STATICS - Lecture Notes / Mehmet Zor
1
23 Agust 2024
FRAMES AND MACHINES
5.
(tvid – 5)
5.1 Definition: They are systems formed by connecting (assembling) various solid parts (elements) to each other. Unlike truss systems, elements can be connected to each other at more than 2 points and external forces can affect the elements not only at the connection points but also at different points. If at least one of the Conditions of Truss Systems mentioned in Article 4.1 is violated, this system becomes a frame system or simple machine and is examined within this scope.
STATICS - Lecture Notes / Mehmet Zor
2
23 Agust 2024
Figure 5.1
Figure 5.2
5- Frame and Machines
STATICS - Lecture Notes / Mehmet Zor
3
23 Agust 2024
5.2 Our aim: To calculate the forces occurring at the connection points of each element (part).
5.3 Solution Method: We have a total of 3 steps to find the forces:
1-Two Force Members (TFM) in the system are determined.
(The concept of TFM will be explained on the next page.)
2- Free Body Diagrams (FBD) of the whole system or a certain part or certain elements are drawn.
3- Unknown forces are determined with the equilibrium equations to be applied to the FBDs.
5- Frame and Machines
Figure 5.3
STATICS - Lecture Notes / Mehmet Zor
4
23 Agust 2024
When solving problems, it is extremely important to identify the Two Force Members at the beginning. Otherwise, the forces cannot be found.
What is a Two Force Member?
Answer: They are elements that are subjected to singular forces from only and only 2 points. These forces can be external forces or reaction forces formed at the connection points.
As a result, two properties of two force members:
Number of connections + number of external forces = 2
Weights and reaction moments at the connections are neglected. (These neglections are made unless otherwise stated.)
5.4 Two Force Member (TFM):
TFM
An example for TFM
5- Frame and Machines
Figure 5.4
STATICS - Lecture Notes / Mehmet Zor
5
23 Agust 2024
TFM
5.4.1 Examples for Two Force Member (TFM):
In the excavator system in the figure, EB, CB, AB, IH are two force members.
Because for each,
Number of connections = 2
Number of external forces outside the connection = 0
+
Total: = 2
Reminder: All bars in truss systems are two force members. Discuss the reason among yourselves.
5- Frame and Machines
Figure 5.5
Figure 5.6
STATICS - Lecture Notes / Mehmet Zor
6
23 Agust 2024
TFM
In the pedal mechanism shown in the figure, the BC arm is a TFM. Because;
Number of connections = 1
Number of external forces outside the connection = 1
+
Total: = 2
400N
5- Frame and Machines
Figure 5.7
..>>(only at point B )
..>>(Force F at point C )
STATICS - Lecture Notes / Mehmet Zor
7
23 Agust 2024
Question: Two single forces will be applied only to points A and B of the L-shaped element. There is no other connection point. Therefore, this is a two force member (TFM). In which of the following cases will the equilibrium of this element be achieved?
A
B
A
B
P
P
A
B
P
P
A
B
P
P
A
B
P
P
(a)
Not in Equilibrium
5.4.2 Equilibrium Condition of TFM:
A
B
P
P
in Equilibrium
Not in Equilibrium
Not in Equilibrium
in Equilibrium
5- Frame and Machines
Figure 5.8
Figure 5.9:
(b)
(c)
(d)
(e)
STATICS - Lecture Notes / Mehmet Zor
8
23 Agust 2024
Condition of Equilibrium for TFM: The acting forces must be on the line connecting the points of action, of equal magnitude and in opposite directions.
Direction of forces in TFM:
In order to provide the above condition, the direction of the forces is placed towards each other (Figure 5.9.d) or outwards (Figure 5.9.e). However, if the sign of the force turns out to be negative as a result of the calculations, it is said to be opposite the direction we chose.
