2.1.1 Centroids Review

Centroid – The centroid of a plane (2D) figure or of a (3D) solid is the geometric center of the object. The centroid of a part may be different than the center of gravity of a part. The center of gravity of a part depends on the material that it is made from and the geometry of the part. The centroid is solely based on the shape of the part. Centroid Symbol –

The distance to the centroid from a given point is called the x bar in the x direction and the y bar in the y direction. The symbols for the x bar and y bar values are shown below.

Centroids for Basic shapes can be found on your formula sheet or as shown below (Note – all dimensions for centroid location are given from the point labeled o.) Rectangle – Right Triangle

Semi-circle

Complex shape Centroids – If there are 2 or more of the simple shapes shown above together, they also have a centroid. We will not cover this in this class.

Unit 2.1.3 Free Body Diagrams

Free Body Diagrams are sometimes referred to as FBD. A free body diagram is a graphical model and visual representation of force and object interactions. Only forces that are exerted on the object are shown. Gravity pulls the object down as exerts a force in the form of weight on an object. Examples are shown below:

The internal forces of the truss members are not shown on the FBD. Only external forces are included. Free Body Diagrams are essential to develop equations used by engineers to describe equilibrium, motion, momentum, strength, and other properties.

What are the components of a FBD? Force - includes magnitude and direction Moment – The tendency of a force to rotate an object As we have already learned

M = Moment F = Force Applied

d(perpendicular)= the distance from the point of rotation to the force perpendicular

If you have a calculator which

is sitting on top of your math book which is sitting on your desk, draw the Free Body Diagram for your math book.

Review Question – Free Body Diagram

1) Visualize the book and the desk and calculator. You should be able to picture how they are oriented with respect to each other. If you can’t visualize the situation, you will have trouble solving the problem.

2) Sketch the book. (Note – this does not have to be a artistic representation, only a line or box representing the book

3) What forces are being exerted on the book? Answer – Gravity – Weight of the book pulling it down. The desk will be pushing up on the book. If not, the book will fall to the floor The Calculator. The weight of the calculator will be pushing down on the book. This must also be resisted by the force the desk exerts upward.

4). Graphically show the forces applied to the book Completed Free Body Diagram (FBD) Fc

Fw Fc = the force exerted by the calculator on the book. ( Note – the force exerted by the book on the calculator is not shown. It is not one of the forces exerted on the book.) Fw = Weight of the book. Gravity. Fd = Force exerted by the desk on the book. Note that the the force Fd has to equal the the sum of forces Fw + Fc. Otherwise the book would fall to the ground or the would lift off of the table. This is known as equilibrium and will be discussed in depth in future chapters.

2.1.4 Calculating Force Vectors

Force Vectors – Forces that have both magnitude and direction See The example below The Force Vector A above has a magnitude of 1200 lbs and a direction of 32˚ clockwise from the negative y axis. (Note - 32˚alone does not define direction since we do not know the axis the angle is measured from just as 1200 does not represent the magnitude without the units, lbs.)

Fd

It is of value to break vectors like Vector A above into forces in the x and y direction. Trigonometry is used to accomplish this task. Trigonometry Introduction Trigonometry is a branch of mathematics that studies relationships between side lengths and angles of triangles. Common relationships include the following equations Sin θ = Opposite Side/Hypotenuse Cos θ= Adjacent Side/Hypotenuse Tan θ= Opposite Side/Adjacent Side These equations relate two of the three sides of a given right triangle to an angle (θ). This diagram shows a right triangle (triangle with one 90˚angle) and the name associated with each side.

