Library / Civil / CIV-12
Civil
Force per connector in a built-up beam
Shear carried by each connector when two parts of a built-up beam are joined at a pitch s.
- Ref
- CIV-12
- Rev
- v1
- Field
- Civil
- Result
- F [N]
The shear flow along an interface is q = VQ/I, the shear force per unit length that the connectors must carry. Multiplying by the pitch s gives the shear over one spacing, and dividing by the n connectors at each station gives the force each one resists. Q is the first moment about the neutral axis of the area that lies on one side of the interface.
Shear carried by each connector when two parts of a built-up beam are joined at a pitch s. The elastic shear flow VQ/I is multiplied by the pitch and shared equally among the n connectors at each pitch point. Q is the first moment of the area on one side of the interface about the neutral axis. Bearing and connector resistance are checked separately.
Calculator
- Shear flow, VQ/I100,000 N/m
- Shear per pitch, VQs/I10,000 N
- Pitch for a capacity of 10 kN per connector0.2 m
What drives the result
Each variable is moved across its typical range while the others stay at the values above. The longest bar is the key driver.
- Each connector carries 5 kN. Halving the pitch would halve that to 2.5 kN.
- Where the shear force is largest, near the supports, the pitch must be closest; it can be widened toward midspan where V falls.
Chart data as tables
Ranked variables
| Variable | Typical range | Lowest result | Highest result | Swing (N) | Local step | Change in result |
|---|---|---|---|---|---|---|
| V Section shear force | 100 to 500,000 N | 25 | 125,000 | 125,000 | +10% | +10 % |
| Q First moment of area | 1e-4 to 0.05 m³ | 250 | 125,000 | 124,800 | +10% | +10 % |
| s Connector pitch | 0.02 to 1 m | 1,000 | 50,000 | 49,000 | +10% | +10 % |
| I Section second moment of area | 5e-5 to 0.01 m⁴ | 200 | 40,000 | 39,800 | +10% | −9.09 % |
| n Connectors per pitch | 1 to 10 | 1,000 | 10,000 | 9,000 | +1 | −33.3 % |
Sweep of Connector pitch
| Connector pitch (m) | Force per connector (N) |
|---|---|
| 0.02 | 1,000 |
| 0.1425 | 7,125 |
| 0.265 | 13,250 |
| 0.3875 | 19,380 |
| 0.51 | 25,500 |
| 0.6325 | 31,630 |
| 0.755 | 37,750 |
| 0.8775 | 43,870 |
| 1 | 50,000 |
Sweep one variable
Variables and how each one contributes
Force per connector
What the equation gives you. Everything else feeds it.
Section shear force
The shear force in the section, in N (the sign is ignored). Raises the force per connector in direct proportion. Raising it by 10 % (to 22,000) moves the result by +10 %, from 5,000 to 5,500 N.
Connectors per pitch
The number of connectors at each pitch station, a whole number. More connectors share the shear, lowering the force on each in inverse proportion. Raising it by 10 % (to 2.2) moves the result by +0 %, from 5,000 to 5,000 N.
First moment of area
The first moment of the area on one side of the interface about the neutral axis, in m³. A larger attached area raises the shear flow in direct proportion. Raising it by 10 % (to 0.0022) moves the result by +10 %, from 5,000 to 5,500 N.
Section second moment of area
The second moment of area of the whole built-up section, in m⁴. A stiffer section lowers the shear flow in inverse proportion. Raising it by 10 % (to 4.4×10−4) moves the result by −9.09 %, from 5,000 to 4,545 N.
Connector pitch
The connector pitch along the beam, in m. A wider spacing means each connector resists the shear of a longer length, raising its force in proportion. Raising it by 10 % (to 0.11) moves the result by +10 %, from 5,000 to 5,500 N.
Worked example with the values above
- Shear flow: q = VQ/I = 20,000 × 0.002 / 4×10−4 = 100,000 N/m.
- Shear over one pitch: q·s = 100,000 × 0.1 = 10,000 N.
- F = 10,000 / 2 = 5,000 N.
Assumptions and limits
- Linear elastic material, plane sections staying plane, and perfect bond between components.
- Sharp-cornered ideal geometry with no fillets, holes or local buckling.
Common mistakes
- Mixing millimetres with metres, which changes I by a factor of 10¹² when it is raised to the fourth power.
- Using the inertia about the wrong axis (width and depth swapped).
- Forgetting that composite properties depend on the modulus of each part, not only its area.
Statistics and uncertainty
These are exact geometric results for the stated shapes, so the uncertainty is in the dimensions. Because the second moment of area scales with depth cubed, a 2 % error in depth gives about a 6 % error in stiffness. Rolled and fabricated sections vary by a few per cent from nominal, and composite sections carry extra uncertainty from the modulus ratio.
Where it is used
- Calculating stiffness, stresses and capacity of beams and built-up members.
- Spacing connectors in composite and built-up sections.
Maths you need
- Second moment of area: I = ∫y² dA about the centroidal axis
- Parallel axis theorem: I = I₀ + Ad² about a parallel axis
- Neutral axis: the line of zero bending stress in an elastic section