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Waqas Haider

Development Length Of Standard Hooks

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Assalam o alaikum.

According to ACI 12.5.2,

development length for fc' = 3000, fy=60000, for normal weight concrete and epoxy less reinforcement, The required development length comes out to be

for #3 = 8.2 inch

for #4 = 10.95 inch

for #6 = 16.42 inch

for #8 = 21.9 inch

 

And if in my case, ACI 12.5.3 is not fulfilled, it means now i have to provide ldh as mentioned above. ldh is STRAIGHT EMBEDMENT LENGTH + RADIUS OF BEND + ONE BAR DIAMETER as shown in figure attached. Now my question is, if in my case, main reinforcement of beam is of #6 and #4, minimum column size required will be 18 inch and 12 inch respectively. Lets say by any means, i can not select #4, #3 bars and size of column where bars are to be terminated is 12 inch, how to fullfil this development length???

post-1702-0-70473600-1420373375_thumb.pn

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You can use headed reinforcement , see section 12.6.

If you cannot use headed reinforcement, and cannot provide the length of embedment for the development of top bars in tension, then consider this joint as PIN, and re-run the analysis to get the correct distribution of moments in mid span and other continuous support.

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Baz bhai, If i make ends of beams as pinned, This member will not be able to redistribute moment and will collapse at the formation of only one plastic hinge. so How REDUCED DESIGN LATERAL FORCE will be calculated and applied to this member? And is such type of beam allowed in IMRF or SMRF which needs special requirements for integrity at joints under lateral deformation of structure?

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It will require two plastic hinges to develop mechanism, as it act as propped beam, and it can still redistribute moment. You are not relying on plastic strength of the structure to carry gravity loads.

If , in a particular frame, your reliability/redundancy factor is less than 1, excluding  the bay which contains propped beam, and your drifts are under control, then you can use a propped beam in one bay of that frame.

If you can afford a non-moment resisting bay, which means your drifts are under control and your structure has adequate redundancy, then you don't have to worry about integrity of that one joint. Other joints and bays of the frame can take care of that.

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Waqas,

 

First note that development length in indicated case is only the straight length as shown in figure and doesn't include bent portion and secondly this requirement can be fulfilled as follows,

 

1,you can use greater no of lesser dia bars complying ldh requirement and if beam width is insufficient to accommodate greater no of bars then you can distribute them in effective flange width (ACI 318-11 section 10.6.6).

 

2. you can multiply ldh by 0.8 if you tie hooked end reinforcing bar with ties parallel or perpendicular to the bar, spaced not greater than 2db (db = hooked bar diametre).(ACI 318-11 figure R12.5.3).

 

3, You can provide excessive reinforcement and multiply ldh by As req/As prov.

 

4, You can multiply ldh by 0.7 by maintaining minimum side cover to the longitudinal bar to 2.5".

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Waqas,

 

First note that development length in indicated case is only the straight length as shown in figure and doesn't include bent portion and secondly this requirement can be fulfilled as follows,

 

1,you can use greater no of lesser dia bars complying ldh requirement and if beam width is insufficient to accommodate greater no of bars then you can distribute them in effective flange width (ACI 318-11 section 10.6.6).

 

2. you can multiply ldh by 0.8 if you tie hooked end reinforcing bar with ties parallel or perpendicular to the bar, spaced not greater than 2db (db = hooked bar diametre).(ACI 318-11 figure R12.5.3).

 

3, You can provide excessive reinforcement and multiply ldh by As req/As prov.

 

4, You can multiply ldh by 0.7 by maintaining minimum side cover to the longitudinal bar to 2.5".

Umair bhai, Thanx for your guidance in this regard.

I have gone for option 2  and 4 i.e. I have specified special requirements of concrete covers and have selected reduced development lengths dat fits in my 12 inch column. Regarding you 1st point, i will study it because it unaware of it and it seems very interesting way. Regarding your 3rd point, please explain it. I also cud not understand it from book/Code. Please explain in a bit detail what is it actually and how it works. Thanx.

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Regarding point # 3, suppose at any support strength requirement shows 3 #6 bars,here to counteract ldh you  provide greater area i.e greater bars for ex if you provide 6 # 6 bars then ldh should be multiplied with As-3#6/As-6#6 i.e 0.5.

 

Concept behind this provision is that if at any support say half ldh is available for stressed bars then these bars will develop half of their strength through bonding in concrete and will be able to take stresses till 0.5xFy,if you add up same no of bars with same ldh then they will also bear stresses upto 0.5xFy and as a whole you will get the required resistance Fy. 

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Regarding point # 3, suppose at any support strength requirement shows 3 #6 bars,here to counteract ldh you  provide greater area i.e greater bars for ex if you provide 6 # 6 bars then ldh should be multiplied with As-3#6/As-6#6 i.e 0.5.

 

Concept behind this provision is that if at any support say half ldh is available for stressed bars then these bars will develop half of their strength through bonding in concrete and will be able to take stresses till 0.5xFy,if you add up same no of bars with same ldh then they will also bear stresses upto 0.5xFy and as a whole you will get the required resistance Fy. 

This bonding stress comes into existence when the bar experience flexural tension or compression. Now If at support, say negative moment of 50 Kip-ft is to be resisted for which i had provided 3-#6 bars, then this 50 Kip-ft moment will be resisted by these 3 bars and these 3 bars will experience corresponding tension and this tension will become cause of required development length of ldh. Now if i have provided 6 bars. This moment will be resisted by 6 bars and share of moment for each bar will be reduced and each bar now will experience less tension, 50% less than before so now 50% less ldh will also work. Is my understading correct bhai?? Kindly confirm me one thing more. We design each bar for 0.9Fy in LRFD. and for development we provide such development that the bar can experience bond stress upto yield before slippage. so each bar can go upto yielding. But after this reduction of ldh, no matter force experience by each bar not gonna increase than share of each bar, but let say, theoratically now bars can't go beyond 0.5Fy if tried to be stressed more than share on calculation basis and bars will slip just after 0.5Fy. Am i getting right??

and My building is in Quetta for which i have assigned zone 4, SDC E. And PCA Commentry says i can not apply this reduction in Zone 4 SDC E. 

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PCA Commentry over 12.2.5 is
Quote

 

Reduction in ld may be permitted by the ratio [(As required)/(As provided)] when excess reinforcement is provided in a flexural member.  Note that this reduction does not apply when the full fy development is required, as for tension lap splices in 7.13, 12.15.1, and 13.3.8.5, development of positive moment reinforcement at supports in 12.11.2, and for development of shrinkage and temperature reinforcement according to 7.12.2.3.  Note also that this reduction in development length is not permitted for reinforcement in structures located in regions of high seismic risk or for structures assigned to high seismic performance or design categories (see 21.11.7.3 and R21.11.7.3).

Reduced ld computed after applying the excess reinforcement according to 12.2.5 must not be less than 12 in.....

 

Now i am confused regarding why this is not allowed? Because in those zones we design for reserve strength beyond yield too?? 

 

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      According to ACI 12.5.2,
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