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Civil
Shear under a linearly increasing span load
Shear force at distance x from the left support of a simply supported span whose downward load rises linearly from zero at the left support to wm at the right.
- Ref
- CIV-14
- Rev
- v1
- Field
- Civil
- Result
- V [N]
A triangular load that rises from zero at the left support to wm at the right has a total of wmL/2 and a resultant at 2L/3, so the left reaction is wmL/6. The load to the left of a section at x is wmx²/(2L), and the shear is the left reaction minus that. It falls to zero at x = L/√3, where the moment is largest.
Shear force at distance x from the left support of a simply supported span whose downward load rises linearly from zero at the left support to wm at the right. The left reaction is wmL/6, and the load to the left of the section is wmx²/(2L). Positive shear is the left reaction less the loads to its left; evaluate just inside the supports.
Calculator
- Left reaction, wmL/612,000 N
- Right reaction, wmL/324,000 N
- Position of zero shear, L/√33.464 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.
- At x = 3 m the shear is 3 kN, still positive, so the section is before the point of maximum moment at 3.464 m.
- The right reaction is double the left (wmL/3 against wmL/6), so the shear just inside the right support is −24 kN.
Chart data as tables
Ranked variables
| Variable | Typical range | Lowest result | Highest result | Swing (N) | Local step | Change in result |
|---|---|---|---|---|---|---|
| L Span | 3 to 30 m | −12,000 | 58,200 | 70,200 | +10% | +67.3 % |
| x Distance from left support | 0 to 6 m | −24,000 | 12,000 | 36,000 | +10% | −63 % |
| wm Peak load intensity | 100 to 100,000 N/m | 25 | 25,000 | 24,980 | +10% | +10 % |
Sweep of Distance from left support
| Distance from left support (m) | Shear force (N) |
|---|---|
| 0 | 12,000 |
| 0.75 | 11,440 |
| 1.5 | 9,750 |
| 2.25 | 6,938 |
| 3 | 3,000 |
| 3.75 | −2,063 |
| 4.5 | −8,250 |
| 5.25 | −15,560 |
| 6 | −24,000 |
Sweep one variable
Variables and how each one contributes
Shear force
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 shear in direct proportion. Raising it by 10 % (to 13,200) moves the result by +10 %, from 3,000 to 3,300 N.
Span
The span, in m. Raises the shear linearly through the reaction wmL/6. Raising it by 10 % (to 6.6) moves the result by +67.3 %, from 3,000 to 5,018 N.
Distance from left support
The distance from the left support, in m. Shear falls with the square of x, passing through zero at 0.577L. Raising it by 10 % (to 3.3) moves the result by −63 %, from 3,000 to 1,110 N.
Worked example with the values above
- Left reaction: wmL/6 = 12,000 × 6 / 6 = 12,000 N.
- Load to the left of the section: wmx²/(2L) = 9,000 N.
- V = 12,000 − 9,000 = 3,000 N.
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