It should be noted that if the angle designated by the symbol θ (pronounced theta), were to be moved to the other non 90˚ angle shown in this triangle, the opposite and adjacent sides would trade places. These sides are determined by the location of the angle. The first 3 letters of the relationships mentioned above can be used to help remember the mathematical expressions. For instance, SOH can remind us that Sin θ = Opposite Side/Hypotenuse and CAH can remind us that Cos θ = Adjacent Side/Hypotenuse. Combining the 3 letters in each expressionyields the following term - SOHCAHTOA . This is a great method used to remind us of the three relationships. In addition to the three equations shown, there is a relationship between all three sides of a

triangle also. It is called the Pythagorean theorem. This theorem states: 𝐚𝟐 + 𝐛𝟐 = 𝐜𝟐 a = the magnitude of the force in either the opposite or adjacent side b = the magnitude of the force in either the opposite or adjacent side not labeled as a. c= the magnitude of the force in the side opposite the right angle. This magnitude is referred to as the Resultant Magnitude The last set of expressions used in vector analysis are what is called inverse functions They allow for the calculation of any angle given the length of two sides to a triangle. Additional Trigonometry Practice is posted in this package of information. See “Review Questions – Trigonometry Practice” Vector addition - Example How do we add two vectors together to arrive at a single force vector?

Lets look at an example:

There are two force vectors shown in the picture above. What if I wanted to replace these vectors with a single force vector?

1) Break each vector into the component x and y vectors using the trig functions from above. See below for Vector C.

See Below for Vector D

2) Add all of the component forces in the x direction together and all of the forces in the y

direction together.

Fx (combined)= 150 lbs + 346.4 lbs = 496.4 lbs to the right Fy(combined) = 259.8 lbs + (-200.0lbs) = 59.8lbs up (Note – a vector going in the negative direction is assigned a negative sign (-). Not assigning a negative sign will result in an error)

3) Draw the component and resultant vector.

4) Solve for the magnitude of the resultant vector using the Pythagorean Theorem.

Resultant Magnitude = 500.0lbs 5) Solve for the direction of the resultant vector

Direction is 7˚ Counterclockwise from the positive x-axis Do a reality check. The direction of the resultant vector should lie between your original two vectors. Does it? The angle of the resultant vector is 7˚ ccw from the positive x axis which is between the two vectors. YES.

The resultant vector magnitude should always be larger than the combined x or the combined y vector magnitude. Is it? 500lbs> 59.8lbs and 496.4lbs YES. If the answer were no for either of these questions, you made a mistake and need to check your work again.

2.1.5 MOMENTS REVIEW SUMMARY SHEET Moments:

➢ The moment of a force (or torque) is a measure of the tendency of the force to rotate the object upon which it acts.

➢ In order to calculate the moment of a force, use the equation below:

𝑴 = 𝒅 × 𝑭

➢ For this equation, it is important to remember that the distance must be

perpendicular to the applied force. The distance from the pivot or fulcrum to the force is called the lever arm or the moment arm.

Units for Moments:

Force Distance Moment

English Customary

Pound (lb) Foot (ft) lb∙ft

SI Newton (N) Meter (m) N∙m

Rotation Direction:

𝐹 = force

𝑑 = distance Pivot

➢ In order to add moments, it is important to know if the direction is clockwise (CW) or counterclockwise (CCW). By convention:

CCW is positive CW is negative

Example Problem 1

A 20-lb force is applied to a wrench with a 9.0 in. long handle. What is the moment of the force applied to the bolt?

Example 1 Calculations:

𝑀 = 𝑑 × 𝐹 = −(0.75 ft × 20 lb) = −15 lb ∙ ft

Example Problem 2

Calculate the moment of the angled force shown below.

Example 2 Calculations:

𝐹𝑦 = 𝐹 ∙ sin 𝜃 = 100 N ∙ sin 60° = 86.6 N

𝑀 = 𝑑 × 𝐹 = 0.50 m × 86.6 N = 43.3 N ∙ m

Static Equilibrium

➢ If the object is not spinning, then the sum of all moments about any point on the object is zero.

𝚺 𝑴 = 𝟎

➢ In other words:

𝑴𝟏 + 𝑴𝟐 + 𝑴𝟑 … = 𝟎

9.0 in.

