Statics Bedford chap 06

Prévia do material em texto

Problem 6.1 In Active Example 6.1, suppose that in
addition to the 2-kN downward force acting at point D,
a 2-kN downward force acts at point C. Draw a sketch of
the truss showing the new loading. Determine the axial
forces in members AB and AC of the truss.
C
5 m5 m
A
D
B
2 kN
3 m
3 m
Solution: The new sketch, a free-body diagram of the entire truss
and a free-body diagram of the joint at A are shown. The angle ˛
between CD and BD is
˛ D tan�1�6/10� D 31.0°
Using the entire truss, the equilibrium equations are
Fx : Ax C B D 0
Fy : Ay � 2 kN� 2 kN D 0
MA : ��2 kN��5 m�� �2 kN��10 m�
C B�6 m� D 0
Solving yields
Ax D �5 kN, Ay D 4 kN, B D 5 kN
Using the free-body diagram of joint A, the equilibrium equations are:
Fx : Ax C TAC cos ˛ D 0
Fy : Ay � TAB � TAC sin ˛ D 0
Solving yields TAB D 1 kN, TAC D 5.83 kN
Because both values are positive, we know that both are in tension
AB : 1 kN (T), AC : 5.83 kN (T)
386
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 6.2 Determine the axial forces in the
members of the truss and indicate whether they are in
tension (T) or compression (C).
C
800 N
0.7 m
0.7 m
A
B
0.4 m
20�
Solution: We start at joint A
∑
Fx : � 7p
65
FAB C 7p
65
FAC � �800 N� sin 20° D 0
∑
Fy : � 4p
65
FAB � 4p
65
FAC � �800 N� cos 20° D 0
Solving we have FAB D �915 N, FAC D �600 N
7 7
44
800 N
A
FACFAB
20°
Next we move to joint C
∑
Fx : � 7p
65
FAC � FBC D 0) FBC D 521 N
C
Cy
FAC
FCB
7
4
In summary we have
FAB D 915 N�C�, FAC D 600 N�C�, FBC D 521 N�T�
Problem 6.3 Member AB of the truss is subjected to a
1000-lb tensile force. Determine the weight W and the
axial force in member AC.
A
B
W
C
60 in
60 in 60 in
Solution: Using joint A
∑
Fx : � 2p
5
�1000 lb�� 1p
2
FAC D 0
∑
Fy : � 1p
5
�1000 lb�� 1p
2
FAC �W D 0
Solving we have FAC D �1265 lb, W D 447 lb
In summary we have
W D 447 lb, FAC D 1265 lb�C�
1000 lb
A2
1
1
1
FAC
W
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.
387
Problem 6.4 Determine the axial forces in members
BC and CD of the truss.
600 lb
D
E
3 ft
3 ft
3 ft
3 ft
A
C
B
Solution: The free-body diagrams for joints E, D, and C are
shown. The angle ˛ is
˛ D tan�1�3/4� D 36.9°
Using Joint E, we have
Fx : ��600 lb�� TCE sin ˛ D 0
Fy : �TCE cos ˛� TDE D 0
Using Joint D, we have
Fx : �TCD � TBD sin ˛ D 0
Fy : TDE � TBD cos ˛ D 0
Finally, using Joint C, we have
Fx : TCD C TCD sin ˛� TAC sin ˛ D 0
Fy : TCE cos ˛� TAC cos ˛� TBC D 0
Solving these six equations yields
TCE D �1000 lb, TDE D 800 lb
TCD D �600 lb, TAC D �2000 lb
TBC D 800 lb, TBD D 1000 lb
A positive value means tension and a negative value means compres-
sion
Thus BC : 800 lb (T), CD : 600 lb (C)
388
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 6.5 Each suspended weight has mass m D
20 kg. Determine the axial forces in the members of
the truss and indicate whether they are in tension (T) or
compression (C).
0.32 m0.16 m0.16 m
0.4 m
A
B
C D
m m
Solution: Assume all bars are in tension. Start with joint D
∑
Fy :
5p
61
TAD � 196.2 N D 0
∑
Fx : � 6p
61
TAD � TCD D 0
Solving: TAD D 306 N, TCD D �235 N
TAD
TCD
5
6
196.2 N
D
Now work with joint C
∑
Fy :
5p
29
TAC � 196.2 N D 0
∑
Fx : � 2p
29
TAC � TBC C TCD D 0
Solving: TAC D 211 N, TBC D �313 N
TAC
5
2
196.2 N
C
TBC TCD
Finally work with joint A
∑
Fy : � 5p
29
�TAB C TAC�� 5p
61
TAD D 0
) TAB D �423 N
T
TAB TAC
TAD
A
2
2
5 5
5
6
In summary:
TAB D 423 N�C�
TAC D 211 N�T�
TAD D 306 N�T�
TBC D 314 N�C�
TCD D 235 N�C�
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.
389
Problem 6.6 Determine the largest tensile and com-
pressive forces that occur in the members of the truss,
and indicate the members in which they occur if
(a) the dimension h D 0.1 m;
(b) the dimension h D 0.5 m.
Observe how a simple change in design affects the
maximum axial loads.
0.6 m
0.4 m
h
1.2 m
0.7 m1 kN 
C
D
B
A
Solution: To get the force components we use equations of the
form TPQ D TPQePQ D TPQXiC TPQYj where P and Q take on the
designations A, B, C, and D as needed.
Equilibrium yields
At joint A:
∑
Fx D TABX C TACX D 0,
and
∑
Fy D TABY C TACY � 1 kN D 0.
At joint B:
∑
Fx D �TABX C TBCX C TBDX D 0,
and
∑
Fy D �TABY C TBCY C TBDY D 0.
At joint C:
∑
Fx D �TBCX � TACX C TCDX D 0,
and
∑
Fy D �TBCY � TACY C TCDY C CY D 0.
At joint D:
∑
Fx D �TCDX � TBDX C DX D 0,
and
∑
Fy D �TCDY � TBDY C DY D 0.
Solve simultaneously to get
TAB D TBD D 2.43 kN,
TAC D �2.78 kN,
TBC D 0, TCD D �2.88 kN.
Note that with appropriate changes in the designation of points, the
forces here are the same as those in Problem 6.4. This can be explained
by noting from the unit vectors that AB and BC are parallel. Also note
that in this configuration, BC carries no load. This geometry is the
same as in Problem 6.4 except for the joint at B and member BC
which carries no load. Remember member BC in this geometry — we
will encounter things like it again, will give it a special name, and will
learn to recognize it on sight.
0.6 m 1.2 m
CY
DY
DX
TBC
−TBC
TBD
TCD
TAB
TAC
−TAB
−TAC−TCD
−TBD
B
y
h
C
D
A
x
1 kN
0.4
m
0.7 m
(b) For this part of the problem, we set h D 0.5 m. The unit vectors
change because h is involved in the coordinates of point B. The new
unit vectors are
eAB D �0.986iC 0.164j,
eAC D �0.864i� 0.504j,
eBC D 0i� 1j,
eBD D �0.768i� 0.640j,
and eCD D �0.832iC 0.555j.
We get the force components as above, and the equilibrium forces at
the joints remain the same. Solving the equilibrium equations simul-
taneously for this situation yields
TAB D 1.35 kN,
TAC D �1.54 kN,
TBC D �1.33,
TBD D 1.74 kN,
and TCD D �1.60 kN.
These numbers differ significantly from (a). Most significantly,
member BD is now carrying a compressive load and this has reduced
the loads in all members except member BD. “Sharing the load” among
more members seems to have worked in this case.
390
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 6.7 This steel truss bridge is in the Gallatin
National Forest south of Bozeman, Montana. Suppose
that one of the tandem trusses supporting the bridge is
loaded as shown. Determine the axial forces in members
AB, BC, BD, and BE.
17 ft 17 ft 17 ft 17 ft
A
B D F
H
GEC
8 ft
10 kip 10 kip 10 kip
Solution: We start with the entire structure in order to find the
reaction at A. We have to assume that either A or H is really a roller
instead of a pinned support.
∑
MH : �10 kip��17 ft�C �10 kip��34 ft�C �10 kip��51 ft�
� A�68 ft� D 0) A D 15 kip
17 ft
A 10 kip 10 kip 10 kip H
17 ft 17 ft 17 ft
Now we examine joint A
∑
Fy :
8p
353
FAB C A D 0) FAB D �35.2 kip
17
8
A
FAB
FAC
Now work with joint C
∑
Fy : FBC � 10 kip D 0) FBC D 10 kip
FAC FCE
FBC
C
10 kip
Finally work with joint B
∑
Fx : � 17p
353
FAB C 17p353
FBE C FBD D 0
∑
Fy : � 8p
353
FAB � 8p
353
FBE � FBC D 0
Solving we find FBD D �42.5 kip, FBE D 11.74 kip
17
8
17
8
FBE
FBC
FAB
B FBD
In Summary we have
FAB D 35.2 kip�C�, FBC D 10 kip�T�,
FBD D 42.5 kip�C�, FBE D 11.74 kip�T�
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.
391
Problem 6.8 For the bridge truss in Problem 6.7,
determine the largest tensile and compressive forces that
occur in the members, and indicate the members in
which they occur.
Solution: Continuing the solution to Problem 6.7 will show the
largest tensile and compressive forces that occur in the structure.
Examining joint A we have
∑
Fx :
17p
353
FAB C FAC D 0) FAC D 31.9 kip
Examining joint C
∑
Fx : �FAC C FCE D 0) FCE D 31.9 kip
Examining joint D
∑
Fy : �FDE D 0) FDE D 0
DFBD FDF
FDE
The forces in the rest of the members are found by symmetry. We have
FAB D FFH D 35.2 kip�C�
FAC D FGH D 31.9 kip�T�
FBC D FFG D 10 kip�T�
FBD D FDF D 42.5 kip�C�
FBE D FEF D 11.74 kip�T�
FCE D FEG D 31.9 kip�T�
FDE D 0
The largest tension and compression members are then
FAC D FEG D FCE D FGH D 31.9 kip�T�
FBD D FDH D 42.5 kip�C�
392
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 6.9 The trusses supporting the bridge in
Problems 6.7 and 6.8 are called Pratt trusses. Suppose
that the bridge designers had decided to use the truss
shown instead, which is called a Howe truss. Determine
the largest tensile and compressive forces that occur
in the members, and indicate the members in which
they occur. Compare your answers to the answers to
Problem 6.8.
17 ft 17 ft 17 ft 17 ft
A
B D F
H
GEC
8 ft
10 kip 10 kip 10 kip
Solution: We start with the entire structure in order to find the
reaction at A. We have to assume that either A or H is really a roller
instead of a pinned support.
∑
MH : �10 kip��17 ft�C �10 kip��34 ft�C �10 kip��51 ft�
� A�68 ft� D 0) A D 15 kip
A H10
kips
10
kips
10
kips
Now we examine joint A
∑
Fy :
8p
353
FAB C A D 0) FAB D �35.2 kip
∑
Fx :
17p
353
FAB C FAC D 0) FAC D 31.9 kip
A
FAC
FAB
17
8
Now work with joint B
∑
Fx : � 17p
353
FAB C FBD D 0) FBD D �31.9 kip
∑
Fy : � 8p
353
FAB � FBC D 0) FBC D 15 kip
FBD
FBC
FAB
B
17
8
Next work with joint C
∑
Fy : FBC C 8p
353
FCD � 10 kip D 0) FCD D �11.74 kip
∑
Fx : FCE C 17p
353
FCD � FAC D 0) FCE D 42.5 kip
FCD
FBC
FCE
10 kip
FAC
C
17
8
Finally from joint E we find
∑
Fy : FDE � 10 kip D 0) FDE D 10 kip
E
FCE
FDE
FEG
10 kip
The forces in the rest of the members are found by symmetry. We have
FAB D FFH D 35.2 kip�C�
FAC D FGH D 31.9 kip�T�
FBD D FDF D 31.9 kip�C�
FBC D FFG D 15 kip�T�
FCD D FDG D 11.74 kip�C�
FCE D FEG D 42.5 kip�T�
FDE D 10 kip�T�
The largest tension and compression members are then
FCE D FEG D 42.5 kip�T�
FAB D FFH D 35.2 kip�C�
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.
393
Problem 6.10 Determine the axial forces in members
BD, CD, and CE of the truss.
F
400
mm
400
mm
400
mm
300 mm
300 mm
6 kN
A
B
C
D
E
G
Solution: The free-body diagrams of the entire truss and of joints
A, B, and C are shown. The angle
˛ D tan�1�3/4� D 36.9°
From the free-body diagram of the entire truss
Fy : Ay � 6 kN D 0
MG : �6 kN��400 mm�C Ax�600 mm�
� Ay�1200 mm� D 0
Solving, Ax D 8 kN, Ay D 6 kN
Using joint A,
Fx : Ax C TAB C TAC cos ˛ D 0
Fy : Ay C TAC sin ˛ D 0
Solving we find
TAB D 0, TAC D �10 kN
Because joint B consists of three members, two of which are parallel,
and is subjected to no external load, we can recognize that
TBD D TAB D 0 and TBD D 0
Finally we examine joint C
Fx : TCE C TCD cos ˛� TAC cos ˛ D 0
Fy : �TAC sin ˛� TCD sin ˛� TBC D 0
}
) TCD D 10 kN, TCE D �16 kN
In summary BD : 0, CD : 10 kN (T), CE : 16 kN (C)
394
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 6.11 The loads F1 D F2 D 8 kN. Determine
the axial forces in members BD, BE, and BG.
3 m
A
B
D
E
G
F2
F1
4 m
C
4 m
3 m
Solution: First find the external support loads and then use the
method of joints to solve for the required unknown forces. (Assume
all unknown forces in members are tensions).
External loads:
y
x
B
A
E
G
GY
C
D
AX
AY
F1 = 8 kN
F2 = 8 kN
3 m
8 m
3 m
∑
Fx : Ax C F1 C F2 D 0 (kN)
∑
Fy : Ay CGy D 0
C
∑
MA : 8Gy � 3F2 � 6F1 D 0
Solving for the external loads, we get
Ax D �16 kN �to the left�
Ay D �9 kN �downward�
Gy D 9 kN �upward�
Now use the method of joints to determine BD, BE, and BG.
Start with joint D.
Joint D :
BD
DE
D
x
y
F1 = 8 kN
θ
cos � D 0.8
sin � D 0.6
� D 36.87°
∑
Fx : F1 � BD cos � D 0
∑
Fy : � BD sin � � DE D 0
Solving, BD D 10 kN �T�
DE D �6 kN �C�
Joint E :
BE
DE
EG
x
y
F2 = 8 kN
DE D �6 kN
∑
Fx D DE� EG D 0
∑
Fy D �BEC F2 D 0
Solving: EG D �6 kN �C�
BE D 8 kN �T�
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.
395
6.11 (Continued )
Joint G :
EG
y
x
CG
BG
GY
θ
�EG D �6 kN �C��
Gy D 9 kN
∑
Fx : �CG� BG cos � D 0
∑
Fy : BG sin � C EGCGy D 0
Solving, we get
BG D �5 kN �C�
CG D 4 kN �T�
Thus, we have
BD D 10 kN �T�
BE D 8 kN �T�
BG D �5 kN �C�
Problem 6.12 Determine the largest tensile and
compressive forces that occur in the members of the
truss, and indicate the members in which they occur if
(a) the dimension h D 5 in;
(b) the dimension h D 10 in.
Observe how a simple change in design affects the
maximum axial loads.
20 in 20 in 20 in
30�
800 lb
A
B
CE
D
h
Solution: Starting at joint A
∑
Fx : � 20p
h2 C 202 FAB � FAC C �800 lb� sin 30
° D 0
∑
Fy :
hp
h2 C 202 FAB � �800 lb� cos 30
° D 0
800 lb
A
20
h
FAB
FAC
Next joint B
∑
Fx : �FBD � 20p
h2 C 202 FBC C
20p
h2 C 202 FAB D 0
∑
Fy : � hp
h2 C 202 FBC �
hp
h2 C 202 FAB D 0
20
h h
B
FBD
FABFBC
20
Finally joint C
∑
Fx : � 20p
h2 C 202 FCD C
20p
h2 C 202 FBC � FCE C FAC D 0
∑
Fy :
hp
h2 C 202 FCD C
hp
h2 C 202 FBC D 0
2020
h h
FCD FBC
C FACFCE
(a) Using h D 5 in we find:
FAB D 2860 lb�T�, FAC D 2370 lb�C�, FBD D 5540 lb�T�
FBC D 2860 lb�C�, FCD D 2860 lb�T�, FCE D 7910 lb�C�
)
FBD D 5540 lb�T�
FCE D 7910 lb�C�
(b) Using h D 10 in we find:
FAB D 1549 lb�T�, FAC D 986 lb�C�, FBD D 2770 lb�T�
FBC D 1549 lb�C�, FCD D 1549 lb�T�, FCE D 3760 lb�C�
)
FBD D 2770 lb�T�
FCE D 3760 lb�C�
396
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 6.13 The truss supports loads at C and E.
If F D 3 kN, what are the axial forces in members BC
and BE? A B
C
D
E
G
1 m
F
2F
1 m 1 m 1 m
Solution: The moment about A is
∑
MA D �1F� 4FC 3G D 0,
from which G D 5
3
F D 5 kN. The sums of forces:
∑
FY D AY � 3FC G D 0,
from which AY D 4
3
F D 4 kN.
∑
FX D AX D 0,
fromwhich AX D 0. The interior angles GDE, EBC are 45°,
from which sin ˛ D cos ˛ D 1p
2
.
Denote the axial force in a member joining I, K by IK.
(1) Joint G :
∑
Fy D DGp
2
CG D 0,
from which
DG D �p2G D � 5
p
2
3
F D �5p2 kN �C�.
∑
Fx D �DGp
2
� EG D 0,
from which
EG D �DGp
2
D 5
3
F D 5kN �T�.
(2) Joint D :
∑
Fy D �DE� DGp
2
D 0,
from which
DE D 5
3
F D 5 kN �T�.
∑
Fx D �BDC DGp
2
D 0,
1 m
1 m 1 m 1 m
AY
AY
AX
F G2F
DG
DE
BD DG
EG
AC
AB AC
BC
CE
F
G
Joint G
Joint A Joint C
Joint D Joint E
45°
45°
45°
45°
45°
CE EG
DE
BE
from which
BD D � 5
3
F D �5 kN �C�.
(3) Joint E :
∑
Fy D BEp
2
� 2FC DE D 0,
from which BE D 2p2F�p2DE D
p
2
3
F D p2 kN �T�.
∑
Fx D �CE� BEp
2
C EG D 0,
from which
CE D EG� BEp
2
D 4
3
F D 4 kN �T�.
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.
397
6.13 (Continued )
(4) Joint A:
∑
Fy D Ay � ACp
2
D 0,
from which AC D 4
p
2
3
F D 4p2 kN �T�.
∑
Fx D ABC ACp
2
D 0,
from which AB D �4
3
F D �4 kN �C�.
(5) Joint C :
∑
Fy D BCC ACp
2
� F D 0,
from which BC D F� ACp
2
D � 1
3
F D �1 kN �C�.
Problem 6.14 If you don’t want the members of the
truss to be subjected to an axial load (tension or compres-
sion) greater than 20 kN, what is the largest acceptable
magnitude of the downward force F? 12 m
3 m
A
F
C
D
B
4 m
Solution: Start with joint A
∑
Fx : �FAB cos 36.9° � FAC sin 30.5° D 0
∑
Fy : �FAB sin 36.9° � FAC cos 30.5° � F D 0
A
36.9°
30.5°
F
FAB
FAC
Now work with joint C
∑
Fx : �FCD � FBC sin 36.9° C FAC sin 30.5° D 0
∑
Fy : FBC cos 36.9° C FAC cos 30.5° D 0
36.9°
30.5°
CFCD
FBC FAC
Finally examine joint D
∑
Fy : FBD D 0
FBD
DDx FCD
Solving we find
FAB D 1.32F, FAC D �2.08F, FCD D �2.4F,
FBC D 2.24F, FBD D 0
The critical member is CD. Thus
2.4F D 20 kN) F D 8.33 kN
398
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 6.15 The truss is a preliminary design for a
structure to attach one end of a stretcher to a rescue
helicopter. Based on dynamic simulations, the design
engineer estimates that the downward forces the stretcher
will exert will be no greater than 1.6 kN at A and at B.
What are the resulting axial forces in members CF, DF,
and FG?
300
mm
290
mm
390
mm
200 mm
480 mm
150 mm
AB
D
C
G
F
E
Solution: Start with joint C
∑
Fy :
48p
3825
FCF � 1.6 kN D 0) FCF D 2.06 kN
FCF
39
48
C
1.6 kN
FCD
Now use joint F
∑
Fx : � 59p
3706
FFG � 29p
3145
FDF C 39p
3825
FCF D 0
∑
Fy :
15p
3706
FFG � 48p
3145
FDF � 48p
3825
FCF D 0
Solving we find FDF D �1.286 kN, FCF D 2.03 kN
FDF
FCF
FFG
59
15
F
39
48
48
29
In Summary
FCF D 2.06 kN�T�, FDF D 1.29 kN�C�, FCF D 2.03 kN�T�
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.
399
Problem 6.16 Upon learning of an upgrade in the heli-
copter’s engine, the engineer designing the truss does
new simulations and concludes that the downward forces
the stretcher will exert at A and at B may be as large as
1.8 kN. What are the resulting axial forces in members
DE, DF, and DG?
Solution: Assume all bars are in tension.
Start at joint C
∑
Fy :
16p
425
TCF � 1.8 kN D 0) TCF D 2.32 kN
∑
Fx : � 13p
425
TCF � TCD D 0) TCD D �1.463 kN
C
TCF
TCD
13
16
1.8 kN
Next work with joint F
∑
Fx : � 59p
3706
TFG � 29p
3145
TDF C 13p
425
TCF D 0
∑
Fy :
15p
3706
TFG � 48p
3145
TDF � 48p
425
TCF D 0
Solving TDF D �5.09 kN, TFG D 4.23 kN
TFG F
TDF
TCF
29
48
13
16
15
59
Next work with joint B
∑
Fx : � 3p
13
TBE D 0) TBE D 0
∑
Fy :
2p
13
TBE C TBD � 1.8 kN D 0) TBD D 1.8 kN
B
TBDTBE
3
2
1.8 kN
