Equilibrium in Three Dimensions

1
Lecture 16

• Equilibrium in Three Dimensions
• Bar supported by cables
• Shaft with levers and thrust bearing
• Plate supported by hinges and a cable

2
Bar Supported by Cables

350 mm

• A bar is supported by a ball,
• a socket joint and two cables
• as shown.

z
600
mm
400
mm
A

• Determine the reaction at support A (the ball
  and socket joint) and the tensions in the two
  cables.

y
B
600 mm
550
C
mm
x
400
1600 N
mm
450
mm
Lecture 16
3
Bar Supported by Cables
1) Determine the cartesian vectors for the
force in cables B and C.
z
600
mm
400
mm
A
y
T T e
B
B
B
B
-0.55 i 0.60 j 0.40 k
600 mm
e

550
B
C
2
2
2
mm
(-0.55) (0.60) (0.40)
400
x
1600 N
mm
T
- 0.606 T i 0.662 T j 0.441 T k
450

B
B
B
B
mm
Lecture 16
4
Bar Supported by Cables

350 mm
z
600
mm
400
T T e
C
mm
C
C
A
y
B
-1.40 i - 0.35 j 0.60 k
e

600 mm
C
550
2
2
2
C
(-1.40) (-0.35) (0.60)
mm
400
1600 N
mm
x
- 0.896T i - 0.224 T j 0.384T k
450
T

C
C
C
C
mm
Lecture 16
5
Bar Supported by Cables
2) Draw FBD and solve the equations of
statics.
z
A
z
FBD
y
T
0.441
A
x
B
A
y
T
0.606
B
T
0.662
B
T
0.384
C
1600 N
T
0.224
550
C
400
T
0.896
C
450
x
Lecture 16
6
Bar Supported by Cables
z
A
z
FBD
y
T
0.441
A
x
B
A
y
0.606
T
B
T
0.662
B
T
0.384
equation of statics
C
1600 N
SFx 0
T
0.224
550
C
Ax - 0.606TB - 0.896 TC 0 (1)
400
T
0.896
C
Ax 2887 N
450
x
Lecture 16
7
Bar Supported by Cables
z
A
z
FBD
y
T
0.441
A
x
B
A
y
0.606
T
B
T
0.662
B
T
0.384
equation of statics
C
SFy 0
1600 N
T
0.224
550
C
Ay 0.662 TB - 0.224 TC 0 (2)
400
T
0.896
C
Ay 704 N
450
x
Lecture 16
8
Bar Supported by Cables
z
Az
FBD
y
T
0.441
A
x
B
A
y
T
0.606
B
T
0.662
B
T
0.384
C
equation of statics
1600 N
T
0.224
550
C
SFz 0
400
T
0.896
C
450
x
Az 0.441TB 0.384 TC - 1600 0 (3)
Az 45.1 N
Lecture 16
9
Bar Supported by Cables
Lecture 16
z
A
z
FBD
y
T
0.441
A
x
B
A
y
T
0.606
B
T
0.662
B
T
0.384
C
1600 N
T
0.224
550
C
equation of statics
SMy(A) 0
400
T
0.896
C
450
x
-0.441TB(550) - 0.384TC(1400) 1600(950) 0
242.6TB 537.6TC 1.52 x 10 (5)
6
10
Bar Supported by Cables
z
A
z
FBD
y
T
0.441
A
x
B
A
y
0.606
T
B
T
0.662
B
T
0.384
C
equation of statics
1600 N
T
0.224
550
C
SMz(A) 0
0.662TB(550) - 0.224TC(1400) 0
400
T
0.896
C
x
450
313.6
TB TC 0.861TC (6)
364.1
Lecture 16
11
Bar Supported by Cables Summary
SFx 0
Ax 2887 N (1)
SFy 0
Ay 704 N (2)
SFz 0
Az 45.1 N (3)
SMx(A) 0
(4)
6
SMy(A) 0
242.6TB 537.6TC 1.52 x 10 (5)
313.6
SMz(A) 0
TB TC 0.861TC (6)
364.1
substitute TB from (6) into (5)
TC 2036 N
TB 1753 N
Lecture 16
12
Shaft with Levers and Thrust Bearings

• The shaft with two levers shown is used to
  change the
• direction of a force.

• Determine the force P required for
  equilibrium and the reactions at supports A and
  B. The support at A is a ball bearing and the
  support at B is a thrust bearing. The bearings
  exert only force reactions on the shaft.

750 N
200
200
A
B
250
200
P
200
Lecture 16
13
Shaft with Levers and Thrust Bearings
750 N
z
Az
200
200
Bz
A
Ax
x
B
y
Bx
P
250
200
200
FBD
Lecture 16
14
Shaft with Levers and Thrust Bearings
750 N
Az
200
200
Bz
A
Ax
x
B
y
Bx
250
200
P
200
FBD
SMy(A) 0

P 750 N
P(200) - 750(200) 0
Lecture 16
15
Shaft with Levers and Thrust Bearings
2) equations of statics
750 N
Az
200
200
Bz
A
Ax
x
B
750 N
y
Bx
250
200
200
FBD
Ax Bx 750 (1)
SFx 0
Ax - 750 Bx 0,
SFy 0
By 0 (2)
Az - 750 Bz 0,
SFz 0
Az Bz 750 (3)
Ax, Bx,
Az, Bz
unknowns
Lecture 16
16
Shaft with Levers and Thrust Bearings
750 N
Az
200
200
Bz
A
Ax
x
B
y
750 N
Bx
250
200
200
FBD
SMx(A) 0
Bz 230.8 N (4)
-750 (200) Bz (650) 0,

Bx 519.2 N (5)
SMz(A) 0

750 (450) - Bx (650) 0,
substitute Bx and Bz into (1) and (3)
Lecture 16
17
Shaft with Levers and Thrust Bearings Summary
reactions at A and B supports
Bz 230.8 N (4)
Bx 519.2 N (5)
SFx 0
Ax - 750 Bx 0,
Ax Bx 750 (1)
SFy 0
By 0 (2)
SFz 0
Az - 750 Bz 0,
Az Bz 750 (3)
Ax 230.8 N (1)
Az 519.2 N (3)
Lecture 16
18
Plate Supported by Cables

• The plate shown has a mass of 100 kg. The hinges
  at support A and B exert only force reactions on
  the plate. Assume that the hinge at B resists any
  force along the axis of the hinge pins.
• Determine the reactions at supports A and B and
  the tension in the cable.

z
500
mm
300 mm
800 mm
D
300 mm
y
A
750 mm
30
B
x
C
1400 mm
1000 mm
19
Plate Supported by Cables
z
500
mm
300 mm
800 mm
D
300 mm
y
A
750 mm
30
B
x
C
1400 mm
1000 mm
plate weight 100 (9.806) 981 N
-0,50 i - 1,400 cos(30) j (0,75 1,400
sin(30)) k
e

CD
2
2
2
(-0,50) (-1,212) (1,45)
20
Plate Supported by Cables
z
500
mm
300 mm
800 mm
D
300 mm
y
A
750 mm
30
B
x
C
1400 mm
1000 mm
plate weight 100 (9.806) 981 N
e
-0.256 i - 0.620 j 0.742 k

