A3 - Kyle Hayes

Ends

   ΣF_X = 0:   F_AX = 0

   ΣF_Y = 0:   -10lbs x 1ft + F_EY x 2ft = 0

F_EY = 10/2 = 5lbs

F_EY = 5lbs

F_AY = -10lbs + F_EY = 0:   F_AY = 5lbs

F_AY = 5lbs

Joint A

   ΣF_Y = 0:   T_AB sin(45) + F_AY = 0:   T_AB = -5/sin(45) = -7.07lbs

                T_AB = -7.07lbs

   ΣF_X = 0:   T_AB cos(45) + T_AC = 0:   T_AC = 7.07 cos(45) = 5lbs

                T_AC = 5lbs

Joint B

   ΣF_Y = 0:   -T_AB sin(45) + T_BC sin(45) = 0:   T_BC = -T_AB sin(45) / sin(45) = 7.07lbs

                T_BC = 7.07lbs

   ΣF_X = 0:   -T_AB cos(45) + T_BC cos(45) + T_BD = 0:   T_BD = -7.07 cos(45) - 7.07 cos(45) = -10lbs

                T_BD = -10lbs

Joint C

   ΣF_X = 0:   T_BC sin(45) + T_CD sin(45) -10lbs = 0:   T_CD = [10 – 7.07 sin(45)] / sin(45) = 7.07lbs

                T_CD = 7.07lbs

   ΣF_Y = 0:   -T_AC – T_BC cos(45) + T_CD cos(45) + T_CE = 0:   T_CE = 5 – 7.07 cos(45) – 7.07 cos(45) = -5lbs

                T_CE = -5lbs

Joint D

  ΣF_Y = 0:    -T_CD sin(45) – T_DE sin(45) = 0:   T_DE = -7.07 sin(45) / sin(45) = -7.07lbs

                T_DE = -7.07lbs

 

Joint E

   All tensions around joint E are already solved for.

My Analysis Diagram

Bridge Designer Analysis



To make sure the hand analysis corresponds to the Bridge Designer, the lengths of the members and the angles must scale to each other. So that all angle are the same between the hand and Bridge Designer analysis. For the members it is just important the relative size to one another is kept the same. If two pieces are the same length as each other than those two corresponding pieces on the Bridge Designer must be the same length. If one is twice the size of the other, than the corresponding piece on the Bridge Designer must be twice the size of the other.



K'NEX

Bridge Designer Analysis

The K’NEX joint test page showed that the pull out for required to remove a member from a joint increase with the more members attached to that joint, it also increase more if it is symmetrical. This test tells us that the average max limit of tension of a member can be 37lbs before the connecter is almost guaranteed to fail, useful information as any member nearing this tension amount must be adjusted and will most likely fail first. We can use the fact the more members per connector increases the tension required to remove the member to strengthen our connection. Where ever there is a spot nearing this maximum tension amount we can add a member that will have a vector force in the same direction as the member nearing the max tension to increase capacity. To explain a bit more clearly, the example only had three of the five slots of the connector use, the other to slots, the ones on the end would not contribute to increasing the strength though as they were only in the x direction and thus can only hold and x direction vector force, but the three that were used either only had a y vector or had a component of them that was in the y direction. This is why adding members increases the tension needed to pull it out, because it is not just that member being pulled in the y direction but some of the pull is being sent to the other member whose vector is in both the x and direction.

    - Kyle Hayes