Forces in all members of truss with 10.8 kN loads

Data and assumptions

A(0,0), B(22.5,0) and C(57.5,0) lie on the top chord (panels of 22.5 m and 35 m); D is 12 m below B. A is a pin and C a roller (vertical reaction); D is a free joint. Members: AB, BC, BD, CD and AD (m = 5, j = 4, r = 3, so m + r = 2j: determinate). Lengths: AB = 22.5 m, BC = 35 m, BD = 12 m, CD = √(35² + 12²) = 37 m (sin = 12/37 = 0.3243, cos = 35/37 = 0.9459), AD = √(22.5² + 12²) = 25.5 m (sin = 12/25.5 = 0.4706, cos = 22.5/25.5 = 0.8824). Loads: 10.8 kN ↓ at B and 10.8 kN ↓ at C.

 A(pin)      B              C(roller)
  o---------o--------------o
    \       |          .
      \     |      .
        \   |  .
            D

Reactions

• ΣM_A = 0: 57.5C_y = 10.8(22.5) + 10.8(57.5) = 864, so C_y = 15.03 kN ↑.
• ΣFy: A_y = 21.6 − 15.03 = 6.57 kN ↑; ΣFx: A_x = 0.

Method of joints (tension +, compression −)

Joint A (AB along +x; AD down-right):

• ΣFy: 6.57 − 0.4706F_AD = 0, so F_AD = +13.97 kN (T).
• ΣFx: F_AB + 0.8824(13.97) = 0, so F_AB = −12.33 kN (C).

Joint B (AB and BC collinear, BD vertical, 10.8 kN ↓):

• ΣFy: −10.8 − F_BD = 0, so F_BD = −10.80 kN (C).
• ΣFx: F_BC = F_AB, so F_BC = −12.33 kN (C).

Joint C (reaction 15.03 kN ↑, load 10.8 kN ↓; CD goes down-left):

• ΣFy: 15.03 − 10.8 − 0.3243F_CD = 0, so F_CD = +13.03 kN (T).
• Check ΣFx: 12.33 − 0.9459(13.03) = 0 ✓.

Joint D (check): AD pulls D with (12.33 ←, 6.57 ↑), CD pulls D with (12.33 →, 4.23 ↑), and BD (compression) pushes D down with 10.80 kN: 6.57 + 4.23 − 10.80 = 0 ✓.

Result

The top chord and the vertical BD are in compression, and the lower members AD and CD are in tension. The 10.8 kN load at C acts directly over the roller, so it only adds to C_y and does not change any member force.