Required Moment of Inertia for a Deflection Limit
The moment of inertia a simple-span beam needs to hold the uniform-load midspan deflection to a limit.
Example
You enter
- Uniform (service) load (kip/ft) 1
- Simple span (ft) 40
- Allowable deflection (in; e.g. span-in / 360) 1
- Modulus E (ksi) 29000
You get
- Required moment of inertia Ix 1986 in^4
- Deflection limit (span/delta) span/480
Details, formula, and sources
I = 5 w L^4 / (384 E delta_allow). A 1.0 kip/ft, 40 ft beam held to 1.0 in needs Ix about 1,986 in^4; pick a rolled shape with at least that Ix, then verify strength. Common limits are span/360 (LL) and span/240 (total). Sizes for stiffness only; the entered load is the service load. A design aid; the structural drawings govern.
I = 5 w L^4 / (384 E delta_allow), L = span x 12, the inverse of delta = 5 w L^4 / (384 E I); span_over_defl = L / delta_allow.
Simple-span uniform-load midspan deflection (AISC / mechanics of materials), solved for the moment of inertia, by name.
The simple-span deflection relation is a standard published structural result; the deflection limits (span/240, span/360) are in the building code.
Estimate. AHJ and licensed professional govern.
Field names used by the API: w_kip_ft, span_ft, allow_defl_in, e_ksi, required_moi_in4, span_over_defl
- Required inertia I = 5 w L^4 / (384 E delta_allow) simple span, uniform loadbeam theory
- Deflection limit delta_allow from the code (span/360 LL, span/240 total) at service loadbuilding code
- Stiffness only verify flexure and shear strength separately with the factored loadscope of this tile