6161103 6.6 frames and machines

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


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


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


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


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


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


� 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


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


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


� 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


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


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.


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


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


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.


Solution

� Member AB is a two force member

� FBD of the parts� FBD of the parts


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


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


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


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


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.


Solution

� Idealized model of the assembly

� Members AB, BC, BE and HI are two � Members AB, BC, BE and HI are two force members


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


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.


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)


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


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


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


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


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.


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


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


Solution

Method 2

� Fail to identify member AB as two force memberforce member

Solution

Member AB

;0=∑ AM


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

=−

=∑


NCNCNBNB

CNB

F

CB

F

mBmN

yxxy

yy

y

xx

x

y

1000;577;577;1000

02000

;0

0

;0

0)4()2(2000

====

=+−

=∑↑+

=−

=∑→+

=−


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.


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


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

=+

−

=∑→+


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(

======

=−

−

=∑↑+

=−

−

=∑

=+

−


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.


Solution

� Member BC is a two force member


FBD of each member� FBD of each member

Solution

Entire Frame

mDmNM 0)8.2()2(981;0 =+−=∑


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


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


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


Solution


� FBD of the members� FBD of the members

Solution

Entire Frame

cmCcmNM 0)5.3()3(20;0 =+−=∑


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 =−=∑→+


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=∑→+


ND

DNN

F

D

F

y

y

y

x

x

20

02040

;0

0

;0

=

=−−

=∑↑+

=

=∑→+


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.


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

=

=−=∑↑+


Pulley B

Pulley C

NR

TPRF

NT

PTF

NP

y

y

800

02;0

400

02;0

200

=

=−−=∑↑+

=

=−=∑↑+

=


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.


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 =−+=∑↑+


Seat

Pulley C

NNNTT

TTF

NNTF

EEA

AEy

SEy

200;275;550

02;0

075;0

===

=−=∑↑+

=−+=∑↑+


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

=

=−−=∑↑+


NNT

TTF

NNTF

EA

AEy

SEy

200;550

02;0

075;0

==

=−=∑↑+

=−+=∑↑+


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.


Solution

� FBD for parts DC and ABG

Solution

Lever ABG

NF

mmNmmFM EAB

140

0)100(35)25(;0

=

=−=∑


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=∑


NF

mmmmF

s

s

C

62.60

0)75(30cos140)150(

=

=+− o


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


Solution

� FBD of each member of the frame

� Ad and CB are two force members

Solution

Pulley B

NBF 045cos5.490;0 =−=∑→+ o


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

=

=−−=∑↑+


� 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

=

=−−=∑→+

=

6161103 6.6 frames and machines

Business

force members

collinear forces

force act

force member fbd

linkthese forces

theresultant forces

force of theboompin

force of themember bc