Irregular i Dad Es

7/30/2019 Irregular i Dad Es

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-,':

Introduction to Vertical Irregularities

'.

12.3 .2.2

12.3.2.2Table 12.3-2 defines vertical structural irregularities and assigns analysis and designprocedures to each type and seismic design category. These irregularities can be divided intotwo categories. The first, dynamic force-distribution irregularities, which are Types Ia, Ib, 2,and 3. The second, irregularities in load path or force transfer, which are Types 4 and 5. Thevertical irregularities areIa. Stiffness Soft Story IrregularityIb. Stiffness Extreme Soft Story Irregularity2. Weight (mass) irregularity3. Vertical geometric irregularity4. In-plane discontinuity in vertical lateral-force-resisting elementSa. Discontinuity in Lateral Stength - Weak Story Irregularity5b. Discontinui ty in Lateral Strength - Extreme Weak Story IrregularityStructures in Seismic Design Categories D, E, and F possessing dynamic force distributionirregularities shall be analyzed using the dynamic analysis procedure (or modal analysisprocedure) prescribed in 12.7. (Refer to Table 12.6.1) Structure Description 3. The verticalforce distribution provided by 12.8.3 may be assumed to be adequate for structures lackingvertical irregulari ty Types Ia, Ib, 2, and 3. However, stiffness and mass discontinuities maysignificantly affect the vertical distribution of forces and, for this reason the modal analysisprocedure, which can account for these discontinuities, is necessary.Although designers may opt to use the dynamic analysis procedure and bypass checks forirregularity Types Ia, Ib, 2, and 3, the reference sections listed in Table 12.3-2 should still bechecked for limitations and design requirements. Note that 12.3.3.1 prohibits structures withvertical irregularity Types Ib, Sa, or 5b for Seismic Design Categories E and F.Regular structures are assumed to have a reasonably uniform distribution of inelasticbehavior in elements throughout the lateral- force-resisting system. When vertical irregularityTypes 4 and 5 exist. there is the possibility of having localized concentrations of excessiveinelastic deformations due to the irregular load path or weak story. In this case, the codeprescribes additional strengthening to correct the deficiencies for structures in certain seismicdesign categories (SDCs). In the case of vertical irregularity Type 5b, limits are placed on thebuilding height for all SDCs exceptS DC A.

2006 lac Structural/Seismic Design Manual, Vol. I 41

irregularidades de distribucion de fuerzas dinamicas

irregularidades en traslaci

de fuerzas

piso debil


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12.3 .2.2 Example 4 Vertical1rregula rlty Typ e 1

~ ,&ample4.Vertical I rregulari ty Type 1a and Type 1b 12.3.2.2

A Seismic Design Category D five-story concrete special moment-resisting frame is shownbelow. The code-prescribed lateral forces F, from Equation 12.8- 11 have been applied andthe corresponding floor level displacements Oe at the floors' centers-of-mass have beendetermined as shown below.

Ft + Fs10

10'

10'

' 0'

' 2'

[';==::;-;::=::::::;11- - - - - --;,. 0" . 2.02i:F, - . . / :

DD . 0', _1.75Triangular .'"shape ; IF, -.. " ' : !DD

/ 0;\... 1.45F . ' /, -.. ! !DD ./ , 02" , .0BF i "'-.. /.' 0.I,' It _ 0.71!"r - r - - - r 7 Y - . , - " ' 7 7 ' , * " Actual shape

[!J Determine if a Type 1a vertical irregularity from Table 12.3-2 (StiffnessSoft Story Irregularity) exists in the first storyCalculations and Discussion CodeReference[L] To determine if this is a Type 1a vertical irregularity (Stiffness-Soft StoryIrregularity) there are two tests

I. The lateral story stiffness is less than 70 percent of that ofthe story above.2. The lateral story stif fness is less than 80 percent of the average stiffness of the threestories above.

42 2006 IBC Structural/Seismic Design Manual, Vol. I

el desplazamiento no es lineal

por que h del primer piso es mas

grande y la rigides es mas peque

rigidez-suave

arriba

promedio

irregularidad en la rigidez de piso suave.


