Example: A three-member truss with a vertical and a horizontal load
Given
- A triangular truss with joints A at (0, 0), C at (4 m, 0) and apex B at (2 m, 1.5 m). Members AB, BC and AC.
- A is a pin support and C is a roller on a horizontal surface.
- Joint B carries 12 kN downward and 4 kN horizontally to the right.
Find
- The support reactions and the force in each member, stating tension (T) or compression (C).
Solution
- Geometry: AB and BC each have a run of 2 m and a rise of 1.5 m, so L = 2.5 m and the direction ratios are 0.8 (horizontal) and 0.6 (vertical).
L = √(2² + 1.5²) = 2.5 m - Moments about A for the whole truss give the roller reaction at C (counterclockwise positive):
ΣM_A: Cy(4) − 12(2) − 4(1.5) = 0 → Cy = 7.5 kN ↑ - Force balance gives the pin reactions at A:
ΣFy: Ay + 7.5 − 12 = 0 → Ay = 4.5 kN ↑; ΣFx: Ax + 4 = 0 → Ax = −4 kN (4 kN ←) - Joint A, with F_AB and F_AC assumed in tension. ΣFy isolates F_AB:
4.5 + 0.6·F_AB = 0 → F_AB = −7.5 kN - ΣFx at A then gives F_AC:
−4 + 0.8(−7.5) + F_AC = 0 → F_AC = 10 kN - Joint C. ΣFy isolates F_BC (BC points up and to the left from C):
7.5 + 0.6·F_BC = 0 → F_BC = −12.5 kN - Check ΣFx at C, then both equations at B:
C: −10 − 0.8(−12.5) = 0 ✓; B: 4 + 6 − 10 = 0 ✓, −12 + 4.5 + 7.5 = 0 ✓
Answer: Ax = 4 kN ←, Ay = 4.5 kN ↑, Cy = 7.5 kN ↑. F_AB = 7.5 kN (C), F_BC = 12.5 kN (C), F_AC = 10 kN (T). The horizontal load makes the two rafters carry different forces even though the truss is symmetric.
Example: A five-joint truss solved joint by joint
Given
- Bottom chord joints A (0, 0), C (4 m, 0) and E (8 m, 0); top chord joints B (2 m, 3 m) and D (6 m, 3 m).
- Members AB, AC, BC, BD, CD, CE and DE. A is a pin, E is a roller.
- Loads: 10 kN downward at C and 6 kN downward at D.
Find
- The force in every member, as tension (T) or compression (C).
Solution
- Check determinacy: 7 members and 5 joints.
m = 2j − 3 = 2(5) − 3 = 7 ✓ - Each diagonal has a run of 2 m and a rise of 3 m, so L = √13 m and the direction ratios are 2/√13 = 0.555 and 3/√13 = 0.832.
L = √(2² + 3²) = √13 = 3.606 m - Reactions from moments about A and vertical balance:
Ey(8) − 10(4) − 6(6) = 0 → Ey = 9.5 kN ↑; Ay = 16 − 9.5 = 6.5 kN ↑ - Joint A (two unknowns):
6.5 + 0.832·F_AB = 0 → F_AB = −7.81 kN; 0.555(−7.81) + F_AC = 0 → F_AC = 4.33 kN - Joint E, the same way:
9.5 + 0.832·F_DE = 0 → F_DE = −11.42 kN; F_CE = −0.555(−11.42) = 6.33 kN - Joint B (no load). ΣFy: the two diagonals must cancel; ΣFx then gives the top chord.
F_BC = −F_AB = 7.81 kN; F_BD = −0.555(7.81) − 0.555(7.81) = −8.67 kN - Joint D. ΣFy gives F_CD; ΣFx is the check.
−0.832(F_CD + F_DE) − 6 = 0 → F_CD = −7.21 + 11.42 = 4.21 kN; 8.67 − 0.555(4.21) + 0.555(−11.42) = 0 ✓ - Joint C is a full check (vertical components of the diagonals against the 10 kN load):
0.832(7.81) + 0.832(4.21) − 10 = 6.5 + 3.5 − 10 = 0 ✓
Answer: F_AB = 7.81 kN (C), F_AC = 4.33 kN (T), F_BC = 7.81 kN (T), F_BD = 8.67 kN (C), F_CD = 4.21 kN (T), F_CE = 6.33 kN (T), F_DE = 11.42 kN (C). The top chord is in compression and the bottom chord in tension, as in any simply supported truss with downward loads.