5.5. The differential relationships in bending
The relationship between the load intensity, the shearing force and the bending moment can be easily derived (Fig. 5.7).
Let us consider the beam loaded by an arbitrary load. Determine the shearing force at the section at a distance z to the left support. Projecting the forces to the left section in the vertical direction we get Analogously determine the shearing force for the next section at distance z+dz to the left support
we get
,
(5.1)
I.E. The intensity of the distributed load is equal to the derivative of the shearing force with respect to the bar section abscissa.
Fig. 5.7.
Let us
determine the bending moment at the section with the abscissa z
by taking the sum of the moments of the forces located to the left
from the section. For it replace the distributed load by its
resultant on the region of the
z
length which is equal to
as applied in the region middle at a distance
to the section
By analogy determine the bending moment at the next section located from the left support on the distance z+dz:
(5.2)
Subtracting the bending moment at two sections we get the increment of the bending moment
(5.3)
The
expression in the parentheses is the shearing force
.
Consequently
(5.4)
from where
I.E. The shearing force is equal to the derivative of the bending moment with respect to the bar section abscissa.
Taking the derivative of both parts of the equality (5.4), we get
(5.5)
I.E. The second derivative of the bending moment with respect to the bar section abscissa is equal to the intensity of the distributed load.
5.6. Drawing bending moment and shearing force diagrams
The diagrams M and Q are drawn to clearly imagine the character of the bending moment and the shearing force change along the bar length and to find the dangerous sections. The drawing technique of the diagrams is considered by the following examples.
Example. Draw the diagram M and Q for the cantilever beam (Fig. 5.8 a).
Solving.
There are two regions (
and BC),
differing by the load character, hence, also by and the change law
and
Q.
а)
b)
c)
Fig. 5.8.
The bending moment at the section of the region a distance z1 to the right of А is found by the moment of the left force. For that purpose the distributed load to the left of the section is replaced by the resultant force qz1 as applied in the middle of the region over the length z1. We get
(5.6)
The minus sign is necessary because the beam tends to bend concave downward. It is the parabola equation. The parabola is drawn by approximately three points:
(5.7)
Do a
section on the region ВС
at a distance z2
to the free bar end. Replace the distributed load along the length а1
by the resultant force
applied in the middle of the region АВ.
The moment at the section is
(5.8)
This is the
equation of the straight line. Determine
for two values
z:
and
Then we get
(5.9)
The diagram М is given in Fig. 5.8 b.
The
shearing force at the section /
— /
is equal to
as the sum of the projection on the vertical line of the force
located to the left of the section.
The
shearing force at the section //—//
is
equal to
The diagram is given in Fig. 5.8 c.