Advanced Methods of Structural Analysis

11.8 Analysis of Statically Indeterminate Trusses 413

Vector displacements

ZED

2 6 6 6 6 6 6 6 6

Z 1

Z 2

Z 3

Z 4

Z 5

3 7 7 7 7 7 7 7 7

DK^1 PED

d

EA

2 6 6 6 6 6 6 6 6

2:2071

8:6594

0:2071

6:8665

1:7929

3 7 7 7 7 7 7 7 7

The negative sign forZ 1 indicates that the jointAaccordingZ-Pdiagram has a
negative displacement, i.e., downwards.
Vector of internal forces

ESD

2 6 6 6 6 6 6 6 6 6 6

S 1

S 2

S 3

S 4

S 5

S 6

3 7 7 7 7 7 7 7 7 7 7

DkAQ

TE

ZD

2 6 6 6 6 6 6 6 6 6 6

0:2071

2:2071

2:5360

3:1217

1:7929

1:7929

3 7 7 7 7 7 7 7 7 7 7

The negative sign forS 4 accordingS-ediagram means that the diagonal member 4
is compressed. The units for all internal forces are kN.
Control:ASDPE. In our case

AESD

2 6 6 6 6 6 4

0  10 0:707 0 0

0000:70710

1000:70700

00 0:707 0  10

0 0 0:707 0 0 1

3 7 7 7 7 7 5



2 6 6 6 6 6 6 6 6 6 6

0:2071

2:2071

2:5360

3:1217

1:7929

1:7929

3 7 7 7 7 7 7 7 7 7 7

D

2 6 6 6 6 6 6 6 6

0

 4

 2

0

0

3 7 7 7 7 7 7 7 7

For calculation of reaction of supports, we need to consider equilibrium condi-
tions for rolled and pinned supports. For example, the free-body diagram for pinned
supportDand corresponding equilibriumequations are shown in Fig.11.28e.

X

XD 0 !XDDS 1 C0:707S 3 D2:0kN;

X

YD 0 !YDDS 2 C0:707S 3 D4:0kN:

Of course, for thisexternallystatically determinate structure all reactions may be
determined considering the structure in whole. However, for calculation of reactions
we used typical approach as for any statically indeterminate structure.