Internal forces in depth: beam element, area rule, cantilever and point load
The equilibrium of the beam element, the area rule for drawing shear and moment by hand, the cantilever with a tip load and the simply supported beam with a point load proved, the method in five steps.
https://www.electronics-engineering.com/en/internal-forces-in-depth-area-rule/
This page goes deeper into the lesson on internal forces: it draws the equilibrium of the beam element, derives the area rule, the quickest way to draw the diagrams by hand, proves the diagrams of the cantilever and of the beam with a point load at an arbitrary position and sums up the method in five steps.
The beam element and the area rule
The proof of the differential relations is easier to follow by looking at the element with all forces and moments in their positive directions.
The two relations can also be integrated between two sections 
and 
:
![\[ V(x_2) - V(x_1) = -\int_{x_1}^{x_2}q(x)\,dx,\qquad M(x_2) - M(x_1) = \int_{x_1}^{x_2}V(x)\,dx \]](https://www.electronics-engineering.com/wp-content/ql-cache/quicklatex.com-8a813b41e2d71bdeb3307fdc0eb4c683_l3.png "Rendered by QuickLaTeX.com")
This is the area rule, the quickest way to draw the diagrams by hand:
- the change in shear between two sections equals (with the sign changed) the total load applied between them;
- the change in moment between two sections equals the area of the shear diagram between them.
A concentrated load 
is the limit of a very narrow distributed load: its integral is over a length of zero, so the shear jumps by 
at the point of application, while the moment, being the integral of a bounded function, stays continuous but changes slope (a kink). In the same way a concentrated couple makes the moment jump.
The cantilever with a load at the tip
Theorem. For a cantilever of span 
fixed at the left, with a load at the free end, at a distance 
from the fixed end:
![\[ V(x) = P,\qquad M(x) = -P(L - x),\qquad M_{max} = PL\ \text{(at the fixed end)} \]](https://www.electronics-engineering.com/wp-content/ql-cache/quicklatex.com-fd70dda355b84b0d1fd4920afd5cd16b_l3.png "Rendered by QuickLaTeX.com")
Proof. It is convenient to write the equilibrium of the right-hand part, which does not contain the fixed end and therefore needs no reactions. Cutting at , the right-hand part carries downwards at its end and the internal actions on its left face. Vertical equilibrium gives a shear equal to ; moment equilibrium about the section gives a moment 
that stretches the top fibres: with the adopted convention it is negative. At 
it equals 
. ∎
Check with the area rule. The area of the shear diagram from to is ; the moment at the right end is zero, so 
. Same result, without writing any equilibrium equation.
The simply supported beam with a point load
Theorem. A simply supported beam of span with a load at a distance 
from the left support (and 
from the right one) has reactions 
, 
, and the maximum moment under the load:
![\[ M_{max} = \frac{Pab}{L} \]](https://www.electronics-engineering.com/wp-content/ql-cache/quicklatex.com-38f908bdc2a33f70efb2c0bd40edba9e_l3.png "Rendered by QuickLaTeX.com")
Proof. Moment equilibrium about 
gives 
; vertical equilibrium gives 
. The shear is 
to the left of the load and 
to the right: it changes sign under the load, where the moment is maximum. By the area rule the moment there equals the area of the shear rectangle between 
and the load: 
. ∎
With the load at midspan 
and we recover the 
of the table in the lesson. It is also the worst position: the product 
with 
fixed is maximum for 
.
The method, step by step
For any statically determinate beam the procedure is always the same:
- Support reactions from the three equilibrium equations of the whole beam (lesson on statically determinate structures). It pays to write the moment equations about the supports, so that each equation has a single unknown.
- Segments: split the beam at the points where the load changes (concentrated loads, start and end of distributed loads, supports).
- Shear: start from one end and proceed with the area rule, adding the jumps of the concentrated loads and the slopes
of the distributed loads. - Moment: integrate the shear, again with the area rule; the maxima are where the shear vanishes or changes sign.
- Checks: the moment must return to zero at free ends and at end supports; at the end the shear must close with the reaction of the last support.
Common mistakes
- Changing convention halfway through: choose a convention for shear and moment (here: moment positive when it stretches the bottom fibres) and keep it, always writing the equilibrium of the same part, right or left.
- Looking for the maximum moment only where
: with point loads the shear jumps without passing through zero, and the maximum is at the jump where the shear changes sign. - Ignoring the sign of the reactions: a negative reaction is not a calculation error, but information about the support (see the beam with an overhang).
- Confusing the area of the load with the area of the shear: the area of the load gives the change in shear, the area of the shear the change in moment.
- Units: the moment is measured in kN·m (force times length), the distributed load in kN/m; mixing them up is the most common mistake in exams.
Exercises on the topic, with simply supported beams, cantilevers and overhangs, are on the exercise page linked at the top.
