Continuous Steel Beam Calculator - Direct Stiffness Method Analysis
This professional Continuous Steel Beam Calculator uses the Direct Stiffness Method (matrix analysis) to deliver engineering-grade results for indeterminate multi-span beams (up to 8 spans) with cantilevers, support settlements, and complex loading.
Quickly compute support reactions, shear force diagrams (SFD), bending moment diagrams (BMD), and deflection curves. Includes full AISC 360 section checks for flexure, shear, and lateral-torsional buckling (LRFD/ASD), plus live load patterning.
Perfect for steel designers needing fast, reliable analysis of continuous beams without expensive FEA software. Free, instant, and mobile-friendly.
Continuous Steel Beam Calculator
Analyze multi-span indeterminate beams — reactions, shear, moment & deflection — using the Direct Stiffness Method. Free, instant, no signup.
● No Signup | ● AISC / Eurocode | ● Up to 8 Spans| Span # | Length (ft) | Left Support | Right Support | Settlement (in) |
|---|
Support settlement causes moment redistribution in continuous beams — enter 0 if no settlement.
Diagram updates automatically as you change inputs.
When checked, the calculator generates all skip-loaded cases and reports the governing envelope of moments, shears, and reactions.
For a prismatic beam element of length $L$, flexural rigidity $EI$, the element stiffness matrix (4×4, degrees of freedom: $v_1, \theta_1, v_2, \theta_2$) is:
$$[k_e] = \frac{EI}{L^3}\begin{bmatrix} 12 & 6L & -12 & 6L \\ 6L & 4L^2 & -6L & 2L^2 \\ -12 & -6L & 12 & -6L \\ 6L & 2L^2 & -6L & 4L^2 \end{bmatrix}$$This is assembled into the global stiffness matrix $[K]$ by superposition over all beam elements.
After applying boundary conditions (pinned: $v=0$; fixed: $v=0, \theta=0$; roller: $v=0$), the reduced system is:
$$[K_r]\{d\} = \{F_r\}$$where $\{d\}$ is the vector of unknown displacements/rotations, $\{F_r\}$ is the reduced load vector. Solved by Gaussian elimination.
For a full-span uniformly distributed load $w$ (force/length) on a span of length $L$:
$$R_{A,fixed} = R_{B,fixed} = \frac{wL}{2}$$ $$M_{A,fixed} = +\frac{wL^2}{12}, \quad M_{B,fixed} = -\frac{wL^2}{12}$$For a partial UDL from $a$ to $b$, fixed-end reactions are integrated accordingly.
For a point load $P$ at distance $a$ from the left end of a span $L$ (where $b = L - a$):
$$R_A = \frac{Pb^2(3a+b)}{L^3}, \quad R_B = \frac{Pa^2(a+3b)}{L^3}$$ $$M_A = +\frac{Pab^2}{L^2}, \quad M_B = -\frac{Pa^2 b}{L^2}$$The demand/capacity ratio (DCR) for flexure:
$$\text{DCR}_{\text{flex}} = \frac{M_u}{\phi_b M_n}, \quad \phi_b = 0.90 \text{ (LRFD)}$$For compact sections (LTB not governing), $M_n = M_p = F_y Z_x$
Plastic (no LTB) when $L_b \le L_p$:
$$L_p = 1.76 r_y \sqrt{\frac{E}{F_y}}$$Inelastic LTB when $L_p < L_b \le L_r$:
$$M_n = C_b \left[M_p - (M_p - 0.7F_y S_x)\frac{L_b - L_p}{L_r - L_p}\right] \le M_p$$Elastic LTB when $L_b > L_r$:
$$M_n = F_{cr} S_x \le M_p, \quad F_{cr} = \frac{C_b \pi^2 E}{(L_b/r_{ts})^2}\sqrt{1 + 0.078 \frac{Jc}{S_x h_o}\left(\frac{L_b}{r_{ts}}\right)^2}$$where $A_w = d \times t_w$ and $C_{v1} = 1.0$ for $h/t_w \le 2.24\sqrt{E/F_y}$.
$$\text{DCR}_{\text{shear}} = \frac{V_u}{\phi_v V_n}$$Deflection is computed by double integration of the moment diagram:
$$EI \frac{d^2 y}{dx^2} = M(x)$$ $$EI \frac{dy}{dx} = \int M(x)\, dx + C_1$$ $$EI\, y = \int\!\!\int M(x)\, dx + C_1 x + C_2$$Constants $C_1, C_2$ are determined by boundary conditions. For a simply supported reference span:
$$\delta_{max} = \frac{5wL^4}{384EI} \quad \text{(UDL, simply supported)}$$The serviceability check requires: $\delta_{max} \le \dfrac{L}{360}$ (or selected limit)
Support settlement $\Delta_s$ at an interior support introduces additional fixed-end moments. For a propped cantilever or interior support settling by $\Delta_s$:
$$M_{settlement} = \frac{6EI \Delta_s}{L^2}$$These are incorporated directly into the fixed-end force vector $\{F_0\}$ before solving $[K]\{d\} = \{F\}$.
