) and solve for the unknown forces.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.
) 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)
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.
Return to: Topic 2.1 - Frames
or select:
Topic 2: Statics II - Topic Table of Contents
Strength of Materials Home Page