Transcript 5_statics_studentx

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Equilibrium of Non-Concurrent Force Systems
support
reactions
support
reactions
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Concurrent and Non-Concurrent Force Systems
non-concurrent force systems
concurrent force systems
lines of action of forces do not intersect at single point
lines of action of forces intersect at single point
Σ𝐹𝑦 = 0
Σ𝐹𝑥 = 0
Σ𝐹𝑥 = 0
Σ𝐹𝑦 = 0
Σ𝑀 = 0
y
FA
A
x
B
W
45°
qA
D
FB
B
y
FA
C
C
FB
A
qB
Ax
x
Ay
W = 100 lb
D
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STEPS:
Free Body Diagrams
pinned joint
resists motion in x and y directions
1.
Choose bodies to include on FBD
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
3000 lbs
A
C
B
D
roller support
resists motion in y direction
B=1500 lbs
C=1500 lbs
Assume center of mass of car is half
way between the front and rear wheels
y
C
B
Ax
A
D
12 ft
Ay
8 ft
x
20 ft
D
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Solve for Unknown Forces
B=1500 lbs
C=1500 lbs
y
B
Ax
C
A
Ay
D
12 ft
8 ft
20 ft
x
Σ𝐹𝑥 = 0
Σ𝐹𝑦 = 0
Σ𝑀 = 0
D
Strategically choosing the order in which the three equilibrium equations are applied can make
the problem easier to solve.
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.
+ Σ𝑀𝐴 = −𝐷 ∙ 40𝑓𝑡 + 1500𝑙𝑏𝑠 ∙ 20𝑓𝑡 + 1500𝑙𝑏𝑠 ∙ 12𝑓𝑡 = 0
one equation
one unknown
𝐷 = 1200𝑙𝑏𝑠
Now we can sum forces in x and y. The order doesn’t matter in this case.
Σ𝐹𝑥 = 𝐴𝑥 = 0
Σ𝐹𝑦 = 𝐴𝑦 + 𝐷 − 1500𝑙𝑏𝑠 − 1500𝑙𝑏𝑠 = 0
𝐴𝑦 = 1800𝑙𝑏𝑠
Should Ay be larger than D? Why? Think critically to evaluate the solution.
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Free Body Diagram Tip
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.
20°
20°
20°
C
B
D
Ax
Ay
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Class Problem
A 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 Problem
A stunt motorcycle driver rides a wheelie across a bridge. The combined weight of the
rider 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 C 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, . . . )
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