CivilSolve

Statics & Mechanics of Materials

Truss Analysis (Method of Joints)

Member forces (tension/compression/zero) and support reactions for a statically determinate plane truss, solved from joint equilibrium with a joint-by-joint walkthrough and zero-force member identification. Checks determinacy (m + r = 2j) and refuses unstable or indeterminate trusses. Use for: 'find the force in each member', 'identify zero-force members', pin-jointed truss reactions. Not for: space trusses, frames with multi-force members, or indeterminate trusses.

Direct solving is free and needs no account. Have a word problem instead? Submit it as text.

Inputs

Load a sample problem:
jointsTruss joints (nodes) with their positions
joints 1
membersStraight two-force members connecting pairs of joints
members 1
supportsSupports; a determinate plane truss needs m + r = 2j
supports 1
loadsJoint loads as signed x/y components (downward = negative fy)
loads 1

Worked examples

These are the solver's own reference problems — the answers come from published sources or independent hand computation, never from the solver itself. The solution below is the live solver output for each.

Example 1

SI equilateral triangle: 4 m side, 10 kN down at the apex

Given

joints
  1. 1.id Ax 0 my 0 m
  2. 2.id Bx 4 my 0 m
  3. 3.id Cx 2 my 3.4641016 m
members
  1. 1.from Ato B
  2. 2.from Ato C
  3. 3.from Bto C
supports
  1. 1.joint Atype pin
  2. 2.joint Btype roller
loads
  1. 1.joint Cfx 0 kNfy -10 kN

Assumptions

  • Ideal pin-jointed truss: members are straight two-force members, loads act only at joints, member self-weight neglected.
  • Member forces are reported tension-positive; negative values are compression. Load components fx/fy are signed: +x right, +y up (downward loads are negative fy).

Solution steps

  1. Determinacy check

    A plane truss is statically determinate when the member count plus reaction components exactly matches the two equilibrium equations available at each joint.

    members + reaction components = 2 × joints

    m = 3, r = 3, j = 3: m + r = 6 = 2j — determinate

  2. Support reactions from global equilibrium

    The reaction components balance the applied joint loads; with the truss determinate they follow from the same equilibrium system the joints satisfy.

    ΣFx = 0, ΣFy = 0, ΣM = 0 for the whole truss

    A: Rx = -4.55e-16 kN; A: Ry = 5 kN; B: R = 5 kN

  3. Equilibrium at joint A

    This joint has at most two unknown member forces left, so its two equilibrium equations (ΣFx = 0, ΣFy = 0) resolve them directly.

    ΣFx = 0 and ΣFy = 0 at this joint

    A–B = 2.89 kN (T); A–C = -5.77 kN (C)

  4. Equilibrium at joint B

    This joint has at most two unknown member forces left, so its two equilibrium equations (ΣFx = 0, ΣFy = 0) resolve them directly.

    ΣFx = 0 and ΣFy = 0 at this joint

    B–C = -5.77 kN (C)

Truss member forces (blue tension, orange compression)2.887 kN T5.774 kN C5.774 kN CABC
blue = tension · orange = compression · dashed = zero-force

Results

Member A–B force (tension +, tension)

2.89kN

Member A–C force (tension +, compression)

-5.77kN

Member B–C force (tension +, compression)

-5.77kN

Reaction at A, x-component

-4.55e-16kN

Reaction at A, y-component

5kN

Roller reaction at B (along its normal)

5kN

Where this answer was checked
source
Classic 3-member equilateral truss, solved by hand at each joint — pin at A(0,0), roller at B(4,0), apex C(2, 2√3); P = 10 kN down at C
verified by
hand-recomputed
derivation
Symmetry: RA = RB = 5 kN up, RAx = 0. Joint C: û(C→A) = (−1/2, −√3/2), û(C→B) = (1/2, −√3/2). ΣFx: (−NCA + NCB)/2 = 0 → NCA = NCB. ΣFy: −(√3/2)(NCA + NCB) − 10 = 0 → NCA = NCB = −10/√3 = −5.7735 kN (C). Joint A: ΣFy: NCA·(√3/2) + 5 = 0 ✓; ΣFx: NCA/2 + NAB = 0 → NAB = +10/(2√3) = +2.8868 kN (T).

Example 2

US 6-8-10 triangle: 20 kip down at the 8 ft apex

Given

joints
  1. 1.id Ax 0 fty 0 ft
  2. 2.id Bx 12 fty 0 ft
  3. 3.id Cx 6 fty 8 ft
members
  1. 1.from Ato B
  2. 2.from Ato C
  3. 3.from Bto C
supports
  1. 1.joint Atype pin
  2. 2.joint Btype roller
loads
  1. 1.joint Cfx 0 kipfy -20 kip

Assumptions

  • Ideal pin-jointed truss: members are straight two-force members, loads act only at joints, member self-weight neglected.
  • Member forces are reported tension-positive; negative values are compression. Load components fx/fy are signed: +x right, +y up (downward loads are negative fy).

Solution steps

  1. Determinacy check

    A plane truss is statically determinate when the member count plus reaction components exactly matches the two equilibrium equations available at each joint.

    members + reaction components = 2 × joints

    m = 3, r = 3, j = 3: m + r = 6 = 2j — determinate

  2. Support reactions from global equilibrium

    The reaction components balance the applied joint loads; with the truss determinate they follow from the same equilibrium system the joints satisfy.

    ΣFx = 0, ΣFy = 0, ΣM = 0 for the whole truss

    A: Rx = -1.64e-15 kip; A: Ry = 10 kip; B: R = 10 kip

  3. Equilibrium at joint A

    This joint has at most two unknown member forces left, so its two equilibrium equations (ΣFx = 0, ΣFy = 0) resolve them directly.

