Forces in Three Dimensions

IntroductionIn previous chapters, we learned how to handle a variety of problems, parallel forces, non-concurrent forces, concurrent forces, trusses, and finally frames. But all of these were coplanar systems. That is, they existed in and were appropriately analyzed in two-dimensions. However, not all structures can be analyzed as such

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IntroductionThere are a number of force systems that are non-coplanar. From the standpoint of architecture, space frames are such a structure and cannot be analyzed in two-dimensions

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IntroductionFrom the standpoint of mechanical design, an example might be a right angle gearbox or perhaps a shaft transmitting power to a series of belt drives. To handle these systems we must look at them in all three dimensions

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Dealing with force systems in three dimensions is virtually identical to dealing with coplanar force systems. The only difference is you will take a three dimensional view and project it into two or three 2-dimensional views, then solve them simultaneously. Before we do so, let's do a quick review of a three dimensional coordinate axis‑

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Introduction

The positive x-axis to the right, the positive y-axis in the positive upward direction and the positive z-axis projecting out of the surface of the paper (or should I say computer screen?). Of course, on a two-dimensional sheet of paper, the positive z-axis is drawn downward and to the left at a 45o angle

The 3-D Axis

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When dealing with moments in three dimensions, we will always maintain a counterclockwise moment as a positive moment regardless of which plane into which we project the structure. For example, looking at the coordinate system above, we can draw the following 2-dimensional projections:

The 3-D Axis

Resolving Parallel Forces

Resolving Parallel ForcesNow let's take a look at how we might resolve a parallel force system into a single equivalent force. To do so, the resultant force must have a magnitude equivalent to the system we are resolving and it must produce the same moment. Consider the following example

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R = 90 ↓

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What we need to do is determine the magnitude of 'R' and the location of 'R'. The location of 'R' is determined by determining distances 'x' and 'z'. The shown location and sense of 'R' is assumed. The algebraic signs of the calculated distances and magnitude will validate our assumption. The resultant is determined by summing all forces in the system:

Resolving Parallel Forces

Caution: In this problem, we are finding a force resultant, not reactions. In the case of finding a resultant, the negative sign is absolute. That is, it indicates actual direction in accordance with our sign convention.

F=−5060−100 =−90

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The location of 'R' must be such that it causes a moment equal to the moment of the system. To do so, it is easier to visualize the problem if we redraw the system in a two dimensional view. First, let's look down the x-axis into the y-z plane and draw an appropriate free body diagram

Resolving Parallel Forces

z = -4.55

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Resolving Parallel Forces

Note: The negative sign on the 90·z term indicates the resultant causes a negative moment. It does not reflect the sense of the resultant.

However, the negative sign on 'z' indicates that our assumption of R being located behind the x-axis is incorrect. It is actually located in the first quadrant with a z-dimension of 4.55 units.

−R⋅z =−50⋅3 − 60⋅4 100⋅8

−90⋅z =−50⋅3 − 60⋅4 100⋅8

Take a moment about the x-axis:

Since we know the magnitude of 'R' is 90:

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Now let's look at a FBD as viewed along the z axis.‑

Resolving Parallel Forces

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Take a moment about the z-axis

Since we know the value of the resultant 'R' is 90:

x = -4.55

Resolving Parallel Forces

As before, the negative sign on the 90·z term indicates the resultant causes a negative moment. It does not reflect the sense of the resultant.

However, the negative sign on 'z' indicates that our assumption of R being located right of the x-axis is incorrect. It is located 4.55 units left of the z-axis.(The fact x=z is purely coincidental)

−R⋅x = 60⋅6−100⋅250⋅5

−90⋅x = 60⋅6−100⋅250⋅5

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Resolving Parallel ForcesThe resultant force would be located as shown. This single force is equivalent to the parallel force system with which we started.