91.111 MechanicsStatics - PowerPoint PPT Presentation

Title: 91.111 MechanicsStatics


1
Lecture 18
FRAMES and MACHINES (continued)
2
Example

• A simplified sketch of the mechanism is used to
  rise the bucket of a bulldozer. The bucket and
  its contents weight 10 kN and have a center of
  gravity at H. Arm ABCD has a weight of 2 kN and a
  center of gravity at B.
• Determine the forces in EI and BF and the
  forces acting on DEFG and ABCD.

D
1 kN
E
C
P
F
H
B
G
10 kN
2 kN
A
Lecture 18
3
Example

• Solution
• 1) Draw a FBD of ABCDEFGHI. Solve for the
• reaction at A and force P.
• 2) Draw a FBD of the bucket GHI. Write the
• equations of statics.
• 3) Draw a FBD of arm DEFG and ABCD. Write the
• equations of statics.
• 4) Solve for the forces.

Lecture 18
4
Example Draw FBD-1 of ABCDEFGHI
0.6 m
D
0.6 m
0.9 m
E
I
0.6 m
C
0.9 m
1 kN
P
H
F
y
0.9 m
G
B
FBD -1
10 kN
2 kN
A
o
1.35 m
0.9 m
60
Ax
x
0.3 m
Ay
o
60
Lecture 18
5
Example Geometry of Member BF
D
1.039
1.559
F
o
19.1
0.520
B
0.900
0.600
Lecture 18
6
Example Equation of Statics
?Fx 0
0.6 m
D
0.9 m
0.6 m
-P Ax 0 (1)
E
I
0.6 m
C
0.9 m
1 kN
P
H
F
0.9 m
y
B
G
FBD-1
10 kN
2 kN
A
o
1.35 m
60
Ax
0.9 m
x
0.3 m
Ay
o
60
Lecture 18
7
Example Equation of Statics
?Fy 0
0.6 m
D
0.6 m
Ay - 2 - 1 - 10 0 (2)
0.9 m
E
I
0.6 m
C
0.9 m
1 kN
P
H
F
y
0.9 m
G
B
FBD -1
10 kN
2 kN
A
o
1.35 m
0.9 m
60
Ax
x
0.3 m
Ay
o
Lecture 18
60
8
Example Equation of Statics
Lecture 18
0.6 m
D
0.6 m
0.9 m
E
I
0.6 m
C
0.9 m
1 kN
P
H
F
y
0.9 m
G
B
FBD -1
10 kN
2 kN
A
o
1.35 m
0.9 m
60
0.3 m
Ax
x
?Mz(A) 0
Ay
o
-2(0.45) - 1(1.65) - 10(2.55) P(1.559) 0
(3)
60
9
Example Equations of Statics

• Summary

?Fx 0
-P Ax 0 (1)
?Fy 0
Ay 13 kN
Ay - 2 - 1 - 10 0 (2)
?Mz(A) 0
-2(0.45) - 1(1.65) - 10(2.55) P(1.559) 0
(3)
P 18 kN
Ax 18 kN
Lecture 18
10
Example Draw FBD-2 of IGH Bucket
Equations of statics
F
I
EI
F - Gx 0 (4)
?Fx 0
EI
?Fy 0
Gy
Gy - 10 0 (5)
H
1.039
Gy 10 kN
Gx
?Mz(A) 0

G
-1.039F - 10(0.3) 0 (6)
10 kN
EI
0.3
F -2.89 kN
EI
FBD - 2
from (4)
Gx 2.89 kN
Lecture 18
11
Example Draw FBD-3 of DEFG
o
?Fx 0
30
0.6 m
Dx
D
Dx 0.9449F - 2.89 2.89 0 (7)
FB
E
Dx - 0.9449 F
0.6 m
FB
Dy
2.89 kN
0.6 m
F
?Fy 0
o
1 kN
30
F
cos(19.1)
Dy 0.3275F - 10 - 1 0 (8)
FB
o
FB
G
10.9
2.89 kN
Dy 11 - 0.3275 F
F
sin(19.1)
FB
FBD - 3
FB
10 kN
0.9 m

?Mz(D) 0
0.9449F (1.039) 0.3275F (0.6)
2.89(0.5196) - 2.89(1.559) - 10(0.9) -
1(0.3) 0 (9)
FB
FB
Lecture 18
12
Example Draw FBD-3 of DEFG
Summary
SFx 0
Dx 0.9449F - 2.89 2.89 0 (7)
FB
Dx - 0.9449 F
FB
SFy 0
Dy 0.3275F - 10 - 1 0 (8)
FB
Dy 11 - 0.3275 F
FB
SMz(D) 0

0.9449F (1.039) 0.3275F (0.6)
2.89(0.5196) - 2.89(1.559) - 10(0.9) -
1(0.3) 0 (9)
FB
FB
Dy 7.58 kN
Dx 9.87 kN
F 10.4 kN
FB
Lecture 18
13
Example
4) Resolve the forces perpendicular and parallel
to the member (shears and axial forces).
11.5 kN
D
4.8 kN
Equations of statics
2.0 kN
E
SFx 0
2.32 kN
F
1.97 kN
SFy 0
G
SMz 0
7.50 kN
10.3 kN
7.22 kN
Lecture 18
14
Frames and Machines

• Suggested Problems
• 6-107, 109, 128, 137

Lecture 18