# Lesson: Forces in a Roof

Authors: Peter Anderson, Wade Goodridge, Sarah Lopez, Natalie Shaheen

The goal is to introduce students to basic principles and calculations of statics, from the field of structural engineering. Specifically truss analysis using the method of joints. The lesson also provides students with an opportunity to investigate a specific branch of engineering and its applications through a creative design experience.

## Class Size

• Written for 3-5 students. Taught workshop style as the instructor works with students individually to help them develop the diagrams and equations needed to analyze their structure.

## Lesson Structure

150 minutes

• Tension and Compression - PASCO Model: 15 minutes
• Truss Analysis - Method of Joints: 10 minutes
• Workbook - Point A: 30 minutes
• Workbook - Point B: 20 minutes
• Workbook - Point C: 15 minutes

## Objectives

Students will be able to:

• Create free body diagrams (FBD) of joints on a roof or truss system.
• Develop equilibrium equations to solve for forces in a rafter and brace beam system from the FBD.
• Determine if a truss member is in compression or tension and what magnitude of force is placing it in those conditions.
• Develop an understanding of how forces flow through a structure from roof to foundation.

## Prerequisite Knowledge

• Students should have developed tactile drawing techniques (Engineering Drafting and Multiview Drawing).
• Students should understand what a vector is (i.e., how it designates the magnitude and direction of a force, and how to represent it) (Fun with Forces).
• Students should conceptually understand how to add vectors representing forces (Fun with Forces).
• Students should know the dimensions of their structure’s roof.

## Accessibility

• The tactile PASCO and patented truss components with adapted tension and compression indicating piece are provided for students to observe compression and tension nonvisually.
• Accessible calculators
• Braillers or 20/20 markers for recording calculations

## Materials

Note: Refer to Accessible Lab Equipment & Instructional Materials for additional information regarding specialized tools/materials.

## Preparation

1. Set up workstations on trays with the appropriate drawing and calculating tools.
2. Assemble PASCO models in advance.
• Attach two medium rods to a joint at 90 degrees from each other to form the peak of the gable end.
• Attach both ends of the long rod to these medium rods as a crossbeam to form a triangle shape.
• Attach the last two medium rods to the joint pieces at each end of the long rod, pointing down as columns (i.e., 90 degrees from the long rod).
• The final model should be roughly shaped like an “A,” but with vertical legs.

Note: The image shows short rods used for columns; however, medium rods are preferable, as they will make the forces slightly more apparent.



## Procedure

### Tension and Compression - PASCO Model: 15 minutes

1. Introduce the lesson.
• Tell. “Houses experience forces from their own weight, wind, people, snow, earthquakes, etc. Successful ones don’t move in response. The trick is that the force going in is resisted by the material and the structure.”
2. Introduce the models.
• Tell. “We have a model truss here to explore. You’ll recognize part of its shape because it is the same as the end of the pavilion model we just worked with. The truss is composed of three members. Two make the angle of the roof. The other is at the bottom connecting them in a horizontal orientation. These three pieces are referred to as truss members. Placing a force on one part of it can cause other truss members of the model to squeeze in (compression) or stretch out (tension).”
3. Use truss model.
• Do. Pass out the prebuilt truss models. Direct students to experiment with placing forces of different direction on the truss model and observing where the structure changes. Note: Do this on a smooth table surface so that the legs can slip. Direct students to pay particular attention to what happens to the flexible part of the crossbeam.
4. Observations.
• Ask. “Is the horizontal member in compression or tension?”
5. Student observations.
• Students do. Have student groups summarize what they tried and what happened.
6. Conclusion.
• Tell. “As forces are applied to real-world structures, similar things happen. Houses don’t usually deflect (bend) when snow falls on them, but beams definitely bear the weight and can deflect under load. Our models emphasize these changes. In a real house these changes are smaller in size, but can still cause real problems like causing doors or windows to get stuck in place or causing cracks to form.”

