System Reactions Example Problem

Lecture 16
Example Problem

System Reactions

Given:
the system and loading as shown.
Determine:
a. the FBD
b. the reactions at A and B analytically

Solution:
a. The system FBD will include the known concentrated and distributed loads, 2 reactions at A (Ax and Ay) and a single reaction at B, perpendicular to the surface supporting the roller. Note that this surface has a 3-4-5 geometry.

b. Write the equations of equilibrium breaking the reaction at B into its rectangular components.

Sum Fx = 0
Ax - Bx + 24
Sum Fy = 0
Ay + By - 24
Sum MA = 0
24k(18ft) + 1k/ft(24ft)(12ft) - By(24ft) - Bx(18ft)

Watch the signs! Realize, in this case, if By is assumed to be positive, Bx must be negative and vice versa because both components must either act toward the diagonal surface or away from the diagnonal surface.

At this point, all three equations have two variables making them impossible to solve. However, remember the 3-4-5 geometry of the reaction at B. Therefore, Bx = 3/5 B and By = 4/5 B. By replacing these values in the moment equation, it is possible to solve for B.

Sum MA = 0
24k(18ft) + 1k/ft(24ft)(12ft) - 4/5B (24ft) - 3/5B (18ft)
432kft + 288kft -19.2B - 10.8B
720kft - 30B
30B = 720 kip-feett
B = 24 kips

Now use the force equations to solve for Ax and Ay.

Ax - 3/5B + 24 = 0
Ax - 14.4 + 24 = 0
Ax = -9.6 kips

Ay + 19.2 - 24 = 0
Ay = 4.8 kips

Finally, find the total reaction at A and its directional angle.

A = SQRT (-9.6^2 + 4.8^2) = 10.73k
ø = arctan (-4.8/9.6) = 26.6 degrees up and to the right.

In summary, the reaction at A equals 10.73 kips acting at an angle of 26.6 degrees up and to the right and the reaction at B equals 24 kips acting perpendicular to the diagonal surface.

Lecture 16 Example Problem

Lecture 16
Example Problem