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6.6 Frames and Machines6.6 Frames and Machines
� Composed of pin-connected multi-force members (subjected to more than two forces)
� Frames are stationary and are used to � Frames are stationary and are used to support the loads while machines contain moving parts, designated to transmit and alter the effects of forces
� Apply equations of equilibrium to each member to determine the unknown forces
6.6 Frames and Machines6.6 Frames and Machines
Free-Body Diagram� Isolate each part by drawing its outlined
shape
- show all the forces and the couple - show all the forces and the couple moments that act on the part
- label or identify each known and unknown force and couple moment with reference to the established x, y and z coordinate system
6.6 Frames and Machines6.6 Frames and Machines
Free-Body Diagram- indicate any dimension used for taking moments
- equations of equilibrium are easier to - equations of equilibrium are easier to apply when the forces are represented in their rectangular coordinates
- sense of any unknown force or moment can be assumed
6.6 Frames and Machines6.6 Frames and Machines
Free-Body Diagram� Identify all the two force members in the
structure and represent their FBD as having two equal but opposite collinear having two equal but opposite collinear forces acting at their points of application
� Forces common to any contracting member act with equal magnitudes but opposite sense on the respective members
6.6 Frames and Machines6.6 Frames and Machines
Free-Body Diagram- treat two members as a system of connected members
- these forces are internal and are not - these forces are internal and are not shown on the FBD
- if the FBD of each member is drawn, the forces are external and must be shown on the FBD
6.6 Frames and Machines6.6 Frames and Machines
Example 6.9
For the frame, draw the free-body diagram of
(a) each member, (a) each member,
(b) the pin at B and
(c) the two members
connected together.
SolutionPart (a)� members BA and BC are not two-force
members� BC is subjected to 3 forces, the resultant force
6.6 Frames and Machines6.6 Frames and Machines
� BC is subjected to 3 forces, the resultant force from pins B and C and the external P
� AB is subjected to the resultant forces from the pins at A and B and the external moment M
Solution
Part (b)
� Pin at B is subjected to two forces, force of the member BC on the pin and the force of
6.6 Frames and Machines6.6 Frames and Machines
member BC on the pin and the force of member AB on the pin
� For equilibrium, these
forces and respective
components must be
equal but opposite
Solution
Part (b)
� But Bx and By shown equal and opposite
on members AB ad BC results from the
6.6 Frames and Machines6.6 Frames and Machines
on members AB ad BC results from the
equilibrium analysis of
the pin rather from
Newton’s third law
Solution
Part (c)
� FBD of both connected members without the supporting pins at A and C
B and B are not shown since
6.6 Frames and Machines6.6 Frames and Machines
� Bx and By are not shown since
they form equal but
opposite collinear pairs
of internal forces
Solution
Part (c)
� To be consistent when applying the equilibrium equations, the unknown force components at A
6.6 Frames and Machines6.6 Frames and Machines
equations, the unknown force components at A and C must act in the same sense
� Couple moment M can be
applied at any point on
the frame to determine
reactions at A and C
6.6 Frames and Machines6.6 Frames and Machines
Example 6.10
A constant tension in the conveyor belt is
maintained by using the device. Draw the
FBD of the frame and FBD of the frame and
the cylinder which
supports the belt.
The suspended black
has a weight of W.
6.6 Frames and Machines6.6 Frames and Machines
Solution
� Idealized model of the device
� Angle θ assumed known
Tension in the belt is the same on � Tension in the belt is the same on each side of the cylinder since it is free to turn
6.6 Frames and Machines6.6 Frames and Machines
Solution
� FBD of the cylinder and the frame
� Bx and By provide equal but � Bx and By provide equal but opposite couple moments on the cylinder
� Half of the pin reactions at A act on each side of the frame since pin connections occur on each side
6.6 Frames and Machines6.6 Frames and Machines
Example 6.11
Draw the free-body diagrams of each part of
the smooth piston and link mechanism used
to crush recycled cans.to crush recycled cans.
6.6 Frames and Machines6.6 Frames and Machines
Solution
� Member AB is a two force member
� FBD of the parts� FBD of the parts
6.6 Frames and Machines6.6 Frames and Machines
Solution
� Since the pins at B and D connect only two parts together, the forces are equal but opposite on the separate FBD of their opposite on the separate FBD of their connected members
� Four components of the force act on the piston: Dx and Dy represent the effects of the pin and Nw is the resultant force of the floor and P is the resultant compressive force caused by can C
6.6 Frames and Machines6.6 Frames and Machines
Example 6.12
For the frame, draw the free-body diagrams of (a)
the entire frame including the pulleys and cords, (b)
the frame without the pulleys and cords, and (c) the frame without the pulleys and cords, and (c)
each of the pulley.
