At a joint where two bars carry a load, the bar forces follow from the force triangle: S₁ = G·sin α₂/sin(α₁+α₂). Calculate the force in bar 1 from load and bar angles – or the permissible load from a given limiting bar force. The force in bar 2 follows by swapping the two angles.
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At a joint where two bars carry a load, the bar forces follow from the force triangle: S₁ = G·sin α₂/sin(α₁+α₂). Calculate the force in bar 1 from load and bar angles – or the permissible load from a given limiting bar force. The force in bar 2 follows by swapping the two angles.
A load G = 10 kN hangs at a joint. Bar 1 includes α₁ = 30° with the load direction, bar 2 on the other side α₂ = 45°. This gives S₁ = 10 kN·sin 45°/sin 75° = 10·0.7071/0.9659 = 7.32 kN. The force in bar 2 follows by swapping the angles: S₂ = 10 kN·sin 30°/sin 75° = 5.18 kN. Check via equilibrium: S₁·sin α₁ + S₂·sin α₂ = 7.32·0.5 + 5.18·0.7071 = 3.66 + 3.66 = 7.32 – the two bar forces form a closed force triangle with G.
Idealised two-bar system as a concurrent force system: both bars are pin-connected at their ends, therefore transmit forces only along their axes, and meet the load at one common point. Self-weight of the bars, flexibility of the connections, gusset plates and eccentricities are excluded. The calculator returns only the magnitude of the bar force; distinguishing tie from strut follows from the geometry and is a design decision. A strut additionally requires a buckling check, a tie a check of its section and connection capacity. As α₁ + α₂ → 180° both bar forces grow without bound; such shallow arrangements are practically unusable.
A load hanging from two bars does not split evenly but according to the bar angles – and both bar forces can be considerably larger than the load itself. This calculator determines the bar force from load and angles, and can be rearranged to find the permissible load from a limiting bar force.
A two-bar system is the simplest concurrent force system in statics: two pin-connected bars meet the load at one joint. Pin-connected bars can transmit forces only along their own axis – so the directions of both bar forces are fixed in advance and only their magnitudes are unknown. The three forces G, S₁ and S₂ must form a closed force triangle; the sine rule applied to that triangle gives S₁ = G·sin α₂/sin(α₁+α₂). The force in the second bar follows from the same formula with the angles swapped.
Like a hammock between two trees: hanging with a deep sag, the rope forces are moderate. Pulled nearly horizontal, the rope forces rise dramatically although the person has not got heavier. The same applies here – the flatter the bar triangle, the larger the bar forces at the same load.
S₁ = G · sin α₂ / sin(α₁ + α₂)
Bar 1: S₁ = G·sin α₂/sin(α₁+α₂)Bar 2: S₂ = G·sin α₁/sin(α₁+α₂)Symmetric (α₁ = α₂ = α): S₁ = S₂ = G/(2·cos α)Equilibrium check: S₁·sin α₁ + S₂·sin α₂ = GValidity condition: 0 < α₁ + α₂ < 180°| Symbol / input | Meaning |
|---|---|
| Bar force S₁ | The magnitude of the force in bar 1, i.e. the bar opposite angle α₂. Whether it is tension or compression follows from the geometry, not from the magnitude: in a typical bracket the upper bar leading away from the load is a tie in tension and the lower one a strut in compression. The strut additionally requires a buckling check. For the force in bar 2, swap the two angles and calculate again. |
| Load G | The external load acting at the joint, usually a weight (G = m·g). Only this single load may act at the joint; if several act, combine them into a resultant first. The self-weight of the bars themselves is not included. |
| Angle α₁ of bar 1 | The angle between the axis of bar 1 and the line of action of the load G, i.e. the angle to the vertical for a hanging load. For a wall bracket this is the angle between the bar and the vertical load direction – not the angle to the wall or to the horizontal. The two bars must lie on opposite sides of the load's line of action, and α₁ + α₂ < 180° must hold. |
| Angle α₂ of bar 2 | The angle between the axis of bar 2 and the same line of action of the load, measured on the opposite side. The smaller the sum α₁ + α₂ – i.e. the more acute the bar triangle – the larger both bar forces become, because the bars increasingly load each other without contributing to carrying the load. |
G is the load acting at the joint, usually a weight G = m·g; if several loads act, combine them into a resultant first. α₁ and α₂ are the angles of the two bar axes against the line of action of the load – for a hanging load, therefore, against the vertical. Crucially: not the angle to the wall or to the horizontal, but to the load direction. Both bars must lie on opposite sides of that line of action and α₁ + α₂ < 180° must hold.
First confirm on a free-body diagram that only two bars and one load actually meet at the joint. Then read the angles against the load direction and enter the load. The result is the magnitude of the force in bar 1, i.e. the bar opposite angle α₂. For the second bar, swap the angles and calculate again. Then decide from the geometry which bar takes tension and which compression: the bar leading upwards away from the load is the tie, the supporting lower bar the strut. The strut additionally needs a buckling check, the tie a section and connection check.
A load G = 10 kN hangs at a joint. Bar 1 includes α₁ = 30° with the load direction, bar 2 on the opposite side α₂ = 45°. This gives S₁ = 10 kN·sin 45°/sin 75° = 10·0.7071/0.9659 = 7.32 kN and – after swapping the angles – S₂ = 10 kN·sin 30°/sin 75° = 5.18 kN. The check via the force triangle: 7.32·sin 30° + 5.18·sin 45° = 3.66 + 3.66 = 7.32. An instructive second case: at α₁ = α₂ = 60° the result is S₁ = S₂ = 10 kN – each bar carries the full load although there are two of them. At α₁ = α₂ = 89° it would be over 280 kN per bar.
The sum α₁ + α₂ is the real lever: it sits in the denominator as sin(α₁+α₂). The closer that sum comes to 180° – the flatter the bar triangle – the larger both bar forces become, theoretically without bound. Conversely the forces are smallest when the sum is near 90°. The widespread expectation that two bars halve a load is wrong: already at α₁ = α₂ = 60° each bar carries the full load. What matters for an economical design is therefore not the number of bars but their angle.
The calculation runs in newtons and radians internally. Load and bar force may be entered in N, kN, lbf or kip, and the angles in degrees or radians. Since load and bar force appear in the same equation, the result is independent of the force unit system chosen.
Wall brackets and jibs with a tie rod and a compression strut; tripod and stand loads; guy wires for masts and antennas; lifting points with two unequally inclined legs; the method of joints on simple trusses; designing rope runs and deflection points; back-calculating the permissible load from a known limiting bar force.
Idealised two-bar system: both bars are pin-connected, transmit forces only along their axes, and meet the load at one point. Self-weight of the bars, flexibility and eccentricity of the connections, gusset plates, bending moments from fixed connections, and spatial arrangements with three or more bars are not covered. The calculator returns magnitudes only – distinguishing tension from compression must be taken from the geometry and implemented in the design. As α₁ + α₂ → 180° the bar forces grow without bound; such arrangements are practically unusable and the calculator reports them as inadmissible.
Common mistake: The most consequential error is assuming two bars halve the load – at shallow angles each bar can carry a multiple of it. The second classic is the wrong angle reference: angles are measured against the line of action of the load, not against the wall or the horizontal; mixing them up swaps sine for cosine and therefore the two bar forces. The compression member is also frequently checked only against compressive stress while the buckling check is forgotten, although slender struts almost always fail by buckling rather than yielding. And finally the formula holds only for pinned connections: fixed bars additionally transmit bending moments.
Wall brackets and jibs with a tie rod and a compression strut; tripod and stand loads; guy wires for masts and antennas; lifting points with two unequally inclined legs; the method of joints on simple trusses; designing rope runs and deflection points; back-calculating the permissible load from a known limiting bar force.
G is the load acting at the joint, usually a weight G = m·g; if several loads act, combine them into a resultant first. α₁ and α₂ are the angles of the two bar axes against the line of action of the load – for a hanging load, therefore, against the vertical. Crucially: not the angle to the wall or to the horizontal, but to the load direction. Both bars must lie on opposite sides of that line of action and α₁ + α₂ < 180° must hold.
Idealised two-bar system: both bars are pin-connected, transmit forces only along their axes, and meet the load at one point. Self-weight of the bars, flexibility and eccentricity of the connections, gusset plates, bending moments from fixed connections, and spatial arrangements with three or more bars are not covered. The calculator returns magnitudes only – distinguishing tension from compression must be taken from the geometry and implemented in the design. As α₁ + α₂ → 180° the bar forces grow without bound; such arrangements are practically unusable and the calculator reports them as inadmissible.