### Truss Analysis - Method of Joints: 10 minutes

1. Introduction.
• Tell. “One of the first courses that mechanical/structural/civil engineers take after math and physics is statics. In statics, engineers analyze structures that are at rest (or moving with a constant velocity). This means we are considering situations where acceleration must be equal to zero.”
2. Newton’s second law.
• Tell. “Newton’s second law of motion states that the force acting on an object is equal to the object’s mass times it’s acceleration. This is described with the equation F = ma. For static situations, where the total acceleration is equal to 0, this equation tells us that the total force must also equal 0.”
3. Static limitations.
• Tell. “Because of this, all the forces acting on a static object must add up to equal zero. This can be thought of like a tug of war where no team is winning, they are in a stalemate. The force pulling from one side is equal to the force pulling from the other, so nothing is moving.”
4. Consequences of imbalances.
• Tell. “Your structure could fail if the posts separate outward, or if the crossbeam pulls apart, or if the posts crumple under the weight. This would happen if some of the forces acting on the structure are not balanced out, and, therefore, one side is ‘winning the tug of war.’ We need to analyze the rafter system to be sure all forces acting in it are truly balanced in all directions.”
5. Method of joints.
• Tell. “To balance out all of our forces, we are going to use a truss analysis method called ‘Method of Joints.’ With this method, we will look at each joint where beams come together, one at a time, and make sure all the forces at each joint add up to zero.”
6. Components.
• Tell. “Remember from a previous lesson that all forces can be split up into components in the X and Y directions. We can use this to simplify the addition of forces at each joint. Instead of trying to account for diagonal forces, we will break everything down into X and Y components, and then make sure all the X components add to 0, and all the Y components add to 0.”
7. Assumptions.
• Tell. “In order to use this method, we will assume that the joints themselves do not exert any forces on the beams. Although this isn’t technically true, this assumption is close enough for most analyses, even in the real world, and allows us to avoid a lot of very complex math and physics.”
8. The Truss.
• Do. Pass out The Truss handout.
• Tell. “To begin our analysis of a truss we will focus on the geometry of the two rafters and the beam on the gable end of your structure. This is a triangle that would look like the figure on this handout. The connection points of the rafters and crossbeam are labeled as B and C and we’ll look at each of these joints in turn. Label A designates the connection of two rafters together at the top.”
9. Overview of work.
• Do. Pass out Truss Analysis Overview
• Tell. “This document gives a step-by-step process for analyzing the truss. You can follow along in it as we go through the steps together. We will begin our analysis looking at connection A.”
10. Prepare materials.
• Do. Make sure each student has materials ready for drawing, calculating, and recording. This includes a calculator, blank paper, Sensational Blackboard and pen, brailler, and/or 20/20 markers, etc.

### Analysis - Point A: 30 minutes

Note: If students have chosen a flat roof for their design, the math will drastically simplify, as there will be no horizontal components to the forces, and there is no point A. These students should stay at the station and attend to Point A, but do no calculations.

