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
Department of Mechanical Engineering
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 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.
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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.
Department of Mechanical Engineering
Sample Problem
SOLUTION:
• Create a free-body diagram for the complete
frame and solve for the support reactions.
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.
• 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.
Department of Mechanical Engineering
Sample Problem 6.4
SOLUTION:
• Create a free-body diagram for the complete frame
and solve for the support reactions.
∑ Fy = 0 = Ay − 480 N
Ay = 480 N ↑
∑ M A = 0 = −(480 N )(100 mm ) + B(160 mm )
B = 300 N →
∑ Fx = 0 = B + Ax
Ax = −300 N ←
Note:
80 = 28.07°
α = tan −1 150
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.
∑ M C = 0 = (FDE sin α )(250 mm ) + (300 N )(60 mm ) + (480 N )(100 mm )
FDE = −561 N
FDE = 561 N C
• Sum of forces in the x and y directions may be used to find the force components at C.
∑ Fx = 0 = C x − FDE cos α + 300 N
0 = C x − (− 561 N ) cos α + 300 N
∑ Fy = 0 = C y − FDE sin α − 480 N
0 = C y − (− 561 N ) sin α − 480 N
C x = −795 N
C y = 216 N
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Sample Problem
• With member ACE as a free-body, check
the solution by summing moments about A.
∑ M A = (FDE cos α )(300 mm ) + (FDE sin α )(100 mm ) − C x (220 mm )
= (− 561 cos α )(300 mm ) + (− 561sin α )(100 mm ) − (− 795)(220 mm ) = 0
(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,
∑ M A = 0 = aP − bQ
Q=
a
P
b
Department of Mechanical Engineering
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:
∑M = 0
∑F = 0
∑F = 0
A
x
y

Results:
– Ax = -120.0 lb
– Ay = 75.0 lb
– C = 120.0 lb
Department of Mechanical Engineering
Internal forces
2-force member
 Break
apart the
structure
Not a 2-force member
2-force member
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Example
2-force member
Not a 2-force member
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
Department of Mechanical Engineering
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
∑M = 0
∑F = 0
∑F = 0
∑M = 0
A
x
y
A
− E x (2) − 981(5.5) = 0
E x = −2698 N
∑F
x
=0
Ax + E x = 0
Ax = −(−2698 N ) = 2698 N
∑F
y
=0
Ay + E y − 981N = 0
Department of Mechanical Engineering
Example


Consider the free-body-diagram
of the members
For the Pulley:
C x = 981N
C y = 981N
∑F
∑F
x
=0
y
=0
∑M = 0
∑M = 0
∑F = 0
A

For Bar ABC
B
x
B y (2.5) − C y (5) = 0 = 0
B y = 1962 N
B = 1962 2 + 1717 2 = 2607 N
− Ay (2.5) − C y (2.5) = 0
Ay = −981N
− Bx − C x + 2698 = 0
Bx = 1717 N
 By 
θ x = tan   = 48.81o
 Bx 
−1
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Example

For the free-body-diagram of
EDB
∑F
y
=0
E y − By = 0
E y = 1962 N

The shearing stress on a cross
section of pin at B
τ=
B
2607
=
= 5.31MPa
2
Apin π (0.025 2)
Department of Mechanical Engineering
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 2force 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
Department of Mechanical Engineering
Solving the problem

Equilibrium equations
∑ M = − P.5 + F .1.5 + F .4 = 0
∑ F = −C = 0
∑F = F + F −P = 0
C
B
x
x
y

A
B
A
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
δA
4
It can be used to relate FA to FB.
=
δB
1.5
=
δD
5
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