EGR 231 Engineering Statics:
Lecture 27: Frames and Machines
Spring 2014
Today:
Structures composed of multi-force members.
Frames: Structures designed to support loads
Machines: structures designed to transmit force or power.
Steps To Analyze a Frame or Machine:
1) Disassemble frame or machine into simplest parts. Isolate each part from
environment and from each other.
2) Show and label all forces that act on each part.
3) Identify any 2 force members
4) Common forces must act with equal magnitude,
but opposite direction. Make sure
these are shown correctly on all FBDs.
5) Apply rigid body equilibrium to each part
and solve for unknowns.
Prob. 6:79,110,116,124
Homework 27:
Problem 6.79
For the frame and loading shown, determine the
Components of all forces acting on member ABD.
Problem 6.110
Two rods are connected by a frictionless collar B.
Knowing that the magnitude of the couple MA is 500 lb-in,
determine a) the couple MC required for equilibrium
b) the corresponding components of the reaction at C.
MA
MC
Problem 6.117
Determine the magnitude of the gripping force produced
when two 60 lb forces are applied as shown.
Problem 6.124
A 500 lb concrete slab is supported by a chain and sling attached to the bucket of the
front end loader shown. The action of the bucket is controlled by two identical
mechanisms, only one of which is shown. Knowing that the mechanism shown supports
one-half of the 500 kg slab, determine the force exerted a) by cylinder CD, b) by
cylinder FH.
Example 1:
Break the machine down into its individual parts, draw the forces on the FBDs, and find
the reactions at each of the pins.
Solution:
1.5 in
Equilibrium of each member:
 F  0 and  M  0
0.75 in
B
A
0.5 in
Start by showing FBDs
P
C
2 in
2.136
2
0.75
D
2 in
10 lb
10 lb
FBy
P
FAy
FBx
FAx
Then apply equilibrium to
each body
FC
FC
FAx
FD
FBx
FBy
P
FAy
FD
10 lb
10 lb
Notice that CD is a 2 force member:
FD  FC
FBy
Start with body BC:
M
B
0
FBx
 0.75

0  0.5 
FC   4.5(10lb)
 2.136 
FC  256.3 lb
F
x
0
0  FBx 
F
y
FC
0.75
FC  10
2.136

FBx  80 lb
0
0  FBy 
10 lb
2
FC
2.136

FBy   240 lb
FAx
Next for Body AB:
FBx
FBy
MA  0
P
FAy
0  (1.5) P  (0.75) FBy
(0.75)( 240)
P
 120 lb
1.5
F
x
0
0   FAx  FBx
F
y
FAx   FBx  80 lb
0
0  P  FAy  FBy
So:

FA 

FAy  P  FBy  120  (240)  360 lb
FAx  FAy  369 lb
FB 
FBx  FBy  253 lb
FC  FD  256.3 lb
2 ft
4 ft
Example 2:
Break the machine down into its
individual parts, draw the forces
on the FBDs, and find the reactions
at each of the pins and supports.
1 ft
C
200lb/ft
D
B
3 ft
Free Body Diagrams:
A
FDy
1000 lb
FB
FC
FDx
FAy
FB
FAx
Start with body AB:
M
A
0
0  2.5(1000)  5( FB )
FB  500 lb
For body BCD:
M
D
0
4
0  (1) FC  (3) FB
5
FC  1200 lb
F
x
0
F
y
0
3
3
0  FAx  1000  FB
5
5
3
3
FAx   1000  (500)
5
5
FAx  300 lb
4
4
0  FAy  1000  FB
5
5
4
4
FAy  (1000)  (500)
5
5
FAy  400 lb
F
F
x
0
3
0  FDx  FB
5
3
FDx   (500)
5
FDx  300 lb
y
0
4
0  FDy  FB  FC
5
4
FDy  (500)  (1200)
5
FDy  800 lb
A
2 ft
Example 3:
Break the machine down into
its individual parts, draw the
forces on the FBDs, and find
3 ft
the reactions at each of the
wheels. (Assume there are four
wheels.)
3 ft
B
3 ft
800 lb
E
G
D
Geometry:
Before setting up the equations
for equilibrium, the geometry
needs to be determined first.
6 ft
1 ft
A
LAD  3
LAB  2
LBG  3
LDG  1
LBH  3
LGE  6
ε
β
2 ft
B
3 ft
α
D
H
3 ft
then
LAG  12  32  3.162
3
  arctan( )  71.56o
1
from law of cosines
A2  B 2  C 2
cos  
2 AB
and
3 ft
θ
1 ft G
E
ρ
6 ft
 32  3.1622  22 
o
  37.76
 2(3.162)(3) 
   arccos 
 22  3.1622  32 
o
  arccos 
  66.72
2(3.162)(2)


then
  180o      180o  71.56o  37.76o  70.68o
and
      71.56o  66.72o  4.84o
Draw FBDs
FB
Body BG:
FAy
Note that body BG is a 2 force member:
FAx
FB  FG
4.84o
FG
Body AB
MA  0
FB
70.86o
800 lb
0  (5cos 4.84o )(800 lb)  (2 cos 4.84o )( FB sin 70.86o )  (2sin 4.84o )( FB cos 70.86o )
(5cos 4.84o )(800 lb)
FB 
2cos(4.84o )sin(70.86o )  2sin(4.84o ) cos(70.86o )
F
x
and FG = 2057 lb
0
0  FAx  FB cos 70.86 o
F
y
= 2057 lb

FAx   FB cos 70.86o  674.3 lb
0
0  FAy  FB sin 70.86o  800
 FAy  FB sin 70.86o  800  1143 lb
Body ADE:
MA  0
FAy
FAx
0  (7)( FE )  (3)( FG cos 70.86o )  (1)( FG sin 70.86o )
(3)(2057 cos 70.86o )  (1)(2057sin 70.86o )
FE 
7
= 567 lb
F
y
0
FD
FG
0   FAy  FD  FE  FG sin 70.86o
FD  FAy  FE  FG sin 70.86o
FD  (1143)  567  (2057)sin 70.86 o  233 lb
The forces are shown for two wheels but in actual device these forces will be
split over two wheel on each side.
FD _ per _ wheel  233 / 2  117 lb
FE _ per _ wheel  567 / 2  284 lb
FE
Workout Problem
Problem 6.149
The telescoping arm ABC is used to provide
an elevated platform for construction workers.
The workers and the platform together have a
mass of 240 kg and have a combined center of
gravity located directly above C. For the position
shown θ = 24 degrees, determine a) the force
exerted at B by the hydraulic cylinder BD,
b) the force exerted on the supporting carriage at A.