CD
-0.256T i - 0.620T j 0.742T k
T T e
C
C
C
C
C
CD
21
Plate Supported by Cables
note hinge transmit shear normal to
the plate
z
693
260
FBD
0.866An
y
150
0.866Bn
Ax
550
0.5An
Bx
GC
At
350
0.5Bn
x
0.742T
Bt
C
981
700
606
0.256T
C
0.620T
1000
C
1212
22
Plate Supported by Cables
z
693
260
FBD
0.866An
y
150
0.866Bn
Ax
550
0.5An
Bx
GC
At
350
0.5Bn
x
0.742T
Bt
C
981
700
606
0.256T
C
0.620T
1000
C
1212
equation of statics
SFx 0
-Ax - Bx - 0.256T 0 (1)
C
23
Plate Supported by Cables
z
z
693
260
FBD
0.866An
y
150
0.866Bn
Ax
550
0.5An
Bx
GC
At
350
0.5Bn
x
0.742T
Bt
C
981
700
606
0.256T
C
0.620T
1000
C
1212
equation of statics
SFy 0
0.5An 0.5Bn - 0.620T 0.866At 0.866Bt 0
(2)
C
24
Plate Supported by Cables
z
z
693
260
FBD
0.866An
y
150
0.866Bn
Ax
550
0.5An
Bx
GC
At
350
0.5Bn
x
0.742T
Bt
C
981
700
606
0.256T
C
0.620T
1000
C
1212
equation of statics
SFz 0
0.866An 0.866Bn 0.742T - 981 - 0.5At -
0.5Bt 0 (3)
C
25
Plate Supported by Cables
z
z
z
693
260
FBD
0.866An
y
150
0.866Bn
Ax
550
0.5An
Bx
GC
At
350
0.5Bn
x
0.742T
Bt
C
981
700
606
0.256T
C
0.620T
1000
C
1212
equation of statics
SMx(A) 0

-981(346) - 0.620T (550) 0.742T (952)
0.866Bn(693) 0.5Bn(400) 0
C
C
-339,426 365.4T 800.1Bn 0 (4)
C
26
Plate Supported by Cables
z
z
z
693
260
FBD
0.866An
y
150
0.866Bn
Ax
550
0.5An
Bx
GC
At
350
0.5Bn
x
0.742T
Bt
C
981
700
606
0.256T
C
0.620T
1000
C
1212
equations of statics
SMy(A) 0

-981(500) 0.256T (550) - 0.742T (1000)
Bx(400) 0
C
C
490,500 - 601.2T 400Bx 0 (5)
C
27
Plate Supported by Cables
z
693
260
FBD
0.866An
y
150
0.866Bn
Ax
550
0.5An
Bx
GC
At
350
0.5Bn
x
0.742T
Bt
C
981
700
606
0.256T
C
0.620T
1000
C
1212
equation of statics
SMz(A) 0

- 0.620T (1000) 0.256T (952)
Bx(693) 0
C
C
- 376.3T 693Bx 0 (6)
C
28
Plate Supported by Cables Summary
-Ax - Bx - 0.256T 0 (1)
SFx 0
C
SFy 0
-0.5An 0.5Bn - 0.620T 0.866At 0.866Bt 0
(2)
C
SFz 0
0.866An 0.866Bn 0.742T - 981 - 0.5At -
0.5Bt 0 (3)
C
SMx(A) 0
-339,426 365.4T 800.1Bn 0 (4)
C
SMy(A) 0
490,500 - 601.2T 400Bx 0 (5)
C
SMz(A) 0
- 376.3T 693Bx 0 (6)
C
from (6) Bx 376.3 / 693T 0.543T
C
C
substitute into (5)
490,500 - 601.2T 400(0.543)T 0 are
obtained
C
C
Bx 694 N
T 490,500 / 384 1,277 N
C
Bn -159 N
Ax -1,021 N
solve for Bn from (4)
solve for Ax from (1)
,
,
from (2) and (3)
An 582 N
Bt 670 N
At 0 N
29
Frames and Machines

• Suggested Problems
• 5-65, 68, 85, 87, 100

30
FBD and Equations of Statics
Bonus Problem 3
31
Bonus Problem 3
3
500 N
4
E

• Determine the reactions,
• the force in the spring and
• in the rigid link DF.

200
F
D
y
300 N
200
B
C
A
x
800 N
k
200
300
300
32
Bonus Problem 3Solution
400 N
300
E
300 N
D
F
Dx
Fx
200
FBD-2
Dx
D
FBD-1
300 N
200
C
B
A
800 N
Cy
By
200
300
33
Bonus Problem 3Solution

• Equations of Statics

Dx 500 N
SFx 0 -800 Dx 300 0
SME 0 -500(200) By(300) -
300(500) 800(400) 0
By -233.333 N
SFy 0 -300 - 233.3 Cy - 400 0
Cy 933.333 N
34
Bonus Problem 3Solution

• Control

SMA 0 -(-233.333)(200) - 933.333(500)
500(200) 300(400) 400(500) 0