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[Plesha, Gray, Costanzo, Engineering Mechanics: Statics]
Sirajuddin, David

Reaction forces: one or two unknowns needed?


How to draw FBDs of the machine? Dissemble into convenient parts to analyze. Draw an FBD of each
Each part is pin-connected 
 reaction forces R in general: R = R x
i + R y j

The reaction R points in some direction, we can always write R = R x
R y j
i +

But, these are two unknowns (R x
, R y
)! It is in our best interest to avoid doing this unless we have to. We would prefer only one unknown.


When do we need two unknowns?
 only when we have no idea which way the “resultant” R actually points
When may we use only one unknown?

2-force members!

Normal forces (e.g. rollers)
Itcanbeshown.com
Sirajuddin, David

Reaction forces: one or two unknowns needed?
 When may we use only one unknown?

2-force members!

Normal forces (e.g. rollers)
Itcanbeshown.com
Sirajuddin, David

Reaction forces: one or two unknowns needed?
 When may we use only one unknown?



2-force members!
Normal forces (e.g. rollers)
For example, remember this problem [P5.85, Plesha, Gray, Costanzo]?
Itcanbeshown.com
Sirajuddin, David

Reaction forces: one or two unknowns needed?
 When may we use only one unknown?



2-force members!
Normal forces (e.g. rollers)
For example, remember this problem [P5.85, Plesha, Gray, Costanzo]?
Itcanbeshown.com
30 o
A
Sirajuddin, David

Reaction forces: one or two unknowns needed?
 When may we use only one unknown?



2-force members!
Normal forces (e.g. rollers)
For example, remember this problem [P5.85, Plesha, Gray, Costanzo]?
One Unknown
A = -Acos30 o i + Asin30 o j
30 o
OR
A
A x
Itcanbeshown.com
Two Unknowns
A = -A x i + A y j
(No need to model the force like this if we can only use one unknown!)
30 o
A y
Sirajuddin, David

Reaction forces: one or two unknowns needed?
 When may we use only one unknown?


2-force members!
Normal forces (e.g. rollers)
2-Force member!, one unknown,
We know the direction
Itcanbeshown.com
Never do this. This is a 2-force member, but there are 4 unknowns. On this HW assignment, you can model them like this and it works out, but it will not always. You must make an insight. This gives a lot of unknowns. If you do not have a lot of leeway, this can put you in trouble of not being able to solve problems
Sirajuddin, David

 By the way, remember zero-force members in trusses?
 That is also key to solving some problems, and you must use physical insight to determine these.
 Like zero-force members, if you cannot identify a two-force member you may not be able to solve the problem
 Friendly off-topic analog:


(1) When given a “frames and machines” problem, your first step is find 2-force members
(2) When given a truss, your first step is to find zero-force members
Itcanbeshown.com
Sirajuddin, David




What do we mean by “reactions”?
For rollers and normal forces it is obvious
For pins, it can sometimes be confusing when many parts are all connected at one point…
Itcanbeshown.com
Sirajuddin, David




What do we mean by “reactions”?
For rollers and normal forces it is obvious
For pins, it can sometimes be confusing when many parts are all connected at one point…

“Which reactions go on which part?” Is the major question (see point
C)
Itcanbeshown.com
[Plesha, Gray, Costanzo, Engineering Mechanics: Statics]
Sirajuddin, David

Reaction forces occur between each part and the pin to which it is connected
 To answer this question, we need to understand exactly where the forces come from:
 Reaction forces are present because a particular part is pushing on a pin AND because the pin is pushing back on the part

This is facilitated by Newton’s third law
 For now, all we need to know is that if these forces are present on an FBD of a machine part, then those same force will be present pushing back on the pin
(meaning, wherever we put the pin physically in the picture [our choice] is the
only place where we draw the equal and opposite forces “connecting,” nowhere else)

The example problem will clarify
Itcanbeshown.com
Sirajuddin, David

 General procedure

Identify 2-force members first
 Definition of a 2-force member

A part that has only two connection points (where pins were)

The FBD of the part itself cannot have a pin on it if one end is connected to more than one member

For this problem:

Cannot have a pin at point C of member CD (we will see why later, C is a trisection of parts)

Can have a pin at point D on member CD (only two parts for the one pin at D, no problem)

If the above is true, then the lines of actions of each force acting at one connection point, intersects the other connection point.