1. Conceptual model of forces - identify the forces at Point A.
• Do. Have students feel the PASCO models. Have them apply a push and feel the result as directed below.
• “What forces did you make when you pushed down on the peak of the PASCO truss or the point of rafter connection at A?” The push down.
• “What forces did you feel back on your hand or finger when you did that”? The point pushed back into my hand.
• “Where did the force come from?” The diagonal rafter parts resisting my push down.
2. Draw the free body diagram together.
• Do. Have students draw a point, or a small circle in the middle of the paper to represent point A. Have students label the point.
• Teach. Facilitate this section to try to get students to suggest these forces: At point A we have three forces. 1) The load from our hand or finger pushing down, 2) a rafter pushing diagonally up and left, and 3) the other rafter pushing diagonally up and right.
• Tell.
• “Draw a force arrow (i.e. a vector) going down into A representing your applied load or push. Then draw two vectors on diagonals pointing into A and angled like the rafters are on the models. One is pointing up and to the left into A and the other is pointing up and to the right into A. These vectors represent the resistance in the rafters to the push you applied. Make sure all your arrows are pointing into point A, since all of these forces are pushing into the joint.”
• “This drawing shows all the basic forces on point A, but to set up the equations we need, we will have to redraw this with more detail.”
3. XY Separation.
• Tell. “Engineers check that their forces are balanced up and down and side to side separately. We’ll check the vertical direction first, then the horizontal. The next figure in your workbook shows a quick reminder of how a diagonal force can be split into vertical and horizontal components for this analysis.”
4. Elaborate on the free body diagram together.
• Tell. “Start a new free body diagram by marking point A in the middle of a new page. We already know the total force coming down on the roof (Total Roof Force, from Columns of Calculation), but that force is spread out across the whole span of the rafters, and not all concentrated at point A. To determine the force at point A, we need to break it down more.”
5. Splitting the force.
• Tell. “In our last lesson The Roof System, we calculated the Total Rafter Force, which represents the force pushing down on each side of the roof. Each rafter will split this force and propagate it evenly to the joints at each end of the rafter. Because of this, the force on the left rafter will be split between points C and A, and the force on the right rafter will be split between points A and B.”
6. Illustrating the force split.
• Tell. “Knowing this, we can say that the vertical force pushing down on point A, due to the load, is ½ Total Rafter Force from the left rafter, and ½ Total Rafter Force from the right rafter. To represent this, draw 2 arrows near each other going down into A and label them both ½ F_load.”
7. X and Y Components.
• Do. Pass out Vector Components handout
• Tell. “Now, let’s look at the diagonal forces from the rafters pushing up into point A. Each of these rafter forces can be split into a horizontal and vertical component. Each rafter force has a vertical part called F_rafter-y and a horizontal force called F_rafter-x. Let’s start with the forces from the left rafter. Draw an arrow going up into A labeled F_rafter-y, and an arrow coming in from the left labeled F_rafter-x.”
8. Draw forces from the right rafter.
• Tell. “To include the forces from the right rafter, draw another arrow going up into A (next to the first one), labeled F_rafter-y, and an arrow coming in A from the right labeled F_rafter-x.”
• Note: Because we assume a symmetric rafter structure, we can assume that the two F_rafter forces have the same magnitude. If a student has an asymmetric structure, the forces should be labeled differently (i.e. F_rafterC-x and F_rafterB-x, etc.)
9. Limit the complexity.
• Tell. “So we’ve gone from 3 forces to 6, which may look more complex, but now that all the forces are horizontal or vertical we can more easily add them together to make them balance.”
10. Equilibrium review.
• Teach. Remember from a previous lesson that an equilibrium equation is when we add up all the forces on an object, and set the sum equal to zero. Since we have split all the forces into vertical and horizontal components, we can make one equilibrium equation to add up all the vertical vectors, and another one to add up all the horizontal vectors.
