Example 1: Frames (non-truss, rigid body) problems
In this first example, we will proceed very carefully and methodically. It is important to get the method and concepts we need to keep in mind firmly established.
In this problem we wish to determine all the external support forces (reactions) acting on the structure shown in Diagram 1 below. Once again our procedure consists basically of three steps.
- Draw a Free Body Diagram of the entire structure showing and labeling all external load forces and support forces, include any needed dimensions and angles.
- Resolve (break) all forces into their x and y-components.
- Apply the Equilibrium Equations (
) and solve for the unknown forces.
Step 1: Free Body Diagram (FBD). Making the FBD is probably the most important part of the problem. A correct FBD usually leads to a quick solution, while an inaccurate FBD can leave a student investing frustrating unsuccessful hours on a problem. With this in mind we will discuss in near excruciating detail the process of making a good FBD.
We note that the structure is composed of members ABC, and CD. These two members are pinned together at point C, and are pinned to the wall at points A and D. Loads of 4000 lb. and 2000 lb. are applied to member ABC as shown in Diagram 1.
In our example, the load forces are already shown by the downward arrows. We next look at the forces exerted on the structure by the supports. Since each support is a pinned joint, the worst case we could have is an unknown x and y-force acting on the structure at each support point. We also must choose directions for the x and y support forces. In some problems the directions of the support forces are clear from the nature of the problem. In other problems the directions the support forces act is not clear at all. However, this is not really a problem. We simple make our best guess for the directions of the support reactions. If our guess is wrong, when we solve for the value of the support forces, that value will be negative. This is important. A negative value when solving for a force does not mean the force necessarily acts in the negative direction, rather it means that the force acts in the direction OPPOSITE to the one we initially chose.
Thus, in our first FBD on the right (Diagram 2), we have shown unknown x and y support forces acting on the structure at each support point.
This is an accurate FBD, but it is not the best. The difficulty is that for our problem, we have three equilibrium conditions ( ), but we have four unknowns (Ax, Ay, Dx, Dy) in this FBD. And as we are well aware, we can not solve for more unknowns than we have independent equations.
We can draw a better FBD by reflecting on the concept of axial and non-axial members. Notice in our structure that member ABC is a non-axial member (since forces act on it at more than two points), while member CD is an axial member (since if we drew a FBD of member CD we would see forces act on it at only two points, D and C). This is important. Since CD is an axial member the force acting on it from the wall (and in it) must act along the direction of the member. This means that at point D, rather than having two unknown forces, we can draw one unknown force acting at a known angle (force D acting at angle of 37o, as shown in Diagram 3). This means we have only three unknowns, Ax, Ay, and D. In Diagram 3, we have also completed Step II, breaking any forces not in the x or y-direction into x and y-components. Thus, in Diagram 3, we have shown the two components of D (which act at 37o), D cos 37o being the x-component, and D sin 37o being the y-component. [Please notice that there are not three forces at point D, there is either D acting at 37o or its two equivalent components, D cos 37o and D sin 37o. In Diagram 3 at this point we really should cross out the D force, which has been replaced by its components.]
Now before we proceed with the final step and determine the values of the support reactions, we should deal with several conceptual questions which often arise at this point. First, why can't we do at point A what we did at point D, that is put in one force acting at a known angle. Member ABC is a horizontal member, doesn't the wall just push horizontally on member ABC, can't we just drop the Ay force? The answer is NO, because member ABC is not an axial member, it is not simply in compression or tension, and the wall does not just push horizontally on member ABC (as we will see in our solution). Thus the best we can do at point A is unknown forces Ax and Ay.
A second question is often, what about the wall, aren't there forces acting on the wall that we should consider? Well, yes and no. YES, there are forces acting on the wall (as a matter of fact they are exactly equal and opposite to the forces acting on the members, in compliance with Newton's Third Law). But NO we should not consider them, because we are making a FBD of the STRUCTURE, not of the wall, so we want to consider forces which act on the structure due to the wall, not forces on the wall due to the structure.
Now Step III. Apply the Equilibrium conditions.
Here we sum the x-forces, keeping track of their direction signs, forces to right, +, to left, -
Sum of y-forces, including load forces. Again keeping track of direction signs.
Sum of Torque about a point. We choose point A. Point D is also a good point to sum torque about since unknowns act through both points A and D, and if a force acts through a point, it does not produce a torque with respect to that point. Thus our torque equation will have less unknowns in it, and will be easier to solve. Notice that with respect to point A, forces Ax, Ay, and D sin 37o do not produce torque since their lines of action pass through point A. Thus in this problem the torque equation has only one unknown, D. We can solve for force D, and then use it in the two force equations to find the other unknowns, Ax and Ay. (Completing the calculations, we arrive at the following answers.) D = +7500 lb. Ax = +6000 lb. Ay = +1500 lb.
Note that all the support forces we solved for are positive, which means the directions we choose for them initially are the actual directions they act. We have now solved our problem. The support force at point D is 7500 lb. acting at 37o. The support forces at A can be left as the two components, Ax = 6000 lb. and Ay = 1500 lb., or may be added (as vectors) obtaining one force at a known angle, as shown in Diagram 4.
Thus the force at point A is 6185 lb. acting at 14o, as shown. This information would help us purchase the correct size hinge (able to support 6185 lb. at A, and able to support 7500 lb. at D), or estimate if the wall is strong enough to support the structure. All very useful and interesting information.
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