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Civil

Reactions to a partial uniformly distributed load

Support reactions of a simply supported span (pin at x = 0, roller at x = L) carrying a downward load of intensity w from x = u to x = v.

Ref
CIV-13
Rev
v1
Field
Civil
Result
RA [N]

A partial uniform load acts as a single resultant W = w(v − u) at its centroid x̄ = (u + v)/2. Taking moments about the left support gives RB = Wx̄/L, and vertical equilibrium gives RA = W − RB. The nearer a load is to a support, the more of it that support carries.

Support reactions of a simply supported span (pin at x = 0, roller at x = L) carrying a downward load of intensity w from x = u to x = v. W = w(v − u) is the resultant and x̄ = (u + v)/2 its position. The page returns the left reaction RA; the right reaction RB is in the derived rows. Reactions are positive upwards.

Calculator

Solve for
N
N/m
10050,000
m
04.9
m
2.16
m
530
  • Right reaction RB10,500 N
  • Resultant load W18,000 N
  • Centroid of the load, x̄3.5 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.

Left reaction (N) current value 7,500 N
wLoad intensityKey driver
12562,500
uLoad start
10517,500
LSpan
5,40015,900
vLoad end
3958,000
  • The left support takes 41.67 % of the 18 kN total, and the right support the remaining 58.33 %.
  • The load centroid at 3.5 m is nearer the right support, so that support carries the larger share.
Chart data as tables

Ranked variables

VariableTypical rangeLowest resultHighest resultSwing (N)Local stepChange in result
w Load intensity100 to 50,000 N/m12562,50062,370+10%+10 %
u Load start0 to 4.9 m10517,50017,390+10%−10.4 %
L Span5 to 30 m5,40015,90010,500+10%+12.7 %
v Load end2.1 to 6 m3958,0007,605+10%+5 %

Sweep of Load start

Load start (m)Left reaction (N)
017,500
0.612514,010
1.22510,900
1.8388,163
2.455,801
3.0633,814
3.6752,203
4.288966.3
4.9105

Sweep one variable

Left reaction (N)Hover, touch, or use the arrow keys on the chart
0500010k15k 01234 Load start (m) now 7,500

Variables and how each one contributes

RA

Left reaction

Result · N

What the equation gives you. Everything else feeds it.

w

Load intensity

N/m · typical 100 to 50,000

The load intensity, in N/m, acting downward. Both reactions scale in direct proportion. Raising it by 10 % (to 6,600) moves the result by +10 %, from 7,500 to 8,250 N.

u

Load start

m · typical 0 to 4.9

The start of the loaded length, measured from the left support, in m. Moving the start toward the right reduces the total load and shifts the resultant right. Raising it by 10 % (to 2.2) moves the result by −10.4 %, from 7,500 to 6,720 N.

v

Load end

m · typical 2.1 to 6

The end of the loaded length, in m, with u < v ≤ L. Extending the end raises the total load and moves its centroid to the right. Raising it by 10 % (to 5.5) moves the result by +5 %, from 7,500 to 7,875 N.

L

Span

m · typical 5 to 30

The span between supports, in m. A longer span makes the load act closer to the left support in relative terms, so RA rises. Raising it by 10 % (to 6.6) moves the result by +12.7 %, from 7,500 to 8,455 N.

Worked example with the values above

  1. Resultant: W = w(v − u) = 6,000 × 3 = 18,000 N at x̄ = 3.5 m.
  2. Right reaction: RB = Wx̄/L = 10,500 N.
  3. RA = W − RB = 7,500 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

Related formulas

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