11. Equilibrium equation of point A in the Y direction.
• Tell. “We are going to make our first equilibrium equation based on the vertical forces in this picture. We know that the sum of these forces must equal 0 for this structure to remain stationary. So 0 = 2 * F_rafter-y - 2 * ½ * F_load.”
12. Explain the 1st equation.
• Tell. “We have just written in our picture what we drew before. The F_load part is negative because it is pushing downward. This equation says that the load down balances with the upward push from the rafters.”
13. Calculate F_rafter-y.
• Students do. Direct students to insert their numbers into the calculation at this point and solve for F_rafter-y. (F_Load is the same thing as Total Rafter Force from The Roof System F_rafter-y should be the same as ½ F_load).
14. Convert F_rafter-y to F_rafter.
• Tell. “Now we know how much upward force the rafters need to provide, but the rafters point diagonally. We need to convert our upward force into this diagonal force to know how much force the rafter exerts in compression or tension along its length. We’ll be using sines and cosines again.”
15. Record the angle of your roof truss.
• Students do. Direct students to measure the angle of their roof using a tactile protractor or calculate it from their structure dimensions with the following equation (roof_angle = arctan(roof_height/tributary_width). Record this value.
• Note: If they are measuring, they should measure the angle between the beam and rafter at point C, not the angle at point A.
16. Remind students of the trigonometry.
• Tell. “When we first learned about vectors, we knew the diagonal force and figured out the up/down and left/right projections. This time we know the up/down and need to figure out the diagonal. We use the same math, but solve for it differently.”
17. Calculate F_Rafter.
• Tell. “Use the equation given in your workbook to solve for F_rafter.” Note: Students' equations should end up as F_rafter = F_rafter-y/sin(roof angle).
18. X component resolution.
• Tell. “Now it is time to calculate our horizontal forces. Let’s start by figuring out what they are.”
• Do. Refer students back to their diagram.
19. Equilibrium equation of Point A in the X direction.
• Tell. “So we have two forces of the same size going in opposite directions. 0 = F_rafter-x - F_rafter-x. Just like we assigned downward forces as negative, we will assign left-pointing forces as negative, and right-pointing forces as positive.”
20. Elaborate on equation.
• Tell. “This equation is well and good, it says that the two horizontal forces cancel out in this equation. It ought to make sense that neither side of the building wins the tug of war. But we don’t know what F_rafter-x is, only that it should be the same on both sides.”
21. Calculation of F_rafter-x.
• Tell. “However, we can use trigonometry to calculate F_rafter-x since we already know F_rafter and the angle.”
22. Calculations.
• Teach. Lead students through these steps again.  They require finding numbers from earlier calculations.
• Setup the equation. F_rafter-x = F_rafter * cos(Roof_angle)
• Calculate cos(Roof_angle). Direct students to type in their angle, then press cos. Stop students at this point to collect their numbers. Cosine should be a number between 0 and 1. Note: Reality check, calculators which may function differently. Be aware if calculators are in degree or radian mode, and whether they need to type “angle, cos” or “cos(, angle, )”.
• Calculate F_rafter-x. Multiply the F_rafter by the cosine number. This should generate a number smaller than F_rafter but in the same order of magnitude. Write this number down on paper. Label this F_rafter-x.
23. Why we would do this.
• Tell. “These are the numbers engineers calculate to figure out what size of lumber to use. Your value of F_rafter is the force that the rafter must be able to resist without crumpling if it’s in compression or pulling apart if it’s in tension.”
24. Review what has been accomplished.
• Tell. “So now we know about all the forces at point A.  Starting with the total load we figured out how much the truss members need to hold, then used those numbers to figure out how strong the rafters needed to be at that angle.  The calculations get easier after this.  We will need the roof angle and F_rafter again for the next two points, so keep or rewrite those numbers somewhere convenient.”

### Analysis - Point B: 20 minutes

Note: For flat roofed structures, point B had a downward force equal to the total rafter load and a vertical force from the column equal to it. Point C has the same forces at play. Adapt the math for these students as you go along.

1. Conceptual model of forces.
• Do.
• Have students find point B on their Truss handouts.
• Hand out the PASCO models.  Direct Students to think about point B which is in the lower right of the triangle.
• Ask. “When you pushed down what forces did point B feel? What forces did the rest of the building make as a response?” A push out from the rafter, a force from the crossbeam, a push up from the post.
2. Draw the free body diagram.
• Teach.
• On a new sheet of paper, label a place in the middle of the page as Point B.
• Draw a force going straight down into B. Label this force Load. This is the other half of the load that is supported by the right side of the roof. (We took care of the first half when we analyzed Point A.)
• Draw a force going diagonally down and to the right into B. Label this force Rafter.
• Draw a force going into B from the left. Label this force Beam. This is the force of the beam pushing on point B.
• Draw a force going up into B. Label this force Column.
3. XY Separation.
• Tell. “We will calculate in the vertical direction first, then in the horizontal direction.”
4. Elaborate on the free body diagram - vertical.
• Tell. “To help us set up our calculations, we will draw a new diagram that shows all the forces as either vertical or horizontal. To do this, we will have to split any diagonal forces into a vertical component and a horizontal component.”
• Ask. “What forces go up and down in this diagram?” The column goes up, the load goes down, the rafter has a part that goes down.
• Teach.
• Get a new sheet of paper, and start a new diagram by drawing point B in the middle of the page.
• Draw an arrow going up into B.  Label it F_column.
• Draw an arrow going down into B.  Label it ½ F_load.
• Draw a second arrow going down into B right next to the last one. Label it F_rafter-y.
5. Elaborate on the Free Body Diagram - horizontal.
• Ask. “What forces go left and right in this diagram?” The rafter and the beam.
• Teach.
• Draw an arrow from left to right going into B. Label it F_rafter-x.
• Draw another arrow from left to right going into B. Label it F_beam.
6. Equilibrium equation of Point B in the Y direction.
• Tell. “We’re going to make an equation now based on the vertical parts of our picture like before. 0 = F_column - ½ F_load - F_rafter-y.”
7. Explain the equation.
• Tell. “This equation says that the column pushes up enough to balance out part of the load, as well as the force of the rafter pushing down. The forces of the load and rafter in the y direction are negative because they are going down.”
• Note: If a student realizes that they have calculated F_rafter-y already in the 1st problem, skip the next section.
8. Convert F_rafter to F_rafter-y.
• Tell. “So we know the diagonal force and need to change it into our y force.”
• Teach.
• Setup the equation. F_rafter-y = F_rafter * sin(angle of roof).
• Calculate sin(angle of roof).  Direct students to type in their angle, then press sin. Stop students at this point to collect their numbers. Sine should be a number between 0 and 1.
• Calculate the F_rafter-y. Multiply F_rafter by the sin(angle of roof). This should generate a number smaller than F_rafter but in the same order of magnitude. Write this number down on paper.
9. Calculate F_column.
• Tell. “So if we know how much the rafter pushes down, and how much of the load is at Point B, we can determine how much force F_column needs to counteract.”
• Teach. Solve for F_column. Should be equal to F_rafter-y + ½ F_load. Direct students to label and write this result.
10. Equilibrium equation of Point B in the X direction.
• Tell. “Now, we’re going to make an equation based on the horizontal parts of our picture like before. 0 = F_rafter-x + F_beam.”
11. Explain the equation.
• Tell. “This equation says that the force from the rafter must be countered by the crossbeam.”
12. Calculate F_beam.
• Teach. Point out that students have calculated F_rafter-x previously. Help students find this number. The value of F_beam should be the same as F_rafter-x, but negative.
• Tell. “This number is negative because we initially described F_beam as pushing on point B to the right, but in reality, it is actually pulling on point B to the left.
13. Review what has been accomplished.
• Tell. “So now we know all the forces at point B. Starting with the force coming down from the rafter we figured out how much force the column needs to be able to resist and how much force our beam needs to be able to pull inward.”

### Analysis - Point C: 15 minutes

1. Symmetry.
• Teach. If students notice that all the forces on point C are the same as point B based on symmetry, affirm their observation, and have them write down, or at least tell, the final values (F_rafter, F_column, F_beam). If you are short on time, you may also point this out, and move to the final answer quickly. Skip to the conclusion of the lesson. If students do not jump to this conclusion, and you have time, this exercise is a good opportunity to get more practice with the ideas of free body diagrams and equilibrium equations.
2. Conceptual model of forces.
• Do. Have students find point C on both their Truss handout and the PASCO model. Have students think about the forces acting on point C when they push down on the model.
• Ask. “When you pushed down what forces did point C feel? What forces did the rest of the building make as a response?” A push down/out from the rafter, a pull from the crossbeam, a push up from the column.
3. Draw the Free Body Diagram.
• Teach.
• Start a free body diagram for point C.
• Draw a force going straight down into C. Label this force Load. This is the other half of the load that is supported by the left side of the roof.
• Draw a force going down and to the left into C. Label this force Rafter.
• Draw a force going to the left into C. Label this force Beam.
• Draw a force going up into C. Label this force Column.
4. XY Separation.
• Tell. “We will calculate in the vertical direction first, then in the horizontal direction.”
5. Elaborate on the Free Body Diagram.
• Ask. “What forces go up and down in this diagram? “ The column goes up, the load goes down, and the rafter has a part that goes down.  These forces must be balanced.  “What forces go side to side in this diagram?  F_rafter points left, F_beam points left.
• Teach.
• On a new page start another free body diagram for point C.
• Draw an arrow going up into C  Label it F_column.
• Draw an arrow going down into C. Label it ½ F_load.
• Draw another arrow going down into C. Label it F_rafter-y.
• Draw an arrow from right to left going into C. Label it F_rafter-x.
• Draw another arrow from right to left going into C. Label it F_beam.
6. Equilibrium Equation of Point C in the Y Direction.
• Tell. “We’re going to make an equation now based on our picture like before. 0 = F_column - ½ F_load - F_rafter-y.”
7. Explain the equation.
• Tell. “This equation says that the column is pushing up with the same force as the rafter and the load are pushing down. The force of the rafter in the y direction and the load are negative because they are going down.”
8. Convert F_rafter to F_rafter-y.
• Tell. “So we know the diagonal force and need to change it into our y force.” Note: If students realize that they have already calculated F_rafter-y skip these steps.
• Teach.
• Setup the equation. F_rafter-y = F_rafter * sin(angle of roof).
• Calculate sin(angle of roof). Direct students to type in their angle, then press sin. Stop students at this point to collect their numbers. Sine should be a number between 0 and 1.
• Calculate the F_rafter-y. Multiply F_rafter by the sin(angle of roof). This should generate a number smaller than F_rafter but in the same order of magnitude. Write this number down on paper.
9. Calculate F_column.
• Tell. “So if we know that the rafter and the load push down and the column must push up with an equal force, what is the F_column?” Sum of ½ F_load and F_rafter-y. Direct students to label and write this result.
10. Equilibrium Equation of Point C in the X Direction.
• Tell. “We’re going to make an equation for the horizontal components based on our picture. 0 = - F_beam - F_rafter-x. Notice that both terms in this equation are negative because they point left.”
11. Explain the equation.
• Tell. “This equation says that the spreading force from the rafter is exactly countered by the crossbeam.”
12. Calculate F_beam.
• Teach. Point out that students have calculated F_rafter-x previously in point A. Help students find this number. The final result should be the same as F_rafter-x and negative.
• Tell. “Like when we looked at point B, the number for F_beam is negative because we initially described F_beam as pushing on point C to the left, but in reality, it is actually pulling on point C to the right.
13. Review what has been accomplished.
• Tell. “So now we know all the forces at point C. Starting with the force coming down from the rafter we figured out how much force the column needs to be able to resist and how much force our beam needs to be able to pull inward.”
14. Conclusion.
• Tell. “So if you imagine the forces moving through this structure they go down through each member and are resisted by the columns and ultimately the ground or by the crossbeam. The reason roofs have crossbeams is so that the crossbeam can resist the spreading force of the diagonal rafters.”

## Standards Alignment

NGSS Standards Alignment:

• SEP 5 - Using mathematics and computational thinking
• CCC 7 - Stability and change
• HS-ETS1-3

CCSS Standards Alignment:

• CC.9-10.R.ST.3, CC.9-10.R.ST.4, CC.9-10.R.ST.5, CC.9-10.R.ST.7, CC.11-12.R.ST.3, CC.11-12.R.ST.4, CC.11-12.R.ST.5, CC.11-12.R.ST.7
• CC.9-12.N.VM.1, CC.9-12.N.VM.2, CC.9-12.N.VM.3, CC.9-12.N.VM.4, CC.9-12.N.VM.5, CC.9-12.A.CED.1, CC.9-12.A.CED.2, CC.9-12.G.SRT.6, CC.9-12.G.SRT.7, CC.9-12.G.SRT.7, CC.9-12.G.GMD.4, CC.9-12.G.MG.1