Triangular Load Maximum Calculator
Largest deflection of a simply supported span with a triangular load. The peak is not at midspan. Runs locally.
Trust summary CVP VERIFIED · production STALE · CVP protocol 1.0.0-proposed · Engineering assurance · one-sided triangle max |δ| = w0 L⁴ √(1−√(8/15)) (2+5√(8/15))/(450 E I); + O3 mpmath tabulated δ.
- Input interpretation
- Enter values to calculate.
- Result
- —
- Verified scope
- one-sided triangle max |δ| = w0 L⁴ √(1−√(8/15)) (2+5√(8/15))/(450 E I); + O3 mpmath tabulated δ.
- Assurance
- Engineering
- Declared partition coverage
- PASS · 5/5 declared partitions (main, alias, signed, awkward, invalid-domain) · Matrix
- Deferred
- Not midspan under the same triangle, and not a station curve.
- Numerical scope
- O2: delta vs a separate-module identity (≤2 ULP). Not midspan under the same triangle, and not a station curve. ≤2 ULP vs O3 applies only to the published tabulated triangular-load-maximum vectors.
- Known limitations
- Core CVP does not include live graph, viewport, or pointer interaction.
- Model
- Largest deflection of a simply supported span under a triangular load that is zero at the left support.
- Scope
- Calculation runs locally in the browser; values are not uploaded.
- Verification
- Engine tested · Source checked · v1.0.0 · CVP VERIFIED · production STALE · CVP protocol 1.0.0-proposed · Engineering assurance · one-sided triangle max |δ| = w0 L⁴ √(1−√(8/15)) (2+5√(8/15))/(450 E I); + O3 mpmath tabulated δ.· View Manifest · CVP overview · Specification
- Versions
- Calculation 1.0.0 · CVP protocol 1.0.0-proposed · Evidence 2026-09-27.o2-o3
- Verification revision
- 2026-09-27.o2-o3 · 1/1 property · digest b84f0c9fe61e
- Legacy regression
- 2/2 tests · Production surface contract 6/6
- Trust layers
- Verification VERIFIED · Production STALE · overall VERIFIED_STALE
- CVP status
- STALE · Capability production binding: STALE · Site report: STALE · Evidence changed after the last successful production attestation. Re-attestation required.
- Reference
- O1 model · O3 expected_values · O3 numerical_behavior · O2 expected_values · O2 numerical_behavior
- Interfaces
- PASS · UI (SSR) / REST / MCP
- Supplemental domain review
- Not performed
- Named expert review
- Not performed
- CVP suite
- 4/4 golden · 1/1 CVP boundary · 3/3 invalid · 1/1 property · 1/1 metamorphic · 8/8 O3 · 2/2 cross-interface · 6/6 CVP contract · Manifest
- Sources
- Hibbeler, Mechanics of Materials
- Gere and Goodno, Mechanics of Materials
- Evidence
- 1 legacy golden · 1 legacy boundary · legacy regression suite · 4/4 oracle-backed golden · 3/3 invalid · Artifact integrity PASS
- Semantic contract
- PASS
Full verification
Formulas
Core equations used by this calculator.
How to use
Enter the peak intensity, the span, the modulus, and the area moment
L, E, and I must be positive.
Read the largest deflection and its position
x is measured from the left support, where the intensity is zero.
Example calculations
Common configurations with formula and result.
Peak of 24 on a span of 2
w0 = 24, L = 2, E = 1, I = 1
Triangular Load Maximum calculator specification
Version 1.0.0 · Engine tested
- Engine tested 2/2 tests · Production surface contract 6/6
- Named expert review Not performed
- Calculation version 1.0.0
- Definition
- A simply supported span of length L carries a triangular load that is 0 at the left support and w0 at the right. The largest deflection is δ = w0 L⁴ √(1−√(8/15)) (2+5√(8/15)) / (450 E I), at x = L √(1−√(8/15)) from the left.
- What it calculates
- Largest deflection of a simply supported span under a triangular load that is zero at the left support.
- Inputs
- w0
- L
- E
- I
- Outputs
- delta
- x
- Formula
δ = w0 L⁴ √(1−√(8/15)) (2+5√(8/15)) / (450 E I).- Assumptions
- Calculation runs locally in the browser; values are not uploaded.
- Keep units consistent with the labels on each field.
- x/L is about 0.519. The midspan value 5 w0 L⁴/(768 E I) is a different page.
- Units
- w0, L, E, and I set the deflection unit.
- Boundary conditions
- missing w0, L, E, or I → MISSING_REQUIRED_INPUT
- L ≤ 0, E ≤ 0, or I ≤ 0 → VALUE_MUST_BE_POSITIVE
- Example
- w0=24 L=2 E=1 I=1 → δ just above 2.5
- Validation cases
2 published on this page · 2/2 tests · Production surface contract 6/6 · View evidence
- w0=24 L=2 E=1 I=1 → δ just above 2.5
- missing I → MISSING_REQUIRED_INPUT
- Sources
- Hibbeler, Mechanics of Materials — Deflection of beams — simply supported beam with a triangular loadSupports: The largest deflection is not at midspan. It sits near 0.519L from the zero-intensity end.
- Gere and Goodno, Mechanics of Materials — Deflection of beams — triangular loadSupports: Supports the closed form δ = w0 L⁴ √(1−√(8/15)) (2+5√(8/15)) / (450 E I).
- Hibbeler, Mechanics of Materials — Deflection of beams — simply supported beam with a triangular load
- Calculation version
- 1.0.0
Background
Interpretation and common distinctions.
Find the largest deflection of a simply supported span under a triangular load. Intensity is 0 at the left support and w0 at the right.
Supported and not supported
Supported: the maximum and its position x from the unloaded end.
Not supported: the midspan value alone, a cantilever, and a point load. /calc/mechanical is not open.
Agent / API notes
Capability id: mechanical.statics.triangular_load_max_deflection · tool id: triangular-load-maximum · pin 1.0.0.
{ "w0": 24, "L": 2, "E": 1, "I": 1 }
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Frequently asked questions
Key distinctions behind the calculation.
Where is x measured from?
From the left support, the end where the intensity is zero. x/L = √(1−√(8/15)).