Portal Method
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3.2.1 Solvin Solvin& & the determ determinate inate stru structure cture 3.2.2 3.2 .2 A#i A#ial al Force' Force' Shear Shear Force Force and (endin& )oment *ia&rams
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.2.1 Solvin Solvin& & the determ determinate inate stru structure cture
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Introduction: A portal rame is oten used in a structure to transer the laterall4 directed loads applied alon& the sides' to the supports at the %ase o the rame.[1] rame.[1] $ortal $ortal rames are oten desi&ned such that the4 are a%le to conidentl4 withstand lateral loads. 5his results in man4 portal rames %ein& staticall4 indeterminate e#ternall4 6i#ed supports or several columns at the %ases7 8%ecause o the rames a%ilit4 to support hori9ontal loadin&' this t4pe o rame is commonl4 used in structures lie %uildin&s' actories' and %rid&es. [2]
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5he %ehavior o these rames under lateral loads c an %e o%served in the General Overview. 5he appro#imate anal4sis o portal rames can %e investi&ated throu&h the portal method. (eore the anal4sis' there are necessar4 assumptions to %e made:
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A point o inlection is located at the center o each mem%er o the portal rame' For each stor4 o the rame' the interior columns %ear twice as much shear as the e#terior columns' ateral orces resisted %4 rame action' Inlection points at mid,hei&ht o columns' Inlection points at mid,span o %eams' olumn shear is %ased on tri%utar4 area' Overturn is resisted %4 e#terior columns onl4.[1]
General Overview 5o solve or a multi,storied %uildin& under lateral loads we are a%le to use what is called an Assumption )ethod. 5he Assumption )ethod consists o two su% cate&ories8 the $ortal and antilever )ethod [1]. (oth these )ethods have there own assumptions to help solve or all unnowns and reactions. On this wii we will %e ocusin& on understandin& the assumptions o a $ortal Frame and how to solve a $ortal Frame e#ample. (ut' %eore we can solve an4 e#amples we must have a &eneral understandin& or the assumptions made in a $ortal Frame and o%tain a &eneral nowled&e or solvin& a $ortal Frame e#ample.
In the $ortal )ethod there are a two main assumptions that must %e made %eore we are a%le to solve or an4 unnowns or reactions. 5he irst assumption is' there must %e hin&es at mid, hei&ht o ever4 column and mid,len&th o ever4 %eam [1][2][3]. 5he second assumption is that the interior column must have twice the %ase shear orce than the two e#terior columns [1][2] [3]. (oth assumptions will %e e#plained with &reater detail in the su% sections to ollow.
5he irst assumption that will %e made applies to the antilever )ethod as well. It states that under lateral loadin&s' a portal rame will have delect in such a manner that its moment dia&rams that will resem%le the ollowin& [1]:
5hat is to sa4 that it is %ein& assumed that there is 9ero moment at the mid,span and mid,hei&ht o each mem%er. om%inin& this with our previous nowled&e o anal49in& structures' we are a%le to mae the assumption that since the rame has multiple 9ero moments we are a%le to replace all 9ero moments with an internal hin&e [1][2]. 5his assumption is true or all $ortal Frames re&ardless o the amount o store4s [1]. For e#ample' a two,store4 %uildin&' as can %e seen %elow' there will %e 1@ hin&es present:
5hereore' the &eneral assumption or all $ortal Frames is that there is a hin&e at mid,hei&ht o ever4 column and mid,len&th o ever4 %eam [1][2][3]. "ven with the irst assumption the $ortal Frame is still indeterminate to 2 de&rees. 5hereore' we must have another assumption that maes our rame determinate. 5his second assumption is uniue to the $ortal method' and is essentiall4 the main dierence %etween the two appro#imate methods or indeterminate structures su%Bected to lateral loads. 5his assumption will %e that the interior columns %ase shear will have dou%le the reaction orce to the two e#terior
columns [1] [2]. Since the column in the middle &enerall4 taes more shear orce and is &enerall4 more sti' this results in dou%le the %ase shear orce. eer to the dia&ram %elow or a visual.
;ith this inal assumption in place our rame now %ecomes determinate and it is a%le to %e solved or in a couple o &eneral' ver4 eas4 steps.
