Advanced Methods of Structural Analysis

11.2 Ancillary Diagrams 373

q=2kN/m 1

l 1 = 8m l 2

1

q M 1 =16kNm

16kNm

M^0

P

1-state 2-state (^1) J-L
Mj 1 = 16kNm
Fig. 11.4 Transformation of external load to the equivalent joint load
moment (FEM) equals toM 1 D
ql^21
8 D16 .kNm/. We can show the joint 1 with
joint moment 16 kNm counterclockwiseaccording a location of extended fibers.
This moment should be transported on the joint-load diagram counterclockwise
(second state).
Figure11.5a presents design diagram of the frame;P 1 D 10 kN,P 2 D 16 kN,
qD 3 kN=m. Degree of kinematical indeterminacy equals 4, wherenr D 3 and
ndD 1. The primary system is shown in Fig.11.5b; all introduced constraints are
labeled as 1–4 and the specified sections as 5–8. Also this figure contains the bend-
ing moment diagram in primary systemMP^0 (first state). Bending moments which
act on joints 1, 2, 3 are shown in Fig.11.5c.
(^23)
1
Joint 1 Joint 2 Joint 3
M1–5
M2–3 M3–2
c
a
5m
q
4m 4m
4m
3m
3m
P 1
P 2
b
P 1
1
2
3
4
q
6
5
(^78) M^0 P
M1–5
M7–2
M2–3 M3–2
Cross bar 1-2-3-4
d
1 23
R1–5
R2–7
Mj 4 =6kNm
Mj 2 =10kNm Mj 3 =10kNm
Mj 4 =12.5kN
Joint-load
diagram
e
Fig. 11.5 (a–c) Design diagram of the frame and computation of the fixed-end moments; (d)Com-
putation of equivalent joint load Pj4;(e) Joint-load diagram (state 2)