6.69 (Continued)
Carry out the indicated operations on the moments to obtain the vectors
defining the moments:
ijk
0.3842 0.2852 0.8781
DjTAFj⊲0.7605i0.4391j0.4753k⊳
0.3842 0.2852 0.8781
DjTBEj⊲0.7605i0.4391j0.4753k⊳
DjTCEj⊲0iC0.8781j0.4753k⊳
The six equations in six unknowns are:
jTADjeADx CjTAFjeAFx CjTBDjeBDx CjTBEjeBEx CjTCEjeCEx
jTADjuADx CjTAFjuAFx CjTBDjuBDx CjTBEjuBEx CjTCEjuCEx
CjTCFjuCFx CMWx D0
jTAFjDjTCFjD16272.5N⊲T⊳ ,
Problem 6.68, obtained by another method. check.
Check: The solution of a six-by-six system by iteration has risks, since
446
Problem 6.70 In Active Example 6.6, suppose that in
addition to being loaded by the 200 N-m couple, the
frame is subjected to a 400-N force at Cthat is hori-
zontal and points toward the left. Draw a sketch of the
frame showing the new loading. Determine the forces
and couples acting on members AB of the frame.
400 mm
600 mm
C
200 N-m
400 mm
AB
Solution: The sketch of the frame with the new loading is shown.
We break the frame into separate bars and draw the free-body diagram
of each bar.
Problem 6.71 The object suspended at Eweighs
200 lb. Determine the reactions on member ACD at A
and C.
3 ft
5 ft
6 ft4 ft
A
BC
D
E
448
Problem 6.72 The mass of the object suspended at G
is 100 kg. Determine the reactions on member CDE at
Cand E.
A
BE
D
C
FG
200 mm
800 mm
Problem 6.73 The force FD10 kN. Determine the
forces on member ABC, presenting your answers as
shown in Fig. 6.25.
1 m 1 m 2 m 1 m
A
D
BC
EG
F
Solution: The complete structure as a free body: The sum of the
Element DEG: The sum of the moments about D
MDFC3EC4GyD0,
F
1m
8 kN
2 kN
B
AC
450
c
2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they
currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Problem 6.74 In Example 6.7, suppose that the frame
is redesigned so that the distance from point Cto the
attachment point Eof the two-force member BE is
increased from 8 in to 10 in. Determine the forces acting
at Con member ABCD.
8 in 8 in
C
6 in
6 in
6 in
W
E
D
B
A
3 in
G
Solution: The analysis of the free-body diagram of the entire struc-
ture as presented in Example 6.7 is unchanged.
body. The angle ˛is now
Problem 6.75 The tension in cable BD is 500 lb.
Determine the reactions at Afor cases (1) and (2).
G
6 in
6 in
8 in 8 in
E
ABC
300 lb
D
(1)
G
6 in
6 in
8 in 8 in
E
ABC
300 lb
D
(2)
Solution: Case (a) The complete structure as a free body: The sum
of the moments about G:
MGD16⊲300⊳C12AxD0,
(a) 12 in
Gx
GxEx
Gy
Ey
Gy
Ay
452
Problem 6.76 Determine the reactions on member
ABCD at A, C, and D.
B
C
0.4 m
0.4 m
600 N
0.6 m 0.4 m 0.4 m
E
A
D
Solution: Consider the entire structure first
Problem 6.77 Determine the forces exerted on member
ABC at Aand C.
C
D
BA
100 lb
E
400 lb
2 ft
1 ft
1 ft
2 ft 2 ft 2 ft
Solution: We start with the free-body diagram of the entire frame.
Two of the equilibrium equations for the whole frame are
454
Problem 6.78 An athlete works out with a squat thrust
machine. To rotate the bar ABD, she must exert a vertical
force at Athat causes the magnitude of the axial force in
the two-force member BC to be 1800 N. When the bar
ABD is on the verge of rotating, what are the reactions
on the vertical bar CDE at Dand E?
0.6 m0.6 m
E
D
B
A
C
1.65 m
0.42 m
Solution: Member BC is a two force member. The force in BC is
along the line from Bto C.
C
0.6D34.99°.
Fx:DxFBC cos D0
Problem 6.79 The frame supports a 6-kN load at C.
Determine the reactions on the frame at Aand D.
0.4 m0.8 m
AB
DE
F
C
0.5 m
0.4 m 1.0 m
6 kN
Solution: Note that members BE and CF are two force members.
Consider the 6 kN load as being applied to member ABC.
Ay
1.0 m
6 kN
0.2D68.20°
Dx
Dy
FE
0.8 m 0.4 m
θφ
Equations of equilibrium:
Member ABC:
Fx:AxCFBE cos FCF cos D0
Solving, we get
AyD11.25 kN
DxD16.3kN
456
Problem 6.80 The mass mD120 kg. Determine the
forces on member ABC, presenting your answers as
shown in Fig. 6.25.
ABC
D
E
200 mm 200 mm
300 mm
m
m
FXDAXCCXD0,
FYDAYBYCCYD0,
4708 N
4708 N
4708 N
2354 N
2354 N
1177 N
B
C
Problem 6.81 Determine the reactions on member
BCD.
C
E
AB
DF
8 in.
8 in.
18 in. 12 in. 8 in.
30 lb
40 lb
G
8 in.
Solution: We will use frame ADG, bar DFG and bar BCD. The
And finally from BCD we have
DxD5.83 lb,D
yD48.9 lb
458
c
2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they
currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Problem 6.82 The weight of the suspended object is
WD50 lb. Determine the tension in the spring and the
reactions at F. (The slotted member DE is vertical.)
A
8 in 8 in 10 in 10 in
10 in
6 in
4 in
B
W
E
C
D
F
Solution: Start with member AB
Finally examine DCE
Problem 6.83 The mass mD50 kg. Bar DE is
horizontal. Determine the forces on member ABCD,
presenting your answers as shown in Fig. 6.25.
AF
DE
B
C
1 m 1 m
1 m
1 m
1 m
m
from which AyD490.5N,
from which FxD981 N, and from above, AxD981 N ,
above. The components at Band Chave the magnitudes
490.5 N
981 N
1387 N
D
460
Problem 6.84 Determine the forces on member BCD.
A
6 ft 400 lb
4 ft
4 ft
8 ft
D
B
E
C
FyDEyCDy400 D0,
from which DyD100 lb. Element AB: The sum of the moments
about A:
MAD8By⊲6⊳400 D0,
where negative sign means that the force is reversed from the direction
shown on the free body diagram.
Ay
Cy
Ax
Ax
By
Bx
400 lb
Problem 6.85 Determine the forces on member ABC.
1 m
DC
E
AB
1 m
2 m 2 m 1 m
6 kN
Member ABC: The equations of equilibrium are
and summing moments about A,
MAD2BY4CYD0.
462
Problem 6.86 Determine the forces on member ABD.
8 in
8 in 8 in
60 lb 60 lb
8 in
8 in
A
CD
B
E
Member ABD: The equilibrium equations for this member are:
FXDAXBXDXD0,
Problem 6.87 The mass mD12 kg. Determine the
forces on member CDE.
200 mm
100 mm
200 mm
200 mm 400 mm
A
CD
B
m
E
Solution: Start with a free-body diagram of the entire frame.
Eq. for entire frame:
⊲C
!⊳FxD0:
Ay
Ax
Dx
464
Problem 6.88 The weight WD80 lb. Determine the
forces on member ABCD.
E
8 in
11 in 12 in5 in
3 in
F
W
AC
DB