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Example 4 Venicallrregularity Type 1 12. 3.2.2

If the stiffness of the story meets at least one of the two criteria above, the structure isdeemed to have a soft story, and a modal analysis (12.9) is generally required by Table12.6- I.The definition of soft story in the code compares values of the lateral stiffness of individualstories. Generally, it is not practical to use stiffness properties unless these can be easilydetermined. There are many structural configurations where the evaluat ion of story stiffnessis complex and is often not an available output from computer programs. Recognizing thatthe basic intent of this irregularity check is to determine if the lateral-force distribution willdiffer significantly from the pattern prescribed by 12.8.3, which assumes a prescribed shapefor the first dynamic mode of response, this type of irregularity can also be determined bycomparing values of drift ratios due to the prescribed lateral forces. This deformat ioncomparison may even be more effective than the stiffness comparison because the shape ofthe first mode shape is often closely approximated by the structure displacements due to thespecified 12.8.3 force pattern . Floor level displacements and corresponding story-drift ratiosare directly available from computer programs. To compare displacements rather thanstiffness, it is necessary to use the reciprocal of the limiting percentage ratios of 70 and 80percent as they apply to story stiffness, or reverse their applicability to the story or storiesabove. The following example shows this equivalent use of the displacement properties.From the given displacements, story drifts and the story-drift rat io's values are determined.The story-drift ratio is the story drift divided by the story height. These story-drift ratios willbe used for the required comparisons because they better represent the changes in the slopeof the mode shape when there are significant differences in interstory heights. (Note: storydisplacements can be used if the story heights are nearly equal.)In terms of the calculated story-drift ratios, the soft story occurs when one of the followingconditions exists.

or

t., 0,. - 0" = (1.08- 0.71) = 0.00308- '=h, h, 120~ Ii,. - 0,. = (1.45 - 1.08) = 0.00308h, h, 120

s s - IiWhen 70 percent of ---"- exceeds ,. ,.h, h,Ii, I [(0, -I i,.) (0 , - Ii, ) (0 , -Ii ,)]When 80 percent of - ' exceeds - . , + _ + .h, 3 h, h, h,

= 0.004930.71- 0)

= 144~ = ..h, h,

the story-drift ratios arc determined as

2006 IBC Structural/Seismic Design Manual. Vol. I 4 3

la rigidez de piso suave ocurre cuando existen estas condiciones


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12. 3.2 .2 Example 4 Vertical Irregularity Type 1

_:J._ = Ii" - Ii" =Iz, s,

(1.75-1.45) = 0.00250120

44

. 0 0 3 0 8 + 0.00308 + 0.00250) = 0.002893Checking the 70-percent requirement:

O.70(1i,,) =0.70(0 .00493) =0.00345 > 0.00308 . . . NGhi: . Soft story exists. . .Note that 70 percent of first story drift is larger than second story drift. Alternately:0.00493 > (0.00308 x 1.30= 0.0040) . . . thus soft story.Also note that structural irregularities ofTypes Ia, Ib, or 2 in Table 12.3-2 do notapply where no story-drift ratio under design lateral force is greater than 130 percentof the story-drift ratio of the next story above, 12.3.2.2, Exception 1.

Checking the 80-percent requirement:

0 . 8 0 ( ~ ) = 0.80(0.00493) = 0.00394 > 0.00289 . . . NGh,.', Soft story exists. . . condition IaAlternately: 0.00493 > (0.00289 x 1.20 = 0.00347) . . . thus soft story.

Check for extreme soft story, (Vertical Structural Irregularity, Type lb)Checking the 60-percent requirement:

0.60(0.00493) = 0.002958 < 0.00308 . . . o.k.Alternately: 0.00493 > (0.00308 x 1.4 = 0.004312) . . . o.k.

Checking the 70-percent requirement:0.70 (0.00493) = 0.003451 > 0.00289 . . . NGAl ternately: 0.00493 > (0.00289 x 1.3 = 0.00375) . . . NG

Thus: Stiffness-Extreme Soft Story exists - condition lb.

2006 IBC Structural/Seismic Design Manual , Vol. I

piso suave existe


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Examp le 4 tI Vertical Irregular it y Type 1 1 2.3.2.2

Recall from Table 12.3-2 for Ib , extreme soft story, reference 12.3.3.1.This building isSOC 0 , and is permitted, Structures having SDC E or F and also having vertical irregularityType Ib shall not be permitted.

CommentarySection 12.8.6 requires that story drifts be computed using the maximum inelastic responsedisplacements b.r , which include the deflection amplification factor Cd

s = Cdb. rx I (Eq 12.8-15)However, for the purpose of the story drift, or story-drift ratio, comparisons needed for softstory determination, the displacement bxe due to the design seismic forces can be used as inthis example. In the example above, only the first story was checked for possible soft-storyvertical irregularity. In practice, all stories must be checked, unless a modal analysis isperformed. It is often convenient to create tables to facilitate this exercise, see Tables 4.1and 4.2.

