Truss member forces; assumptions of perfect truss

Data

Take A(0,0) pin and B(6,0) roller on the bottom; F(6,7.5), E(12,7.5), D(18,7.5), C(24,7.5) on the top (6 ft panels, 7.5 ft high); the lower chord runs in a straight line from B to C, so G(12,2.5) and H(18,5). Loads: 15 kN ↓ at E, 10 kN ↓ at C. The members marked in the figure are taken as BE, EG, ED, DH and GH. (m = 13, r = 3, j = 8: determinate.)

Reactions

• ΣM_A = 0: 6B_y − 15(12) − 10(24) = 0, so B_y = 70 kN ↑.
• ΣFy: A_y = 25 − 70 = −45 kN (45 kN ↓); ΣFx: A_x = 0.

Joint C (cut at the right-hand end)

Member CH has direction (−6, −2.5)/6.5 from C, so its vertical component is 2.5/6.5 = 0.3846.

• ΣFy: −10 − 0.3846F_CH = 0, so F_HC = −26.0 kN (C).
• ΣFx: −0.9231F_CH − F_CD = 0, so F_CD = +24.0 kN (T).

Joint H

Members HG and HC lie on the same straight line and DH is the only other member with no load at H, so F_DH = 0 and F_GH = F_HC = −26.0 kN (C).

Joint D

ED and DC are collinear and equal (24 kN T); with DH = 0, DG must also be zero: F_ED = +24.0 kN (T), F_DG = 0.

Joint G

BG and GH are collinear (26 kN C each) and DG = 0, so the vertical member F_EG = 0.

Joint E (cos = 6/9.605 = 0.6247, sin = 0.7809 for EB)

• ΣFy: −15 − 0.7809F_BE = 0 (E→B points down-left, EG = 0), so F_BE = −19.21 kN (C) (it pushes E upward with 15 kN).
• ΣFx: −F_FE + 24 + 0.6247(19.21) = 0 → F_FE = +36 kN (T) (check by section: ΣM_B on the left part: 7.5F_FE = 6×45 = 270, so F_FE = 36 kN).

Result (indicated members)

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.