NONCONCURRENT FORCES
Forces that do not intersect at a single point can create rotation (moment). Equilibrium requires X and Y resultants to be zero AND the total moment to be zero.
Lecture Notes: NonConcurrent.pdf NonConcurrent.one
Image  Video Lesson Description and Link  Duration  Date  Download 
NonConcurrent Forces  10:27 min  20140331 
Video: NonConcurrent Forces
1. General Definition of Equilibrium
In 2 dimensions, if all the forces are NOT concurrent then you must have equilibrium both forces and moments.Mathematically this can be written as;
Notes;
1. Clockwise is a positive moment.
2. Upwards is positive Y, to the right is positive X.
2. Take care with units here  especially with mm. It is best to convert everything to m first.
3. Moments can be taken around any point, although we usually select a point at the most unknown forces (e.g. A pin joint), or at the intersection of forces. This eliminates as many unknowns as possible in the first calculation.
Simple Balancing BeamComplete a FBD for the plank.(i.e. Find all the forces acting on it) + (5 × 0·50)  ( F x 0.25) = 0 So force F = 2·50 Nm / 0·25 m = 10 N Now check the Y direction:  5N  10N + F_{pivot} = 0 So F_{pivot} = 15N There are no forces in the X direction: 
If in equilibrium, the anticlockwise turning effect of force F must equal the clockwise turning effect of the 5N load. 
2. Resultant of Nonconcurrent Forces
If we want to replace a set of forces with a single resultant force we must make sure it has not only the total Fx, Fy but also the same moment effect (about any chosen point).It turns out that when we add up the moment of several forces we get the same answer as taking the moment of the resultant.
To obtain the total moment of a system of forces, we can either...
 Calculate each moment (from each force separately) and add them up, keeping in mind the CW and CCW sign convention.
 Calculate the moment caused by the resultant of the system of forces about that point (So long as the resultant is in the RIGHT PLACE to create the right rotation).
Resultant Force (Non Concurrent)Sum all Forces in the Y direction:Now find total moment about A: Now place the resultant to give same moment: 
From Ivanoff (ex 6.8 Old Edition) 
3. Beam Reactions
A
common problem in statics is to calculate the beam reactions developed
by a set of nonconcurrent forces. The most common example is the
Simply Supported Beam which is a model of a typical bridge. In a simply
supported beam, one support is a Roller joint and other is a Pin Joint.
The roller may take many forms, some of which are shown below (first
row of table).
This all looks like structural engineering, but keep in mind that in mechanical engineering these connections can exist in other forms.
For example a simply supported beam might appear as a shaft, where a locating bearing acts as a pin joint, and a nonlocating (axially free) bearing acts as a roller joint.
1.Simply supported beams
A simply supported beam has the absolute minimum to hold the beam in place. One side has a pin joint (allows it to rotate but prevents movement), and the other side is a roller joint (up/down force only). The other advantage is that any expansion/contraction or deflection of the beam can be tolerated at the roller joint because it allows horizontal movement.To solve a Simply Supported Beam:
1. SPLIT INTO COMPONENTS.
Convert any forces at an angle to X and Y components.
2. MOMENT EQUILIBRIUM:
Take moment about the PIN joint A using unknown force F_{B} at roller joint.
NOTE: Make sure you give a negative sign if moment is anticlockwise.
The best way to do this is to put the moment in brackets, THEN look back at the diagram and see which way it is trying to spin around point A. This is the safest way because there are many possibilities: +/ Fy, +/ Fx, & +/ dy, +/ dx.
3. Solve to find F_{B }4. VERTICAL FORCE EQUILIBRIUM:
Add all vertical forces (Y components) to find F_{AY }
5. HORIZONTAL FORCE EQUILIBRIUM:
_{ }Add all horizontal forces (X components) to find F_{AX }
Make sure you are consistent with units. Usually best to work in m, but force is sometimes more convenient in kN. You can use anything though, so long as you are CONSISTENT. i.e. Convert everything over to mm, or kN etc FIRST (before you begin any work).
Find Reactions at A and B1.
COMPONENTS. (Not needed)
Take moment about the PIN joint A using unknown force R_{B} at roller joint.
3. Solve to find F_{B}_{}
We can check the answer by
testing moment equilibrium at the other end B.
There are no horizontal
forces, so equilibrium is complete. 
2. Cantilever beams
A cantilever beam has the absolute minimum to hold the beam in place from one side only. One side has more than a pin joint  it has a fixed support (solid or encastered joint). This prevents movement in X and Y and also rotation.To solve a Cantilever Beam:
1. SPLIT INTO COMPONENTS.
Convert any forces at an angle to X and Y components.
2. MOMENT EQUILIBRIUM:
Take moment about the fixed joint A. The moment applied by the forces must be resisted by the fixed joint.
3. VERTICAL FORCE EQUILIBRIUM:
Add all vertical forces (Y components) to find F_{AY. }This must be taken at the fixed joint too.
4. HORIZONTAL FORCE EQUILIBRIUM:
_{ }Add all horizontal forces (X components) to find F_{AX. } This must be taken at the fixed joint as well! _{ }
Basically, the fixed joint takes everything, so you simply work out the resultant Fx, Fy and M_{A} for the poor old fixed support  it must everything!
Find Reactions at A1.
SPLIT INTO COMPONENTS
Taking
moment about the fixed joint A...
3. VERTICAL FORCE EQUILIBRIUM
Add
all horizontal forces (X components) to find F_{AX. }

Worked Example
In
this example the problem looks complicated, but it isn't. It is really
just a simply supported beam. If we take the FBD (Free Body Diagram) of
the WHOLE truss we can forget about the fact that it is made up of 7
pieces bolted together.
Now it is just another plain old
simply supported beam.
This is why Free Body
Diagrams are so powerful and essential to engineering.
Find Reactions at L and R where...

Solving Reactions (Animation)
Animation showing how equilibrium of moments can be used to find the reaction forces at the supports.Questions:
Homework Assignment:Do all questions 6.10 to 6.13 (page 100101: Resultant of nonconcurrent forces)
Do all questions 6.14 to 6.21 (page 105107: Equivalent force/moment systems)
Do all questions 7.1 to 7.4 (Beams)