Table 4. I Soft-Story Status lnSto ry Story Story-drift 0 .8x (S tory- 0 .7x (Story - Avg . of Story-drift Ratio

Leve l Displacement Drift Rat io drift Ratio ) drift Ratio) of Next 3 Stories5 2.02 in 0.27 in 0.00225 0.00180 0.00 1584 1.75 0.30 0 .00250 0.00200 0.001753 1.45 0.37 0 .00308 0.00246 0.0021 62 1.08 0.37 0 .00308 0.0024 6 0.00 216 0.00261

0.71 0.71 0 .00493 0.00394 0.00345 0.00289

Tab le 4.2 Soft-Story Status IbStory Story Story-drift 0 .7x (S tory- 0.6x (Story- Avg, of Story-drift Ratio

Level Disp lacement Drift ratio drift Ratio) dr ift Ratio) of Next 3 Stories5 2.02 in 0.27 in 0.00225 0.001 58 0.001354 1.75 0.30 0 .00250 0.00175 0.001503 1.45 0.37 0.0030 8 0.002 16 0.001852 1.08 0.37 0.00308 0.002 16 0.00185 0.0026 1

0.71 0 .71 0.0049 3 0.00345 0.00296 0.00289

Soft StoryStatus la

NoNoNoNoYes

Soft StoryStatus lb

NoNoNoNoYes

2006 IBC Structural/Seismic Design Manual, Vol. I 45

no hay tres pis

superiores

80% 70


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12.3.2.2 Example 5 Vertical Irregularity Type 2

...' a!nple 5; ~ r t c a r r e g u a r t y Type 2 1 2 . 3 ~ 2 . 2

The five-story special moment frame office building has a heavy utility equipmentinstallation at Level 2. This results in the floor weight distribution shown below.Ws =90 k

W,= 110k

W, = 110 k

W,= 100k

[!J Determine if there is a Type 2 vertical weight (mass) irregularity. , .:Calci!li!tions andDiscussion Code Reference

A weight, or mass, vertical irregularity is considered to exist when the effective mass of anystory is more than 150 percent of the effective mass of an adjacent story. However, thisrequirement does not apply to the roof if the roofis lighter than the floor below. Note that itdoes apply if the roof is heavier than the floor below.Checking the effective mass of Level 2 against the effective mass of Levels 1 and 3At Levell

1.5 X WI = 1.5(100 kips) = 150 kipsAt Level 3

1.5 X W3 = 1.5(110 kips) = 165 kipsWz = 170 kips > 150 kips

.. Weight irregularity exists.


sale en el reglamento en el inciso 7 pero sale 110%

no se aplica para el techo

arriba y abajo


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Example 5 Ver tical Ir regularity Type 2 12.3 .2 .2

I , ..,.CommentaryAs in the case of vertical irregularity Type la or Ib, this Type 2 irregularity also results in aprimary mode shape that can be substantially different from the triangular shape and lateralload distribution given by 12.8.3. Consequently, the appropriate load distribution must bedetermined by the modal analysis procedure of 12.9, unless the irregular structure is notmore than two stories and is Occupancy Category l or II (see Table 12.6-1).

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12.3 .2 .2 Ex ample 6 Vertical Ir regularity Type 3

ample 6Vertical Irregularity Type 3 12.3.2.2

The lateral-foree-resisting system of the five-story special moment frame building shownbelow has a 25-foot setback at the third, fourth, and fifth stories.

4@2S' -100'I ..... - ......... ...-

5 DDDDDDDDDD.DDD

' / "/ //"/ / / / / / , / / / , /

Level

4

[!J Determine if a Type 3 vertical irregularity (vertical geometric irregularity)existsC a l c u J ~ t i C ? n s andDiscussion Code Reference

A vertical geometric irregularity is considered to exist where the horizontal dimension of thelateral-foree-resisting system in any story is more than 130 percent of that in the adjacentstory. One-story penthouses are not subject to this requirement.In this example, the setback of Level 3 must be checked. The ratios of the two levels are

Width of Level 2 = (lOa ft) = 1.33Width of Level 3 (75 ft)133 percent > 130 percent.'. Vertical geometric irregulari ty exists.

48 2006 IBC Structural/Seismic Des ign Manual, Vol. I


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Example 6 Vertical Irregularity Typ e 3 12.3 .2 .2

, .CommentaryThe more than l30-percent change in width of the lateral-force-resisting system betweenadjacent stories could result in a primary mode shape that is substantially different from theshape assumed for proper applications of Equation 12.8-11. If the change is a decrease inwidth of the upper adjacent story (the usual situation), the mode shape difference can bemitigated by designing for an increased stiffness in the story with a reduced width.Similarly, if the width decrease is in the lower adjacent story (the unusual situation), theType la soft-story irregularity can be avoided by a proportional increase in the stiffness ofthe lower story. However, when the width decrease is in the lower story, there could be anoverturning moment-load-transfer discontinuity that would require a dynamic analysis perTable 12.6-1.Note that if the frame elements in the bay between lines 4 and 5 were not included as part ofthe designated lateral-force-resisting system, the vertical geometric irregularity would notexist.

