Simply supported beam with two point loads

Two equal loads placed symmetrically on a supported beam create something useful: between them the bending moment is constant and the shear force is zero. That middle region is in pure bending, which is why this arrangement is the basis of the four-point bending test.

It also describes plenty of real structures — a beam picked up by two clamps, a shelf loaded by two brackets, a shaft carried on two bearings with two pulleys, or a truss chord panel.

The calculator below is set to a DN50 steel tube spanning 2400 mm, loaded with 400 N at 600 mm from each support.

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-two-loads#v1-ARsAABZFMzNxQgAAUkKUAQIAAAAWRAAAyEMAAAAAAQAAFkQAAMhDAAAAAA

How the beam is modelled

  • The left support is a pin, the right one a roller.
  • Two equal vertical loads of 400 N act at 600 mm from the ends, measured from the left and from the right respectively.
  • The beam is a hollow round tube: 60.3 mm outer diameter, 52.5 mm inner diameter.
  • The span is 2400 mm and the material is construction steel.

How to read the results

Each support carries exactly one load, so the reactions are simply 400 N each.

The bending moment diagram rises linearly from the supports and stays flat between the two loads, at a value of F * a. That flat plateau is the pure bending region, and it is also where the beam is most heavily stressed.

The maximum deflection is at midspan and is smaller than for a single load of the same total magnitude placed at the centre — spreading the load over two points is a cheap way to reduce bending.

Because the section is a tube, the beam is equally stiff in both lateral directions, so the same numbers apply if the loads are applied sideways.

Formulas

Formula What it gives
a = 600 mm Distance from each support to the nearest load.
M_max = F * a Bending moment, constant across the region between the two loads.
delta_mid = F * a * (3 * L^2 - 4 * a^2) / (24 * E * I) Deflection at midspan, the largest one.
R_left = R_right = F Support reactions; each support carries one of the loads.

FAQ

Why put two loads instead of one in the middle?

It reduces the maximum bending moment and the deflection for the same total load, and it creates a region of constant moment that is much easier to measure or to inspect than a single peak.

What exactly is pure bending?

A state where the bending moment is constant and the shear force is zero, so the section only rotates and does not slide. Material tests use it because shear would otherwise distort the measurement.

Is the deflection really smaller than with one central load?

Yes. For the same total load applied at one point in the middle, the midspan deflection would be higher; moving the material out towards the supports shortens the effective bending span.

Can I use unequal loads?

Yes, set a different value for each load in the calculator. The pure bending region then disappears, and the moment diagram becomes a trapezoid instead of a flat plateau.

More beam examples

Beam calculation examples