IBC SEAOC STRUCTURAL SEISMIC DESIGN MANUAL VOLUME 1 CODE APPLICATION

Table of contents :
2020-08-12_14-26-54......Page 1
2020-08-12_14-27-13......Page 2
2020-08-12_14-27-22......Page 3
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@Seismicisolation @Seismicisolation

2018 IBC®SEAOC STRUCTURAL/SEISMIC DESIGN MANUAL VOLUME 1 CODE APPLICATION EXAMPLES

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i @Seismicisolation • @Seismicisolation

Copyright Copyrtght 0 2020 Structuta1 Engtneers Assocaatton of California. All t1ghts reserved. This publication

or any partlhereof must not be reproduced in any form wtlhoutlhe writte.n permission of the Structw'al Engineers Associatton of California.

Publisher Sttucrurol Engineers As.c;ociatton of California (SEAOC) 921 lhhStre:c"_ _ = 400 kops

TOlal design base shear

DETERMINE THE FOLLOWING: I. Design crite'rta for lhe moment fmme system. 2. Seismtc design mome1u at point A

1. Design Criteria for the Moment Frame System According to the two listed tequlrements. the moment frame must be designed for lhe greater value of enher the Q£ value due to the design base shear V loadmg oo the combined moment frame-shear wall system, or the Q~ \•alue result1ng froan at least 25 percent of the design forces. This 25 percent requ1rement may be interpreted in two v.>ays.

a.

Q~ nl3)' be found by an equivalen1 lateral-force analys1s of the independent moment frame using 25 percent oflhe design base shear V.

b. Q~ mOl)' be found by factormg the oombmed moment frame-shear \\'311 system Q~ value such that Q~ coJTesponds to the moment that would occur tf the portion of the ba.~ shear reststed by lhe moment frame V,:- were w be equal to at lea~ 25 percent of the design base shear V.

@Seismicisolation 2018 IBC SEAOC SlructiXai/Seismic Design Manual, \All 1 @Seismicisolation

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Desig'J Example 10 • Dual S)'siems

§12.2.5. 1

2. Seismic Design Moment at Point A It lS elected to detemune Q£ per optton b descrtbed Oil page 29, beeause thts procedure tndudes the mteraction eftects between the moment frame and the shear wall.

From the combt.ned moment frame-she.ar wall analysis with fOrces due to the design base shear r·-= 400 kips, the ponion , ~r oflhe base shear reststed by the mome-nt frame 1sequaJ to the sum of the first-stOI')' tiame co1wnn shears tn the direction oflooding. For thiS example, it has been determined that

IF= rvro~= 45 ktps < 0.25 x r• 100 kips ll1e required \'aJues QE corresponding to a frame base shear reststance equal to 25 percent of V lS gtven by

and the seismic design momem a t A ls

M(.c= 2.22(53.0) = 11 7.8 kap-ft Commentary The use of a dual system has the advantage of pfOviding the structure with an lndependent ven.tcalloadcarrying system capable of resis6ng 25 percent of tJ'Ie des1gn bac;e shear~ while m the same tune tJte prtmary system. e1lher shear waJI or braced frame. carries us propomol'lal share of the design base shear. For this oonfiguration. lhe code permits use of a larger R-value for the pnmary system than \lw"Ould be pennined wtthout the 25 percent frame system. Per Table 12.2-1 , R = 7 for spec tal retnforced concrete shear walls wnh spec-ial moment frames capable of resisung at Jeast 25 petcent of the prescribed seismic fOI"Ces. Per Table 12.2-1. the R-value for special reinforced concrete shear walls alone would be 5 or6.

Design criteria in[e.rpretanon Ia uwolvtng the design of the momeru frame independent from the shear wall or bracing system for 25 percent of the design base shear should be COtlsidered for high-rise buildmgs. The slender configuration of lhe shear waJLs or brncsng systems can acruaJiy load tJ:.e moment frame at lhe upper levels of the combtned model, and excesssve.ly large moment frame design ac.tions would resul! from the use of destg.n criteria Interpre-tation Ib. whe.re these large actions would be multiplied by 0·25 " .

v,

Dual systems are often used in structutes with tJte primaJ)• lateml-force-resjsungsyste.m located 10 a structural core. The moment frames are placed at the perimetl!r of the btulding to cowuerac.t torsional eftkts. Where lht.s configuration tS present, destgn criteria lnterpretatJon lb shoukl be ullhzed tn order to envelope potential torsional eftkts. For addnional mformation. refer to SEAOC Blu1! Book article 4.02.020 ..Dual Systems'" a\'atlable at seaoc.orglstore.