Finally work with joint D
∑
Fx : �TDE � 10p
541
TDG C 29p
3145
TDF C TCD D 0
∑
Fy :
21p
541
TDG C 48p
3145
TDF � TBD D 0
Solving: TDG D 6.82 kN, TDE D �7.03 kN
TDE TCD
TBD
TDG
D
TDF
21
10
48
29
In summary:
TDE D 7.03 kN�C�, TDF D 5.09 kN�C�, TDG D 6.82 kN�T�
400
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 6.17 Determine the axial forces in the
members in terms of the weight W.
A
B E
D
C
1 m
1 m
0.8 m 0.8 m 0.8 m
W
Solution: Denote the axial force in a member joining two points
I, K by IK. The angle between member DE and the positive x axis
is ˛ D tan�1 0.8 D 38.66°. The angle formed by member DB with the
positive x axis is 90° C ˛. The angle formed by member AB with the
positive x axis is ˛.
Joint E :
∑
Fy D �DE cos ˛�W D 0,
from which DE D �1.28W �C� .
∑
Fy D �BE� DE sin ˛ D 0,
from which BE D 0.8W �T�
Joint D :
∑
Fx D DE cos ˛C BD cos ˛�CD cos ˛ D 0,
from which BD �CD D �DE.
∑
Fy D �BD sin ˛C DE sin ˛�CD sin ˛ D 0,
from which BD CCD D DE.
Solving these two equations in two unknowns:
CD D DE D �1.28W �C� , BD D 0
Joint B :
∑
Fx D BE� AB sin ˛� BD sin ˛ D 0,
from which AB D BE
sin ˛
D 1.28W�T�
∑
Fy D �AB cos ˛� BC D 0,
from which BC D �AB cos ˛ D �W�C�
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.
401
Problem 6.18 The lengths of the members of the truss
are shown. The mass of the suspended crate is 900 kg.
Determine the axial forces in the members.
12 m
12 m
5 m
13 m
13 m
C
D
B
A
40�
Solution: Start with joint A
∑
Fx :� FAB cos 40° � FAC sin 27.4° D 0
∑
Fy :� FAB sin 40° � FAC cos 27.4° � �900 kg��9.81 m/s2� D 0
A
FAC
FAB
8829 N
40°
27.4°
Next work with joint C
∑
Fx :� FCD cos 40° � FBC cos 50° C FAC sin 27.4° D 0
∑
Fy :� FCD sin 40° C FBC sin 50° C FAC cos 27.4° D 0
27.4°
50°
40°
FAC
FCD
C
FBC
Finally work with joint B
∑
Fy : FAB cos 50° � FBC sin 50° � FBD cos 27.4° D 0
50°
50°
27.4°
FAB
FBC
FBD
T B
Solving we find
FAB D 10.56 kN D 10.56 kN�T�
FAC D �17.58 kN D 17.58 kN�C�
FCD D �16.23 kN D 16.23 kN�C�
FBC D 6.76 kN D 6.76 kN�T�
FBD D 1.807 kN D 1.807 kN�T�
402
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 6.19 The loads F1 D 600 lb and F2 D
300 lb. Determine the axial forces in members AE, BD,
and CD.
F2
F1
D
A
B
C
4 ft
3 ft
G
E
6 ft
4 ft
Solution: The reaction at E is determined by the sum of the
moments about G:
MG D C6E� 4F1 � 8F2 D 0,
from which
E D 4F1 C 8F2
6
D 800 lb.
The interior angle EAG is
˛ D tan�1
(
6
8
)
D 36.87°.
From similar triangles this is also the value of the interior angles ACB,
CBD, and CGD. Method of joints: Denote the axial force in a member
joining two points I, K by IK.
Joint E :
∑
Fy D EC AE D 0,
from which AE D �E D �800 lb �C� .
∑
Fy D EG D 0,
from which EG D 0.
Joint A:
∑
Fy D �AE� AC cos ˛ D 0,
from which AC D �AE
0.8
D 1000 lb�T�.
∑
Fy D AC sin ˛C AB D 0,
from which AB D �AC�0.6� D �600 lb�C�.
Joint B :
∑
Fy D BD sin ˛� AB� F1 D 0,
GX GY
6 ft
4 ft 4 ft
F1
F2
E
EG
E AE AE
AC AB
BD
BC AB
DG
CD
F2 F1
BD
α
αα
Joint E JointA Joint B Joint D
from which BD D F2 C AB
0.6
D �300
0.6
D �500 lb�C� .
∑
Fx D �BC� BD cos ˛ D 0,
from which BC D �BD�0.8� D 400 lb�T�.
Joint D :
∑
Fy D �BD sin ˛�CD � F1 D 0,
from which CD D �F1 � BD�0.6� D �300 lb�C�
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.
403
Problem 6.20 Consider the truss in Problem 6.19. The
loads F1 D 450 lb and F2 D 150 lb. Determine the axial
forces in members AB, AC, and BC.
Solution: From the solution to Problem 6.19 the angle ˛ D 36.87°
and the reaction at E is E D 4F1 C 8F2
6
D 500 lb. Denote the axial
force in a member joining two points I, K by IK.
Joint E :
∑
Fy D EG D 0.
∑
Fx D AEC E D 0,
from which AE D �E D �500 lb�C�.
Joint A:
∑
Fx D �AE� AC cos ˛ D 0,
from which AC D �AE
0.8
D 625 lb�T� .
∑
Fy D AC sin ˛C AB D 0,
from which AB D �AC�0.6� D �375 lb�C�
Joint B:
∑
Fy D BD sin ˛� F2 � AB D 0,
from which BD D F2 C AB
0.6
D �375 lb�C�
∑
Fx D �BC� BD cos ˛ D 0,
from which BC D �BD�0.8� D 300 lb�T�
EG
E AE AE
BC AB
BDABAC
Joint E Joint A Joint B
F2
α
α
404
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 6.21 Determine the axial forces in members
BD, CD, and CE of the truss.
C E
G
FD
HA
B
4 ft
4 ft
4 ft4 ft4 ft
12 kip
Solution: The free-body diagrams for the entire truss as well as
for joints A, B and C are shown.
From the entire truss:
Fx : Ax D 0
FH : �12 kip��8 ft�� Ay�12 ft� D 0
Solving, yields Ax D 0, Ay D 8 kip
From joint A:
Fx : Ax C TAD cos 45° D 0
Fy : Ay C TAB C TAD sin 45° D 0
Solving yields TAB D �8 kip, TAD D 0
From joint B:
Fx : TBD C TBC cos 45° D 0
Fy : TBC C sin 45° � TAB D 0
Solving yields TBD D 8 kip, TBC D �11.3 kip
From joint C:
Fx : TCE � TBC cos 45° D 0
Fy : �TBC sin 45° � TCD D 0
Solving yields TCD D 8 kip, TCE D �8 kip
Thus we have
BC : 11.3 kip (C), CD : 8 kip (T), CE : 8 kip (C)
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.
405
Problem 6.22 The Warren truss supporting the
walkway is designed to support vertical 50-kN loads at
B, D, F, and H. If the truss is subjected to these loads,
what are the resulting axial forces in members BC, CD,
and CE?
6 m6 m6 m6 m
A C E G I
B D F H
2 m
Solution: Assume vertical loads at A and I Find the external loads
at A and I, then use the method of joints to work through the structure
to the members needed.
3 m 3 m
6 m 6 m 6 m
50 kN 50 kN 50 kN 50 kN
x
AY IY
∑
Fy : Ay C Iy � 4�50� D 0 (kN)
∑
MA : � 3�50�� 9�50�� 15�50� � 21�50�C 24 Iy D 0
Solving Ay D 100 kN
Iy D 100 kN
Joint A:
y
x
AB
AC
A
AY
θ
tan � D 23
� D 33.69°
∑
Fx : AB cos � C AC D 0
∑
Fy : AB sin � C Ay D 0
Solving, AB D �180.3 kN �C�
AC D 150 kN �T�
Joint B :
50 kN
BD
BC
AB
y
B
x
θ θ
AB D �180.3 kN
� D 33.69°
∑
Fx : BC cos � C BD � AB cos � D 0
∑
Fy : � 50� AB sin � � BC sin � D 0
Solving, BC D 90.1 kN �T�
BD D �225 kN �C�
Joint C :
BC
AC CE
CD
y
C
θ θ
x
� D 33.69°
AC D 150 kN �T�
BC D 90.1 kN �T�
∑
Fx : CE� ACCCD cos � � BC cos � D 0
∑
Fy : CD sin � C BC sin � D 0
Solving,
CE D 300 kN �T�
CD D �90.1 kN �C�
Hence BC D 90.1 kN �T�
CD D �90.1 kN �C�
CE D 300 kN �T�
406
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 6.23 For the Warren truss in Problem 6.22,
determine the axial forces in members DF, EF, and FG.
Solution: In the solution to Problem 6.22, we solved for the forces
in AB, AC, BC, BD, CD, and CE. Let us continue the process. We
ended with Joint C. Let us continue with Joint D.
Joint D :
D
BD
CD DE
θ θ
DF
x
y
50 kN
� D 33.69°
BD D �225 kN �C�
CD D �90.1 kN �C�
∑
Fx : DF� BD C DE cos � � CD cos � D 0
∑
Fy : � 50�CD sin � � DE sin � D 0
Solving, DF D �300 kN �C�
DE D 0
At this point, we have solved half of a symmetric truss with a
symmetric load. We could use symmetry to determine the loads in
the remaining members. We will continue, and use symmetry as a
check.
Joint E :
CE E EG
x
y
DE EF
θ θ
� D 33.69°
CE D 300 kN �T�
DE D 0
∑
Fx : EG�CEC EF cos � � DE cos � D 0
∑
Fy : DE sin � C EF sin � D 0
Solving, we get
EF D 0
EG D 300 kN �T�
Note: The results are symmetric to this point!
Joint F :
50 kN
EF FG
DF F FH
x
y
θ θ
� D 33.69°
DF D �300 kN �C�
EF D 0
∑
Fx : FH� DFC FG cos � � EF cos � D 0
∑
Fy : � 50� EF sin � � FG sin � D 0
Solving: FH D �225 kN �C�
FG D �90.1 kN �C�
Thus, we have
DF D �300 kN �C�
EF D 0
FG D �90.1 kN �C�
Note-symmetry holds!
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.
407
Problem 6.24 The Pratt bridge truss supports five
forces (F D 300 kN). The dimension L D 8 m. Deter-
mine the axial forces in members BC, BI, and BJ.
A
B C D E G
I J K L M
H
LLL L L L L L
F F F F F
LL
Solution: Find support reactions at A and H. From the free body
diagram,
∑
Fx D AX D 0,
∑
Fy D AY CHY � 5�300� D 0,
and
∑
MA D 6�8�HY � 300�8C 16C 24C 32C 40� D 0.
From these equations, AY D HY D 750 kN.
From the geometry, the angle � D 45°
Joint A: From the free body diagram,
∑
Fx D AX C TAB cos � C TAI D 0,
∑
Fy D TAB sin � C AY D 0.
From these equations,
TAB D �1061 kN
and TAI D 750 kN.
Joint I: From the free body diagram,
∑
Fx D TIJ � TAI D 0,
∑
Fy D TBI � 300 D 0.
From these equations,
TBI D 300 kN
and TIJ D 750 kN.
Joint B: From the free body diagram,
∑
Fx D TBC C TBJ cos � � TAB cos � D 0,
∑
Fy D �TBI � TBJ sin � � TAB sin � D 0.
From these equations,
TBC D �1200 kN
and TBJ D 636 kN.
B G
I J K L M H
L L L L L L
L
F F F F F
HYAY
L = 8 m F = 300 kN
AY
A I
y
x
x
x
y
TAI
TAB TBI
TBC
TBJ
TBITAB
TIJTAI
θ
θ
θ θ
F
Joint B
Joint A Joint I
y
A
8 8 8 8 8 8
408
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 6.25 For the roof truss shown, determine the
axial forces in members AD, BD, DE, and DG. Model
the supports at A and I as roller supports.
A
B
C F
H
I
E
3 m 3 m 3 m 3 m 3 m 3 m
D G
6 kN
6 kN
8 kN 8 kN
10 kN
3.6 m
Solution: Use the whole structure to find the reaction at A.∑
MI : �6 kN��3 m�C �8 kN��6 m�C �10 kN��9 m�
C �8 kN��12 m�C �6 kN��15 m�
C A�18 m� D 0) A D 19 kN
6 kN 8 kN
10 kN
8 kN
6 kN
IA
Now work with joint A
∑
Fy : FAB sin 21.8° C A D 0) FAB D �51.2 kN
∑
Fx : FAD C FAB cos 21.8° D 0) FAD D 47.5 kN
A
A
FAB
FAD
21.8°
Next use joint B
∑
Fx : ��FAB C FBC C FBD� cos 21.8° D 0
∑
Fy : ��FAB C FBC � FBD� sin 21.8° � �6 kN� D 0
Solving: FBC D �43.1 kN, FBD D �8.08 kN
6 kN
B
FBC
FBDFAB
Next go to joint C
∑
Fy : ��8 kN�� FCD C �FCE � FBC� sin 21.8° D 0
∑
Fx : �FCE � FBC� cos 21.8° D 0
Solving: FCD D �8 kN, FCE D �43.1 kN
8 kN
C
FCD
FCDFBC
Finally examine joint D
∑
Fx : �FAD C FDG � FBD cos 21.8° C FDE cos 50.19° D 0
∑
Fy : FBD sin 21.8° C FCD C FDE sin 50.19°D 0
Solving: FDE D 14.3 kN, FDG D 30.8 kN
D
FCD
FDE
FDGFAD
FBD
50.19°
In Summary
FAD D 47.5 kN�T�, FBD D 8.08 kN�C�,
FDE D 14.32 kN�T�, FDG D 30.8 kN�T�
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.
409
Problem 6.26 The Howe truss helps support a roof.
Model the supports at A and G as roller supports. Deter-
mine the axial forces in members AB, BC, and CD.
4 ft 4 ft 4 ft 4 ft 4 ft 4 ft
800 lb
8 ft
A
B
C
G
F
E
D
H I J K L
600 lb600 lb
400 lb400 lb
Solution: The strategy is to proceed from end A, choosing joints
with only one unknown axial force in the x- and/or y-direction, if
possible, and if not, establish simultaneous conditions in the unknowns.
The interior angles HIB and HJC differ. The pitch angle is
˛Pitch D tan�1
(
8
12
)
D 33.7°.
The length of the vertical members:
BH D 4
(
8
12
)
D 2.6667 ft,
from which the angle
˛HIB D tan�1
(
2.6667
4
)
D 33.7°.
CI D 8 8
12
D 5.3333 ft,
from which the angle
˛IJC D tan�1
(
5.333
4
)
D 53.1°.
The moment about G:
MG D �4C 20��400� C �8C 16��600� C �12��800�� 24A D 0,
from which A D 33600
24
D 1400 lb. Check: The total load is 2800 lb.
From left-right symmetry each support A, G supports half the total
load. check.
The method of joints: Denote the axial force in a member joining two
points I, K by IK.
Joint A:
∑
Fy D AB sin ˛P C 1400 D 0,
from which AB D � 1400
sin ˛p
D �2523.9 lb �C�
∑
Fx D AB cos ˛Pitch C AH D 0,
from which AH D �2523.9��0.8321� D 2100 lb �T�
400 lb
600 lb
800 lb
600 lb
400 lb
A G
4 ft 4 ft 4 ft 4 ft 4 ft 4 ft
AB
AH
CD
CI CJBC
BI
HI IJ
CI
BCBH
AH AB
BH BI
HI1400 lb
400 lb
αPitch
αPitch αPitch αIJC
αPitch αPitch
Joint A
Joint I Joint C
Joint H Joint B
600 lb
Joint H :
∑
Fy D BH D 0, or, BH D 0.
∑
Fx D �AHCHI D 0,
from which HI D 2100 lb �T�
Joint B :
∑
Fx D �AB cos ˛Pitch C BC cos ˛Pitch
C BI cos ˛Pitch D 0,
from which BCC BI D AB
410
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.
6.26 (Continued )
∑
Fy D �400� AB sin ˛Pitch C BC sin ˛Pitch
� BI sin ˛Pitch D 0,
from which BC� BI D ABC 400
sin ˛Pitch
.
Solve the two simultaneous equations in unknowns BC, BI:
BI D � 400
2 sin ˛Pitch
D �360.56 lb �C�,
and BC D AB� BI D �2163.3 lb �C�
Joint I :
∑
Fx D �BI cos ˛Pitch �HIC IJ D 0,
from which IJ D 1800 lb �T�
∑
Fy D CBI sin ˛Pitch CCI D 0,
from which CI D 200 lb (T)
Joint C:
∑
Fx D �BC cos ˛Pitch CCD cos ˛Pitch CCJ cos ˛IJC D 0,
from which CD�0.8321� CCJ�0.6� D �1800
∑
Fy D �600�CI� BC sin ˛Pitch CCD sin ˛Pitch
�CJ sin ˛IJC D 0,
from which CD�0.5547� �CJ�0.8� D �400
Solve the two simultaneous equations to obtain CJ D �666.67 lb �C�,
and CD D �1682.57 lb �C�
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.
411
Problem 6.27 The plane truss forms part of the
supports of a crane on an offshore oil platform. The
crane exerts vertical 75-kN forces on the truss at B, C,
and D. You can model the support at A as a pin support
and model the support at E as a roller support that can
exert a force normal to the dashed line but cannot exert
a force parallel to it. The angle ˛ D 45°. Determine the
axial forces in the members of the truss.
3.4 m3.4 m 3.4 m3.4 m
1.8 m
2.2 m A E
F G H
C
DB
α
Solution: The included angles
� D tan�1
(
4
3.4
)
D 49.64°,
ˇ D tan�1
(
2.2
3.4
)
D 32.91°,
� D tan�1
(
1.8
3.4
)
D 27.9°.
The complete structure as a free body: The sum of the moments about
A is
MA D ��75��3.4��1C 2C 3�C �4��3.4�Ey D 0.
with this relation and the fact that Ex cos 45° C Ey cos 45° D 0, we
obtain Ex D �112.5 kN and Ey D 112.5 kN. From
∑
FAx D Ax C Ex D 0, AX D �EX D 112.5 kN.
∑
FAy D Ay � 3�75�C Ey D 0,
from which Ay D 112.5 kN. Thus the reactions at A and E are symmet-
rical about the truss center, which suggests that symmetrical truss
members have equal axial forces.
The method of joints: Denote the axial force in a member joining two
points I, K by IK.
Joint A:
∑
Fx D AB cos � C Ax C AF cos ˇ D 0,
∑
Fy D AB sin � C Ay C AF sin ˇ D 0,
from which two simultaneous equations are obtained.
Solve: AF D �44.67 kN �C� ,
and AB D �115.8 kN �C�
Joint E:
∑
Fy D �DE cos � C Ex � EH cos ˇ D 0.
∑
Fy D DE sin � C Ey C EH sin ˇ D 0,
from which two simultaneous equations are obtained.
AX
AX
AY EY
EX
AY EY
EX
75 kN 75 kN 75 kN
3.4
m
3.4
m
3.4
m
3.4
m
AB
AB BF
EH
AF
γ
γ γθ θ
γβ β
ββ
DE BF
AF
FG GH
DH
EH
BG DG DH
CG
CDBC
DE
BC
75 kN 75 kN 75 kN
CD
Joint A Joint E Joint F
Joint B Joint D Joint C
Joint H
Solve: EH D �44.67 kN�C� ,
and DE D �115.8 kN�C�
Joint F :
∑
Fx D �AF cos ˇ C FG D 0,
from which FG D �37.5 kN �C�
∑
Fy D �AF sin ˇ C BF D 0,
from which BF D �24.26 kN �C�
Joint H:
∑
Fx D EH cos ˇ �GH D 0,
412
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.
6.27 (Continued )
from which GH D �37.5 kN �C�
∑
Fy D �EH sin ˇ C DH D 0,
from which DH D �24.26 kN �C�
Joint B:
∑
Fy D �AB sin � � BFC BG sin � � 75 D 0,
from which BG D 80.1 kN �T�
∑
Fx D �AB cos � C BCC BG cos � D 0,
from which BC D �145.8 kN �C�
Joint D:
∑
Fy D �DE sin � � DH� DG sin � � 75 D 0,
from which DG D 80.1 kN �T�
∑
Fx D DE cos � �CD � DG cos � D 0,
from which CD D �145.8 kN �C�
Joint C :
∑
Fx D CD� BC D 0,
from which CD D BC Check.
∑
Fy D �CG� 75 D 0,
from which CG D �75 kN �C�
Problem 6.28 (a) Design a truss attached to the
supports A and B that supports the loads applied at points
C and D.
(b) Determine the axial forces in the members of the
truss you designed in (a)
A B
C
D
2 ft
1000 lb
2000 lb
4 ft
5 ft 5 ft5 ft
Problem 6.29 (a) Design a truss attached to the
supports A and B that goes over the obstacle and
supports the load applied at C.
(b) Determine the axial forces in the members of the
truss you designed in (a). A B
C4 m
Obstacle
6 m 3.5 m 4.5 m
1 m
2 m
10 kN
Solution: This is a design problem with many possible solutions.
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.
413
Problem 6.30 Suppose that you want to design a truss
supported at A and B (Fig. a) to support a 3-kN down-
ward load at C. The simplest design (Fig. b) subjects
member AC to 5-kN tensile force. Redesign the truss so
that the largest force is less than 3 kN.
A
B
C
A
B
C
3 kN
1.2 m
1.6 m
(a) (b)
3 kN
Solution: There are many possible designs. To better understand
the problem, let us calculate the support forces in A and B and the
forces in the members in Fig. (b).
Ax
Ay
Bx
C
xB
1.6 m
3 kN
1.2 m
A
θ
tan � D 1.2
1.6
� D 36.87°
sin � D 0.6
cos � D 0.8
∑
Fx: Ax C Bx D 0
∑
Fy : Ay � 3 kN D 0
C
∑
MA: 1.2Bx � 1.6�3� D 0
Solving, we get Ax D �4 kN
Bx D 4 kN
Ay D 3 kN
Note: These will be the external reactions for every design that we
produce (the supports and load do not change).
Reference Solution (Fig. (b))
Joint C :
θ
BC
AC
3 kN
� D 36.87°
∑
Fx : � BC� AC cos � D 0
∑
Fy : AC sin � � 3 kN D 0
Solving: BC D �4 kN �C� AC D5 kN �C�
Thus, AC is beyond the limit, but BC (in compression) is not,
Joint B :
BX
AB
BC
∑
Fx : Bx C BC D 0
∑
Fy : AB D 0
Solving, BC and Bx are both already known. We get AB D 0
Thus, we need to reduce the load in AC. Consider designs like that
shown below where D is inside triangle ABC. Move D around to adjust
the load.
B C
D
A
However, the simplest solution is to place a second member parallel
to AC, reducing the load by half.
414
