## Problem 821

8.16 Solutions: Plastic Analysis—Rigid-Jointed Frames 3

 Solution topic: I'LiMi: Analysis - Ki^iil JoluIciI frame* ) l'rûhltm Number: Kiiicmilic Mcthùd l'açc No, 2 Mechanism II: Ileam BCD i -fi * H tN rT v- ^ S YC = 8.0£-8 .Cff ■A = 0 sr Internal Work Done- l:\ier.i:il Work Done + M, (û+ffi * M. (flj = (75 v Jv^ 8) * M„ (tf+ ■ * 6000 150,0 kNm Mcchaiibra [II- Ku ay 9 Ti IN D J qy H H1 ■ r 'I l J! h 6 Mifa f \ i S « lî JQj? SjOfl p> i CUG70 J5 Internal Work Dene M Ism-mill Worfi Done (H (°> +A/j ( \$ * ,mo.i7i?>]=(-is xiae/i 7] .91 kN'm Media □ ism III: Cninbined Henni UCl> »nd Jinny In mechanism II the rxnm ion at joint]? = -Û I n meehamsnt 111 lite naiitw at ji>i at !l ■ + 0 Adding ctiuutirms for Mechiinisrïii |l + ItJ A.b\il.0=6Dd0 2&7Mt0= - ill lowing for the hinge eliminated at ioinl IÏ: 12 ■1.6WP0 -7920 Mr- kNm
 Solution Topic: rlaslic Ani1>ïis - Kî;i ,-i ffiEB u I Bttml Work Ine = Extendi Vdik Dtfnv 2.5.Wt,0-4O0 IÛ.C hiNui Mtthanism 111: Canibincd Ilium l>tF and Sway In metdudium 1 [ the rotation at joint f = — G In nicch.nnisni 111 (he s mat ion at jni (it F ™ + ff Adding equation for M Initowuifl Jor 1!« Inline eliminated al ioim B: ¿2 k iftj 4.5Mfl- SStf UW fcNm

Solution

Topic: n.iMit Anil^is - Hi ¡¡id JuLnlccC frames 3

I'rDblcm ¡Numherj - Sialic iVIclhod I'msc >"a. «

\fc = + (20 x 5.5} + (15-5 * 3.0) - (I0.O * 1.5} - * S i) -+1413 -Ma = - (12 * 1.0) + (10 * 3.0) + (J7.S * 2.0) - (HL, * 7.0) = + 113.0 - ?JGift .Mfe - + (tO k 1.5) + (47.5 * 1.0) - (/An * 5.5) = + 62,5 -Mt = 0 - {!!(, * 4 0) - 0 - 4.0//,;

Assume the vullupse mechanism as indicated previously, i.e. piasiie hinges developing under llie point load 31C (+ 2.0.1^) 31 and joint !■' (- -Up).

0= 141.5 - 13SHa lf(; =+ 10.JRkN snd M, = 41IS™h befor*

Chf(k tlu vil lit of tin? bg nd i ng ninmenl at otht r jmssible hi u^c potil ion«

Mt = + 60.0 + 4.0//,, - + SO.O - (4.0 " 10.4\$) = 3S.0& tMin i 1.5 3/n Mo = + 113-0 - ?.0l/(i =+ 113.0 - (7.0 K 10.48) = 3^-64 liNrn £ Mv Mt - + «.5 - 5.Siftj - + 62.5 - (5 5 * 10-4«)-4.S6 kNltl S Mp

Solution

Topic: rijiülic Analysis - Kigjil Joiitlcri trames 3 l'tübkm Kumíiir; S.ll - Kinematic Mtlliod

I'LIJC 3

ExtÉnill wflrii l>(mç = |<lí v Jvc) +- t'S * ¿vil) + ( lí¿vk) +1">* íhb)) [fiS * + +

internai W(uk Done = liJilcnwl Work Ojnc i.OhtfB = 2 JO0 ,tfp - 48.0 hiN m

Miilisniim V; Cuirtltinn! lh',uu DEF üml G:lI>Ic.

Iwuvnamviri KHH*ofia(atiíii fiw link BCDF, intend Work IJonc li ,Wp(íf+/J> + A/p {P+ít)J = 11SM^.Qty + Jt/p{4,OÍ7)J - 7 fíSít0 Eslífnal Work Dow

te.tC» * &.[>£\$ + (]S X + H 6.00) + (¡5 * O.OÍÜ) ■ 3300

interna] Woík = External Work

(.'[lie rendar slunild ccnlinn ihís ansner by ;i<3dijiij the ihmrfc equation!;).

l'his value is less llian tEi n 1 obtained for ÜW ¡¡able median ism. Assume llic gable inccliajiiíin (i.e. !i hites al I? and D) lo be Hie critical medunfem ami clicek iht bcndijig. moments at either possible hinge pusiliüns do not eseeed the values.

Miilisniim V; Cuirtltinn! lh',uu DEF üml G:lI>Ic.

Solution

Topic: l'Lihlk A.n:dysis - Hinitt .liimlnl Francs 3

i'mhlcin Number: it,2 I - Khi vin ¡Hic Method I'^eNti, ?

Afc = + <3S.S * 3.0)-(12.0» 7.i)+(IOUÛ* 1.3)-+ 10.5 Wm S 1.5¡1/, Mk - "(10 « l.î)- 24.Û + (S.O *4.S) + (16.5 « 3.0)-+ 1(1.5 kN ni iA/p Ma = - 24.0 kTJin s; M,

This frame can also be readily analysed using the sialic method of uns lysis ¡is fallows

Sol ul ion

Td|iii: Piaille A il :i Iyils — Ri^id Join led Pramcs 3

Pmhltm Number S.ll - Sialic Melhod l'une Nu. 1

A^-+<22.5* J.0) + (10x _ (7.5 * HÙ + 3(0.5/;/,,) ^ + S2.S -

Alii = + (2Ï.S * 0.0) + ( 10 * ¿,0}-(1S * 3.0)- 9Wa + (6,0xG.SHà

Mt + (22,i * i.o) + (IOx |j)-(1i * irO)-tti * 3,0) - 1ri/f% - (<>.Q x rt.if/o = + US - 3.0Ha

Assume [he collapse mechanism jih indicated previously, i.e. piaslie hinges developing a[ jriim l! (- IJA^) tt jnint 13

Sublniclhig equation ( I ) fnun équation (2) gives:

-IShtf - - 110 . - Mr - 4H.0 kNm as before add J?A 12.0 kN

C'liKt tht value of ilie bendiug moment ai other possible hinge positions

Mç = + Sï.i - 6.0//A-+ S3J - (6.0 * I2.Û) - + lO.i kisin s 1 5.1/? Alt = + SÎ.5 - 3.0JJ.H = + S2.5 - (3JÛ W 1J.0) = + Jfi.5 kNm S M, Mi ; - - 60,0 + S.OJft = - 60.(1 - (3,0 * 12,0) - - 2-4.0 kNm £ M?