| Application | Deflection Limit | Code Reference |
|---|---|---|
| Office/residential floor — live load | L / 360 | AISC 360, IBC |
| Floor — total load | L / 240 | AISC 360 |
| Roof — live load | L / 180 | AISC 360 |
| Sensitive equipment | L / 480 to L / 600 | Project specific |
| Steel (A36, A992, A572) | E = 29,000 ksi (200,000 MPa) | AISC Steel Manual |
| Aluminum 6061-T6 | E = 10,000 ksi (69,000 MPa) | ADM |
| Concrete (fc'=4000 psi) | E ≅ 3,605 ksi (24,855 MPa) | ACI 318 Eq. 19.2.2.1 |
| Douglas Fir-Larch #2 | E = 1,600 ksi (11,030 MPa) | NDS Supplement |
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Continuous Steel Beam Calculator: Complete User Guide
Step-by-step instructions, all calculation formulas, worked examples, and code compliance checks for multi-span indeterminate steel beam analysis using the Direct Stiffness Method.
Accuracy Statement: This calculator uses the Direct Stiffness Method (matrix stiffness analysis) — the same analytical approach used in commercial software like SAP2000 and ETABS for elastic beam elements. Results are analytically exact for prismatic beam elements under static loading. Shear deformation is not included (negligible for standard wide-flange beams with span-to-depth ratios above 10). All results are suitable for preliminary design and professional verification. Always confirm final designs with a licensed structural engineer and the applicable building code.
What Is a Continuous Steel Beam Calculator?
A Continuous Steel Beam Calculator is a structural analysis tool that determines the support reactions, shear forces, bending moments, and deflections in a steel beam that spans continuously over three or more supports without internal hinges. Unlike a simply supported beam, a continuous beam is statically indeterminate, meaning standard statics equations alone cannot solve it — you need to account for the elastic stiffness of each span.
This tool solves the indeterminate system using the Direct Stiffness Method, assembles a global stiffness matrix from individual beam elements, applies boundary conditions for each support type, and solves for all unknown displacements and rotations simultaneously. It then back-calculates every internal force, diagram, and deflection at 200 evenly spaced points across the entire beam.
Where Engineers and Designers Apply Continuous Beam Analysis
- Multi-story steel framing: Floor beams that run continuously over steel columns or girders to reduce deflections and increase load capacity.
- Bridge girders and highway overpasses: Continuous girders over multiple piers reduce mid-span moments by as much as 30% versus simple spans.
- Industrial platform beams: Equipment support frames where vibration control and deflection limits are tight (L/480 or stricter).
- Parking structure framing: Long-span PT or steel beams that must carry heavy vehicle loads across multiple bays.
- Roof and mezzanine framing: Continuous purlins or secondary framing over interior support walls.
- Retrofit and renovation checks: Verifying that an existing continuous beam can carry new or increased loads.
Key User Pain Points — and How This Calculator Solves Them
Manual math is too slow
Solving a 3-span beam by hand using the Three-Moment Equation or Moment Distribution takes 30–60 minutes per load case — more for unequal spans.
No free code-compliance checks
Most free tools stop at reactions and moments. Engineers still need to verify flexural capacity, shear, LTB, and deflection limits against AISC 360 or Eurocode 3 separately.
Can’t visualize behavior
Without shear force, bending moment, and deflection diagrams, engineers can’t quickly identify critical locations or explain results to non-engineers.
Pattern loading is always skipped
Checking alternating-span (skip) loading per ASCE 7 / ACI 318 manually means building 2n load cases. Almost everyone skips it — and misses the true critical moment.
Expensive software for routine checks
SAP2000 and ETABS cost hundreds to thousands of dollars per year — too much for a quick preliminary check or a student project.
Settlement and stiffness effects ignored
Differential support settlement redistributes moments across a continuous beam in ways that a simple-span analysis completely misses. Most free calculators ignore it.
Understanding the Continuous Beam Diagram
The diagram below shows the key components of a multi-span continuous steel beam as modeled in the calculator. Study this before entering your inputs.
Figure 1: 3-span continuous beam with UDL on spans 1–2, point load on span 3, pin-roller-roller-pin supports. Bending moment diagram shows sagging (+) in spans and hogging (−) over interior supports.