2006 IBC Structural/Seismic Design Manual, Vol.J 49


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12.3.2.2 Examp le 7 11 Vertical Ir regular i ty Type 4

,Ex, mple 7Vertica l I rregularity Type 4 12.3 .2.2

A concrete building has the building frame system shown below. The shear wallbetween lines A and B has an in-plane offset from the shear wall between lines Cand D.

3@25'=75'r r r T

Level IE- - - - - ~12'

12'

12'

12'

12'

5 'DO' - L - -4 0--:3 0 05'2 50'1

/ / '/ / , ' / / / '/ / / " /

Shear wall

Shear wall

[!J Determine if there is a Type 4 vertical irregularity (in-plane discontinuity)in the verticallateral-force-resisting element

Calculations andDiscussion Code ReferenceA Type 4 vertical irregularity exists when there is an in-plane offset of the lateral-forceresisting elements greater than the length of those elements . In this examp le, the leftside of the upper shear wall (between lines A and B) is offset 50 feet from the leftside of the lower shear wall (between lines C and D) .This 50-foot offset is greaterthan the 25-foot length of the offset wall clements.

: . In-plane discontinuity exists .

50 2006 IBC Structural/Seismic Design Manual, Vol . I


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Examp le 7 Vertical Irregularity Type 4 12.3.2.2

s: '"C;ommentaryThe intent of this irregularity check is to provide correction offorce transfer or load-path deficiencies. It should be noted that any in-plane offset, even those less than or equalto the length or bay width of the resisting element, can result in an overturning momentload-transfer discontinuity that requires the application of 12.3.3.3. When the offsetexceeds the length of the resisting element, there is also a shear transfer discontinuitythat requires application of 12.3.3.4 for the strength of collector elements along theoffset. In this example, the columns under wall A-B are subject to the prov isions of12.3.3.3, and the collector element between lines Band C at Level 2 is subject to theprovisions of 12.3.3.4.

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12.3.2 .2 Example 8 Vert Ic al Ir regu lar it y Type 5a

IE. ampleBVer fi ca l l r regular ity Type 5a 12.3.2,,2

A concrete bearing-wall building has the typical transverse shear-wall configurationshown below. All walls in this direction are identical, and the individual piers havethe shear contribution given below. Then, V, is the nominal shear strength calcu lated inaccordance with Chapter 19, and Vm is defined herein as the shear corresponding to thedevelopment of the "nominal flexure strength also calculated in accordance with Chapter19." Note that VII/ is not defined in ACI or Chapter 19.

Level J

PIER \I V",1 20 kips 30 kips2 30 403 15 104 80 1205 15 10

[!J Determine if a Type 5 vert ical irregularity (discontinuity in capacity- weakstory) condition exists

Calculations andDiscussion Code ReferenceA Type Sa weak-story discontinuity in capacity exists when the story strength is lessthan 80 percent of that in the story above. The story strength is the total strength of allseismic-force-resisting elements shari ng the story shear for the direction under consideration.Using the smaller values of VI and VI/ given for each pier, the story strengths are

First story strength = 20 + 30 + 10 = 60 kipsSecond story strength = 80 + 10 = 90 kips

Check if first-story strength is less than 80 percent of that of the second story.60 kips < 0.8(90) = 72 kips:. Weak story condition exists.



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Example 8 VerlicallrregularJty Type Sa 12.3 .2.2

Check if first-story strength is less than 65 percent of that of the second story (IrregularityType 5b).60 kips < 0.65(90 kips) = 58.5 kips

:. 60 kips > 58.5 kips. . Therefore the lower story is not an extreme soft story, Irregularity Type 5b.

CommentaryThis irregularity check is to detect any concentration of inelastic behavior in onesupporting story that can lead to the loss of vertical load capacity. Elementssubject to this check are the shear-wall piers (where the shear contribution isthe lower of either the shear at development of the flexural strength, or theshear strength), bracing members and their connections, and frame columns.Frame columns with weak column-strong beam conditions have a shearcontribution equal to that developed when the top and bottom of the columnare at flexural capacity. Where there is a strong column-weak beam condition,the column shear resistance contribution should be the shear corresponding tothe development of the adjoining beam yield hinges and the column baseconnection capacity. In any case, the column shear contribution shall notexceed the column shear capacity.An extreme weak story is prohibited (under 12.3.3.1) for structures more than twostories or 30 feet in height if the "weak story" has a calculated strength oflessthan 80 percent of the story above . A weak-story condition is absolutely prohibited inSDC E and F.