Solution
Part (a)
� Consider the entire frame, interactions at
the points where the pulleys and cords
6.6 Frames and Machines6.6 Frames and Machines
the points where the pulleys and cords
are connected to the frame
become pairs of internal
forces which cancel
each other and not
shown on the FBD
6.6 Frames and Machines6.6 Frames and Machines
Solution
Part (b) and (c)
� When cords and pulleys are removed, their effect on the frame must be their effect on the frame must be shown
6.6 Frames and Machines6.6 Frames and Machines
Example 6.13
Draw the free-body
diagrams of the bucket and
the vertical boom of the back the vertical boom of the back
hoe. The bucket and its
content has a weight W.
Neglect the weight of the
members.
6.6 Frames and Machines6.6 Frames and Machines
Solution
� Idealized model of the assembly
� Members AB, BC, BE and HI are two � Members AB, BC, BE and HI are two force members
6.6 Frames and Machines6.6 Frames and Machines
Solution
� FBD of the bucket and boom
� Pin C subjected to 2 forces, force of the link BC and force of the of the link BC and force of the boom
� Pin at B subjected to three forces, force by the hydraulic cylinder and the forces caused by the link
� These forces are related by equation of force equilibrium
6.6 Frames and Machines6.6 Frames and Machines
Equations of Equilibrium� Provided the structure is properly
supported and contains no more supports and members than necessary to prevent collapse, the unknown forces at the collapse, the unknown forces at the supports and connections can be determined from the equations of equilibrium
� The selection of the FBD for analysis are completely arbitrary and may represent each of the members of the structure, a portion or its entirety.
6.6 Frames and Machines6.6 Frames and Machines
Equations of Equilibrium
� Consider the frame in fig (a)
� Dismembering the frame in fig (b), equations of equilibrium can be usedof equilibrium can be used
� FBD of the entire frame in fig (c)
6.6 Frames and Machines6.6 Frames and Machines
Procedures for AnalysisFBD� Draw the FBD of the entire structure, a portion
or each of its members� Choice is dependent on the most direct solution � Choice is dependent on the most direct solution
to the problem� When the FBD of a group of members of a
structure is drawn, the forces at the connected parts are internal forces and are not shown
� Forces common to two members which are in contact act with equal magnitude but opposite sense on their respective FBD
6.6 Frames and Machines6.6 Frames and Machines
Procedures for AnalysisFBD
� Two force members, regardless of their shape, have equal but opposite collinear forces acting at the ends of the memberhave equal but opposite collinear forces acting at the ends of the member
� In many cases, the proper sense of the unknown force can be determined by inspection
� Otherwise, assume the sense of the unknowns
� A couple moment is a free vector and can act on any point of the FBD
6.6 Frames and Machines6.6 Frames and Machines
Procedures for AnalysisFBD� A force is a sliding vector and can act at any
point along its line of action
Equations of Equilibrium� Count the number of unknowns and compare
to the number of equilibrium equations available
� In 2D, there are 3 equilibrium equations written for each member
6.6 Frames and Machines6.6 Frames and Machines
Procedures for AnalysisEquations of Equilibrium
� Sum moments about a point that lies at the intersection of the lines of action of as many unknown forces as possibleintersection of the lines of action of as many unknown forces as possible
� If the solution of a force or couple moment magnitude is found to be negative, it means the sense of the force is the reserve of that shown on the FBD
6.6 Frames and Machines6.6 Frames and Machines
Example 6.14
Determine the horizontal and vertical
components of the force which the pin C components of the force which the pin C
exerts on member CB
of the frame.