Cow that we have an understandin& and some &eneral nowled&e a%out the assumptions and what a $ortal Frame is. ;e will now o%tain the sills to solve an e#ample. For ever4 $ortal Frame there are a %asic couple o steps that 4ou must ollow to solve or all unnown and reaction orces. I 4ou ollow these ! &eneral steps than 4ou will %e a%le to solve or most $ortal Frame uestions. 5o %e&in a $ortal )ethod uestion' 4ou must now what has to %e solved or' in most cases 4ou will %e ased to solve or all unnowns and reaction orces in the entire rame. In that case 4ou will ollow all ! steps' i 4ou are not to solve or all unnowns and reaction orces it is up to 4ou to decide whether to deviate rom the ! steps or to shorten 4our method. In this &eneral case we will %e solvin& all unnowns and reaction orces. It shall %e nown that or these &eneral ! steps the method was derived rom the steps in the A. /assimali Structure Anal4sis te#t%oo [1]. For 4our irst step 4ou must solve the %ase shear reactions in all columns' eepin& in mind assumption num%er two 6interior column has dou%le %ase shear orce than the e#terior7.
Once 4our irst step is complete' 4ou will tae the rame &iven to 4ou and place hin&es at all mid,hei&ht and mid,len&ths o the rame.
Once all %ase shear orces have %een calculated and hin&es placed' 4ou must then split the rame up at all hin&e locations.
Once the Frame is split into pieces 4ou must then tae the moment a%out the hin&es' preera%l4 4ou should start with the piece where the a#ial orce was applied. 5his same process must %e repeated on the other side o the Frame. Once this is solved or 4ou are a%le to %alance out ever4 piece %4 euili%rium and solve or the moments at the i#ed column ends.
Once all values have %een ound 4ou are a%le to ill in and %ac solve an4 unnowns remainin& within 4our rame. Once this is completed 4ou have successull4 complete a $ortal Frame e#ample.
"#ample Pro$lem 1
5he portal method will %e used as an appro#imate method to &enerate the a#ial' shear and %endin& moment dia&rams or the %uildin& rame shown %elow. 5he %uildin& is 2 store4s tall' and is divided into eual si9ed %a4s' each with dimensions o !m # !m. 5he %uildin& is e#posed to two lateral loadin&s o @ C and +@ C' actin& at the top o the second store4 and irst store4 respectivel4.
5o %e&in anal49in& this 12 de&ree indeterminate structure' we must irst mae use o our simpli4in& assumptions. ;e will %e&in %4 placin& hin&es at the mid,span and mid,hei&ht o each mem%er' as this has %een determined to %e the appro#imate location o 9ero moment. 5his irst assumption has reduced the de&ree o indeterminac4 to 2. 5he second assumption that must now %e made is tain& the stiness o the interior columns to %e twice that o the e#terior columns. 5his assumption allows us to tae the hori9ontal reaction orce o the middle column as %ein& dou%le the orce at either o the letmost or ri&htmost column. Cow we have a relationship which %inds 3 o our unnowns to a sin&le unnown' which has removed our once indeterminate structure' leavin& a staticall4 determinate one in its place.
Cow that the issue o resolvin& the %uildin&Ds indeterminac4 has %een overcome' all that remains is solvin& a comple#' %ut determinate s4stem. 5o do so' the irst step is to sum the orces in the xdirection' or &lo%al euili%rium to solve or the hori9ontal reactions at the %ase o the structure' which are all &iven in terms o the varia%le F 1.
E F x@@ F 1@@kN +@kN H6 F 12 F 1 F 171@@kN H F 12!kN
Ater this is done' a similar procedure will %e used to anal4se the second store4 o the %uildin&. 5he two store4 rame will %e separated at an ar%itrar4 location throu&h the cross sections o the columns to 4ield somethin& that resem%les the i&ure to the %ottom let. In this case' the assumption stands that the interior columns will %ear twice the orce o the e#terior ones' so we can mae a new euation in terms o F 2.
E F x@@ F 2@@kN H6 F 22 F 2 F 27@kN H F 21@kN
At this point' we will %e&in to disassem%le the entire structure at the hin&es. 5he implied condition that there is no moment at the location o the hin&es' still stands' and allows us to solve the orces in each mem%er o the structure %4 separatin& it into individual sections. 5his is shown in the i&ure to the ri&ht.
*ependin& on which piece we are looin& at' there ma4 %e an4where rom 1 to 3 unnown orces actin& on it' so our three euations o euili%rium will %e suicient to ind each o them.
5his e#ample will &o throu&h the process e#plicitl4 or the three sections which contain the let column o the i&ure to the ri&ht. 5he procedure will %e the e#act same or the remainin& + sections. 5he i&ure to the let shows the pieces that we will %e looin& at now.
