10.2 Construction of Influence Lines by the Displacement Method 349
13452689107 11
k
0.081610.17280.08320.0224
+
R
bInfluence line for Mk^0
Inf. line Mk^0
(factorl l )
Mk^0 depends of position P
the left or right at k
Mk^0 = 0 for any position P
LoadP =1
in the left span
Load P =1
in the right span
P = 1 P^ =^1
P = 1
1345268910711
k
ul R =u^2 ( 3 −u)
2
ul
ul
ul
ul
11
P = 1
13452689107
k
(All ordinates
must be multiplied
by factor l )
–1.2EI
l
⋅IL(Z 1 )
0.01920.0336
0.0288
0.03840.0288
0.03840.03360.0192
−
+
a
Final
Inf. line Mk
(factor l )
+
−
0.10080.2064 0.0512
0.02880.03840.0336
0.1216
c 0.0192
Fig. 10.12 (a–c) Construction of influence line for bending momentMk
10.2.1.3 Influence Line for Shear ForceQk
This influence line should be constructed by formula (10.19).
According to Fig.10.11c, the reaction at the support 1 due toZ 1 D 1 is directed
downward, so the shear in primary system at sectionkdue to primary unknown
Z 1 D 1 isQkD 3 EI=l. Therefore, (10.19) becomes
IL.Qk/D
3 EI
l^2
IL.Z 1 /CIL
Q^0 k
: (10.23)
The first term
3 EI=l^2
IL.Z 1 /of the (10.23) is presented in Fig.10.13a. The
second term of the (10.23) presents influence line of shear at sectionkin the primary
system. For calculation of this term we need to consider location of the loadPD 1
in both spans separately.