Step-by-Step Guide to Using the Calculator
Step 1 — Choose Your Unit System
The unit toggle at the top of the calculator switches between two fully consistent unit systems. All inputs, outputs, and formulas automatically convert:
| Quantity | Imperial (US) | SI (Metric) | Common Mistake |
|---|---|---|---|
| Span length | ft | m | Entering mm instead of m in SI mode — results in wildly large EI |
| Point load | kips (1 kip = 1000 lb) | kN | Entering lb instead of kips in Imperial mode |
| Distributed load | kip/ft | kN/m | Entering lb/ft — divide by 1000 to convert to kip/ft |
| Modulus E | ksi (29,000 for steel) | MPa (200,000 for steel) | Entering GPa (200) instead of MPa (200,000) in SI mode |
| Moment of inertia I | in⁴ | cm⁴ | Entering mm⁴ — too large by a factor of 10,000 |
| Deflection output | in | mm | Checking L/360 against deflection in wrong units |
Step 2 — Define Beam Geometry (Geometry Tab)
The Geometry tab captures all dimensional and structural parameters for the beam model.
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Select number of spans (2–8) Each span is the distance between two adjacent supports. Interior supports (rollers) are added automatically. A 3-span beam has 4 supports: 2 end supports + 2 interior supports.
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Enter individual span lengths Spans do not need to be equal. Enter each span length separately in the Span Lengths & Support Conditions table. Unequal spans are fully supported — this is where the matrix method is required over simplified hand calculations. Minimum valid span: 0.1 ft (or 0.03 m).
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Set support types per span Assign each support as Pin, Roller, Fixed, or Free. The left column shows the left support of each span; the right column shows the right support. For most floor and roof beams: set the two end supports as Pin and all interior supports as Roller. Use Fixed only for built-in (fully restrained) end conditions — not for standard column connections.
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Enter support settlement values (optional) If any support is expected to settle (e.g., on compressible soil or a flexible support beam), enter the settlement in inches or mm. Even small settlements (0.25″) can significantly redistribute moments in stiff systems. Leave as 0 if no settlement is expected or if supports are on concrete or steel framing.
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Enter cantilever overhangs (optional) If the beam extends beyond the end supports (a cantilever), enter the overhang length in the Left/Right Cantilever fields. The beam diagram will show the overhang as a dashed extension. Loads on the overhang are entered in the Loads tab under the appropriate span number.
Support Type Reference
Step 3 — Enter Material & Section Properties
These values define the elastic stiffness \(EI\) of the beam, which controls how moments distribute between spans and how much the beam deflects.
| AISC Section | Iₓ (in⁴) | Sₓ (in³) | Zₓ (in³) | d (in) | tᴥ (in) | A (in²) |
|---|---|---|---|---|---|---|
| W12×26 | 204 | 33.4 | 37.2 | 12.22 | 0.230 | 7.65 |
| W14×30 | 291 | 42.0 | 47.3 | 13.84 | 0.230 | 8.85 |
| W14×48 | 485 | 70.2 | 78.4 | 13.79 | 0.340 | 14.1 |
| W16×40 | 518 | 64.7 | 73.0 | 16.01 | 0.305 | 11.8 |
| W18×35 | 510 | 57.6 | 66.5 | 17.70 | 0.300 | 10.3 |
| W18×46 | 712 | 78.8 | 90.7 | 18.06 | 0.360 | 13.5 |
| W21×44 | 843 | 81.6 | 95.4 | 20.66 | 0.350 | 13.0 |
| W24×55 | 1350 | 114 | 130 | 23.57 | 0.395 | 16.2 |
Step 4 — Enter Loads (Loads Tab)
The calculator accepts four load types. All loads are span-referenced — you assign each load to a specific span by selecting the span number.
| Load Type | Inputs Required | Units (Imperial) | Units (SI) | Typical Use |
|---|---|---|---|---|
| UDL (Uniformly Distributed) | Intensity w, start position a, end position b (0 = full span) | kip/ft | kN/m | Floor self-weight, snow load, uniform live load |
| Point Load | Magnitude P, position from left end of span | kips | kN | Column reaction, equipment load, concentrated post load |
| Trapezoidal Load | Intensity at left w₁, intensity at right w₂ | kip/ft | kN/m | Soil pressure, hydrostatic load, tributary area triangular load |
| Applied Moment | Magnitude M, position from left end | kip-ft | kN-m | Eccentric connection moment, cantilever tip reaction |
Load Type Tagging (Dead / Live / Snow / Wind)
Tag each load with its load type to enable AISC 360 LRFD automatic load combinations. Dead loads are permanent (self-weight, superimposed dead); Live loads are occupancy-dependent and pattern-loadable. For pattern loading analysis, only Live (L) loads are alternated across spans per ASCE 7.
