Moment Diagram by Parts

The moment-area method of finding the deflection of a beam will demand the accurate computation of the area of a moment diagram, as well as the moment of such area about any axis. To pave its way, this section will deal on how to draw moment diagram by parts and to calculate the moment of such diagrams about a specified axis.

Basic Principles

  1. The bending moment caused by all forces to the left or to the right of any section is equal to the respective algebraic sum of the bending moments at that section caused by each load acting separately.
    $M = ( \, \Sigma M \, )_L = ( \, \Sigma M \, )_R$


  2. The moment of a load about a specified axis is always defined by the equation of a spandrel
    $y = kx^n$

    where n is the degree of power of x.


The graph of the above equation is as shown below

Area and Centroid of General Spandrel


and the area and location of centroid are defined as follows.

$A = \dfrac{1}{n + 1}bh$


$X_G = \dfrac{1}{n + 2}b$


Cantilever Loadings
A = area of moment diagram
Mx = moment about a section of distance x
barred x = location of centoid
Degree = degree power of the moment diagram

Couple or Moment Load
cantilever-moment-load.jpg$A = -CL$

$M_x = -C$

$\bar x = \frac{1}{2} L$

Degree: zero

Concentrated Load
cantilever-concentrated-load.jpg$A = -\frac{1}{2} PL^2$

$M_x = -Px$

$\bar x = \frac{1}{3} L$

Degree: first

Uniformly Distributed Load
cantilever-uniform-load.jpg$A = -\frac{1}{6} w_o L^3$

$M_x = -\frac{1}{2} w_o x^2$

$\bar x = \frac{1}{4}L$

Degree: second

Uniformly Varying Load
cantilever-triangular-load.jpg$A = -\frac{1}{24} w_o L^3$

$M_x = -\dfrac{w_o}{6L} x^2$

$x = \frac{1}{5}L$

Degree: third


Saan po galing yung L sa Mx=−[wox2]/6L under Uniformly Varying Load? Pagkaintindi ko po sa moment is Mx = -1/2 * (wo) * x * (1/3)x = −[wox2]/6