Simply supported beam with an eccentric load

Loads rarely sit exactly in the middle. A bracket bolted to one side, a wheel passing along a beam, or a machine hanging off centre all produce an eccentric point load, and the first thing that changes is how the two supports share the load.

This page models that case: a solid rectangular steel bar, 10 mm thick and 60 mm deep, spanning 1800 mm between a pin and a roller, with a 400 N load applied 600 mm from the left support.

Loading at the third point is also the standard arrangement of a four-point bending test, where the middle section is left in pure bending and can be measured without shear interference.

Calculator

This calculator is already set up for the problem above. Change the length, the load, the profile or the material and it recalculates immediately; the address bar keeps a shareable link to your exact configuration.

Beam
l=1,000mm
fixation type
0F = 101000
Input data:
Beam:
Length: 1000mm
Left: Fixed end
Right: None
Material: Construction steel (Generic)
Density: 7850 kg/m3
E: 200,000 MPa
G: 77,000 MPa
Profile:
Ax: 78.54mm2 Jxx: 981.75mm4
Asy: 78.54mm2 Iyy 490.87mm4
Asz: 78.54mm2 Izz: 490.87mm4
Simulation results:
Displacements (mm):
# x y zz
Reactions (newtons):
# x y zz

Direct link to this setup: https://beamcalc.net/en/solve/simply-supported-third-point-load#v1-ASsAAOFEAAAgQQAAcEKUAQEAAAAWRAAAyEMAAAAA

How the beam is modelled

  • The left support is a pin, the right one a roller.
  • A vertical point load of 400 N acts 600 mm from the left support, which is one third of the span.
  • The beam is a solid rectangular bar: 10 mm wide and 60 mm deep, so it bends about its strong axis.
  • The span is 1800 mm and the material is construction steel.

How to read the results

The support closer to the load carries more of it: the left reaction is twice the right one here, because the load sits at one third of the span.

The bending moment peaks under the load rather than at midspan, and its value is F * a * b / L — with a shorter loaded side, the peak grows.

The deflection maximum no longer sits under the load exactly; it shifts slightly towards midspan. The formula below gives the deflection at the load itself, and the calculator reports the deflection at every node, so you can see the whole shape and read off the peak.

Because the bar is loaded in its strong direction, the section is deep rather than wide: laying it flat instead would make the beam 36 times more flexible.

Formulas

Formula What it gives
a = 600 mm, b = 1200 mm Distances from the load to the left and right supports, with a + b = L.
M_max = F * a * b / L Maximum bending moment, located under the load.
R_left = F * b / L, R_right = F * a / L Support reactions; the nearer support takes the larger share.
delta_load = F * a^2 * b^2 / (3 * E * I * L) Deflection at the load point. The maximum is a little larger and sits slightly towards midspan.

FAQ

Which support is more loaded?

The one closer to the load. The reactions split inversely to the distance from the load, so at the third point the near support carries twice the far one.

Why is the flat bar placed on edge?

Because the moment of inertia of a rectangle grows with the cube of the depth. Standing the 10 mm bar up to a 60 mm depth makes it 36 times stiffer in bending than laying it flat, since the width shrinks at the same time.

Where does the maximum bending moment occur?

Directly under the load, not at midspan. For an eccentric point load the moment diagram is a triangle that peaks at the loading point.

How is this related to a four-point bending test?

In a four-point test two symmetric loads create a region of constant bending moment between them, which lets you measure material properties without shear stresses. Moving the load to a third point is the same family of arrangement.

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