--------------------------------- | ---------- | ------------ | -------------- | ------------- | | Point load P at midspan | PL/4 | P/2 | PL³/48EI | Midspan | | UDL w over full span | wL²/8 | wL/2 | 5wL⁴/384EI | Midspan | | Point load P at distance a from left | Pab/L | P×max(a,b)/L | See formula | Under load | | Two equal P at third points | PL/3 | P | 23PL³/648EI | Between loads |

Cantilever beam

Fixed-fixed beam

Propped cantilever (fixed-pinned)

Worked Example — Simply Supported W16x36 Under Floor Load

Problem: A W16x36 floor beam (ASTM A992, Fy = 50 ksi) spans 24 ft and carries a uniformly distributed service dead load of 0.80 klf and service live load of 1.00 klf. Calculate the reactions, moment, shear, and live load deflection.

Step 1 — Determine factored and service loads

Service loads:
  w_D = 0.80 klf (dead)
  w_L = 1.00 klf (live)
  w_total = 1.80 klf (service, unfactored)

LRFD factored load (ASCE 7 LC2: 1.2D + 1.6L):
  w_u = 1.2(0.80) + 1.6(1.00) = 0.96 + 1.60 = 2.56 klf

Step 2 — Calculate reactions

For symmetric UDL on simply supported beam:

Service reactions:
  R_A = R_B = w_total × L / 2 = 1.80 × 24 / 2 = 21.6 kips (each)

Factored reactions:
  R_u = w_u × L / 2 = 2.56 × 24 / 2 = 30.72 kips (each)

Step 3 — Calculate maximum bending moment

Service moment at midspan:
  M_max = w_total × L² / 8 = 1.80 × 24² / 8 = 1.80 × 576 / 8 = 129.6 kip-ft

Factored moment:
  M_u = w_u × L² / 8 = 2.56 × 24² / 8 = 2.56 × 576 / 8 = 184.3 kip-ft

Step 4 — Calculate maximum shear

Service shear at supports:
  V_max = w_total × L / 2 = 1.80 × 24 / 2 = 21.6 kips

Factored shear:
  V_u = w_u × L / 2 = 2.56 × 24 / 2 = 30.72 kips

Step 5 — Calculate live load deflection

W16x36 properties: I_x = 448 in⁴, E = 29,000 ksi

w_L = 1.00 klf = 1.00/12 = 0.0833 kip/in
L = 24 ft = 288 in

δ_L = 5 × w_L × L⁴ / (384 × E × I)
    = 5 × 0.0833 × 288⁴ / (384 × 29,000,000 × 448)

Numerator: 5 × 0.0833 × 6,879,707,136 = 2,864,867,904
Denominator: 384 × 29,000,000 × 448 = 4,988,928,000

δ_L = 2,864,867,904 / 4,988,928,000 = 0.574 in

Deflection limit L/360: 288/360 = 0.800 in
δ_L = 0.574 in < 0.800 in → OK

Deflection ratio: L/501 (well within L/360)

Step 6 — Check beam capacity (quick verification)

W16x36: Z_x = 64.0 in³

φM_n = 0.90 × 50 × 64.0 = 2,880 kip-in = 240 kip-ft
M_u = 184.3 kip-ft

D/C ratio = 184.3 / 240 = 0.768 → 77% utilization → OK

Shear: φV_n ≈ 126 kips (W16x36, compact web)
V_u = 30.72 kips → D/C = 0.244 → OK

Summary

Supported Beam Types and Load Cases

This calculator supports the following boundary conditions and load types:

Boundary conditions

• Simply supported — pin at one end, roller at the other. Zero moment at both supports.
• Cantilever — fixed at one end, free at the other. Full moment at the fixed support.
• Fixed-fixed — both ends fully restrained. Moment at both supports and at midspan.
• Propped cantilever — fixed at one end, pinned at the other. Moment only at the fixed end.

Load types

• Point load — concentrated force at any location along the span
• Uniform distributed load (UDL) — constant load intensity over the full span
• Partial UDL — constant load intensity over a portion of the span
• Triangular load — linearly varying load (zero at one end, maximum at the other)
• Applied moment — concentrated moment at any location

Limitations

• All analyses assume linear elastic material behavior (Hooke’s Law applies)
• All sections are assumed prismatic (constant cross-section along the span)
• Small deflection theory applies (deflections are small relative to span length)
• Superposition is valid for combining multiple load cases
• Does not account for connection stiffness, semi-rigid behavior, or partial fixity

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Frequently Asked Questions

What is the maximum moment in a simply supported W12×26 under 2 kip/ft over a 20-foot span? For a uniformly distributed load w on a simply supported span L, the maximum moment occurs at midspan and equals M = wL²/8. With w = 2 kip/ft and L = 20 ft: M = 2 × 20² / 8 = 2 × 400 / 8 = 100 kip-ft. The end reactions are each wL/2 = 2 × 20 / 2 = 20 kips. The W12×26 has φMn ≈ 119 kip-ft (compact section, Fy = 50 ksi, fully braced), so the section passes flexure at 84% utilization. Always confirm with the actual unbraced length — if Lb > Lp, LTB reduces φMn below this value.

How does the shear force diagram relate to the applied loading? The shear force diagram is the integral of the distributed load along the span: a uniform load (w) produces a linearly varying shear diagram, while a point load produces a step discontinuity equal in magnitude to that load. The shear is zero at midspan for a symmetrically loaded simply supported beam, and the location where shear passes through zero corresponds to the point of maximum bending moment. Reading the shear diagram first is the fastest way to identify where the beam is most critical in flexure.

Where does maximum bending moment occur under a single point load versus a uniform load? For a simply supported beam with a single midspan point load P and span L, the maximum moment is PL/4 and occurs at midspan. For a uniformly distributed load w (force per unit length), the maximum moment is wL²/8, also at midspan. If the point load is off-center at distance a from one support, the maximum moment is Pab/L where b = L − a, and it occurs directly under the load. These moment values are the primary inputs when sizing a beam for flexural capacity.

What is tributary width and how do I use it to get the beam line load? Tributary width is the floor or roof area width that drains load to a particular beam, typically taken as half the spacing to each adjacent beam. Multiplying the tributary width (in feet or meters) by the applied surface load (psf or kN/m²) gives the distributed line load (lb/ft or kN/m) to enter into the beam calculator. For example, a beam at 10 ft spacing with a 50 psf floor load receives 10 ft × 50 psf = 500 lb/ft as its tributary dead plus live load.

How do I use beam end reactions to design the connections? The vertical reaction at each support is the required shear demand on the connection at that end. For a bolted or welded shear tab, the connection must develop at least the full factored reaction (applying your LRFD or ASD load factors). For a simply supported beam with no moment transfer, the connection is designed for vertical shear only; if the beam is continuous or has partial fixity, a moment component must also be designed into the connection detail.

What sign convention does the calculator use for shear and moment? The standard beam sign convention treats a positive shear as one where the left face of a cut section acts downward (or the resultant of forces to the left of a section is upward). A positive bending moment produces tension on the bottom fiber (sagging). These conventions mean a simply supported beam under gravity load will have positive moment throughout its span, and a cantilever under a downward tip load will have negative moment (hogging) along its full length. Confirm the sign convention in the tool output before using the values in a connection or section capacity check.

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