London Bridge Tutorial
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London Bridge Tutorial

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TutorialPage 1 London Bridge Construction Manual November 17, 2001IntroductionP ROBLEM Dalso be not too difficult to construct.CPage 2 London Bridge Construction Manual November 17, 2001and each student will benefit from the experience!Olympics. This event will enhance their knowledge of the world of engineering and technology,I urge you to encourage your students to take part in this event of the London District Sciencethat the students will learn what types of bridges are effective, and what types of designs are not.participants will have the opportunity to witness their creations in action. It is through this processbut also through participating in the testing of the other bridges. On the competition day,for their creativity. The student will learn not only through the construction of their own bridge,The London Bridge Competition provides innovative and self motivated students with an outletrather than on the design.this competition. Success relies more upon the construction of the bridge,to build their bridge. This is probably the most important part of the project forAfter the design process has been completed the student must figure out howONSTRUCTIONdesign restrictions. The structure must not only meet these restrictions, butrequirements, students must design a structure which lies within the givenAfter devising the method by which they are going to meet the competitionESIGNthe competition. the student must devise a method by which they ...

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November 17, 2001
Tutorial
London Bridge Construction Manual
Page 1
Introduction
The purpose of the London Bridge popsicle stick bridge design competition is to give high-school students the opportunity to exhibit the skills which they have learned in their mathematics, physics and technology classes. In the class room, students are given ample opportunity to display their understanding of a concept, but are seldom required to demonstrate their ability to put this knowledge to practical use. As students will soon discover, the ability to apply the concepts which they are learning in class is as important as the comprehension of the concepts themselves. The London Bridge competition teaches students three vital aspects of the modern engineering world:
P ROBLEM S OLVING When assigned the task of designing and constructing a popsicle stick bridge the student must devise a method by which they will fulfill the requirements of the competition. D ESIGN After devising the method by which they are going to meet the competition requirements, students must design a structure which lies within the given design restrictions. The structure must not only meet these restrictions, but also be not too difficult to construct. C ONSTRUCTION After the design process has been completed the student must figure out how to build their bridge. This is probably the most important part of the project for this competition. Success relies more upon the construction of the bridge, rather than on the design.
The London Bridge Competition provides innovative and self motivated students with an outlet for their creativity. The student will learn not only through the construction of their own bridge, but also through participating in the testing of the other bridges. On the competition day, participants will have the opportunity to witness their creations in action. It is through this process that the students will learn what types of bridges are effective, and what types of designs are not. I urge you to encourage your students to take part in this event of the London District Science Olympics. This event will enhance their knowledge of the world of engineering and technology, and each student will benefit from the experience!
November 17, 2001
London Bridge Construction Manual
Page 2
HOW TO RESOLVE A TRUSS
Recall from OAC Statics: Find the tension in each rope.  3 F = 0 1. 3 F X = 0 = T 1 cos45°-T 2 cos45°  T 1 = T 2  sub in 2. 2. 3 F Y = 0 = T 1 sin45°+T 2 sin45°-1  1 = T 2 sin45°+T 2 sin45° =  0.71 N T 2 Therefore: T 1 = 0.71 N
Figure 23: OAC statics problem.
Although the principle of drawing a free body diagram and summing all the force vectors to zero is the same, a truss is very different. A truss possesses not only tension members, but compression members as well. To account for this, the following steps must be taken:
1. Treat the truss as a solid and determine the forces at the supports A and B. 2. Hypothesize(guess) which members are in tension, and which are in compression. 1  3. In your free body diagram, draw compression members as ones pushing on the pin, and tension members as ones pulling on the pin. 4. Draw a free body diagram for all pins. 5. Use Newton's Second Law: 3 F X = 0 3 F Y = 0                    6. Solve for the unknown forces.
     1 If after calculating the forces, one or more result in a negative, this means that your guess as to whether the member was in tension or compression, was wrong. November 17, 2001 London Bridge Construction Manual Page 3
Eg 1) What are the forces in truss?
Solution: AB and AC are in compression. BC is in tension. Since the equal.
Forces at
FBD at B:
  November 17, 2001
each member of the following
truss and loading are symmetrical, forces F 1 and F 2 will be
Supports: 3 F = 0     3 F Y = 0 = F 1 + F 1 -1  1 = 2F 1  F 1 = 0.5 N
       3 F = 0 3 F X = 0 = -ABcos60°+BC  BC = Abcos60°  0.5AB = BC                                                3 F Y = 0 = 1/2-ABsin60°  1/2 = ABsin60° London Bridge Construction Manual
Page 4
    
 / 3/2AB = 1/2  AB = 0.58N  BC 0.5(1/ / 3) =  BC = 0.29 N Note:  Because this truss is symmetrical the forces in member "AC" will be equal to those found in member "AB". Therefore we do not need to draw another free body diagram at pin "C". All we have left to do is check our answers by evaluating the forces acting in pin "A". They should sum to zero if we are correct.
FBD at A:
3 F = 0 3 F = 0 = 1/ / 3cos60°-1/ / 3cos60° X    0 = 0
3 F Y = 0 = 1/ / 3sin60°+1/ / 3sin60°-1  0 0 =
Since the forces at the last pin examined summed to zero, then the truss is in equilibrium, and all forces have been determined.
NOTES:                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                             
November 17, 2001
London Bridge Construction Manual
Page 5
                                                                                                                                                                                                                                                                                                                                                                                                                                                                    