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 6.31 The bridge structure shown in Example
6.2 can be given a higher arch by increasing the 15°
angles to 20°. If this is done, what are the axial forces
in members AB, BC, CD, and DE?
2b
F F F F F
bbbb
(1)
2b
F F F
(2)
b b b b
B
C
D
EA
15�15�
G JI KH
F F
a a
Solution: Follow the solution method in Example 6.3. F is known
Joint B :
α
y
F
x
20°
TBC
TAB
Joint C :
F
TBC TCD
20°20°
C
For joint C,
∑
Fx : � TBC cos 20° C TCD cos 20° D 0
∑
Fy : � F� TBC sin 20° � TCD sin 20° D 0
TBC D TCD D �1.46F �C�
For joint B.
∑
Fx : TBC cos 20� TAB cos ˛ D 0
∑
Fy : TBC sin 20° � F� TAB sin ˛ D 0
Solving, we get ˛ D 47.5° and TAB D �2.03F �C�
For the new truss (using symmetry)
Members Forces
AG, BH, CI, F
DJ, EK
AB, DE 2.03F (C)
BC, CD 1.46F (C)
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.
415
Problem 6.32 In Active Example 6.3, use the method
of sections to determine the axial forces in members BC,
BI and HI.
A B C D E F
G H I J K L
100 kN
M
1 m
Solution: The horizontal members of the truss are each 1 m in
length. We cut through the relevant members and draw a free-body
diagram of the section to the right of the cut.
We will use equilibrium equations for this section that are designed to
allow us to easily solve for the unknowns.
The equilibrium equations
MI : TBC�1 m�� �100 kN��4 m� D 0) TBC D 400 kN
MB : �THI�1 m�� �100 kN��5 m� D 0) THI D �500 kN
Fy : TBI sin 45° � 100 kN D 0) TBI D 141 kN
In summary we have
BC : 400 kN (T), BI : 141 kN (T), HI : 500 kN (C)
Problem 6.33 In Example 6.4, obtain a section of the
truss by passing planes through members BE, CE, CG,
and DG. Using the fact that the axial forces in members
DG and BE have already been determined, use your
section to determine the axial forces in members CE
and CG.
K
L
L
D
L L L L
G J
IC
B E H
F F2F
A
Solution: From Example 6.4 we know that
TDG D �F, TBE D F
Ax D 0, Ay D 2F
We make the indicated cuts and isolate the section to the left of the
cuts. The equilibrium equations are
Fx : TDG C TBE C TCG cos 45° C TCE cos 45° D 0
Fy : Ay � FC TCG sin 45° � TCE sin 45° D 0
Solving yields TCE D Fp
2
, TCG D �Fp
2
We have CE :
Fp
2
�T�, CG :
Fp
2
�C�
416
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 6.34 The truss supports a 100-kN load at J.
The horizontal members are each 1 m in length.
(a) Use the method of joints to determine the axial
force in member DG.
(b) Use the method of sections to determine the axial
force in member DG.
A B C D
E F G H
100 kN
J
1 m
Solution:
(a) We draw free-body diagrams of joints J, H, and D.
From joint J we have
Fy : TDJ sin 45° � �100 kN� D 0
) TDJ D 141 kN
From joint H we have Fy : TDH D 0
From joint D we have
Fy : �TDG sin 45° � TDH � TDJ sin 45° D 0
Solving yields TDG D �141 kN
(b) We cut through CD, DG and GH. The free-body diagram of
the section to the right of the cut is shown. From this diagram
we have
Fy : �TDG sin 45° � �100 kN� D 0
) TDG D �141 kN
In summary (a), (b) DG : 141 kN (C)
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.
417
Problem 6.35 For the truss in Problem 6.34, use the
method of sections to determine the axial forces in
members BC, CF, and FG.
Solution:
∑
Fx: � BC�CF cos 45� FG D 0
∑
Fy : �CF sin 45° � 100 D 0
∑
MC: � �1�FG� 2�100� D 0
Solving BC D 300 kN �T�
CF D �141.4 kN �C�
FG D �200 kN �C�
1 m
45°
F FG
CF
G H1 m 1 m
J
D
BC C
100 kN
Problem 6.36 Use the method of sections to determine
the axial forces in members AB, BC, and CE.
A B
C
D
E
G
1 m 1 m 1 m
1 m
F
2F
Solution: First, determine the forces at the supports
AX
AY
GY
B
F
2F
D
C E
θ
1 m1 m 1 m
1 m
Θ = 45°
∑
Fx: Ax D 0
∑
Fy : Ay CGy � 3F D 0
C
∑
MA: � 1�F�� 2�2F�C 3Gy D 0
Solving Ax D 0 Gy D 1.67F
Ay D 1.33F
Method of Sections:
AX = 0
AY
BC
AB
CE1 m
1 m
y
B
C
F
x
AY = 1. 33 F
AX = 0
∑
Fx : CEC AB D 0
∑
Fy : BCC Ay � F D 0
C
∑
MB: ��1�Ay C �1�CE D 0
Solving, we get
AB D �1.33F �C�
CE D 1.33F �T�
BC D �0.33F �C�
418
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 6.37 Use the method of sections to determine
the axial forces in members DF, EF, and EG.
A
B
C
D
E
F
G H
300 mm
400 mm 400 mm 400 mm 400 mm
18 kN 24 kN
Solution: We will first use the free-body diagram of the entire
structure to find the reaction at F.
MB : �18 kN� �400 mm�
� �24 kN� �1200 mm�
C F �800 mm� D 0
) F D 27 kN
Next we cut through DF, EF, EG and look at the section to the right
of the cut. The angle ˛ is given by
˛ D tan�1�3/4� D 36.9°
The equilibrium equations are
MF : TEG �300 mm�� �24 kN� �400 mm� D 0
ME : �TDF �300 mm�� �24 kN� �800 mm�
C F�400 mm� D 0
Fy : F� �24 kN�C TEF sin ˛ D 0
Solving yields TDF D �28 kN, TEF D �5 kN, TEG D 32 kN
Thus DF : 28 kN (C), EF : 5 kN (C), EG : 32 kN (T)
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.
419
Problem 6.38 The Pratt bridge truss is loaded as
shown. Use the method of sections to determine the axial
forces in members BD, BE, and CE.
17 ft 17 ft 17 ft 17 ft
A
B D F
H
GEC
8 ft
10 kip 30 kip 20 kip
Solution: Use the whole structure to find the reaction at A.∑
MH : �20 kip��17 ft�C �30 kip��34 ft�
C �10 kip��51 ft�� A�68 ft� D 0
) A D 27.5 kip
Now cut through BD, BE, CE and use the left section
∑
MB : �A�17 ft�C FCE�8 ft� D 0) FCE D 58.4 kip
∑
ME : �10 kip��17 ft�� A�34 ft�� FBD�8 ft� D 0
) FBD D �95.6 kip
∑
Fy : A� 10 kip� 8p
353
FBE D 0) FBE D 41.1 kip
In Summary
FCE D 58.4 kip�T�, FBD D 95.6 kip�C�, FBE D 41.1 kip�T�
A
H10
kip
30
kip
20
kip
A
C
B
8
17
A
10 kip
FCE
FBE
FBD
Problem 6.39 The Howe bridge truss is loaded as
shown. Use the method of sections to determine the axial
forces in members BD, CD, and CE.
17 ft 17 ft 17 ft 17 ft
A
B D F
H
GEC
8 ft
10 kip 30 kip 20 kip
Solution: Use the whole structure to find the reaction at A (same
as 6.38) A D 27.5 kip
Now cut through BD, CD, and CE and use the left section.
∑
MC : �A�17 ft�� FBD�8 ft� D 0) FBD D �58.4 kip
∑
MD : �A�34 ft�C �10 kip��17 ft�C FCE�8 ft� D 0
) FCE D 95.6 kip
∑
Fy : A� 10 kipC 8p
353
FCD D 0) FCD D �41.1 kip
In SummaryFBD D 58.4 kip�C�, FCE D 95.6 kip�T�, FCD D 41.1 kip�C�
FBD
FCD
FCE
10 kip
A
A
B
C
17
8
420
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 6.40 For the Howe bridge truss in Problem
6.39, use the method of sections to determine the axial
forces in members DF, DG, and EG.
Solution: Same truss as 6.39.
Cut through DF, DG, and EG and use left section
∑
MD : �A�34 ft�C �10 kip��17 ft�C FEG�8 ft� D 0
) FEG D 95.6 kip
∑
MG : �A�51 ft�C �10 kip��34 ft�C �30 kip��17 ft�� FDF�8 ft�
D 0) FDF D �69.1 kip
∑
Fy : A� 10 kip� 30 kip� 8p
353
FDG D 0) FDG D �29.4 kip
In summary
FEG D 95.6 kip�T�, FDF D 69.1 kip�C�, FDG D 29.4 kip�C�
FDF
D
8
17
FDG
FEG
E
30 kip10 kipA
Problem 6.41 The Pratt bridge truss supports five
forces F D 340 kN. The dimension L D 8 m. Use the
method of sections to determine the axial force in
member JK.
A
B C D E G
I J K L M
H
LLL L L L L L
F F F F F
LL
Solution: First determine the external support forces.
L L L L L L
F F F F F
AX
AY
HY
F = 340 kN, L = 8 M
∑
Fx : Ax D 0
∑
Fy : Ay � 5FCHy D 0
C
∑
MA: 6LHy � LF� 2LF� 3LF� 4LF� 5LF D 0
Solving: Ax D 0,
Ay D 850 kN
Hy D 850 kN
Note the symmetry:
Method of sections to find axial force in member JK.
B
A
AY
L L
JI
JK
K
CK
D
CDC
F F
θ
� D 45°
L D 8M
F D 340 kN
Ay D 850 kN
∑
Fx : CDC JKCCK cos � D 0
∑
Fy : Ay � 2F�CK sin � D 0
C
∑
MC: L�JK�C L�F�� 2L�Ay� D 0
Solving, JK D 1360 kN �T�
Also, CK D 240.4 kN �T�
CD D �1530 kN �C�
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.
421
Problem 6.42 For the Pratt bridge truss in Prob-
lem 6.41, use the method of sections to determine the
axial force in member EK.
Solution: From the solution to Problem 6.41, the support forces
are Ax D 0, Ay D Hy D 850 kN.
Method of Sections to find axial force in EK.
DE
EK
E G
KL
F F HY
L
θ
∑
Fx : � DE� EK cos � �KL D 0
∑
Fy : Hy � 2F� EK sin � D 0
∑
ME: � �L��KL�� �L��F�C �2L�Hy D 0
A
B C D E G
I J K L M
H
L L L L L L
F F F F F
L
Solution: EK D 240.4 kN �T�
Also, KL D 1360 kN �T�
DE D �1530 kN �C�
Problem 6.43 The walkway exerts vertical 50-kN
loads on the Warren truss at B, D, F, and H. Use
the method of sections to determine the axial force in
member CE.
6 m6 m6 m6 m
A C E G I
B D F H
2 m
Solution: First, find the external support forces. By symmetry,
Ay D Iy D 100 kN (we solved this problem earlier by the method of
joints).
B
BD
A
y
x
AY
CD
D
CEC
50 kN
2 m
6 m
θ
tan � D 2
3
� D 33.69°
∑
Fx: BD CCD cos � CCE D 0
∑
Fy : Ay � 50CCD sin � D 0
∑
MC: � 6Ay C 3�50�� 2BD D 0
Solving: CE D 300 kN �T�
Also, BD D �225 kN �C�
CD D �90.1 kN �C�
422
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 6.44 Use the method of sections to determine
the axial forces in members AC, BC, and BD.
600 lb
D
E
3 ft
4 ft
4 ft
3 ft
A
C
B
Solution: Obtain a section by passing a plane through members
AC, BC, and BD, isolating the part of the truss above the planes. The
angle between member AC and the horizontal is
˛ D tan�1�4/3� D 53.3°
The equilibrium equations are
MC : �600 lb� �4 ft�� TBD cos ˛ �3 ft� D 0
MB : �600 lb� �8 ft�C TAC sin ˛ �4 ft� D 0
Fy : �TBC � TAC cos ˛� TBD cos ˛ D 0
Solving yields
TBD D 1000 lb, TAC D �2000 lb, TBC D 800 lb
Thus
BD : 100 lb (T), AC : 2000 lb (C), BC : 800 lb (T)
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.
423
Problem 6.45 Use the method of sections to determine
the axial forces in member FH, GH, and GI.
I
C
A
B D F
H
E G
400 mm 400 mm
6 kN 4 kN
400 mm400 mm
300 mm
300 mm
Solution: The free-body diagram of the entire truss is used to find
the force I.
MA : I�600 mm�� �4 kN� �1200 mm�
� �6 kN� �800 mm� D 0
) I D 16 kN
Obtain a section by passing a plane through members FH, GH, and
GI, isolating the part of the truss to the right of the planes. The angle
˛ is
˛ D tan�1�3/4� D 36.9°
The equilibrium equations for the section are
MH : TGI cos ˛ �300 mm�C I�300 mm� D 0
MG : I�300 mm�� TFH cos ˛ �400 mm� D 0
Fx : �TGH � TGI sin ˛� TFH sin ˛ D 0
Solving yields TGI D �20 kN, TFH D 20 kN, TGH D �16 kN
Thus GI : 20 kN (C), FH : 20 kN (T), GH : 16 kN (C)
424
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 6.46 Use the method of sections to determine
the axial forces in member DF, DG, and EG.
I
C
A
B D F
H
E G
400 mm 400 mm
6 kN 4 kN
400 mm400 mm
300 mm
300 mm
Solution: The free-body diagram of the entire truss is used to find
the force I.
MA : I�600 mm�� �4 kN� �1200 mm�
� �6 kN� �800 mm� D 0
) I D 16 kN
Obtain a section by passing a plane through members DF, DG, and
EG, isolating the part of the truss to the right of the planes. The angle
˛ is
˛ D tan�1�3/4� D 36.9°
The equilibrium equations for the section are
MG : I �300 mm�� TDF�300 mm� D 0
MD : TEG�300 mm�C I�600 mm�
� �4 kN��400 mm� D 0
Fy : �TDG sin ˛� �4 kN� D 0
Solving yields TDF D 16 kN, TEG D �26.7 kN, TDG D �6.67 kN
Thus DF : 16 kN (T), EG : 26.7 kN (C), DG : 6.67 kN (C)
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.
425
Problem 6.47 The Howe truss helps support a roof.
Model the supports at A and G as roller supports.
(a) Use the method of joints to determine the axial
force in member BI.
(b) Use the method of sections to determine the axial
force in member BI.
2 m 2 m 2 m 2 m 2 m 2 m
2 kN
4 m
A
B
C
G
F
E
D
H I J K L
2 kN2 kN
2 kN2 kN
Solution: The pitch of the roof is
˛ D tan�1
(
4
6
)
D 33.69°.
This is also the value of interior angles HAB and HIB. The complete
structure as a free body: The sum of the moments about A is
MA D �2�2��1C 2C 3C 4C 5�C 6�2�G D 0,
from which G D 30
6
D 5 kN. The sum of the forces:
∑
FY D A� 5�2�CG D 0,
from which A D 10� 5 D 5 kN.
The method of joints: Denote the axial force in a member joining I, K
by IK.
(a) Joint A:
∑
Fy D AC AB sin ˛ D 0,
from which AB D �A
sin ˛
D �5
0.5547
D �9.01 kN (C).
∑
Fx D AB cos ˛C AH D 0,
from which AH D �AB cos ˛ D 7.5 kN (T).
Joint H :
∑
Fy D BH D 0.
Joint B :
∑
Fx D �AB cos ˛C BI cos ˛C BC cos ˛ D 0,
∑
Fy D �2� AB sin ˛� BI sin ˛C BC sin ˛ D 0.
Solve: BI D �1.803 kN �C� , BC D �7.195 kN �C�
(b) Make the cut through BC, BI and HI. The section as a free body:
The sum of the moments about B:
MB D �A�2�CHI�2 tan ˛� D 0,
from which HI D 3
2
A D 7.5 kN�T�. The sum of the forces:
∑
Fx D BC cos ˛C BI cos ˛CHI D 0,
∑
Fy D A� FC BC sin ˛� BI sin ˛ D 0.
Solve: BI D �1.803 kN �C� .
F
F
F
F
F = 2 kN
GA
2 m 2 m 2 m 2 m 2 m 2 m
(a)
AB
A AH
HI
BI
BCF
AH HI AB
BH
α α α
Joint A Joint H Joint B
BH
BI
BC
2 kN
(b)
A
2 m
α
α
αB
426
c� 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rightsreserved. 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 6.48 Consider the truss in Problem 6.47. Use
the method of sections to determine the axial force in
member EJ.
Solution: From the solution to Problem 6.47, the pitch angle is
˛ D 36.69°, and the reaction G D 5 kN. The length of member EK is
LEK D 4 tan ˛ D 166 D 2.6667 m.
The interior angle KJE is
ˇ D tan�1
(
LEK
2
)
D 53.13°.
Make the cut through ED, EJ, and JK. Denote the axial force in a
member joining I, K by IK. The section as a free body: The sum of
the moments about E is
ME D C4G� 2�F�� JK�2.6667� D 0,
from which JK D 20� 4
2.6667
D 6 kN �T�.
The sum of the forces:
∑
Fx D �DE cos ˛� EJ cos ˇ � JK D 0.
∑
Fy D DE sin ˛� EJ sin ˇ � 2FCG D 0,
from which the two simultaneous equations:
0.8321DE C 0.6EJ D �6,
0.5547DE � 0.8EJ D �1.
Solve: EJ D �2.5 kN �C� .
DE F
E
β
α
EJ
JK
F
G
2 m 2 m
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.
427
Problem 6.49 Use the method of sections to determine
the axial forces in member CE, DE, and DF.
C E
G
FD
HA
B
4 ft
4 ft
4 ft4 ft4 ft
12 kip
Solution: The free-body diagrams for the entire structure and the
section to the right of the cut are shown.
From the entire structure:
MA : ��12 kip� �4 ft� H �12 ft� D 0
) H D 4 kip
Using the section to the right of the cut we have
ME : H�4 ft�� TDF�4 ft� D 0
MD : H�8 ft�C TCE�4 ft� D 0
Fy : H� TDE sin 45° D 0
Solving yields
TDF D 4 kip, TCE D �8 kip, TDE D 5.66 kip
Thus we have
DF : 4 kip (T)
CE : 8 kip (C)
DE : 5.66 kip (T)
428
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 6.50 For the bridge truss shown, use the
method of sections to determine the axial forces in
members CE, CF, and DF. D F H J
I
200 kN 200 kN 200 kN 200 kN 200 kN
B
A
C
E
G
3 m 4 m
7 m
5 m 5 m 5 m 5 m
Solution: From the entire structure we find the reactions at A∑
Fx : Ax D 0
∑
MI : �200 kN��5 m�C �200 kN��10 m�C �200 kN��15 m�
C �200 kN��20 m�� Ay�20 m� D 0) Ay D 500 kN
200 kN
I
200 kN 200 kN 200 kN 200 kN
Ax
Ay
Now we cut through DF, CF, and CE and use the left section.
∑
MC : �200 kN��5 m�� Ay�5 m�C Ax�3 m�� FDF�4 m� D 0
) FDF D �375 kN
∑
MF : �200 kN��10 m�C �200 kN��5 m�� Ay�10 m�C Ax�7 m�
C 5p
26
FCE�4 m�� 1p
26
FCE�5 m� D 0) FCE D 680 kN
∑
Fx : Ax C FDF C 5p
26
FCE C 5p
41
FCF D 0
) FCF D �374 kN
FDF
FCF
Ay
Ax
4
5
5
1
FCE
200 kN 200 kN
D
C
Summary:
FDF D 375 kN�C�, FCE D 680 kN�T�, FCF D 374 kN�C�
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.
429
Problem 6.51 The load F D 20 kN and the dimension
L D 2 m. Use the method of sections to determine the
axial force in member HK.
Strategy: Obtain a section by cutting members HK,
HI, IJ, and JM. You can determine the axial forces in
members HK and JM even though the resulting free-
body diagram is statically indeterminate.
A B C
D
H
K
G
J
M
E
I
F
F
L
L
L
L
L
Solution: The complete structure as a free body: The sum of the
moments about K is MK D �FL�2C 3�CML�2� D 0, from which
M D 5F
2
D 50 kN. The sum of forces:
∑
FY D KY CM D 0,
from which KY D �M D �50 kN.∑
FX D KX C 2F D 0,
from which KX D �2F D �40 kN.
The section as a free body: Denote the axial force in a member joining
I, K by IK. The sum of the forces:
∑
Fx D Kx �HIC IJ D 0,
from which HI� IJ D Kx . Sum moments about K to get MK D
M�L��2�C JM�L��2�� IJ�L�CHI�L� D 0.
Substitute HI� IJ D Kx , to obtain JM D �M� Kx
2
D �30 kN �C�.
∑
Fy D Ky CMC JMCHK D 0,
from which HK D �JM D 30 kN�T�
F 2L
2L
2L
F
MKX
KX
KY
KY
L
HI IJ
HK JM
M
L
430
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 6.52 The weight of the bucket is W D
1000 lb. The cable passes over pulleys at A and D.
(a) Determine the axial forces in member FG and HI.
(b) By drawing free-body diagrams of sections, explain
why the axial forces in members FG and HI are
equal. 3 ft 6 in
3 ft
3 ft
3 ft 3 in
35°
L
J
H
F
C
K
I
G
E
B
AD
W
Solution: The truss is at angle ˛ D 35° relative to the horizontal.
The angles of the members FG and HI relative to the horizontal are
ˇ D 45° C 35° D 80°. (a) Make the cut through FH, FG, and EG,
and consider the upper section. Denote the axial force in a member
joining, ˛, ˇ by ˛ˇ.
