Example 54 Rigid Jointed Frame with Sway Frame

A rigid-jointed frame is fixed at support A, pinned at support H and supported on a roller at F as shown in Figure 5.23. For the relative EI values and loading given:

i) sketch the bending moment diagram, ii) determine the support reactions and iii) sketch the deflected shape (assuming axially rigid members) and compare with the shape of the bending moment diagram, (the reader should check the answer using a computer analysis solution). EI=10xl03 kNm2

12 kN 12 kN

10 kN

10 kN

Support ii sillies- bv in in -3.0 m i 3,0 m i 3.0 m

Figure 5.23

Consider the frame analysis as the superposition of two effects: t'iiiLil Forces ™ 'No-Sway Fnrcts1 4- 'Sway Forces*

Figure 25.5

Consider the No-Sway Frame: Fixed-end Moments Member BCD

Fixed-end Moments Member DEF

Fixed-end Moments Member DGH

Since support H pinned, the fixed-end moments are (MDH—0.5MHD) at D and zero at H.

Distribution Factors : Joint B

Distribution Factors : Joint D

No-Sway Moment Distribution Table:

Jnim

a

I!

D

F

M

as

ha

111!

mi

nu

13-f

FD

lilt

Drtlributln Kate«»

OLM

M3

IJ.-H.

txt

u

1,4

r-'iu-iS-rmt McmcnU

-J J.I)

T».0

*t.s

it

j" lfrs^

-■flJil

-ÎKÏÎ

Ctrmnr

f Ji.lT ■

r*

- ÎKI ■

-+ li.U

Sl^IlfF

* ÎjW.

_-Ml

- 3.W)

Cirr^mif

1 I.mi

f*

- '

1 + IJSU

Hjljnfc

4- 1.17

i L.'JO

-O.SJ

-0.-iï

HVSÎ-

- O.J I «

«

Bjbrift

HklS

* ftlfi

-0.44

-0LÎÎ

- DÜ5

irarrv-ah rr

1-o.oî-

Tslul

+ J.SJ

* HUT

-I Ml

t ÎO.frfi

- liJD

0

0

Determine the value of the reactions and prop force P:

Consider member DEF:

Consider member DGH:

Consider member BA and a section to the left of D:

:0u06 tNm

:0u06 tNm

Equation

Solve equations (1) and (2) simultaneously:

-VSb+lSiHi16 67 + 0.44 f^ Vxm* 16.67 +(0.44 *2<>.!W}

Consider the equilibrium of the complete frame:

Since the direction of the prop force is right-to-left the sway of the frame is from left-

toright as shown.

Apply an arbitrary sway force P' to determine the ratio of the fixed-end moments.
Sway Moments

The fixed-end moments in each member are related to the end-displacements (5) in each case. The relationship between 5ab, 5ad and 5DH can be determined by considering the displacement triangle at joint B and the geometry of the frame.

Displacement triangle:

Rigid-jointed frames 439 Assume arbitrary fixed-end moments equal to:

{- 24.0 : - 24.0 : + 20.0 : + 20.0 : - ! 5.0} * (¿/¿¡W100

Sway-Qnly Moment Distribution Table:

-lOiBÏ

A

ii

F

Fl

All

SA

BD

19»

1) II

FD

EID

CJi«iCr ili iiehio F*Cl»rs

O

«us

U3

(M It

O.ÎT

l.û

1.4

Flitd^+fid Moments

- 24.0

- M.O

+ 2ftO

-

- li.O

<}

0

15¡1 la ne*

1.«

» Î.-IS .

2..W

- 1.3.1

- IJJ

- 0.76 "

- 1.1 Î

~t J.II

13il In Mte

u.J-I

L - DSt

- OJ3

tiiTy-frier

• y.22 ■

■ * (J.Ji

Jin la nee

_> 0.1 1

■ U.IX _

(J. Ii

- LI. 1E)

0.10

C»iry--ovfr

-rt.iis

— 4»JCI» '

fl.LW

lï:i la nc«

„+ o.wl

'd.di

- UM

-LI.!«

-1MB

farrv-fïicr

* O.W2

**

Ti>t*ï

- 12S)5

-IIA

+ 11.94

* ULtiO

- Ifi.Kfl

- liU

V

it

Determine the value of the arbitrary sway force P:

Consider member DEF:

Consider member DGH:

Consider member AB and a section to the left of D:

2J.95 fcNm

2J.95 fcNm

II ,00 kNn

tSiii tSiii

J&6Q kMni t

■ in (!>■-. :■■ 2:.i>\> i) .■ !''-. ■ : ■ .:-■?/,'\ Equation

■■ ■ (.■ ■■ , y.!>i ■■. -.■::) ■ -■ <\ :. ¡"- -■ ■ ■::.■ ■:: ■■■I//', Equation

Solve equations (3) and (4) simultaneously:

+ 14 95 + L UK'/, p 1- 0.4S +■ OMHa J"A-+(We-<a44x 16.26)

Consider the equilibrium of the complete frame:

Stairs Moment Distribution

The multiplying factor for the sway moments=+2.02 Final Moments Distribution Table:

Joint

A

B

D

F

H

AB

BA

BD

DB

DH

DF

FD

HD

No-Sway Moments

+9.93

+19.87

-19.87

+20.06

-6.66

-13.40

0

0

Sway Momentsx2.02

-46.36

-44.24

+44.24

+37.57

-33.93

-3.64

0

0

Final Moments (kNm)

-36.43

-24.37

+24.37

+57.63

-40.59

-17.04

0

+1 -1

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