Fixed End Moments

The horizontal deflection at the rafter level=5DH=0.85AB=(0.8x20)=16 mm Final values of support reactions:

Continuity Moments:

Free bending moment member BCD:

Free bending moment member DEF:

Free bending moment member DGH:

* The maximum value along the length of members BCD can be found by identifying the point of zero shear as follows:

Examples in structural analysis 444 tciwim fliMMe

f^------ - A ~ IT

7m f i®

jxiinl hio.11Lir;i\ijr^"

G

1

"J

l

Deflected Shape

5.3.2 Problems: Moment Distribution - Rigid-Jointed Frames with

Sway

A series of rigid-jointed frames are indicated in Problems 5.13 to 5.18 in which the relative EI values and the applied loading are given. In each case:

i) sketch the bending moment diagram and ii) sketch the deflected shape (assuming axially rigid members) and compare the shape of the bending moment diagram with a computer analysis solution of the deflected shape.

25 kni

ej ef 21

Ma PA 4,0m

Problem 5.13

Problem 5.14

Problem 5.15

Problem 5.16

Problem 5.17

Problem 5.18

5.3.3 Solutions: Moment Distribution—Rigid-Jointed Frames with

Solution

Topic: M anient Dislrilmliun- Rigirt-Jtiintrd Pnmks wilh Sivny IVnlilviH ¡MMffltefl 5,13 Plig^

li llx

Consider ilit fttne analysis as the swporiwiiiion of two effects

Kin j I Kortts =• '[VoSwpy F'orccs" + LSway t'orefs'

Ii INI

Prop l y P

p* arbilnuy svi ;iv f orcc

it

/

irS

SA

Nn-H«'iiy Kraut? Consider lhr>'i-S» jv hrjimc fistd-traJ Moments:

Member BC*

J fa

-2S.0 kKnl

* Since support C is pinned, the fiwd-eJid momçnls îire (^g - Sfct/2) St B and. *çn3 m C.

(Mu* - MnQ) - J-25.0 - {Û.S X Î5.0J1 - - 3Î.5 kNm.

Solution

Topic: Moment IKslrilniliun- Di^id-Joinlcd Fniinisi)ilh Swav i'rulilrin Number; f.1 5 I'm«« No. 4

Tie muttiplying factor for the jmjr momenta = 3,41S

Fjggj M allium IllHlrilmlinii TaTilc.

Joint

A

It

C

Att

l!C

cu

Nu-Swav Moinnils

+ 1 8.44

+ 3&W

- 36.85

u

Moments« 3.4] S

- 77,25

-55.0S

+ 52.08

0

final Moment*

-5$JM

- 15.20

+ 15.20

0

The horizontal deflection at B = (Ah Sinfl - (34|,y/;/) * 0.S94 -ViSJSfEl The vertical dgftxlfen nl - Cosi^ - (34l.ifEI) * 0.J47 = 152.7W

1-in :i I Vfilikc (1 tn% = + 37.S3 - (I0.W « 3-41 J) = 0

Final va lite of Ft-+ 32JS1 4 (3J0S « 3j*1 5) - + kN f

Finalviiueof - + 47JS-(3.05 * 3.415) - + 37.0 LN f a r:

1-in :i I Vfilikc (1 tn% = + 37.S3 - (I0.W « 3-41 J) = 0

Final va lite of Ft-+ 32JS1 4 (3J0S « 3j*1 5) - + kN f

Finalviiueof - + 47JS-(3.05 * 3.415) - + 37.0 LN f a r:

Solution

T<h|»fc; Muntcnl DislHbution — Rip;id*Joi filed Frunir* with Sw-uj" Problem Tiuiiilï^r; f.IS E'n^t 3

Assume srhilnry rixeri-end tiwnieiiK

Moment l)i*lrtl)ulion Tihk:

Jiinl

A

B

B

1}

Alt

tB

IIA

BD

lie

CB

DC

Dbtrltalbn

1J0

L,

OJ

U

04

OJT

e.«

1.1)

im-il-fini

+

-lîf.O

l^linci'

- 9M

- 5^.1

- 'Î.Î ,

, -TIJÎ

-M.ÎJ

ùirwtr

-ÏSM

Ihilincr

+ lO.W

* 14.Ï* .

+ 11.43

(tro-tvir

* t<1.7> J

' - T.U

iLiUnrt

-iî!

-331

,

-3J»

Cinymr

-Ï.<H J

-114

Babil»

-161

-Mf.fi I

+ 181

* IJ3

-ll.vî

Ttlil

a

0

-4SJ1

+ IJMS

- tus

+ ÄS.3Ü

0

Dclcrminc Cht Vulur uf the ¡irWtnirJ' twoy f p< JSJÎtNnij, El^l.KLVm rV

B H

1

CöUsidtr Menlbcr b1>; •tn - 0 +1 WAt - {fffcx 4,i Consider MlnbtfCEi +rtpZMç-Q + 35 33 - {tf t * 6.0) Tor Lhe complete Ihunec

+YT —ÛÎ; = 0 +34.«+14,21 t"=(>

liUl

H» [jj ■■ H'b-H

.13 tN

0 0

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