1. Determine the force in members GM and MN.

2. Determine the forces in members BC and FG.

 Zero‐Force Members:
EK, EF

Cut (Upper Side)

1200 N
FCJ FFJ
Cut
FBC FFG
800 N C

FCJ FFJ
FBC FFG

+
M C  0  FFG 4   12002   0  FFG  600 N C 
Fy  0   FBC  FFG  0  FBC  600 N T 

 600
3. If it is known that the center pin A supports one-half of the vertical
loading shown, determine the force in member BF.
 Gy Hy
Ax

Ay
Reactions at the supports

From equilibrium of whole truss; Fx  0  Ax  0

48  210
Center pin A supports one-half of the vertical loading. Ay   26 kN
2

Because of symmetry, G y  H y  13 kN
 I. Cut

DE

DF
BF

AF
Gy
Hy=13 kN

Ay
DE I. Cut (Right side)
DF There are four unknowns.
BF

AF

Hy=13 kN
 Joint A
AB AF

45o 45o

Ay=26 kN

Fx  0   AB cos 45  AF cos 45  0  AB  AF

Fy  0  AB sin 45  AF sin 45  26  0  AB  AF  18.38 kN C 
DE

DF
I. Cut (Right side) 
BF
45o
AF
+ Hy=13 kN
M D  0
10(12)  8(24)  8(36)  13(48)  BF (16)  
AF cos 45(16)  
AF sin 45(12)  0
18.38 18.38

BF  24.24 kN T 
4. Determine the forces in members FG, CG, BC, and EF for the loaded crane truss.

25 kN 25 kN D
25 kN
E

25 kN
50 kN

F C
25 kN

 25 kN
G
B
25 kN

Zero‐Force Members: BG
 25 kN 25 kN
D 1st Cut 25 kN 25 kN
25 kN
E (Upper part) 25 kN D
E
25 kN 50 kN
50 kN 25 kN

F C F C
25 kN
25 kN
1st Cut CG
25 kN
G FG BC
B
25 kN
G
Fx  0  CG  0
+
M F  0 50(8)  BC (4)  0  BC  100 kN C 

Fy  0  25  FG  
BC  50  0  FG  25 kN T 
100

Joint F: EF
Fy  0
25 kN
FC  25  25 cos 45  EF cos 45  
FG  0
F 25

25 kN  EF  45.7 kN T 
FG
5. Determine the forces in members DE, EI, FI, and HI of the arched roof truss.
Zero‐Force Members:
BK, HF

Reactions at the supports

From equilibrium of whole truss;

Fx  0  Gx  0

Because of symmetry of the truss:

Ay  G y  150 kN

Gx
Ay Gy
+
M A  0  G y (40)  25(4)  75(10)  100(20)  75(30)  25(36)  0  G y  150 kN 
 
Fy  0  G y  Ay  225  275  100  0  Ay  150 kN 
1st Cut (Right side)

EF 25 kN
I FI F

HI
H

G 1st Cut

Gy=150 kN
Ay=150 kN Gy=150 kN
+
4
M I  0   EF 12  2512  15016  0  EF  315.48 kN C 
4 6
2 2

4 14
Fy  0  EF
  HI  150  25  0  HI  75.93 kN T 
 315.48 4 6
2 2
14  16
2 2

6 16
Fx  0  
EF  FI  
HI 0  FI  205.356 kN T 
315.48 4 62 2
75.93 14  16
2 2
2nd Cut (Right side)
75 kN
DE
E 25 kN
EI
IF
I F

IH

25 kN 2nd Cut
G

+ Gy=150 kN
Ay=150 kN Gy=150 kN
M I  0
3 10
 DE 6  DE 4  2512  756  15016  0  DE  297.008 kN C 
3  10
2 2
3  10
2 2

4 3 14
Fy  0   EI  
DE 
HI  75  25  150  0
4 62 2
 297.008 3  10
2 2
75.93 14  16
2 2

 EI  26.4 kN T 
6. Determine the forces acting in members DE, DI, KJ, AJ.

4m 4m 4m

B C D E F

3m
M I G
3m
L 37o
K J H
20 kN
6m

A
 4m 4m 4m
Zero‐Force Members:
EF, FG B T C D E F

3m
M I
G
3m
L
K J H 37o

20 kN
6m

Ax A

Ay
Reactions at the supports

From equilibrium of whole truss; Fy  0 Ay  20 sin 37  0  Ay  12 kN

+
M A  0  T (12)  20 cos 37(6)  20 sin 37(12)  0  T  20 kN

Fx  0  T  Ax  20 cos 37  0  Ax  4 kN
 4m 4m 4m
Joint A:
BT C D E F
AL AJ 3m
Ax=4 kN M I
G
3m
L
Ay=12 kN K J H 37o
Fx  0
20 kN
8
Ax  AJ 0 1st Cut 6m
6 2  82
Ax
AJ  5 kN C 
A

Ay

E 1st Cut (Right side)

DE +
EI M J  0
 DE 6  20 sin 374   0  DE  8 kN T 
IJ
KJ
J 37o
AJ 20 kN
 4m 4m 4m
2nd Cut (Right side)
T C D E F
DE E
3m
DI M I
G
I 3m
KI
L
K J 37o
KJ H

K J 37o 20 kN
J 6m
20 kN
AJ 2nd Cut
Ax A

+ Ay
M I  0
DE 3  KJ 3  
 AJ cos 373  20 cos 373  20 sin 374  0  KJ  12 kN T 
8 5

+
M K  0
DE 6  
 AJ sin 374  20 sin 378  DI cos 373  DI sin 374  0  DI  7.5 kN T 
8 5
7. The hinged frames ACE and DFB are connected by two hinged bars, AB and CD,
which cross without being connected. Compute the force in AB.
 I. Cut (Left Side) I. Cut (Right Side)
A
 AB 

2 m
 B CD
CD 3.5 m AB
2
tan  
3.5
  29.74 o
Ex

Ey

I. Cut (Left Side)

+
M E  0  AB cos 6   AB sin 1.5  CD cos 4   CD sin 3  0
CD  3 AB
I. Cut (Right Side)
+
M F  0  106   AB cos 4   AB sin 3  CD cos 6   CD sin 1.5  0
60  5.95CD  1.98 AB  0  AB  3.78 kN C 