Truss ideal assumptions and member forces

Ideal assumptions of a truss

1. All members are straight and joined at their ends by smooth (frictionless) pins.
2. Loads and support reactions act only at the joints.
3. The self-weight of members is negligible (or is lumped at the joints).
4. Every member is a two-force member, so it carries only axial force (tension or compression), with no bending or shear.
5. Members are rigid enough that the truss geometry does not change appreciably under load, and all members lie in one plane.

Data

Take A(0,0) pin, C(20,0) roller, apex B(10,6) and internal joint D(10,3) (the 3 m + 3 m heights in the figure). Members: AB, BC, AD, DC, BD (m = 5, j = 4, r = 3, so m + r = 2j and the truss is determinate and stable). Loads at B: 40 kN ↓ and 30 kN →. Lengths: AB = BC = √(10² + 6²) = 11.66 m, AD = DC = √(10² + 3²) = 10.44 m, BD = 3 m.

          30kN→ B  (40 kN ↓)
              / | \
            /   |   \
          /     D     \
        /  .    |   .    \
      A(pin) ------------- C(roller)

Support reactions

• ΣM_A = 0: C_y(20) − 40(10) − 30(6) = 0, so C_y = 29 kN ↑.
• ΣFy = 0: A_y = 40 − 29 = 11 kN ↑.
• ΣFx = 0: A_x + 30 = 0, so A_x = 30 kN ←.

Method of joints (tension +, compression −)

Joint C (cos = 10/11.66 = 0.8575, sin = 6/11.66 = 0.5145 for CB; 0.9578 and 0.2874 for CD):

• ΣFy: 29 + 0.5145F_CB + 0.2874F_CD = 0.
• ΣFx: −0.8575F_CB − 0.9578F_CD = 0, so F_CD = −0.8953F_CB.
• Substituting: 29 + 0.5145F_CB − 0.2573F_CB = 0, so F_CB = −112.73 kN (C) and F_CD = +100.92 kN (T).

Joint D (members DC, DA, DB; no load):

• ΣFx: 0.9578(100.92) − 0.9578F_DA = 0, so F_DA = 100.92 kN (T).
• ΣFy: −0.2874(100.92) − 0.2874(100.92) + F_DB = 0, so F_DB = 58.00 kN (T).

Joint A: AD pulls A with (96.67, 29.00); AB gives (0.8575, 0.5145)F_AB. ΣFx: −30 + 96.67 + 0.8575F_AB = 0, so F_AB = −77.75 kN (C). Check ΣFy: 11 + 29.00 − 0.5145(77.75) = 11 + 29.00 − 40.00 = 0 ✓.

Result

Joint B closes the check: ΣFy = −40 + 0.5145(77.75) + 0.5145(112.73) − 58 = −40 + 40 + 58 − 58 = 0 ✓, and ΣFx = 30 + 0.8575(77.75) − 0.8575(112.73) = 30 + 66.67 − 96.67 = 0 ✓.