Advanced Methods of Structural Analysis

182 6 Deflections of Elastic Structures

Fig. 6.22 Design diagram
of the beam and bending
moment diagrams for actual
and unit states

Unit

state

P =1

b =l/8 d =l/4

l/8

M

(m)

a =Pl/8 c =Pl/4

Pl/8

MP

(kN⋅m)

yC

Ω

0.25l 0.25l 0.25l 0.25l

P

EI kEI

Actual

state

AB

DEC

Solution.Bending moment diagrams in actual and unit states are presented in
Fig.6.24. For computationC D.MpM/=N EI, we have to subdivide bending
moment diagrams on the parts, within which the bending stiffness is constant. These
parts for the left half of the beam areADDl=4andDCDl=4. The Vereshcha-
gin rule for multiplication of diagramsMpandMN within portionADleads to the
following result

C1D

1

EI



1

2



Pl

8



l

„ ƒ‚^4 ...

̋



2

3



l

„ƒ‚...^8

yc

D

Pl^3

768 EI

: (a)

For portionCDthe trapezoid rule is applied. According to (6.22)weget

C2D

l=4

6 kEI

2

6

6

42 

Pl

8



l

„ ƒ‚^8 ...

2ab

C 2

Pl

4



l

„ƒ‚^4 ...

2cd

C

Pl

8



l

„ƒ‚^4 ...

ad

C

Pl

4



l

„ƒ‚^8 ...

cb

3

7

7

5 D

7

768

Pl^3

kEI

: (b)

Finally, the vertical displacement atCbecomes

CD 2



Pl^3

768 EI

C

7

768

Pl^3

kEI



D

Pl^3

48 EI

;

where D.1=8/C.7=8k/; factor 2 takes into account symmetrical partCEBof the
beam. IfkD 1 ,then D 1.

Example 6.16.A portal frame is subjected to horizontal forceP as shown in
Fig.6.23. The bending stiffness for each member is shown on design diagram. Cal-
culate (a) the horizontal displacement at the rolled supportBand (b) the angle of
rotation at the same pointB.