[email protected] ENGR-36_Lec-17_Frames.pptx 1 Bruce Mayer, PE Engineering-36: Engineering...

[email protected] • ENGR-36_Lec-17_Frames.pptx1

Bruce Mayer, PE Engineering-36: Engineering Mechanics - Statics

Bruce Mayer, PELicensed Electrical & Mechanical Engineer

[email protected]

Engineering 36

Chp 6:

Machines



Introduction: MultiPiece Structures• For the equilibrium of structures made of several

connected parts, the internal forces as well the external forces are considered.

• In the interaction between connected parts, Newton’s 3rd Law states that the forces of action and reaction between bodies in contact have the same magnitude, same line of action, and opposite sense.

• Three categories of engineering structures are considered:

• Frames: contain at least one multi-force member, i.e., a member acted upon by 3 or more forces.

• Trusses: formed from two-force members, i.e., straight members with end point connections

• Machines: structures containing moving parts designed to transmit and modify forces.



Analysis of Machines• Frames and Machines are structures with at least one

multiforce member. Frames are designed to support loads and are usually stationary. Machines contain moving parts and are designed to transmit & modify forces.

• A free body diagram of the complete frame is used to determine the external forces acting on the frame.

• Internal forces are determined by dismembering the frame and creating free-body diagrams for each component.

• Forces between connected components are equal, have the same line of action, and opposite sense.

• Forces on two force members have known lines of action but unknown magnitude and sense.

• Forces on multiforce members have unknown magnitude and line of action. They must be represented with two unknown components.



Pin Equilibrium Consider Actions of

Members AD & CF on Pin-C

Member CF• Pulls Pin RIGHT• Pushes Pin UP

Member AD• Pulls Pin LEFT• Pushes Pin Down

FBD

xAD xCF

yAD

yCF



Pin Equilibrium Note that Pin-C is in

Equilibrium

• By ΣFx = ΣFy = 0

The forces on the MEMBERS caused by the Pin are Equal and Opposite as Predicted by Newton’s 3rd Law

yy

xx

CFAD

CFAD

xCF

yAD

yCF

xAD

http://van.hep.uiuc.edu/van/demos/Action-Reaction%20and%20Fire-Extinguisher%20Cart/TamaraPushingOnDan3.jpg



Machines• Machines are structures designed to transmit

and modify forces. Their main purpose is to transform input forces into output forces.

• Given the magnitude of P, determine the magnitude of Q.

• Create a free-body diagram of the complete machine, including the reaction that the wire exerts.

• The machine is a nonrigid structure. Use one of the components as a free-body.

• Taking moments about A,

Pb

aQbQaPM A 0



Example: CenterPull Brake For the Center-Pull

Bicycle Brake shown at right find Normal Force Exerted by the BrakePad on the Rim• Neglect the brake-

opening Spring Force



Example: CenterPull Brake There is a Single

cable connecting Pts B, G, & C• Thus TGB = TGC

The FBD for the Cable Connector

The Forces at the connector are CONCURRENT → Particle Applies

Take the ΣFy = 0

• Note:727.6

7433.06.727cm

cm 5

Hyp

Oppsin

0 N 200 BGCyBGCy TT

sinBGCBGCy TT



Example: CenterPull Brake The ΣFy = 0 at G

Use Trig & Geometry to find TBCG

The FBD for Brake Arm BAE

N 1002N 200

0 N 2002

0 N 200

BGCy

BGCy

BGCyBGCy

T

T

TT

N 5.1345

727.6N 100

sin

BGC

BGC

BGCyBGC

T

T

TT



Example: CenterPull Brake Find Fnormal by

ΣMA = 0• Note

Use F∙d to find Moments about A

N 90272.6

5.4N 5.134

cos

BGCx

BGCx

BGCBGCx

T

T

TT

lbs) (15.7 N 70

N9

630

cm 9

cmN018450

cm 9cm 2cm 5.4

normal

normal

normalBGCxBGCy

F

F

FTT



Example: CenterPull Brake

70 N



WhiteBoard Work

Let’s WorkThis NiceProblem

For the Cutters with Applied Hand Force, P, Find the Force Applied to the Gripped Bolt

Big Ol’ Bolt Cutter



Pliers (4-pc Structure) AB is TWO-Force Member



Double Toggle Machine



Bruce Mayer, PERegistered Electrical & Mechanical Engineer

[email protected]

Engineering 36

Appendix



Friction-Filled Bearing Consider Non-

Frictionless Bearing From the Diagram

• r ≡ Brg Radius• F ≡ Brg Support

Force as determined by Static Analysis

If the Brg has non-negligible Friction the Moment that OPPOSES rotation

O

r

F FrMO



Friction-Filled Bearing The Rotation

Resisting Moment:

Where• µ ≡ the Bearing’s

COEFFICIENT of FRICTION

µ Discussed in Detail in Chp08

FrMO

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