For a straight member, the forces follow the geometry of the member

For any curved or odd shaped member, draw a line connecting the two connection points, this is the line of action of the forces at each connection point. (This tends to seem strange to students. Just try and trust yourself using the methods).
[Plesha, Gray, Costanzo, Engineering Mechanics: Statics]
Itcanbeshown.com
Sirajuddin, David

 General procedure



Identify 2-force members first
Connect these forces to where you choose to put the pins in your FBD (pins can be left on some of the parts themselves, or a point FBD can be drawn of any pin)

I draw circles to help me keep track of where the pins are
Fill in the remaining forces in any order you wish
Itcanbeshown.com
[Plesha, Gray, Costanzo, Engineering Mechanics: Statics]
Sirajuddin, David

Itcanbeshown.com
[Plesha, Gray, Costanzo, Engineering Mechanics: Statics]
Sirajuddin, David

2-force members!
Itcanbeshown.com
[Plesha, Gray, Costanzo, Engineering Mechanics: Statics]
Sirajuddin, David

(3) Draw the reaction forces on each two-force member
These forces “connect” wherever we choose to put the pins on these diagrams (I put circles to show this)
Itcanbeshown.com
[Plesha, Gray, Costanzo, Engineering Mechanics: Statics]
Sirajuddin, David

These forces “connect” wherever we choose to put the pins on these diagrams (I put circles to show this)
Itcanbeshown.com
[Plesha, Gray, Costanzo, Engineering Mechanics: Statics]
Sirajuddin, David

(5) “Connect” reactions from 2-force members to the pins
These forces “connect” wherever we choose to put the pins on these diagrams (I put circles to show this)
Itcanbeshown.com
F
CF
F
CD [Plesha, Gray, Costanzo, Engineering Mechanics: Statics]
Sirajuddin, David

(6) Fill in any other forces whose direction is known
These forces “connect” wherever we choose to put the pins on these diagrams (I put circles to show this)
Itcanbeshown.com
F
CF
Fill in the roller force next if you like, it is convenient
F
CD
[Plesha, Gray, Costanzo, Engineering Mechanics: Statics]
Sirajuddin, David

(6) Fill in any other forces whose direction is known
These forces “connect” wherever we choose to put the pins on these diagrams (I put circles to show this)
F
G
F
CF
Fill in the roller force next if you like, it is convenient
Itcanbeshown.com
F
CD [Plesha, Gray, Costanzo, Engineering Mechanics: Statics]
Sirajuddin, David

(7) Choose where the remaining pins are physically located
The rest of the forces, we no nothing about which way they actually point
 x,y components
Remember where your pins are!
We have more to figure out
F
G
F
CF
Itcanbeshown.com
F
CD [Plesha, Gray, Costanzo, Engineering Mechanics: Statics]
Sirajuddin, David

(7) Choose where the remaining pin A is physically located
The rest of the forces, we no nothing about which way they actually point
 x,y components
Remember where your pins are!
We have more to figure out
F
G
F
CF
Itcanbeshown.com
F
CD [Plesha, Gray, Costanzo, Engineering Mechanics: Statics]
Sirajuddin, David

A y
The rest of the forces, we no nothing about which way they actually point
 x,y components for all remaining forces
Remember where your pins are! We have more to figure out
E y
E x
A x
F
G
Itcanbeshown.com
F
CF
F
CD
B y
A y
By definition (see the figure), there is no pin at E, or B (no connecting part present)
A x
Can choose pin locations for B x
A on the dump or part ABC
(no difference in forces)
[Plesha, Gray, Costanzo, Engineering Mechanics: Statics]
Sirajuddin, David

Choice of pin position of A does not change anything
The rest of the forces, we no nothing about which way they actually point
 x,y components for all remaining forces
Remember where your pins are! We have more to figure out
E y
E x
A x
F
G
A y
Itcanbeshown.com
F
CF
F
CD
B y
A y
By definition (see the figure), there is no pin at E, or B (no connecting part present)
A x
Can choose pin locations for B x
A on the dump or part ABC
(no difference in forces)
[Plesha, Gray, Costanzo, Engineering Mechanics: Statics]
Sirajuddin, David

