Statics:Hibbeler 1 Smithfield Bridge, Pittsburgh 1882 Smithfield Other Lindenthal Projects Royal Albert Bridge at Saltash 1859 Isambard Kingdom Brunel, Engineer Saltash Brunel Balance of Forces in Equilibrium Balance of Forces in Equilibrium PULLEYS, FRAMES, AND MACHINES Compute the forces in pulleys, ropes, beam, and frame members that require multiple free-body diagrams . Compute the reactions for compound beams, frames, or similar devices. Show why it is advantageous to: Separate elements at pins and draw multiple FBDs & identify 2-force members
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Royal Albert Bridge at Saltash 1859 Balance of Forces in ...Statics:Hibbeler 1 Smithfield Bridge, Pittsburgh 1882 Smithfield Other Lindenthal Projects Royal Albert Bridge at Saltash
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Statics:Hibbeler 1
Smithfield Bridge, Pittsburgh 1882
Smithfield
Other Lindenthal Projects
Royal Albert Bridge at Saltash 1859 Isambard Kingdom Brunel, Engineer
SaltashBrunel
Balance of Forces in Equilibrium
Balance of Forces in EquilibriumPULLEYS, FRAMES, AND MACHINES
Compute the forces in pulleys, ropes, beam, and frame members that require multiple free-body diagrams.
Compute the reactions for compound beams, frames, or similar devices.
Show why it is advantageous to: Separate elements at pins and draw multiple FBDs & identify 2-force members
Statics:Hibbeler 2
Pulley Systems Pulley Systems – Mechanical Advantage
W
T
Pulley Systems – Mechanical Advantage
W
T
Free Body Diagrams of Pulley System
FBD Top
T
Reaction
T
T
FBD A
W
T T
Pulley Systems – Other Cuts and FBDs
Fb
FBD
W
TFbar
T T
Pulley Systems – Other Cuts and FBDs
Fbar T
FBD
W
T
W
Statics:Hibbeler 3
Pulleys – Method of Joints Approach Pulleys – Method of Joints Approach
Pulleys – Method of Joints ApproachAy
P
Cy
RR
R
10 lb
FBD 2
P P R
P
10 lb 10 lb
100 lbFBD 1
FBD 3
Pulleys – Method of Sections Approach
Pulleys – Method of Sections Approach
RP
R
10 lb
100 lb
10 lb
10 lb
Statics:Hibbeler 4
QUICK PROBLEM SOLVING
Given: A frame and loads as shown.
Find: The reactions that the pins exert on the frame at A, B, and C.
Plan:
a) Draw a FBD of members AB and BC.
b) Apply the equations of equilibrium to each FBD to solve for the six unknowns.
QUICK PROBLEM SOLVING (continued)
FBDs of members AB and BC:
BY
BB
X
0.4m
500NCA X
A
B
BY
BX
1000N
45º
+ MA = BX (0.4) + BY (0.4) – 1000 (0.2) = 0
+ MC = -BX (0.4) + BY (0.6) + 500 (0.4) = 0
BY = 0 and BX = 500 N
Summing moments about A and C on each member, we get:
0.2m 0.4m
CY
AY
0.2m 0.2m
FBDs of members AB and BC:
BY
BB X
0.4m
500NC
0.2m 0.4m
C
A XA
B
BY
BX
1000N
45º
0.2m 0.2m
QUICK PROBLEM SOLVING (continued)
+ FX = AX – 500 = 0 ; AX = 500 N
+ FY = AY – 1000 = 0 ; AY = 1,000 N
For FBD of BC:
+ FX = 500 – CX = 0 ; CX = 500 N
+ FY = CY – 500 = 0 ; CY = 500 N
For FBD AB: CY
AY
WORKING EXAMPLE
Given: The wall crane supports an external load of 700 lb.
Find: The force in the cable at the winch motor W and the horizontal and vertical components of the pin reactions at A, B, C, and D.
Plan:
a) Draw FBDs of the frame’s members and pulleys.
b) Apply the equations of equilibrium and solve for the unknowns.