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@Seismicisolation @Seismicisolation

Des;gn Example 11

• Introduction Jo Horizontal Irregularities

Design Example 11 Introduction to Horizontal Irregularities

§12.3.2.1

§12.3.2.1

Hortzontal structural 1rreguJartties are ide-nttfied 10 Table 12.3-1 . There are five types of horizontal itregularities: I a. Torsional Irregularity-to be considered where diaphragms are rigid or semi·rtgid (diaphragms

not deemed flexible per Secttlt 12.3. 1. 1 or calculated as flexible per Secnon 12.3. 1.3). I b.

E.~treme tors tonal trregulamy- to be constdered where dtaphmgms are rigid or sema-rtg1d (diaphragms not deemed He:tible per Section 12.3.1. 1 or calc-ulated as ftexable per Secuon

12.3.1.3). 2. Re..e.ntrant comer trregulamy. 3. Diaphragm discontinuity arregularuy.

4. Out-of-plane offsets ii'J'eg.ularity. 5. Nonparallel systents 1rreguJamy.

These arregulartties can be categom:ed as bemg eitJ'ler speciaJ.response co.nditioos or cases of t.rregular load path. Types Ia, I b, 2, 3, and 5 are special-response condttions. Types Ia tmd I b. Where the ratio of ma.x1mum. stOI)' dnft 10 avemge stOI)' drift exceeds the given limtt. there IS the-potentjaJ tOr an unbalance in the tneJastic deformation demands at the 1\\'0 ex'treme sides of a SUM)'. As a conseque:nc~ the equivalent stiffness oftJ'le side havtng maximum deformation wlll be reduced. Md the eccentricny betw~n the cenrers of ma~ and ngidity will be utcrea.sed along with the COITtsponding torsions. An amplificauon factor A_. ts to be applied 10 lhe acctde-ntal tors1on ."1111 to represent the effectS of thtsunbalanced st1ffness. as requ1red in Section 12.8.4.3. Type 1. The response of plan projeettons adjocent 10 re-e-ntrant comers ("flapping.. behavior acung 10 open and clo.o:oe the angle ofthe re-e-ntrant comer) can resuh 1n concentrated forces at the oorner point. Elements must be provaded to uansfer lhese forces 1010 the djaphragm.s. Type 3. Excessive ope.nings in a diaphragm can result in a flexible dJaphragm response along_ wtth force concentrations and load path deficiencies at the boundartes of the openings. EJeme.nt~ must be provided 10 transfer the forces 1nto the diaphragm and the structural system. 7j~ 4. The out~f.plane offSet trr~ulamy represents a d1sconnnuity in the lood path. Shears and overrumtng moments must be tranSferred from the level above the ofl'!iet 10 the level below, wttJ\ a horizontal offset in the load path fOr the she.ars.

7jpe 5. Systents wath nonparallel lateml-force.resisung elements require speciaJ consideratiOn to determine appropriate design forces for elements.

@Seismicisolation 2018 IBC SEAOC SlructiXai/Seismic Design Manual, \All 1 @Seismicisolation

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Desig'J Example 12 • Horizontal Irregularity Type 1a and Type 1b

§12.3.2. 1

Design Example 12 Horizontal Irregularity Type 1a and Type 1 b

§12.3.2.1

PROBLEM STATEMENT A lhree-story spec tal moment-resisung frame buildlng has ngid floor diaphragms. Unde-r code-prescrtbed seismtc tOrcest including lhe eftb:-ts of accidenml lotSton. it has the followtng elastic: displacemetus 8~ at LeveJs I and 2:

81..2 = 1.20 in

s.., = 1.90 in

8t,t = 1.00 in

8R. 1 = 1.20 in

--------------)>6.v ------> ~~ 2

Figure J]. J.

DETERMINE THE FOLLOWING: I. If a Type Ia or Type lb torstonaltrregulanty e.' 1.2 .. . thus, torsional trregulartty exast~Type Ia t....

