Transcript ENGR-36_Lec-16_Frames_H13e

Engineering 36
Chp 6:
Frames
Bruce Mayer, PE
Licensed Electrical & Mechanical Engineer
[email protected]
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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.
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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 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.
• Forces between connected components are equal,
have the same line of action, and opposite sense.
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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.
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Example: Pin & Roller Frame
 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.
 Members ACE and BCD are
• With the force on the link DE known,
connected by a pin at C and
the sum of forces in the x and y
by the link DE. For the
directions may be used to find the force
loading shown, determine
components at C.
the force(s) in link DE and
• With member ACE as a free-body,
the components of the
check the solution by summing
force exerted by the Pin at
moments about A
C on member BCD.
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Example: Pin & Roller Frame
 SOLUTION
• Create a free-body diagram for the
COMPLETE FRAME and solve for the
support reactions.
F
M
 0  A y  480 N
y
A y  480 N 
 0    480 N 100 mm   B 160 mm
A

B  300 N 
F
x
A x   300 N 
 0  B  Ax
 Note
  tan
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1 80
150
 28 . 07 
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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.
M
C
 0    F DE sin   250 mm   300 N 8 0 mm    480 N 100 mm
F DE   561 N
(DE in Compression)

F DE  561 N C
• Use the Sum of forces in the x and y directions to find the force
components at C.
 F x  0  C x  F DE cos   300 N
0  C x    561 N  cos   300 N
 F y  0  C y  F DE sin   480 N
0  C y    561 N  sin   480 N
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C x   795 N
C y  216 N
Bruce Mayer, PE
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Example: Pin & Roller Frame
 With member ACE as a freebody, check the solution by
summing moments about A
M
A
  F DE cos  300 mm    F DE sin  100 mm   C x  220 mm
   561 cos  300 mm     561 sin  100 mm     795

 220

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mm
 0
Example: Pin & Pin Frame
 For the Frame
Shown Below Find
• The RCNs at points
A&C
 Draw FBDs noting
that the forces at B
are Equal & Opp
• FBD for AB
• The SHEAR
FORCES acting on
the PINS at A & C
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Example: Pin & Pin Frame
• FBD for Member BC
 For AB take ΣMB = 0
M
B
 0  150 lb 3 ft   F Ay 6 ft 
 F Ay   450 ft  lb  6 ft   75 . 0 lb
F Ay  75 . 0 lb
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Example: Pin & Pin Frame
 For AB take ΣFy = 0
• Recall FAy = 75 lbs
 For AB take ΣFx = 0
F
x
 0  F Ax  F Bx
 F Bx  F Ax
• Will use Later:
FBx = FAx
F
y
 0  F Ay  F By  150 lb
 F By  150 lb   F Ay
F By  150 lb   75 lb   75 lb
F By  75 . 0 lb
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Example: Pin & Pin Frame
 For BC take ΣMC =
0
M
C
0
0  150 ft  lb  F By  2 ft   F Bx 4 ft  sin 60  
Recall : F By  75 . 0 lb
 F Bx  150 ft  lb  150 ft  lb
 3 . 46 ft 
 F Bx  86 . 6 lb
F Bx  86 . 6 lb
 For BC take ΣFx = 0
F
x
 0  F Bx  FCx
 FCx  F Bx  86 . 6 lb
FCx  86 . 6 lb
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Bruce Mayer, PE
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Example: Pin & Pin Frame
 For BC take ΣFy = 0
F
y
 0  FCy  F By
 FCy  F By  75 lb
FCy  75 . 0 lb
 Recall from Before:
FAx = FBx, and
FBx = 86.6 lbs
• Thus
F Ax  86 . 6 lb
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Example: Pin & Pin Frame
 Now Find SHEAR
FORCE on Pins at
A&C
 Find the Shear Force
Magnitude by
F 
F x F
2
2
y
 In this Case
 The Forces at A & C
F
A
 F Ax iˆ  F Ay ˆj
F C  FCx iˆ  FCy ˆj
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FA 
86 . 6 lb 2  7 5 lb 2
 115 lb
FC 
86 . 6 lb 2  7 5 lb 2
 115 lb
 Thus the connecting
pins must resist a
shear force of 115 lbs
Bruce Mayer, PE
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WhiteBoard Work
Let’s Work
This Nice
Problem
30cm
 Find the Forces Acting on Each of the
Members, and on the Frame at Pts A & D
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Engineering 36
Appendix
Bruce Mayer, PE
Registered Electrical & Mechanical Engineer
[email protected]
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Bruce Mayer, PE
[email protected] • ENGR-36_Lec-16_Frames.pptx