66 3 Multispan Beams and Trusses
Reaction of Supports and Internal Forces
ReactionsRAandRBfor any load can be calculated using following equilibrium
conditions:
RA!
X
MBD 0 IRB!X
MAD0:For calculation of the internal forces that arise in the members of the hinged chain,
we need to show the free-body diagram for any jointn(Fig.3.23). The equilibrium
equation
P
XD 0 leads to the relationshipSncos ̨DSn 1 cosDH: (3.2)Thus, for any vertical load acting on the given truss, the horizontal component of the
forces, which arise in the all members of the hinged chain, are equal. The horizontal
componentHof the forcesSn,Sn 1 is called a thrust.
Now we will provide an analysis for the case of a moving load. The influence
lines for reactionsRAandRBare the same as for a simply supported beam. How-
ever, the construction of an influence line for thrustHhas some special features.
Let us consider them.
ThrustH(Section 1-1, SPLC Is Panel 7-C; Ritter’s Point isC)
Internal forceS, which arises in the elementm-kof the hinged chain, is denoted as
SleftandSright. The meaning of the subscript notation is clear from Fig.3.23.
If loadPD 1 is located to the left of joint 7, then thrustHcan be calculated
by considering therightpart of the structure. The active forces are reactionRB
and internal forcesS 7 - C,S 8 - C,andSright. The last forceSrightcan be resolved into
two components: a horizontal component, which is the required thrustH,anda
vertical component, which acts along the vertical lineC-k. Now we form the sum
of the moment of all forces acting on the right part of the truss around pointC, i.e.,
H!
P
MCrightD 0. In this case, the vertical component of forceSrightproduces no
moment, while the thrust produces momentHf.
If loadP D 1 is located right at jointC, then thrustH can be calculated by
considering theleftpart of the structure. The active forces are reactionRAand
internal forcesS 7 - C,S 8 - C,andSleft. The forceSleft, which is applied at jointm,
can be resolved into a horizontal componentHand a vertical component. The latter
component acts along vertical linem-7. Now we find the sum of the moment of
all the forces, which act on theleftpart of the truss, around pointC. In this case,
the vertical component of forceSleftproduce the nonzero moment around jointC
and thrustHhas a new arm (m-7) around the center of momentsC.Inorderto
avoid these difficulties we translate the forceSleftalong the line of its action from
jointminto jointk. After that we resolve this force into its vertical and horizontal
components. This procedure allows us to eliminate the moment due to the vertical
component ofS, while the moment due to the horizontal component ofSis easily
calculated asHf.