0.45

Checktng fOr e.xtreme torsional irregularity

A_,_ = 0 ·70 =- 1.56 . . . th~ extreme torsion irregulattty exists-Type 1b t..., 0.45

@Seismicisolation 2018 IBC SEAOC SlructiXai/Seismic Design Manual, \All 1 @Seismicisolation

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Desig'J Example 12 • Horizontal Irregularity Type 1a and Type 1b

§12.3.2. 1

2. Compute Amplification Factor Ax for Level 2

§12.8.4.3

Where torsional irregularity exists at a Level.t. lhe acctdental torstonaJ moment ,\ /,.must be tncreas:ed b)• an amplification factor A,.. Thts must be done for each level. and each levei may h:we a different A_. value. In thts example, A.~~ IS computed for Level2. Nore that A.~~ is a funcuon of the displacements. not the story drafts. A .. -

(s_ J

•

o_,'¥-

A,= ( -

Eq 12.8- 14

1.26'"'

S., +S,,

1.20+ 1.90

2

2

....

·· -

90

1.

1.2(1.55)

-

I •• .

.)) In

J

= 1.04 > 1.0 ... Note illat A, shall be not less than

1.0.

Therefore,useA:%== 1.04.

Commentary In Sec.tion 12.8.4.3, there is the provision that the more severe loading shall be considered. The L01erpretatJon of this f« the case of the.story drift and displacementS to be used f« the average ' ·aJues .6...,.8 and Sm8 is as fOllows. The mOSt severe condnion is when both SN.,.t'and 8~\'are compured for the same accidental center-of-mass displaceme:JU that causes the maxtmum displaceme.nt s_. For lhe c.olldt1ion shown in thtsexample whete SR,X==8_.lhec.enters of mass ar all levels should be displac.ed by the accidenta1 eccerllticny to the right side R. and both s.-,x and 84-f should be evaluated for this load coodition. Table 12.3- 1 triggers a number of special design requirements for torstonally trregular structures. In fact, if irregulamy Type Ib (extreme torsional trregulanty) ls present. Section 12.3.3.1 is U"igge.red, which prohtbits such structures f"Of SOC E or F. It i.s i.m.pon.ant to recogmze that torsional irregularity as defined in tenns of sroty drift .d.r• while the evaJuatiott of Ax by Equation 12.8-14 is, in terms of dtsplacements. (\,.There can be instances where the .story-dnft values tndtcate torstonalt.rregulamy and where the related dt.splaceme.m \'3.lues produce an Ax value less lhan 1.0. llus result is not the intent of the provision. and the ''alue ofA_. used to de.temu.ne acc.idemal totsion should not be less lhan 1.0. The displacement and story-d.nft values should be obtained by the equivalent lateral-force method with the oode-prescnbed late.ral forces. Theore.tically. 1f the dynam te analysis ptoced ure \\'ere 10 be used, lhe \1al ues of a_ and a;M8 would ha''e to be fOund for each dynamtc mode. then combined by the appropriate SRSS or CQC procedures, and then scaled to the code-prescribed base she.ar. Howt\•er. an vtew of the conlplexuy of thts derennuwion and the judgmental ruuure of the 1.2 factot, u is reasoned that the equivalent static. force t'l'k!lhod IS sufficJently accurate 10 detect torstOnal trregularity and a •aluar.e the Ax factOI'".

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@Seismicisolation @Seismicisolation

Design Example 12 • Horizontal Irregularly Type f a and Type 1b

§12.3.2.1

Section 12.7.3 reqmres lhe use of a three-dimensional mode-l ifirr~ularuy Type laorl)rpe lb ls present For eases of large eccemricity and low lorsiooaJ ngidity,lhe stauc fotce procedure can resuh in a negative dJspfacement on one stde and a posatrveon the olhe-r. For example,lhJS occurs if841 = - 0.40 inches and 8R.1 = 1.80 inches. The ''aJue of8'"" an Equation 12.8- 14 should be calculated a~ the aJgebmic average. 8

- ~'-l+8.a:J -0.40+1 .80 1.40 _ 070 ' 2 2 - 2 - · In

mg-

Where modal analysis is used, the algebraic average value 8.._ should be found for each mode, and the andividuaJ modal results must be properly combined to detennine the total response \1alue for 8"',. Alternatively, static analysis may be used to compute lhe effects of acC-identaJ torsion., ancluding amplification. The results may be added to those from modalrutalysis.

@Seismicisolation 2018 IBC SEAOC SlructiXai/Seismic Design Manual, \All 1 @Seismicisolation

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Desig'J Example 13 • Horizontal Irregularity Type 2

§12.3.2. 1

Design Example 13 Horizontal Irregularity Type 2

§12.3.2.1

PROBLEM STATEMENT The plan configumtion of a 10-story speciaJ moment frame butldmg is as shown in Ftgure 13- I.

