60 APPLIED MECHANICS 
example, the beams shown in Fig. 66 are acted on by forces P, Q, and R 
to the right of the transverse section XY, and the bending moment at 
XY is equal to Px/+Qxm-Rxn. 
The loads on a beam also tend to shear the beam transversely, and 
the shearing action at any transverse section is equal to the resultant of 
the transverse forces on one side of the section. For example, the 
shearing action at the section XY of the beams shown in Fig. 66 is 
equal to the resultant of the forces P, Q, and R which act to the right 
of the section, and this resultant is equal to P+ Q-—R. 
The drawing of the bending moment diagram for a beam is simply 
the application of the construction explained in Article 59. In Fig. 67 
is shown a horizontal cantilever carrying vertical loads AB, BC, and CD. 
abed is the line of loads, or polygon of forces. A pole o is chosen so 
that the pole distance of is a simple multiple or sub-multiple of the 
linear unit. The funicular polygon a’d’n’ is then drawn. It is easy to 
show, as in Article 60, that the bending moment at any section XY is 
equal to a,d, x oh, z.e. the depth of the funicular polygon under the 
section multiplied by the pole distance. The depth of the funicular 
polygon is measured by the force scale, and the pole distance by the 
linear scale. It follows that, since the pole distance is the same for 
all parts of the funicular polygon, the depth of the funicular polygon 
under any section of the beam is a measure of the bending moment on 
the beam at that section, the scale for measuring the bending moment 
being found as explained in Article 59. 
The shearing force diagram is constructed by drawing horizontals 
across the spaces A, B, C, and D at ; 
the levels a, 6, c, and d respectively. A B c D E 
The depth of this diagram under any 
section of the beam, measured with 
the force scale, ‘gives the vertical 
shearing force on the beam at that 
section. For example, at the section 
XY the shearing force is the resultant 
of the forces to the right of XY, and 
is equal to BC+ CD=bc+cd=bd. 
Another example is illustrated in 
Fig. 68. The beam in this case is 
supposed to rest on supports at its 
ends. There are three forces AB, 
CD, and DE acting downwards, a 
force BC acting upwards, and the 
reactions EF and FA at the sup- 
ports acting upwards. The bending 
moment and shearing force diagrams Fic. 68 
are drawn as already explained. It sa 
will be noticed that the forces DE and EF are equal, and therefore 
there is no shearing force on that part of the beam in the space D; also, 
the bending moment on that part of the beam is uniform. The thick 
line HKLM shows roughly how the beam will bend ; the points K and L 
where the bending moment changes its sign are points of inflexion. 
In the examples illustrated by Figs, 67 and 68 the loads acting on 