6.6 Frames and Machines6.6 Frames and Machines
Solution
Method 1
� Identify member AB as two force member
FBD of the members AB and BC� FBD of the members AB and BC
Solution
NF
mFmN
M
AB
AB
C
7.1154
0)4(60sin)2(2000
;0
=
=−
=∑
o
6.6 Frames and Machines6.6 Frames and Machines
NC
CNN
F
NC
C
F
NF
y
y
y
x
x
x
AB
1000
0200060sin7.1154
;0
577
060cos7.1154
;0
7.1154
=
=−−
=∑↑+
=
=−
=∑→+
=
o
o
6.6 Frames and Machines6.6 Frames and Machines
Solution
Method 2
� Fail to identify member AB as two force memberforce member
Solution
Member AB
;0=∑ AM
6.6 Frames and Machines6.6 Frames and Machines
0
;0
0
;0
0)60cos3()60sin3(
;0
=−
=∑↑+
=−
=∑→+
=−
=∑
yy
y
xx
x
yx
A
BA
F
BA
F
mBmB
Moo
Solution
Member BC
mBmN
M
y
C
0)4()2(2000
;0
=−
=∑
6.6 Frames and Machines6.6 Frames and Machines
NCNCNBNB
CNB
F
CB
F
mBmN
yxxy
yy
y
xx
x
y
1000;577;577;1000
02000
;0
0
;0
0)4()2(2000
====
=+−
=∑↑+
=−
=∑→+
=−
6.6 Frames and Machines6.6 Frames and Machines
Example 6.15
The compound beam is pin connected at B.
Determine the reactions at its support. Determine the reactions at its support.
Neglect its weight and thickness.
6.6 Frames and Machines6.6 Frames and Machines
Solution
� FBD of the entire frame
� Dismember the beam into two segments since there are 4 unknowns but 3 equations since there are 4 unknowns but 3 equations of equilibrium
Solution
Segment BC
0
;0
=
=∑→+ x
B
F
6.6 Frames and Machines6.6 Frames and Machines
08
;0
0)2()1(8
;0
0
=+−
=∑↑+
=+−
=∑
=
yy
y
y
B
x
CkNB
F
mCmkN
M
B
Solution
Member AB
BkNA
F
xx
x
03
)10(
;0
=+
−
=∑→+
6.6 Frames and Machines6.6 Frames and Machines
kNCkNBBmkNMkNAkNA
BkNA
F
mBmkNM
M
BkNA
yyxAyx
yy
y
yA
A
xx
4;4;0;.32;12;6
054
)10(
;0
0)4()2(54
)10(
;0
053
)10(
======
=−
−
=∑↑+
=−
−
=∑
=+
−
6.6 Frames and Machines6.6 Frames and Machines
Example 6.16
Determine the horizontal and vertical
components of the force which the pin at C components of the force which the pin at C
exerts on member ABCD of
the frame.
6.6 Frames and Machines6.6 Frames and Machines
Solution
� Member BC is a two force member
� FBD of the entire frame
FBD of each member� FBD of each member
Solution
Entire Frame
mDmNM 0)8.2()2(981;0 =+−=∑
6.6 Frames and Machines6.6 Frames and Machines
NA
NAF
NA
NAF
ND
mDmNM
y
yy
x
xx
x
xA
981
0981;0
7.700
07.700;0
7.700
0)8.2()2(981;0
=
=−=∑↑+
=
=−=∑→+
=
=+−=∑
Solution
Member CEF
mFmNM 0)6.1)(45sin()2(981;0 =−−=∑ o
6.6 Frames and Machines6.6 Frames and Machines
NC
NNCF
NC
NCF
NF
mFmNM
y
yy
x
xx
B
BC
245
0981)45sin2.1734(;0
1226
0)45cos2.1734(;0
2.1734
0)6.1)(45sin()2(981;0
−=
=−−−=∑↑+
=
=−−−=∑→+
−=
=−−=∑
o
o
o
6.6 Frames and Machines6.6 Frames and Machines
Example 6.17
The smooth disk is pinned at D and has a weight of
20N. Neglect the weights of others member,
determine the horizontal and vertical components determine the horizontal and vertical components
of the reaction at pins B and D
6.6 Frames and Machines6.6 Frames and Machines
Solution
� FBD of the entire frame
� FBD of the members� FBD of the members
Solution
Entire Frame
cmCcmNM 0)5.3()3(20;0 =+−=∑
6.6 Frames and Machines6.6 Frames and Machines
NA
NAF
NA
NAF
NC
cmCcmNM
y
yy
x
xx
x
xA
20
020;0
1.17
01.17;0
1.17
0)5.3()3(20;0
=
=−=∑↑+
=
=−=∑→+
=
=+−=∑
Solution
Member AB
BNF 01.17;0 =−=∑→+
6.6 Frames and Machines6.6 Frames and Machines
NB
BNNF
NN
cmNcmNM
NB
BNF
y
yy
D
DA
x
xx
20
04020;0
40
0)3()6(20;0
1.17
01.17;0
=
=+−=∑↑+
=
=+−=∑
=
=−=∑→+
Solution
Disk
Fx ;0=∑→+
6.6 Frames and Machines6.6 Frames and Machines
ND
DNN
F
D
F
y
y
y
x
x
20
02040
;0
0
;0
=
=−−
=∑↑+
=
=∑→+
6.6 Frames and Machines6.6 Frames and Machines
Example 6.18
Determine the tension in the cables
and also the force P required to and also the force P required to
support the 600N force using the
frictionless pulley system.
