11.2 Ancillary Diagrams 375
adbcEI, a –24° h^ =^ 0.4ma M1–0 M M3–4
M1–2 3–204123bcM1–0 M1–2M3–2 M
3–4d1Joint 13Joint 3
Joint-load diagram30 aEI 30 aEIFig. 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 = 4mjD1 2
0j = 0.05rad D = 0.032maM^0 Pb M1–0 M1–2c1Joint 1M1–0 M1–21Joint-load
diagramMj 1 = 0.014EI (kNm)dFig. 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).