Similar to trusses, frames are generally fixed, load

carrying structures.

The main difference between a frame and a truss is

that in a frame at least one member is a “multi

force member” (çoklu kuvvet elemanı).

A multi force member supports three or more forces

or at least two forces and one or more couples.

In contrast with a truss, the force or moment can be

exerted to any point on the frame member; it does

not have to be applied the joint as in the trusses.

Machines are structures which contain moving parts and are

designed to transmit input forces or couples to output forces or

couples.

Therefore a machine is an assembly of rigid and sturdy members

that are capable of generating work by means of some kind of

motion.

A mechanism is a term used to describe the physical devices which

enable the parts of a machine to conduct the intended movements.

A machine may contain several mechanisms.

Machines are designed to change forces, enhance and

amplify their magnitudes and transmit them.

Whether a machine is as simple as a hand tool or as

complex as an airplane, the main aim is to convert input

forces into output forces.

The main difference between a frame and a

machine is that although frames are rigid

structures, machines are not. Machines may be

fixed to some supporting surface or body, but

they will always consist moving parts.

The forces acting on each member of a connected system are

found by isolating the member with a FBD and applying the

equations of equilibrium.

The principle of action and reaction must be carefully observed

when we represent the forces of interaction on the separate FBDs.

It would be appropriate to identify the two forces members in the

frame, if there is any, before starting with the solution.

In order to determine the forces in a frame or machine,

it is divided into a sufficient number of members or

groups of members, but initially the support forces to

be used in the analysis must generally be determined

from the equilibrium of the whole frame.

The structure is then, dismembered and the equilibrium of

each member is considered separately.

The equilibrium equations for the several parts will be related

through the terms involving the forces of interaction.

It should be kept in mind that when going from one member to

the other, the direction of the interaction force must be

changed in accordance with Newton’s third law.

In general the FBDs of pins in the structures are not drawn;

pins are considered as a complementary part of one of the

two members it connects. It must be clearly decided which

member the pin will belong to. Whereas, the FBD of a pin

will be considered if:

It connects three or more members,

It connects a support and two or more members,

A load is directly applied to the pin.

Machines are considered as “ideal machines” when the work

output is equal to the energy input. It is impossible to build such

machines. In a real machine friction forces always generate

useless work which causes loss of energy, therefore, work output

is always less than the energy input. In other words, the

mechanical efficiency is always less than one, h<1.

inputenergy

utputo workh

Mechanical advantage is the ratio of the output force of a

machine to the input force necessary to work the machine.

This concept is totally different from the mechanical

efficiency and should not be mixed with it. Mechanical

advantage is generally greater than one.

ForceInput

ForceOutputAdvantageMechanical

Small bolt cutter operated by hand for cutting small bolts and rods is shown. For

a hand grip P=150 N. Determine the force Q developed by each jaw on the rod

to be cut.

Mechanical advantage

The forces acting on the two parts of the bolt cutter behave as mirror images of each other with respect to

x-axis. Thus, we can not have an action on one member in the +x direction and its reaction on the other

member in the –x direction. Consequently the forces at E and B have no x-components and CD is a two-

force member.

1. The clamp shown in the figure is frequently used in welding operations.

Determine the clamping force on the two metal pieces at E and the magnitudes of

the forces supported by pins A, B and D.

2. The elements of a rear suspension for a front-wheel-drive car are

shown in the figure. Determine the magnitude of the force at each joint

if the normal force F exerted on the tire has a magnitude of 3600 N.

3. Determine the force P exerted on the twig G. Note that there is a

horizontal line of symmetry for the handles, but there is no line of

symmetry for the jaws.

4. Calculate the x- and y-components of all forces acting on each member

of the loaded frame.

5. The truck shown is used to deliver food to aircraft. The elevated unit weighs

1000 kg with center of gravity at G. Determine the required force in the hydraulic

cylinder AB.

6. A basketball hoop whose rim

height is adjustable is shown.

The supporting post ABCD

weighs 400 N with the center of

gravity at point C, and

backboard-hoop assembly

weighs 220 N with the center of

gravity at point G. The height of

the rim is adjustable by means of

the screw and hand crank IJ,

where the screw is vertical. If a

person with 800 N weight hangs

on the rim, determine the support

reactions at D and the forces

supported by all members. Hint:

Member IJ is a two-force

member.