Choice of pin position of D does not change anything
The rest of the forces, we no nothing about which way they actually point
 x,y components for all remaining forces
Remember where your pins are! We have more to figure out
E y
E x
A x
F
G
A y
Itcanbeshown.com
F
CF
F
CD
B y
A y
By definition (see the figure), there is no pin at E, or B (no connecting part present)
A x
Can choose pin locations for B x
A on the dump or part ABC
(no difference in forces)
[Plesha, Gray, Costanzo, Engineering Mechanics: Statics]
Sirajuddin, David

Changes due to choosing a different location for pin C
A y
We could have chosen pins to be at different locations. How does this change things? Try moving the pin at
C on FBD ABC to FBD CD
New unknowns arise, and CD ≠ 2force member anymore (4-force member)
E y
E x
A x
F
G
Itcanbeshown.com
F
CF
F
CD
B y
A y
A x
B x
[Plesha, Gray, Costanzo, Engineering Mechanics: Statics]
Sirajuddin, David

Changes due to choosing a different location for pin C
We could have chosen pins to be at different locations. How does this change things? Try moving the pin at
C on FBD ABC to FBD CD
New unknowns arise, and CD ≠ 2force member anymore (4-force member)
F
CF
E y
E x
A x
F
D
F
G
A y
Itcanbeshown.com
C x
C y
F
CF
C x
F
CF
C y
B y
A y
A x
B x
[Plesha, Gray, Costanzo, Engineering Mechanics: Statics]
Sirajuddin, David

Changes due to choosing a second different location for pin C
Move pin at point C to FBD CF
New unknowns arise, and CD != 2force member anymore (4-force member)

Leaving the pin at FBD ABC is probably the best strategy
F
F
E y
F
CD
E x
A x
F
G
A y
C x
Itcanbeshown.com
C y
F
CD
C x
C y
B y
A y
A x
B x
[Plesha, Gray, Costanzo, Engineering Mechanics: Statics]
Sirajuddin, David

Can also isolate pins, draw point FBDs
If we take off pin C, then…
C y
F
CF
E y
C x
F
CF
F
CD
Pin C
F
CD
E x
A x
F
G
A y
[Plesha, Gray, Costanzo, Engineering
Mechanics: Statics]
Itcanbeshown.com
F
CD
F
CF
C x
C y
B y
A y
B x
But now we have one more
FBD to work with to find the additional C reactions
A x
Sometimes this actually is useful
Note: point FBD of pin C  no distances
 only force eqns. can be used
Sirajuddin, David

CF
A y
Note: we may use 3 equations for each FBD
E y
A x
E x
F
G
Itcanbeshown.com
F
CF
F
CD
B y
A y
A x
B x
[Plesha, Gray, Costanzo, Engineering Mechanics: Statics]
Sirajuddin, David

CF
E y
A y
Start here, find A x
, A y
,
F
G
A x
E x
F
G
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F
CF
F
CD
B y
A y
A x
B x
[Plesha, Gray, Costanzo, Engineering Mechanics: Statics]
Sirajuddin, David

CF
A y
FBD of EDG

F
CD
E y
A x
E x
F
G
Itcanbeshown.com
F
CF
F
CD
B y
A y
A x
B x
[Plesha, Gray, Costanzo, Engineering Mechanics: Statics]
Sirajuddin, David

CF
A y
FBD of EDG

F
CD
E y
A x
E x
F
G
Itcanbeshown.com
F
CF
F
CD
B y
A y
A x
B x
[Plesha, Gray, Costanzo, Engineering Mechanics: Statics]
Sirajuddin, David

CF
A y
E y
A x
E x
F
G
Itcanbeshown.com
F
CF
F
CD
B y
A y
Then, FBD of ABC has three unknowns, Find
F
CF
A x
B x
[Plesha, Gray, Costanzo, Engineering Mechanics: Statics]
Sirajuddin, David

CF
A y
E y
A x
E x
F
G
Itcanbeshown.com
F
CF
F
CD
B y
B x
A y
Then, FBD of ABC has three unknowns, Find F
CF,
 F
CF
= 27,172 lb
A x
[Plesha, Gray, Costanzo, Engineering Mechanics: Statics]
Sirajuddin, David


M. E. Plesha, Gray, Costanzo. Engineering
Mechanics: Statics. McGraw-Hill Higher
Education. 2010.