Member forces of truss with 20 and 10 kN loads

Data

Top joints A(0,8), B(12,8), C(24,8); bottom joints D(6,0), E(18,0). Members: AB, BC, AD, DB, BE, EC, DE (m = 7, j = 5, r = 3: determinate). Loads: 20 kN ↓ at A and 10 kN ↓ at B. Supports: pin at C and roller at E (vertical reaction). The inclined members have length √(6² + 8²) = 10 m, so sin = 0.8 and cos = 0.6.

Reactions

• ΣM_C = 0 (anticlockwise +): 20(24) + 10(12) − 6E_y = 0, so E_y = 100 kN ↑.
• ΣFy: C_y = 30 − 100 = −70 kN (70 kN ↓); ΣFx: C_x = 0.

Method of joints

Joint A: ΣFy: −20 − 0.8F_AD = 0, so F_AD = −25.00 kN (C). ΣFx: F_AB + 0.6(−25) = 0, so F_AB = +15.00 kN (T).

Joint D (AD compression pushes D with 15 kN → and 20 kN ↓):

• ΣFy: −20 + 0.8F_DB = 0, so F_DB = +25.00 kN (T).
• ΣFx: 15 + 0.6(25) + F_DE = 0, so F_DE = −30.00 kN (C).

Joint B (AB = 15 T pulls left, DB = 25 T pulls down-left; BE goes down-right):

• ΣFy: −10 − 0.8(25) − 0.8F_BE = 0, so F_BE = −37.50 kN (C).
• ΣFx: −15 − 0.6(25) + 0.6(−37.50) + F_BC = 0, so F_BC = +52.50 kN (T).

Joint E: BE pushes E with (22.5 →, 30 ↓) and DE (compression) pushes with 30 kN →. ΣFy: 100 − 30 + 0.8F_EC = 0, so F_EC = −87.50 kN (C). Check ΣFx: 22.5 + 30 + 0.6(−87.5) = 0 ✓.

Result