Advanced Methods of Structural Analysis

6.3 Maxwell–Mohr Method 165

Ta b l e 6. 3 Calculation of vertical displacement of the joint 6 of the truss

Length of the

bar (m)

Axial forces

Bar Axial rigidity Actual state (Np) Unit stateNN

NpNlN

EA

O 1  3 4 2EA 0 0 0

O 3  5 4 2EA 4P=3 4=3 64P=18EA

O 5  7 4 2EA 4P=3 4=3 64P=18EA

O 7  9 4 2EA 0 0 0

U 2  4 4 2EA 4P=3 2/3 32P=18EA

U 4  6 4 2EA 4P=3 2/3 32P=18EA

U 6  8 4 2EA 4P=3 2/3 32P=18EA

U 8  10 4 2EA 4P=3 2/3 32P=18EA

V 1  2 3 EA 0 0 0

V 3  4 3 EA^000

V 5  6 3 EA 0 0 0

V 7  8 3 EA 0 0 0

V 9  10 3 EA 0 0 0

D 2  3 5 EA 5P=3 5=6 125P=18E

D 3  6 5 EA 0 5/6 0

D 6  7 5 EA 0 5/6 0

D 7  10 5 EA 5P=3 5=6 125P=18E

6.3.2 Deflections Due to Change of Temperature

It is often in engineering practice that the members of a structure undergo ther-
mal effects. In case of statically determinate structures, the change of temperature
leads to displacements of points of a structurewithoutan appearance of temperature
internal forces, while in case of statically indeterminate structures the change of
temperature causes an appearance of temperature internal forces. Often these inter-
nal forces may approach significant values. Analysis of any statically indeterminate
structure subjected to change of temperature is based on calculation of displacement
of statically determinate structure. So,calculation of displacements due to change of
temperature is a very important problem for analysis of both statically determinate
and indeterminate structures.
Two first terms of Maxwell–Mohr’s integral (6.11) may be rewritten as follows

kpD

X

Zl

0

MNkpC

X

Zl

0

NNkxp; (6.14)

wherep D .Mp=EI/dxis the mutual angular displacement of both sections
faced apart at a distance dxdue to the given load andxpD.Np=EA/dxis the
mutual axial displacement of both sections faced apart at a distance dxdue to the
given load.
These terms may be easily computed for the case of temperature change. Let
us consider elementary part of a structure with length dx. The height of the cross