[email protected] ENGR-36_Lec-16_Frames.pptx 1 Bruce Mayer, PE Engineering-36: Engineering...

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

Bruce Mayer, PELicensed Electrical & Mechanical Engineer

[email protected]

Engineering 36

Chp 6:

Frames



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 Frames• 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.



Not Fully Rigid Frames• Some frames may collapse if removed from

their supports. Such frames can NOT be treated as rigid bodies.

• A free-body diagram of the complete frame indicates four unknown force components which can not be determined from the three equilibrium conditions.

• The frame must be considered as two distinct, but related, rigid bodies.

• With equal and opposite reactions at the contact point between members, the two free-body diagrams indicate 6 unknown force components.

• Equilibrium requirements for the two rigid bodies yield 6 independent equations.



Example: Pin & Roller Frame

Members ACE and BCD are connected by a pin at C and by the link DE. For the loading shown, determine the force(s) in link DE and the components of the force exerted by the Pin at C on member BCD.

SOLUTION PLAN• Create a free-body diagram for the

complete frame and solve for the support reactions. (won’t collapse)

• Define a free-body diagram for member BCD. The force exerted by the link DE has a known line of action but unknown magnitude (2-frc member); determined by summing moments about C.

• With the force on the link DE known, the sum of forces in the x and y directions may be used to find the force components at C.

• With member ACE as a free-body, check the solution by summing moments about A



Example: Pin & Roller Frame SOLUTION

• Create a free-body diagram for the COMPLETE FRAME and solve for the support reactions.

N 4800 yy AF N 480yA

mm 160mm 100N 4800 BM A N 300B

xx ABF 0 N 300xA

07.28tan150801

Note



Example: Pin & Roller Frame• Define a free-body diagram for member

BCD. The force exerted by the 2-force link DE has a known line of action but unknown magnitude. It is determined by summing moments about C.

N 561

mm 100N 480mm 08N 300mm 250sin0

DE

DEC

F

FM

C N 561DEF

• Use the Sum of forces in the x and y directions to find the force components at C.

N 300cosN 561 0

N 300cos0

x

DExx

C

FCF N 795xC

N 480sinN 5610

N 480sin0

y

DEyy

C

FCFN 216yC

(DE in Compression)



Example: Pin & Roller Frame

0mm 220795mm 100sin561mm 300cos561

mm 220mm 100sinmm 300cos

xDEDEA CFFM

With member ACE as a free-body, check the solution by summing moments about A



Example: Pin & Pin Frame For the Frame

Shown Below Find• The RCNs at points

A & C• The SHEAR

FORCES acting on the PINS at A & C

Draw FBDs noting that the forces at B are Equal & Opp• FBD for AB



Example: Pin & Pin Frame• FBD for Member BC For AB take ΣMB = 0

lb 0.75ft 6lbft 450

ft 6ft 3lb 1500

Ay

AyB

F

FM

lb 0.75AyF



Example: Pin & Pin Frame For AB take ΣFy = 0

• Recall FAy = 75 lbs

For AB take ΣFx = 0

• Will use Later: FBx = FAx

lb 0.75ByF

lb 75lb 75lb 150

lb 150

lb 1500

By

AyBy

ByAyy

F

FF

FFF

AxBx

BxAxx

FF

FFF

0



Example: Pin & Pin Frame For BC take ΣMC = 0

For BC take ΣFx = 0

lb 6.86

ft 46.3 lbft 150 lbft 150

lb 0.75 :Recall

60sinft 4ft 2 lbft 1500

0

Bx

Bx

By

BxBy

C

F

F

F

FF

M

lb 6.86BxF

lb 6.86

0

BxCx

CxBxx

FF

FFF

lb 6.86CxF



Example: Pin & Pin Frame For BC take ΣFy = 0

Recall from Before: FAx = FBx, and FBx = 86.6 lbs• Thus

lb 75

0

ByCy

ByCyy

FF

FFF

lb 0.75CyF

lb 6.86AxF



Example: Pin & Pin Frame Now Find SHEAR

FORCE on Pins atA & C

The Forces at A & C

Find the Shear Force Magnitude by

In this Case

Thus the connecting pins must resist a shear force of 115 lbsjFiFF

jFiFF

CyCxC

AyAxA

ˆˆ

ˆˆ

22yx FF F

lb 115lb 57lb 6.86

lb 115lb 57lb 6.86

22

22

C

A

F

F



WhiteBoard Work

Let’s WorkThis NiceProblem

Find the Forces Acting on Each of the Members, and on the Frame at Pts A & D

30cm





Bruce Mayer, PERegistered Electrical & Mechanical Engineer

[email protected]

Engineering 36

Appendix

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