6.6 Frames and Machines6.6 Frames and Machines
Solution
� FBD of each pulley
� Continuous cable and frictionless pulley = frictionless pulley = constant tension P
� Link connection between pulleys B and C is a two force member
Solution
Pulley A
NP
NPFy
200
06003;0
=
=−=∑↑+
6.6 Frames and Machines6.6 Frames and Machines
Pulley B
Pulley C
NR
TPRF
NT
PTF
NP
y
y
800
02;0
400
02;0
200
=
=−−=∑↑+
=
=−=∑↑+
=
6.6 Frames and Machines6.6 Frames and Machines
Example 6.19
A man having a weight of 750N supports
himself by means of the cable and
pulley system. If the seat has a pulley system. If the seat has a
weight of 75N, determine the force
he must exert on the cable at A and
the force he exerts on the seat.
Neglect the weight of the cables
and pulleys.
6.6 Frames and Machines6.6 Frames and Machines
Solution
Method 1
� FBD of the man, seat and pulley C� FBD of the man, seat and pulley C
Solution
Man
Seat
NNTF SAy 0750;0 =−+=∑↑+
6.6 Frames and Machines6.6 Frames and Machines
Seat
Pulley C
NNNTT
TTF
NNTF
EEA
AEy
SEy
200;275;550
02;0
075;0
===
=−=∑↑+
=−+=∑↑+
6.6 Frames and Machines6.6 Frames and Machines
Solution
Method 2
� FBD of the man, seat and pulley C as � FBD of the man, seat and pulley C as a single system
Solution
NT
NNTF
E
Ey
275
0750753;0
=
=−−=∑↑+
6.6 Frames and Machines6.6 Frames and Machines
NNT
TTF
NNTF
EA
AEy
SEy
200;550
02;0
075;0
==
=−=∑↑+
=−+=∑↑+
6.6 Frames and Machines6.6 Frames and Machines
Example 6.20
The hand exerts a force of 35N on the grip of the
spring compressor. Determine the force in the
spring needed to maintain equilibrium of the spring needed to maintain equilibrium of the
mechanism.
6.6 Frames and Machines6.6 Frames and Machines
Solution
� FBD for parts DC and ABG
Solution
Lever ABG
NF
mmNmmFM EAB
140
0)100(35)25(;0
=
=−=∑
6.6 Frames and Machines6.6 Frames and Machines
Pin E
NF
NFF
FFF
FFF
NF
x
EFED
EFEAy
EA
140
014060cos2;0
060sin60sin;0
140
=
=−=∑→+
==
=−=∑↑+
=
o
oo
Solution
Arm DC
M C ;0=∑
6.6 Frames and Machines6.6 Frames and Machines
NF
mmmmF
s
s
C
62.60
0)75(30cos140)150(
=
=+− o
6.6 Frames and Machines6.6 Frames and Machines
Example 6.21
The 100kg block is held in equilibrium by
means of the pulley and the continuous
cable system. If the cable is cable system. If the cable is
attached to the pin at B,
compute the forces which this
pin exerts on each of its
connecting members
6.6 Frames and Machines6.6 Frames and Machines
Solution
� FBD of each member of the frame
� Ad and CB are two force members
Solution
Pulley B
NBF 045cos5.490;0 =−=∑→+ o
6.6 Frames and Machines6.6 Frames and Machines
NB
NNBF
NB
NBF
y
yy
x
xx
3.837
05.49045sin5.490;0
8.346
045cos5.490;0
=
=−−=∑↑+
=
=−=∑→+
o
o
Solution
Pin E
NF
NNFF
CB
CBy
1660
05.4903.8375
4;0
=
=−−=∑↑+
6.6 Frames and Machines6.6 Frames and Machines
� Two force member BC subjected to bending as caused by FBC
� Better to make this member straight so that the force would only cause tension in the member
NF
NNFF
NF
AB
ABx
CB
1343
08.346)1660(5
3;0
1660
=
=−−=∑→+
=