0--

ce2S' .-100'

I< -r ". ~

" ~

"' ~

0-

0- -'Figure /J-1.

DETERMINE THE FOLLOWING: I. If 3 Type 2 re-enuant corner irregularny exists.

1. Determine If a Type 2 Re-entrant Comer Irregularity Exists

§12.3.2.1

A Type 2 re-entrant comer l!regul3nt)' exists when lhe plan COilfiguraUOil of 3 suucture and its lateralforce-resisting system contain re-e.ntrant corners. where both projection.~ of the sm.tcture beyond a re-enuant cor-ner are greater lhan I5 percent of the plan dtmens.on of the suucrure 1n the dtreclion considered(seeTable

12 . 3-1~

The plan configumtion of this bwlding and tl'i lateral-force-rtsisung system has re-entra.Jn corner dimensions as shown. For the sadeson line 1. the proJeCtion beyond the re-e.ntrant corner as

100ft - 75 ti=2S ft

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@Seismicisolation @Seismicisolation

Design Example 13 • HorizontalltregU/arity Type 2

llus is

§12.3.2.1

25 or 25 petcent of the 100-foot plan dimension .. . More than 15 percent 100

For the stdes on line E. the ptojeclion tS

60ft - 40ft=20ft ThiS

1S

20

60

or 33.3 perc.em oflhe 60-foot plan dime-nston ... More lhan IS pe-rcenL

Since both projecttons e.xeeed 15 percem,lhere is a re-entrant c (1.0 + 0.105S,..,)D + 0.525!"1,(1£ + 0.15L 8. (0.6 - 0. 14S,.)D+ 0.7!"l,Q,

Jn this shear waJI. the timber column carries only axial loads. The awopnate dead, live. and seismic loads were given as 0:6.0 kipS L= 3.0 k1ps

with !"l, = 3.0

For the required strength des1gn check, both load combinations must be checked. P. =(l.2 + 0.2SDS)D+ !"l,Q,+ O.SL

P = (1.2 + 0.2( 1.10)](6.0) + 3.0(7.0) + 0.5(3.0) = 3 1.0 kips (compression) P =(0.9 - 0.2SIJS)D +!loQ,,

P = [0.9 - 0.2( 1.10)](6.0) + 3.0(- 7.0) = -16.9 kips (ns.

ASCE 7-16 permits lhe redw'ldancy fact« to be taken as 1.0 in the tOIIowmgc:ircwnstances (see Secuon

12.3.4.1): I . Structures ass1gned to Seism1c Design Category B or C. (Note that the lood combmauons that

1nclude the redundancy factor are not used for Seismtc Design Category A.) 2. Dnft calculation and P-deha effects. 3. Oestgn of nonstrucmral components.

4. Design of nonbuilding structures that are nm similar to buildmg.'i. 5. Design of oollectot element~. splices, and their connections for which lhe load combt.nauons

walh overstrength factor of Section 12.4.3 ate used.

6. Design of members or connec.tions where the load eombtnations \Vtth 0\'erstrength of SectiOil I2.4.3 are required for desJgn.

7. Dtaphrag.m loads determined using Equation 12.10- 1, including the limtts tmpo. c, •• = 0.044S..,J, (Equation 12.8-5) = 0.044(0.935)(1.5) = 0.062

V. = C, W = (0.088)(86,050 ktps) = 7572 ktps l.OOV, = ( 1.00)(7572 ktps) = 7572 kips> Va = 6163 ktps

Therefore, a scale factor of: 7572/6163; 1.23 shall be applied to aJI sei.~mic forces derived from the modal analysis in this direction. YDirtttlon

r,. = 0.70 sec< c. T, = ( 1.4)(0.61 sec)= 0.85 sec Therefore, for they d1rec.tion, the go"em1ng pertod T,. = 0. 70 se-conds. Use Equatton 12.8-3 to detennine c1 for this directioo. since~ is less than TL: C, =S0 .f[7{Ril,)J, where S01 = 0.401 C, = 0.401/(0.70(8/1.5)) = 0. 107 > c, •• =0.044S0 sf, (Equation 12.8-5)= (0.044)(0.935)1.5 = 0.062

v, = C, W= (0.107)(86,050 ktps) = 9207 ktps l.OOV,= (1.00)(9207 kips)= 9207 kips> v.,=6159 ktps

The-refore. a scale factor of: 920716759 = 1.36 shall be applied to all 5e-lS:mic forcesde,l'r\'ed ti'om the modal analysts ln lh1s dtrect.ion.