Step 5 — Run the Analysis
Click “Analyze Beam” on any tab to run the full Direct Stiffness Method solution. The calculator:
- Assembles the global stiffness matrix from all beam elements
- Calculates fixed-end forces for every load type
- Applies boundary conditions (constrains DOFs at each support)
- Solves the reduced system by Gaussian elimination with partial pivoting
- Back-calculates reactions, internal forces, and deflections at 200 points
- Performs equilibrium verification (ΣR = ΣP, tolerance <1%)
- Auto-populates Mu and Vu in the Section Check tab
Step 6 — Read and Interpret Results (Results Tab)
Results are presented in six sections:
- Equilibrium check: Green = solver converged and reactions balance total applied load. Red = check your inputs (usually a missing support or conflicting support types).
- KPI summary: Six key numbers at a glance: Max+M, Max−M, Max|V|, Max deflection, total load, and equilibrium status.
- Support reactions table: Vertical reaction at each support, with upward/downward flag. Negative reactions mean the support would need to resist uplift — check your anchorage.
- Shear force diagram (SFD): Shows positive/negative shear across the full beam length. Shear changes sign at maximum moment locations.
- Bending moment diagram (BMD): Shows sagging (+) moments in span interiors and hogging (−) moments over interior supports. Both peaks are labeled.
- Deflection diagram: Shows the deflected shape. Maximum value is labeled with the span length to deflection ratio (e.g., “L/420”).
All Calculation Formulas — Complete Reference
The following formulas are used directly in the calculator. Click any heading to expand the full derivation and notes.
For a prismatic Euler-Bernoulli beam element of length \(L\) and flexural rigidity \(EI\), the 4×4 element stiffness matrix relates the four end forces (two shear forces and two moments) to the four end displacements (two vertical displacements \(v\) and two rotations \(\theta\)):
\[ [k_e] = \frac{EI}{L^3} \begin{bmatrix} 12 & 6L & -12 & 6L \\ 6L & 4L^2 & -6L & 2L^2 \\ -12 & -6L & 12 & -6L \\ 6L & 2L^2 & -6L & 4L^2 \end{bmatrix} \]where the DOF vector is \(\{d_e\} = \{v_1,\, \theta_1,\, v_2,\, \theta_2\}^T\) and the force vector is \(\{f_e\} = \{V_1,\, M_1,\, V_2,\, M_2\}^T\).
This matrix is assembled into the global stiffness matrix \([K]\) by direct superposition (the Direct Stiffness Method). For \(n\) spans with \(n+1\) nodes, the global matrix is \(2(n+1) \times 2(n+1)\) in size.
The global system before applying boundary conditions is:
\[ [K]\{d\} = \{F\} \]where \(\{F\} = \{F_{ext}\} + \{F_{0}\}\) is the sum of external nodal forces and fixed-end forces from distributed loads.
Boundary conditions are applied by removing the rows and columns corresponding to constrained DOFs:
- Pin or Roller: Vertical displacement \(v = 0\) (constrain the vertical DOF). Rotation \(\theta\) is free.
- Fixed support: Both \(v = 0\) and \(\theta = 0\) (constrain both DOFs).
After applying boundary conditions, the reduced system is:
\[ [K_r]\{d_r\} = \{F_r\} \]This is solved for the free DOFs \(\{d_r\}\) (the unknown rotations at all nodes) by Gaussian elimination with partial pivoting. The calculator uses 200-point integration for smooth diagrams.
For a full-span uniformly distributed load of intensity \(w\) (force per unit length) on a span of length \(L\), the fixed-end reactions (treating both ends as fixed) are:
\[ R_{A} = R_{B} = \frac{wL}{2} \quad \text{(shear)} \] \[ M_{A} = +\frac{wL^2}{12}, \quad M_{B} = -\frac{wL^2}{12} \quad \text{(moment)} \]These fixed-end forces are assembled into the global load vector \(\{F_0\}\). After solving, the solver recovers the actual moments at each node accounting for continuity (not the fixed-end values).
Partial Span UDL (start at \(a\), end at \(b\))
For a partial UDL, the calculator converts it to an equivalent point load at the centroid of the loaded region:
\[ P_{eq} = w(b-a), \quad \text{applied at } x_{centroid} = a + \frac{b-a}{2} \]Then the point load fixed-end force formulas are applied (see Formula 4 below).
