Beam Span Calculator

Bending moment, required section modulus and deflection for steel, LVL and sawn timber beams — step by step
Find the maximum bending moment and required section modulus for a simply supported 20 ft beam carrying 500 lb/ft
Check the deflection of a W10x15 steel beam spanning 20 ft under 500 lb/ft
Size an LVL beam for a 14 ft span carrying 400 lb/ft
Convert a 55 psf floor load on a 12 ft tributary width to a line load in lb/ft

The Two Checks Every Beam Must Pass

A simply supported beam under a uniform load ww over a span LL has:

Mmax=wL28,Vmax=wL2,Δmax=5wL4384EIM_{\max} = \frac{wL^2}{8}, \qquad V_{\max} = \frac{wL}{2}, \qquad \Delta_{\max} = \frac{5wL^4}{384EI}

Strength is the first check. Bending stress is fb=M/Sf_b = M/S, so the section modulus you need is

Sreq=MmaxFbS_{req} = \frac{M_{\max}}{F_b}

with FbF_b the allowable bending stress for the material. For a rectangle, S=bd2/6S = bd^2/6 and I=bd3/12I = bd^3/12; for a rolled steel shape, SxS_x and IxI_x come from the shape tables.

Stiffness is the second, and on residential spans it usually governs. Deflection is limited by a fraction of the span — commonly L/360L/360 for live load and L/240L/240 for total load on floors — and because ΔL4\Delta \propto L^4, a 10% longer span deflects about 46% more.

Getting ww right: a floor load in psf becomes a line load by multiplying by the tributary width — half the joist span on each side. 5555 psf on a 12 ft tributary width is 660660 lb/ft.

Steel, LVL and Sawn Lumber

The equations above are identical for all three; the material properties and the design method are not.

MaterialTypical FbF_bTypical EE
A36 steel (ASD, 0.66Fy0.66F_y)24,000\approx 24{,}000 psi29×10629 \times 10^6 psi
A992 steel, Fy=50F_y = 50 ksi33,000\approx 33{,}000 psi29×10629 \times 10^6 psi
LVL / engineered lumber2,6002{,}6003,1003{,}100 psi1.91.92.0×1062.0 \times 10^6 psi
No. 2 Douglas fir-larch, sawn900\approx 900 psi1.6×1061.6 \times 10^6 psi
  • Steel is designed to AISC, and bending capacity depends on how the compression flange is braced — an unbraced beam can fail by lateral-torsional buckling well below M=FbSM = F_b S. Web crippling and bearing at supports need checking too.
  • LVL values are product-specific: the published FbF_b is adjusted for depth, and the binding numbers sit in the manufacturer's ICC-ES evaluation report, not a generic table.
  • Sawn lumber is designed to the NDS, where the tabulated FbF_b is multiplied by adjustment factors for load duration, wet service, size and repetitive members.

These figures are for learning the calculation, not for specifying a beam. A real member must be selected from the applicable code and span tables — AISC for steel, the NDS and IRC tables for wood, the manufacturer's data for LVL — with the engineer of record responsible for the final design.

Common Mistakes to Avoid

  • Mixing feet and inches. M=wL2/8M = wL^2/8 with ww in lb/ft and LL in ft gives lb·ft; multiply by 12 before dividing by an FbF_b in psi.
  • Checking strength and stopping. Deflection governs most residential spans. A section that passes bending can still bounce far past L/360L/360.
  • Using the wrong support condition. wL2/8wL^2/8 is the simply supported case. A cantilever of length LL carries wL2/2wL^2/2 — four times the moment — and deflects wL4/8EIwL^4/8EI.
  • Confusing SS and II. Section modulus SS sizes for stress; moment of inertia II sizes for deflection. They are not interchangeable.
  • Forgetting tributary width, point loads and self-weight. A beam under a bearing wall or a post above carries a concentrated load that a uniform-load formula ignores entirely.
  • Treating a rule of thumb as a specification. Depth-to-span ratios and "one size up" habits are not design; the governing code is.

