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Civil
Peak sagging moment from a triangular span load
Largest bending moment in a simply supported span carrying a downward load that rises linearly from zero at the left support to wm at the right.
- Ref
- CIV-15
- Rev
- v1
- Field
- Civil
- Result
- Mmax [N·m]
For a triangular load the shear is zero at x = L/√3, and the moment there is wmL²/(9√3), about 0.0642 wmL². This is slightly above the midspan value of 0.0625 wmL² for the same shape and well below the 1/8 coefficient that applies to a uniform load of the same peak intensity. Use the true peak intensity wm in N/m.
Largest bending moment in a simply supported span carrying a downward load that rises linearly from zero at the left support to wm at the right. The shear is zero at x = L/√3 ≈ 0.577L, and that is where the moment peaks, a little beyond midspan toward the heavy end.
Calculator
- Position of the maximum, L/√33.464 m
- Moment at midspan, wmL²/1627,000 N·m
- Moment if the same wm acted uniformly, wmL²/854,000 N·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.
- The peak moment of 27.71 kN·m is 51.32 % of what a uniform load of the same peak intensity would give.
- It occurs 0.464 m beyond midspan, toward the heavily loaded end.
Chart data as tables
Ranked variables
| Variable | Typical range | Lowest result | Highest result | Swing (N·m) | Local step | Change in result |
|---|---|---|---|---|---|---|
| L Span | 1 to 30 m | 769.8 | 692,800 | 692,100 | +10% | +21 % |
| wm Peak load intensity | 100 to 100,000 N/m | 230.9 | 230,900 | 230,700 | +10% | +10 % |
Sweep of Span
| Span (m) | Peak sagging moment (N·m) |
|---|---|
| 1 | 769.8 |
| 4.625 | 16,470 |
| 8.25 | 52,390 |
| 11.88 | 108,600 |
| 15.5 | 184,900 |
| 19.13 | 281,600 |
| 22.75 | 398,400 |
| 26.38 | 535,500 |
| 30 | 692,800 |
Sweep one variable
Variables and how each one contributes
Peak sagging moment
What the equation gives you. Everything else feeds it.
Peak load intensity
The peak load intensity at the right support, in N/m. Scales the moment in direct proportion. Raising it by 10 % (to 13,200) moves the result by +10 %, from 27,710 to 30,480 N·m.
Span
The span, in m. Raises the moment with the square of the span. Raising it by 10 % (to 6.6) moves the result by +21 %, from 27,710 to 33,530 N·m.
Worked example with the values above
- Position of zero shear: x = L/√3 = 3.4641 m.
- Coefficient: 1/(9√3) = 0.06415.
- Mmax = 12,000 × 6² × 0.06415 = 27,712.8 N·m.
Assumptions and limits
- Statically determinate beam on two supports with ideal pin and roller conditions.
- Loads are static and vertical, and the beam weight is included in the load or neglected.
Common mistakes
- Dropping the sign of a reaction that is negative, which means uplift, and not providing a hold-down.
- Measuring the load position from the wrong support.
- Using a total load where the formula needs intensity per unit length, or the reverse.
Statistics and uncertainty
Reactions and moments follow from equilibrium alone, so they are exact for the idealised loading. The real uncertainty is in the loads: a variable load of ±20 % or a mis-measured overhang length passes straight through, and a 5 % error in a lever arm gives a 5 % error in the reaction.
Where it is used
- Finding support reactions before drawing shear and bending moment diagrams.
- Checking hold-downs, bearings and cantilever roots.
Maths you need
- Static equilibrium: ΣF = 0 and ΣM = 0
- Shear force: the vertical force on one side of a section
- Bending moment: the moment of forces on one side of a section