Advanced Methods of Structural Analysis

304 8 The Displacement Method

Ta b l e 8. 6 Symmetric frame and corresponding half-frame for symmetric and antisymmetric
loading
Number
of spans Symmetrical loading Antisymmetrical loading

1,3,5,... Δv≠^0
Δh= 0
φ= 0

Half-frame

A

A

Half-frame Δv= 0

Δh≠^0

j≠ 0

A

A

Number of unknowns Number of unknowns

Entire frame Half-frame Entire frame Half-frame

FM: 9 FM: 5 FM: 9 FM: 4

DM: 9 DM: 4 DM: 9 DM: 5

2,4,6,...

Δv= 0

Δh= 0

j= 0

Half-frame

A

A

Half-frame Δv= 0

Δh≠ 0

0.5EI j≠ 0

A

A

EI

Number of unknowns Number of unknowns
Entire frame Half-frame Entire frame Half-frame
FM: 12 FM: 6 FM: 12 FM: 6
DM: 10 DM: 4 DM: 10 DM: 6
Internal
force
diagrams

Bending moment diagram – symmetrical

Normal force diagram – symmetrical

Shear force diagram – antisymmetrical

Bending moment diagram – antisymmetrical

Normal force diagram – antisymmetrical

Shear force diagram – symmetrical

In the case of antisymmetrical load, the horizontal and angular displacements at
the pointAexist, while a vertical displacement is zero. It means that an equivalent
half-frame at pointAmust contain a support, which would allow such displace-
ments, i.e., which would allow the horizontal and angular displacements and does
not allow a vertical one. Only roller support corresponds to these types of displace-
ments. Table8.6also contains the mathematical conditions for replacing the given
frame by its equivalent half-frame. In all cases, the support at the pointAfor equiv-
alent half-frame simulates the displacement at the pointAfor entire frame.