This is a truss problem using the method of joints. The truss is shown above. The horizontal and vertical struts have a length a. The three forces F1 = F2 = F3 = 20 kN. Find the strut loads for all of the struts.

We begin by taking a free body diagram of the whole truss, the equations of equilibrium are as follows:

F1 + F2 + F3 − XA = 0 (1a, x-direction)
N − YA = 0 (1b, y-direction)
a F2 + 2 a F3 − 2 a N = 0 (1c, moments around Point V)

Solving Equation (1a) for XA,

XA = F1 + F2 + F3 (2)

Solving Equation (1b) for YA,

YA = N (3)

Substituting Equation (3) into Equation (1c) and solving for N,

N = YA = F2/2 + F3 (4)

For convenience, we note that sin(45) = cos(45) = 2-1/2, and write (and solve when advantageous) the strut equations as follows:

F2 + S3 = 0 (5, Joint II, x-direction)

S3 = -F2 (5, solving Equation (5))

S7 = 0 (6, zero force member, not included in future equations)

− F1 − F2 − F3 − S8 − 2-1/2 S9 = 0 (7, Joint V, x-direction)

− 2-1/2 S2 + F2 − 2-1/2 S5 + 2-1/2 S9 = 0 (8, Joint VI, x-direction)

−S6 + S8 = 0 (9, Joint IV, x-direction)

F3 + S6 + 2-1/2 S5 = 0 (10, Joint III, x-direction)

− 1/2 F2 − F3 − 2-1/2 S9 = 0 (11, Joint V, y-direction)

S9 = 2-1/2 ( F2 + 2 F3 ) (12, solving Equation (11) for S9)

S8 = − F1 − F2/2 (13, solving Equation (7) for S8)

S6 = – F1 − F2/2 (14, solving Equation (9) for S6)

S5 = 2-1/2 ( −2 F3 + 2 F1 + F2 ) (15, solving Equation (1) for S5)

S2 = – 21/2 F1 (16, solving Equation (8) for S2)

−S4 + F3 − F1 − F2/2 = 0 (17, Joint III, y-direction)

S4 = F3 − F1 − F2/2 (18, solving Equation (17) for S4)

F3 − F1 − F2/2 − S1 = 0 (19, Joint II, y-direction)

S1 = F3 − F1 − F2/2 (20, solve Equation (19) for S1)

With all of these computed and solved, the results (in kN) are as follows:

  • Reactions
    • N = 30
    • XA = 60
    • YA = 30
  • Struts (by number)
    1. 10
    2. -28.28
    3. -20
    4. -10
    5. 14.14
    6. -30
    7. 0
    8. -30
    9. -42.42

The problem is taken from Theoretical Mechanics – A Short Course -Targ.