Department of Mechanical Engineering

Frames and Machines

Structure that has at least one non 2-force member Frames

– The members cannot be moved relative to each other rigid structure

Machines– The members can be moved relative to each other non

rigid structure The analysis method for both is similar:

– Take apart the members and perform analysis individually on each member

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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 and 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 between connected components are equal, have the same line of action, and opposite sense.

• 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.

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Frames Which Cease To Be Rigid When Detached From Their Supports

• 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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Members ACE and BCD are connected by a pin at C and by the link DE. For the loading shown, determine the force in link DE and the components of the force exerted at C on member BCD.

SOLUTION:

• Create a free-body diagram for the complete frame and solve for the support reactions.

• Define a free-body diagram for member BCD. The force exerted by the link DE has a known line of action but unknown magnitude. It is determined by summing moments about C.

• With the force on the link DE known, the sum of forces in the x and y directions may be used to find the force components at C.

• With member ACE as a free-body, check the solution by summing moments about A.

Sample Problem

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Sample Problem 6.4SOLUTION:

• Create a free-body diagram for the complete frame and solve for the support reactions.

N 4800 −==∑ yy AF ↑= N 480yA

( )( ) ( )mm 160mm 100N 4800 BM A +−==∑→= N 300B

xx ABF +==∑ 0 ←−= N 300xA

°== − 07.28tan 150801α

Note:

Department of Mechanical Engineering

Sample Problem• Define a free-body diagram for member

BCD. The force exerted by the link DE has a known line of action but unknown magnitude. It is determined by summing moments about C.

( )( ) ( )( ) ( )( )N 561

mm 100N 480mm 06N 300mm 250sin0−=

++==∑DE

DECF

FM α

CFDE N 561=• Sum of forces in the x and y directions may be used to find the force components at C.

( ) N 300cosN 561 0 N 300cos0

+−−=+−==∑

αα

x

DExxC

FCF

N 795−=xC

( ) N 480sinN 5610

N 480sin0

−−−=

−−==∑α

α

y

DEyy

C

FCF

N 216=yC

Department of Mechanical Engineering

Sample Problem

• With member ACE as a free-body, check the solution by summing moments about A.

( )( ) ( )( ) ( )( )( ) ( )( ) ( )( ) 0mm 220795mm 100sin561mm 300cos561

mm 220mm 100sinmm 300cos=−−−+−=

−+=∑αααα xDEDEA CFFM

(checks)

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Machines• Machines are structures designed to transmit and

modify forces. Their main purpose is to transform input forces into output forces.

• Given the magnitude of P, determine the magnitude of Q.

• Create a free-body diagram of the complete machine, including the reaction that the wire exerts.

• The machine is a nonrigid structure. Use one of the components as a free-body.

• Taking moments about A,

PbaQbQaPM A =−==∑ 0

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Example

Given:– A frame– The weight = 75 lb– Properties of the members– Type of the pins

Question:– Internal forces on all of the

members– Shearing stress on pin B– Change of length of

member AC

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Determine the reactions

Support reactions (after FBD) – Ax– Ay– C

Equations to be used:

Results:– Ax = -120.0 lb– Ay = 75.0 lb– C = 120.0 lb

0

0

0

=

=

=

∑∑∑

y

x

A

F

F

M

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Internal forces

Break apart the structure

2-force member

2-force member

Not a 2-force member

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Example2-force member

2-force member

Not a 2-force member

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Example

6-25 A lever is loaded and supported as shown. Determine(a) The reactions at A and C.(b) The normal stress in the

½-in.-diameter rod CD.(c) The shearing stress in the

½-in.-diameter pin at A, which is in double shear.

(d) The change in length of rod CD, which is made of a material with a modulus of elasticity of 30(106) psi.

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Example

Draw FBD

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Example

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Example

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Example The hoist pulley structure of Figure

is rigidly attached to the wall at C. A load of sand hangs from the cable that passes around the 1-ft-diameter, frictionless pulley at D. The weight of the sand can be treated as a triangular distributed load with a maximum intensity of 70 lb/ft. Determine– (a) All forces acting on member

ABC.– (b) The shearing stress on the

cross section of the ½-in.-diameter pin D, which is in double shear.

– (c) The change in length of the ¼ × 1-in. member BE [E = 29(106) psi].

Department of Mechanical Engineering

Example The frictionless pulley and

frame structure of Figure is used to support a 100-kg mass m. Determine(a) Reactions at the supports

A and E.(b) The force exerted on bar

ABC by the pin at B.(c) The shearing stress in the

25mm –diameter pin at B, which is in single shear.

Department of Mechanical Engineering

Department of Mechanical Engineering

Example Consider the free-body-diagram

of the entire from

NNAEA

F

x

xx

x

2698)2698(0

0

=−−==+

=∑NE

EM

x

x

A

26980)5.5(981)2(

0

−==−−

=∑0

0

0

=

=

=

∑∑∑

y

x

A

F

F

M

0981

0

=−+

=∑NEA

F

yy

y

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Example Consider the free-body-diagram

of the members

For the Pulley:

For Bar ABC

02698

0)5.2()5.2(

00)5()5.2(

=+−−

=−−

==−

xx

yy

yy

CBCA

CB

NCNC

y

x

981981

==

0

0

=

=

∑∑

y

x

F

F

NBNANB

x

y

y

1717

981

1962

=

−=

=

0

0

0

=

=

=

∑∑∑

x

B

A

F

M

M

o

x

yx B

BNB

81.48tan

260717171962

1

22

=

=

=+=

−θ

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Example For the free-body-diagram of

EDB

NEBE

F

y

yy

y

1962

0

0

=

=−

=∑

The shearing stress on a cross section of pin at B

MPaAB

pin

31.5)2025.0(

26072 ===

πτ

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Structurally Indeterminate problems

Number of unknowns > number of balance equations– 3 in 2D and 6 in 3D problems

Must employ the compatibility condition of the structure– Usually coming from the displacements

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Example Given

– P = 150 kN Note:

– Members A and B are 2-force member

– C is a pin– After FBD, 4 unknowns

Determine– Reactions at C– Forces on members A and

B Assumptions:

– Member CD is rigid (does not bend) it rotates

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Solving the problem

Equilibrium equations

Compatibility condition– Assume:

» the member CD rotates» Pins B, A, and D move in a

circular arc» Small displacement B, A, and

D in straight motion» Use similar triangle rule to

determine the displacement ratio

0

0

04.5.1.5.

=−+=

=−=

=++−=

∑∑∑

PFFF

CF

FFPM

ABy

xx

ABC

55.14DBA δδδ

==

It can be used to relate FA to FB.