Example 3: Frame (non-truss, rigid body) Problems
This third example is somewhat similar to example one, but we will extend the problem by not only determining the external support reactions acting on the structure, but we will also determine the internal force in member CD of the structure shown in Diagram 1.
We first calculate the support forces. The procedure to find the external support reactions consists of our basic statics procedure..
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, CD, AD, and cable DE. These members are pinned together at several points as shown in Diagram 1. A load of 12,000 lb. is acting on member ABC at point B, and a load of 8000 lb. is applied at point C. 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 simply 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: Draw a Free Body Diagram of the entire structure. In the FBD (Diagram 2), we have shown unknown x and y support forces acting on the structure at point A, however, at point E we have shown one unknown force 'E' acting at a known angle (37o).
We can do this at point E since we know that ED is a cable, and a cable is an axial member which can only be in tension. Since the cable pulls axially on the wall, the wall pulls equally and in the opposite direction on the structure., as shown in Diagram 2.
In Diagram 2, we have also included Step II, Resolve any forces not in the x or y-direction into x and y-components. Thus, we have shown the two components of E (which act at 37o) - E cos 37o being the x-component, and E sin 37o being the y-component. We have also used given angles and dimensions to calculate some distance, as shown, which may be needed when we apply the equilibrium equations.
We also note that at point A we have two members pinned together to the wall, axial member AD, and non-axial ABC. Because of these two members (as opposed to a single axial member, such as at point E), the best we can do at point A is to replace the hinged joint by an unknown Ax and Ay support forces acting on the structure as shown in Diagram 2. However, we have a good FBD since we have only three unknown forces, and we have three independent equations from our equilibrium conditions.
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 E 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 E 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 - E. We can solve for E, and then use it in the two force equations to find the other unknowns, Ax and Ay. Doing the mathematics we arrive at the following answers.) E = +10,800 lb. Ax = +8620 lb. Ay = +13500 lb.
We see that all the support forces we solved for are positive, which means the directions we chose for them initially are the actual directions they act. We have now solved part one of our problem. The support force at point E is 10,800 lb. acting at 37o. The support force(s) at A can be left as the two components, Ax = 8,620 lb. and Ay = 13,500 lb., or may be added (as vectors) obtaining one force at a known angle.
The second part of the problem is to determine the force in axial member CD. (We know member CD is axial as there are only two points at which forces acts on CD, point C and point D.) To determine the force in an internal member of a structure we use a procedure similar to that used to find the external support reactions. That is, we draw a FBD, not of the entire structure, but of a member of the structure, (choosing not the member we wish to find the force in, but a member it acts on). Thus, if we wish to find the force in member CD, we draw a FBD - not of member CD, but a member CD acts on, such as member ABC, or member AD. In this example we will use member ABC to find the force in member CD.
To find the force in a member of the structure we will use the following steps:
First, determine the external support reactions acting on the structure (as we did in the first part of this example). Then continue with steps below
- I. Draw a Free Body Diagram of a member of the structure showing and labeling all external load forces and support forces acting on that member, include any needed dimensions and angles. (The member selected should be one acted on by the member in which we wish to find the force.)
- II. Resolve (break) all forces into their x and y-components.
- III. Apply the Equilibrium Equations () and solve for the unknown forces.
Step 1: FBD of Member ABC. There are actually two good FBD for member ABC. In Diagram 3 we have shown the first of these. Notice at the left end we have shown both the wall support reactions at A, and also the force from axial member AD which acts on member ABC. At point C we have shown the force from axial member CD which acts on member ABC. That is, we have isolated member ABD and indicated the forces on it due to the other members (and the wall) attached to it.
The second good FBD of member ABC is shown in Diagram 4. What we have done in this diagram is to look more closely at the left end of member ABC and observe that the effect of the wall forces and the effect of member AD, is to give some net x and y-force acting on member ABC. Thus, rather than show both the wall forces and the force due to AD on ABC, we simple show an ACx and an ACy force which is the net horizontal and vertical force acting on ABC at the left end. This is fine to do, as we are looking for force CD, and that is still present in our FBD. This second FBD is slightly easier than the first in that it will result in one less force (AD) in the equilibrium equations. We will use the second FBD in the rest of the problem.
Step 2: Resolve forces into their x and y-components (This is done in Diagram 5.) Notice we have chosen directions earlier for the forces. These may not be the correct directions, but our solution will tell us if we have the right or wrong directions for the unknown forces.
Sum Fx: ACx + CD cos 53.8o = 0
Sum Fy: ACy -12,000 lb. + CD sin 53.8o = 0
Sum TA: -12,000 lb. (4 ft) + CD sin 53.8o (8 ft ) =0
Solving: CD = 7,440 lb., ACx = - 4390 lb. (- sign shows force acts opposite direction chosen), ACy = 6000 lb.
We have determined the force in CD to be 7,440 lb. Since the force is positive, this indicates that we have chosen the correct direction for the force CD (which indicates it is in tension). This solves our problem. (For a little further analysis of forces at point A, select MORE.)
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