For a concentrated load \(P\) at distance \(a\) from the left end of a span of length \(L\) (with \(b = L - a\)):
\[ R_A = \frac{Pb^2(3a+b)}{L^3}, \quad R_B = \frac{Pa^2(a+3b)}{L^3} \] \[ M_A = +\frac{Pab^2}{L^2}, \quad M_B = -\frac{Pa^2 b}{L^2} \]Check: \(R_A + R_B = P\) (vertical equilibrium). \(M_A - M_B = P \cdot a \cdot b / L\) (moment equilibrium about A).
Special case: midspan point load (\(a = b = L/2\)):
\[ R_A = R_B = \frac{P}{2}, \quad M_A = -M_B = \frac{PL}{8} \]For a trapezoidal load varying linearly from \(w_1\) (at the left end) to \(w_2\) (at the right end) over the full span \(L\):
\[ R_A = \frac{L(7w_1 + 3w_2)}{20}, \quad R_B = \frac{L(3w_1 + 7w_2)}{20} \] \[ M_A = +\frac{L^2(3w_1 + 2w_2)}{60}, \quad M_B = -\frac{L^2(2w_1 + 3w_2)}{60} \]Special case: triangular load (zero at one end). If \(w_2 = 0\) (zero at right end):
\[ R_A = \frac{7w_1 L}{20} = 0.35w_1 L, \quad R_B = \frac{3w_1 L}{20} = 0.15w_1 L \]Note the asymmetry: the reaction is larger at the “heavy” end as expected.
After solving \([K_r]\{d_r\} = \{F_r\}\) for all free DOFs, the full displacement vector \(\{d\}\) is reconstructed (setting all constrained DOFs to zero). The reaction vector is then:
\[ \{R\} = [K]\{d\} - \{F\} \]The support reactions are the vertical components of \(\{R\}\) at the constrained vertical DOFs (every node position where \(v = 0\)).
Equilibrium Verification
\[ \sum_{i=1}^{n+1} R_i = \sum_{\text{loads}} P_j \quad \text{(must balance)} \]The calculator checks this with a tolerance of 1% of the total load. If equilibrium fails, the model has an error (check support types and load inputs).
Once end moments \(M_A, M_B\) and end shears are known for each element from the solved displacements, the internal forces at any position \(x\) within a span are computed by equilibrium:
\[ M(x) = M_A + V_{start} \cdot x - \int_0^x q(s)\,(x-s)\,ds \] \[ V(x) = V_{start} - \int_0^x q(s)\,ds \]where \(V_{start}\) is the shear at the left end of the span, and \(q(s)\) is the distributed load intensity at position \(s\).
For a UDL of intensity \(w\) from 0 to \(L\):
\[ M(x) = M_A + V_{start} \cdot x - \frac{w x^2}{2} \] \[ V(x) = V_{start} - wx \]The element end moments are recovered from the solved nodal displacements using the element stiffness relationship:
\[ M_A = EI\left(\frac{6v_0}{L^2} + \frac{4\theta_0}{L} - \frac{6v_1}{L^2} + \frac{2\theta_1}{L}\right) \] \[ M_B = EI\left(\frac{-6v_0}{L^2} - \frac{2\theta_0}{L} + \frac{6v_1}{L^2} - \frac{4\theta_1}{L}\right) \]Deflection at any position \(x = \xi L\) within a span is interpolated from the four nodal DOFs using cubic Hermite shape functions:
\[ y(\xi) = N_1(\xi)\,v_0 + N_2(\xi)\,\theta_0 + N_3(\xi)\,v_1 + N_4(\xi)\,\theta_1 \]where:
\[ N_1 = 1 - 3\xi^2 + 2\xi^3 \] \[ N_2 = L\,\xi(1-\xi)^2 \] \[ N_3 = 3\xi^2 - 2\xi^3 \] \[ N_4 = L\,\xi^2(\xi - 1) \]and \(\xi = x/L\) is the normalized position along the span (\(0 \le \xi \le 1\)).
This is the exact elastic deflection curve (the Bernoulli beam assumption gives a quartic polynomial for UDL which is captured exactly by the cubic shape functions through the nodal rotations).
Reference formula for simply supported span under UDL
\[ \delta_{max} = \frac{5wL^4}{384EI} \quad \text{(occurs at midspan)} \]For a continuous beam, actual midspan deflections are significantly smaller because the end rotations \(\theta_0, \theta_1\) are restrained by adjacent spans, reducing the effective span.
When support \(i\) settles by an amount \(\Delta_s\) (downward), this is equivalent to applying a prescribed displacement. For the adjacent spans of length \(L\), this generates additional fixed-end forces:
\[ V_{settlement} = \frac{12EI\,\Delta_s}{L^3} \quad \text{(equivalent shear)} \] \[ M_{settlement} = \frac{6EI\,\Delta_s}{L^2} \quad \text{(equivalent moment)} \]These are added to the fixed-end force vector before solving. The resulting moment redistribution can be significant: for a stiff beam (large \(EI\)) over a short span \(L\), even small settlements produce large moments.
