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

Parallel Chord Pratt Truss Example

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)

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

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

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

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)

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)

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)

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)

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)

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)

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)

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)

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)

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)

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 G11-10-9F, 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 (G11-10-9F)

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 11-12-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)

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