Advanced Methods of Structural Analysis

6.2 Initial Parameters Method 153

Fig. 6.4 Design diagram of
cantilevered beam and its
deflected shape x

y

Initial

parameters:

y 0 =0 q 0 =0

q

MA=

qa^2

2 l

a

RA=qa

x

y(x)

ymax

Solution.Since a distributed load is interrupted atxDathen an additional dis-
tributed load should be applied from the pointxDauntil pointxDland this
additional load must be compensated by the load of same intensity in the opposite
direction. Expression for the elastic curve is

EIy.x/DEIy 0 CEI 0 x







MA.x0/^2

2Š

C

RA.x0/^3

3Š



q.x0/^4

4Š

C

q.xa/^4

4Š

;

where the initial parameters arey 0 D^0 and 0 D^0 ; the third and fourth terms in
parentheses take into account the uniformly distributed load over theall lengthof
the beam and the compensated distributed load. The reactions areMADqa^2 =2and
RADqa. So the equations of elastic curve and slope become

EIy.x/D

qa^2

2

x^2

2

qa

x^3

6

C

qx^4

24



q.xa/^4

24

;

EI.x/D

qa^2 x

2



qax^2

2

C

qx^3

6



q.xa/^3

6

:

The beam with compensated load has two portions so the last term in both equations
must be taken into account only for second portion.axl/, i.e., for positive
values ofxa.
The transversal deflections of the beam at the pointxDaand at the free end are:

EIy.a/D

qa^2

2

a^2

2

qa

a^3

6

C

qa^4

24

D

qa^4

8

;

EIy.l/D

qa^2 l^2

4



qal^3

6

C

ql^4

24



q.la/^4

24

:

The right part of the beam is unloaded and has no constraints, therefore the second
portion.axl/is not deformable and the slopes for any point at this portion
are equal

EI.a/DEI.l/D

qa^3

6

: