Find reactions and member forces in a determinate plane truss. Follow joint equilibrium and identify tension, compression, and zero-force members.
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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.id Ax 0 my 0 m
2.id Bx 4 my 0 m
3.id Cx 2 my 3.4641016 m
members
1.from Ato B
2.from Ato C
3.from Bto C
supports
1.joint Atype pin
2.joint Btype roller
loads
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
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
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
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)
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.
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
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
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
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)
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.
5-member truss with a zero-force vertical: 20 kN down at the apex
Given
joints
1.id Ax 0 my 0 m
2.id Bx 6 my 0 m
3.id Cx 3 my 4 m
4.id Dx 3 my 0 m
members
1.from Ato D
2.from Dto B
3.from Ato C
4.from Cto B
5.from Cto D
supports
1.joint Atype pin
2.joint Btype roller
loads
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
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
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
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
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)
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)
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.
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.)