1. Truss Analysis Using the Graphical Method (“Shaping Structures: Statics” by Waclaw Zalewski and Edward Allen) (“Form & Forces” by Allen & Zalewski)

2. Parallel Chord Pratt Truss Example

4. 1600 lbs
5600 lbs
10’-0”
10’-0”
10’-0”
10’-0”
10’-0”
10’-0”
10’-0”
5600 lbs
FORM DIAGRAM (Drawn to Scale, here 1/8” = 1’-0”) (includes reactions already solved)

5. Step 1: Label the FORM DIAGRAM
1600 lbs
5600 lbs B C D E F G A H J
5600 lbs
Step 1: Label the FORM DIAGRAM
Capital Letters Between Forces and Reactions, Clockwise, Start at Left Side

6. Step 2: Label the FORM DIAGRAM
1600 lbs
5600 lbs B C D E F G 4 6 7 9 2 11 A H 1 3 5 8 10 12 J
5600 lbs
Step 2: Label the FORM DIAGRAM
Numbers Between Diagonals, Start at Left Side

7. Step 3: Begin the FORCE POLYGON (Scale 1” = 2850 lbs)D E F G H J 8 2 3 4 5 6 1 7 9 10 11 12 a b c d j e f g h
Step 3: Begin the FORCE POLYGON (Scale 1” = 2850 lbs)
Sum of Forces in ‘Y’ direction, begin with Force A-B

8. Step 4: Begin DRAWING FORCES
1600 lbs
5600 lbs A B C D E F G H J 8 2 3 4 5 6 1 7 9 10 11 12 a 1 b c
At Joint A1J there are two forces: d
A1 in the Y direction j e
1J in the X direction
A1 is compression acting downward f
It starts at ‘a’ and goes vertically down to ‘1’ g
1J is horizontal, but in this case its value is zero
1J is a point at the tip of the arrow h
Step 4: Begin DRAWING FORCES
Like method of Joints, start at joint with 2 unknowns (A1J)

9. Step 5: Continue DRAWING FORCES
1600 lbs
5600 lbs A B C D E F G H J 8 2 3 4 5 6 1 7 9 10 11 12 a 2 b c
At Joint B21A there are two unknown forces: d
B2 in the X direction
j, 1 e
21 in the diagonal direction
B2 is compression acting horizontally f
It starts at ‘b’ and goes horizontally ‘2’ g
21 is diagonal and must go from ‘2’ to ‘1’
Thus we can locate point ‘2’ based on geometry h
Step 5: Continue DRAWING FORCES
Like method of Joints, move to next joint with 2 unknowns (B21A)

10. Step 6: Continue DRAWING FORCES
1600 lbs
5600 lbs A B C D E F G H J 8 2 3 4 5 6 1 7 9 10 11 12 a 2 b c
At Joint 23J1 there are two unknown forces: d 3
23 in the Y direction
j, 1 e
J3 in the X direction
23 is compression acting vertically f
It starts at ‘2’ and goes vertically to ‘3’ g
3J is horizontal and must go from ‘3’ to ‘J’
Thus we can locate point ‘3’ based on geometry h
Step 6: Continue DRAWING FORCES
Like method of Joints, move to next joint with 2 unknowns (23J1)

11. Step 7: Continue DRAWING FORCES
1600 lbs
5600 lbs A B C D E F G H J 8 2 3 4 5 6 1 7 9 10 11 12 a 2 b 4 c
At Joint C432B there are now two unknown forces: d 3
C4 in the X direction
j, 1 e
43 in the diagonal direction
C4 is compression acting horizontally f
It starts at ‘c’ and goes horizontally to ‘4’ g
43 is diagonal and must go from ‘4’ to ‘3’
Thus we can locate point ‘4’ based on geometry h
Step 7: Continue DRAWING FORCES
Like method of Joints, move to next joint with 2 unknowns (C432B)

12. Step 8: Continue DRAWING FORCES
1600 lbs
5600 lbs A B C D E F G H J 8 2 3 4 5 6 1 7 9 10 11 12 a 2 b 4 c
At Joint 5J34, two unknowns: d 5 3
45 in the Y direction
j, 1 e
J5 in the diagonal direction
45 is compression acting vertically f
It starts at ‘4’ and goes vertically to ‘5’ g
J5 is horizontal and must go from ‘5’ to ‘J’
Thus we can locate point ‘5’ based on geometry h
Step 8: Continue DRAWING FORCES
Like method of Joints, move to next joint with 2 unknowns (5J34)

13. Step 9: Continue DRAWING FORCES
1600 lbs
5600 lbs A B C D E F G H J 8 2 3 4 5 6 1 7 9 10 11 12 a 2 b 4 c 6
At Joint D654C, two unknowns: d 5 3
D6 in the X direction
j, 1 e
65 in the diagonal direction
D6 is compression acting horizontally f
It starts at ‘d’ and goes horizontally to ‘6’ g
65 is diagonal and must go from ‘6’ to ‘5’
Thus we can locate point ‘6’ based on geometry h
Step 9: Continue DRAWING FORCES
Like method of Joints, move to next joint with 2 unknowns (D654C)