The section as a free body: The perpendicular distance from point F
is LFW D 3
p
2 sin ˇ C 3.5 D 7.678 ft.
The sum of the moments about F is MF D �WLFW CW�3.25��
jEGj�3� D 0, from which EG D �1476.1 lb �C�.
The sum of the forces:
∑
FY D �FG sin ˇ � FH sin ˛� EG sin ˛�W sin ˛�W D 0,
∑
FX D �FG cos ˇ � FH cos ˛� EG cos ˛�W cos ˛ D 0,
from which the two simultaneous equations:
�0.9848FG � 0.5736FH D 726.9, and �0.1736FG � 0.8192FH D
�389.97.
Solve: FG D �1158.5 lb �C� , and FH D 721.64 lb �T�. Make the
cut through JH, HI, and GI, and consider the upper section.
The section as a free body: The perpendicular distance from point
H to the line of action of the weight is LHW D 3 cos ˛C 3
p
2 sin ˇ C
3.5 D 10.135 ft. The sum of the moments about H is MH D �W�L��
jGIj�3�CW�3.25� D 0, from which jGIj D �2295 lb �C�.
∑
FY D �HI sin ˇ � JH sin ˛�GI sin ˛�W sin ˛�W D 0,
∑
FX D �HI cos ˇ � JH cos ˛�GI cos ˛�W cos ˛ D 0,
from which the two simultaneous equations:
�0.9848HI� 0.5736JH D 257.22,
and �0.1736HI� 0.8192JH D �1060.8.
Solve: HI D �1158.5 lb�C� ,
and JH D 1540.6 lb�T� .
W
W W
W
FH
FG
JH
HI
GI
EG
α
β
3.25 ft
3 ft 3.5 ft
(b) Choose a coordinate system with the y axis parallel to JH. Isolate
a section by making cuts through FH, FG, and EG, and through HJ,
HI, and GI. The free section of the truss is shown. The sum of the
forces in the x- and y-direction are each zero; since the only external
x-components of axial force are those contributed by FG and HI, the
two axial forces must be equal:
∑
Fx D HI cos 45° � FG cos 45° D 0,
from which HI D FG
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.
431
Problem 6.53 Consider the truss in Problem 6.52. The
weight of the bucket is W D 1000 lb. The cable passes
over pulleys at A and D. Determine the axial forces in
members IK and JL.
Solution: Make a cut through JL, JK, and IK, and consider the
upper section. Denote the axial force in a member joining, ˛, ˇ by
˛ˇ. The section as a free body: The perpendicular distance from point
J to the line of action of the weight is L D 6 cos ˛C 3p2 sin ˇ C
3.5 D 12.593 ft. The sum of the moments about J is MJ D �W�L�C
W�3.25�� IK�3� D 0, from which IK D �3114.4 lb�C�.
The sum of the forces:
∑
Fx D JL cos ˛� IK cos ˛
�W cos ˛� JK cos ˇ D 0,
and
∑
Fy D �JL sin ˛� IK sin ˛
�W sin ˛�W� JK sin ˇ D 0,
from which two simultaneous equations:
0.8192JL C 0.1736JK D �1732
and 0.5736JL C 0.9848JK D 212.75.
Solve: JL D 2360 lb�T� ,
and JK D �1158.5lb�C� .
W
W
3.5 ft
3 ft
3.25 ft
β
αJL
JK
IK
Problem 6.54 The truss supports loads at N, P, and R.
Determine the axial forces in members IL and KM.
2 m
2 m
2 m
2 m
1 m
6 m
2 m 2 m 2 m 2 m 2 m
K
I
M
L
O
N
Q
P RJ
H
F
D
G
E
C
BA
1 kN 2 kN 1 kNSolution: The strategy is to make a cut through KM, IM, and
IL, and consider only the outer section. Denote the axial force in a
member joining, ˛, ˇ by ˛ˇ.
The section as a free body: The moment about M is
MM D �IL � 2�1�� 4�2�� 6�1� D 0,
from which IL D �16 kN �C� .
The angle of member IM is ˛ D tan�1�0.5� D 26.57°.
The sums of the forces:
∑
Fy D �IM sin ˛� 4 D 0,
from which IM D � 4
sin ˛
D �8.944 kN (C).
∑
Fx D �KM� IM cos ˛� IL D 0,
from which KM D 24 kN�T�
α
KM
IM
IL
1 kN 2 kN 1 kN
1 m
2 m 2 m 2 m
432
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 6.55 Consider the truss in Problem 6.54.
Determine the axial forces in members HJ and GI.
Solution: The strategy is to make a cut through the four members
AJ, HJ, HI, and GI, and consider the upper section. The axial force
in AJ can be found by taking the moment of the structure about B.
The complete structure as a free body: The angle formed by AJ with the
vertical is ˛ D tan�1
(
4
8
)
D 26.57° . The moment about B is MB D
6AJ cos ˛� 24 D 0, from which AJ D 4.47 kN (T).
The section as a free body: The angles of members HJ and HI relative
to the vertical are ˇ D tan�1
(
2
8
)
D 14.0°, and � D tan�1
(
1.5
2
)
D
36.87° respectively. Make a cut through the four members AJ, HJ,
HI, and GI, and consider the upper section. The moment about
the point I is MI D �24C 2AJ cos ˛C 2HJ cos ˇ D 0. From which
HJ D 8.25 kN �T� . The sums of the forces:
∑
Fx D �AJ sin ˛CHJ sin ˇ�HI sin � D 0,
from which HI D AJ sin ˛�HJ sin ˇ
sin �
D 2� 2
sin �
D 0.
∑
FY D �AJ cos ˛�HJ cos ˇ �HI cos � �GI� 4 D 0,
from which GI D �16 kN �C�
AJ HJ
HI GI
2 m 2 m 2 m
1 kN 2 kN 1 kN
2 m 2 m
1 m
I
γα β
Problem 6.56 Consider the truss in Problem 6.54. By
drawing free-body diagrams of sections, explain why the
axial forces in members DE, FG, and HI are zero.
Solution: Define ˛, ˇ to be the interior angles BAJ and ABJ
respectively. The sum of the forces in the x-direction at the base yields
AX C BX D 0, from which Ax D �Bx . Make a cut through AJ, BD and
BC, from which the sum of forces in the x-direction, Ax � BD sin ˇ D
0. Since Ax D AJ sin ˛, then AJ sin ˛� BD sin ˇ D 0. A repeat of the
solution to Problem 6.55 shows that this result holds for each section,
where BD is to be replaced by the member parallel to BD. For example:
make a cut through AJ, FD, DE, and CE. Eliminate the axial force
in member AJ as an unknown by taking the moment about A. Repeat
the solution process in Problem 6.55, obtaining the result that
DE D AJ sin ˛� DF sin ˇ
cos �DE
D 0
where �DE is the angle of the member DE with the vertical. Similarly,
a cut through AJ, FH, FG, and EG leads to
FG D AJ sin ˛� FH sin ˇ
cos �FG
D 0,
and so on. Thus the explanation is that each member BD, DF, FH and
HJ has equal tension, and that this tension balances the x-component
in member AJ
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.
433
Problem 6.57 In Active Example 6.5, draw the free-
body diagram of joint B of the space truss and use it to
determine the axial forces in members AB, BC, and BD.
1200 lb
B
D (10, 0, 0) ft
C (6, 0, 6) ft
A (5, 3, 2) ft
z
y
x
Solution: From Active Example 6.5 we know that the vertical
reaction force at B is 440 lb.
The free-body diagram of joint B is shown. We have the following
position vectors.
rBA D �5iC 3jC 2k� ft
rBC D �6iC 6k� ft
rBD D �10i� ft
The axial forces in the rods can then be written as
TAB
rBA
jrBAj D TAB�0.811iC 0.487jC 0.324k�
TBC
rBC
jrBCj D TBC�0.707iC 0.707k�
TBD
rBD
jrBDj D TBDi
The components of the equilibrium equations are
Fx : 0.811TAB C 0.707TBC C TBD D 0
Fy : 0.487TAB C 440 lb D 0
Fz : 0.324TAB C 0.707TBC D 0
Solving yields TAB D �904 lb, TBC D 415 lb, TBD D 440 lb
Thus AB : 904 lb (C), BC : 415 lb (T), BD : 440 lb (T)
434
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 6.58 The space truss supports a vertical 10-
kN load at D. The reactions at the supports at joints A,
B, and C are shown. What are the axial forces in the
members AD, BD, and CD?
B (5, 0, 3) m
AyAx
Az
Cy
Cz
By
C (6, 0, 0) m
D (4, 3, 1) m
10 kN
z
y
x
A
Solution: Consider the joint D only. The position vectors parallel
to the members from D are
rDA D �4i� 3j� k,
rDB D i� 3jC 2k,
rDC D 2i� 3j� k.
The unit vectors parallel to the members from D are:
eDA D rDAjrDAj D �0.7845i� 0.5883j � 0.1961k
eDB D rDBjrDBj D 0.2673i� 0.8018j C 0.5345k
eDC D rDCjrDCj D 0.5345i� 0.8018j � 0.2673k
The equilibrium conditions for the joint D are
∑
F D TDAeDA C TDBeDB C TDCeDC � FD D 0,
from which
∑
Fx D �0.7845TDA C 0.2673TDB C 0.5345TDC D 0
∑
Fy D �0.5883TDA � 0.8018TDB � 0.8108TDC � 10 D 0
∑
Fz D �0.1961TDA C 0.5345TDB � 0.2673TDC D 0.
Solve: TDA D �4.721 kN �C� , TDB D �4.157 kN �C�
TDC D �4.850 kN �C�
10 kN
TDC
TDB
TDA
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.
435
Problem 6.59 Consider the space truss in Prob-
lem 6.58. The reactions at the supports at joints A, B,
and C are shown. What are the axial forces in members
AB, AC, and AD?
Solution: The reactions at A are required for a determination of
the equilibrium conditions at A.
The complete structure as a free body: The position vectors are rAB D
5iC 3k, rAC D 6i, rAD D 4iC 3jC k. The sum of the forces:
∑
Fx D Ax D 0,
∑
Fy D Ay C Cy C By � 10 D 0,
and
∑
Fz D Az CCz D 0.
The moments due to the reactions:
M D rAB ð FB C rAC ð FC C rAD ð FD D 0
M D
∣∣∣∣∣∣
i j k
5 0 3
0 By 0
∣∣∣∣∣∣C
∣∣∣∣∣∣
i j k
6 0 0
0 Cy Cz
∣∣∣∣∣∣C
∣∣∣∣∣∣
i j k
4 3 1
0 �10 0
∣∣∣∣∣∣ D 0
D ��3By C 10�i� �6Cz�jC �5By C 6Cy � 40�k D 0.
These equations for the forces and moments are to be solved for the
unknown reactions. The solution:
Ax D Cz D 0,
Ay D 2.778 kN,
By D 3.333 kN,
and Cy D 3.889 kN
The method of joints: Joint A: The position vectors are given above.
The unit vectors are:
eAB D 0.8575iC 0.5145k,
eAC D i,
eAD D 0.7845iC 0.5883jC 0.1961k.
The equilibrium conditions are:
∑
F D TABeAB C TAC C eAC C TADeAD C A D 0,
from which
∑
Fx D 0.8575TAB C TAC C 0.7845TAD D 0
∑
Fy D 0TAB C 0TAC C 0.5883TAD C 2.778 D 0
∑
Fz D 0.5145jTABj C 0jTACj C 0.1961jTADj D 0.
Solve: TAB D 1.8 kN �T� , TAC D 2.16 kN �T�
TAD D �4.72 kN �C�
Ay
Ax
Az TAB
TAC
TAD
436
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 6.60 The space truss supports a vertical load
F at A. Each member is of length L, and the truss rests on
the horizontal surface on roller supports at B, C, and D.
Determine the axial forces in members AB, AC, and AD.
F
A
B
C
D
Solution: By symmetry, the axial forces in members AB, AC, and
AD are equal. We just needto determine the angle � between each of
these members and the vertical:
F
A
TAB
TAC = TAB
TAD = TAB
θ
θ
θ
FC 3TAB cos � D 0,
so TAB D TAC D TAD D � F3 cos � .
From the top view,
L
C
b
60°
30°
L /2
we see that
b(
L
2
) D tan 30°
and
bC c(
L
2
) D tan 60°,
from which we obtain
c D 1
2
L�tan 60° � tan 30°�.
Then � D arcsin
( c
L
)
D 35.26°
and TAB D TAC D TAD D � F
3 cos 35.26°
D �0.408F.
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.
437
Problem 6.61 For the truss in Problem 6.60, deter-
mine the axial forces in members AB, BC, and BD.
Solution: See the solution of Problem 6.60. The axial force in
member AB is TAB D �0.408F, and the angle between AB and the
vertical is � D 35.26°. The free-body diagram of joint B is
TAB
TBC
TBD = TBC
θ
30°
30°
From the equilibrium equation
TAB sin � C 2TBC cos 30° D 0,
we obtain
TBC D TBD D 0.136F.
Problem 6.62 The space truss has roller supports at B,
C, and D and supports a vertical 800-lb load at A. What
are the axial forces in members AB, AC, and AD? 800 lb
B
D (6, 0, 0) ft
C (5, 0, 6) ft
A (4, 3, 4) ft
z
y
x
Solution: The position vectors of the points A, B, C, and D are
rA D 4iC 3jC 4k,
rC D 5iC 6k,
rD D 6i.
The position vectors from joint A to the vertices are:
rAB D rB � rA D �4i� 3j� 4k,
rAC D rC � rA D 1i� 3jC 2k,
rAD D rD � rA D 2i� 3j� 4k
Joint A: The unit vectors parallel to members AB, AC, and AD are
eAB D rABjrABj D �0.6247i� 0.4685j� 0.6247k,
eAC D rACjrACj D 0.2673i� 0.8018j C 0.5345k,
and eAD D rADjrADj D 0.3714i� 0.5570j � 0.7428k.
The equilibrium conditions at point A:
∑
Fx D �0.6247TAB C 0.2673TAC C 0.3714TAD D 0
∑
Fy D �0.4685TAB � 0.8018TAB � 0.5570TAD � 800 D 0
∑
Fz D �0.6247TAB C 0.5345TAC � 0.7428TAD D 0.
800 lb
TAD
TAC
TAB
Solve: TAB D �379.4 lb �C� , TAC D �665.2 lb �C� ,
and TAD D �159.6 lb �C�
438
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 6.63 The space truss shown models an
airplane’s landing gear. It has ball and socket supports
at C, D, and E. If the force exerted at A by the wheel is
F D 40j (kN), what are the axial forces in members AB,
AC, and AD?
B
(1, 0, 0) m
A
(1.1, –0.4, 0) m
0.4 m
0.6 m
y
x
z
E (0, 0.8, 0) m
C
D
F
Solution: The important points in this problem are A (1.1, �0.4,
0), B (1, 0, 0), C (0, 0, 6), and D (0, 0, �0.4). We do not need point
E as all of the needed unknowns converge at A and none involve the
location of point E. The unit vectors along AB, AC, and AD are
uAB D �0.243iC 0.970jC 0k,
uAC D �0.836iC 0.304jC 0.456k,
and uAD D �0.889iC 0.323j� 0.323k.
The forces can be written as
TRS D TRSuRS D TRSXiC TRSYjC TRSZk,
where RS takes on the values AB, AC, and AD. We now have three
forces written in terms of unknown magnitudes and known directions.
The equations of equilibrium for point A are∑
Fx D TABuABX C TACuACX C TADuADX C FX D 0,
∑
Fy D TABuABY C TACuACY C TADuADY C FY D 0,
and
∑
Fz D TABuABZ C TACuACZ C TADuADZ C FZ D 0,
where F D FXiC FYjC FZk D 40j kN. Solving these equations for
the three unknowns, we obtain TAB D �45.4 kN (compression),
TAC D 5.26 kN (tension), and TAD D 7.42 kN (tension).
y
z
x
E
D
C
B
F
A
(0, 0.8, 0) m
0.4
m
0.6
m (1, 0, 0) m
(1.1, −0.4, 0) m
TABTAD
TAC
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.
439
Problem 6.64 If the force exerted at point A of
the truss in Problem 6.63 is F D 10iC 60jC 20k (kN),
what are the axial forces in members BC, BD and BE?
Solution: The important points in this problem are A (1.1, �0.4,
0), B (1, 0, 0), C (0, 0, 0.6), D (0, 0, �0.4), and E (0, 0.8, 0). The
unit vectors along AB, AC, AD, BC, BD, and BE are
uAB D �0.243iC 0.970jC 0k,
uAC D �0.836iC 0.304jC 0.456k,
uAD D �0.889iC 0.323j� 0.323k,
uBC D �0.857iC 0jC 0.514k,
uBD D �0.928iC 0j� 0.371k,
and uBE D �0.781iC 0.625jC 0k.
The forces can be written as TRS D TRSuRS D TRSXiC TRSYjC
TRSZk, where RS takes on the values AB, AC, and AD when dealing
with joint A and AB, BC, BD, and BD when dealing with joint B. We
now have three forces written in terms of unknown magnitudes and
known directions.
Joint A: The equations of equilibrium for point A are,
∑
Fx D TABuABX C TACuACX C TADuADX C FX D 0,
∑
Fy D TABuABY C TACuACY C TADuADY C FY D 0,
and
∑
Fz D TABuABZ C TACuACZ C TADuADZ C FZ D 0,
where F D FXiC FYjC FZk D 10iC 60jC 20k kN. Solving these
equations for the three unknowns at A, we obtain TAB D �72.2 kN
(compression), TAC D �13.2 kN (compression), and TAD D 43.3 kN
(tension).
Joint B: The equations of equilibrium at B are
∑
Fx D �TABuABX C TBCuBCX C TBDuBDX C TBEuBEX D 0,
∑
Fy D �TABuABY C TBCuBCY C TBDuBDY C TBEuBEY D 0,
and
∑
Fz D �TABuABZ C TBCuBCZ C TBDuBDZ C TBEuBEZ D 0.
Since we know the axial force in AB, we have three equations in the
three axial forces in BC, BD, and BE. Solving these, we get TBC D
32.7 kN (tension), TBD D 45.2 kN (tension), and TBE D �112.1 kN
(compression).
y
z
x
E
C
D
B
F
A
(0, 0.8, 0) m
0.4
m
0.6
m
(1, 0, 0) m
(1.1, −0.4, 0) m
TAB
TDETAD
TBC
440
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 6.65 The space truss is supported by roller
supports on the horizontal surface at C and D and a ball
and socket support at E. The y axis points upward. The
mass of the suspended object is 120 kg. The coordinates
of the joints of the truss are A: (1.6, 0.4, 0) m, B: (1.0,
1.0, �0.2) m, C: (0.9, 0, 0.9) m, D: (0.9, 0, �0.6) m,
and E: (0, 0.8, 0) m. Determine the axial forces in
members AB, AC, and AD.
x
y
z
B
A
D
C
E
Solution: The important points in this problem are A: (1.6, 0.4,
0) m, B: (1, 1, �0.2) m, C: (0.9, 0, 0.9) m, and D: (0.9, 0, �0.6) m.
We do not need point E as all of the needed unknowns converge at A
and none involve the location of point E. The unit vectors along AB,
AC, and AD are
uAB D �0.688iC 0.688j� 0.229k,
uAC D �0.579i� 0.331jC 0.745k,
and uAD D �0.697i� 0.398j� 0.597k.
The forces can be written as TRS D TRSuRS D TRSXiC TRSYjC
TRSZk, where RS takes on the values AB, AC, and AD. We now
have three forces written in terms of unknown magnitudes and known
directions. The equations of equilibrium for point A are
∑
Fx D TABuABX C TACuACX C TADuADX C FX D 0,
∑
Fy D TABuABY C TACuACY C TADuADY C FY D 0,
and
∑
Fz D TABuABZ C TACuACZ C TADuADZ C FZ D 0,
where F D FXiC FYjC FZk D �mgj D �1177j N. Solving these
equations for the three unknowns, we obtain TAB D 1088 N (tension),
TAC D �316 N (compression), and TAD D �813 N (compression).
y
x
E B
D
C
z
A
mg
TAB
TAD
TAC
L
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.
441
Problem 6.66 The free-body diagram of the part of the
construction crane to the left of the plane is shown. The
coordinates (in meters) of the joints A, B, and C are (1.5,
1.5, 0), (0, 0, 1), and (0, 0, �1), respectively. The axial
forces P1, P2, and P3 are parallel to the x axis. The axial
forces P4,P5, and P6 point in the directions of the unit
vectors
e4 D 0.640i� 0.640j� 0.426k,
e5 D 0.640i� 0.640j� 0.426k,
e6 D 0.832i� 0.555k.
The total force exerted on the free-body diagram by the
weight of the crane and the load it supports is �Fj D
�44j (kN) acting at the point (�20, 0, 0) m. What is
the axial force P3?
Strategy: Use the fact that the moment about the line
that passes through joints A and B equals zero.
y
x
z
P1
A
B
F
C
P4P5
P3P6
P2
Solution: The axial force P3 and F are the only forces that exert
moments about the line through A and B. The moment they exert about
pt B is
MB D

 i j k�20 0 �1
0 �44 0

C

 i j k0 0 �2
P3 0 0


D �44i� 2P3jC 880k (kN-m).
The position vector from B to A is
rBA D 1.5iC 1.5j� k (m),
and the unit vector that points from B toward A is
eBA D rBAjrBAj D 0.640iC 0.640j� 0.426k.
From the condition that
eBA ÐMB D 0.640��44� C 0.640��2P3�
� 0.426�880� D 0,
we obtain P3 D �315 kN.
442
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 6.67 In Problem 6.66, what are the axial
forces P1, P4, and P5?
Strategy: Write the equilibrium equations for the
entire free-body diagram.
Solution: The equilibrium equations are
∑
Fx D P1 C P2 C P3 C 0.64P4 C 0.64P5 C 0.832P6 D 0,
∑
Fy D �0.64P4 � 0.64P5 � 44 D 0,
∑
Fz D �0.426P4 C 0.426P5 � 0.555P6 D 0,
∑
MB D

 i j k�20 0 �1
0 �44 0

C

 i j k0 0 �2
P3 0 0


C

 i j k1.5 1.5 �1
P1 0 0


C

 i j k1.5 1.5 �1
0.64P4 �0.64P4 �0.426P4


C

 i j k1.5 1.5 �1
0.64P5 �0.64P5 0.426P5

 D 0.
The components of the moment equation are
∑
MBx D �44� 1.279P4 � 0.001P5 D 0,
∑
MBy D �2P3 � P1 � 0.001P4 � 1.279P5 D 0,
∑
MBz D 880� 1.5P1 � 1.92P4 � 1.92P5 D 0.
Solving these equations, we obtain
P1 D 674.7 kN,
P2 D P3 D �315.3 kN,
P4 D P5 D �34.4 kN,
and P6 D 0.
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.
443
Problem 6.68 The mirror housing of the telescope is
supported by a 6-bar space truss. The mass of the
housing is 3 Mg (megagrams), and its weight acts at G.
The distance from the axis of the telescope to points A,
B, and C is 1 m, and the distance from the axis to points
D, E, and F is 2.5 m. If the telescope axis is vertical
(˛ D 90°), what are the axial forces in the members of
the truss?
Mirror housing
A
B
CG
E
D
F
4 m
1 m
A FD
B C
E
60°
60° 60°
60°
60°60°
G
END VIEW y
x
z
y
α
Solution: A cut through the 6-bar space truss leads to six equations
in the unknowns (see Problem 6.59). However for this problem an
alternate strategy based on reasonable assumptions about the equality
of the tensions is used to get the reactions. Assume that each support
carries one-third of the weight, which is equally divided between the
two bars at the support.
The coordinate system has its origin in the upper platform, with the
x axis passing though the point C. The coordinates of the points are:
A�� cos 60°, sin 60°, 0� D ��0.5, 0.866, 0�,
B�� cos 60°,� sin 60°, 0� D ��0.5,�0.866, 0�,
C�1, 0, 0�,
D��2.5, 0,�4�,
E�2.5 cos 60°,�2.5 sin 60°,�4� D �1.25,�2.165,�4�,
F�2.5 cos 60°, 2.5 sin 60°,�4� D �1.25, 2.165,�4�.
Consider joint B in the upper housing. The position vectors of the
points E and D relative to B are
rBD D �2iC 0.866j� 4k,
rBE D 1.75i� 1.299j� 4k.
The unit vectors are
eBD D �0.4391iC 0.1901j� 0.8781k,
and eBE D 0.3842i� 0.2852j � 0.8781k.
The weight is balanced by the z components:
∑
Fz D �W3 � �0.8781�TBD � �0.8781�TBE D 0.
Assume that the magnitude of the axial force is the same in both
members BD and BE, TBE D TBD. The weight is W D 3�9.81� D
29.43 kN. Thus the result: TBE D TBD D �5.5858 kN �C� . From
symmetry (and the assumptions made above) the axial force is the
same in all members.
A
F
D
B C
E
60°
60° 60°
60°
60°60°
G
y
x
z
y
4 m
1 m
α
Mirror housing
A
B
E
D
F
CG
r = 1 m
R = 2.5 m
y
x
C
D
E
A
B
4 m
444
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 6.69 Consider the telescope described in
Problem 6.68. Determine the axial forces in the members
of the truss if the angle ˛ between the horizontal and the
telescope axis is 20°.
Solution: The coordinates of the points are,
A�� cos 60°, sin 60°, 0� D ��0.5, 0.866, 0� �m�,
B�� cos 60°,� sin 60°, 0� D ��0.5,�0.866, 0� �m�,
C�1, 0, 0� �m�,
D��2.5, 0,�4� �m�,
E�2.5 cos 60°,�2.5 sin 60°,�4� D �1.25,�2.165,�4� �m�,
F�2.5 cos 60°, 2.5 sin 60°,�4� D �1.25, 2.165,�4� �m�.
The coordinates of the center of gravity are G (0, 0, 1) (m). Make a
cut through the members just below the upper platform supports, such
that the cut members have the same radial distance from the axis as
the supports. Consider the upper section.
The section as a free body: The strategy is to sum the forces and
moments to obtain six equations in the six unknown axial forces. The
axial forces and moments are expressed in terms of unit vectors. The
position vectors of the points E, D, and F relative to the points A, B,
and C are required to obtain the unit vectors parallel to the members.
The unit vectors are obtained from these vectors. The vectors and their
associated unit vectors are given in Table I. Note: While numerical
values are shown below to four significant figures, the calculations
were done with the full precision permitted (15 digits for TK Solver
Plus.)
Table I
Vector x y z Unit x y z
Vector
rAD �2 �0.866 �4 eAD �0.4391 �0.1901 �0.8781
rAF 1.75 1.299 �4 eAF 0.3842 0.2852 �0.8781
rBD �2 0.866 �4 eBD �0.4391 0.1901 �0.8781
rBE 1.75 �1.299 �4 eBE 0.3842 �0.2852 �0.8781
rCE 0.25 �2.165 �4 eCE 0.0549 �0.4753 �0.8781
rCF 0.25 2.165 �4 eCF 0.0549 �0.4753 �0.8781
The equilibrium condition for the forces is
jTABjeAD C jTAFjeAF C jTBDjeBD C jTBEjeBE C jTCEjeCE
C jTCFjeCF CW D 0.
This is three equations in six unknowns. The unit vectors are given in
Table I. The weight vector is W D jWj��j cos ˛� k sin ˛�, where ˛ is
the angle from the horizontal of the telescope housing. The remaining
three equations in six unknowns are obtained from the moments:
rA ð �TAD C TAF�C rB ð �TBD C TBE�C rC ð �TCE
C TCF�C rG ðW D 0.
D
A
B
C
E
F
y
x
−25000
−100 −50 0 50 100
−20000
−15000
−10000
−5000
0
5000
10000
15000
20000
25000
A
x
i
a
l
F
,
N
Axial Forces in Bars
|AF| & |CF|
|AD| & |BD|
|CE| & |BD|
alpha, deg
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.
445
6.69 (Continued )
Carry out the indicated operations on the moments to obtain the vectors
defining the moments:
rA ð TAD D jTADj
∣∣∣∣∣∣
i j k
�0.5 0.866 0
�0.4391 �0.1901 �0.8781
∣∣∣∣∣∣
D jTADj��0.7605i� 0.4391jC 0.4753�
D jTADj�iuADx C juADy C juADz�
rA ð TAF D jTAFj
∣∣∣∣∣∣
i j k
�0.5 0.866 0
0.3842 0.2852 �0.8781
∣∣∣∣∣∣
D jTAFj��0.7605i� 0.4391j� 0.4753k�
D jTAFj�iuAFx C juAFy C kuAFz�
rB ð TBD D jTBDj
∣∣∣∣∣∣
i j k
�0.5 �0.866 0
�0.4391 0.1901 �0.8781
∣∣∣∣∣∣
D jTBDj�0.7605i� 0.4391j� 0.4753k�
D jTBDj�iuBDx C juBDy C kuBDz�
rB ð TBE D jTBEj
∣∣∣∣∣∣
i j k
�0.5 �0.866 0
0.3842 �0.2852 �0.8781
∣∣∣∣∣∣
D jTBEj�0.7605i� 0.4391j� 0.4753k�
D jTBEj�iuBEx C juBEy C kuBEz�
rC ð TCE D jTCEj
∣∣∣∣∣∣
i j k
1 0 0
0.0549 �0.4753 �0.8781
∣∣∣∣∣∣
D jTCEj�0iC0.8781j� 0.4753k�
D jTCEj�iuCEx C juCEy C kuCEz�
rC ð TCF D jTCFj
∣∣∣∣∣∣
i j k
1 0 0
0.0549 0.4753 �0.8781
∣∣∣∣∣∣
D jTCFj�0iC 0.8781jC 0.4753k�
D jTCFj�iuCFx C juCFy C kuCFz�
rG ðW D jWj
∣∣∣∣∣∣
i j k
0 0 1
�0 � cos ˛ � sin ˛
∣∣∣∣∣∣
D jWj�i cos ˛� j0C k0� D �iMWx�
The six equations in six unknowns are:
jTADjeADx C jTAFjeAFx C jTBDjeBDx C jTBEjeBEx C jTCEjeCEx
C jTCFjeCFx CWx D 0
jTADjeADy C jTAFjeAFy C jTBDjeBDy C jTBEjeBEy C jTCEjeCEy
C jTCFjeCFy CWy D 0
jTADjeADz C jTAFjeAFz C jTBDjeBDz C jTBEjeBEz C jTCEjeCEz
C jTCFjeCFz CWz D 0
jTADjuADx C jTAFjuAFx C jTBDjuBDx C jTBEjuBEx C jTCEjuCEx
C jTCFjuCFx CMWx D 0
jTADjuADy C jTAFjuAFy C jTBDjuBDy C jTBEjuBEy C jTCEjuCEy
C jTCFjuCFy D 0,
jTADjuADz C jTAFjuAFz C jTBDjuBDz C jTBEjuBEz C jTCEjuCEz
C jTCFjuCFz D 0
This set of equations was solved by iteration using TK Solver 2. For
˛ D 20° the results are:
jTADj D jTBDj D �1910.5 N �C� ,
jTAFj D jTCFj D 16272.5 N �T� ,
jTBEj D jTCEj D �19707 N �C� .
Check: For ˛ D 90°, the solution is jTADj D jTAFj D jTBDj D jTBEj D
jTCEj D jTCFj D �5585.8 N �C�, which agrees with the solution to
Problem 6.68, obtained by another method. check.
Check: The solution of a six-by-six system by iteration has risks, since
the matrix of coefficients may be ill-conditioned. As a reasonableness
test for the solution process, TK Solver Plus was used to graph the
axial forces in the supporting bars over the range �90° < ˛ < 90°.
The graph is shown. The negative values are compression, and the
positive values are tension. When ˛ D �90°, the telescope platform is
pointing straight down, and the bars are in equal tension, as expected.
When ˛ D 90° the telescope mount is upright and the supporting bars
are in equal compression, as expected. The values of compression
and tension at the two extremes are equal and opposite in value,
and the values agree with those obtained by another method (see
Problem 6.58), as expected. Since the axial forces go from tension
to compression over this range of angles, all axial forces must pass
through zero in the interval. check.
446
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 6.70 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 C that 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
A B
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.
Starting with bar BC, we have the equilibrium equations
MB : C�400 mm�
� �400 N��400 mm�
� �200 N-m� D 0
Fy : C� By D 0
Fx : �Bx � 400 N D 0
Now using bar AB we have the equilibrium equations
Fx : Ax C Bx D 0
Fy : Ay C By D 0
MA : MA C By �600 mm� D 0
Solving these six equations yields C D 900 N and
Ax D 400 N, Ay D �900 N
Bx D �400 N, By D 900 N
MA D �540 N-m
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.
447
Problem 6.71 The object suspended at E weighs
200 lb. Determine the reactions on member ACD at A
and C.
3 ft
5 ft
6 ft4 ft
A
B C
D
E
Solution: We start with the free-body diagram of the entire frame.
We have the equilibrium equations:
Fx : Ax D 0
Fy : Ay � 200 lb D 0
MA : MA � �200 lb� �6 ft� D 0
Next we use the free-body diagram of the post ACD. Notice that BD
is a two-force body and the angle ˛ is
˛ D tan�1�3/4� D 36.9°
The equilibrium equations are
MC : MA C Ax �5 ft�C TBD cos ˛ �3 ft� D 0
Fx : Ax CCx � TBD cos ˛ D 0
Fy : Ay CCy � TBD sin ˛ D 0
Solving these six equations we find TBD D �500 lb and
Ax D 0, Ay D 200 lb
Cx D �400 lb, Cy D �500 lb
MA D 1200 ft-lb
448
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 6.72 The mass of the object suspended at G
is 100 kg. Determine the reactions on member CDE at
C and E.
A
B E
D
C
F G
400 mm 400 mm 400 mm800 mm
200 mm
800 mm
Solution: The free-body diagram of the entire frame and of mem-
ber CDE are shown. The angle ˛ is
˛ D tan�1�4/8� D 26.6°
The equilibrium equations are
MC : TAB cos ˛ �400 mm�
C TAB sin ˛ �800 mm�
� �981 N��1200 mm� D 0
Fx : Cx � TAB sin ˛ D 0
Fy : Cy � TAB cos ˛� 981 N D 0
The free-body diagram for bar CDE is shown. Note that DF is a
two-force member. The angle ˇ is
ˇ D tan�1�3/4� D 36.9°
The equilibrium equations are
ME : TDF cos ˇ�600 mm�CCx�800 mm� D 0
Fx : TDF cos ˇ C Ex CCx D 0
Fy : TDF sin ˇC Ey CCy D 0
Solving these six equations, we find
TAB D 1650 N, TDF D �1230 N and
Cx D 736 N, Cy D 2450 N
Ex D 245 N, Ey D �1720 N
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.
449
Problem 6.73 The force F D 10 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
B C
E G
F
Solution: The complete structure as a free body: The sum of the
moments about G:
∑
MG D C3F� 5A D 0,
from which A D 3F
5
D 6 kN which is the reaction of the floor. The
sum of the forces:
∑
Fy D Gy � FC A D 0,
from which Gy D F� A D 10� 6 D 4 kN.
∑
Fx D Gx D 0.
Element DEG: The sum of the moments about D
∑
M D �FC 3EC 4Gy D 0,
from which E D F� 4Gy
3
D 10� 16
3
D �2 kN.
The sum of the forces:
∑
Fy D Gy � FC EC D D 0,
from which D D F� E�Gy D 10C 2� 4 D 8 kN.
Element ABC : Noting that the reactions are equal and opposite:
B D �D D �8 kN ,
and C D �E D 2 kN .
The sum of the forces:
∑
Fy D AC BCC D 0,
from which A D 8� 2 D 6 kN. Check
F
A
GY
GX
2 m 3 m
F
E
F
D
A
GY
1m
1m
2m 1m
B = −D C = −E
8 kN
3 m1 m
2 kN
6 kN
BA C
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 C to the
attachment point E of the two-force member BE is
increased from 8 in to 10 in. Determine the forces acting
at C on 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.
From the example we know that
Ax D 42.2 lb, Ay D 40 lb, D D 42.4 lb
The free-body diagram for ABCD is shown. Note that BE is a two-force
body. The angle ˛ is now
˛ D tan�1�6/10� D 31.0°
The equilibrium equations are
MC : TBE cos ˛ �6 in�C D �6 in�C Ax �6 in� D 0
Fx : TBE cos ˛CCx C Ax � D D 0
Fy : TBE sin ˛C Cy C Ay D 0
Solving yields TBE D �124 lb and
Cx D 66.7 lb, Cy D 24 lb
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.
451
Problem 6.75 The tension in cable BD is 500 lb.
Determine the reactions at A for cases (1)and (2).
G
6 in
6 in
8 in 8 in
E
A B C
300 lb
D
(1)
G
6 in
6 in
8 in 8 in
E
A B C
300 lb
D
(2)Solution: Case (a) The complete structure as a free body: The sum
of the moments about G:
∑
MG D �16�300� C 12Ax D 0,
from which Ax D 400 lb . The sum of the forces:
∑
Fx D Ax C Gx D 0,
from which Gx D �400 lb.
∑
Fy D Ay � 300CGy D 0,
from which Ay D 300�Gy . Element GE : The sum of the moments
about E:
∑
ME D �16Gy D 0,
from which Gy D 0, and from above Ay D 300 lb.
Case (b) The complete structure as a free body: The free body diagram,
except for the position of the internal pin, is the same as for case (a).
The sum of the moments about G is
∑
MG D �16�300� C 12Ax D 0,
from which Ax D 400 lb .
Element ABC : The tension at the lower end of the cable is up and to
the right, so that the moment exerted by the cable tension about point
C is negative. The sum of the moments about C:
∑
MC D �8B sin ˛� 16Ay D 0,
noting that B D 500 lb and ˛ D tan�1
(
6
8
)
D 36.87°,
then Ay D �150 lb.
(a) 12 in
16 in
Gx
Gx Ex
Gy
EyGy
Ay
Ax 300 lb
(b)
Ay Cy
CxAx
B
α
300 lb
8 in8 in
452
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 6.76 Determine the reactions on member
ABCD at A, C, and D.
B
C
0.4 m
0.4 m600 N
0.6 m 0.4 m 0.4 m
E
A
D
Solution: Consider the entire structure first∑
MA : Dy�0.6 m�� �600 N��1.0 m� D 0) Dy D 1000 N
∑
Fx : Ax D 0
∑
Fy : Ay C Dy � 600 N D 0) Ay D �400 N
600 N
E
C
Ay
Dy
Ax
Now examine bar CE. Note that the reactions on ABD are opposite to
those on CE.∑
ME : �600 N��0.4 m�CCy�0.8 m� D 0) Cy D �300 N
∑
MB : �Cx�0.4 m�� �600 N��0.4 m� D 0) Cx D �600 N
600 N
E
T
Cy
Cx
In Summary we have
Ax D 0, Ay D �400 N
Cx D �600 N, Cy D �300 N
Dx D 0, Dy D 1000 N
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.
453
Problem 6.77 Determine the forces exerted on member
ABC at A and C.
C
D
BA
100 lb
E
400 lb2 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
Fx : Ax C 100 lb D 0
ME : �Ax �2 ft�� Ay �4 ft�
� �100 lb� �1 ft�
� �400 lb� �2 ft� D 0
Next we examine the free-body diagram of bar ABC. Note that BD
is a two-force body and that the angle ˛ D 45°. The equilibrium
equations are
MC : �Ay �4 ft�� TBD sin ˛ �2 ft�
� �400 lb� �2 ft� D 0
Fx : Ax C TBD cos ˛CCx D 0
Fy : Ay C TBD sin ˛C Cy � 400 lb D 0
Solving, we find that TBD D �70.7 lb and
Ax D �100 lb, Ay D �175 lb
Cx D 150 lb, Cy D 625 lb
454
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 6.78 An athlete works out with a squat thrust
machine. To rotate the bar ABD, she must exert a vertical
force at A that 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 D and E?
0.6 m0.6 m
E
D
BA
C
1.65 m
0.42 m
Solution: Member BC is a two force member. The force in BC is
along the line from B to C.
y
D
C
Dx
Dy
x
Ay
FBC
0.6 m 0.6 m
0.42 mθ
(FBC = 1800 N)
tan Θ = 0.42
0.6
Θ = 34.990
�FBC D 1800 N�
tan � D 0.42
0.6
� D 34.99°.
∑
Fx: Dx � FBC cos � D 0
∑
Fy : Ay � FBC sin � C Dy D 0
C
∑
MD: � 1.2Ay C 0.6FBC sin � D 0
Solving, we get Dx D 1475 N
Dy D 516 N
Ay D 516 N
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.
455
Problem 6.79 The frame supports a 6-kN load at C.
Determine the reactions on the frame at A and D.
0.4 m0.8 m
A B
D E
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.
Ax
Ay
FCFFBE
B C
0.4 m
1.0 m
6 kN
θ φ
tan � D 0.5
0.4
� D 51.34°
tan � D 0.5
0.2
� D 68.20°
Member DEF
Dx
Dy
FCFFBE
FE
0.8 m 0.4 m
θ
φ
Equations of equilibrium:
Member ABC:
∑
Fx: Ax C FBE cos � � FCF cos � D 0
∑
Fy : Ay � FBE sin � � FCF sin � � 6 D 0
C
∑
MA: � �0.4�FBE sin � � �1.4�FCF sin � � 1.4�6� D 0
Member DEF:
∑
Fx : Dx � FBE cos � C FCF cos � D 0
∑
Fy : Dy C FBE sin � C FCF sin � D 0
C
∑
MD: �0.8��FBE sin ��C 1.2FCF sin � D 0
Unknowns Ax, Ay, Dx, Dy, FBE, FCF we have 6 eqns in 6
unknowns.
Solving, we get
Ax D �16.8 kN
Ay D 11.25 kN
Dx D 16.3 kN
Dy D �5.25 kN
Also, FBE D 20.2 kN �T�
FCF D �11.3 kN �C�
456
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 6.80 The mass m D 120 kg. Determine the
forces on member ABC, presenting your answers as
shown in Fig. 6.25.
A B C
D
E
200 mm 200 mm
300 mm
m
m
Solution: The equations of equilibrium for the entire frame are
∑
FX D AX C EX D 0,
∑
FY D AY � 2mg D 0,
and summing moments at A,
∑
MA D �0.3�EX � �0.2�mg� �0.4�mg D 0.
Solving yields AX D �2354 N, AY D 2354 N, and EX D 2354 N.
Member ABC: The equilibrium equations are
∑
FX D AX CCX D 0,
∑
FY D AY � BY CCY D 0,
and
∑
MA D ��0.2�BY C �0.4�CY D 0.
We have three equations in the three unknowns BY, CX, and CY.
Solving, we get BY D 4708 N, CX D 2354 N, and CY D 2354 N. This
gives all of the forces on member ABC. A similar analysis can be made
for each of the other members in the frame. The results of solving for
all of the forces in the frame is shown in the figure.
AX
EX
AY BY
BY
DY
DY
CY
CY
CX
CX
2354 N 4708 N
4708 N
4708 N
4708 N
2354 N
2354 N
2354 N
2354 N
2354 N
1177 N
2354 N
1177 N
A
B C
B
E
D
C
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.
457
Problem 6.81 Determine the reactions on member
BCD.
C
E
A B
D F
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
free-body diagrams ares shown.
The angle ˛ D tan�1�18/24� D 36.9°
From ADG we have
MD : Bx�24 in�C �40 lb��8 in�
� �30 lb� �20 in� D 0
MA : By�18 in�� �40 lb��16 in�
� �30 lb��38 in� D 0
Fy : By � �30 lb�� TAD cos ˛ D 0
From DFG we have
MF : �Dy C TAD cos ˛� �12 in�
� �30 lb��8 in� D 0
And finally from BCD we have
Fy : By CCy C Dy D 0
MD : Bx�24 in�CCx�16 in� D 0
Fx : Dx CCx C Bx D 0
Solving these seven equations, we find TAD D 86.1 lb and
Bx D 11.7 lb, By D 98.9 lb
Cx D �17.5 lb, Cy D �50 lb
Dx D 5.83 lb, Dy D �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
W D 50 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
EC
D
F
Solution: Start with member AB
∑
MA : ��50 lb��8 in�C 1p
2
FB�16 in� D 0) FB D 35.4 lb
50 lb
Ay
Ax
1
1
FB
Now examine BCF∑
MF : FB�20
p
2� in� FC�10 in� D 0) FC D 100 lb
∑
Fx : � 1p
2
FB C FC C Fx D 0) Fx D �75 lb
∑
Fy : � 1p
2
FB C Fy D 0) Fy D 25 lb
Fy
Fx
1
1
FCFB
Finally examine DCE
∑
MD : �T�16 in�C FC�10 in� D 0) T D 62.5 lb
Dy
Dx
T
FC
Summary
Tension in Spring D 62.5 lb
Fx D 25 lb, Fy D �75 lb
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.
459
Problem 6.83 The mass m D 50 kg. Bar DE is
horizontal. Determine the forces on member ABCD,
presenting your answers as shown in Fig. 6.25.
A F
D E
B
C
1 m 1 m
1 m
1 m
1 m
m
Solution: The weight of the mass hanging is W D mg D
50�9.81� D 490.5 N The complete structure as a free body: The sum
of the moments about A:∑
MA D �2WC Fy D 0,
from which Fy D 981 N. The sum of the forces:
∑
Fy D Ay C Fy �W D 0,
from which Ay D �490.5 N,
∑
Fx D Ax C Fx D 0,
from which Ax D �Fx . Element BF: The sum of the moments
about F:∑
MF D �Bx � By D 0,
from which By D �Bx . The sum of the forces:
∑
Fy D By C Fy D 0,
from which By D �981 N, and Bx D 981 N.
∑
Fx D Bx C Fx D 0,
from which Fx D �981 N, and from above, Ax D 981 N ,
Element DE: The sum of the moments about D:∑
MD D �Ey � 2W D 0,
from which Ey D �981 N. The sum of the forces:
∑
Fy D �Dy � Ey �W D 0,
from which Dy D 490.5 N .
∑
Fx D �Dx � Ex D 0,
from which Dx D �Ex . Element CE : The sum of the moments
about C:∑
MC D Ey � Ex D 0,
from which Ex D �981 N, and from above Dx D 981 N .
∑
Fy D Ey CCy D 0,
from which Cy D 981 N.
∑
Fx D Ex CCx D 0,
from which Cx D 981 N, and
Element ABCD: All reactions on ABCD have been determined
above. The components at B and C have the magnitudes
B D C D p9812 C 9812 D 1387 N , at angles of 45°.
Dy
Dy
Cx
Bx
Ay
Ax
By
By
Bx
Cy
Cy
Dx
Dx Ex
Ex
EyEy
Fy
Fx
Cx
W
490.5 N
981 N
1387 N
D
C
B
A
45°
45°
490.5 N
981 N
1387 N
460
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 6.84 Determine the forces on member BCD.
A
6 ft
400 lb
4 ft
4 ft
8 ft
D
B
E
C
Solution: The following is based on free body diagrams of the
elements: The complete structure as a free body: The sum of the
moments about D:
∑
MD D ��6�400C 8Ey D 0,
from which Ey D 300 lb. The sum of the forces:
∑
Fx D Dx D 0.
∑
Fy D Ey C Dy � 400 D 0,
from which Dy D 100 lb. Element AB: The sum of the moments
about A:
∑
MA D �8By � �6�400 D 0,
from which By D �300 lb. The sum of forces:
∑
Fy D �By � Ay � 400 D 0,
from which Ay D �100 lb.
∑
Fx D �Ax � Bx D 0,
from which (1) Ax C Bx D 0 Element ACE: The sum of the moments
about E:
∑
ME D �8Ax C 4Cx � 8Ay C 4Cy D 0,
from which (2) �2Ax CCx � 2Ay CCy D 0. The sum of the forces:
∑
Fy D Ay C Ey � Cy D 0,
from which Cy D 200 lb .
∑
Fx D Ax �Cx D 0,
from which (3) Ax D Cx . The three numbered equations are solved:
Ax D �400 lb, Cx D 400 lb , and Bx D �400 lb .
Element BCD:
The reactions are now known:
By D �300 lb , Bx D �400 lb , Cy D 200 lb ,
Dx D 0 , Dy D 100 lb ,
where negative sign means that the force is reversed from the direction
shown on the free body diagram.
Ay
Ay
Cy
Cy
Cx
Cx
Dx
Dy
Ax
Ax
By
By
Bx
Bx
400 lb
E
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.
461
Problem 6.85 Determine the forces on member ABC.
1 m
D C
E
A B
1 m
2 m 2 m 1 m
6 kN
Solution: The frame as a whole: The equations of equilibrium are
∑
FX D AX C EX D 0,
∑
FY D AY C EY � 6000 N D 0,
and, with moments about E,
∑
ME D 2AX � �5�6000 D 0.
Solving for the support reactions, we get AX D 15,000 N and EX D
�15,000 N. We cannot yet solve for the forces in the y direction at A
and E.
Member ABC: The equations of equilibrium are
∑
FX D AX � BX D 0,
∑
FY D AY � BY � CY D 0,
and summing moments about A,
∑
MA D �2BY � 4CY D 0.
Member BDE: The equations of equilibrium are
∑
FX D EX C DX C BX D 0,
∑
FY D EY C DY C BY D 0,
and, summing moments about E,
∑
ME D �1�DY C �1�DX C �2�BY C �2�BX D 0.
Member CD: The equations of equilibrium are
∑
FX D �DX D 0,
∑
FY D �DY CCY � 6000 D 0,
and summing moments about D,
∑
MD D ��4�6000C 3CY D 0.
Solving these equations simultaneously gives values for all of the
forces in the frame. The values are AX D 15,000 N, AY D �8,000 N,
BX D 15,000 N, BY D �16,000 N, CY D 8,000 N, DX D 0, and
DY D 2,000 N.
EY
EX E DY
DX
DX D
BXBY
DY
BX
AX
AY BY
CY
CY
D
B
BA
C
C
6 kN
462
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 6.86 Determine the forces on member ABD.
8 in
8 in 8 in
60 lb 60 lb
8 in
8 in
A
C D
B
E
Solution: The equations of equilibrium for the frame as a
whole are
∑
FX D AX CCX D 0,
∑
FY D AY � 60� 60 D 0,
and
∑
MA D 16CX � 16�60�� 24�60� D 0.
Solving these three equations yields
AX D �150 lb,
AY D 120 lb,
and CX D 150 lb.
Member ABD: The equilibrium equations for this member are:
∑
FX D AX � BX � DX D 0,
∑
FY D AY � BY � DY D 0,
and
∑
MA D �8BY � 8DY � 8BX � 16DX D 0.
Member BE: The equilibrium equations for this member are:
∑
FX D BX C EX D 0,
∑
FY D BY C EY � 60� 60 D 0,
and
∑
MB D �8�60�� 16�60�C 16EY D 0.
Member CDE: The equilibrium equations for this member are:
∑
FX D CX C DX � EX D 0,
∑
FY D DY � EY D 0,
and
∑
MD D 8EX � 16EY D 0.
Solving these equations, we get BX D �180 lb, BY D 30 lb, DX D
30 lb, DY D 90 lb, EX D 180 lb, and EY D 90 lb. Note that we have
12 equations in 9 unknowns. The extra equations provide a check.
AY
AX
BX EX
EX
EY EYBX
BY
BY
DX
DY
DXCX
DY
B
60 lb 60 lb
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.
463
Problem 6.87 The mass m D 12 kg. Determine the
forces on member CDE.
200 mm
100 mm
200 mm
200 mm 400 mm
A
C D
B
m
E
Solution: Start with a free-body diagram of the entire frame.
Eq. for entire frame:
�
C!�
∑
Fx D 0: � Ax CCx D 0) Ax D Cx �1�
�C"�
∑
Fy D 0: Ay � 117.7 D 0) Ay D 117.7 N
C
∑
Mc D 0: Ax�0.4�� 117.7�0.7� D 0 Ax D 206 N
∴ Ax D Cx D 206 N.
Now look at free-body diagram ABD.
Eq. for ABD:
C
∑
MB D 0:
� Dx�0.2�� 117.7�0.2�C 206�0.2� � 117.7�0.1� D 0
Dx D 29.45 N
�
C!�
∑
Fx D 0:
� 206C 117.7C Bx � 29.45 D 0 Bx D 117.75 N
�C"�
∑
Fy D 0: 117.7� By C Dy D 0
Draw free-body diagram of CDE
Eq. for CDE:
�
C!�
∑
Fx D 0:
206C 29.45C Ex D 0 Ex D �235.45 or Ex D 235.45 
C
∑
MD D 0:
� Ex�0.2�C Ey�0.4� D 0 Ey D Ex�0.2�0.4 D
��235.45��0.2�
0.4
Ey D �117.7 or Ey D 117.7 N
�C"�
∑
Fy D 0:
Ey � Dy D 0 or Dy D �Ey D ���117.7� Dy D 117.7 N#
Ay
Ax
Cx
B
D W = (9.81) (12)
W = 117.7
Ax = 206
Ay = 117.7 T = 117.7
By
Bx 
Dx 
Dy
Cx = 206
Dx = 29.45
Dy
Ey
Ex
464
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 6.88 The weight W D 80 lb. Determine the
forces on member ABCD.
E
8 in
11 in 12 in5 in
3 in
F
W
A C DB
Solution: The complete structure as a free body: The sum of the
moments about A:
∑
MA D �31WC 8Ex D 0,
from which Ex D 310 lb. The sum of the forces:
∑
Fx D Ex C Ax D 0,
from which Ax D �310 lb .
∑
Fy D Ey C Ay �W D 0,
from which (1) Ey C Ay DW.
Element CFE: The sum of the forces parallel to x:
∑
Fx D Ex �Cx D 0,
from which Cx D 310 lb . The sum of the moments about E:
∑
ME D 8F� 16Cy C 8Cx D 0.
For frictionless pulleys, F DW, and thus Cy D 195 lb . The sum of
forces parallel to y:
∑
Fy D Ey �Cy C F D 0,
from which Ey D 115 lb .
Equation (1) above is now solvable: Ay D �35 lb .
Element ABCD: The forces exerted by the pulleys on element ABCD
are, by inspection: Bx DW D 80 lb , By D 80 lb , Dx D 80 lb ,
and Dy D �80 lb , where the negative sign means that the force is
reversed from the direction of the arrows shown on the free body
diagram.
Ay
Ax
Bx
Cx
Cy
Cx
Cy
Dy
Dx
By
Ey
Ex
W
F
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.
465
Problem 6.89 The woman using the exercise machine
is holding the 80-lb weight stationary in the position
shown. What are the reactions at the built-in support E
and the pin support F? (A and C are pinned connections.)
E F
80 lb
A
B
D2 ft 2 in
60�
2 ft
C
1 ft 6 in
6 ft
9 in
Solution: The complete structure as a free body: The sum of the
moments about E:
∑
M D �26W� 68W sin 60° C 50Fy � 81W cos 60° CME D 0
from which (1) 50Fy CME D 10031. The sum of the forces:
∑
Fx D Fx CW cos 60° C Ex D 0,
from which (2) Fx C Ex D �40.
∑
Fy D �W�W sin 60° C Ey C Fy D 0,
from which (3) Ey C Fy D 149.28
Element CF: The sum of the moments about F:
∑
M D �72Cx D 0,
from which Cx D 0. The sum of the forces:
∑
Fx D Cx C Fx D 0,
from which Fx D 0 . From (2) above, Ex D �40 lb
Element AE: The sum of the moments about E:
∑
M D ME � 72Ax D 0, .
from which (4) ME D 72Ax . The sum of the forces:
∑
Fy D Ey C Ay D 0,
from which (5) Ey C Ay D 0.
∑
Fx D Ax C Ex D 0;
from which Ax D 40 lb, and from (4) ME D 2880 in lb D 240 ft lb .
From (1) Fy D 143.0 lb , and from (2) Ey D 6.258 lb . This
completes the determination of the 5 reactions on E and F.
26
in
81 in
50 in
42
W
W
60°
in
ME
Ex
Ey Fy
Fx
ME
Ex
Ey
Cx
Fx
Ay
Ax
Fy
Cy
466
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 6.90 Determine the reactions on member
ABC at A and B.
E
B C
DA
9 in
8 in
4 in13 in
80 lb
Solution: We first examine the entire structure.∑
MD : �Ay �13 in�C �80 lb��21 in� D 0
Solving: Ay D 129.2 lb
80 lb
DyAy
DxAx
Next examine body ABC
∑
MB : Ax �8 in�� Ay �13 in�C �80 lb� 4 in D 0
∑
Fx : Ax C Bx D 0
∑
Fy : Ay C By C 80 lb D 0
Ax
Ay
Bx
By
80 lb
Solving and summarizing we have
Ax D 170 lb, Ay D 129.2 lb
Bx D �170 lb, By D �209 lb
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.
467
Problem 6.91 The mass of the suspended object is
m D 50 kg. Determine the reactions on member ABC.
D E 0.2 m
0.6 m0.8 m
0.2 m
C
0.6 m
A B
m
Solution: Begin with an examination of the pulley at B.
∑
Fx : �Bx C 1p
2
�490.5 N� D 0) Bx D 347 N
∑
Fy : �By � 490.5 N� 1p
2
�490.5 N� D 0) By D �837 N
490.5 N
490.5 N
By
Bx
1
1
Now examine the entire structure∑
MD : ��490.5 N��1.6 m�� Ax�0.6 m� D 0) Ax D �1308 N
490.5 N
Dy
Ay
Dx
Ax
Finally look at member ABC
∑
MC : �Ax�0.6 m�� Ay�1.4 m�� Bx�0.6 m�� By�0.6 m� D 0
) Ay D 771 N
∑
Fx : Ax C Bx CCx D 0) Cx D 961 N
∑
Fy : Ay C By CCy D 0) Cy D 66.6 N
By
Cy
Ay
Bx
Cx
Ax
In Summary
Ax D �1308 N, Ay D 771 N
Bx D 347 N, By D �837 N
Cx D 961 N, Cy D 66.6 N
468
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 6.92 The unstretched length of the string is
LO. Show that when the system is in equilibrium the
angle ˛ satisfies the relation sin ˛ D 2�LO � 2F/k�L.
k
F
1–
2
L
1–
4
L
1–
4
L
α α
Solution: Since the action lines of the force F and the reaction E
are co-parallel and coincident, the moment on the system is zero, and
the system is always in equilibrium, for a non-zero force F. The object
is to find an expression for the angle ˛ for any non-zero force F.
The complete structure as a free body:
The sum of the moments about A
∑
MA D �FL sin ˛C EL sin ˛ D 0,
from which E D F. The sum of forces:
∑
Fx D Ax D 0,
from which Ax D 0.
∑
Fy D Ay C E� F D 0,
from which Ay D 0, which completes a demonstration that F does not
exert a moment on the system. The spring C: The elongation of the
spring is s D 2 L
4
sin ˛� LO, from which the force in the spring is
T D k
(
L
2
sin ˛� LO
)
Element BE: The strategy is to determine Cy , which is the spring force
on BE. The moment about E is
∑
ME D �L4 Cy cos ˛�
L
2
By cos ˛� L2 Bx cos ˛ D 0,
from which
Cy
2
C By D �Bx . The sum of forces:
∑
Fx D Bx D 0,
from which Bx D 0.
∑
Fy D Cy C By C E D 0,
from which Cy C By D �E D �F. The two simultaneous equations
are solved: Cy D �2F, and By D F.
The solution for angle ˛: The spring force is
Cy D T D k
(
L
2
sin ˛� LO
)
,
from which k
(
L
2
sin ˛� LO
)
D �2F.
Solve: sin ˛ D
2
(
LO � 2F
k
)
L
F
E
L
α
α
E
Ay
By
CyBx
Ax
L
4 L
4
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.
469
Problem 6.93 The pin support B will safely support a
force of 24-kN magnitude. Based on this criterion, what
is the largest mass m that the frame will safely support?
C
B
500 mm
100 mm
300 mm
m
300 mm 400 mm 400 mm
D
E
A F
Solution: The weight is given by W D mg D 9.81 g
The complete structure as a free body:
Sum the forces in the x-direction:
∑
Fx D Ax D 0,
from which Ax D 0
Element ABC: The sum of the moments about A:
∑
MA D C0.3Bx C 0.9Cx � 0.4W D 0,
from which (1) 0.3Bx C 0.9Cx D 0.4W. The sum of the forces:
∑
Fx D �Bx �Cx CWC Ax D 0,
from which (2) Bx CCx DW. Solve the simultaneous equations (1)
and (2) to obtain Bx D 56 W
Element BE : The sum of the moments about E:
∑
ME D 0.4W� 0.7By D 0,
from which By D 4
7
W. The magnitude of the reaction at B is
jBj DW
√(
5
6
)2
C
(
4
7
)2
D 1.0104W.
For a safe value of jBj D 24 kN, W D 24
1.0104
D 23.752 kN is the
maximum load that can be carried. Thus, the largest mass that can be
supported is m DW/g D 23752 N/9.81 m/s2 D 2421 kg.
Cy Cy
Cx
Cx
Bx
By
Ay
Ax
Ex
Ex
Ey
Ey
Bx
ByW W
FW
470
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 6.94 Determine the reactions at A and C.
3 ft
A
72 ft-lb
36 lb
C
B
3 ft
8 ft4 ft
18 lb
Solution: The complete structure as a free body:
The sum of the moments about A:
∑
MA D �4�18�C 3�36�C 12Cy � 72 D 0,
from which Cy D 3 lb. The sum of theforces:
∑
Fy D Ay CCy � 18 D 0,
from which Ay D 15 lb.
∑
Fx D Ax CCx C 36 D 0,
from which (1) Cx D �Ax � 36
Element AB: The sum of the forces:
∑
Fy D Ay � By � 18 D 0,
from which By D �3 lb. The sum of the moments:
∑
MA D 6Bx � 4�18�� 4By � 72 D 0,
from which Bx D 22 lb. The sum of the forces:
∑
Fx D Ax C Bx D 0,
from which Ax D �22 lb From equation (1) Cx D �14 lb
Ay
By
Bx
Cy
CxAx
Ax
Ay
3 ft
8 ft
6 ft
4 ft
36 lb
72 ft-lb
72 ft-lb
18 lb
18 lb
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.
471
Problem 6.95 Determine the forces on member AD.
400 N
200 N
C
D
130 mm
400 mm
400 mm 400 mm
B
A
Solution: Denote the reactions of the support by Rx and Ry . The
complete structure as a free body:
∑
Fx D Rx � 400 D 0,
from which Rx D 400 N. The sum of moments:
∑
MA D 800C� 400�930� C 400�530� � 400�200� D 0,
from which C D 300 N.
∑
Fy D CC Ry � 400� 200 D 0,
from which Ry D 300 N. Element ABC : The sum of the moments:
∑
MA D �4By C 8C D 0,
from which By D 600 N. Element BD : The sum of the forces:
∑
Fy D By � Dy � 400 D 0,
from which Dy D 200 N.
Element AD: The sum of the forces:
∑
Fy D Ay C Dy � 200 D 0,
from which Ay D 0: Element AD: The sum of the forces:
∑
Fx D Ax C Dx D 0
and
∑
MA D �400�200� C 800Dy � 400Dx D 0
Ax D �200 N, and Dx D 200 N.
Element BD: The sum of forces:
∑
Fx D Bx � Dx � 400 D 0
from which Bx D 600 N. This completes the solution of the nine
equations in nine unknowns, of which Ax , Ay , Dx , and Dy are the
values required by the Problem.
200 N
400 N
400 N
Dy
Dy
Bx
BxRx
Ax
Ax
Ay
Ay
Ry By
By
Dx
Dx
C
472
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 6.96 The frame shown is used to support
high-tension wires. If b D 3 ft, ˛ D 30°, and W D
200 lb, what is the axial force in member HJ? B
A
C
D
E
FG
J
H
I W
W
W
b b b b
α α
αα
Solution: Joints B and E are sliding joints, so that the reactions
are normal to AC and BF, respectively. Member HJ is supported by
pins at each end, so that the reaction is an axial force. The distance
h D b tan ˛ D 1.732 ft
Member ABC. The sum of the forces:
∑
Fx D Ax C B sin ˛ D 0,
∑
Fy D Ay �W� B cos ˛ D 0.
The sum of the moments about B:
∑
MB D bAy � hAx C bW D 0.
These three equations have the solution: Ax D 173.21 lb, Ay D
�100 lb, and B D �346.4 lb.
Member BDEF: The sum of the forces:
∑
Fx D Dx � B sin ˛� E sin ˛ D 0,
∑
Fy D Dy �WC B cos ˛� E cos ˛ D 0.
The sum of the moments about D:
∑
MD D �2bW� bE cos ˛� hE sin ˛� bB cos ˛C hB sin ˛ D 0.
These three equations have the solution: Dx D �259.8 lb, Dy D
350 lb, E D �173.2 lb.
Member EGHI: The sum of the forces:
∑
Fx D Gx C E sin ˛�H cos ˛ D 0,
∑
Fy D Gy �WC E cos ˛CH sin ˛ D 0.
The sum of the moments about H:
∑
MH D bGy � hGx C bWC 2bE cos ˛� 2hE sin ˛ D 0.
These three equations have the solution: Gx D 346.4 lb, Gy D 200 lb,
and H D 300 lb. This is the axial force in HJ.
Ay
DyGy
Gx
Dx
Ax
B
B
E
H W
b
W
W
h
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.
473
Problem 6.97 Determine the force exerted on the ball
by the bolt cutters and the magnitude of the axial force
in the two-force member AB.
3 in 6 in 4 in
20 in
A
20 lb
20 lb
B
Solution: Free-body diagrams of the top head and the top handle
are shown.
From the head we learn that
Fx : Cx D 0
From the handle we have
MD : ��20 lb��20 in�
CCy�4 in� D 0
) Cy D 100 lb
Now we return to the head
MA : Cy�6 in�� F�3 in� D 0
Fy : F� TAB CCy D 0
Solving yields
Force on the ball D F D 200 lb, Axial force D TAB D 300 lb
474
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 6.98 The woman exerts 20-N forces to the
pliers as shown.
(a) What is the magnitude of the forces the pliers exert
on the bolt at B?
(b) Determine the magnitude of the force the members
of the pliers exert on each other at the pinned
connection C.
20 N
20 N
C
45�
25 mm 80 mm
B
50 mm
Solution: Look at the piece that has the lower jaw of the pliers
(a)
∑
MC : B�25 mm�� �20 N cos 45°��80 mm�
� �20 N sin 45°��50 mm� D 0
B D 73.5 N
(b)
∑
Fx : Cx � 20 N sin 45° D 0) Cx D 14.14 N
∑
Fy : Cy � B� 20 N cos 45° D 0) Cy D 87.7 N
Thus the magnitude is C D
√
Cx2 CCy2 D 88.8 N
45°
20 N
Cy
Cx
B
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.
475
Problem 6.99 Figure a is a diagram of the bones and
biceps muscle of a person’s arm supporting a mass.
Tension in the biceps muscle holds the forearm in the
horizontal position, as illustrated in the simple mechan-
ical model in Fig. b. The weight of the forearm is 9 N,
and the mass m D 2 kg.
(a) Determine the tension in the biceps muscle AB.
(b) Determine the magnitude of the force exerted on
the upper arm by the forearm at the elbow joint C.
(a)
A
B
C
290 
mm 
50 
mm
150 mm
9 N
200 mm
m
(b)
Solution: Make a cut through AB and BC just above the elbow
joint C. The angle formed by the biceps muscle with respect to the
forearm is ˛ D tan�1
(
290
50
)
D 80.2°. The weight of the mass is W D
2�9.81� D 19.62 N.
The section as a free body: The sum of the moments about C is
∑
MC D �50T sin ˛C 150�9� C 350W D 0,
from which T D 166.76 N is the tension exerted by the biceps muscle
AB. The sum of the forces on the section is
∑
FX D Cx C T cos ˛ D 0,
from which Cx D �28.33 N.
∑
FY D Cy C T sin ˛� 9�W D 0,
from which Cy D �135.72. The magnitude of the force exerted by the
forearm on the upper arm at joint C is
F D
√
C2x CC2y D 138.65 N
W
T
200 
mm
150 
mm
50 
mm
9N
Cy
Cx
α
476
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 6.100 The bones and the tendons in a horse’s
rear leg are shown in Fig. a. A biomechanical model of
the leg is shown in Fig. b. If the horse is stationary and
the normal force everted on its leg by the ground is
N D 1200 N, determine the tension s in the superficial
digital flexor BC and the patellar ligament DF.
40 cm
72 cm
8
cm
10
cm
8
cm
A
N
B
C
D
EF
6 cm
6 cm
6 cm
3 cm
(a) (b)
Solution: The free-body diagrams for AB and AE are shown. The
angle
˛ D tan�1�18/58� D 17.2°
The equilibrium equations for AB are
MA : ��1200 N��10 cm�
C TBC cos ˛�8 cm�
C TBC sin ˛�3 cm� D 0
Fx : Ax � TBC sin ˛ D 0
Fy : Ay C TBC cos ˛C 1200 N D 0
One of the equilibrium equations for AE is
ME : �TDF�8 cm�� Ax�49 cm�
� Ay�10 cm� D 0
Solving these four equations yields
Ax D 417 N, Ay D �2540 N
TBC D 1410 N, TDF D 625 N
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 permissionin writing from the publisher.
477
Problem 6.101 The pressure force exerted on the
piston is 2 kN toward the left. Determine the couple
M necessary to keep the system in equilibrium.
A
B
M C
400 mm
350 mm300 mm
45°
Solution: From the diagram, the coordinates of point B are �d, d�
where d D 0.3 cos�45°�. The distance b can be determined from the
Pythagorean Theorem as b D
√
�0.35�2 � d2. From the diagram, the
angle � D 37.3°. From these calculations, the coordinates of points B
and C are B (0.212, 0.212), and C (0.491, 0) with all distances being
measured in meters. All forces will be measured in Newtons.
The unit vector from C toward B is uCB D �0.795iC 0.606j.
The equations of force equilibrium at C are
∑
FX D FBC cos � � 2000 D 0,
and
∑
FY D N� FBC sin � D 0.
Solving these equations, we get N D 1524 Newtons(N), and FBC D
2514 N.
The force acting at B due to member BC is FBCuBC D �2000iC
1524j N.
The position vector from A to B is rAB D 0.212iC 0.212j m, and the
moment of the force acting at B about A, calculated from the cross
product, is given by MFBC D 747.6k N-m (counter - clockwise). The
moment M about A which is necessary to hold the system in equi-
librium, is equal and opposite to the moment just calculated. Thus,
M D �747.6k N-m (clockwise).
A C
B
d
d
b
0.35 m0.3 m
45° θ
A
B
N
x
x
y
y
FBC
FBCX
FBCY
c
2000 N
M
rAB
FBC uCB
478
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 6.102 In Problem 6.101, determine the forces
on member AB at A and B.
Solution: In the solution of Problem 6.101, we found that the force
acting at point B of member AB was FBCuBC D �2000iC 1524j N,
and that the moment acting on member BC about point A was given by
M D �747.6k N-m (clockwise). Member AB must be in equilibrium,
and we ensured moment equilibrium in solving Problem 6.101.
From the free body diagram, the equations for force equilibrium are
∑
FX D AX C FBCuBCX D AX � 2000 N D 0,
and
∑
FY D AY C FBCuBCY D AY C 1524 N D 0.
Thus, AX D 2000 N, and AY D �1524 N.
y
x
AX
AY
A
B
M
FBC uCB
Problem 6.103 In Example 6.8, suppose that the
object being held by the plier’s is moved to the left
so that the horizontal distance from D to the object at
E decreases from 30 mm to 20 mm. Draw a sketch of
the pliers showing the new position of the object. What
forces are exerted on the object at E as a result of the
150-N forces on the pliers?
150 N
70 mm 30 mm30 mm 30 mm
30 mm
B D
150 N
A
C
E
Solution: The analysis of the bottom grip of the pliers (member 3)
is unchanged.
The reactions Dx D �1517 N, Dy D 500 N.
From the free-body diagram of the lower jaw (member 2) we obtain
MC : �E�20 mm�� Dx�30 mm� D 0
Therefore E D 2280 N
2280 N
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.
479
Problem 6.104 The shovel of the excavator is
supported by a pin support at E and the two-force
member BC. The 300-lb weight W of the shovel acts at
the point shown. Determine the reactions on the shovel
at E and the magnitude of the axial force in the two-force
member BC.
W
12 in
15 in
3 in
Shovel
Hydraulic
cylinder
12 in
C
B
A
D
E
7 in 20 in
Solution: The angle
˛ D tan�1�3/15� D 11.3°
The equilibrium equations for the shovel are
Fx : Ex � TBC cos ˛ D 0
Fy : Ey C TBC sin ˛� 300 lb D 0
MC : ��300 lb��20 in�C Ex�12 in�
� Ey�7 in� D 0
Solving yields
Ex D 604 lb, Ey D 179 lb, TBC D 616 lb
Thus Ex D 604 lb, Ey D 179 lb, axial force D 616 lb
480
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 6.105 The shovel of the excavator has a pin
support at E. The position of the shovel is controlled by
the horizontal hydraulic piston AB, which is attached to
the shovel through a linkage of the two-force members
BC and BD. The 300-lb weight W of the shovel acts
at the point shown. What is the magnitude of the force
the hydraulic piston must exert to hold the shovel in
equilibrium?
W
12 in
15 in
3 in
Shovel
Hydraulic
cylinder
12 in
C
B
A
D
E
7 in 20 in
Solution: From the solution to Problem 6.104 we know that the
force in member BC is
TBC D 616 lb
We draw a free-body diagram of joint B and note that AB is horizontal.
The angles are
˛ D tan�1�3/15� D 11.3°
ˇ D tan�1�4/15� D 14.9°
The equilibrium equations for joint B are
Fx : TBC cos ˛� TAB � TBD sin ˇ D 0
Fy : �TBD cos ˇ � TBC sin ˛ D 0
Solving yields TAB D 637 lb, TBD D �125 lb
637 lb
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.
481
Problem 6.106 The woman exerts 20-N forces on the
handles of the shears. Determine the magnitude of the
forces exerted on the branch at A.
BA
C
D
E
25 mm25 mm36 mm
65 mm 20 N
20 N
Solution: Assume that the shears are symmetrical.
Consider the 2 pieces CD and CE
∑
Fx D 0) Dx D Ex
∑
Fy D 0) Dy D Ey
∑
MC D 0) Dx D Ex D 0
20 N
20 N
Ey
Ex
C
Dx
Dy
Now examine CD by itself
∑
MC D ��20 N��90 mm�C Dy�25 mm� D 0) Dy D 72 N
20 N
Cy
Cx
Dx = 0
Dy
Finally examine DBA
∑
MB : A�36 mm�� Dy�50 mm� D 0
Bx
Dx = 0By
Dy
A
Solving we find A D 100 N
482
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 6.107 The person exerts 40-N forces on the
handles of the locking wrench. Determine the magnitude
of the forces the wrench exerts on the bolt at A.
A B
E
D
C
50 mm 30 mm 75 mm
8 mm 40 mm
40 N
40 N
Solution: Recognize that DE is a 2-force member. Examine
part CD
∑
Fx :
75p
5689
FDE CCx D 0
∑
Fy :
8p
5689
FDE C Cy C 40 N D 0
∑
MC :
8p
5689
FDE�30 mm�C �40 N��105 mm� D 0
Solving we find
Cx D 1312.5 N, Cy D 100 N, FDE D �1320 N
Now examine ABC∑
MB : A�50 mm��Cx�40 mm� D 0
∑
Fx : Bx �Cx D 0
∑
Fy : By �Cy � A D 0
Solving: A D 1050 N, Bx D 1312.5 N By D 1150 N
Answer: A D 1050 N
Cx
Cy
FDE
D
75
40 N
8
A
Cx
Cy
Bx
By
Problem 6.108 In Problem 6.107, determine the
magnitude of the force the members of the wrench exert
on each other at B and the axial force in the two-force
member DE.
Solution: From the previous problem we have
B D
√
Bx2 C By2 D 1745 N
FDE D 1320 N�C�
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.
483
Problem 6.109 The device is designed to exert a large
force on the horizontal bar at A for a stamping operation.
If the hydraulic cylinder DE exerts an axial force of
800 N and ˛ D 80°, what horizontal force is exerted on
the horizontal bar at A?
A
90°
250
 mm 25
0 
m
m
250 mm
B
D
C E
400 mm
α
Solution: Define the x-y coordinate system with origin at C. The
projection of the point D on the coordinate system is
Ry D 250 sin ˛ D 246.2 mm,
and Rx D 250 cos ˛ D 43.4 mm.
The angle formed by member DE with the positive x axis is � D
180� tan�1
(
Ry
400� Rx
)
D 145.38°. The components of the forceproduced by DE are Fx D F cos � D �658.3 N, and Fy D F sin � D
454.5 N. The angle of the element AB with the positive x axis is ˇ D
180� 90� ˛ D 10°, and the components of the force for this member
are Px D P cos ˇ and Py D P sin ˇ, where P is to be determined. The
angle of the arm BC with the positive x axis is � D 90C ˛ D 170°.
The projection of point B is Lx D 250 cos � D �246.2 mm, and Ly D
250 sin � D 43.4 mm. Sum the moments about C:
∑
MC D RxFy � RyFx C LxPy � LyPx D 0.
Substitute and solve: P D 2126.36 N, and Px D P cos ˇ D 2094 N is
the horizontal force exerted at A.
Py
Fy
Cy
Fx
Cx
Px
B
D
484
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 6.110 This device raises a load W by
extending the hydraulic actuator DE. The bars AD and
BC are 4 ft long, and the distances b D 2.5 ft and h D
1.5 ft. If W D 300 lb, what force must the actuator exert
to hold the load in equilibrium?
A
C
B
D E
b
h
W
Solution: The angle ADC is ˛ D sin�1
(
h
4
)
D 22.02°. The
distance CD is d D 4 cos ˛.
The complete structure as a free body: The sum of the forces:
∑
Fy D �WCCy C Dy D 0.
∑
Fx D Cx C Dx D 0.
The sum of the moments about C:
∑
MC D �bWC dDy D 0.
These have the solution:
Cy D 97.7 lb,
Dy D 202.3 lb,
and Cx D �Dx.
Divide the system into three elements: the platform carrying the
weight, the member AB, and the member BC.
The Platform: (See Free body diagram) The moments about the
point A:
∑
MA D �bW� dB D 0.
The sum of the forces:
∑
Fy D AC BCW D 0.
These have the solution:
B D �202.3 lb,
and A D �97.7 lb.
Element BC: The sum of the moments about E is
∑
MC D �
(
h
2
)
Cy C
(
d
2
)
Cx C
(
d
2
)
B D 0, from which
(1) dCx � hCy � dB D 0. The sum of the forces:
∑
Fx D Cx � Ex D 0, from which
(2) Ex � Cx D 0,
∑
Fy D Cy � Ey C B D 0,
from which
A
A
B
B
W
E
Ex
Cx
Dx
Dy
Cy
Ex
Ey
(3) Cy � Ey C B D 0
Element AD: The sum of the moments about E:
∑
ME D
(
d
2
)
Dy C
(
h
2
)
Dx �
(
d
2
)
A D 0,
from which
(4) dDy C hDx � dA D 0.
These are four equations in the four unknowns: EX, EY, Dx , CX and
DX
Solving, we obtain Dx D �742 lb.
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.
485
Problem 6.111 The four-bar linkage operates the forks
of a fork lift truck. The force supported by the forks is
W D 8 kN. Determine the reactions on member CDE.
0.7 m
0.15 m
0.2 m
0.15 m
0.2 m
0.2 m
0.3 m
Forks
W
ED
A
F
C
B
Solution: Consider body BC. Note that AB is a 2-force body.
∑
Fx : �Cx D 0
∑
MB : �Cy�0.2 m�� �8 kN��0.9 m� D 0
) Cx D 0, Cy D �36 kN
Now examine CDE. Note that DF is a 2-force body.
∑
ME : �Cy�0.15 m��Cx�0.15 m�C 3p
13
FDF�0.15 m� D 0
∑
Fx : Cx C Ex C 2p
13
FDF D 0
∑
Fy : Cy C Ey � 3p
13
FDF D 0
Solving we find FDF D �43.3 kN, Ex D 24 kN, Ey D 0
Note that Dx D 2p
13
FDF, Dy D � 3p
13
FDF
Summary:
Cx D 0, Cy D �36 kN
Dx D �24 kN, Dy D 36 kN
Ex D 24 kN, Ey D 0
Cx
Cy
FAB
W = 8 kN
2
3
D
FDF
Cy
Cx
Ey
Ex
486
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 6.112 If the horizontal force on the scoop is
F D 2000 lb, what is the magnitude of the axial force
in the hydraulic actuator AC?
10 in 12 in20 in
B
C
D
A
38 in
28 in
10 in
Scoop
F
Solution: We start with the free-body diagram of the scoop. Note
that BC is a two-force body. The angle
˛ D tan�1�38/32� D 49.9°
We have the following equilibrium equation
MA : ��2000 lb� �10 in�
C TBC cos ˛�28 in�
C TBC sin ˛�12 in� D 0
) TBC D 735 lb
Now we work with the free-body diagram of joint C. The angles
ˇ D tan�1�20/66� D 16.9°
� D tan�1�10/38� D 14.7°
The equilibrium equations are
Fx : TBC cos ˛C TAC sin ˇ � TCD sin � D 0
Fy : �TBC sin ˛� TAC cos ˇ � TCD cos � D 0
Solving yields TAC D �1150 lb, TCD D 553 lb
Thus 1150 lb
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.
487
Problem 6.113 A 10-kip horizontal force acts on the
bucket of the excavator. Determine the reactions on
member ACF at A and F.
Bucket
2 ft 3 ft
2 ft
9 ft
2 ft E
F
D
B
A
C
4 ft 4 in
5 ft 6 in
1 ft 8 in
1 ft 4 in
10 kip
Solution: We start with the free-body diagram of the entire struc-
ture. The angle
˛ D tan�1�36/72� D 26.6°
The equilibrium equations are
Fx : Ax � TBC sin ˛C 10 kip D 0
Fy : Ay � TBC cos ˛ D 0
MA : �10 kip� �66 in�
C TBC sin ˛ �52 in�
� TBC cos ˛ �60 in� D 0
Next we examine the free-body diagram of the member on the right.
The angle
ˇ D tan�1�84/4� D 87.3°
The equilibrium equations are
Fx : Fx C 10 kip� TDE sin ˇ D 0
Fy : Fy � TDE cos ˇ D 0
MF : TDE sin ˇ �24 in�C �10 kip� �120 in� D 0
Solving these six equations we find
TBC D 21.7 kip, TDE D �49.2 kip
Ax D �0.294 kip, Ay D 19.4 kip
Fx D 59.2 kip, Fy D 2.34 kip
488
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 6.114 The structure shown in the diagram
(one of the two identical structures that support the
scoop of the excavator) supports a downward force F D
1800 N at G. Members BC and DH can be treated as
two-force members. Determine the reactions on member
CDK at K.
H
L K
100
mm
160
mm
Shaft
260
mm
320
mm
380
mm
200
mm
C
D
1040
mm
1120
mm
260
mm
B
J
180
mm
Scoop
G
F
Solution: Start with the scoop
∑
MJ :
4p
17
FBC�0.44 m�� 1p
17
FBC�0.06 m�
� �1800 N��0.2 m� D 0
) FBC D 873 N
1
4
1800 N
FBC
Jx
Jy
Now examine CDK
∑
MK :
56p
3161
FDH�0.26 m�� 4p
17
FBC�0.52 m� D 0
∑
Fx : � 56p
3161
FDH C 4p
17
FBC CKx D 0
∑
Fy : � 5p
3161
FDH � 1p
17
FBC CKy D 0
Solving we find Kx D 847 N, Ky D 363 N
FDH
FBC
1
4
Kx
Ky
56
5
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.
489
Problem 6.115 The loads F1 D 440 N and F2 D 160 N.
Determine the axial forces in the members. Indicate
whether they are in tension (T) or compression (C).
B
200 mm
400 mm
A
F1
700 mm
F2
C
Solution: The sum of the moments about C is
MC D �0.7BY C 0.7F1 C 0.4F2 D 0,
from which By D 0.7F1 C 0.4F2
0.7
D 531.43 N .
The axial loads at joint B are
AB D �By D �531.4 N �C� ,
and BC D 0 .
Similarly, the sum of the forces at the joint A is
∑
FAx D �F2 C AC cos ˇ D 0,
from which
AC D F2
cos ˇ
D 184.3 N �T�
490
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 6.116 The truss supports a load F D 10 kN.
Determine the axial forces in the members AB, AC,
and BC.
B
3 m
A
4 m 3 m
DC
F
Solution: Find the support reactions at A and D.
∑
Fx : Ax D 0
∑
Fy : Ay C Dy � 10 D 0
C
∑
MA: ��4��10�C 7Dy D 0
Solving,Ax D 0,
Ay D 4.29 kN
Dy D 5.71 kN
Joint A:
tan � D 34
� D 36.87°
�Ay D 4.29 kN�
∑
Fx : FAB cos � C FAC D 0
∑
Fy : Ay C FAB sin � D 0
Solving, FAB D �7.14 kN �C�
FAC D 5.71 kN �T�
Joint C:
∑
Fx : FCD � FAC D 0
∑
Fy : FBC � 10 kN D 0
Solving FBC D 10 kN �T�
FCD D C5.71 kN �T�
4 m 3 m
3 m
10 kN
AX
AY
DY
AY
FAB
FAC x
y
θ
FAC
FBC
FCD
10 kN
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.
491
Problem 6.117 Each member of the truss shown in
Problem 6.116 will safely support a tensile force of
40 kN and a compressive force of 32 kN. Based on this
criterion, what is the largest downward load F that can
safely be applied at C?
Solution: Assume a unit load F and find the magnitudes of the
tensile and compressive loads in the truss. Then scale the load F
up (along with the other loads) until either the tensile limit or the
compressive limit is reached.
External Support Loads:
∑
Fx : Ax D 0 (1)
∑
Fy : Ay C Dy � F D 0 (2)
∑
MA: � 4FC 7Dy D 0 (3)
Joint A:
tan � D 3
2
� D 36.87°
∑
Fx : FAC C FAB cos � D 0 (4)
∑
Fy : FAB sin � C Ay D 0 (5)
Joint C
∑
Fx : FCD � FAC D 0 (6)
∑
Fy : FBC � F D 0 (7)
Joint D
tan � D 3
3
� D 45°
∑
Fx : � FCD � FBD cos � D 0 (8)
∑
Fy : FBD sin � C Dy D 0 (9)
Setting F D 1 and solving, we get the largest tensile load of 0.571 in
AC and CD. The largest compressive load is 0.808 in member BD.
Largest Tensile is in member BC. BC D F D 1
The compressive load will be the limit
Fmax
1
D 32
0.808
Fmax D 40 kN
A
B
DC
F
4 m 3 m
3 m1
4 3
25
AX
AY
DYF
AY
FAC
FAB
y
x
θ
φ
FCD
DY
FBD
x
y
FBC
FAC FCD
F
x
y
492
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 6.118 The Pratt bridge truss supports loads
at F, G, and H. Determine the axial forces in members
BC, BG, and FG.
B C D
4 m
60 kN 80 kN 20 kN
F G H
EA
4 m 4 m 4 m 4 m
Solution: The angles of the cross-members are ˛ D 45°.
The complete structure as a free body:
The sum of the moments about A:
∑
MA D �60�4�� 80�8�� 20�12� C 16E D 0,
from which E D 70 kN. The sum of the forces:
∑
Fx D Ax D 0.
∑
Fy D Ay � 60� 80� 20C E D 0,
from which Ay D 90 kN
The method of joints: Joint A:
∑
FY D Ay C AB sin ˛ D 0,
from which AB D �127.3 kN �C�,
∑
Fx D AB cos ˛C AF D 0,
from which AF D 90 kN �T�. Joint F:
∑
Fx D �AFC FG D 0,
from which FG D 90 kN �T� .
∑
Fy D BF� 60 D 0,
from which BF D 60 kN �C�. Joint B:
∑
Fx D �AB cos ˛C BCC BG cos ˛ D 0,
and
∑
Fy D �AB sin ˛� BF� BG sin ˛ D 0,
from which:
�AB sin ˛� BF� BG sin ˛ D 0.
Solve: BG D 42.43 kN �T� ,
and � AB cos ˛C BCC BG cos ˛ D 0,
from which BC D �120 kN �C�
4 m
60 kN
60 kN
Joint A Joint F Joint B
80 kN 20 kN
4 m 4 m 4 m
4 m
E
Ax
Ay
AB
AB BF
BF
BG
BC
AF AF FG
Ay
α α α
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.
493
Problem 6.119 Consider the truss in Problem 6.118.
Determine the axial forces in members CD, GD,
and GH.
Solution: Use the results of the solution of Problem 6.130:
BC D �120 kN �C�,
BG D 42.43 kN �T�,
and FG D 90 kN �T�.
The angle of the cross-members with the horizontal is ˛ D 45°.
Joint C:
∑
Fx D �BCC CD D 0,
from which CD D �120 kN �C�
∑
FY D �CG D 0,
from which CG D 0.
Joint G:
∑
Fy D BG sin ˛CGD sin ˛CCG� 80 D 0,
from which GD D 70.71 kN �T� .
∑
Fy D �BG cos ˛CGD cos ˛� FGCGH D 0,
from which GH D 70 kN �T�
BC
