Example 4: Frame (non-truss, rigid body) Problems
In our fourth example, we will examine a problem in which it will initially seem that there are too many unknowns to allow us to determine the external support forces acting on the structure, however, by a slight variation of our approach we will find that we can determine all the unknowns. The problem then is this, for the structure shown in Diagram 1 below, determine the external support forces acting on the structure, and additionally, determine the force in member CF. Our usual procedure to find the external support reactions consists of three basic steps.
- I. 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.
- II. Resolve (break) all forces into their x and y-components.
- III. Apply the Equilibrium Equations (
) and solve for the unknown forces.
We note that the structure is composed of members ABC, DEF, DE, and CF. These members are pinned together at several points as shown in Diagram 1. A load of 6000 lb. is acting on member DEF at point E, and a load of 3000 lb. is applied at point F. These 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.
Step 1: Free Body Diagram (FBD).
In the FBD (Diagram 2), we have shown unknown x and y support forces acting on the structure at pinned support points A and D, (Ax, Ay, Dx, Dy). If we think ahead somewhat, we realize that there could be a problem. We have four unknown forces supporting the structure, but there are only three equations in our Equilibrium Conditions, 
.
Normally, one can not solve for more unknowns then there are independent equations. Our first reaction should be to see if we can draw a better FBD. Perhaps we can replace the two unknowns at either point A or D by one unknown acting at a known angle (which is possible if we have a single axial member acting at the support point). However in this case both member ABC and member DEF are non-axial members, and the forces in them (and on them from the wall) do not act along the axis of the member. Thus, we already have the best FBD possible - we can not reduce the number of external unknowns acting on the structure.
At this point we will simply continue with our normal analysis procedure and see what results.
Step II: Resolve any forces not in the x or y-direction into x and y-components. (All forces are already in either the x or y-direction.
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 D. Forces Ay, Dx and Dy do not produce torque since their lines of action pass through point A. Thus, the torque equation has only one unknown, Ax. We solve for Ax, and then use it in the sum of forces in the x-direction equation to find the unknown, Dx . And if we do so, we find: Ax = +18000 lb. Dx = +18000 lb. (The positive signs indicate we initially chose the correct direction for the forces.) However, please notice that while we found Ax and Dx, we can not find Ay and Dy. There are still two unknowns in the y-equation and not enough information to determine then at this point. Thus, analysis of the structure as a whole has enabled us to determine several of the external support forces, but not all of them. What now?
First, an overview. There are problems for which the static equilibrium conditions are not enough to enable one to solve the problem. They are called Statically Indeterminate Problems, and we will be considering these a bit later. Then there are problems which, on first glance, appear to be statically indeterminate, but are not. That is the case here. To find the remaining unknown support forces (and at the same time, determining the force in member CF), we will now take out a member of the structure and apply our statics analysis procedure to the selected member of the structure (rather than the entire structure). We will select member ABC to analyze. (See Diagram 3)
Step 1: Free Body Diagram (FBD)
In Diagram 3, we have drawn a FBD of member ABC, showing and labeling all forces external to member ABC which act on it. That is at point A we have the forces of the wall acting on ABC, at point B we have an axial force on ABC due to member BE, and at point C we have a axial force on ABC due to the member CF. Both member BE and CF are axial members and so we know the directions their forces act - along their axis. We do have to guess if they push or pull on member ABC, and we have chosen those directions as shown, (if the direction chosen is wrong, the force's value will be negative when we solve). We also are happy with the FBD as it has only three unknowns acting on the member, which indicates that we should be able to solve completely for the unknowns.
Step 2: Resolve forces into x/y components.
Here we have resolved force CF into its horizontal and vertical components as shown in Diagram 4. All other forces are already in x or y-direction.
Step 3. Apply the Equilibrium conditions. 
We now solve the first equation for CF, then use that value in the last equation to find BE, and use both values in the middle equation to find Ay, giving us: CF = 30,000 lb., BE = 36,000 lb., Ay = -12, 000 lb. Please note that the value of Ay is negative, which indicates the direction we chose was incorrect. Ay acts downward rather than in the positive y-direction we initial chose. Additionally, we now can return to the equations for the entire structure (see below), and knowing the value for Ay, we can use it in the y-forces equation to solve for the value of Dy.
Equilibrium Conditions for entire structure (from first part of problem)
From the y-forces equation: (-12,000 lb.) - Dy -6000 lb. -3000 lb = 0; Solving Dy = -21,000 lb.
Once again, the negative sign indicates we selected the wrong direction for Dy, rather than acting downward it actually acts upward. See Diagram 5 for final force values and directions.
Ax = 18000 lb.
Ay = 12,000 lb. Dx = 18000 lb. Dy = 21,000 lb. CF = 30,000 lb. BE = 36,000 lb.
No comments:
Post a Comment