Note: Both diagrams are 2d(they might look 3d because I drew them manually.) I have a prismatic frame that looks like this:Frame
I made a section on the half of the first pillar(the red line represents the section), to study the internal forces in it(Normal force, shear force and moment.) Here is a free body diagram at this section:
A is a fixed support so I calculated its reactions as follows:

  • $X_A= 100 KN$
  • $Y_A= 200 KN$
  • $M_A= 300 KN$

Note: $AB=3m$
I have also calculated the normal and the shear force at B, but I have a problem calculating the moment at B and here is why:
now let's apply the equilibrium equation of the moments at B, as:

$$\Sigma M_B=0$$

As you see there are the following moments:

  • $X_A\times 3$
  • $M_B$
  • and at last $M_A$

I have done it like this:
$$ 3X_A + M_B - M_A = 0 $$ I just need to know what to do with $M_A$ should I multiply it with the distance $AB$ or just leave it.(I tried both cases and I think both are wrong). now can somebody till me what to do with the $M_A$?


I guess the diagrams are somehow misleading,(I am not good at drawing!). The axis of moment B is the straight line perpendicular to the plane and passing through B and the same for A. Now please look at the first picture: imagine there is a force just above the red section, this force is causing the bending moment in B and the reaction moment in A. I want to calculate the internal moment B, and that's why I need the sum of moments at B.

  • $\begingroup$ Is $M_B$ moment of force about point A applied by a force at distance say $x$ from A ? $\endgroup$ Jan 14, 2014 at 20:04
  • $\begingroup$ @rijulgupta I guess my diarams were a little misleading, please take a look at the edit. $\endgroup$
    – Mohammad
    Jan 15, 2014 at 12:28

2 Answers 2


Edit with respect to correction/edit in question : First of all lets assume that $M_A$ is caused by force $F$ applied at distance $x$ from A, and distance from A to B is $L$.

To calculate $M_a$
$xF = M_A$
$M_a = \sqrt[2]{x^2 + L^2} F \sin\theta$
$M_a = \sqrt[2]{x^2 + L^2} F x / \sqrt[2]{x^2 + L^2}$
$M_a = M_A $
$M_a$ is moment of the force which provides moment $M_A$ about point A, about point B.

Rest of your equations look correct !

  • $\begingroup$ I want to calculate the moment at B, how can I acheive that without adding the $M_A$ $\endgroup$
    – Mohammad
    Jan 14, 2014 at 20:45
  • $\begingroup$ I sense you are interested in calculating $M_B$ only, so I suggest taking the axis along AB, moment of $X_A$ is not rotating about this axis, so leave it, about this axis the moment of $M_A$ is $M_A$ only so just equate both of them for equilibrium. $\endgroup$ Jan 14, 2014 at 20:50
  • $\begingroup$ Are you talking about total Moment about B ? $\endgroup$ Jan 14, 2014 at 20:51
  • $\begingroup$ No the diagrams are 2d not 3d, $M_A$ is the moment around A .the reaction of A towards bending. $\endgroup$
    – Mohammad
    Jan 14, 2014 at 20:54
  • $\begingroup$ The total moment around B is zero since it is an internal moment, I want to calculate the bending moment around B. $\endgroup$
    – Mohammad
    Jan 14, 2014 at 20:56

$M_a$ is a Moment of Couple and is a free vector. it can be moved any where in space.

whereas Force(X and Y) are sliding vectors they can only be moved along their line of action (principle of transmissibility).


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