11.2 Ancillary Diagrams 375
adbc
EI, a –24° h^ =^ 0.4m
a M1–0 M M3–4
M1–2 3–2
04123
b
c
M1–0 M1–2M3–2 M
3–4
d
1
Joint 1
3
Joint 3
Joint-load diagram
30 aEI 30 aEI
Fig. 11.6 (a,b) Design diagram of the beam and computation of the fixed-end moments;
(c,d) Joint-load diagram in case of temperature change
Figure11.6c presents the bending moments vicinity the joints 1 and 3. The joint
moments equal
Mj1DM 1 0 M 1 2 D90 ̨EI60 ̨EID30 ̨EI.kNm/counterclockwise;
Mj2D0; Mj3D90 ̨EI60 ̨EID30 ̨EI.kNm/clockwise:
The final joint-load diagram is shown in Fig.11.6d.
Example 11.2.The continuous beam is subjectedto the angular displacement'
of fixed support 0 and vertical displacementof rolled support 2 (Fig.11.7a).
Construct the joint-load diagram.
l 1 = 5m l 2 = 4m
j
D
1 2
0
j = 0.05rad D = 0.032m
a
M^0 P
b M1–0 M1–2
c
1
Joint 1
M1–0 M1–2
1
Joint-load
diagram
Mj 1 = 0.014EI (kNm)
d
Fig. 11.7 (a,b) Design diagram of the beam and computation of the fixed-end moments;
(c,d) Joint-load diagram in case of settlements of supports
Solution.The primary system of the displacement method and bending moment
diagram caused by given settlements of supports is shown in Fig.11.7b (first state).