Equilibrium is only achieved in options d and e.Therefore, there is only one condition for two force members to be in equilibrium:
Note the following points in the FBD examples shown starting from the next page:
5- Frame and Machines
A
B
P
P
A
B
P
P
Figure 5.9.d
Figure 5.9.e
STATICS - Lecture Notes / Mehmet Zor
9
23 Agust 2024
E
C
D
FCB
FCB
FBE
FBE
FBDs of TFMs
FBDs of other parts
H
B
C
A
B
I
E
B
FBD Example-1
5- Frame and Machines
Figure 5.10
Figure 5.11
Figure 5.12
Free Body Diagram
Two Force Members (TFM)
STATICS - Lecture Notes / Mehmet Zor
10
23 Agust 2024
In the structure in equilibrium in the figure, draw the Free Body Diagram of the entire system and each element (part) separately.
5- Frame and Machines
FBD Example-2
FBD of Each Parts
ÇKE
FBD of Entire Systems
Figure 5.13
Figure 5.14
Figure 5.15
Figure 5.16
Figure 5.17
STATICS - Lecture Notes / Mehmet Zor
11
23 Agust 2024
Step 1: Determination of TEM(1)s
Step 2: Drawing of FBD(2)s.
Step 3: Application of Equilibrium Equations
5- Frame and Machines
Now Sample Equilibrium Problems Related to Frame Systems and Machines will be solved. First, let's recall our steps: We had 3 steps to find the forces in frame systems and machines:
The following points will also help you in solving and understanding the problems:
FBDs of TFMs are useless when calculating forces.
SCDs of non-TFM elements lead us to the result.
Since the problems we examine are plane problems, scalar solutions with equations 3.2 will be preferred.
(1) TFM: Dual Force Element, (2) FBD: Free Body Diagram
Figure 5.18
Figure 5.19
STATICS - Lecture Notes / Mehmet Zor
12
23 Agust 2024
TFM
BD is a two force member (TFM). Because it is subjected to a single force from only two points. Note that the FBD force is on the BD line.
«Why did we take the direction of the FBD and other forces in this way?» You should be able to answer the question with the information you have learned so far.
Solution:
1st Step: Detection of TFMs:
3rd: Application of Equilibrium Equations
5- Frame and Machines
Accordingly, calculate the force generated on pin D.
Example 5.1:
5- Frame and Machines
A load of W = 200N is hung from end A of the frame system in the figure.
Figure 5.20
Figure 5.21
Figure 5.22
2nd Step:
Drawing FBDs
STATICS - Lecture Notes / Mehmet Zor
13
23 Agust 2024
TFM
Calculate the forces that occur in the A, B and C connections in the frame system shown in the figure.
Example 5.2:
Solution
2nd Step:
Drawing FBDs
3rd: Application of Equilibrium Equations to rod CB
5- Frame and Machines
1st Step: Detection of TFMs:
BD is a two force member (TFM). Because it is subjected to a single force from only two points. Its weight is also neglected.
Figure 5.23
Figure 5.24
Figure 5.25
STATICS - Lecture Notes / Mehmet Zor
14
23 Agust 2024
Example 5.3 In the system shown in the figure, a weight of W=80N is hung from the end of the rope passed through pulley A. Since the system is in equilibrium, calculate the forces that occur on pins B and C.
Equilibrium of the Entire System:
Equilibrium of the Pulley:
Equilibrium of the rod ABC:
Figure 5.26
Figure 5.27
Figure 5.28
Figure 5.29
5- Frame and Machines
STATICS - Lecture Notes / Mehmet Zor
15
23 Agust 2024
Example 5.4 (Midterm Exam -2016) How much clamping force comes to the object at E as a result of the hand force of P=50N applied to the pliers in the figure? Calculate. (Dimensions are in millimeters.)
Solution:
TFM: Rod AB
Lower handle
Compression Jaw
Figure 5.30
Figure 5.31
Figure 5.32
Figure 5.33
5- Frame and Machines
STATICS - Lecture Notes / Mehmet Zor
16
23 Agust 2024
Example 5.5 2010/Final Exam.
The mechanism in the figure maintains the horizontal position of the load while lifting it up. The pin located on the outer perimeter of the 40 cm radius pulley can slide frictionlessly in the channel on the ABC arm. The length of the ABC and DE arms is 80 cm each and the weight of the lifted load is 1200 N. The mechanism is lifted upwards by pulling the rope wrapped around the pulley. Determine the force P that must be applied to the rope and all the forces acting on the ABC arm in order to keep the load on the L-shaped CEF platform 80 cm above the ground as shown in the figure.