3. Scaling of Drifts from Modal Analysts Determine the scale t3ctor for modaJ design drifts 10 each considered onl'lOgonal direction for the structure 11'1: accordance wath Sections 12.9.1.2 and 12.9.1.4.2. The value for displaceme.nt and drift quanuues. based on modal analysis using a ODE Respcmse Specttum already factored by Rllv> shall be multiplied by the quanutyCi/~. Addinonally.1f V,< C,Wand whereC, lS per Equauon 12.8-6,lhen dnftsde,rived from the modal analysis shaJI also be nluluplied by the factor: CJWIJ~. C/1, = 5.011.5 = 3.33 C, = 0.5S11(Ril,) = (0.5)(0.601 Y(S/1.5) = 0.0563

110

20181BC

SEAOC Structu.raVSeismic Design Manual. \ - r-• z.. -r • >T

>T

>T

>T

/

/

/

/

/

/

/

Figun H-1-

DETERMINE THE FOLLOWING : I . Design criteria. 2. Design lateral seismic tbrce on a panelm the founh story. 3. Design lateral seismic fOrce on a panel at the first story.

1. Design Criteria

§13.3.1

For design of exterior element-; such aoe; lhe cladding panels on a butldi.ng. destgn la[eral setsmtc forces are determined from Equanon 13.3-1 or 13.3-4, wtlh upper and lower lunils given by Equattons 13.3-2 and 13.3-3, respec:uvely. The pa.nels are attached out-of-plane at the t\\'0 e.levatioos, =1 and =u- For lhe design of clildding element flexure governed by out-of-plane loading. the intent oflhe code ts to determine a value of F, that representS the average of the acceleration inputs from the 1\lfO attachment locauons. For rectangular panel.s.,. this can be taken as lhe ilve:rage of the two F, values 31 =equal to =t. and =u2018 /BC SEAOC StrocturaVSei.smic Design Manual. 1.01. 1 @Seismicisolation @Seismicisolation

167

Desig'J Example 53 • Exterior NonstructuraiiMIII Element

§13.5

Assum1n_g that a modal response spectrum aoalys•s was not perfocmed fOr this struerure and therefore Equation 13.3-4 was not used, use Equation I 3.3--1 to detennine F, as follows:

_ 0.4a,.SbS[1+2 =] > - IV,-0.3So;I,W,

F, aP

R,IIP

Eq 13.3-1 and Eq 13.3-3

h

=I .o. R, =2.5 for lhe waJI element

Tl3.5-l

F, is oot reqwred 10 be taken greater than F/1 = I .6SN~,WP

Eq 13.3-2

2. Design Lateral Seismic Force on a Panel at the Fourth Story As:sum1ngconnections are I foot above and below the nominal 12-foot panel height as shown 1n F1gure 53-2: =u =471\ =L =37ft

=60ft

h F.

= 0.4(l.O)(I.O)[I+2l•47]rv. =0.41111' 2.5/1.0 60 , ,

F

= OA(I.O)(I.O)[I+2l•J7]rv =0.357JV 2.5/ 1.0 60 , ,

,..,

pL

F,.. = Fe I. ST., C,. dete-rmtned per &juauon 12.8-3, has been multtplted by 1.5. Check T s

r, r, = 12.0 sec

F 22- 12

For nonbuildtng Strucrures located where 5 1 > 0.6g. the mtntmum specified C, \'alue shall be C = 0.8S, = (O.SXI.O) = O. l ' (R/1,)

Eq 15.4-2

f8l ll.ii)

Thus. C,=O.I ... Equation 15.4-2 govems. Eq 12.8-1

I'= C,W=(O.IX300) = 30 kops

178

20181BC SEAOC Structu.raVSeismic Design Manual. \-0.> ~-o~) =1.75 Tab/~

le\•e l

2

45

I

30

Sum

Nme:

Height h,

55-I S1ory[orres and SIOIJ' shrars (k =I 75) Story Force

w,

J~h:

ctt

F,

,.,

781.85 384.56

200

100

156,369.45 38,455.83

0.803 0.197

24.09 5.91

24.09 30

300

194,825.28

1.00

30

k = 1.75

Cn = Wxh: /194825.28

It~ m feet

Fx in kips Vx •n kips

Wxin k1ps

Story

~.:

Weight

Shear

2018/BC SEAOC StrocturaVSeismic Design Manual. 1.01. 1 @Seismicisolation @Seismicisolation