F

7. The pruning mechanism of a pole saw is shown as it cuts a branch S. For the

particular position drawn, the actuating cord is parallel to the pole and carries a

tension of 120 N. Determine the shearing force P applied to the branch by the cutter

and the total force supported by the pin at E. The force exerted by the light return

spring at C is small and may be neglected.

8. In the frame shown determine the forces acting at pins A, B and D.

9. The mechanism in the figure is used to raise the bucket of a bulldozer. The bucket and its contents

weigh 10 kN and have a center of gravity at H. Arm ABCD has a weight of 2 kN and a center of gravity

at B, arm DEFG has a weight of 1 kN and a center of gravity at E. The weights of the hydraulic cylinders

can be neglected. Determine the forces in the hydraulic cylinders CJ, BF and EI and also determine all

the forces acting at arm DEFG.

10. The linkage shown is used on a garbage truck to lift a 9000 N dumpster. Points A-G are

pins, and member ABC is horizontal. For the position shown, when the dumpster just fully

lifts off the ground, determine the force hydraulic cylinder CF. Roller at G is contact with the

dumpster.

Ax

Ay

FG

FBD of dumpster

NAFAF

NAFAF

NFFM

yGyy

xGxx

GGA

6910045sin90000

15910045cos0

225000)600()1500(90000

Ex

Ey

FCD

FG

FBD of member EDG

NEEF

NEEF

NFFM

yyy

xxx

GCDE

5.79549045sin2250045cos1125000

63640045cos11250045cos225000

1125000)300()1500(225000

Bx

By

FCF

Ex

Ey

From equilibrium of whole system;

NF

FEE

M

CF

CFyx

E

68.118724

0)450(45cos)45cos900900(450)45cos900()1950(900

0

11. A hydraulic lift-table is used to raise a 1000 kg crate. It consists of a platform and two

identical linkages on which hydraulic cylinders exert equal forces. In the figure only one

linkage and one cylinder are shown. Members EDB and CG are each of length 2a and

member AD is pinned to the midpoint of EDB. If the crate is placed on the table, so that

half of its weight is supported by the system shown, determine the force exerted by each

cylinder in raising the crate for q=60o, a=0.70 m and L=3.20 m. Show that the result

obtained is independent of the distance d.

FBD of Platform

Two-Force Members: AD, BC, CG, DH.

30o

q=60o

FAD

W/2

FC FB

A B C

BCCBCBy

ADADx

FW

FW

FFW

FFF

FFFx

220

20

000

Pulley C FC

C

FCG

30o

FBC

300300

0300

cosFFFcosFF

FsinFF

CGCCCGy

BCCGx

30o

q=60o

FBD of CG

C

G

FCG

q=60o 30

30

cos

FF

sin

FF

CCG

BCCG

FBD of BC

FCG

FBC FBC

30tanF

F

C

BC

30o

q=60o

N.F

FF.F.

.tanF.FF.

.F.F..sinF..cosFM

DH

/W

CBDH

F

CBDH

BCBDHDHE

BC

635124

070670

02113070670

021170350021261002120

2

FBD of EDB

B

E

FB

q=60o

Ex

o.sin

.

sin

.

m.DHcosEHDEEHDEDH

021260

91270

912602222

D

H

Ey

FBC

FDH

0.7 m

0.61 m

1.4 m

Cosine theorem:

Sine theorem:

12. The mechanism is designed to keep its load level while raising it. A pin on the rim of the 80 cm

diameter pulley fits in a slot on arm ABC. Arm ABC and DE are each 80 cm long and the package

being lifted weighs 80 kN. The mechanism is raised by pulling on the rope that is wrapped around the

pulley. Determine the force P applied to the rope and all the forces acting on arm ABC when the

package has been lifted 80 cm as shown.

SOLUTION

FBD of Platform Two force member: ED

69.28 mm

FBD of DE

FBD of ABC

FBD of the pulley

13. A simple folding chair comprised of two identical frames, one on each side, is shown in

the figure. The half frame shown carries half the weight of a 70 kg person. Determine all

the forces acting on member BF. Neglect the weights of the connecting elements (not

shown) and the seat, and also the friction at points A, B and F.

Detail of contact at F

Dimensions in “mm”

Geometry of the Chair

SOLUTION

Two force member: EG

Equilibrium of the whole system:

FBD of DC

GE two force member

FBD of BF