Advanced Methods of Structural Analysis

256 7 The Force Method

for portion 6-8

tavD

0 C 20

2

D 10 ı tDj 0 .C20/jDC 20 ı:

These parameters for each element of the frame are shown on the primary system
in Fig.7.20b. Detailed calculation of free terms is presented in Table7.13.

Ta b l e 7. 1 3 Calculation of free termsitof canonical equations according to formula (7.18a)
̨tav

R

NNidsfor portion ̨th

R

MNidsfor portion

1  3 4  5 6  8 1  3 4  5 6  8

P

1t ̨ 5  1  5 D
25 ̨

0 0 ̨^30 0:6 10  5 D

2 ,500 ̨

0 ̨0:6^20 ^12 

10  10 D

1 ,666:7 ̨

4,191.7 ̨

2t 0 0 0 ̨^30 0:6^3 C 28 
5 D 1 ,375 ̨

̨0:6^10 ^12 

3  3 D75 ̨

0 1,450 ̨

The first term in (7.18a) is positive if normal force in the unit state and tempera-
ture of the neutral fibertavhas the same sign (for example, for element 1-3, normal
force in the first unit state equalsC 1 and average temperature equalsC 5 ). The sec-
ond term in (7.18a) is positive if the bending moment diagram in the unit state is
located on the side of more heated fibers. For member 1-3, the first component is
1tD ̨ 5  1  5 ,where 1  5 presents the area of the axial force diagram along
this member.
Canonical equations become

666:67X 1 C275X 2 C4; 191:7 ̨EID0;

275X 1 C170:67X 2 C1; 450 ̨EID0: (d)

Roots of these equations areX 1 D8:2988 ̨EI;X 2 D4:8759 ̨EI. Bending mo-
ment diagram is constructed using formula (7.20). The corresponding calculation is
presented in Table7.14. ColumnsMN 1 X 1 ,MN 2 X 2 ,andMthave to be multiplied
by factor ̨EI.

Ta b l e 7. 1 4 Calculation of bending moments
Points MN 1 MN 1 X 1 MN 2 MN 2 X 2 Mt
1  10 C82:998 8:0 39:008 C43:99
3  10 C82:998 3:0 14:628 C68:37
4 0:0 0.0 3:0 14:628 14:628
5 0:0 0.0 0.0 0.0 0.0
6  10 C82:998 0.0 0.0 C82:99
8 0:0 0.0 0.0 0.0 0.0

signs of

bending moments

+ −

+

−

The resulting bending moment diagram is presented in Fig.7.20d.
Static verification.

X

MD.68:37C14:62882:99/ ̨EID.82:99882:99/ ̨EI0: