66.

Determine the force in each member of the truss, and state D 600 N
if the members are in tension or compression.

4m

SOLUTION E 900 N
C
Method of Joints: We will begin by analyzing the equilibrium of joint D, and then
proceed to analyze joints C and E. 4m

Joint D: From the free-body diagram in Fig. a, A B

FDE a b - 600 = 0
+ F = 0; 3 6m
: x
5
FDE = 1000 N = 1.00 kN (C) Ans.

1000 a b - FDC = 0
4
+ c Fy = 0;
5
FDC = 800 N (T) Ans.

Joint C: From the free-body diagram in Fig. b,

+ F = 0; FCE - 900 = 0
: x

FCE = 900 N (C) Ans.

+ c Fy = 0; 800 - FCB = 0
FCB = 800 N (T) Ans.

Joint E: From the free-body diagram in Fig. c,

R+ Fx = 0; - 900 cos 36.87 + FEB sin 73.74 = 0

FEB = 750 N (T) Ans.

Q+ Fy = 0; FEA - 1000 - 900 sin 36.87 - 750 cos 73.74 = 0

FEA = 1750 N = 1.75 kN (C) Ans.

67.

Determine the force in each member of the Pratt truss, and J

state if the members are in tension or compression.
2m
K I
2m
L H
SOLUTION 2m
A
G
Joint A: B C D E F
2m 2m 2m 2m 2m 2m
+ c Fy = 0; 20 - FAL sin 45 = 0
10 kN 10 kN
FAL = 28.28 kN (C) 20 kN

+ F = 0; FAB - 28.28 cos 45 = 0

: x

FAB = 20 kN (T)

Joint B:
+ F = 0; FBC - 20 = 0
: x

FBC = 20 kN (T)
+ c Fy = 0; FBL = 0

Joint L:

R+ Fx = 0; FLC = 0
+QFy = 0; 28.28 - FLK = 0
FLK = 28.28 kN (C)

Joint C:
+ F = 0; FCD - 20 = 0
: x

FCD = 20 kN (T)

+ c Fy = 0; FCK - 10 = 0
FCK = 10 kN (T)

Joint K:

R+ Fx - 0; 10 sin 45 - FKD cos (45 - 26.57) = 0

FKD = 7.454 kN (L)

+QFy = 0; 28.28 - 10 cos 45 + 7.454 sin (45 - 26.57) - FKJ = 0

FKJ = 23.57 kN (C)

Joint J:
+ F = 0; 23.57 sin 45 - FJI sin 45 = 0
: x

FJI = 23.57 kN (L)

+ c Fy = 0; 2 (23.57 cos 45) - FJD = 0
FJD = 33.3 kN (T) Ans.

Due to Symmetry

FAL = FGH = FLK = FHI = 28.3 kN (C) Ans.

FAB = FGF = FBC = FFE = FCD = FED = 20 kN (T) Ans.
FBL = FFH = FLC = FHE = 0 Ans.
FCK = FEI = 10 kN (T) Ans.
FKJ = FIJ = 23.6 kN (C) Ans.
FKD = FID = 7.45 kN (C) Ans.
622.

Determine the force in each member of the double scissors B C

truss in terms of the load P and state if the members are in
tension or compression.
L/3
SOLUTION
E F D
A
L 2L
c + MA = 0; Pa b + P a b - (Dy)(L) = 0
3 3 L/3 L/3 L/3
Dy = P
P P
+ c Fy = 0; Ay = P

Joint F:
+ F = 0; 1 (1)
: x FFD - FFE - FFB a b = 0
22
FFD - FFE = P

+ F = 0; 1
: y FFB a b -P = 0
22

FFB = 22P = 1.41 P (T)

Similarly,
FEC = 22P

Joint C:

+ F = 0; 2 1 1
: x FCA a b - 22P a b - FCD a b = 0
25 22 22
2 1
FCA - FCD = P
25 22
1 1 1
+c Fy = 0; FCA - 22P + FCD =0
25 22 22
225
FCA = P = 1.4907P = 1.49P (C)
3

22
FCD = P = 0.4714P = 0.471P (C)
3

Joint A:

+ F = 0; 22 1 225 2
: x FAE - Pa b - Pa b = 0
3 22 3 25
5
FAE = P = 1.67 P (T)
3
Similarly,
FFD=1.67 P (T)

From Eq.(1), and Symmetry,

FFE = 0.667 P (T) Ans.
FFD = 1.67 P (T) Ans.
FAB = 0.471 P (C) Ans.
FAE = 1.67 P (T) Ans.
FAC = 1.49 P (C) Ans.
FBF = 1.41 P (T) Ans.
FBD = 1.49 P (C) Ans.
FEC = 1.41 P (T) Ans.
FCD = 0.471 P (C) Ans.
625.

Determine the force in each member of the truss in terms of P

the external loading and state if the members are in tension
or compression. B L C
P

L L

SOLUTION
Joint B: A D
L
+ c Fy = 0; FBA sin 2u - P = 0

FBA = P csc 2u (C) Ans.

+ F = 0;
: P csc 2u(cos 2u) - FBC = 0
x

FBC = P cot 2 u (C) Ans.

Joint C:
+ F = 0;
: P cot 2 u + P + FCD cos 2 u - FCA cos u = 0
x

+ c Fy = 0; FCD sin 2 u - FCA sin u = 0

cot 2 u + 1
FCA = P
cos u - sin u cot 2 u

FCA = (cot u cos u - sin u + 2 cos u) P (T) Ans.

FCD = (cot 2 u + 1) P (C) Ans.

Joint D:
+ F = 0;
: FDA - (cot 2 u + 1)(cos 2 u) P = 0
x

FDA = (cot 2 u + 1)(cos 2 u) (P) (C) Ans.

626.

The maximum allowable tensile force in the members of the P

truss is 1Ft2max = 2 kN, and the maximum allowable
compressive force is 1Fc2max = 1.2 kN. Determine the B L C
maximum magnitude P of the two loads that can be applied P
to the truss. Take L = 2 m and u = 30.
u

L L

SOLUTION
(Tt)max = 2 kN A D
L
(FC)max = 1.2 kN

Joint B:

+ c Fy = 0; FBA cos 30 - P = 0

P
FBA = = 1.1547 P (C)
cos 30
+ F = 0;
: FAB sin 30 - FBC = 0
x

FBC = P tan 30 = 0.57735 P (C)

Joint C:

+ c Fy = 0; -FCA sin 30 + FCD sin 60 = 0

sin 60
FCA = FCD a b = 1.732 FCD
sin 30
+ F = 0;
: P tan 30 + P + FCD cos 60 - FCA cos 30 = 0
x

tan 30 + 1
FCD = a b P = 1.577 P (C)
23 cos 30 - cos 60
FCA = 2.732 P (T)

Joint D:
+ F = 0;
: FDA - 1.577 P sin 30 = 0
x

FDA = 0.7887 P (C)

1) Assume FCA = 2 kN = 2.732 P

P = 732.05 N

FCD = 1.577(732.05) = 1154.7 N 6 (Fc)max = 1200 N (O.K.!)

Thus, Pmax = 732 N Ans.

631.

Determine the force in members CD, CJ, KJ, and DJ of the 8000 lb
truss which serves to support the deck of a bridge. State if 4000 lb 5000 lb
these members are in tension or compression.
B C D E F
A G

12 ft

SOLUTION L K J I H
9 ft 9 ft 9 ft 9 ft 9 ft 9 ft
a + MC = 0; - 9500(18) + 4000(9) + FKJ(12) = 0

FKJ = 11 250 lb = 11.2 kip (T) Ans.

a + MJ = 0; -9500(27) + 4000(18) + 8000(9) + FCD(12) = 0

FCD = 9375 lb = 9.38 kip (C) Ans.

+ F = 0; 3
: x - 9375 + 11 250 - FCJ = 0
5

FCJ = 3125 lb = 3.12 kip (C) Ans.

Joint D: FDJ = 0 Ans.

*632.

Determine the force in members EI and JI of the truss 8000 lb

which serves to support the deck of a bridge. State if these 4000 lb 5000 lb
members are in tension or compression.
B C D E F
A G

12 ft

SOLUTION L K J I H
9 ft 9 ft 9 ft 9 ft 9 ft 9 ft
a + ME = 0; - 5000(9) + 7500(18) - FJI(12) = 0

FJI = 7500 lb = 7.50 kip (T) Ans.

+ c Fy = 0; 7500 - 5000 - FEI = 0

FEI = 2500 lb = 2.50 kip (C) Ans.

646.

Determine the force in members CD and CM of the M L K

Baltimore bridge truss and state if the members are in 2m
tension or compression. Also, indicate all zero-force N O P J
members. 2m
A I
B C D E F G H

2 kN 2 kN
5 kN 3 kN

SOLUTION 16 m, 8 @ 2 m

Support Reactions:

a + MI = 0; 21122 + 5182 + 3162 + 2142 - Ay 1162 = 0

Ay = 5.625 kN
+ F = 0;
: Ax = 0
x

Method of Joints: By inspection, members BN, NC, DO, OC, HJ

LE and JG are zero force members. Ans.

Method of Sections:

a + MM = 0; FCD142 - 5.625142 = 0

FCD = 5.625 kN 1T2 Ans.

a + MA = 0; FCM 142 - 2142 = 0

FCM = 2.00 kN T Ans.

647.

Determine the force in members EF, EP, and LK of the M L K

Baltimore bridge truss and state if the members are in 2m
tension or compression. Also, indicate all zero-force N O P J
members. 2m
A I
B C D E F G H

2 kN 2 kN
5 kN 3 kN

SOLUTION 16 m, 8 @ 2 m

Support Reactions:

a + MA = 0; Iy 1162 - 21122 - 31102 - 5182 - 2142 = 0

Iy = 6.375 kN

Method of Joints: By inspection, members BN, NC, DO, OC, HJ

LE and JG are zero force members. Ans.

