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MAAE 2001 Engineering Graphical Design

On Resolving a Given Force Along Three Given Directions in 3D Space Using the Cutting Plane Method

Systems involving an applied load distributed along three supports arise frequently in engineering design. A couple of examples are a landing-gear assembly with the aircraft resting on the ground, or a tripod with legs resting on un-even terrain Consider the tripod resting in the un-even way shown in Figure 1. All the members are two-force members and the lines of action of the applied load, F, and reaction forces in the legs are through the ball joint centre, O. Frequent design problems involve resolving a given force along the directions of the legs. One could proceed with the auxiliary plane method found in most engineering graphics texts. Instead, we will use the cutting plane method. We take advantage of the fact that the tripod is in a state of static equilibrium. We can construct a Force Vector Diagram and determine the magnitudes of the reaction forces in the three legs by using the following steps. STEP 1. First, let’s call the forces that act along legs A, B, and C as FOA, FOB, and FOC, respectively. Now, as we know, two intersecting lines define a plane. In the Space Diagram above, the lines of action of all forces intersect in the centre of the ball joint, O. Define one plane using the given force F and FOA. Define a second plane with FOB, and FOC. We obtain the two coloured plane segments FOA and BOC shown in Figure 2. The line of intersection (LoX) of planes FOA and BOC is the line of action of the resultant of forces F+FOA and the resultant

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of forces FOB+FOC. This must be true because of the following arguments:

1. The resultant of any two forces must lie in the plane spanned by the two forces.

2. Given one applied force and three reaction forces all concurrent and in

equilibrium, the resultant of any pair of the four forces in the given system must be equal in magnitude, but oppositely directed, to the resultant of the remaining pair.

We can express the equilibrium condition algebraically in the following equation:

0

0

=+++⇒=∑

OCOBOA FFFF

F

We can rearrange the system as:

)( OCOBOA FFFF +−=+

Clearly, the resultants are equal in magnitude, but opposite in direction. Both resultants must act through the centre of the ball joint; hence they must be on the same line. That line is the LoX of the planes spanned by the forces in each pair. Because the centre of the ball joint, O, is a point on the LoX, the problem of resolving the given force along the three given directions reduces to finding just one additional point. This is most elegantly done using a cutting plane. STEP 2. Find the LoX of planes FOA and BOC. The cutting plane is established in edge view in either the H or the F view, and cuts the lines of action of FOA, FOB, FOC, and passes through the tip of F in that view. In this example, we use the F view. In Figure 3, line GF is in plane FOA, while HC is in BOC. Point O is common to both FOA and BOC and hence is a point on the LoX. Clearly, GF and HC also meet on the LoX. In the H view, GF and HC intersect in point J. Join OH and JH to establish the horizontal projection of the LoX. Project J into the F view to establish the Front view of the LoX. See Figure 4.

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STEP 3. Now we can construct the Force Vector Diagram. We know the direction and magnitude of the applied force, so we can layout this force to scale. We know the direction and magnitude of F and we know the direction cosines of the resultant F+FOA, but not its magnitude, or sense. However, we do know the line of action of the reaction force in leg A. Thus, we have enough information to layout the force vector polygon of F+FOA. We can draw the direction of the LoX (the resultant of F+FOA) through either the tip, or tail of F in the both the H and F views in the Force Vector Diagram. Draw the direction of FOA through the opposite end of F in both views. In Figure 5, we put FOA through the tail of F. The same results would be obtained had we put FOA through the tip of F.

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This gives the force polygon illustrated in Figure 5. Because F+FOA is the resultant of F and FOA, the tip of vector FOA must join the tail of F. Moreover, the tip of the resultant joins the tip of F. Thus, we have the H and F projections of the reaction force in leg A, both magnitude and direction. The magnitudes are the lengths of the polygonal edges. To determine magnitude from the projections, an auxiliary view is required. The corresponding arrowheads are shown in Figure 6. The arrowhead for the resultant is purposely halved. The reason for this will be clear shortly. Now, we turn our attention to determining the reactions in legs B and C. When we located the arrowhead indicating the direction of F+FOA, we also established that of FOB+FOC. The magnitude is identical, but the direction opposite. So, we proceed by drawing the directions of FOB and FOC through either the tip or tail of the resultant. Because the direction of the resultant is known, those of FOB and FOC must follow the rules of vector addition. Additionally, due to the equilibrium condition, the Force Vector Diagram must close. That is, all arrows representing vectors of the applied force and reaction forces in the legs must be joined tip-to-tail. Inspecting the Force Vector Diagram below, one sees the same line must be used to indicate both resultants F+FOA and FOB+FOC. The half arrowhead is used in the sense that the upper half of the line represents F+FOA, while the lower half represents FOB+FOC.

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Space Diagram Force Vector Diagram Tension or Compression. The final task is to determine whether the reactions are tensile or compressive. The Force Vector Diagram gives us the forces that the joint feels. The same forces are equal in magnitude, but opposite in direction in the respective two-force members. Whether the member is in a state of tension or compression can be determined by considering the direction of the force in the Space Diagram relative to the joint being considered. COMPRESSION: The member is in a state of compression if the force direction in the Force Vector Diagram points towards the joint in the space diagram. TENSION: The member is in a state of tension if the force direction in the Force Vector Diagram points away from the joint in the space diagram. Transferring the directions of the arrows form the Force Vector Diagram onto the members in the Space Diagram, we see that all the forces point towards the joint. Hence, all three reaction forces are compressive. Note, the state should be the same in all views. If it isn’t, this is a clue that an error in construction has likely been made.