Startin& with the top section' we have an e#ternal load' and internal orces' %ein& a hori9ontal and vertical component orce actin& at %oth hin&es. 5he e#ternal load is nown' and as 4ou ma4 recall' so is the orce la%elled F By' which was determined rom the &lo%al euili%rium anal4sis o the top loor to %e F 21@kN . Our procedure or solvin& or the three remainin& unnowns is as ollows: •
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>se the sum o the orces in the x direction to ind the remainin& unnown hori9ontal orce F Ax. Find the sum o the moments a%out one o the hin&es to solve or one o the unnown vertical orces 6we will tae the sum o the moments a%out ( to solve or F Ay. >se the sum o the orces in the y direction to ind the remainin& vertical orce' F By.
E F x@ FAxF AxFAxF Ax@@kN F Ax F BxH@kN H F BxH@kN H6H F 27H@kN H6H1@kN 7H3@kN E M B@@H6@kN 762.!m7H6H F Ax762.!m762.!m76 F Ay7@H6@kN 7 62.!m7H6H3@kN 762.!m762.!m76 F Ay7 F Ay62!kNm7 62.!m7 F Ay1@kN E F y@ FByF By@ F AyH F By F Ay1@kN Cow that we have solved all o the orces at this section we will move on to the ne#t. At this point weDre &oin& to have to decide which section we will anal4se ne#t' and we have some options here. Ideall4 we would pro&ress in some orderl4 manne r' and solve or one o the adBacent sections 6either immediatel4 to the ri&ht or directl4 %elow7 %ut we could &o to an4 section which contains three or less unnown orces. ;e will proceed downwards. 5his section has three hin&es correspondin& to + internal orces' as well as another e#ternal lateral load. From Newton's Third Law of Motion' we now that the orces which we ound at hin&e ( in the a%ove section will have eual and opposite reaction orces on this s4stem at (' thus we alread4 now two o our internal orces' F Bx' and F By. ie the case or the irst section' we also now the hori9ontal orce in the hin&e at * FDx' rom our &lo%al euili%rium o the entire structure' to %e F 22!kN . ;e now have a s4stem with three unnowns as %eore' and we will ollow the same procedure as we are aced with the same issue o one unnown hori9ontal orce and two vertical orces. E F x@@+@kN H F Bx F Cx F DxF CxH+@kN F BxH F DxF CxH+@ kN 6H1@kN 7H6H F 17 F CxH+@kN 6H1@kN 7H6H2!kN 7 F CxH!kN E M D@@6!m76 FBx7H6+@kN 762.!m7H6 F Cx762.!m762.!m7 6 F Cy7@6!m76H1@kN 7H6+@kN 762.!m7H6H!kN 762.!m762.!m7 6 F Cy7 F Cy60-.!kNm762.!m7 F Cy3!kN E F y@ FDyF DyFBy@ F By F CyH F Dy F By FCy1@kN 3!kN !kN Cow we will continue to move downwards to our %ottom section. In the last part we had alread4 used the act that F Dx F 12!kN ' and o course that relationship still stands. (ecause we now the orces at the hin&e' *' we are let with one unnown vertical orce and or the irst time' a moment. In each o the other sections there were no moments to %e calculated' which is the
result o us choosin& to %rea the sections at the hin&e locations. ;e will use our euations o euili%rium to solve or the two remainin& unnowns as alwa4s. E M D@ M E M E ME @62.!m76 FDx7H M E 62.!m76 F Dx762.!m7 6H2!kN 7H+2.!kNm E F y@ FEyF Dy@ F DyH F Ey F Dy!kN ;e would now return to the middle section o the top store4 and ollow wor our wa4 down a&ain' then &o up to the ri&htmost section o the top store4 and &o downwards until all o the unnown orces are resolved. Ater &oin& throu&h all individual sections' all o the pin reactions will have %een ound. 5hese pin reactions' as 4ou ma4 have reali9ed' correspond to the internal shear and a#ial orce that e#ists in the accordin& mem%er. 5hese orces are summari9ed in this ima&e.