2006 IBC Slructural/Seismic Design Manual. Vol. I 53


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12.3.3.1 Example 9 II Vertical Irregularity Type 5a

Example 9Verticallrregulaljty Type Sa 12.3.3.1

A five-story building has a steel special moment-resisting frame (SMRF). The frameconsists ofW24 beams and W14 columns with the following member strengthproperties.

5

4 . -.3 .. -' .. .2

1 .. ......

// "/ / / / // . / ///

Beams at Levels I and 2:Mllb = ZF."= 250 kip-ftColumns on lines Band C at

both levels:M" c=250 kip-ft ataxial loading of 1.2PD + 0.5PLColumn base connections at grade(based on grade-beam strength):Jvf"GB= 100 kip-ftIn addition, assume for the purposes

of illustration only, that the columnshave been designed such that astrong beam-weak column condi tionis permitted.

12'

12'

12"

12'

14'

Level

A o

Determine if a Type 5 vertical irregularity (discontinuity in capac ity-weak story)cond ition exists in the first story.[!J Determine first-story strength[!J Determine second-story strength~ Determine if weak-story exists at first story

Calculations andDiscussion Code ReferenceA Type 5 weak-story discontinuity in capacity exists when the story strength is less than 80percent of that of the story above (where it is less than 65 percent, an extreme weak storyexists) . The story strength is consi dered to be the total strength of all seismic-foree-resistingelements that share the story shear for the direction under consideration.To determine if a weak story exists in the first story, the sums of the column shearsin the first and second stories-when the member moment capacities are developedby lateral loading-must be determined and compared.In this example, it is assumed that the beam moments at a beam-column joint are

54 2006 IBC Stru ctur al/Se ism ic Design Manual, Vol . I


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Example 9 Vertical Irregularity Type Sa 12.3.3.1

distributed equally to the sections of the columns directly above and below the joint.Given below are the calculations for first and second stories.

[!J Determine first story strengthColumns A and D must be checked for strong column-weak beam considerations

2002Mc =400 >M; =250: . Strong column-weak beam condition exists.

....~ ) 250

--... FOR MOMENT200

v

Next, the shear in each column must be determined.Note moment capacity of beam (25012) governs overmoment capacity of column (200) to determine shear

a M. /2 =125kip-ft1 2 5 ~L ) 250125 I

-'

Clear height = 14 ft - 2 ft = 12 ft125+100 187-k '- --- = .) "iPS12 v o Mf=100k ipft FOR SHEAR

Checking columns Band C for strong column-weak beamconsiderations

2Mc = 400 < 2Jvfb = 500

200 .........

250 (+ ) 2 5 0200" -

:. Strong beam-weak column condition exists.Next, the shear in each column must be determined.Note moment capacity of column governs over vmoment capacity of beam to determine shear.

Clear height = 14 ft - 2 ft = 12 ft

FOR MOMENT200 ..........a 200(+) 200

Me = 200 kip-ft 200'-'200 r'

VB= Vc= 200 + 100 = 25.0 kips12 V4

M =100klp-ftG100 .....JFOR SHEAR

First story strength = VA + VB + VD = 2(18.75) + 2(25.0) =87.5 kips2006 IBC Structural/Seismic Design Manual, Vol. I 55


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12.3.3 .1 Example 9 Ver t ical Irregu la rit y Type 5a

~ Determine second story strengthColumns A and D must be checked for st rong column-weak beam at Level 2

:. strong column-weak beam condition exists. FOR MOMENT

'- '125Mb I 2= 125 kip-ftv

""""00

Clear height = 12 ft - 2 ft = 10ft 125r- -V - tr - 125 + 125 -?5 0 k'" - " D - - _. IpS. 10

-J125

FOR SHEAR

vMb I 2= 125 kip-ft

125 " J

Checking columns B and C for strong column-weak beam considerations2Mc =400 < 2Mb= 500: . Strong beam-weak column condition exists .

vFOR MOMENT

Me = 200 kip-ftClear height = 12 ft - 2 ft = IO ft 10'

200 ........

200 (+ )200200 ........

VB = Vc = 200 + 200 = 40.0 kips10 200 J1"""' FOR SHEAR

v Me=200 klp-ft 200 '-"

Second story strength= VI + VB + Vc+ VD + 2(25.0) + 2(40.0) = 130.0 kips

56 2006 IBC Structural/Seismic Design Manual, Vol . I


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Example 9 Vertical Irregularity Type Sa 12.3 .3. 1

~ Determine if weak story exists at first storyFirst story strength = 87.5 kipsSecond story strength = 130.0 kips

87.5 < 0.80(130) = 104:. Weak story condition in first story exists.

