STATICS - Lecture Notes / Mehmet Zor

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The center of gravity of the elbow weighing W = 400 N is at point G. The elbow is connected to a pin that allows rotation at end B, and the 400 N horizontal force applied to end C is balanced by the AG cable. Accordingly, calculate the forces that occur in the cable and connection B. calculate the forces occurring in the cable and connection B.

3- EQUILIBRIUM

Example 3.1 (2016 – 1st MidTerm Exam)

3.8: 2D (Plane) Equilibrium Examples

In these problems we will prefer the scalar solution (equations 3.2).

Solution

We isolate the elbow from the cable and connection B.

1

1st Step (FBD drawing):

1

..>>

Figure 3.31

Figure 3.32

STATICS - Lecture Notes / Mehmet Zor

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3- EQUILIBRIUM

Scalar Equilibrium Equations are written the Plane

From these 3 equations, the 3 unknowns are found as follows:

T= 518.71 N, B_x = -222.6 N, B_y = 887.07 N

• A negative Bx (-) value indicates that it is in the opposite direction to the selected direction.

• If Bx had been selected to the left, the + sign would have appeared.

• The directions of Bx, By can be chosen arbitrarily at the beginning. If the sign ( - ) appears, it is said to be in the opposite direction of the chosen direction.

SCD -1

2nd Step:

A practical tip for those who have difficulty determining the sign and direction of the moment of a force: Think of the object as having only a pin connection (allowing rotation) at the point where the moment is taken. (Ignore other connections and other forces.) Imagine in which direction the force we are examining will rotate the object around this fixed joint. If it rotates counterclockwise, the moment of the force is positive, and in the opposite direction, it is negative (this rule applies if +x is to the right and +y is to the top). Remember that the perpendicular distance is perpendicular to the force and passes through the point where the moment is taken. For example: in the example above, try to see that the 400N force will rotate this object clockwise when there is only a pin connection at point G, and therefore its moment will be negative (-400x300).

Figure 3.33

STATICS - Lecture Notes / Mehmet Zor

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1

In the motionless system shown in the figure, find the forces occurring in the cable passing through the frictionless pulley B and in the connection C.

1st Step : Free Body Diagram (FBD-1 ) is drawn

• Since it is a C pin connection, a reaction force occurs in both directions. Since it is a single cable, the same pulling force (T) occurs in all parts. (If there was friction in the reel, T would change according to the parts.)

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Example 3.2

• In isolation no. 1, since both parts AB and BD of the cable are cut imaginarily, 2 T forces will occur.

Solution:

Figure 3.34

Figure 3.35

Figure 3.36

STATICS - Lecture Notes / Mehmet Zor

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2nd Step: Force calculations from Equilibrium Equations

3- EQUILIBRIUM

Figure 3.37

STATICS - Lecture Notes / Mehmet Zor

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A fixed crane weighing 1 kN is used to lift an object weighing 2.4 kN. The crane is supported by a pin connection at A and a roller support at B. The centre of mass of the crane is point G. Find the reactions at supports A and B.

1st Step : Let's draw the FBD by isolating the crane from the connections.

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2nd Step: Force Calculations from Equilibrium Equations:

Example 3.3

Since the connection at B is a roller support, it allows movement in the y-axis direction and cannot carry force in the y-axis direction. Therefore, only the reaction force in the x-axis direction occurs at the B support.

Solution:

Reaction moments do not occur in A and B because they allow rotation.

Figure 3.38

Figure 3.39

STATICS - Lecture Notes / Mehmet Zor

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Example 3.4

Solution:

We isolated the beam from the fixed connection (built-in) at point A and drew its SCD. At the same time, we shifted the 5kN force at point C along its own line to point B. The 15kNm load at end C is a force couple and has only a turning moment effect (M_c).

Point A is the fixed end. It does not allow rotation or translation in any direction. For this reason, reaction forces (N, V) in the x and y directions and also reaction moment (M) in the z direction occur. (See: topic 3.4.4)

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As shown in the figure, 4 singular forces and a 15 kN-m couple are applied to the beam, whose A end is concreted to a wall.

1st Step:: Free Body Diagram (FBD) of Beam.

Calculate the reactions that will occur at end A.

Figure 3.40

Figure 3.41

STATICS - Lecture Notes / Mehmet Zor

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2nd Step: Equilibrium Equations and Force Calculations

Figure 3.42

STATICS - Lecture Notes / Mehmet Zor

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Solution:

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Neglecting friction,

a-) Calculate the reaction force at the contact point C that the cylinders apply to each other,

b-) Calculate the reaction forces coming from the ground (B) and the side walls (A, D) to the cylinders.

Example 3.5

1st Step: Let's draw the FBDs for each cylinder:..>>

Cylinders a and b, with weights Wa = 100 N and Wb = 200 N respectively, are placed in a fixed case as shown in the figure.

Calculating the angle 𝜃

C

Figure 3.43

Figure 3.44

Figure 3.45

STATICS - Lecture Notes / Mehmet Zor

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2nd Step: Equilibrium Equations

from the equilbrium of cylinder b:

from the equilbrium of cylinder a:

​

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Figure 3.46

STATICS - Lecture Notes / Mehmet Zor

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Or we could solve option b from the equilibrium of the entire system. Like this:

1st. Step

FBD of the entire system

Notice that the system is isolated (i.e., cut off from contact) from the wall and the floor. Since the cylinders are not isolated from their interfaces, no force is shown at point C. (Because the forces in the non-isolated parts remain as internal forces of the system.) However, since, in the previous pages, each cylinder is also isolated from the other cylinder, the R_c force is applied to the FBD of each.

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2nd Step

Equilibrium Equations:

Figure 3.47