Example 2: Frame (non-truss, rigid body) Problems

Again in this second example, we will continue to proceed very carefully and methodically. This second example is reasonably straight forward, but points out some aspects of axial/non-axial forces, and torque. In this second example (Diagram 1, below) we will want to determine the value of the external support reactions. The general procedure to find the external support reactions consists of three basic steps.

We observe, as shown in Diagram 1, that the structure is composed of members AB and BCD. These members are pinned together at point B, and are pinned to the floor at points A and D. Additionally, point B supports a pulley with which a person is hoisting a 200 lb. load. Member BCD has a weight of 160 lb., which may be considered to act at the center of member BCD.

Step 1: Free Body Diagram (FBD).
We now proceed normally, that is we first draw our FBD (Diagram 2), showing and labeling all loads and support reactions acting on the structure. As we do so we will note several items: that member AB is an axial member (only in tension or compression), and that the person / load / pulley combination at point B produces a net 400 lb. downward load at point B. (This results since both sides of the rope have 200 lb. force in them - one side due to the load, and the other side due to the pull of the person. Point B must support both the load and the pull of the person which results in a total force of 400 lb. acting on point B)
In the FBD (Diagram 2), at point A we have shown one unknown support force 'A' acting at a known angle (37o). We can do this at point A since we know member AB is an axial member. In an axial member the force is along the direction of the member, thus the floor must exert a force on the member also along the direction of the member (due to equal and opposite forces principle). However, at point D, since member D is a non-axial member, the best we can do is to show an unknown Dx and Dy support forces acting on the structure at point D
[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 - indicating our original direction was incorrect].
In Diagram 2, we have also included Step II: Resolve (break) any forces not in the x or y-direction into x and y-components. Thus, we have shown the two components of A (which act at 37o) - A cos 37o being the x-component, and A sin 37o being the y-component. .

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 chose Point D to calculate torque. Since two unknown forces (Dx, Dy) are acting at Point 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 fewer unknowns in it, and will be easier to solve. We now proceed through the structure, looking a each force and calculating the torque due to that force with respect to the chosen Point D, and entering it in our torque equation (above) with the correct sign (+ for counterclockwise acting torque, - for clockwise acting torque). In this example, we must be careful to use the correct distance in the torque relationship - Torque = Force x Perpendicular Distance. (See Torque Review if needed.)

Finally, solving for our unknowns we obtain: A = +343 lb. Dx = +274 lb. Dy = +354 lb.
We observe 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. (Notice that means that A acts at 37o as shown, Dx act in the negative x-direction, and Dy acts in the positive y - direction.)

We have now solved our problem - finding the support reactions (forces) acting on the structure. We could add (as vectors) Dx & Dy to find one resultant force acting a some angle on point D, as follows:
D = Square Root [Dx2 + Dy2] = Square Root [(-274 lb)2 + (354 lb)2] = 447.7 lb.
Tangent (angle) = Dy/Dx = 354lb/-274 lb = -1.29, so Angle = ArcTangent (-1.29) = 127.8o (from x -axis)
so support force at Point D could also be expressed as: D = 447.7 lb. @ 127.8o

(Please note that the force at D does not act along the direction of member BCD, which it would do if BCD were an axial member.

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Continue to Example 3
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Topic 2: Statics II - Topic Table of Contents