示例题目

Step 1: Mmax=wL2/8=(500 lb/ft)(20 ft)2/8M_{\max} = wL^2/8 = (500\ \text{lb/ft})(20\ \text{ft})^2 / 8
Step 2: =(500)(400)/8=25,000 lb\cdotpft= (500)(400)/8 = 25{,}000\ \text{lb·ft}
Step 3: Convert to lb·in: 25,000×12=300,000 lb\cdotpin25{,}000 \times 12 = 300{,}000\ \text{lb·in}
Step 4: Sreq=M/Fb=300,000 lb\cdotpin÷24,000 lb/in2=12.5 in3S_{req} = M/F_b = 300{,}000\ \text{lb·in} \div 24{,}000\ \text{lb/in}^2 = 12.5\ \text{in}^3
Step 5: A W10×15 has Sx=13.8 in3>12.5 in3S_x = 13.8\ \text{in}^3 > 12.5\ \text{in}^3, so it satisfies bending — subject to the deflection, shear, bracing and bearing checks, and to confirming the properties against the AISC tables
Answer: Mmax=25,000M_{\max} = 25{,}000 lb·ft and Sreq=12.5S_{req} = 12.5 in³; a W10×15 (Sx=13.8S_x = 13.8 in³) passes the bending check

Step 1: Convert to consistent inch units: w=500 lb/ft÷12=41.67 lb/inw = 500\ \text{lb/ft} \div 12 = 41.67\ \text{lb/in} and L=240 inL = 240\ \text{in}
Step 2: L4=2404=3.318×109 in4L^4 = 240^4 = 3.318 \times 10^{9}\ \text{in}^4
Step 3: Numerator: 5wL4=5(41.67)(3.318×109)=6.912×10115wL^4 = 5(41.67)(3.318 \times 10^{9}) = 6.912 \times 10^{11}
Step 4: Denominator: 384EI=384(29×106)(68.9)=7.673×1011384EI = 384(29 \times 10^{6})(68.9) = 7.673 \times 10^{11}
Step 5: Δ=6.912×1011÷7.673×1011=0.90 in\Delta = 6.912 \times 10^{11} \div 7.673 \times 10^{11} = 0.90\ \text{in}
Step 6: Limits: L/240=240/240=1.00L/240 = 240/240 = 1.00 in and L/360=240/360=0.67L/360 = 240/360 = 0.67 in
Answer: Δ0.90\Delta \approx 0.90 in — inside L/240L/240 but over the L/360L/360 limit, so stiffness governs and a deeper section is needed if L/360L/360 applies

Step 1: Mmax=wL2/8=(400)(14)2/8=(400)(196)/8=9,800 lb\cdotpftM_{\max} = wL^2/8 = (400)(14)^2/8 = (400)(196)/8 = 9{,}800\ \text{lb·ft}
Step 2: Convert: 9,800×12=117,600 lb\cdotpin9{,}800 \times 12 = 117{,}600\ \text{lb·in}
Step 3: Sreq=117,600÷2,600=45.2 in3S_{req} = 117{,}600 \div 2{,}600 = 45.2\ \text{in}^3
Step 4: A single 1¾ in × 11⅞ in ply gives S=bd2/6=(1.75)(11.875)2/6=246.8/6=41.1 in3S = bd^2/6 = (1.75)(11.875)^2/6 = 246.8/6 = 41.1\ \text{in}^3 — short of 45.2
Step 5: Two plies give 2×41.1=82.3 in32 \times 41.1 = 82.3\ \text{in}^3; a single deeper 1¾ in × 14 in ply gives (1.75)(196)/6=57.2 in3(1.75)(196)/6 = 57.2\ \text{in}^3
Step 6: Either satisfies bending, but the depth adjustment factor and the deflection check must still be run against the manufacturer's published values
Answer: Sreq=45.2S_{req} = 45.2 in³ — a single 1¾×11⅞ ply (41.1 in³) is insufficient; two plies (82.3 in³) or one 1¾×14 ply (57.2 in³) satisfy bending, pending the deflection check and the manufacturer's data

常见问题

For a simply supported beam under a uniform load, M = wL²/8, with w the load per unit length and L the clear span. A cantilever of the same length and load carries wL²/2 instead — four times as much — so the support condition must be identified before any formula is applied.

It depends entirely on the load. For a uniform 500 lb/ft the moment is 25,000 lb·ft = 300,000 lb·in, so with an allowable bending stress of 24,000 psi the required section modulus is 12.5 in³ — a W10×15 exceeds that in bending, but the deflection check above shows it is over the L/360 limit. The final section must be selected against the AISC specification and signed off by the engineer of record.

There is no single answer: the span depends on the load, the number of plies, the depth, and the product's published Fb and E. Work it as a calculation — find M = wL²/8, divide by the allowable bending stress for the required section modulus, then check deflection — and confirm the result against the manufacturer's span tables and ICC-ES report, which is the binding document.

Deflection scales with the fourth power of the span while moment scales with the square, so as spans get longer the stiffness demand grows far faster than the strength demand. Serviceability limits such as L/360 also exist to stop bouncy floors and cracked finishes, which occur well below any risk of failure.

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