(T 12.3-2, Item 5a)

2006 IBC Structural/Seismic Design Manual, Vol. I 57


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12.3.2.1 Introduction 10 Ho r/zonla /lrregularitles

Horizontal structural irregularities are identified in Table 12.3-1. There are five types ofhorizontal irregularities:la. Torsional Irregularity - to be considered when diaphragms are not flexible asdetermined in 12.3.1.2lb . Extreme Torsional Irregularity - to be considered when diaphragms are notflexib le as determined in 12.3.1.22. Re-entrant Comer Irregularity.3. Diaphragm Discontinuity Irregularity.4. Out-of-plane Offsets Irregularity.5. Nonparallel Systems - Irregularity.

Introduction toHorizontal Irregularities 12.3.2.1

58

These irregularities can be categorized as being either special response conditionsor cases of irregular load path. Types Ia, Ib, 2, 3, and 5 are special response conditions:Type 1a and 1b. When the ratio of maximum story drift to average story drift exceeds thegiven limit, there is the potential for an unbalance in the inelastic deformation demands at thetwo extreme sides ofa story. As a consequence, the equivalent stiffness of the sidehaving maximum deformation will be reduced, and the eccentr icity between thecenters of mass and rigidity will be increased along with the corresponding torsions.An amplification factor Ax is to be applied to the accidental torsion M'a to represent theeffects of this unbalanced stiffness, I2 .8.4.I to 12.8.4.3.Type 2. The opening and closing deformation response or flapping action of theprojecting legs of the building plan adjacent to re-entrant comers can result inconcentrated forces at the comer point. Elements must be provided to transferthese forces into the diaphragms.Type 3. Excessive openings in a diaphragm can result in a flexible diaphragmresponse along with force concentrations and load path deficiencies at theboundari es of the openings. Elements must be provided to transfer the forcesinto the diaphragm and the structural system.Type 4. The out-of-plane offset irregularity represents the irregular load path category. In thiscase, shears and overturning moments must be transferred from the level above the offset tothe level below the offset, and there is a horizontal offset in the load path for the shears.Type 5. The response deformations and load patterns on a system with nonparallellateral-force-resisting elements can have significant differences from those of a regularsystem. Further analysis of deformation and load behavior may be necessary.

2006 IBC Stru ctural/Seismic Design Manual, Vol. I


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Example 10 a Horizontal Irregularity Type 1a and Type 1b 12.3.2.1

ri3fnple 1oui on a/Irregulari ty Type 1a and ype ' lb 12.3,,2 .

A three-story special moment-resisting frame building has rigid floor diaphragms.Under code-prescribed seismic forces, including the effects of accidental torsion, it hasthe following elastic displacements OX! at Levels I and 2.

OL.: = 1.20 in OR,:!. = 1.90 in

A Type 1a torsional irregularity is considered to exist when the maximum storydrift, includ ing accidental tors ion effects, at one end of the structure transverse toan axis is more than 1.2 times the average of the story drifts of the two ends of thestructure, see 12.8.6 for story drift determination

[}J Determine i f a Type 1a or Type 1b torsional irregularity exists at thesecond storyIf it does:~ Compute the torsional amplification factor Ax fo r Level 2

O,2 = 1.00 in OR,l = 1.20 in

Level3

2

Cl'cu'ations andDiscussion

OR,2----------------->OR,1- - - - - --7

Code Reference

2006 IBC Structural/S eismic Design Manual, Vol. I 59


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12.J.2.1 Example 10 . Horizon'al Irregular ity Type 1a and Typ e 1b

ITI Determine if a Type 1a torsional irregularity exists at the second storyReferring to the above figure showing the displacements bJe due to theprescribed lateral forces, this irregularity check is defined in terms of storydrift D.x=(bx - bx-d at ends R (right) and L (left) of the structu re. Torsionalirregularity exists at Leve l x when

where

Determining story drifts at Level 2D.L,2 = 1.20 - 1.00 = 0.20 inD.R2 = 1.90 - 1.20 = 0.70 inA = 0.20 + 0.70 = 0 45 .U al'g . In2

Checking 1.2 criteriaD.""" = 0.7 = 1.55 > 1.2D. a,- 0.45: . Tors ional irregularity exis ts - Type Ia.

Check for extreme torsional irregularity

D. 0_70 I 55 h . - I - . T Ib---"!!!!. = - - = . . ..t us, extreme torsion lrregu anty exists - ype .D. .", 0.45


T 12.3-1


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Example 10 . Horizontal Irregularity Type 1a and Type 1b

~ Compute amplification factor Ax for Leve l 212.3.2.1

12.8.4.3When torsional irregularity exists at a Level x, the accidental torsional moment M'nmust beincreased by an amplification factor Ax. This must be done for each level, and each level mayhave a different Ax value. In this example, A., is computed for Level 2.Note thatAx is a function of the displacements as opposed to/versus the drift.

