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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

TT

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:

BYB

BX

0.4m500N

CA 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 0.4mCY

AY

0.2m 0.2m

FBDs of members AB and BC:BYBB X

0.4m500N

C

0.2m 0.4mC

A X A

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: CYAY

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.

Statics:Hibbeler 5

FBD of the Pulley E

T T

E

EXAMPLE (continued)

+↑ Fy = 2 T – 700 lb = 0

T = 350 lb

700 lb

Equations of Equilibrium:

+Fx = Cx – 350 = 0 Cx = 350 lbFBD of pulley C

C

350 lb

CY

CX

350 lb

+↑ Fy = Cy – 350 = 0 Cy = 350 lb +FX = – Bx + 350 – 350 sin 30° = 0

Bx = 175 lb

+↑ Fy = By – 350 cos 30° = 0By = 303.1 lbFBD of pulley B

BY

BX30°

350 lb

350 lb

B

Note that member BD is a two-force member

FBD of member ABC

AX

AY

A 45°

TBDB

175 lb303.11 lb 700 lb

350 lb

4 ft 4 ft

+MA = TBD sin 45° (4 ft) – 303.1 (4 ft) – 700 (8 ft) = 0

TBD = 2409 lb+FY = AY + 2409 sin 45° – 303.1 – 700 = 0

AY = – 700 lb +FX = AX – 2409 cos 45° + 175 – 350 = 0

AX = 1880 lb

EXAMPLE (continued)

A FBD of member BD

45

2409 lb

B

D

At D, the x and y component are

+ DX = –2409 cos 45° = –1700 lb

+ DY = 2409 sin 45° = 1700 lb

45°

B

2409 lb

Why are trusses like donuts?

Donuts are not good in compression eitherAfter the 5th one you feel kinda sick

Both = fun!You can’t drive to school without them

If they fall to the ground they are not any good anymore

Analysis of Machines

Statics:Hibbeler 6

Pliers, Cutters, and GripsFind vertical clamping force at E

APPLICATIONS

Frames are commonly used to support external loads.

How is a frame different than a t ?truss?

How can you determine the forces at the joints and supports of a frame?

Find all reaction forces and forces at pinned connections in press

Statics:Hibbeler 7

Elevation View of Cairo Bridge

(Source: S. Nam, University of Illinois)

2D Schematic of Cairo Bridge

(Source: S. Nam, University of Illinois)

View of Finite Element Model

-1

-0.5

0

0.5

1

0 10 20 30 40

Time (sec)

Acc

. (g)

Statics:Hibbeler 8

Animation of Bridge Deflection

(Source: S. Nam, University of Illinois)BACK

Graph of Maximum Deflection

Vertical Displacement

0

2

4

6

8

10

0 500 1000 1500 2000 2500 3000 3500 4000

bridge cood. (ft)

TEDA Stadium Tianjin, China