1. Equilibrium of Non-Concurrent Force Systems
support reactions
weight of bridge deck on beam
BEAM

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3. Concurrent and Non-Concurrent Force Systems
lines of action of forces intersect at single point
lines of action of forces do not intersect at single point
Σ 𝐹 𝑥 =0
Σ 𝐹 𝑦 =0
Σ 𝐹 𝑥 =0
Σ 𝐹 𝑦 =0
Σ𝑀=0 FB FA x y W
45° A B C D x y qB qA FB FA
W = 100 lb A D Ay Ax B C

4. Free Body Diagrams STEPS: pinned joint Choose bodies to include on FBD
resists motion in x and y directions
Choose bodies to include on FBD
STEPS:
3000 lbs
Draw the body of interest A
Show loads exerted by interacting bodies; name the loads B C D
Define a coordinate system
roller support
resists motion in y direction
Label distances and angles
Assume center of mass of car is half way between the front and rear wheels B
B=1500 lbs C
C=1500 lbs x y A Ax Ay
12 ft
8 ft
20 ft D

5. Solve for Unknown Forcesx y
12 ft
8 ft
20 ft B
B=1500 lbs C
C=1500 lbs Ay D A Ax
Σ 𝐹 𝑥 =0
Σ 𝐹 𝑦 =0
Σ𝑀=0
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.
𝐷=1200𝑙𝑏𝑠
one equation
one unknown
Σ 𝑀 𝐴 =𝐷∙40𝑓𝑡−1500𝑙𝑏𝑠∙20𝑓𝑡−1500𝑙𝑏𝑠∙12𝑓𝑡=0 +
Now we can sum forces in x and y. The order doesn’t matter in this case.
Σ 𝐹 𝑥 = 𝐴 𝑥 =0
Σ 𝐹 𝑦 = 𝐴 𝑦 +𝐷−1500𝑙𝑏𝑠−1500𝑙𝑏𝑠=0
𝐴 𝑦 =1800𝑙𝑏𝑠
Should Ay be larger than D? Why? Think critically to evaluate the solution.

6. 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. A B C D
20°
20° A D Ay Ax B C
20° x y
20 ft
8 ft
12 ft

7. Sum forces in the x‐direction to find Ax
Class Problem: A man who weighs 890 N stands on the end of a diving board as he plans his dive.
Draw a FBD of the beam.
Sum forces in the x‐direction to find Ax
Sum moments about point A to find the reaction at B (By).
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.
Solution: a. 𝑦
𝑊=890𝑁
1.5m 3m
𝐴 𝑥 𝑥
𝐴 𝑦
𝐵 𝑦

8. 𝐹 𝑥 = 𝐴 𝑥 =0
Form good habits early! b.
1.5m 3m 𝑥 𝑦
𝐴 𝑥
𝐴 𝑦
𝐵 𝑦
𝑊=890𝑁
ALWAYS put “= 0” on the right hand side.
𝑀 𝐴 =−890𝑁∙4.5𝑚+ 𝐵 𝑦 ∙1.5𝑚=0 +
ALWAYS write this when summing moments, including the point that moments are being summed about. c.
𝐵 𝑦 = 890𝑁∙4.5𝑚 1.5𝑚 =2670𝑁
𝐹 𝑦 = 𝐴 𝑦 + 𝐵 𝑦 −890𝑁=0
Tip: Always assume the reactions act in a positive coordinate reaction. The signs of the answers will set things straight in the end. d.
𝐴 𝑦 +2670𝑁−890𝑁=0
𝐴 𝑦 =−2670𝑁+890𝑁
𝐴 𝑦 =−1780𝑁
Negative means it points downward.

9. Class Problem: The catapult has a horizontal beam that pivots about point P. A massive boulder (mass = 720kg ) is permanently mounted at B. Before loading a smaller rock (for launch) into the scoop at M, the rope connected between the ground and R holds the system at rest. If the weight of the machine itself (including the scoop at M) is neglected, what is the tension in the rope? 𝑥 𝑦
0.6m
2.6m
0.8m
𝑃 𝑥
𝑃 𝑦 𝑅
720𝑘𝑔
𝐵=720𝑘𝑔∙9.81 𝑚 𝑠 2
𝐵=7063.2𝑁
𝑀 𝑃 =0 +
=𝑅∙2.6𝑚−7063.2𝑁∙0.8𝑚
𝑅∙2.6𝑚=7063.2𝑁∙0.8𝑚
𝑅= 𝑁