MDME: MANUFACTURING, DESIGN, MECHANICAL ENGINEERING

# NON-CONCURRENT 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 Non-Concurrent.pdf Non-Concurrent.one

 Type Video Lesson Description and Link Duration Date Lecture Non-Concurrent Forces 10:27 min 20140331 Online Worked Problem: Beam with vertical force 17:02 min 20200504 Online Worked Problem: Beam with angled force 6:31 min 20200504 Online Worked Problem: Truss Reactions 10:27 min 20200504

## 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 Beam

Complete 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 + Fpivot = 0
So Fpivot = 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 Non-concurrent 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...
1. Calculate each moment (from each force separately) and add them up, keeping in mind the CW and CCW sign convention.
2. 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).
(Of course, if the body is in equilibrium then there is no resultant - it should be zero. We only do this when we are replacing a set of forces for some reason).

#### 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 non-concurrent 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 non-locating (axially free) bearing acts as a roller joint. (Above) A shaft mounted in bearings acting as a simply supported beam. (Above) The left bearing acts like a PIN support and the right bearing acts like a ROLLER support - allowing axial movement.

## 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.

There are many ways to show a simply supported beam, but they always have a PIN support on one end and a ROLLER on the other. (Above) The roller joint allows the beam to change length without fighting against it, such as when the beam deflects or changes temperature (expansion/contraction).

Pin supports can often be seen in real life; (Above) Pin joint on the Sydney Harbour Bridge (Above) A roller joint on a bridge

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 FB at roller joint.
NOTE: Make sure you give a negative sign if moment is anti-clockwise.
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 FB

4. VERTICAL FORCE EQUILIBRIUM: Add all vertical forces (Y components) to find FAY

5. HORIZONTAL FORCE EQUILIBRIUM: Add all horizontal forces (X components) to find FAX

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).

### 1. COMPONENTS.  (Not needed)

2. MOMENT EQUILIBRIUM: Take moment about the PIN joint A using unknown force RB at roller joint. Taking moments about point A to find reaction at B....

0    =  + (3*8) - (5 * FB)
kNm  =  m * kN  -  m * kN   (Checking units)

3. Solve to find FB

5*FB = 24
FB = 4.8 kN Reaction at B solved.

4. VERTICAL FORCE EQUILIBRIUM 0  =  - 8 +  4.8  +   FAY
FAY = 3.2 kN The completed Free Body Diagram.

We can check the answer by testing moment equilibrium at the other end B.
MB = -(2*8) + (5*3.2) = -16 + 16 = 0

5. HORIZONTAL FORCE EQUILIBRIUM

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 FAY. This must be taken at the fixed joint too.

4. HORIZONTAL FORCE EQUILIBRIUM: Add all horizontal forces (X components) to find FAX.  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 MA for the poor old fixed support - it  must everything!

### 1. SPLIT INTO COMPONENTS
Convert the 9kN force @ 300 degrees...
Fy = 9 * sin(300) = -7.794 kN
Fx = 9 * cos(300) = +4.5 kN Split all forces into components.

2. MOMENT EQUILIBRIUM Taking moment about the fixed joint A...
0      =  (4.5*0)    +  (7.794*3)  -   (3*5)     +  (MA)
(kNm)   =   (kN*m)     +     (kN *m)     -   (kN*m)    +   (kNm)
-MA = 23.383 - 15
MA = -8.383 kNm  (Negative = anticlockwise!) 3. VERTICAL FORCE EQUILIBRIUM Add all vertical forces (Y components) to find FAY.
0  =  -7.794  +  3  + FAY
FAY  = 4.794 kN Vertical reaction force solved.

4. HORIZONTAL FORCE EQUILIBRIUM: Add all horizontal forces (X components) to find FAX.

0     =   4.5  +  FAX
FAX  = - 4.5 kN   (Negative = to-the-left) Horizontal reaction force solved.

## 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... Forces A= 2.2kN, B=1.3kN, C=6kN 1. COMPONENTS.

Forces are already in components (no work to do).

2. MOMENT EQUILIBRIUM: Take moment about the PIN joint L using unknown force RR at roller joint. Taking moments about point A to find reaction at R....

Before you do the moment equation, you should draw all the unknown forces - guessing their direction.

(Note: If you guess wrong, you will simply get a negative answer).

Here are my guesses for RLX, RLY and RR. Guess the direction of reaction forces, then do the moment equation...

So the moment equation is: (in kNm)

0 = + (2.2*0.8) + (6*1.6) - (1.3*1.385) - (3.2 * RR)

3. Solve to find RR:

0 = + (2.2*0.8) + (6*1.6) - (1.3*1.385) - (3.2 * RR)

0 = + 1.76 + 9.6 -1.8005 - 3.2 * RR

3.2 * RR = 9.5595

RR = 2.987 kN

4. VERTICAL FORCE EQUILIBRIUM: 0  =  - 2.2 - 6 + 2.987  +   RLY
RLY = 5.213 kN Vertical Reaction L solved.

5. HORIZONTAL FORCE EQUILIBRIUM 0  =  0  +  0  - 1.3  +   RLX
RLX = 1.3 kN The completed Free Body Diagram.

Check

We can check the answer by testing moment equilibrium at the other support at R.
MR = -(2.2*2.4) - (6*1.6) - (1.3*1.385) + (RLY*3.2) = 0
So RLY = 16.6805 / 3.2  =  5.213 kN

Extra Step

Now we can resolve these components at the pin joint (L) into a resultant reaction force;

RL  = ( RLX2  +  RLY2 )0.5
= (1.32  +  5.2132 )0.5
=  5.372 kN

Angle of  RL  = ATAN ( RLY  /  RLX )
= 76.00o

### 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 100-101: Resultant of non-concurrent forces)
Do all questions 6.14 to 6.21 (page 105-107: Equivalent force/moment systems)
Do all questions 7.1 to 7.4 (Beams)