20%
(  reinforcement as worked out in a member)

The lap length shall be = 0.80 + ( 2*0.20) =1.20*Ld

Generally column pedestals supporting steel columns of shed /PEB
the column pedestals are under direct tension & bending .
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PostPosted: Sat May 15, 2021 2:43 pm    Post subject: Reply with quote

Laps - in water retaining structures

Structures like cylinderical water tanks , overground / underground , wherever , hoop tension reinforcement is provided, this reinforcement bars are under direct tension and LAP LENGTH shall be 2*Ld.

However Lap length shall be based on tensile stress of maximum 130 Mpa as sections are uncracked and for strength calculations the maximum permissible stress is 130 Mpa.

For M25 mix , and Hysd bars
Normal Ld = {1300/(4*9) }/1.4=~ 26D (But minimum 30D for flexural tension)
For Lap length in direct tension = 2*26D =52 D
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PostPosted: Sun May 16, 2021 5:33 am    Post subject: Reply with quote

Welding of deformed bars cold worked
Reference - BIS Code 9417 on cold worked bars welding

Welding length
As per IS 9417, the bars to be lapped in fillet lapping , the effective weld is 0.2D thickness as per sketches of lap weld given in this code.

Design
I think design of welded lap length must account for eccentric moment of magnitude T*D on the lap weld , here T is the maximum tensile force in bar & D is the bar dia., IN conjunction with  T causing shear stress. In weld .
Since two stresses are acting simultaneously, e equivalent stress needs to be worked out as per IS 816.
Equivalent stress fe = √[ f^2 +(1.8*q^2)]
f is max axial stress as result of bending  & q is stress due to shear
Permissible stress = 1100 kg/cm2 , at shop welding
Permissible sressst= 0.80* 1100 = 880 kg/cm2 for site welding
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PostPosted: Sun May 16, 2021 1:07 pm    Post subject: Reply with quote

Welding length at laps

The effective weld thickness is 0.20D

For 25mm dia bars Beng lapped with 25 mm diameter

The weld thickness = 0.20*25=5 mm, If 8 mm weld is specified in details, only 5 mm will be effective for design

Maximum permissible tension in bar = 2750*4.91= 13500 kg
For single V weld
Moment on weld in one direction = 13500*2.5= 33750 kgcm
Since ,single weld , moment in orthogonal direction=13500*2.5/2=16875 kgcm
Taking 50cm single V  weld length
Area of weld = 0.5*50 = 25 cm2
Modulus Z = (0.5*50^2)/6 =208.33 cm3

Axial stress (max) due to Moment = 33750/208.33=162 kg/cm2
Axial stress max due to single V ecc mo=16875/208.33=81 kg/cm2
Shear stress = 13500/25= 540kg/cm2
Resultant axial stress due to moment=√[162^2+81^2]=199.2kg/cm2

Equivalent stress  = √[ 199.2^2  +  (1.8*540^2)]= 752 kg/cm2

Permissible stress for site weld = 0.80*1100=880 kg/cm2

Hence 50 cm length is ok

For both sides welding i.e.double V weld on both sides, half of length can be taken

Lap length needed is 20D for single V weld
If it is shop weld , the weld length for lap is 80%=16D forsingle V weld
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