Library / Civil / CIV-17
Civil
Hogging moment at the root of a loaded overhang
Bending moment at support B for a uniform downward load w acting along an overhang of length a beyond the support.
- Ref
- CIV-17
- Rev
- v1
- Field
- Civil
- Result
- MB [N·m]
Cut the beam just to the free-end side of support B and take moments of the overhang load: the uniform load w over length a has resultant wa acting a/2 from the cut, giving a moment of wa²/2. It hogs the beam, so with sagging positive the value is negative. This moment is independent of the span between the supports and of what else loads the beam.
Bending moment at support B for a uniform downward load w acting along an overhang of length a beyond the support. A section cut at B isolates the overhang, which behaves as a cantilever, so the moment is wa²/2 and hogging (negative when sagging is positive). The free end carries no applied moment.
Calculator
- Shear at B from the overhang, w·a12,000 N
- Moment if the overhang were twice as long−36,000 N·m
- Equivalent point load at the tip giving the same moment, wa/26,000 N
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 hogging moment of 9 kN·m at the support is the largest negative moment in the overhang, and tension is on the top face there.
- Doubling the overhang length would quadruple it to 36 kN·m, so overhangs should be kept short or reinforced at the top.
Chart data as tables
Ranked variables
| Variable | Typical range | Lowest result | Highest result | Swing (N·m) | Local step | Change in result |
|---|---|---|---|---|---|---|
| a Overhang length | 0.1 to 10 m | −400,000 | −40 | 400,000 | +10% | +21 % |
| w Load intensity | 100 to 100,000 N/m | −112,500 | −112.5 | 112,400 | +10% | +10 % |
Sweep of Overhang length
| Overhang length (m) | Moment at support B (N·m) |
|---|---|
| 0.1 | −40 |
| 1.338 | −7,156 |
| 2.575 | −26,520 |
| 3.813 | −58,140 |
| 5.05 | −102,000 |
| 6.287 | −158,100 |
| 7.525 | −226,500 |
| 8.762 | −307,100 |
| 10 | −400,000 |
Sweep one variable
Variables and how each one contributes
Moment at support B
What the equation gives you. Everything else feeds it.
Load intensity
The uniform load intensity on the overhang, in N/m. Scales the moment in direct proportion. Raising it by 10 % (to 8,800) moves the result by +10 %, from −9,000 to −9,900 N·m.
Overhang length
The overhang length, in m. Raises the moment with the square of the length. Raising it by 10 % (to 1.65) moves the result by +21 %, from −9,000 to −10,890 N·m.
Worked example with the values above
- Resultant of the load: w·a = 8,000 × 1.5 = 12,000 N, acting 0.75 m from support B.
- Moment at B: 12,000 × 0.75 = 9,000 N·m, hogging.
- MB = −9,000 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