For the truss with 25, 50 and 10 kN downward loads, which equation gives the roller reaction E_y?

Moments about C: 25(12) + 50(6) - E_y(3) = 0, so E_y = 200 kN up

Sum of vertical forces only: E_y = 25 + 50 + 10 = 85 kN up

Moments about E: 25(12) + 50(6) = 10(3), so E_y = 100 kN

Moments about C with the 10 kN load included at 6 m, giving 125 kN

From the vertical equilibrium, C_y = 85 - 200 = -115 kN. What does this mean?

The hinge reaction is 115 kN acting downward (holding the truss down)

The hinge pushes up with 115 kN

The calculation must be repeated with the opposite sign convention for loads

At joint D, AD (compression) pushes D with 25 kN to the right and 25 kN down. How is F_DE found and what is it?

From ΣFx with DB's horizontal part: DE = 50 kN compression

From ΣFx with DB's horizontal part: DE = 50 kN tension

From ΣFy alone: DE = 25 kN tension

From ΣFx ignoring DB: DE = 25 kN compression

2070 Asar · 8 marks

Truss member forces with 25, 50, 10 kN loads

Short answer

Reactions: 200 kN ↑ at E, 115 kN ↓ at C, no horizontal. Using method of joints: AB 25 kN T, BC 125 kN T, AD 35.36 kN C, DB 35.36 kN T, BE 106.07 kN C, EC 176.78 kN C, DE 50 kN C.

Determine the member forces for given truss loaded as shown in figure below. [Figure: truss with joints A, B, C on top (spans 6 m and 6 m), D and E at bottom (3 m, 6 m, 3 m); loads 25 kN at A, 50 kN at B and 10 kN at C, downward; a hinge support at E (hinge) and roller at E side, depth 3 m.]

Data

Top joints A(0,3), B(6,3), C(12,3); bottom joints D(3,0), E(9,0). Members: AB, BC, AD, DB, BE, EC, DE (m = 7, j = 5, r = 3: determinate). Loads: 25 kN at A, 50 kN at B and 10 kN at C, all downward. Support: hinge (pin) at C and roller at E (vertical reaction). All sloping members are at 45° (3 m rise over 3 m run, length 4.243 m).

Reactions

ΣM_C = 0: 25(12) + 50(6) − E_y(3) = 0, so E_y = 200 kN ↑.
ΣFy: C_y = 85 − 200 = −115 kN (115 kN ↓); ΣFx: C_x = 0.

Method of joints

Joint A: ΣFy: −25 − 0.7071F_AD = 0, so F_AD = −35.36 kN (C). ΣFx: F_AB + 0.7071(−35.36) = 0, so F_AB = +25.00 kN (T).

Joint D: AD is in compression and pushes D with 0.7071 × 35.36 = 25 kN to the right and 25 kN downward.

ΣFy: −25 + 0.7071F_DB = 0, so F_DB = +35.36 kN (T).
ΣFx: 25 + 0.7071(35.36) + F_DE = 0, so F_DE = −50.00 kN (C).

Joint B (AB = 25 T pulls left, DB = 35.36 T pulls down-left):

ΣFy: −50 − 0.7071(35.36) − 0.7071F_BE = 0, so F_BE = −106.07 kN (C).
ΣFx: −25 − 25 + 0.7071(−106.07) + F_BC = 0, so F_BC = +125.00 kN (T).

Joint E: BE is in compression and pushes E with 75 kN to the right and 75 kN downward; DE (compression) pushes E with 50 kN to the right. ΣFy: 200 − 75 + 0.7071F_EC = 0, so F_EC = −176.78 kN (C). Check ΣFx: 50 + 75 − 0.7071(176.78) = 0 ✓.

Result

Member	Force (kN)	Nature
AB	25.00	tension
BC	125.00	tension
AD	35.36	compression
DB	35.36	tension
BE	106.07	compression
EC	176.78	compression
DE	50.00	compression

✓ Answer checked by Claude Opus 5.5 (high effort)

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Truss member forces with 25, 50, 10 kN loads

Data

Reactions

Method of joints

Result

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