Beam Load Calculator

Beam load formulas and step-by-step examples

Declare the model and bending axis first

The beam is prismatic with constant elastic modulus $E$, second moment of area $I$ and span $L$. Both load magnitudes are nonnegative and act downward: a concentrated force $P$ and full-span uniform intensity $w$. Linear elastic Euler–Bernoulli bending assumes small deflections and neglects shear deformation. The flexural rigidity is $EI$; it is not determined by area alone.

Simple supports are an ideal pin and roller at the span ends. A cantilever has a perfectly fixed left end and a free right end. The support choice changes the boundary conditions and coefficients substantially. Neither model includes finite support flexibility, a second span, a partial uniform load or an arbitrary point-load position. Self-weight is included only if it has already been added to the entered uniform intensity.

Use the matching peak-deflection formula

For the simple beam, the point force acts at midspan and the maximum downward deflection is also at midspan. For the cantilever, the point force is at the free end and the maximum is at that free end. Same-direction combinations can be added because the adopted model is linear and their deflection maxima coincide.

Set the inactive load component to zero for a pure load case. A distributed intensity contributes total force $W=wL$; substituting $W$ directly for $w$ would introduce the wrong dimensions. The calculator keeps force and force-per-length unit families separate and converts the fourth-power inertia factor before evaluating either expression.

Work the retained simply-supported point example

The original inputs are $L=3\,\mathrm{m}$, $P=5\,\mathrm{kN}$, $E=200\,\mathrm{GPa}$ and $I=8\times10^6\,\mathrm{mm}^4$, with no uniform component. First normalize the load, modulus and inertia. The fourth power matters: $1\,\mathrm{mm}^4=10^{-12}\,\mathrm{m}^4$.

The corresponding reactions are $R_A=R_B=2.5\,\mathrm{kN}$ and the maximum sagging moment is $3.75\,\mathrm{kN}\,\mathrm{m}$. With a separate full-span intensity of $w=2\,\mathrm{kN}/\mathrm{m}$, its own peak deflection is $1.318359375\,\mathrm{mm}$. Combining both downward loads gives $3.076171875\,\mathrm{mm}$, with each simple support carrying $5.5\,\mathrm{kN}$.

Separate internal moment from the support reaction couple

For the cantilever, $C_A$ is the external counterclockwise reaction-couple magnitude. Internal bending moment is sagging-positive, so the fixed-end internal moment under these downward loads is $M(0)=-C_A$. With both original load components active, the cantilever gives a free-end deflection of $40.78125\,\mathrm{mm}$ and a reaction couple of $24\,\mathrm{kN}\,\mathrm{m}$. The free end is not a second support.

Read signed stations with the declared convention

The coordinate $x$ runs from the left support or clamp. Bending moment $M$ is positive for sagging, internal shear satisfies $V=dM/dx$, and deflection $v$ is positive downward. Accordingly, the model’s curvature equation is $EI\,v''=-M$. The reported slope is $v'$, a dimensionless derivative; treating it as a rotation angle is a small-slope approximation.

At the simple-beam point-load location $a=L/2$, the pure point example has shear limits $V(a^-)=2.5\,\mathrm{kN}$ and $V(a^+)=-2.5\,\mathrm{kN}$. Moment and deflection remain continuous. Simple-beam endpoint shear values are internal span-side values and exclude support reaction jumps. At a cantilever’s free-end point load, the reported shear jump to zero is included.

Invert one load without inventing a combined-load split

A specified nonnegative maximum deflection and positive stiffness determine one pure load for the selected support model. A single deflection cannot determine two independent load magnitudes. Inverse mode therefore offers pure point or uniform loading and does not present an arbitrary combined answer as unique.

Using the original simple-beam stiffness with target $\delta=1.7578125\,\mathrm{mm}$ recovers $P=5\,\mathrm{kN}$. This is the load satisfying the adopted equation, not a rated capacity. A zero target yields a zero load and zero response; the maximum location is then not unique and $L/\delta$ is undefined.

Keep optional comparisons narrow

You choose the positive divisor $N$ or a positive absolute allowance. For example, $L=3\,\mathrm{m}$ with an adopted divisor of $240$ gives an allowance of $12.5\,\mathrm{mm}$. A calculated deflection equal to that allowance has $D=1$. The comparison uses exact unrounded values and says only below, equal to or above the adopted limit; it does not select a code requirement or certify a beam.

The optional extreme-fiber distance $c$ must match the same neutral axis as $I$. In the original point-load example, $c=50\,\mathrm{mm}$ gives $S=0.00016\,\mathrm{m}^3$ and a stress magnitude of $23.4375\,\mathrm{MPa}$. That elastic magnitude is not a material allowable and does not check shear stress, lateral stability or local section behavior.