Method of Sections:

a + MK = 0; 3122 + 6.375142 - FEF142 = 0

FEF = 7.875 = 7.88 kN 1T2 Ans.

a + ME = 0; 6.375182 - 2142 - 3122 - FLK 142 = 0

FLK = 9.25 kN 1C2 Ans.

+ c Fy = 0; 6.375 - 3 - 2 - FED sin 45 = 0

FED = 1.94 kN T Ans.

671.

Determine the support reactions at A, C, and E on the 10 kN

9 kN
compound beam which is pin connected at B and D.

10 kN m
E
B C D
SOLUTION A

Equations of Equilibrium: First, we will consider the free-body diagram of 1.5 m 1.5 m 1.5 m 1.5 m 1.5 m 1.5 m
segment DE in Fig. c.

+ MD = 0; NE(3) - 10(1.5) = 0
NE = 5 kN Ans.
+ ME = 0; 10(1.5) - Dy(3) = 0
Dy = 5 kN
+ F = 0; Dx = 0 Ans.
: x

Subsequently, the free-body diagram of segment BD in Fig. b will be considered

using the results of Dx and Dy obtained above.

+ MB = 0; NC(1.5) - 5(3) - 10 = 0
NC = 16.67 kN = 16.7 kN Ans.
+ MC = 0; By(1.5) - 5(1.5) - 10 = 0
By = 11.67 kN
+ F = 0; By = 0
: x

Finally, the free-body diagram of segment AB in Fig. a will be considered using the
results of Bx and By obtained above.
+ F = 0; Ax = 0 Ans.
: x

+ c Fy = 0; 11.67 - 9 - A y = 0
A y = 2.67 kN Ans.
+ MA = 0; 11.67(3) - 9(1.5) - MA = 0
MA = 21.5 kN # m Ans.
6102.

The tractor boom supports the uniform mass of 500 kg in

the bucket which has a center of mass at G. Determine the
force in each hydraulic cylinder AB and CD and the G A B
resultant force at pins E and F. The load is supported 0.25 m
equally on each side of the tractor by a similar mechanism. E C

1.5 m
0.3 m
0.1 m
SOLUTION 0.2 m 1.25 m

a + ME = 0; 2452.510.12 - FAB10.252 = 0
F
FAB = 981 N Ans.
+ F = 0;
: -Ex + 981 = 0; Ex = 981 N
0.6 m
x
D
+ c Fy = 0; Ey - 2452.5 = 0; Ey = 2452.5 N
0.4 m 0.3 m
FE = 2198122 + 12452.522 = 2.64 kN Ans.

a + MF = 0; 2452.512.802 - FCD1cos 12.2210.72 + FCD1sin 12.2211.252 = 0

FCD = 16 349 N = 16.3 kN Ans.

+ F = 0;
: Fx - 16 349 sin 12.2 = 0
x

Fx = 3455 N

+ c Fy = 0; -Fy - 2452.5 + 16 349 cos 12.2 = 0

Fy = 13 527 N

FF = 21345522 + 113 52722 = 14.0 kN Ans.

6109.

The symmetric coil tong supports the coil which has a mass
of 800 kg and center of mass at G. Determine the horizontal
and vertical components of force the linkage exerts on H
plate DEIJH at points D and E. The coil exerts only vertical
reactions at K and L. 300 mm D J
E I
400 mm
C
45 30 30
SOLUTION 100 mm
B
A 45
F

Free-Body Diagram: The solution for this problem will be simplified if one realizes 50 mm
100 mm
that links BD and CF are two-force members.

Equations of Equilibrium : From FBD (a), K G L

a + ML = 0; 78481x2 - FK12x2 = 0 FK = 3924 N

From FBD (b),

a + MA = 0; FBD cos 4511002 + FBD sin 4511002 - 39241502 = 0

FBD = 1387.34 N
+ F = 0;
: A x - 1387.34 cos 45 = 0 A x = 981 N
x

+ c Fy = 0; A y - 3924 - 1387.34 sin 45 = 0

A y = 4905 N

From FBD (c),

a + ME = 0; 4905 sin 4517002 - 981 sin 4517002

- FCF cos 1513002 = 0

FCF = 6702.66 N
+ F = 0;
: Ex - 981 - 6702.66 cos 30 = 0
x

Ex = 6785.67 N = 6.79 kN Ans.

+ c Fy = 0; Ey + 6702.66 sin 30 - 4905 = 0

Ey = 1553.67 N = 1.55 kN Ans.

At point D,

Dx = FBD cos 45 = 1387.34 cos 45 = 981 N Ans.

Dy = FBD sin 45 = 1387.34 sin 45 = 981 N Ans.