### ENGR-36_Lec-16_Frames_H13e

```Engineering 36
Chp 6:
Frames
Bruce Mayer, PE
[email protected]
Engineering-36: Engineering Mechanics - Statics
1
Bruce Mayer, PE
[email protected] • ENGR-36_Lec-16_Frames.pptx
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.
Engineering-36: Engineering Mechanics - Statics
2
Bruce Mayer, PE
[email protected] • ENGR-36_Lec-16_Frames.pptx
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.
Engineering-36: Engineering Mechanics - Statics
3
Bruce Mayer, PE
[email protected] • ENGR-36_Lec-16_Frames.pptx
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.
Engineering-36: Engineering Mechanics - Statics
4
Bruce Mayer, PE
[email protected] • ENGR-36_Lec-16_Frames.pptx
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
 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
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
C on member BCD.
Engineering-36: Engineering Mechanics - Statics
5
Bruce Mayer, PE
[email protected] • ENGR-36_Lec-16_Frames.pptx
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
Engineering-36: Engineering Mechanics - Statics
6
1 80
150
 28 . 07 
Bruce Mayer, PE
[email protected] • ENGR-36_Lec-16_Frames.pptx
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
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
Engineering-36: Engineering Mechanics - Statics
7
C x   795 N
C y  216 N
Bruce Mayer, PE
[email protected] • ENGR-36_Lec-16_Frames.pptx
Example: Pin & Roller Frame
 With member ACE as a freebody, check the solution by
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

Engineering-36: Engineering Mechanics - Statics
8
Bruce Mayer, PE
[email protected] • ENGR-36_Lec-16_Frames.pptx
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
Engineering-36: Engineering Mechanics - Statics
9
Bruce Mayer, PE
[email protected] • ENGR-36_Lec-16_Frames.pptx
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
Engineering-36: Engineering Mechanics - Statics
10
Bruce Mayer, PE
[email protected] • ENGR-36_Lec-16_Frames.pptx
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
Engineering-36: Engineering Mechanics - Statics
11
Bruce Mayer, PE
[email protected] • ENGR-36_Lec-16_Frames.pptx
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
Engineering-36: Engineering Mechanics - Statics
12
Bruce Mayer, PE
[email protected] • ENGR-36_Lec-16_Frames.pptx
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
Engineering-36: Engineering Mechanics - Statics
13
Bruce Mayer, PE
[email protected] • ENGR-36_Lec-16_Frames.pptx
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
Engineering-36: Engineering Mechanics - Statics
14
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
[email protected] • ENGR-36_Lec-16_Frames.pptx
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
Engineering-36: Engineering Mechanics - Statics
15
Bruce Mayer, PE
[email protected] • ENGR-36_Lec-16_Frames.pptx
Engineering-36: Engineering Mechanics - Statics
16
Bruce Mayer, PE
[email protected] • ENGR-36_Lec-16_Frames.pptx
Engineering 36
Appendix
Bruce Mayer, PE
Registered Electrical & Mechanical Engineer
[email protected]
Engineering-36: Engineering Mechanics - Statics
17
Bruce Mayer, PE
[email protected] • ENGR-36_Lec-16_Frames.pptx
```