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Equilibrium of Non-Concurrent Force Systems
supportreactionssupport
reactionssupport
reactionssupport
reactions
weight of bridge deck on beamweight of bridge deck on beam
BEAMBEAM
Concurrent and Non-Concurrent Force Systems
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FB
FA
x
y
W
45°
x
y
BA
FBFA
W = 100 lb
AB C D
A
DAy
Ax
B C
concurrent force systems non-concurrent force systemslines of action of forces intersect at single point lines of action of forces do not intersect at single point
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Free Body Diagrams
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AB C D
x
y
2. Draw the body of interest
3. Show loads exerted by interacting bodies; name the loads
4. Define a coordinate system
5. Label distances and angles
12 ft 8 ft 20 ft
3000 lbs
B
B=1500 lbs
C
C=1500 lbs
Ay
D
D
A
Ax
pinned jointresists motion in x and y directions
roller supportresists motion in y direction
1. Choose bodies to include on FBD
STEPS:
Assume center of mass of car is halfway between the front and rear wheels
3
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Solve for Unknown Forces
Strategically choosing the order in which the three equilibrium equations are applied can make the problem easier to solve.
x
y
12 ft 8 ft 20 ft
B
B=1500 lbs
C
C=1500 lbs
Ay
D
D
A
Ax
+
Now we can sum forces in x and y. The order doesn’t matter in this case.
When solving for the “reaction forces” on a beam like this, it is usually easiest to sum moments about the point where there are the most unknowns.
Point A has two unknowns, so let’s begin by summing moments about that point.
one equation
one unknown
Should Ay be larger than D? Why? Think critically to evaluate the solution. 4
5
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A
D
Ay
Ax
BC
20°20°
x
y
AB
C
D
20°
Free Body Diagram Tip
12 ft8 ft
20 ft
Sometimes it’s helpful to strategically align your coordinate system; here, the coordinate system is aligned with the beam. In this case, it really doesn’t make that much difference in solution difficulty (compared with a horizontal / vertical alignment), but it may be a little easier since the distances used in moment calculations are clearly labeled on the beam.
6
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Class ProblemA man who weighs 890 N stands on the end of a diving board as he plans his dive.
a. Draw a FBD of the beam.b. Sum forces in the x direction to find Ax‐c. Sum moments about point A to find the reaction at B (By).d. Sum forces in the y direction to find Ay.‐
Assumptions:• Ignore dynamic effects.• Ignore deflection (bending) of the diving board.• Assume the weight of the man can be lumped exactly 3m horizontally from point B.
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Class ProblemA stunt motorcycle driver rides a wheelie across a bridge. The combined weight of therider and the motorcycle is 2.45 kN (about 550 lbs).
a. Draw a FBD of the beam for x = 2 m.b. Determine the reactions at A and C for x = 2 m.c. Derive an equation for the reactions at A and B as a function of x.d. Enter the equation from (c) into Mathcad, and plot Ay and Cy versus x on the same plot.
Assumptions:• The tire will have frictional forces with the road that could lead to a non zero value for Ax.‐• Ignore these forces when computing reactions.• Ignore dynamic effects (bumps, bouncing, change in motorcycle angle, . . . )