    ΣFx = 0 and ΣFy = 0 at this joint

    A–B = 7.5 kip (T); A–C = -12.5 kip (C)

  4. Equilibrium at joint B

    This joint has at most two unknown member forces left, so its two equilibrium equations (ΣFx = 0, ΣFy = 0) resolve them directly.

    ΣFx = 0 and ΣFy = 0 at this joint

    B–C = -12.5 kip (C)

Truss member forces (blue tension, orange compression)33.36 kN T55.6 kN C55.6 kN CABC
blue = tension · orange = compression · dashed = zero-force

Results

Member A–B force (tension +, tension)

7.5kip

Member A–C force (tension +, compression)

-12.5kip

Member B–C force (tension +, compression)

-12.5kip

Reaction at A, x-component

-1.64e-15kip

Reaction at A, y-component

10kip

Roller reaction at B (along its normal)

10kip

Where this answer was checked
source
Classic 3-4-5 proportioned triangular truss, solved by hand at each joint — pin at A(0,0), roller at B(12,0) ft, apex C(6,8) ft; P = 20 kip down at C
verified by
hand-recomputed
derivation
AC = BC = 10 ft (6-8-10). Symmetry: RA = RB = 10 kip. Joint C: û(C→A) = (−0.6, −0.8), û(C→B) = (0.6, −0.8). ΣFx → NCA = NCB; ΣFy: −0.8(NCA + NCB) = 20 → NCA = NCB = −12.5 kip (C). Joint A: ΣFx: 0.6·(−12.5) + NAB = 0 → NAB = +7.5 kip (T). Check ΣFy at A: 0.8·(−12.5) + 10 = 0 ✓.

Example 3

5-member truss with a zero-force vertical: 20 kN down at the apex

Given

joints
  1. 1.id Ax 0 my 0 m
  2. 2.id Bx 6 my 0 m
  3. 3.id Cx 3 my 4 m
  4. 4.id Dx 3 my 0 m
members
  1. 1.from Ato D
  2. 2.from Dto B
  3. 3.from Ato C
  4. 4.from Cto B
  5. 5.from Cto D
supports
  1. 1.joint Atype pin
  2. 2.joint Btype roller
loads
  1. 1.joint Cfx 0 kNfy -20 kN

Assumptions

  • Ideal pin-jointed truss: members are straight two-force members, loads act only at joints, member self-weight neglected.
  • Member forces are reported tension-positive; negative values are compression. Load components fx/fy are signed: +x right, +y up (downward loads are negative fy).

Solution steps

  1. Determinacy check

    A plane truss is statically determinate when the member count plus reaction components exactly matches the two equilibrium equations available at each joint.

    members + reaction components = 2 × joints

    m = 5, r = 3, j = 4: m + r = 8 = 2j — determinate

  2. Support reactions from global equilibrium

    The reaction components balance the applied joint loads; with the truss determinate they follow from the same equilibrium system the joints satisfy.

    ΣFx = 0, ΣFy = 0, ΣM = 0 for the whole truss

    A: Rx = -9.09e-16 kN; A: Ry = 10 kN; B: R = 10 kN

  3. Zero-force members by inspection

    Two textbook rules: at an unloaded, unsupported joint with exactly two non-collinear members, both are zero-force; with three members of which two are collinear, the third is zero-force. Spotting these first shortens the joint-by-joint work.

    zero-force members carry no load under this loading

    Zero-force: C–D

  4. Equilibrium at joint A

    This joint has at most two unknown member forces left, so its two equilibrium equations (ΣFx = 0, ΣFy = 0) resolve them directly.

    ΣFx = 0 and ΣFy = 0 at this joint

    A–D = 7.5 kN (T); A–C = -12.5 kN (C)

  5. Equilibrium at joint B

    This joint has at most two unknown member forces left, so its two equilibrium equations (ΣFx = 0, ΣFy = 0) resolve them directly.

    ΣFx = 0 and ΣFy = 0 at this joint

    D–B = 7.5 kN (T); C–B = -12.5 kN (C)

  6. Equilibrium at joint C

    This joint has at most two unknown member forces left, so its two equilibrium equations (ΣFx = 0, ΣFy = 0) resolve them directly.

    ΣFx = 0 and ΣFy = 0 at this joint

    C–D = 0 kN (zero)

Truss member forces (blue tension, orange compression)7.5 kN T7.5 kN T12.5 kN C12.5 kN C0ABCD
blue = tension · orange = compression · dashed = zero-force

Results

Member A–D force (tension +, tension)

7.5kN

Member D–B force (tension +, tension)

7.5kN

Member A–C force (tension +, compression)

-12.5kN

Member C–B force (tension +, compression)

-12.5kN

Member C–D force (tension +, zero-force)

0kN

Reaction at A, x-component

-9.09e-16kN

Reaction at A, y-component

10kN

Roller reaction at B (along its normal)

10kN

Where this answer was checked
source
Classic zero-force-member configuration, solved by hand at each joint — pin A(0,0), roller B(6,0), D(3,0) on the bottom chord, apex C(3,4); members AD, DB, AC, CB, CD; 20 kN down at C
verified by
hand-recomputed
derivation
Joint D carries no load and AD, DB are collinear → CD is zero-force (rule 2), and NAD = NDB. Reactions by symmetry: 10 kN each. AC = CB = 5 m (3-4-5). Joint A: û(A→C) = (0.6, 0.8). ΣFy: 0.8·NAC + 10 = 0 → NAC = −12.5 kN (C). ΣFx: 0.6·(−12.5) + NAD = 0 → NAD = +7.5 kN (T) = NDB. NCB = −12.5 by symmetry. (CD ≈ 0 is asserted separately in the test file — a 0-expected value has no relative tolerance.)