Advanced Methods of Structural Analysis

336 10 Influence Lines Method

SinceX 3 is antisymmetrical unknown, for the right part of the arch it is necessary

to change sign on the opposite and make the changeu! 1 u. Therefore, if unit

loadPis located on the right part of the arch then reactionRAis

RADX 3 D12 .1u/^2





1

4

C

1 u

6



for0:5u1:0:

Corresponding influence line is presented in Fig.10.6d.

Moment at Support A:

MAD 1 ulCX 1 CX 2 fX 3

l

2

Du



 1 C

9

2

u 6 u^2 C

5

2

u^3



l for 0 u0:5

MADX 1 CX 2 fX 3

l

2

D.1u/^2



5

2

u^2 u



l for0:5u1:0:

Corresponding influence line is presented in Fig.10.6e.

10.1.2.4 Bending Moment at CrownC

MCDX 1 Du^2





3

4

C

5

2

u

5

4

u^2



l for 0 u0:5:

SinceX 1 is symmetrical unknown, for the right part of the arch it is necessary to

make the changeu! 1 u. Therefore, if unit loadPis located on the right part of

the arch, then bending moment at crown is

MCDX 1 D.1u/^2





3

4

C

5

2

.1u/

5

4

.1u/^2



l for0:5u1:0:

Influence line forMCis presented in Fig.10.6f.

ConclusionsIf loadPis placed in the portion of0:132lin both sides from crown

C, then the extended fibers atCare located below the neutral line of the arch. The

direction of the support momentMAdepends on the location of the load: if loadP

is placed within o:4lfrom the left support, then extended fibers in vicinity of theA

are located outside of the arch.

Discussion

1.For given parabolic nonuniform arch we obtained the precise results. It happens
because a cross moment of inertia is increasing from crown to supports according
to formulaIxDIc=cos'x.SincedxDdscos'x,ds=EIxDdx=EIcand all
integrals are presented in exact form.
2.In arch with clamped supports subjected to distributed load along half-span the
maximum bending moments arise at supports. For this case, the following law