20 lb

50 cm

60°

60° F = 100 N

Fy

Example Problem 3:

Example 3 Calculations: (establish the sign of each moment)

Force #

Magnitude Newtons (N)

Distance from Pivot (Meters)

SIGN CCW + CW -

1 10 8 -

2 5 5 +

3 50 2 -

4 4 2.5 +

5 F? 10 ? assume positive

Example 3 Calculations approach #1

−𝑀1 + 𝑀2 + −𝑀3 + 𝑀4 + 𝑀5 = 0

-(8*10) + (5*5) + -(50*2) + (4*2.5) + (F*10) = 0

-145 + 10*F=0

10*F = 145

F = 145 / 10

F = 14.5N*M

Example 3 Calculations approach #2

CCW=CW

𝑀2 + 𝑀4 + 𝑀5 = 𝑀1 + 𝑀3

5(5)+4(2.5)+F(10)=8(10)+50(2)

10F = 145

F = 14.5N*M

POE REACTION FORCES REVIEW SUMMARY SHEET Forces applied to an object (like pushing an pulling), there are also the forces that result from supporting that object. We often call the support forces “reactions” or “reaction forces”. The direction of the reaction forces depends on the type of support that the external object provides. When an object simply rests on a surface (applying its weight (mass times gravity) to the support), a normal force reaction is applied to the object. If the

1

2

3 4 5

0M = 0xF =0yF =

surface is inclined, an additional friction force may also result, which keeps the object from sliding down the incline.

All of the forces {external applied forces, loads, friction,..} we have discussed in PLTW Unit 2 generate reaction forces to keep objects in static equilibrium. If an object is in static equilibrium the following MUST all be true: the object is not rotating, not moving vertically - raising or falling, or moving horizontally – right or left.

Sum of all moments = 0; Sum of all y forces = 0; Sum of all x forces = 0

For beams, levers or trusses the reaction forces are at the pivot point or pin, and also at the other support, such as a roller. Two reaction forces are at point A the pin, one in the y direction (vertical) and potentially one in the x direction (horizontal). One reaction force will be at point B in the y direction as a partial counterbalance to Load F. Review the example below:

PLTW 2.1.7 TRUSSES REVIEW SUMMARY SHEET

STATICS Understandings

1. Static – still; Dynamic – changing/moving.

2. Laws of motion describe the interaction of forces acting on a body.

3. Structural member properties including centroid location, moment of inertia, and modulus of elasticity are

important considerations for structure design.

4. Static equilibrium occurs when the sum of all forces acting on a body are equal to zero.

5. Applied forces are vector quantities with a defined magnitude {length}, direction {angle}, and sense

{arrowhead}, and can be broken into x and y component vectors.

6. Forces acting at a distance from an axis, pivot point, pin or fulcrum attempt or cause an object to rotate. These

forces generate a moment. These also generate a y or x force.

7. In a statically determinate truss, translational and rotational equilibrium equations can be used to calculate

external and internal forces.

8. Free body diagrams are used to illustrate and calculate forces acting upon a given body.

A statically determinate truss will always have one solution for reaction forces and member tensions/compressions for a given load, and will always obey this relationship among J,M, and R.

If 2J is not equal to M+R, the truss has the potential to have “built-in” tension or compression, so there is not enough information to determine the tensions; it depends on the tensions members were under, for instance, when they were bolted together. Even if bolted/welded together without tension / compression, later temperature changes might cause internal forces in the truss, and an indeterminate truss does not have the freedom to expand or contract to relieve those forces. All of the forces {external applied forces, loads, friction,..} we have

discussed in PLTW Unit 2 generate reaction forces to keep objects in static equilibrium. If an object is in static equilibrium the following MUST all be true: the object is not rotating, not moving vertically - raising or falling, or moving horizontally – right or left.

Trusses tend to be designed around the triangle. Below we will walk through solving for all external forces acting on a Truss, made of 1 triangle:

Below we will walk through solving for all external forces acting on a Truss made of 2 triangles:

Another Example:

Method of Joints is the process for determining the internal forces of trusses, identifying Compression or Tension, and

their magnitude. Please review the concept, however, know that this concept was only demonstrated, not taught for

students to be able to execute this process.

Method of Joints

A method of analysis of trusses which constructs free body diagrams of each joint and determines the forces acting in that joint by considering equilibrium of the joint pin.