The maximum elastic bending stress at any section is:
\[ f_b = \frac{M}{S_x} \]where \(S_x = I_x / c\) and \(c = d/2\) for a symmetric section.
For LRFD design, the demand-to-capacity ratio (DCR) for flexure is:
\[ \text{DCR}_{flex} = \frac{M_u}{\phi_b M_n} \le 1.0 \]where \(\phi_b = 0.90\) (AISC LRFD) and \(M_n\) is the nominal flexural strength.
For compact sections not subject to LTB:
\[ M_n = M_p = F_y Z_x \]The plastic moment \(M_p\) represents the fully yielded cross-section state where the entire cross-section carries \(F_y\) in either tension or compression.
For ASD design: \(M / \Omega_b \le M_n\), where \(\Omega_b = 1/0.90 \approx 1.67\).
LTB reduces the available flexural strength when the compression flange is not adequately braced. The limit lengths are:
\[ L_p = 1.76\, r_y \sqrt{\frac{E}{F_y}} \quad \text{(plastic limit)} \] \[ L_r = 1.95\, r_{ts} \frac{E}{0.7F_y} \sqrt{\frac{Jc}{S_x h_0} + \sqrt{\left(\frac{Jc}{S_x h_0}\right)^2 + 6.76\left(\frac{0.7F_y}{E}\right)^2}} \]Case 1 — No LTB: If \(L_b \le L_p\), then \(M_n = M_p\).
Case 2 — Inelastic LTB: If \(L_p < L_b \le L_r\):
\[ M_n = C_b \left[M_p - (M_p - 0.7F_y S_x)\frac{L_b - L_p}{L_r - L_p}\right] \le M_p \]Case 3 — Elastic LTB: If \(L_b > L_r\):
\[ F_{cr} = \frac{C_b \pi^2 E}{(L_b/r_{ts})^2}\sqrt{1 + 0.078\frac{Jc}{S_x h_0}\left(\frac{L_b}{r_{ts}}\right)^2} \] \[ M_n = F_{cr}\, S_x \le M_p \]where \(C_b\) is the moment gradient factor (1.0 is conservative; calculated automatically when the moment diagram is known).
where:
- \(\phi_v = 1.00\) for hot-rolled I-shapes with \(h/t_w \le 2.24\sqrt{E/F_y}\) (most standard W-shapes qualify)
- \(A_w = d \times t_w\) (web area using overall depth \(d\) and web thickness \(t_w\))
- \(C_{v1} = 1.0\) for most W-shapes where the web slenderness is below the shear yielding limit
Note: \(\phi_v V_n\) uses \(\phi_v = 1.0\) (not 0.9) for qualifying sections per AISC 360-22 Section G2.1. This is a recent code clarification — older references may show \(\phi_v = 0.9\).
The deflection serviceability check requires that the maximum computed deflection in each span does not exceed the allowable limit:
\[ \delta_{max} \le \frac{L}{n_{limit}} \]where \(L\) is the span length and \(n_{limit}\) is the deflection limit ratio selected in the settings (240, 360, 480, or 600).
| Limit | Application | Code Reference |
|---|---|---|
| L / 360 | Live load deflection, floor beams supporting plaster or brittle finishes | AISC DG-3, IBC 1604.3 |
| L / 240 | Total load deflection, general construction | AISC 360 Commentary |
| L / 480 | Live load, floors with sensitive equipment or glass | Project specific / AISC |
| L / 600 | Precision or vibration-sensitive applications | Project specific |
| L / 180 | Roof members, live load only | AISC 360 Commentary |
For a 20-ft span with a L/360 limit, the allowable deflection is:
\[ \delta_{allow} = \frac{20 \times 12}{360} = \frac{240}{360} = 0.667 \text{ in} \]Fully Worked Example — 3-Span Office Floor Beam
This example follows a complete design check for a W14×30 beam supporting an office floor across three equal 20-ft spans. All steps use the calculator's exact formulas.