Eg2) What are the forces in each member of the following truss?   
Solution: AC and AF are in compression. CF is in tension. Since the truss and loading are symmetrical, forces F 1 and F 2 will be equal. BD, AD, AE, and EG are all zero force members. This means that there only function is to preserve the structural integrity of the truss. The forces on them are negligible. .
Forces at Supports:
November 17, 2001
3 F 0 = 3 F Y = 0 = F 1 + F 1 -1 1 = 2F  1 London Bridge Construction Manual
Page 6
FBD at C:
FBD at B:
F 1 = 0.5 N
     3 F = 0 3 F X = 0 = DC-BCcos30 °  DC = BCcos30 °
3 F Y = 0 = 1/2-BCsin30° =  1/2 BCsin30°  BC = 1.0 N ˆ  DC = 1cos30°  DC = 0.87 N
3 F = 0 3 F X = 0 = BCcos30°-ABcos30°+ BDcos60°  0.87 = 0.87AB + 0.5BD 3 F Y  0 = BCsin30°-ABsin30°-= BDsin60°  0.5 = 0.5AB + 0.87BD         Solving these equations we find: AB = 1.0N, and BD = 0.  
FBD at D:
November 17, 2001
3 F = 0 3 F X = 0 = -DC +DE-ADcos60°-BDcos60°  0.87 = DE - 0.5AD       3 F Y = 0 = -ADsin60°+BDsin60°  AD = 0   
London Bridge Construction Manual
Page 7
Note : Because of the we do not need to pins "F", "G" and "E". "AF", "AE", "EG" will members "AC", "BD" We can now check our forces at the last pin, they should sum to zero.
    ˆ DE = 0.87N
symmetry of this truss, evaluate the forces at The forces in members be equal to those in and "AD" respectively. answers by summing "A". If we are correct
FBD at A: Since AE and AD are zero force members, they do not appear in our equations. 3 F = 0 3 F X = 0 = -AGcos30°+ABcos30° 0 = 0 3 F Y = 0 AGsin30°+ABsin30°-1 =  0 = 0
Eg3) What are the forces in each member of the following truss?
Solution: AB and ED are in compression. AE and BD are in tension. AC and EC are zero force members. Since the truss and loading are symmetrical, forces F 1 and F 2 will be equal.
November 17, 2001
London Bridge Construction Manual
Page 8
Forces at Supports:
FBD at B:
FBD at A:
   
3 F = 0 3 F Y = 0 = F 1 + F 1 -2 2 = 2F 1 F 1 = 1.0 N
3 F = 0 3 F X = 0 = BC-ABcos60°   1/2AB = BC = 3 / F3 Y  A 0B  ==  11-ABsin60°  /2  AB = 1.15 N  BC = 1/2(1.15)  = 0.58 N
3 F = 0 3 F X = 0 = ABcos60°-AE-ACcos60°  ACcos60°= 2/ / 3(1/2)-AE  1/2AC = / 3/3-AE 3 F Y = 0 = ACsin60°+ABsin60°-1 1 = / 3/2AC+2/ / 3( / 3/2) 1 = / 3/2AC+1  AC = 0  1/2AC = / 3/3-AE  AE = 0.58 N Note:  Because of the symmetric nature of the truss we do not need to draw free body diagrams at points "E" and "D". The forces in members "EC" and "ED" are equal to those in members "AB" and "AC" respectively. All we have left to do is check our answers by November 17, 2001 London Bridge Construction Manual Page 9
                                              
Page 10
3 F Y = 0 = -ECsin60°-ACsin60° 0 = 0
                                                                                                                                                   
Since the sum of the forces at the last point evaluated equal zero, the truss is in equilibrium and all forces are known. NOTES:                                                                                                                                                                                                                                                                                                                                                                                                                                                        
November 17, 2001
1. Resolve the truss. (All unmarked angles = 60°)
summing the forces at pin "C". If we are correct the forces at "C" will sum to zero. FBD at C: 3 F = 0 3 F X = 0=BD+ACcos60°-ECcos60°-BD  1/2EC = 1/2AC 0 = 0
London Bridge Construction Manual
SUPPLEMENTARY PROBLEMS
                                                  
2. Resolve the truss. (All unmarked angles = 45°)
3. Resolve the truss.  
Answers: AE = CE = BF = BD N 2. AG = CG = 0.5 HE = = = AE = 0   AC = KI CI = 2.6 = KF = AK =  LF = 1.0 = BE = AE = JH = JG = KG 0 N =
November 17, 2001
London Bridge Construction Manual
1. AC = 0.58 N, 0.29 N, = DF = 0   AC =  0.71 N, N, BE BD = HF N 3. 3.0 N, = N, AF 0.5 N, 2.85 N, N, BD
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