Advanced Methods of Structural Analysis

6.2 Initial Parameters Method 155

Thus, displacement at pointBequals

EIy.l/D



RA

P

2



l^3

2



RAl^3

6

C

P

6



l

2

 3

: (e)

2.Deflection at pointBand stiffnesskare related asyBD.RB=k/D.PRA/= k,
therefore
EIy.l/DEI

PRA

k

: (f)

Solving system of equations (e, f) leads to an expression for the reaction at

supportA

RADP

1 C

11

48

kl^3

EI

1 C

kl^3

3E I

:

Substitution of the last expression into (d) leads to the equation of elastic curve

for the given system.

Special cases:

1.IfkD 0 (cantilever beam), thenRADP

2.IfkD1(clamped-pinned beam), thenRAD11P =16andRBD5P =16.

Example 6.4.The continuous beam in Fig.6.6is subjected to a uniform loadqin

the second span. Flexural stiffnessEIis constant. Derive the equation of the elastic

curve of the beam. Calculate the reactions of supports.

y( l ) = 0

x

Initial

parameters:

y 0 = 0, q 0 ≠ 0

q

l

y

RA

q 0

x

y(x)

RB RC

A B C

y( 2 l ) = 0

l

Fig. 6.6 Design diagram of two-span continuous beam and its deflected shape

Solution.The universal equation of elastic curve of a beam is

EIy.x/DEIy 0 CEI 0 x



RA.x0/^3

3Š

C

RB.xl/^3

3Š



q.xl/^4

4Š

:

Since the initial parametery 0 D 0 then

EIy.x/DEI 0 x

RAx^3

6



RB.xl/^3

6

C

q.xl/^4

24

: (a)