Advanced Methods of Structural Analysis

154 6 Deflections of Elastic Structures

As can be seen in this example, the initial parametersy 0 and 0 for the given struc-

ture are known right away, so there is no necessity to use the boundary conditions

at the right end.

Example 6.3.A uniform beam with clamped left support and elastic support at the

right end is subjected to a concentrated forcePat pointC, as shown in Fig.6.5;

stiffness parameter of the elastic support isk. Derive the equation of elastic curve.

Compute the reaction of a clamped support and consider the special caseskD 0

andkD1.

Fig. 6.5 Design diagram of

one-span beam with elastic

support and its deflected

shape

Initial

parameters:

y 0 =0, q 0 =0

l

x

y

MA

RA

Parameters

at x=l:

y≠0, q≠ 0

RB

0.5l

k

P

Solution.The general expression for elastic curve according to (6.3)and(6.5)is

EIy.x/DEIy 0 CEI 0 xC

MA.x0/^2

2Š



RA.x0/^3

3Š

C

P.x0:5l/^3

3Š

: (a)

The last term should be taken into account only for positive.x0:5l/.

Initial parameters arey 0 D 0 and 0 D 0. These conditions lead to the equation

EIy.x/D

MAx^2

2



RAx^3

6

C

P.x0:5l/^3

6

; (b)

which contains the unknown reactionRAand reactive momentMAat the clamped

supportA. For their determination we have two additional conditions:

1.The bending moment at supportBis zero, therefore

M.l/DMACRAlP



l

l

2



D 0

and the reactive momentMAin terms ofRAbecomes

MADRAl

Pl

2

: (c)

This expression allows us to rewrite (b) for elastic curve of a beam as follows:

EIy.x/D



RAl

Pl

2



x^2

2



RAx^3

6

C

P

6



x

l

2

 3

: (d)