BEAMS AND BENDING 101 
_ positive or negative shear. This is important in the case of open web 
or braced girders, and is referred to again in Art. 207, p. 233. 
108. Bending and Shearing by Forces in different Planes.—Suppose 
_ a beam AB (Fig. 130) to be acted on by a force P, at B and a force P, at 
D, the lines of action of P, and P, being in different planes, but per- 
_ pendicular to AB. Consider the bending action at a section C at a 
_ distance 2,, from P, and x, from P,. So far as the bending action of P, is 
concerned P, may be replaced by a force Q acting at B in a direction 
parallel to P,, the magnitude of Q being such that Qz,=P,7,. There 
_ are now two forces acting at B, namely, P, and Q, and their resultant R 
sy he 
A D 
ees = Op 
Fig. 1380. 
7 may be found by the parallelogram of forces. The resultant bending 
moment at C is then Rz,, and the plane of bending is ABR. 
If the plane ABP, be perpendicular to the plane ADP,, then 
R= /Pi+Q?, and Ra, = /{(P\7)?+(P,a,)*t. 
The following alternative method will in general be more convenient. 
Through C (Fig. 131) draw CF parallel to P, and equal to M,=P,z,. 
Draw CE parallel to P, and equal to M,=P,«,. Complete the parallelo- 
m CEYF. Then 6Y=M will be the resultant bending moment at 
©. Or, after drawing CF, draw FY parallel to P, and equal to P,z,, 
then CY, the closing side of the triangle CEY, is the resultant bending 
moment at C. 
It is understood, of course, that the actual scale drawing of the 
parallelogram or triangle must be made on a plane perpendicular to the 
length of the beam, and not in oblique projection, as shown. 
Any number of forces at right angles to the beam and in different 
planes may be dealt with in a similar manner. If there are more than 
two forces, a polygon will take the place of the parallelogram or triangle. 
If any of the given forces are not perpendicular to the beam, resolve 
them parallel and perpendicular to the beam, and deal with the com- 
ponents perpendicular to the beam, as above, to find the resultant bending 
moment at any section. 
To find the resultant shearing force at any section, consider the forces 
to one side of the section. Find the shearing forces at the section due to 
_ these forces separately. The resultant of these shearing forces is the 
_ resultant shearing force at the section. 
Exercises VIIa. 
Draw the bending moment and shearing force diagrams for the examples 
given. In this set of exercises the bending moments and shearing forces at 
@ suflicient number of sections should be found by calculation. 
