Truss member forces; assumptions of perfect truss

Data

A(0,0) roller, B(3,0) and C(6,0) pin on the bottom; D(1,2) and E(5,2) on the top (top chord 1 m + 4 m + 1 m, height 2 m). Members: AB, BC, AD, DE, EC, DB, BE (m = 7, j = 5, r = 3: determinate). Loads: 20 kN ↓ at D; 30 kN at E at 45° pointing down-left (21.21 kN ← and 21.21 kN ↓). Lengths: AD = EC = √(1² + 2²) = 2.236 m (sin = 0.8944, cos = 0.4472); DB = BE = 2.828 m (45°).

Reactions

• ΣM_C = 0 (anticlockwise +): −6A_y + 20(5) + 21.21(1) + 21.21(2) = 0, so A_y = 27.27 kN ↑.
• ΣFx: C_x = 21.21 kN →. ΣFy: C_y = 20 + 21.21 − 27.27 = 13.94 kN ↑.

Method of joints

Joint A: ΣFy: 27.27 + 0.8944F_AD = 0, so F_AD = −30.49 kN (C). ΣFx: F_AB + 0.4472(−30.49) = 0, so F_AB = +13.64 kN (T).

Joint D: AD pushes D with (13.64 →, 27.27 ↑).

• ΣFy: 27.27 − 20 − 0.7071F_DB = 0, so F_DB = +10.29 kN (T).
• ΣFx: 13.64 + F_DE + 0.7071(10.29) = 0, so F_DE = −20.91 kN (C).

Joint B: AB pulls B to the left (13.64 kN); DB goes up-left and BE up-right at 45°. No external load acts at B, so ΣFy: 0.7071(10.29) + 0.7071F_BE = 0 gives F_BE = −10.29 kN (C). ΣFx: −13.64 − 7.27 − 7.27 + F_BC = 0, so F_BC = +28.18 kN (T).

Joint C: ΣFy: 13.94 + 0.8944F_EC = 0, so F_EC = −15.59 kN (C). Check ΣFx: 21.21 − 28.18 + 0.4472(15.59) = 0 ✓.

Result

Assumptions of a perfect 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.