Advanced Methods of Structural Analysis

(Jacob Rumans) #1

226 7 The Force Method


Displacement in the primary system due to applied load is


1PD


MN 1 MP^0
EI

D

1
EI



1
3



qa^2
2

a

3
4

a

1
EI



qa^2
2

aaD

5
8

qa^4
EI

.m/:

The negative sign at each term means that ordinates of the two bending moment
diagramMN 1 andMP^0 are located on different sidesof the neutral line of the cor-
responding members of the frame. The primary unknown isX 1 D1P=ı 11 D
15
32 qa .kN/. The positive sign shows that thechosendirection for the primary un-
known coincides with itsactualdirection.


Construction of bending moment diagram The final bending moment diagram will
be constructed based on the superposition principleMDMN 1 X 1 CMP^0 .Thefirst
term presents the bending moment diagram due to actual primary unknownX 1 D
.15=32/ qa; the procedure above for construction of bending moment diagram is
presented in Fig.7.9e.
The ordered calculation of bending moments at specified points of the frame is
presented in Table7.3. Signs of bending moments are chosen arbitrarily and used
only for convenience of calculations; these rules do not influence the final bending
moment diagram. In our case, signs are accepted as shown below.


Ta b l e 7. 3 Calculation of bending moments
Points MN 1 MN 1 X 1 MP^0 MN 1 X 1 CMP^0
1 0 0 0 0
2 a=2 15=64 C1=8 7=64
30 a 15=32 C1=2 C1=32
300 Ca C15=32 1=2 1=32
4 Ca C15=32 1=2 1=32
Factor qa^2 qa^2 qa^2


signs of
bending moments
+ −

+

The bending moment diagram is presented in Fig.7.9e. This diagram allows us
to trace a corresponding elastic curve of the frame; this curve is shown on Fig.7.9e
by dotted line. Rolled support at point 1 does not prevent horizontal displacement
of the cross-bar. Therefore, this structure presents the frame with sidesway (cross-
bar translation). In this case, all ordinates of bending moment diagram for vertical
member 3–4 are located on one side and therefore, elastic curve for this member has
no point of inflection.


Kinematical verificationDisplacement in the direction of constraint 1 in the origi-
nal system has to be zero. This displacement may be calculated by multiplication of
two bending moment diagrams, i.e., one isfinal bending moment diagram for given
structureMP(Fig.7.9e), and the second is bending moment diagramMN 1 in unit
state (Fig.7.9c).

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