7.6 Temperature Changes 255
ı 11 D666:67
EI;ı 12 Dı 21 D275
EI;ı 22 D170:67
EI: (b)h = 0.6mb= 0.2mt 1 =− 10 °C10
m5m3mC
B
1 2Ahh t 3 =+20°Ct 2 =0h= 0.6mb = 0.4mhtin= +20°tout= − 10 ° tav=+5°Dt=30°Design diagram Primary systemX 1X 215
4
36 8tav=+10°
Δt=+ 20 °tav=+5°
Δt= 30 °tav= –5°
Δt= 10 °abd 21d 11d 22d 12X 1 =110
10M 1X 2 =1
38N 1 M 2+1.0
X 1 =1c82.9914.628Mt kNm 68.37
(factor aEI)82.99
68.37
14.62843.99d X
2 =1(^6) X 1 =1
18
3
13
10
MΣ
e
Fig. 7.20(a)Design diagram and temperature distribution; (b) Primary system. (c) Bending
moment and axial force diagrams in unit states. (d) Final bending moment diagram and static
verification. (e) Summary unit bending moment diagram
For calculation of free terms we will use expressions (7.18a). These expressions
contain the average temperaturetavand temperature gradienttfor each member;
they are
For portion 1-3
tavD
20 C.10/
2
D 5 ıI tDj 10 .C20/jD 30 ı;
for portion 4-5
tavD
0 C.10/
2
D 5 ı tDj 10  0 jD 10 ı; (c)
