Simply Supported Beam Formulas: Shear, Moment, and Deflection for Common Load Cases

A simply supported beam — resting on a pin at one end and a roller at the other, free to rotate at both ends — is the most common statically determinate member in civil engineering practice. Floor beams, purlins, and secondary framing members behave this way to a good approximation. Because the system is determinate, the reactions, shear V(x), moment M(x), and deflection δ(x) all have closed-form solutions in terms of the span L and flexural rigidity EI.

This article gives the standard formulas for five load cases: a full-span UDL, a central point load, a point load at any location, two symmetric point loads, and a triangular load — plus superposition for combining cases and one worked numeric example in kN, m, and MPa covering the unit-conversion step most students get wrong.

Sign Convention and Assumptions

Take x = 0 at support A and x = L at support B; upward reactions and sagging moment (tension at the bottom) are positive. Formulas assume Euler-Bernoulli beam theory: linear-elastic material, small deflections, constant EI, and static loads. Shear deformation is neglected, reasonable for span-to-depth ratios of about 10 or more.

Case 1: Uniformly Distributed Load Over the Full Span

A load w (kN/m or N/mm) spread over the full span — self-weight, finishes, or a uniform live load.

Case 2: Concentrated Load at Midspan

A single load P at midspan — a column, a beam reaction, or equipment framing in at the center.

Case 3: Concentrated Load at Any Point Along the Span

Most point loads are not centered. Let P sit at distance a from support A and b from support B, with a + b = L.

When a ≠ b the true maximum lies between the load and midspan, on the longer segment. The load-point value is within about 3% of δ_max for loads between 0.4L and 0.6L, but it underestimates the true maximum by roughly 25% when the load sits at a quarter point (a = 0.75L) and by more as the load nears a support — use the δ_max expression for serviceability checks.

Case 4: Two Equal, Symmetrically Placed Point Loads

Two identical loads P, each a distance a from its nearest support — as in third-point loading for flexural beam tests.

Because both the load and the geometry are symmetric about midspan, the center deflection given above is also the maximum.

Case 5: Triangular (Linearly Varying) Load

A load rising linearly from zero at support A to a peak w₀ at support B — hydrostatic pressure on a wall, or a varying wind or earth pressure diagram.

This deflection maximum does not fall at a tidy fraction of L; it comes from integrating the M/EI curve twice. If the load peaks at the left support instead, mirror x about midspan.

Summary Table of Standard Load Cases

For quick verification, the simple beam calculator runs these cases automatically, and the beam deflection tool isolates a serviceability check.

Combining Load Cases by Superposition

Real beams typically combine self-weight (UDL) with one or two point loads. Superposition states that for a linear-elastic, small-deflection system, the shear, moment, and deflection at any x can be found by adding the contributions of each load case, provided all cases share the same support conditions.

Two pitfalls: you can only add two tabulated M_max values directly if both occur at the same x — true for a UDL plus a centered point load, not for off-center combinations, which need the full combined M(x) diagram. Also, superposition only holds in the linear-elastic range; once a section cracks or yields, simple addition of deflections becomes an approximation.

Worked Example: UDL Plus a Midspan Point Load

Simply supported RC beam: L = 6 m, section 300 × 500 mm, f’_c = 28 MPa, carrying uniform service load w = 25 kN/m plus a service point load P = 40 kN at midspan. Find V_max, M_max, midspan deflection.

Step 1 — Section and material properties

I = bh³/12 = (300)(500)³/12 = 3.125 × 10⁹ mm⁴. Per ACI 318-19, E_c = 4700√f’_c (MPa) = 4700√28 = 4700 × 5.2915 ≈ 24,870 MPa. EI = 24,870 × 3.125 × 10⁹ = 7.7719 × 10¹³ N·mm².

Step 2 — Shear and moment by superposition

Both cases peak in shear at the supports and in moment at midspan, so the maxima add directly. UDL: V_max,w = wL/2 = 25(6)/2 = 75 kN; M_max,w = wL²/8 = 25(36)/8 = 112.5 kN·m. Point load: V_max,P = P/2 = 20 kN; M_max,P = PL/4 = 40(6)/4 = 60 kN·m. Combined: V_max = 95 kN; M_max = 172.5 kN·m.

Step 3 — Deflection, with units converted carefully

Converting: L = 6000 mm. Since 1 kN/m = 1 N/mm numerically, w = 25 N/mm with no scaling needed; P = 40,000 N.

δ_w = 5wL⁴/384EI = 5(25)(6000)⁴/(384 × 7.7719 × 10¹³) = 1.62 × 10¹⁷/2.9844 × 10¹⁶ ≈ 5.43 mm.

δ_P = PL³/48EI = 40,000(6000)³/(48 × 7.7719 × 10¹³) = 8.64 × 10¹⁵/3.7305 × 10¹⁵ ≈ 2.32 mm.

Both maxima occur at midspan: δ_total ≈ 5.43 + 2.32 = 7.75 mm — well within an illustrative L/360 = 16.7 mm limit, though a real check should separate dead/live components and use an effective (cracked) I rather than the gross I_g used here for simplicity.

Assumptions & Limitations

• Assumes a prismatic beam with constant EI; stepped or tapered sections need a numerical solution.
• Deflections here use gross I_g. ACI 318-19 requires a smaller effective I_e for cracked RC members, giving larger real deflections.
• Pin and roller supports are idealizations; real bearings provide some restraint, reducing actual results.
• Shear deformation is neglected, valid for span-to-depth ratios of about 10 or more.
• Loads are static; dynamic or impact loads need separate amplification factors.
• These are elastic, service-load formulas, not a substitute for the factored-load strength checks in NSCP 2015 and ACI 318-19.

For the cracked-section I_e calculation, RHCES's beam deflection calculator pairs well with the elastic analysis here.

Frequently Asked Questions

Do these formulas apply to continuous or cantilever beams?

No. Every case assumes one pin, one roller, no intermediate or fixed supports. A cantilever is still statically determinate but has its own set of formulas (for a full-span UDL, M_max = wL²/2 at the fixed end and δ_max = wL⁴/8EI at the free end). Continuous, propped, and fixed-end beams are indeterminate and need moment distribution, the three-moment equation, or a stiffness/matrix analysis.

Should I use gross or cracked moment of inertia for deflection?

For steel or timber, gross I is usually fine. For reinforced concrete at service load, ACI 318-19 calls for an effective I_e reduced from I_g for cracking — I_g alone under-predicts deflection.

Can I just add the M_max values from two load-case tables?

Only if both maxima fall at the same location, as with a UDL plus a centered point load. Otherwise, sum the full M(x) functions across the span and find the peak of the combined diagram.

These cases and superposition cover most simply supported beam checks. The simple beam calculator and beam deflection tool automate the arithmetic, part of the full catalog of free web tools. Offline spreadsheets are on the downloads page.

Written by Engr. Ruel H. Cepeda, Structural Engineer, Philippines.