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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

2

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.

7

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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, . . . )