Beam Load Calculator — Shear Force & Bending Moment (Simply Supported)
Inputs
Point Loads (downward +)
Load P (kN) | Position x (m) | Actions |
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Uniformly Distributed Loads (UDL)
Intensity w (kN/m) | Start a (m) | End b (m) | Actions |
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Results
What This Calculator Does
We assume a simply supported beam (pin at the left end, roller at the right). You can add point loads and uniformly distributed loads (UDLs). Reactions are found by static equilibrium: ∑F = 0 and ∑MA = 0. The shear function \(V(x)\) is evaluated piecewise across all load breakpoints; the bending moment is obtained by integrating shear numerically along the span.
Shear Force & Bending Moment — Concepts, Examples, and Tips
This tool helps you sketch the shear force diagram (SFD) and bending moment diagram (BMD) for a simply supported prismatic beam (pin at the left, roller at the right). It supports point loads and uniformly distributed loads (UDLs), computes support reactions, and draws SFD/BMD live. All calculations run privately in your browser.
When to Use This
- Quick checks during early sizing or coursework.
- Understanding how loads change shear/moment shape before a detailed design package.
- Exploring “what if” scenarios (moving a point load, adjusting a UDL length, etc.).
Inputs & Units
- Beam length \(L\): metres (m)
- Point load \(P\): kilonewtons (kN), downward taken as positive here
- UDL \(w\): kilonewtons per metre (kN/m)
- Shear \(V\): kN | Moment \(M\): kN·m
How the Calculations Work
Reactions from equilibrium with origin at the left support:
\( R_B = \dfrac{\sum_i P_i x_i + \sum_j w_j(b_j-a_j)\,\dfrac{a_j+b_j}{2}}{L}, \quad R_A = \sum_i P_i + \sum_j w_j(b_j-a_j) - R_B. \)
Shear at position \(x\):
\( V(x) = R_A - \sum_{x_i \le x} P_i - \sum_j w_j\,\max\!\bigl(0,\,\min(x,b_j)-a_j\bigr). \)
Moment from \( dM/dx = V \): \( M(x) = \int_0^x V(s)\,ds \) with \(M(0)=0\).
Worked Example (Step-by-Step)
Beam: \(L=6\,\text{m}\), \(P_1=10\,\text{kN}\) at \(x=2\,\text{m}\), UDL \(w=3\,\text{kN/m}\) from \(3\) to \(6\,\text{m}\).
- Total load \(=10+3(6-3)=19\,\text{kN}\).
- Moment about A \(=10\cdot2 + 3\cdot3 \cdot \tfrac{3+6}{2} = 60.5\,\text{kN·m}\).
- \(R_B=60.5/6=10.083\,\text{kN}\), \(R_A=19-10.083=8.917\,\text{kN}\).
- Shear: jump at point load; linear drop under UDL with slope \(-w\).
Interpreting the Diagrams
- Point load → jump in SFD; slope change in BMD.
- UDL → linear SFD; parabolic BMD.
- Where \(V=0\) → local extremum of \(M\).
Common Mistakes (and Fixes)
- Keep all loads within \(0\le x\le L\).
- Downward loads positive; reactions positive upward.
- Ensure UDL \(a\le b\); the tool warns if flipped.
- Negative reactions → uplift: re-check inputs/supports.
Assumptions & Limits
- Statically determinate, simply supported beam; prismatic section.
- Linear statics only; no deflection/stress checks.
- No axial or torsion; vertical loads only.
Disclaimer: Educational tool. Not a substitute for professional structural analysis or code-compliant design checks.