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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
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 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.
Department of Mechanical Engineering
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
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
Department of Mechanical Engineering
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)
Department of Mechanical Engineering
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
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
Department of Mechanical Engineering
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
Department of Mechanical Engineering
Internal forces
Break apart the structure
2-force member
2-force member
Not a 2-force member
Department of Mechanical Engineering
Example2-force member
2-force member
Not a 2-force member
Department of Mechanical Engineering
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.
Department of Mechanical Engineering
Example
Draw FBD
Department of Mechanical Engineering
Example
Department of Mechanical Engineering
Example
Department of Mechanical Engineering
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
Department of Mechanical Engineering
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
Department of Mechanical Engineering
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
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
Department of Mechanical Engineering
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
Department of Mechanical Engineering
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.