179

Desig'J Example 56 • Flexible Nonbuilding Structure

§15. 4 and §15.5

Design Example 56 Flexible Nonbuilding Structure

§15.4 and §15.5

PROBLEM STATEMENT A mil S[etl bin tower is suppotted by a conctete foundmjon. llle towe.r sits on symmetrtcally broced legs. The tbllowtng 1nformauon 1s gwen:

Weight of ttm•er and max.tmum notmal operaung contents; 150 ktps Stiffness of supponi1tg tO\\'er; 8.30 kiplan

Risk Caregory Ill I,

:c

1.25

Site Class 0

S, = 1.70,S, =0.65 S0 s = 1.20, Sm = 0.65

Seismic Design Category 0

Figu,.. 56-/.

DETERMINE THE FOLLOWING: I. Pt-nod of VJbtation.

2. Desogn base shear.

180

20181BC SEAOC Structu.raVSeismic Design Manual. \ns of Section 1605 shall nm apply to this requirement. Instead. design shall be based on 0.7 times nominal eanhquake loads. 1.0 limes other nominal loads and tnvesrigat1on wtth one or more of the \'3ftable loads se.t to :ze.ro.'· The exc.epuon tn Secuon 1807.2.3 tn the2018 IBC states: .. W11e-re eanhquake loads are included, lhe mtnimum safe-ty of factor for retainutg wall sliding and ovenuming shall be I. I:• Per IBC Sectton 1803.5. I2. Irem I. for struc-tures assigned to Setsmic Design categories D through F. the geotechmcal repon shall mclude lhe '"dynamic setsmtc lateral earth pressures on tOundauon wall.s and re-truning walls supporting more than 6 feet of bad.1ill heighL'" Consequent!)'. for the design of foundatJon and retainmg walls supporung less than 6 feet of back:fi.ll. n is interpreted th31 seismtc forces need not be constdered.

Accordtng to the SEAOC Blue Book arude. the k>cauon of the resultant of the acuve and seismic earth pressures may both be taken at the one-t.htrd potnt from the base of the cantile\'et wall~ refer to Figure 58- 1.

- -·---·-

-~ I

,....,

,.,.,....,..

""""-

I

~ ~

I

I

~

c·:

~

!

f•

I

I Figure 58- I.

186

20181BC SEAOC Structu.raVSeismic Design Manual. \cluding E (2 and 3). The load combinauons ofASCE 7 Chapter 2 shall not apply to thts requtrement.

Instead. design shall be based on 0.7 umes nominal earthquake Joads and 1.0 urnes other nominal loads.

F.-.,. =0.7(750 lb)+ 0.7(1250 lb) + 1.0(!1,)(525 pll)(l5 ft) = 53381b Mor= 0. 7(750 lb)(6.67) + 0.7( 1250 lb)(8.75 ft) + 1.0(!1,)(525 pll)( 151/3) = 30.85 kip-ft

Commentary ASCE 7 Section 11.8.3 requtr~ that the proJecl georechntcal repon mcludes recommendations for dynamic seismtc lateral eanh pressure~ and IBC Section 1803.5.11 funher statesthm geotechnical reportS shall provtde dynamic seismic lateral eanh pressures on foundation waHs and retainmg walls supportutg more lhan 6 feet of bad-fill. In order to provtde an efficte.nt yet code-compltam retatn.inglba.:;ement wall desjgn. the structutal enginetoJ mu. = (0.2)( I.0)( 162.3) = 32.5 ktps

£_.= n,o,= (2)(50.6)= 101.2 ktps

(1·- 100%)

£.,.= O,Q,= (2)(771.7) = 1543.4 ktp-1\ £.,. = O,Q, = (2)(382.7) = 765.4 ktps

(M - 100"/o)

£.,. = n,o, = (2)(6.8) = 13.6 k•P• £_. = n,o, = (2)(64.5) = 129.o ktp-ft

(V - 30%)

Eq 12.4-7

(P- 100%)

(M- 30%)

3. Column Design (Strength Design Forces Acting on Column)

§12.4

11te seismic load effect,. E. will be crttical fOf the compression (down\\'afd) 001td11ion OOI'tsidertng overstrength; therefore.• using Equation 12.4-Stsappropriate.

For design a.xta1 forces: J.2D+ E,+ E.. = (1.2)(162.3) + 32.5 + 765.4 =992.7 ktps For des~gn shear forces (strong axisf. 101.2 ktps For design moments (suong a.-xis): I543.4 kip-ft For design shear forces (weak axis): 13.6 kips For des~gn mornen