328 10 Influence Lines Method
Influence line for bending MomentMk
Should be constructed using formula (10.4). Bending moment at sectionkin the pri-
mary system due to primary unknownX 1 D 1 isMkD0:4(Fig.10.2a), therefore
IL.Mk/D0:4IL.X 1 /CIL
Mk^0
: (10.10)
Construction of IL.Mk/step by step is presented in Fig.10.3.
k
P= 1
1345268910711
0.01920.0336 0.0288 0.4⋅IL X
1
− − (factor l)
0.0384 0.02880.03840.03360.0192
a
Load P=1 in the left span Load P=1 in the right span
Mk=0 (see primary system
in Fig. 10.1.b)
0
0.120.240.160.08
Inf. line Mk
(factor l)
+
b
+
− Inf. line (factor l)Mk
0.10080.20640.12160.0512
0.02880.03840.03360.0192
IL(Mk)=0.4IL(X 1 )+IL(Mk)
c^0
0
Fig. 10.3 Construction of influence line for bending moment at sectionk
The first term0:4IL.X 1 /of (10.10) is presented in Fig.10.3a. The second
term IL
Mk^0
is the influence line of bending moment at sectionkin the pri-
mary system. If loadP D 1 is located in the left span, then the influence line
presents the triangle with maximum ordinate at the sectionk; this ordinate equals to
ab= lD.0:4l0:6l/ = lD0:24l.IfloadPD 1 is located in the right span, then
the bending moment at sectionkdoes not arise, and the influence line has zeros
ordinates. It happens because the primarysystem presents two separate beams. In-
fluence line forMk^0 is shown in Fig.10.3b. Summation of two graphs0:4IL.X 1 /and
IL
Mk^0
leads to the final influence line for bending moment at sectionk; this influ-
ence line is presented in Fig.10.3c. The maximum bending moment at the sectionk
occurs when the loadPis located at the same section. The positive sign means that
if a load is located within the left span, then extended fibers at the sectionkwill be
located below the neutral line. It is easy to show the following important property
of influence line forMk. The sum of the absolute values of the slopes at the left and
right of sectionkis equal to unity.