5he ne#t step is to ind the a#ial' shear and %endin& moment dia&rams. Once a&ain this will %e done e#plicitl4 or the two mem%ers which mae up the let column o the structure' and the remainder will %e summari9ed %elow. 5his part is uite simple. 5o ind the shear and a#ial orce in a mem%er' one would normall4 %e reuired to mae a cut alon& the mem%er and then solve or these internal orces' however since this procedure reuired us to place hin&es at the mid,spans and mid,hei&hts o the mem%ers' we can tae the reaction orces at these hin&es as the internal orces. All loads are assumed to %e applied to the Boints and thus the shear orce is constant alon& the len&th o the mem%er' and accordin&l4 the slope o the moment will also %e constant. 5his &ives simpl4 that or the top letmost column' the shear orce is simpl4 &iven %4 1@ C' and the a#ial orce is 1@ C 6in tension7. ecall that these reaction orces were ound at a pin so that there would %e no internal moment at that point' and thus simpli4in& our anal4sis. Since we now that the shear is constant over the mem%er' the moment at the mem%erDs end can %e calculated %4 multipl4in& the shear %4 the hal len&th o the mem%er. 5his would result in a moment o 2!Cm or at the top o the mem%er in uestion and ,12! Cm at the %ottom. 5he same procedure is used to ind the a#ial orce' shear orce and %endin& moment in the %ottom let column. Once a&ain we ind that the a#ial orce in the mem%er is ! C 6in tension7 shear in the mem%er is 2! C and accordin&l4 the internal moment at the mem%erDs ends are o ma&nitude +2.! Cm. 5his %endin& moment can %e conirmed to %e correct %4 comparin& it with the support moment reaction at the %ase o the column' which was o%tained in our anal4sis o the determinate structure. 5his corresponds to the ollowin& a#ial orce dia&ram' shear dia&ram and %endin& moment dia&ram. Axial Force Diagram
Shear Force Diagram
Bending Moment Diagram
"#ample Pro$lem 2
5he portal method will %e used to construct the shear orce and moment dia&ram or &irder "FGJ. 5he %uildin& structure is two stories hi&h' with 3 %a4s located on irst loor and one su%seuent loor on second level' each with dimensions 2@m # 12m. 5he %uildin& is e#posed to two lateral loadin&s o 2@ C and 1@ C' actin& at the top o the second store4 and irst store4 respectivel4.
5o anal4se this indeterminate structure' we will calculate the internal loads at the inluence points. ;e will place hin&es at the mid wa4 o each %eam where it has 9ero moment. Similar to pro%lem 1 a%ove the same assumptions o tain& the interior column stiness to %e twice o the e#terior. 5his assumption allows us to have one unnown in the structure and thereore the other internal orces can easil4 %e calculated.
;e can now solve the determinate structure' we d o this %4 summin& all the x orces or euili%rium to solve or hori9ontal reaction at the %ase o the structure. ;e do this or the entire structure to ind our varia%le F 1. In this case we have two interior columns with %ear twice the orce o the e#teriors. E F x@@ F 1@2@kN 1@kN H6 F 12 F 12 F 1 F 273@kN H+ F 1!kN
Cow since we have ound the hori9ontal orces at the %ase F 1' we can ocus on the second level store4. 5he same method is used to calculate the hori9ontal orce at the %ase cut o the second store4 to ind varia%le F 2. In this case there is onl4 one %a4 located at the second level and thereore there is onl4 e#terior columns. 5hereore a new euation in terms o F 2 will %e ormed.
E F x@@ F 2@2@kN H6 F 2 F 272@kN H2 F 21@kN
;ith an4 structure 4ou alwa4s want to start at the top to %e&in solvin& 4our unnowns. On the top loor we have an e#ternal load o 2@ C' and internal orces o Fy and Fx'. 5he e#ternal load o FBx is nown as F 21@kN . Cow we can solve or the three unnows as ollows:
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5he sum o all orces in the x direction to ind the remainin& unnown hori9ontal orce F Ax.
E F x@ FAxF AxFAxF Ax@2@kN H F AxH F Bx2@kN H F Bx2@kN H 6 F 272@kN H61@kN 71@kN
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alculate the moment a%out one o the hin&es to solve or one o the unnown vertical orces 6we will tae the sum o the moments a%out ( to solve or F By.
E M A@@H61@kN 76+m761@m76 F Ay7 F By6+@kNm7 61@m7 F By+kN
•
>se the sum o the orces in the y direction to ind the remainin& vertical orce' F Ay.
E F y@ FAyF Ay@ FAyH F By F By+kN
Cow we will continue to solve or another section. Ideall4 4ou want to solve the section with e#ternal orces on them %ecause 4ou can easil4 calculate 4our 3 internal orces. In this case 4ou can solve section with a e#ternal orce o 1@ C to calculate 4our 3 unnowns. >sin& the same steps a%ove with 4our e#ternal load F 1!kN . E F x@ FCxF CxFCxF Cx@1@kN H F CxH F Dx1@kN H F Dx1@kN H 6 F 171@kN H6!kN 7!kN E M C @@6!kN 76+m7H61@m76 F Dy7 F Dy63@kNm761@m7 F Dy3kN
E F y@ FCyF Cy@ F DyH F Cy F Dy3kN Cow we will continue to proceed downwards at section ". From CewtonDs 5hird aw o )otion' we now that the internal orces at the hin&es * are eual and opposite reactions orces on section ". Since we alread4 calculated the internal orces or hin&e * we can calculate hori9ontal' vertical and moment at point ".