( )

24 = (jmI.'' . 1.260 ' 8bma., = 1.90 in... (bR.2)b = bL,] + bR., = 1.30 + 1.90 = 1.60 inavg 2 2A , = ( 1.90 )2= 0.98 < 1.0 . . . Note Ax shall not be less than 1.0- 1.2(1.60):. use Ax = 1.0.

Commentary

(!BC Eq 16-44)

In 12.8.4.3, there is the provision that the more severe loading shall be considered. Theinterpretation of this for the case of the story drift and displacements to be used for theaverage values I'l.b",.gand bmg is as follows . The most severe condition is when both bR,X andbL,Xare computed for the same accidental center-o f-mass displacement that causes themaximum displacement bmax. For the condition shown in this example where bRX= bma.n thecenters-of-mass at all levels should be displaced by the accidental eccentricity to the rightside R, and both bR,Xand bL..rshould be evaluated for this load condition.Table 12.3-1 triggers a number of special design requirements for torsionally irregularstructures. In fact, if irregularity Type Ib (Extreme Torsional Irregularity) is present,12.3.3.1 is triggered, which prohibits such structures for SOC E or F. It is important torecognize that torsionai irregularity is defined in terms of story drift I'l.." while the evaluationofA.r by Equation 12.8-14 is, in terms of displacements bxc There can be instances where thestory-drift values indicate torsional irregularity and where the related displacement valuesproduce an Ax value less than 1.0. This result is not the intent of the provision, and the valueof Ax used to determine accidental torsion should not be less than 1.0.The displacement and story-drift values should be obtained by the equivalent lateral-forcemethod with the code-prescribed lateral forces. Theoret ically, if the dynamic analysisprocedure were to be used, the values of I'l.ma.' and I'l.m.gwould have to be found for eachdynamic mode, then combined by the appropriate SRSS or CQC procedures, and then scaledto the code-prescribed base shear. However, in view of the complexity of this determinationand the judgmental nature of the 1.2 factor, it is reasoned that the equivalent static forcemethod is sufficiently accurate to detect torsional irregularity and evaluate the Ax factor.

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12.3 .2.1 Example 10 . Horizontal Irregular ity Typ e 1a and Type 1b

If the dynamic analysis procedure is either elected or required, then 12.7.3 requires the useof a three-dimensional model if there are any irregularities.For cases oflarge eccentricity and low torsional rigidity, the static force procedure can resul tin a negative displacement on one side and a positive on the other. For example, this occurs ifDu =- DAD in. and DR .3 = 1.80 in. The value of Dmg in Equation 12.8-14 should be calculatedas the algebraic average.

= (- 40)+ 1.80 = lAO =0.70 in2 2When dynamic analysis is used, the algebraic average value Dmg should be found for eachmode, and the individual modal results must be properly combined to determine the totalresponse value for Dm .g

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Example 11 Horizontal Irregularity Type 2 12.3.2.1

Example 11orizontallrregularity T}'J e 2 12.3.2..1

The plan configuration of a ten-story special moment frame building is as shown below.

0 -0 - ~ oN@)

c


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1 2.3.2.1 Example 11 Horiz ontal Irregu larity Typ e 2

This is 20 or 33.3 percent of the 60-ft plan dimension . . . More than 15 percent.~ .Since both projections exceed 15 percent, there is a re-entrant comer irregularity.

Re-entrant comer irregularity exists.Commentary

Whenever the Type 2 re-entrant comer irregularity exists, see the diaphragm designrequirements of 12.3.3.4 for SDC D, E, and F.

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Examp le 12 IZ Horizon ta l Irregular ity Typ e 3 12.3.2.1

a;8Vple 12Horizonfal lrregularity ype 3 12.3.2 .1

A five-story concrete building has a bearing wall system located around the perimeterof the buil ding. Lateral forces are resisted by the bearing walls acting as shear wa lls.The floor plan of the second floor of the building is shown below. Th e symmetricallyplaced open area in the diaphragm is for an atrium, and has dimensions of 40 feet by 75 feet.All diaphragms above the second floor are without significant openings.? ~ ~ ?r 125' 1E 75'

- --

bCD

- "-

Second floor plan

[!J Determine if a Type 3 diaphragm discontinuity irregularity exists at the secondfloor level

Calculations and Discussion Code ReferenceA Type 3 diaphragm discontinuity irregularity exists when diaphragms have abruptdiscontinuities or variations in stiffness, including cutout or open areas comprising more than50 percent of the gross enclosed area of the diaphragm, or changes in effective diaphragmstiffness ofmore than 50 percent from one sto ry to the next.The first check is for gross area

Gro ss enclosed area of the diaphragm is 80 ft x 125 ft = 10,000 sq ftArea of opening is 40 ft x 75ft = 3000 sq ft50 percent of gross area = 0.5(10,000) = 5000 sq ft3000 < 5000 sq ft" No diaphragm discontinuity irregularity exists.

