Advanced Methods of Structural Analysis

152 6 Deflections of Elastic Structures

account according to the universal equation (6.3) are reactionRAof the supportA
and uniformly distributed loadq. The expression of elastic curve is

EIy.x/DEIy 0 CEI 0 x



P.xaP/^3

3Š



q.xaq/^4

4Š

; (a)

wherePDRAD0:5ql,aP D 0 ,aqD 0 ; the initial parameters arey 0 D 0 and
 0 ¤ 0.
The negative sign before the brackets means that they-axis is directed downward.
The signs before the first and second terms in the brackets indicate that bending
moments atxdue to a reactionPDRAand distributed loadqare positive and
negative, respectively. Equation (a) contains one unknown parameter 0 , which can
be calculated taking into account the boundary condition at the right support. Since
the vertical displacement atBis zero, i.e.,y.l/D 0 ,then

EIy.l/DEI 0 l

ql

2

l^3

6

C

ql^4

24

D 0 (b)

This equation leads toEI  0 Dql^3 =24. Therefore, the expression (a) of the elastic
curve becomes

EIy.x/D

ql^3

24

x

ql

2

x^3

6

C

qx^4

24

D

ql^4

24



x

l

 2

x^3

l^3

C

x^4

l^4



: (c)

An expression for slope of the beam may be derived from (c) by differentiation

EI.x/DEI

dy

dx

D

ql^3

24



1  6

x^2

l^2

C 4

x^3

l^3



:

The slopes at the left and right supports areEI.0/D ql^3 =24andEI.l/D
ql^3 =24.
Since the elastic curve is symmetrical with respect to the middle point of the
beam, then the maximum displacement occurs at the pointxD0:5l. Thus, from (c)
we can obtain the following result

EIymaxD

5

384

ql^4 :

The positive sign indicates that the displacement occurs in positive direction of the
y-axis.

Example 6.2.A uniform cantilevered beam has a uniform loadqover the interval
aof the beam, as shown in Fig.6.4. Derive the equation of the elastic curve of the
beam. Determine the slope and deflection of the beam at the free end.