CG
CG
GD
GH
80 kN
Joint C Joint G
BG
CD α α
494
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 6.120 The truss supports loads at F and H.
Determine the axial forces in members AB, AC, BC, BD,
CD, and CE.
A I
200 lb
F
100 lb
H
J
E
D
C G
B
4 in
6 in 6 in 6 in 6 in
4 in
4 in
Solution: The complete structure as a free body: The sum of the
moments about I:
∑
MA D 100�6� C 200�12� � 24AY D 0,
from which AY D 125 lb. The sum of forces:
∑
Fx D Ax D 0.
The method of joints: The angles of the inclined members with the
horizontal are
˛ D tan�1�0.6667� D 33.69°
Joint A:
∑
Fx D AC cos ˛ D 0,
from which AC D 0.
∑
Fy D Ay C ABC AC sin ˛ D 0,
from which AB D �125 lb �C�
Joint B :
∑
Fyt D �ABC BD sin ˛ D 0,
from which BD D �225.3 lb �C� .
∑
Fx D BD cos ˛C BC D 0,
from which BC D 187.5 lb �T�
Joint C :
∑
Fx D �BC� AC cos ˛CCE cos ˛ D 0,
from which CE D 225.3 lb �T�
∑
Fy D �AC sin ˛CCD CCE sin ˛ D 0,
from which CD D �125 lb �C�
Ax Ay
I
200 lb
100 lb
12
in
Joint A Joint B Joint C
6
in
6
in
AB
Ay AB AC
CD
CEAC BD
BC
BCα α
α
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.
495
Problem 6.121 Consider the truss in Problem 6.120.
Determine the axial forces in members EH and FH.
Solution: Use the results from the solution to Problem 6.132:
CE D 225.3 lb �T�,
CD D �125 lb �C�,
BD D �225.3 lb �C�.
The method of joints: The angle of inclined members with the hori-
zontal is ˛ D 33.69°.
Joint D :
∑
Fy D �BD sin ˛�CD C DF sin ˛ D 0,
from which DF D �450.7 lb �C�.
∑
Fx D �DF cos ˛C DE� BD cos ˛ D 0,
from which DE D 187.5 lb �T�
Joint F :
∑
Fx D �DF cos ˛C FH cos ˛ D 0,
from which FH D �450.7 lb �C�
∑
Fy D �200� DF sin ˛� FH sin ˛� EF D 0,
from which EF D 300 lb �T�
Joint E :
∑
Fy D �CE sin ˛C EF� EG sin ˛ D 0,
from which EG D 315 lb �T�
∑
Fx D �DEC EH�CE cos ˛C EG cos ˛ D 0,
from which EH D 112.5 lb �T�
BD
CD
DE
DF 200 lb
DF
EF
EF
EH
EG
FH
DE
CE
Joint D Joint F Joint E
α
α α αα
496
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 6.122 Determine the axial forces in members
BD, CD, and CE.
2 m
2 m
2 m
2 m
A
C
E
G
I
6 m
D
B
F
H
10 kN
14 kN
Solution: Use the method of sections
10 kN
14 kN
2 m
1.5 m
FBD FCD
FCE
A
B
D
C
x
y
Θ Θ
Θ
tan � D 2
1.5
� D 53.13°
∑
Fx : FCE cos � � FCD cos � C 24 D 0
∑
Fy : � FBD � FCD sin � � FCE sin � D 0
∑
MB: � 2�10�� 1.5FCD sin � � 1.5FCE sin � D 0
3 eqns-3 unknowns.
Solving FBD D 13.3 kN,
FCD D 11.7 kN,
FCE D �28.3 kN
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.
497
Problem 6.123 For the truss in Problem 6.122, deter-
mine the axial forces in members DF, EF, and EG.
Solution: Use method of sections
10 kN
14 kN
3
3
2 m
2 m
2
D E
A
1.5
FDF
FEF
FEG
φ Θ
tan � D 2
1.5
� D 53.13°
tan � D 2
3
� D 33.69°
∑
Fx : 24C FEG cos � � FEF cos � D 0
∑
Fy : � FDF � FEF sin � � FEG sin � D 0
∑
ME: 3FDF � 2�14�� 4�10� D 0
Solving, FEG D �32.2 kN �C�
FDF D 22.67 kN �T�
FEF D 5.61 kN �T�
2 m
2 m
2 m
2 m
A
C
E
G
I
6 m
D
B
F
H
10 kN
14 kN
498
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 6.124 The truss supports a 400-N load at G.
Determine the axial forces in members AC, CD, and CF.
400 N
300 mm
B
D
F
H
GECA
300 mm300 mm300 mm
600 mm
Solution: The complete structure as a free body: The sum of the
moments about A:
∑
MA D �900�400� C 600B D 0,
from which B D 600 N. The sum of forces:
∑
Fx D Ax C B D 0,
from which Ax D �600 N.
∑
Fy D Ay � 400 D 0,
from which Ay D 400 N.
The method of joints: The angle from the horizontal of element BD is
� D tan�1
(
300
900
)
D 18.43°.
The angle from the horizontal of element AD is
˛AD D 90� tan�1
(
300
600� 300 tan �
)
D 59.04°.
The angle from the horizontal of element CF is
˛CF D 90� tan�1
(
300
600�1� tan ��
)
D 53.13°.
Joint B:
∑
Fx D BC BD cos � D 0,
from which BD D �632.5 N �C�
∑
Fy D ABC BD sin � D 0,
from which AB D 200 N �T�
Joint A:
∑
Fy D Ay � AD sin ˛AD � AB D 0,
from which AD D 233.2 N �T�
∑
Fx D Ax C ACC AD cos ˛AD D 0,
from which AC D 480 N �T�
600 mm
900 mm
Joint D Joint C
Joint B Joint A
400 N
B
B
AB
BD
BD
DF
AC
AD
AB
AC CE
CD
CDAD
CF
Ax
AX
AY
Ay
θ
θ
ADα
ADα
CFα
Joint D :
∑
Fx D �AD cos ˛AD � BD cos � C DF cos � D 0,
from which DF D �505.96 N �C�
∑
Fy D AD sin ˛AD CCD � BD sin � C DF sin � D 0,
from which CD D �240 N �C�
Joint C :
∑
Fy D �CD �CF sin ˛CF D 0,
from which CF D 300 N �T�
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.
499
Problem 6.125 Consider the truss in Problem 6.124.
Determine the axial forces in members CE, EF,
and EH.
Solution: Use the results of the solution of Problem 6.124:
AC D 480 N �T�,
CF D 300 N �T�,
DF D �505.96 N �C�,
� D 18.4°,
˛CF D 53.1°.
The method of joints: The angle from the horizontal of element EH is
˛EH D 90� tan�1
(
300
600� 900 tan �
)
D 45°
Joint C:
∑
Fx D �ACC CECCF cos ˛CF D 0,
from which CE D 300 N �T�
AC CE CE EG
CF DF
FH
EHEF
EFCF
CD
Joint C Joint F Joint E
CF
α CF
α
EH
α
θ
Joint F :
∑
Fy D �CF cos ˛CF � DF cos � C FH cos � D 0,
from which FH D �316.2 N �C�
∑
Fy D EFCCF sin ˛CF � DF sin � C FH sin � D 0,
from which EF D �300 N �C�
Joint E :
∑
Fy D �EH sin ˛EH � EF D 0,
from which EH D 424.3 N �T�
Problem 6.126 Consider the truss in Problem 6.124.
Which members have the largest tensile and compressive
forces, and what are their values?
Solution: The axial forces for all members have been obtained in
Problems 6.124 and 6.125 except for members EG and GH. These are:
Joint E:
∑
Fx D �CEC EGC EH cos ˛EH D 0,
from which EG D 0
Joint G:
∑
Fy D �GH� 400 D 0,
from which GH D �400 N �C�.
This completes the determination for all members. A comparison of
tensile forces shows that AC D 480 N �T� is the largest value, and a
comparison of compressive forces shows that BD D �632.5 N �C�
is the largest value.
CE
EF EH
EG
EG
GH
400 N
EH
α
Joint E Joint G
500
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 6.127 The Howe truss helps support a roof.
Model the supports at A and G as roller supports. Use
the method of joints to determine the axial forces in
members BC, CD, CI, and CJ.
2 m 2 m 2 m 2 m 2 m 2 m
6 kN
4 m
A
B
C
G
F
E
D
H I J K L
4 kN4 kN
2 kN2 kN
Solution: The free body diagrams for the entire truss and the
required joints are shown.
The whole truss: The equations of equilibrium for the entire truss are:
∑
FX D 0,
∑
FY D AY CGY � 18 kN D 0.
Instead of using the moment equation here (it would work), we see
that the loading is symmetric. Thus, AY D GY D 9 kN.
We need unit vectors along AB, BC, CD, (note that these are the same),
and along BI, and CJ. We get
uAB D uBC D uCD D 0.832iC 0.555j,
uBI D 0.832i� 0.555j,
and uCJ D 0.6i� 0.8j.
Joint A:
The equations of equilibrium are
∑
FX D TABuABX C TAH D 0
and
∑
FY D TABuABY C AY D 0.
Joint H: The equations of equilibrium are
∑
FX D �TAH C THI D 0,
and
∑
FY D TBH D 0.
Joint B:
∑
FX D �TABuABX C TBCuBCX C TBIuBIX D 0,
∑
FY D �TABuABY C TBCuBCY C TBIuBIY � TBH � 2 D 0,
Joint I:
∑
FX D �THI C TIJ � TBIuBIX D 0,
and
∑
FY D TCI � TBIuBIY D 0,
12 m
6 kN
4 m
A
B
C
G
GYAY
F
ED
H I J K L x
y
4 kN4 kN
2 kN 2 kN
TAB
TAH
TDF
TFH
TFG
TEF
TCX
TCI
TCD
TCJ
TBX
TBC
TAH
TBH
THI
THI TIJ
x x
xx
y y
yy
A
F
E
C
I
H
4 kN
Joint C:
∑
FX D �TBCuBCX C TCJuCJX C TCDuCDX D 0,
∑
FY D �TBCuBCY C TCJuCJY C TCDuCDY � TCI � 4 D 0.
Solving these equations in sequence (we can solve at each joint before
going to the next), we get
TAB D �16.2 kN, TAH D 13.5 kN, TBH D 0 kN,
THI D 13.5 kN, TBC D �14.4 kN, TBI D �1.80 kN,
TIJ D 12.0 kN, TCI D 1.00 kN, TCJ D �4.17 kN,
and TCD D �11.4 kN.
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.
501
Problem 6.128 For the roof truss in Problem 6.127,
use the method of sections to determine the axial forces
in members CD, CJ, and IJ.
Solution: The free body diagram of the section is shown at the
right. The support force at A is already known from the solution to
Problem 6.139. The equations of equilibrium for the section are
∑
FX D TCDuCDX C TCJuCJX C TIJ D 0,
∑
FY D TCDuCDY C TCJuCJY C AY D 0,
and
∑
MC D yCTIJ � 4AY D 0.
Solving, we get
TIJ D 12.0 kN,
TCJ D �4.17 kN,
and TCD D �11.4 kN.
Note that these values check with the values obtained in
Problem 6.139.
2 kN
4 kN
C
D
I J
AY
TCD
TCJ
TIJ
502
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 6.129 A speaker system is suspended from
the truss by cables attached at D and E. The mass of
the speaker system is 130 kg, and its weight acts at G.
Determine the axial forces in members BC and CD.
G
E
C
A
B D
0.5 m 0.5 m 0.5 m0.5 m
1 m
1 m
Solution: The speaker as a free body: The weight of the speaker
is W D 130�9.81� D 1275.3 N. Make a cut through the suspension
cables D, E, the sum of the moments about cable D is
∑
MD D ��1�WC �1.5�E D 0,
from which E D 850.2 N. The sum of the forces:
∑
Fy D DC E�W D 0,
from which D D 425.1 N.
The structure as a free body: The sum of the moments about C is
∑
MC D C�1�A� 0.5D� �2�E D 0,
from which A D 1912.95 N. The sum of the forces:
∑
Fy D �ACCy �W D 0,
from which Cy D �637.65 N and
∑
Fx D Cx D 0.
The method of joints: The angle of member DE relative to the hori-
zontal is ˛ D tan�1
(
1
1.5
)
D 33.69° . The angles of members AB, BC,
and CD are ˇ D 90� tan�1�0.5� D 63.43°.
Joint E :
∑
Fy D �E� DE sin ˛ D 0,
from which DE D �1532.72 N �C�.
∑
Fx D �CE� DE cos ˛ D 0,
from which CE D 1275.3 N �T�
A
B D
D
E
E
1 m
0.5 m
0.5 m1 m
Joint E Joint D Joint C
2 mCy
CY
Cx
W
CE
DE
DE
BC
AC
CD
CECD
BD
E
D
β β βα
α
Joint D :
∑
Fy D CD sin ˇC DE sin ˛� D D 0,
from which CD D 1425.8 N �T�
Joint C :
∑
Fy D �CD sin ˇ � BC sin ˇ CCy D 0,
from which BC D 2138.7 N �T�
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 portionof this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
503
Problem 6.130 The mass of the suspended object is
900 kg. Determine the axial forces in the bars AB
and AC.
Strategy: Draw the free-body diagram of joint A.
y
z
x
D (0, 4, 0) m
A (3, 4, 4) m
C (4, 0, 0) mB (0, 0, 3) m
Solution: The free-body diagram of joint A is.
TAD
TAB TAC
(900) (9.81) N
The position vectors from pt A to pts B, C, and D are
rAB D �3i� 4j� k (m),
rAC D i� 4j� 4k (m),
rAD D �3i� 4k (m).
Dividing these vectors by their magnitudes, we obtain the unit vectors
eAB D �0.588i� 0.784j� 0.196k,
eAC D 0.174i� 0.696j� 0.696k,
eAD D �0.6i� 0.8k.
From the equilibrium equation
TABeAB C TACeAC C TADeAD � �900��9.81�j D O,
We obtain the equations
� 0.588TAB C 0.174TAC � 0.6TAD D 0,
� 0.784TAB � 0.696TAC � �900��9.81� D 0,
� 0.196TAB � 0.696TAC � 0.8TAD D 0.
Solving, we obtain
TAB D �7200 N,
TAC D �4560 N,
TAD D 5740 N.
504
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 6.131 Determine the forces on member ABC,
presenting your answers as shown in Fig. 6.25. Obtain
the answer in two ways:
(a) When you draw the free-body diagrams of the
individual members, place the 400-lb load on the
free-body diagram of member ABC.
(b) When you draw the free-body diagrams of the
individual members, place the 400-lb load on the
free-body diagram of member CD.
A
C
B
200 lb
400 lb
1 ft
1 ft
1 ft
D
E
F
1 ft 1 ft
Solution: The angle of element BE relative to the horizontal is
˛ D tan�1 ( 12 ) D 26.57° .
The complete structure as a free body: The sum of the moments
about A:
∑
MA D �3�400�� 1�200� C 2Fy D 0
from which Fy D 700 lb. The sum of forces:
∑
Fy D Ay C Fy � 200 D 0,
from which Ay D �500 lb .
∑
Fx D Ax C Fx C 400 D 0.
(a) Element CD: The sum of the moments about D:
∑
MD D 200C 2Cy D 0,
from which Cy D �100 lb .
∑
Fy D �Dy �Cy � 200 D 0,
from which Dy D �100.
∑
Fx D �Cx � Dx D 0,
from which Dx D �Cx .
Element DEF: The sum of the moments about F:
∑
MF D �3Dx C B cos ˛ D 0,
from which Dx D B
( cos ˛
3
)
.
∑
Fy D Fy C B sin ˛C Dy D 0,
from which B D �700C 100
sin ˛
D �1341.6 lb , and Dx D
�400 lb
Element ABC:
∑
MA D �2B cos ˛� 3�400�� 3Cx D 0.
The sum of the forces
∑
Fy D Cy � B sin ˛C Ay D 0,
400 lb
200 lb
100 lb
400 lb
400 lb
500 lb
400 lb
1341
 lb
B
B
Cx
Cx Dx
Dx
FxAx
FyAy
Dy
Dy
Cy
Cy
B
B
26.6°
from which Check:
B D 500C 100
sin ˛
D �1341.6 lb.
check. From above: Cx D �Dx D 400 lb .
∑
Fx D 400CCx C B cos ˛C Ax D 0,
from which Ax D 400 lb .
(b) When the 400 lb load is applied to element CD instead, the
following changes to the equilibrium equations occur: Element
CD:
∑
Fx D �Cx � Dx C 400 D 0,
from which Cx C Dx D 400. Element ABC:
∑
Fx D Cx C Ax � B cos ˛ D 0.
Element DEF : No changes. The changes in the solution for
Element ABC Cx D 800 lb when the external load is removed,
instead of Cx D 400 lb when the external load is applied, so that
the total load applied to point C is the same in both cases.
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.
505
Problem 6.132 The mass m D 120 kg. Determine the
forces on member ABC.
A B C
D
E
m
200 mm200 mm
300 mm
Solution: The weight of the hanging mass is given by
W D mg D 120 kg
(
9.81
m
s2
)
D 1177 N.
The complete structure as a free body: The equilibrium equations are:
∑
FX D AX C EX D 0,
∑
FY D AY �W D 0,
and
∑
MA D 0.3EX � 0.4W D 0.
Solving, we get
AX D �1570 N,
AY D 1177 N,
and EX D 1570 N.
Element ABC: The equilibrium equations are
∑
FX D Ax CCX D 0,
∑
FY D AY CCY � BY �W D 0,
and:
∑
MA D �0.2BY C 0.4cY � 0.4W D 0.
Solution gives BY D 2354 N (member BD is in tension),
CX D 1570 N,
and CY D 2354 N.
Ay
Ax Cx
Cx
Ex
B
B
B W
B
Cy
Cy
506
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 6.133 Determine the reactions on member
ABC at B and C.
0.2 m 0.2 m
A
B
C
D
E
4 kN
0.2 m
0.2 m
2 kN-m
Solution: We draw free-body diagrams for the entire structure, and
for members BD and ABC.
From the entire structure:
Fx : Cx C 4 kN D 0
ME : �Cy�0.4 m�� �4 kN��0.4 m�
� �2 kN-m� D 0
From body ABC
MA : Bx�0.2 m�CCx�0.4 m� D 0
And from body BD
MD : By�0.2 m�� �2 kN-m� D 0
Solving these four equations yields
Bx D 8 kN, By D 10 kN
Cx D �4 kN, Cy D �9 kN
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.
507
Problem 6.134 The truck and trailer are parked on a
10° slope. The 14,000-lb weight of the truck and the
8000-lb weight of the trailer act at the points shown.
The truck’s brakes prevent its rear wheels at B from
turning. The truck’s front wheels at C and the trailer’s
wheels at A can turn freely, which means they do not
exert friction forces on the road. The trailer hitch at D
behaves like a pin support. Determine the forces exerted
on the truck at B, C, and D.
14 ft
A
B
C
D
y
x
4 ft
3 ft
5 ft 6 in
6 ft
8 kip
14 kip
10�
3 ft
2 ft 9 ft
Solution: We separate the two vehicles and draw a free-body
diagram of each. Starting with the trailer we have
MA : ��8 kip� cos 10°�4 ft�C �8 kip� sin 10°�6 ft�
C Dx�5.5 ft�� Dy�16 ft� D 0
Fx : ��8 kip� sin 10° � Dx D 0
Now we use the free-body diagram for the truck
MB : C�11 ft�C Dy�2 ft�� Dx�5.5 ft�
� �14 kip� cos 10°�8 ft�C �14 kip� sin 10°�3 ft� D 0
Fx : Bx C Dx � �14 kip� sin 10° D 0
Fy : By C Dy CC� �14 kip� cos 10° D 0
Solving yields
Bx D 3820 lb, By D 6690 lb
Dx D �1390 lb, Dy D �1930 lb, C D 9020 lb
508
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 6.135 The 600-lb weight of the scoop acts at
a point 1 ft 6 in to the right of the vertical line CE. The
line ADE is horizontal. The hydraulic actuator AB can
be treated as a two-force member. Determine the axial
force in the hydraulic actuator AB and the forces exerted
on the scoop at C and E.
1 ft 6 in
2 ft 6 in
1 ft
5 ft
2 ft
C
B
D E
A
Scoop
Solution: The free body diagrams are shown at the right. Place the
coordinate origin at A with the x axis horizontal. The coordinates (in
ft) of the points necessary to write the needed unit vectors are A (0,
0), B (6, 2), C (8.5, 1.5), and D (5, 0). The unit vectors needed for
this problem are
uBA D �0.949i� 0.316j,
uBC D 0.981i� 0.196j,
and uBD D �0.447i� 0.894j.
The scoop: The equilibrium equations for the scoop are
∑
FX D �TCBuBCX C EX D 0,
∑
FY D �TCBuBCY C EY � 600 D 0,
and
∑
MC D 1.5EX � 1.5�600 lb� D 0.
Solving, we get
EX D 600 lb,
EY D 480 lb,
and TCB D 611.9 lb.
Joint B: The equilibrium equations for the scoop are
∑
FX D TBAuBAX C TBDuBDX C TCBuBCX D 0,
and
∑
FY D TBAuBAY C TBDuBDY C TCBuBCY D 0.
Solving, we get
TBA D 835 lb,
and TBD D �429 lb.
TCB
TBA
TBD
TCB
EX E
G
C
EY
1.5 ft 1.5 ft
600 lb
x
y
c� 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright lawsas 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.
509
Problem 6.136 Determine the force exerted on the bolt
by the bolt cutters.
100 N
100 N
90 mm 60 mm 65 mm
55 mm
75
mm
40 mm
300 mm
A
C
D
B
Solution: The equations of equilibrium for each of the members
will be developed.
Member AB: The equations of equilibrium are:
∑
FX D AX C BX D 0,
∑
FY D AY C BY D 0,
and
∑
MB D 90F� 75AX � 425�100� D 0
Member BD: The equations are
∑
FX D �BX C DX D 0,
∑
FY D �BY C DY C 100 D 0,
and
∑
MB D 15DX C 60DY C 425�100� D 0.
Member AC: The equations are
∑
FX D �AX CCX D 0,
∑
FY D �AY CCY C F D 0,
and
∑
MA D �90FC 125CY C 40CX D 0.
Member CD: The equations are:
∑
FX D �CX � DX D 0,
∑
FY D �CY � DY D 0.
Solving the equations simultaneously (we have extra (but compatible)
equations, we get F D 1051 N, AX D 695 N, AY D 1586 N, BX D
�695 N, BY D �435 N, CX D 695 N, CY D 535 N, DX D �695 N,
and Dy D �535 N
75 mm
40 mm
100 N
55 mm
90 mm 60 mm 65 mm 300 mm
F
AX
BX
BA
BY
AY
λ
75 mm
40 mm
55 mm
90 mm 60 mm
60 mm
65 mm
65 mm
300 mm
300 mm
100 N
AX
BX
BY
DX
DY
DX
CX
DY
CY
AY CY
CXC
D
D
B
C
D
F
Problem 6.137 For the bolt cutters in Problem 6.136,
determine the magnitude of the force the members exert
on each other at the pin connection B and the axial force
in the two-force member CD.
Solution: From the solution to 6.136, we know BX D �695 N,
and BY D �435 N. We also know that CX D 695 N, and CY D 535 N,
from which the axial load in member CD can be calculated. The load
in CD is given by TCD D
√
C2X CC2Y D 877 N
510
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.