14. Step 10: Continue DRAWING FORCES
1600 lbs
5600 lbs A B C D E F G H J 8 2 3 4 5 6 1 7 9 10 11 12 a 2 b 4 c 6
At Joint E76D, two unknowns: d 5 3
E7 in the X direction
j, 1 e
76 in the vertical direction 7
E7 is compression acting horizontally f
It starts at ‘e’ and goes horizontally to ‘7’ g
67 is vertical and must go from ‘6’ to ‘7’
Thus we can locate point ‘7’ based on geometry h
Step 10: Continue DRAWING FORCES
Like method of Joints, move to next joint with 2 unknowns (E76D)

15. Step 11: Continue DRAWING FORCES
1600 lbs
5600 lbs A B C D E F G H J 8 2 3 4 5 6 1 7 9 10 11 12 a 2 b 4 c 6
At Joint 78J56, two unknowns: d 5 8 3
8J in the X direction
j, 1 e
78 in the diagonal direction 7
8J is tension acting horizontally f
It starts at ‘8’ and goes horizontally to ‘j’ g
78 is diagonal and must go from ‘7’ to ‘8’
Thus we can locate point ‘8’ based on geometry h
Step 11: Continue DRAWING FORCES
Like method of Joints, move to next joint with 2 unknowns (78J56)

16. Step 12: Continue DRAWING FORCES
1600 lbs
5600 lbs A B C D E F G H J 8 2 3 4 5 6 1 7 9 10 11 12 a 2 b 4 c 6
At Joint F987E, two unknowns: d
5,8 3
F9 in the X direction
j, 1 e
98 in the vertical direction 7
F9 is compression acting horizontally f
It starts at ‘f’ and goes horizontally to ‘9’ 9 g
98 is vertical and must go from ‘8’ to ‘9’
Thus we can locate point ‘9’ based on geometry h
Step 12: Continue DRAWING FORCES
Like method of Joints, move to next joint with 2 unknowns (F987E)

17. Step 13: Continue DRAWING FORCES
1600 lbs
5600 lbs A B C D E F G H J 8 2 3 4 5 6 1 7 9 10 11 12 a 2 b 4 c 6
At Joint 9-10J8, two unknowns: d
5,8 3 10
10J in the X direction
j, 1 e
9-10 in the diagonal direction 7
10J is tension acting horizontally f
It starts at ‘10’ and goes horizontally to ‘j’ 9 g
9-10 is diagonal and must go from ‘9’ to ‘10’
Thus we can locate point ‘10’ based on geometry h
Step 13: Continue DRAWING FORCES
Like method of Joints, move to next joint with 2 unknowns (9-10J8)

18. Step 14: Continue DRAWING FORCES
1600 lbs
5600 lbs A B C D E F G H J 8 2 3 4 5 6 1 7 9 10 11 12 a 2 b 4 c 6
At Joint G F, two unknowns: d
5,8
3,10
G11 in the X direction
j, 1 e
10-11 in the vertical direction 7
G11 is compression acting horizontally f
It starts at ‘g’ and goes horizontally to ‘11’ 9 g
10-11 is vertical and must go from ‘10’ to ‘11’ 11
Thus we can locate point ‘11’ based on geometry h
Step 14: Continue DRAWING FORCES
Like method of Joints, move to next joint with 2 unknowns (G F)

19. Step 15: Continue DRAWING FORCES
1600 lbs
5600 lbs A B C D E F G H J 8 2 3 4 5 6 1 7 9 10 11 12 a 2 b 4 c 6
At Joint J10, two unknowns: d
5,8
3,10
12J in the X direction
j, 1 12 e
11-12 in the diagonal direction 7
12J is zero force f
It starts at ‘J’ and goes horizontally to ‘12’ a distance of zero, so it is a point at ’12’. 9 g 11
11-12 is a diagonal and must go from ‘11’ to ‘12’
Thus we can locate point ‘12’ based on geometry h
Step 15: Continue DRAWING FORCES
Like method of Joints, move to next joint with 2 unknowns (11-12-J10)

20. Step 16: Completed Force Polygon
1600 lbs
5600 lbs A B C D E F G H J 8 2 3 4 5 6 1 7 9 10 11 12 a 2 b 4 c 6
The completed force Polygon: d
5,8
3,10
j, 1,12
Indicates member forces. e
To find the magnitude of a force, 7
measure the diagram; it is drawn to scale. (in this case 1” = 2850 lbs) f 9 g 11 h
Step 16: Completed Force Polygon
This is sometimes called a Maxwell-Cremona Diagram