## 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

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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

 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 J ' - T.U iLiUnrt -iî! -331 , -3J» Cinymr -Ï.; •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