📌 Problem Setup
Beam: W14×30 (ASTM A992, Fy = 50 ksi, E = 29,000 ksi)
Spans: 3 equal spans, L = 20 ft each — Pin-Roller-Roller-Pin supports
Loads: Dead load (D) = 0.8 kip/ft UDL (all spans) + Live load (L) = 1.2 kip/ft UDL (all spans)
Section properties (W14×30): Ix = 291 in⁴, Sx = 42.0 in³, Zx = 47.3 in³, d = 13.84 in, tw = 0.230 in
Unbraced length: Lb = 6 ft (braced at third-points)
Design method: LRFD | Deflection limit: L/360
Step 1: LRFD Load Combination
Governing combination per ASCE 7: \(1.2D + 1.6L\)
\[ w_u = 1.2(0.8) + 1.6(1.2) = 0.96 + 1.92 = \mathbf{2.88 \text{ kip/ft}} \]Step 2: Analysis Results (from calculator)
Step 3: Flexural Capacity Check (AISC F2)
\[ M_p = F_y Z_x = 50 \times 47.3 = 2365 \text{ kip-in} = \mathbf{197.1 \text{ kip-ft}} \] \[ L_p = 1.76 \times 1.74 \sqrt{\frac{29000}{50}} = 73.4 \text{ in} = 6.12 \text{ ft} \]Since \(L_b = 6.0 \text{ ft} \le L_p = 6.12 \text{ ft}\): No LTB, \(M_n = M_p = 197.1 \text{ kip-ft}\)
\[ \phi_b M_n = 0.90 \times 197.1 = \mathbf{177.4 \text{ kip-ft}} \] \[ \text{DCR}_{flex} = \frac{96.0}{177.4} = \mathbf{0.54} \quad \Rightarrow \text{ PASS} \]Step 4: Shear Capacity Check (AISC G2)
\[ A_w = d \times t_w = 13.84 \times 0.230 = 3.18 \text{ in}^2 \] \[ \phi_v V_n = 1.00 \times 0.6 \times 50 \times 3.18 \times 1.0 = \mathbf{95.5 \text{ kips}} \] \[ \text{DCR}_{shear} = \frac{36.8}{95.5} = \mathbf{0.39} \quad \Rightarrow \text{ PASS} \]Step 5: Deflection Check
\[ \delta_{allow} = \frac{L}{360} = \frac{20 \times 12}{360} = \mathbf{0.667 \text{ in}} \] \[ \delta_{max} = 0.28 \text{ in} \le 0.667 \text{ in} \quad \Rightarrow \text{ PASS (L/857)} \]Summary Table
| Check | Demand | Capacity | DCR | Status |
|---|---|---|---|---|
| Flexure (AISC F2) | 96.0 kip-ft | 177.4 kip-ft | 0.54 | PASS |
| LTB (AISC F2) | Lb=6.0 ft | Lp=6.12 ft | N/A | No LTB |
| Shear (AISC G2) | 36.8 kips | 95.5 kips | 0.39 | PASS |
| Deflection (L/360) | 0.28 in | 0.667 in | 0.42 | PASS (L/857) |
Conclusion: W14×30 (A992) is adequate for this 3-span 20-ft office floor beam at 54% flexural utilization. The beam is deflection-controlled: utilization of 42% on the deflection check is the practical limit here given the generous section. A lighter section could be trialed — try W14×26 or W12×26 in the calculator.
Continuous vs Simply Supported Beams — Side-by-Side Comparison
Understanding the structural advantage of continuity is essential for justifying its use. The table below compares a 3-span 20-ft beam (same beam, same load) under two different modeling assumptions.
| Result | Three Simply Supported Spans (independent) | Continuous 3-Span Beam | Change |
|---|---|---|---|
| Max positive moment (midspan) | w L² / 8 = 2.0 × 400/8 = 100 kip-ft | 57.6 kip-ft | ↓ 42% reduction |
| Max negative moment (at supports) | 0 kip-ft (no continuity) | 96.0 kip-ft (over interior supports) | New check required |
| Max deflection (service load) | 5wL⁴/384EI = 0.71 in | 0.28 in | ↓ 61% reduction |
| End support reactions | wL/2 = 20 kips each | ≈ 22 kips each | ↓ 10% increase |
| Interior support reactions | wL/2 + wL/2 = 40 kips (sum of 2 simple beams) | 54.7 kips | ↑ 37% increase — design column/footing for this |
| Required beam size (W-shape) | W14×38 minimum for moment | W14×30 adequate (54% DCR) | Lighter section = cost saving |
Key Insight: Interior Support Reactions Are Always Higher in Continuous Beams
While midspan moments and deflections decrease dramatically in a continuous beam, interior column/footing loads increase significantly. Always re-check the supporting structure (columns, beams, foundations) when switching from simple to continuous framing. This is one of the most commonly missed consequences of continuity in preliminary design.