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12. 3.2.1 Example 12 Horizonral lrregu/arit y Type 3

The second check is for stiffness.The sti ffuess of the second floor diaphragm with its opening must be compared withthe stiffness of the solid diaphragm at the third floor. If the change in stiffness exceeds50 percent, a diaphragm discontinuity irregularity exists for the structure.This comparison can be performed as follows.Find the simple beam mid-span deflections L12 and L1 J for the diaphragm s at Levels2 and 3, respectively, due to a common distri buted load IV such as I kif.

w = 1kif

-.

" ...., ------- ~ - - . >- ----- - - - - - - -- .. ..... ~ Deflected shapew=1 kif

t : . ~ - - - - - - - - - - - ~

If L1 2> 1.5L1J, there is diaphragm discontinuity.

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Examp le 13 Horizontal Ir regularity Type 4 12.3.2.1

.... an Ie -1 3rizontallrregularity TjIj e 4 12.3.2.1

A four-story building has a concrete shear wall lateral-force-resisting system in abui lding frame system configuration. The plan configuration of the shear walls is.shown below.

10'

10'

10'

10'

Elevation Line EoII I0 ) - "in'"'"

oI. ITypical floor plan

0) @ 0I I


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12.3.2 .1 Exampl e 14 Horizontal Irregularit y Type 5

Example 14Hor izonta l Irregularity Type 5 12.3.2.1

A ten-story building has the floor plan shown below at all levels. Special momentresisting frames are located on the perimeter of the building on lines 1,4, A, and F.

0-0 - inN~~

Typical floor plan

[!J Determine if a Type 5 nonparallel system irregularity existsCalculations and Discussion Code Reference

A Type 5 nonparallel system irregularity is considered to exist when the verticallateral-force-resisting elements are not parallel to or symmetric about the majororthogonal axes ofthe build ing's lateral-foree-resisting system.The vertical lateral-foree-resisting frame elements located on line F are notparallel to the major orthogonal axes of the building (i.e., lines 4 and A).Therefore a nonparallel system irregularity exists, and the referenced sectionin Table 12.3-1 applies to the design, see 12.5.3, 12.7-3, and Table 12.6-1.:. A nonparallel system irregularity exists.A 3-dimensional dynamic analysis is recommended.

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Example 15 Reliability/Redundancy Coefficientp 12.3 .4

E.rample 15Redundancy Factor p 12.3.4

The calculation of the redundancy factor p has changed considerably between earliercodes (1997 UBC; 2000 and 2003 1BC; ASCE/SEI 7-02) and the ASCE/SEI 7-05. Thecalculation is in some ways simpler, although it nevertheless requires some effort forconditions that do not compl y with prescriptive requirements (unless the full penalty istaken, as described below).ASCE/SEI 7-05 permits the redundancy factor to be taken as 1.0 in the followingcircumstances (12.3.4.1):

I. Struc tures assigned to Seismic Design Category B or C. (Note that the loadcombinations that include the redundancy factor are not used for Seismic DesignCategory A.)

2. Drift calculation and P-delta effects.3. Design of nonstructural components.4. Design of nonbuilding structures that are not similar to buildings.5. Design of collector elements, splices and their connections for which the loadcombinations with overstrength factor of 12.4.3.2 are used.6. Design ofmembers or connections where the load combinat ions with overstrength

of 12.4.3.2 are required for des ign.7. Diaphragm loads determined using Eq. 12.10-1 (note that this does not apply to

forces transferred through a diaphragm, such as due to an out-o f-plane offset inthe seismic load resisting system, and the higher p factor may apply as otherwiserequired).

8. Structures with damping systems designed in accordance with 18.Additionally, 12.3.4.2 ident ifies two other conditions in which p may be taken as 1.0.Note that the criteria for these condi tions need only be met at floor levels in which morethan 35-percent of the base shea r is being resisted; for the top level or levels of tallerstructures, the cond itions need not be met. The factor may be taken as 1.0 when either ofthe conditions listed below is met. In all other conditions, p is taken as 1.0. There is nolonger a calcu lated p factor between the minimum and maximum values.

Irregular i Dad Es

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