[email protected] ENGR-36_Lec-16_Frames.pptx 1 Bruce Mayer, PE Engineering-36: Engineering...
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Transcript of [email protected] ENGR-36_Lec-16_Frames.pptx 1 Bruce Mayer, PE Engineering-36: Engineering...
[email protected] • ENGR-36_Lec-16_Frames.pptx1
Bruce Mayer, PE Engineering-36: Engineering Mechanics - Statics
Bruce Mayer, PELicensed Electrical & Mechanical Engineer
Engineering 36
Chp 6:
Frames
[email protected] • ENGR-36_Lec-16_Frames.pptx2
Bruce Mayer, PE Engineering-36: Engineering Mechanics - Statics
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.
[email protected] • ENGR-36_Lec-16_Frames.pptx3
Bruce Mayer, PE Engineering-36: Engineering Mechanics - Statics
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.
[email protected] • ENGR-36_Lec-16_Frames.pptx4
Bruce Mayer, PE Engineering-36: Engineering Mechanics - Statics
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.
[email protected] • ENGR-36_Lec-16_Frames.pptx5
Bruce Mayer, PE Engineering-36: Engineering Mechanics - Statics
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
[email protected] • ENGR-36_Lec-16_Frames.pptx6
Bruce Mayer, PE Engineering-36: Engineering Mechanics - Statics
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
[email protected] • ENGR-36_Lec-16_Frames.pptx7
Bruce Mayer, PE Engineering-36: Engineering Mechanics - Statics
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)
[email protected] • ENGR-36_Lec-16_Frames.pptx8
Bruce Mayer, PE Engineering-36: Engineering Mechanics - Statics
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
[email protected] • ENGR-36_Lec-16_Frames.pptx9
Bruce Mayer, PE Engineering-36: Engineering Mechanics - Statics
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
[email protected] • ENGR-36_Lec-16_Frames.pptx10
Bruce Mayer, PE Engineering-36: Engineering Mechanics - Statics
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
[email protected] • ENGR-36_Lec-16_Frames.pptx11
Bruce Mayer, PE Engineering-36: Engineering Mechanics - Statics
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
[email protected] • ENGR-36_Lec-16_Frames.pptx12
Bruce Mayer, PE Engineering-36: Engineering Mechanics - Statics
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
[email protected] • ENGR-36_Lec-16_Frames.pptx13
Bruce Mayer, PE Engineering-36: Engineering Mechanics - Statics
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
[email protected] • ENGR-36_Lec-16_Frames.pptx14
Bruce Mayer, PE Engineering-36: Engineering Mechanics - Statics
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
[email protected] • ENGR-36_Lec-16_Frames.pptx15
Bruce Mayer, PE Engineering-36: Engineering Mechanics - Statics
WhiteBoard Work
Let’s WorkThis NiceProblem
Find the Forces Acting on Each of the Members, and on the Frame at Pts A & D
30cm
[email protected] • ENGR-36_Lec-16_Frames.pptx16
Bruce Mayer, PE Engineering-36: Engineering Mechanics - Statics
[email protected] • ENGR-36_Lec-16_Frames.pptx17
Bruce Mayer, PE Engineering-36: Engineering Mechanics - Statics
Bruce Mayer, PERegistered Electrical & Mechanical Engineer
Engineering 36
Appendix