Advanced Methods of Structural Analysis

184 6 Deflections of Elastic Structures

moment diagrams is located one under the other and vertical portionACis located
one besides the other. For vertical memberAC, the area of the bending moment
diagram in actual state is ̋ 2 D .1=2/hPh. Corresponding ordinate from unit
bending moment diagram isy 2 D.2=3/ 1 h(Fig.6.23a). Using Vereshchagin rule
and taking into account the different flexuralrigidities for the vertical and horizontal
members, we find the required displacement:

BD

X 1

EI

Zs

0

MpMNdsD

1

EI 1



1

2

hPh

„ƒ‚...

̋



2

3

 1 h

„ƒ‚...

yc

„ ƒ‚ ...

ACelement



1

EI 2



1

2

Phl

„ƒ‚...

̋

„ƒ‚... 1 h

yc

„ ƒ‚ ...

CDelement

D

Ph^3

3 EI 1



Plh^2

2 EI 2

: (a)

Each term of the expression for horizontal displacement has negative sign, because
the bending moment diagrams for actual and unit states are located on different
sides of the basic line of the frame. The result of multiplication of diagrams within
the vertical memberBDequals to zero, since in actual state the bending moments
within the memberBDare zeros. A final negative sign means that assumed unit
force produces a negative work along the real horizontal displacementB.
(b)Angle of rotation at B.The bending moment diagram for actual state is shown
in Fig.6.23b; this diagram for problems (a) and (b) is same. For required displace-
mentB, the unit state presents the same structure with concentrated coupleMD 1 ,
which acts at pointB; direction of the unit couple is chosen in arbitrary way. In the
unit state, only vertical reactions1= larise. The extended fibers are located indoor of
the frame. The centroid and area ̋of bending moment diagram in actual state and
corresponding ordinateycfrom bending moment diagram for unit state are shown
in Fig.6.23b.
The result of multiplication of diagrams within two vertical membersACandBD
equals to zero. Indeed, for these portions the procedureMpMN D 0 becauseMND 0
for memberAC,andMpD^0 for memberBD. For memberCD, the Vereshchagin
rule leads to the following result

BD

X 1

EI

Zs

0

MpMNdsD

1

EI 2



1

2

Phl

„ƒ‚...

̋



1

3

 1

„ƒ‚...

yc

„ ƒ‚ ...

CDelement

D

Plh

6 EI 2

: (b)

The positive sign is adopted because the bending moment diagrams for actual and
unit states are located on one side of the basic lineCDof the frame. A final positive
sign means that assumed unit couple produce a positive work along the real angular
displacementB. Or by other words, the actual direction of angular displacement
coincides with the chosen direction for unit coupleM, i.e., the section at supportB
rotates counterclockwise.