Common Mistakes When Using a Continuous Beam Calculator
Real-World Applications of Continuous Steel Beam Analysis
Multi-Span Floor Framing in Commercial Buildings
In a typical office building with 20-ft bay spacing, secondary floor beams frame between girders and run continuously when moment connections are used. The reduction in mid-span deflection (approximately 60% compared to simple spans) allows architects to use shallower floor depths, reducing overall building height while meeting L/360 serviceability limits for the finished floor system. Engineers use a continuous beam calculator to quickly select the lightest adequate W-shape and confirm that interior girder reactions are within column capacity before detailed connections are designed.
Industrial Mezzanines and Equipment Platforms
Equipment platforms carry concentrated point loads from machinery, HVAC units, and storage racks that rarely align with support points. A continuous beam calculator allows engineers to position these point loads at any location within each span and immediately see the resulting moment and deflection at critical sections. Pattern loading is critical here: a forklift on one span while the adjacent span is empty creates a very different moment diagram than two forklifts operating simultaneously.
Bridge Girder Preliminary Design
Highway overpass girders are classic multi-span continuous beams. Bridge engineers use continuous beam analysis to determine girder reactions (for pier and abutment footing design), midspan moments (for section sizing), and negative moments over piers (which may require top-flange reinforcement or cover plates). The influence line concept applies here: a truck load moving across the bridge creates different critical load cases depending on truck position, which is the engineering basis behind the influence line output available in advanced continuous beam tools.
Structural Retrofit and Existing Building Assessment
When assessing whether an existing continuous beam can carry new mechanical equipment or increased live loads, engineers input the known section properties from as-built drawings, enter existing dead loads plus the new proposed load, and check the Section Check tab for demand-to-capacity ratios. A DCR above 1.0 flags an inadequate section that requires either strengthening (cover plates, supplemental framing) or load redistribution.
Frequently Asked Questions — Continuous Steel Beam Calculator
Understanding the Section Check Tab
The Section Check tab runs AISC 360 LRFD design checks automatically using the maximum forces from the analysis. Here is what each check means and how to interpret the results:
Flexural Capacity
Compares the maximum design moment Mu to the LRFD design flexural strength φbMn. For compact sections in the no-LTB range: Mn = Mp = FyZx. If LTB governs, Mn is reduced per the AISC F2 formulas shown above. A DCR of 1.0 means the beam is exactly at its code-permitted limit.
Lateral-Torsional Buckling (LTB)
LTB check determines whether the beam can achieve its full plastic moment capacity or if it buckles laterally before yielding. The critical input is the unbraced length Lb — the distance between points where the compression flange is restrained from lateral movement. For composite beams with a concrete slab on top, Lb = 0 and LTB never governs. For non-composite beams, Lb equals the brace spacing.
Shear Capacity
Shear governs in short spans and near concentrated loads. The web area Aw = d × tw controls shear capacity. Thicker webs or deeper sections have higher shear capacity. If shear governs your design, consider a section with a thicker web (MC shapes, or W-shapes with higher d/tw ratios) rather than simply increasing the depth.
Serviceability and Deflection
The deflection check is performed at service load levels (unfactored D + L), not LRFD factored loads. If you entered LRFD factored loads in the Loads tab, divide your deflection result by approximately 1.45 (the typical load factor for 1.2D + 1.6L with equal dead and live loads) to get the service-load deflection for the L/360 check. Alternatively, enter service-level loads in a separate analysis run specifically for deflection checking.
Exporting and Sharing Results
The Results tab generates a complete plain-text engineering report containing all inputs, support reactions, maximum forces, and section check results. Two export options are available:
- Copy Report to Clipboard: Copies the formatted text report for pasting into Word, Excel, email, or a calculation package. The report includes the date, unit system, span geometry, section properties, load summary, reaction table, governing maximum forces, and equilibrium verification.
- Print / Save PDF: Opens the browser’s print dialog. Select “Save as PDF” to generate a professional PDF calculation. The print stylesheet expands all formula accordions automatically so no information is hidden.
Calculator Scope and Limitations
This calculator is designed for elastic analysis of prismatic beam elements under static loading. The following conditions are outside its current scope:
- Non-prismatic sections (stepped or tapered beams) — use constant-I approximation or divide into separate elements
- Axial loads (beam-column interaction) — P-M interaction not included
- Dynamic loading, seismic base shears, or vibration analysis
- Composite action (steel beam + concrete deck acting together)
- Non-linear geometric effects (P-Delta)
- In-plane frame analysis (horizontal loads on a portal frame)
- Section classification for plastic design (moment redistribution per AISC Appendix 1)
For these scenarios, use full-frame FEA software such as SAP2000, ETABS, RISA-3D, or STAAD.Pro. Use this calculator for beam-only preliminary analysis and for verification of output from larger models.
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