Three-Force Member Example Problem

Lecture 15
Example Problem

Three-Force Members

Given:
the system with the loading indicated below.
Determine:
a. the FBD of the horizontal element using the images below
b. the reactions at B and C algebraically
c. the reactions at B anc C graphically

Solution:
a. The free body diagram of the horizontal member will include the 12k force at D and the reactions at B and C. The support at B will have a horizontal and a vertical component although the angle at which the total force B acts is unknown at this point. The direction of C is already known because it must act in the direction of the wire, but it is more easily solved when it is broken into its rectangular components.

b. Apply the familiar equations of equilibrium to algebraicly solve for the reactions at B and C.

Sum Fx = 0 = Bx + Cx
Sum Fy = 0 = -12k + By + Cy
Sum MB = 0 = -Cy(6ft) + 12k(14ft)

Solve for Cy using the moment equation:

6Cy = 168kft
Cy = 28k

Solve for By using Cy in Sum Fy = 0:

-By = -12k + 28k
-By = 16k
By = -16k

Now there are two more unknowns, Bx and Cx, but only one other equation. However, from observation, C is a 3-4-5 triangle. Using this information, solve for C and Cx:

Cy = 28k = 4/5 C
C = 22.4k
Cx = -3/5 (22.4k)
Cx = -13.44k

Now solve for Bx:

Bx = -Cx
Bx = 13.44k

Finally, solve for B using the Pythagorean Theorem:

B = SQRT (13.44k^2 + (-16k)^2) = SQRT (436) = 20.9k acting down and to the right

The angle for the reaction at B would be:

ų = arctan (-16 / 13.44)
ų = 53.5 degrees below the x axis

c. To solve for the reactions graphically, remember that any one force in a three-force system must be the resultant of the other two forces. This must be so for the element to remain in equilibrium. Additionally, if the system is in equilibrium and if the forces are not parallel, they must be concurrent.

Now, extend the lines of action of the two forces that have a known direction until they intersect. This is the point of origin for the three-force system. It is a VERY important point! Now, the ine of action and the direction of the reaction at B can be found by simply drawing a line from point B to the point of origin. This is the line of action of the force. The direction of the resulting third force is also known when inspection determines what kind of force must be applied to put the system into equilibrium. The magnitude of B and C can be found using the triangle method and making assumptions about the sense of the forces.

Lecture 15 Example Problem

Lecture 15
Example Problem