92 APPLIED MECHANICS 
rected on the assumption that the load Q is distributed over the surface 
AB. Generally the distribution of the load on Q 
‘AB would not be uniform. The point just dis- A\B 
cussed is of little practical importance, but some- dx} 
times students find it to be a difficulty. = <= 
104. Relations between Bending Moment and cD 
1 
Shearing Force Diagrams.—Let M be the bending R am ¥eat 
moment and F the shearing force at a section AC ef 
of a beam (Fig. 122). The section AC may be NE 
anywhere except at a point where there is a M a 
concentrated load, but 7t may be as near to that 
point as is desired. Take a section BD at an X ee 
indefinitely small distance dx from AC, and let f N 
M+dM be the bending moment at BD. Let Q F 
be the resultant of any loads which there may D 
be on the beam between the sections AC and BD, x x 
and let its distance from BD be nda where n is Fra. 122 
a fraction. aes 
Let R be the resultant of all the external forces acting on the beam 
to the left of the section AC, and let its distance from AC be Then, 
M=Ra, F=R, M+dM=R(#+dz) — Qndx= Ra + Rdx — Qndz, there- 
fore, dM = Rdw — Qnda, and Be =R-Qn=F- Qn. 
Tf Q is the résultant of a load which is distributed over AB, then, 
since dx is indefinitely small, Q will be so small that it may be neglected, 
and then ae sR. 
dx 
Again, if Q is the resultant of loads concentrated at points in AB, 
these loads may be avoided by taking da small enough, and then as 
before Say 
x 
The shearing force F at the section AC is therefore a measure of the 
slope of the bending moment line at the point corresponding to AC. In 
other words, the shearing force at any section is equal to the rate of 
increase of the bending moment at that section. 
Again, dM=Fdz, therefore the difference between the bending 
moments at two sections indefinitely near to one another is equal to the 
area of the shearing force diagram between these sections, and, in passing 
from one section to any other section, it is obvious that the sum of all the 
increments dM will be equal to the sum of all the increments Fdz, and 
therefore the difference between the bending moments at any two sections 
is equal to the area of the shearing force diagram between these sections. 
These results are very interesting and very useful. For example, 
referring to a horizontal beam, if the bending moment line is horizontal 
at any point its slope is nil, and there can therefore be no shearing force 
at the corresponding section. Again, at the highest point of the bending 
moment line the slope changes from positive to negative, and therefore 
where the maximum bending moment occurs the shearing force must 
change its sign, and the shearing force line will cross the base line. The 
converse of this is not always true, and all that can be said about the 
