Advanced Methods of Structural Analysis

7.3 Analysis of Statically Indeterminate Structures 241

MP^0 DR^0 Axq

x^2

2

D24xx^2

QP^0 D



R^0 Aqx



cos'D.242x/cos'

NP^0 D



RA^0 qx



sin'D.242x/sin': (c)

Internal forces at specified points in the loaded state of the primary system are pre-
sented in Table7.9.

Ta b l e 7. 9 Internal forces of
the arch in the loaded state

Points MP^0 Q^0 P NP^0

0 0:0 16.968 16:968

1 63 14.400 10:800

2 108 10.7328 5:178

3 135 5.8206 1:455

4 144 0.0 0:0

5 135 5:8206 1:455

6 108 10:7328 5:178

7 63 14:400 10:800

8 0:0 16:968 16:968

Computation of unit and loaded displacements For calculation of displacements,
the Simpson’s formula is applied.Unit and loaded displacements are

ı 11 D

MN 1 MN 1

EI

D

Xn

1

li

6 EI



a^21 C4c 12 Cb 12



;

1PD

MN 1 M^0

P

EI

D

Xn

1

li

6 EI

.a 1 aPC4c 1 cPCb 1 bP/; (d)

whereliis the length of theith straight portion of the arch (Table7.7);nnumber of
the straight portions of the arch;a 1 ,aPordinates of the bending moment diagrams
MN 1 andMP^0 at the extreme left end of the portion;b 1 ,bP ordinates of the same
bending moment diagrams at the extreme right end of the portion; andc 1 ,cPare the
ordinates of the same bending moment diagrams at the middle point of the portion.
Calculation of the unit and loaded displacements is presented in Table7.10.
Section “Unit state,” columnsa 1 , c 1 ,andb 1 contain data from columnMN 1
(Table7.8). Section “Loaded state,” columnsaP,cP,andbP contain data from
columnMP^0 (Table7.9). As an example for portion 1 – 2.l 1  2 =6D 3:5377=6D
0:5896/, the entries of columns 6 and 10 are obtained by following way

0:5896

EI

h

.2:625/^2 C4.3:5625/^2 C.4:50/^2

i

D45:9335=EI;

0:5896

EI

Œ.2:625/ 63 C4.3:5625/85:5C.4:50/108

D1;102:40=EI: