Truss building project: of straws to a desired length and predict the maximum amount of weight of weight it will hold (for a Objective: Build a truss made of straws minimum of 20 of 20 seconds) when the weight is placed in the middle of the of the truss, and which member(s) will fail first when max weight is exceeded. The basic design of the of the truss is seen in the following figure:
The length is 4 times the height. This will require the following steps 1. Cutting, measuring, and testing the compression strength of the of the straws. 2. Writing the Buckling Formula as a function of the of the length of the of the straws being used 3. Decide on the span or total length of the of the truss based on the length of each of each member of the of the truss. 4. Do the calculations necessary to predict how much weight the truss will hold based on the length of the of the truss and which member will fail when maximum weight capacity is exceeded. Note: The nodes are to be constructed using a hot glue gun so they will not fail before the members. 5. Build the trusses being careful that the hot glue does not inadvertently add strength to a given member due to too much overlap etc. 6. Test the truss to see how much weight it will actually hold and which of its of its members fail when maximum weight is exceeded.
Materials needed: hot glue gun, straw purchased from a local grocery store, scissors, string, weights and hangers or a
bucket with some sand and a scale, compression/tension testers, truss tester. Photos of the of the compression tester and the truss tester are found below. More detailed descriptions of each of each tester are included in the body of this of this lesson.
LESSON:
Step 1. Intro: We are all going to be civil and material science engineers. Define: Civil Engineer/ Material Science engineer Discuss: Today we are going to build trusses from drinking straws. Ask: if anyone knows what a truss is. Define: Truss by showing pictures found on Google images. Show: the design of truss we are going to use. Discuss: If you are an engineer designing a truss you are going to need to know the design parameters. The two parameters we are going to focus on in this project are the length and the amount of weight it can support. The length a truss can span and the amount of weight it can support not only depends on the design of the truss but also the materials used to build the truss. For this project drinking straws will be used, so the first thing that needs to be done is to figure out how strong a drinking straw is. Note: for this project it is assumed the tensile strength of the straw is greater than the compression strength so only the compression strength will be tested.
Compression Test:
Demo: Adjust the bolt at the back of the tester‘s “arm” until it is balanced and sits horizontally on its own. Cut a piece of straw to the proper length for use in the compression tester. At the proper length the “arm” of the tester will be horizontal. Next add weight at the end of the arm slowly. Keep an eye on the straw. Make a note of the amount of weight on the arm when the straw begins to buckle or bend under the pressure. Wait 20sec. then add more weight and let it sit for 20sec more. Continue this process until the straw “fails” and the arm of the tester falls or is no longer supported by the straw, and record the weight on the arm.
Note: The design of this tester is such that the weight (force) applied on the straw is twice that of the weight (force) on
the end of the straw because the weight is twice the distance from the fulcrum.
Step 2. Intro: the Buckling Formula (found on Wikipedia.com)
where
F = maximum or critical force (vertical load on column), E = modulus of elasticity, I = area moment of inertia, L = unsupported length of column, K = column effective length factor, whose value depends on the conditions of end support of the column, as follows. For both ends pinned (hinged, free to rotate), K = 1.0. For both ends fixed, K = 0.50. For one end fixed and the other end pinned, K = 0.699.... For one end fixed and the other end free to move laterally, K = 2.0. Explain: Our Goal is to write this equation as F(L) or Force(weight as a function of the length L of the straw where E, I and K are constants for the material (straw ) being used. Note: If two or more types of straws are used all of these constants would need to be re-calculated.
F can be considered weight for this project although there is no problem converting the weight to a force. This is what is determined by the compression tester. Don’t forget to double the amount of weight as represented by
F due to the design of the tester. I I can be calculated using the following formula found on wikipedia.com. A straw has a hollow cylindrical cross section
Hollow Cylindrical Cross Section:
•
•
•
•
DO = outside diameter D I = inside diameter r O = outside radius r I = inside radius
K = 0.50 for this project because the tester used fixes both ends of the straw. L is the length of the straw used in the tester. Data collection and calculation of E and I : Next the students will get into group of three or four at the most. They will cut test and measure (using a micrometer)at least two straws in each group. Students will record the following for each straw: Length, inside diameter, outside diameter, and the weight at which it buckled and failed (again don’t forget to double the weight on the end of the straw). Collect the data from each group and write it all on the board. Have the students calculate the average for each of the measurements collected above. Have the students use these averages to calculate the constants E,I found in the formulas above. The students should all now have the Buckling Formula written as a function of the length of the straw.
Step 3. What happens when a force is applied to the truss: Discuss: Forces Draw a picture of a house with truss on top.
Ask: how much weight is each wall holding?? (assuming the force is in the middle of the truss.)
Discuss: physics, for each action there is an equal and opposite reaction so the walls are actually pushing up on the truss with a force of 100 lbs each. Ask: how are these forces transferred through each of the members of the truss? Discuss the possibilities. Define: tension (pulling apart) and compression (smashing or pushing ends together) Ask: When you build a house you want it to be “safe” structurally. What does safe mean?? Ans: it means the house won’t move, the trusses won’t move or buckle! Define: Equilibrium‐ (do an arm wrestle if necessary) A truss is considered to be safe when is in a state of equilibrium, meaning that the sum of the opposing forces is 0. In the picture above there is a force of 200lbs pushing down on the truss and the walls are pushing back with a force of 100lbs each creating a state of equilibrium. Discuss: So the goal of truss design is to make it is strong enough to maintain a state of equilibrium, or in other words ensure that none of its members moves or buckles under the pressure! Question: how much force is acting on each member of the truss? Note: remember as we go through these calculations the sum of the forces in the y direction need to be zero as well as the sum of the forces in the x direction.
Explain: The forces acting on the members of the truss can be represented using vectors. In the beginning it is assumed that the forces at each node are going away form the node, but calculations may prove otherwise. The forces acting at each node will be calculated by setting up a system of equations that are derived using the idea of equilibrium or
.
As the students will need to be able to set up systems of equations and solve for the forces acting at each node and thus the forces applied to each member of the truss demontrate the following example.
Explain picture: lets start with the easy part Fy = 100 + Asin45 = 0. So member A must be pushing back down on “the wall” to maintain a state of equilibrium. But there is a problem; A is angled at 45 degrees which creates forces in the x direction. So there is a force on member B which must clearly be in the x direction and again because of the angle of A there is a force in the x direction there as well so B + Acos45 = 0. Again the sum of the forces must equal zero or the truss will move which we don’t want. Demo: solve for A then B. Notice A =‐141.4 because it applies a force in the opposite direction so it is in compression rather than tension as assumed in the beginning. Show:
Symmetry can be used to figure the forces on the other side of the truss.
Note: The extent of the detail used in the example above depends on the students. I tried not to calculate all the forces
for each node for my students. I demonstrated nodes 1 and 2 in full and then helped them set up the system of equations needed to solve for the forces at node 3. I then had my students do the calculations for nodes 4 and 5 on there own and I checked them.
Step 4. Have each group decide on the length of their trusses. Make sure each group chooses a different length so that results are not duplicated. Note: Remember the length of the truss is determined by the height because the length of the truss is 4 times the height. They now have all the information necessary to predict how much weight their truss will hold using the Buckling formula and the force calculations shown above. Have each group complete their calculations and predictions.
Step 5.
Check each group’s calculations and have them build their trusses.
Step 6. Test the trusses to see how much weight they hold. Note: there could be a contest to see which group’s predictions were the most accurate as a percentage of weight.
Reflection and possible changes to this project:
‐Instead of using a truss tester, have the students build two trusses and attach them together using cross‐members and just use to sets of weights, one for each truss.
‐Increase the difficulty level by moving the weight from the middle of the truss to eliminate some of the symmetry involved in this project. In the example, load P is at the center so by 'inspection' we can determine that each support carries half of the load (P/2). If the load is at any other location, then the reactions are proportional to the distance. In the example, we said that the beam was 10‐ft long and the force was at 5 ft from the edge, 5/10 = 1/2 so the reaction is 200*(1/2) = 100 lbs. If the load was applied 2‐ft from the left edge, then the left reaction would be (8/10)*200 = 160 lbs and the reaction on the right would be (2/10)*200 = 40 lbs (it should be apparent that the side closer to the load carries a higher force). Alternatively we can sum moments (which was actually your question), calling CCW positive we set up the equation: R1*0 ‐ 200*2 + R2*10 = 0 (this is the same as equation 1 on page 2 but now the 200‐lb force is applied 2‐ft from the left edge). Solving for R2 we get 40‐lbs so R1 = 160‐lbs. The same process can be applied if you have more than one force. Note that this is for a beam; in a truss, the forces are applied at the nodes only, otherwise you need to analyze the bending of each member.
‐Use different angles between members of the truss or use a different truss design altogether.