Shear dia&ram
)oment dia&ram
E F y@ FEyF Ey@ F DyH F Ey F Dy3kN E F x@ FExF Ex@ F DxH F Ex F Dx!kN E M E @@6!kN 76+m7H6 M E 7E M E 3@kNm >sin& the same steps we can continue to the ri&ht o the structure to calculate the hori9ontal and vertical interior columns orces and the moment at point ). E M K @@H63kN 762@m761@kN 76+m76+kN 761@m761@kN 76+m7 H61@m76 FFy7 F Fy612@kNm761@m7 F Fy12kN E K y@ FKy@ F KyH63kN 7H612kN 76+kN 7kN E F x@ FLxF Lx@ FFxH F Lx F Fx1@kN E F y@ FLyFLy@ F FyH F Ly F Fy12kN E M L@@61@kN 76+m7H6 M E 7E M E +@kNm
Cow with all the orces and moments calculated we can ind the shear and moment dia&ram or "FGJ.
"#ample Pro$lem 3
5he $ortal )ethod is an appro#imate anal4sis used or anal49in& %uildin& rames su%Bected to lateral o and vertical loadin& o !@ C and 2! C' actin& at the top o the second store4 and irst store4 respectivel4. 5he two store4 %uildin& divided into eual si9ed %a4s' each with dimensions o m # 2m. *etermine the appro#imate values o moment' shear and a#ial orce in each mem%er o the rame.
In order to solve such pro%lem usin& the portal method the ollowin& assumptions are made:
1. $lacin& hin&es 6appro#imate location o 9ero moment7 at mid, hei&ht o each column and centre o each %eam. 2. 5he hori9ontal shear is divided amon& all the columns on the %asis that each interior column taes twice as much as e#terior column
First' consider the upper part and place hin&es at mid,hei&ht o each column and centre o each %eam. O%tain the shear in each column rom a ree %od4 dia&ram %4 assumin& shear o the interior column eual to twice the shear in e#terior column.
>pper part o the structure
For simplicity, each node is given a number from 1 to 10
>se the three euations o euili%rium to solve or the unnown orces %4: •
•
•
>se the sum o the orces in x direction to ind the remainin& hori9ontal orces. alculate the moment a%out one o the hin&es to solve or one o the unnown vertical orces. >se the sum o the orces in the y direction to ind the rest o the vertical orces.
E M !@ y6H2712.!617 @ y+.2!E Fy@+.2!H y!@ y!+.2!E Fx@12. !H x!@ x!12.!
E Fy@ y3! @
E M 1@12.!617H y2627 y2 +.2! M E Fy@H+.2! y1 @ y1+.2!E Fx@ x112. !H!@@ x13-.!
Cow' consider the %ottom part and place hin&es at mid,hei&ht o each column and centre o each %eam. O%tain the shear in each column rom a ree %od4 dia&ram %4 assumin& shear o the interior column eual to twice the shear in e#terior column.
ower part o the structure
E M -@H y+62710.-!61712.! E M @ y062712.-!6 E Fy@1!. 617+.2!627@ y+21.0-!E Fy 1712.!617+.2!627 +2!H!@ y @21.0-!H @ y021.0-!E Fy@ y 1@H1!.+2! H+.2!H y-@ y-1!.+2!E Fx@ 1!.+2!E Fx@ x @ y1@!@ 10.-!H12.!H x2@ x2+.2! +.2! 5he sum o the orces on the %ase o the structure shown in the dia&ram %elow:
Axial Force Diagram
Shear Force Diagram
Bending Moment Diagram
,e-erences K ?ump up to:1.@@ 1.@1 1.@2 1.@3 1.@ 1.@! 1.@+ 1.@- 1.@0 1.@ 1.1@ /assimali' A. 62@117. Strutura! Ana!ysis" S# Edition $%th ed&. Stamord' 5: en&a&e earnin&. ?ump up to:2.@ 2.1 2.2 2.3 2. 2.! 2. K *r. Itehar Anam 62@117. LAppro#imate ateral oad Anal4sis %4 $ortal )ethodL. Cov 2!th' 2@13. Mhttp:<
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