EGR 231 Engineering Statics:
Lecture 26: 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 the
environment and from other frame or machine parts.
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.
Homework 26:
Problem 6.50
Determine the force in member BD
and the components of the reaction at C.
60 lb
Problem 6.54
An aircraft tow bar is positioned by means of a single hydraulic cylinder CD which is
connected to two identical arm and wheel units DEF. The entire tow bar has a mass of
200 kg and its center of gravity is located at G. For the position shown, determine a) the
force exerted by the cylinder on bracket C, b) the force exerted on each arm by the pin at
E.
Problem 6.61
Knowing that P = 60 lb and Q = 90 lb, determine the components of all forces acting on
member BCDE of the assembly shown.
Problem 6.69
Knowing that each pulley has a radius of 250 mm, determine the components of the
reaction at D and E.
4.8 kN
D
10ft
Example 1:
Find the forces in the pins that at
points A, B, and C.
150 lb
9 ft
r = 1ft
B
C
E
30o
Solution:
Equilibrium of each member:
 F  0 and  M  0
A
FBy
Start by showing FBDs
FCy
FBx
x
F
y
E
FCx
0
B
r = 1ft
 FCy  285 lb
0  (10) FCy  (19)150
F
150 lb
D
Body BCE:
M
150 lb
0

0  FBx  FCx
FBx   FCx
150 lb
FCx
FAy
10tan30o
o
30
0
20tan30o+1
FCy
FAx

0  FBy  FCy  150
10 ft
FBy   FCy  150
19 ft
FBy  (285)  150  135 lb
Body ACD:
M
A
0
0  (10) FCy  (19)150  (10 tan 30o ) FCx  (20 tan 30o 1)150
0  (10)( 285)  2850  5.77 FCx 1882
F
x
y

FAx  FCx  150  176.2 lb

FAy  FCy  150  135 lb
0
0  FAy  FCy  150
So:
and FBx  326.2 lb
0
0  FAx  FCx  150
F
 FCx  326.2 lb
FA 
FAx 2  FAy 2  222.0 lb
FC 
FB 
FCx 2  FCy 2  353 lb
FBx 2  FBy 2  353 lb
Example 2:
Find the forces acting in each of the pins shown on the body
3 in
3 in
A
B
E
3.5 in
1500 lb
C
2.5 in
D
4 in
Draw FBDs
A
FB
A FA
B
FA
1500
1500 lblb
FCy
C
FCx
FB
B
EE
1500 lb
lb
1500
C
FCx
x
FCy
Equilibrium of each member:
 F  0 and  M  0
Body AB is a 2 force member:
FB
FA
FA  FB along the horizontal
FA
Body BCD:
M
C
F
x
F
y
0
 FA  1714 lb
0  (3.5) FA  (4)1500
0
0  FA  FCx
FCy
1500 lb
FCx

FCx   FA  1714 lb
0
0  FCy  1500

FCy  1500 lb
Or you could have analyzed CBE instead and found the same forces in C:
Therefore the pin forces are:
So: FA  FB  1714 lb
and
FC 
FD  FE  1500 lb
FCx 2  FCy 2  2278 lb
Workout Problem:
For the frame and loading shown, determine the components of all forces acting on
member ABD.
4kN
3kN
FAy
Solution:
Start with support reactions:
A
MA  0
B
FAx
0  625FF  750(4) 1250(3)
FF  10.8kN
F
x
0
FF
F
y
0  FAx  FF
0
3kN
4kN
F
0  FAy  4  3
FAy  7 kN
FAx   FF  10.8kN
FBy
For individual components:
FBx
FBy
FAy
FAx
FBx
FF
FDy
FCy
FDx
For Body ABC:
M
B
FCx
FDx
FCx
0
0  500 FAy  500 FDy  250 FDx
F
x
4
FDy
FCy
3
F
0
y
0
0  FAy  FBy  FDy
0  FAx  FBx  FDx
0  2 FAy  2 FDy  FDx
For Body CD:
M
C
0
F
x
F
0
y
0
0  500 FDy  250(4)  750(3)
0   FCx  FDx
0   FCy  4  FDy  3
0  2 FDy  13
0  FCx  FDx
0  FCy  FDy  7
For Body CD:
M
B
0
0  2 FCx  5FF
Solve using FAx  10.8kN
FDy  13 / 2  6.5kN
F
x
F
0
y
0  FF  FCx  FBx
FAy  7 kN
FCy  7  FDy  7  (6.5)  0.5kN
FDx  2 FAy  2 FDy  2(7)  2( 6.5)  27 kN
0
0   FBy  FCy
FF  10.8kN
FBx   FDx  FAx  (27)  (10.8)  16.2kN
FBy   FAy  FDy  7  (6.5)  0.5kN
FCx   FDx  27kN