Advanced Methods of Structural Analysis

168 6 Deflections of Elastic Structures

Solution.Since the required displacement isC, then a structure in the unit state
is loaded by unit forceXD 1 at pointC. The bending moment and axial force di-
agrams are shown in Fig.6.13. General expression for displacements due to change
of temperature is

ktD

X

Zs

0

̨

t 1 Ct 2

2

NNkdxC

X

Zs

0

̨

jt 1 t 2 j

h 0

MNkdx: (a)

For memberABof the structure integral

R
NNkdxDh 1 , while for memberBC
integral

R

NNkdxD 0.

For membersBCandABof the frame, the integral

R
MNkdxequals to.1=2/ll
andlh, respectively. Therefore, for the given structure the required displacement
becomes

CD ̨

0 C 20

2

h 1

„ ƒ‚ ...

first term6:15

 ̨

j 0  20 j

d



1

2

ll ̨

j 0  20 j

b

lh

„ ƒ‚ ...

second term6:15

: (b)

The first term of (b), which takes into account axial forces, is negative, since the
strains of the elementABinduced by the variation in temperature is positive (elon-
gation) and by the unit load is negative (compressed). The second and third terms of
(b), which take into account bending moment, are negative, since the tensile fibers of
the elementsABandBCinduced by the variation in temperature are located inside
of the frame, and by the unit load are located outside of the frame.
The final result for the required displacement is

CD10 ̨



hC

l^2

d

C 2

lh

b



:

Negative sign shows that the actual displacementCdue to the variation in tem-
perature is opposite with induced unit loadX.

Example 6.10.Determine the horizontaldisplacement of pointBof the uniform
semicircular bar in Fig.6.14a, if the indoor and outdoor temperature rises byt 1 ıC
andt 2 ıC, respectively. The height of cross section bar ish 0.

Solution.A temperature effect related to curvilinear bar, therefore the general ex-
pression for temperature displacement should be presented in terms of curvilinear
coordinatesinstead ofxas for straight member

BtD

X

Zs

0

̨

t 1 Ct 2

2

NNkdsC

X

Zs

0

̨

jt 1 t 2 j

h 0

MNkds: (a)