5- Frame and Machines
AC=DE=80cm
Figure 5.34
40cm
P
F
STATICS - Lecture Notes / Mehmet Zor
17
23 Agust 2024
F
80cm
40cm
69.28cm
SOLUTION
Platform + Load
Rod ED is TFM (two force member)
FBD of Rod ED
1st Step: Detection of TFMs
2nd Step: Drawing FBDs
3rd Step :
Application of Equilibrium Equations for the Platform+Load
5- Frame and Machines
It is calculated from the geometry of the system.
Figure 5.35
Figure 5.36
STATICS - Lecture Notes / Mehmet Zor
18
23 Agust 2024
40cm
69.28cm
40cm
FBD of Rod ABC
FBD of the Pulley
5- Frame and Machines
Application of Equilbirium Equations to the Rod ABC
Figure 5.37
Figure 5.38
STATICS - Lecture Notes / Mehmet Zor
19
23 Agust 2024
In the system shown, the excavator lifts a load of 10 kN. Point H is the center of gravity of the load and the system is balanced in the position shown. Accordingly; calculate the forces in the hydraulic cylinders BC and JK and the reaction force components in the pin A.
Example 5.6 (Final Exam)
1st Step : TFMs are elements BC, JK and GI
Solution:
2nd Step : Let's separate the escavator from the A and B connections and draw its FBD.
3rd Step: Equilbrium Equations
Figure 5.39
Figure 5.40
5- Frame and Machines
STATICS - Lecture Notes / Mehmet Zor
20
23 Agust 2024
For bucket
For element EKI :
If we examine for other elements :
Figure 5.41
Figure 5.42
Figure 5.43
Figure 5.44
5- Frame and Machines
bucket
STATICS - Lecture Notes / Mehmet Zor
21
23 Agust 2024
Questions 5.1 (*): If θ = 30o in the system shown in the figure, calculate the force occurring on pin C.
5- Frame and Machines
Figure 5.44
STATICS - Lecture Notes / Mehmet Zor
22
23 Agust 2024
Calculate the forces that occur in the D and B pins in the frame system shown in the figure. Answer: B=888N, D=942.8N
5- Frame and Machines
Question 5.2 (*) :
Figure 5.45
STATICS - Lecture Notes / Mehmet Zor
23
23 Agust 2024
How much clamping force occurs on object A as a result of the 50N hand force applied to the pliers in the figure? Calculate.
(Neglect the dimensions of object A)
Question 5.3 (*)
θ
Answer: 483.7N
Figure 5.46
5- Frame and Machines
STATICS - Lecture Notes / Mehmet Zor
24
23 Agust 2024
As a result of the 240N hand force applied to the pliers in the figure, what is the tightening force on the nut tightened by the pliers? Calculate by neglecting friction.
(Answer: 1680N)
Question 5.4
5- Frame and Machines
Figure 5.47
STATICS - Lecture Notes / Mehmet Zor
25
23 Agust 2024
P
Question 5.5 (*)
5- Frame and Machines
Figure 5.48
STATICS - Lecture Notes / Mehmet Zor
26
23 Agust 2024
The trailer connection mechanism in the figure carries a load of 9000 N. In this case, calculate the force occuring in the CF hydraulic cylinder. Neglect friction. The pins allow rotation but not translation. Pin G is connected to a wheel. (Answer= 109.4kN )
5- Frame and Machines
Question 5.6 (*)
Figure 5.49
STATICS - Lecture Notes / Mehmet Zor
27
23 Agust 2024
Question 5.7 (*) (asked in a Final Exam): Question 5.7 (*) (Final Question): The lifting mechanism in the figure lifts the 15 kN piece A upwards. Thanks to the thrust force coming from the CD hydraulic cylinder, the EDF piece and the FH arm transmit the power to the B platform. The 20 kN B platform, which is in contact with the A piece, can move on the fixed vertical column by means of its wheels on the left. When it reaches the position in the figure, the system is stopped and remains in balance. Calculate the forces in the CD hydraulic cylinder, the FH arm and the E pin in this position. Neglect the weights of all other parts and friction. The E pin only allows rotation. Dimensions are in meters.
5- Frame and Machines
Figure 5.50