Simply Supported Beam (Center Load + UDL)

stressmass-cost

Verified build 8 relations · 3 identities proven · 2 modeling steps · 6 parity samples

Every floor you have ever stood on is this THING: a beam resting on two supports, carrying both a crowd of distributed weight and the occasional concentrated piano. The cantilever taught the flexure machinery; this page teaches the move that makes beam tables useful — superposition:

\delta_P = \frac{P L^3}{48\,EI} \qquad \delta_w = \frac{5\,w L^4}{384\,EI} \qquad \delta = \delta_P + \delta_w

Three things to notice, one per formula:

Each load case is a table row. Engineers almost never integrate the beam equation at the desk; they look up the case (Shigley’s Table A-9 holds sixteen) and trust the row. The widget shows both rows separately — $\delta_P$ and $\delta_w$ are live readouts, not hidden intermediates — so you can watch which load owns the deflection as you trade $P$ against $w$ .
The span exponents disagree. The point load deflects with $L^3$ but the distributed load with $L^4$ — stretch the span and the UDL always wins eventually, which is why long roofs fear snow (a $w$ ) more than equipment (a $P$ ). (Newton for newton, spread load actually deflects only 5/8 as much as a centered point load — the UDL wins on span, not on efficiency.)
Adding answers is a theorem, not a habit. $\delta = \delta_P + \delta_w$ works because $EI\,v'' = M(x)$ is linear — double the cause, double the effect, causes add. The moment the beam yields, buckles, or deflects enough to change its own geometry, superposition dies with the linearity (the warn banners are exactly that boundary). And a subtlety the widget’s $M_{max}$ relation flags: the maxima added only because both load cases peak at midspan — fields always superpose, peak values only sometimes do.

The size-depth configuration runs it backwards: name the loads, span, and the margin a code demands, and the widget returns the joist depth — then immediately reports the deflection that depth costs, because strength-sized beams are routinely stiffness-governed in practice.

Try it

Configuration

Material

T3, bare flat sheet 0.010-0.128 in. thick, AMS 4037 / AMS-QQ-A-250/4 (MIL-HDBK-5J Table 3.2.3.0(b1), p. 3-71)

Bound properties of 2024-T3 aluminum sheet (bare)
E	10.5 Msi	typical	mil-hdbk-5j
sigma_y	47 ksi	design min.	mil-hdbk-5j
rho	0.1 lb/inch**3	typical	mil-hdbk-5j

3 materials in the database are not listed here: no published value in our cited sources for every property this THING needs.

Materials modeled here: 2024-T3 aluminum sheet (bare) 304 stainless steel 6061-T6 aluminum 7075-T6 aluminum AISI 1045 medium-carbon steel AISI 4340 low-alloy steel (Ni-Cr-Mo) ASTM A36 structural steel (hot-rolled) C26000 Cartridge Brass (70/30) Nylon 6/6 (PA66), unfilled Ti-6Al-4V

Governing relations

I = \frac{b h^3}{12}

Assumes: solid rectangular cross-section, bending about the strong axis

Source: Gere, J. M., & Goodno, B. J., Mechanics of Materials, 9th ed., Cengage, 2018 — flexure formula (§5.5), shear-deflection and small-deflection limits, App. E section properties.

\delta_P = \frac{P L^3}{48\,E I}

Assumes: simply supported (pinned-roller), point load exactly at midspan; linear elastic, small deflections, Euler–Bernoulli kinematics

Source: Budynas, R. G., & Nisbett, J. K., Shigley's Mechanical Engineering Design, 10th ed., McGraw-Hill, 2015 — ch. 4 (Deflection and Stiffness): deflections by integration and by superposition, and Table A-9 (beam deflection cases; the simply-supported center-load and uniform-load rows used here).

\delta_w = \frac{5\,w L^4}{384\,E I}

Assumes: uniform load over the full span, same support and kinematic assumptions

\delta = \delta_P + \delta_w

Assumes: THE method relation — the beam equation EI·v″ = M(x) is linear in the loads, so the response to both loads together is exactly the sum of the responses to each alone; legal only while everything stays linear (elastic material, small deflections, no contact or buckling) · Valid while: Total midspan deflection exceeds L/10 — the small-deflection assumption is breaking down, and with it the linearity that superposition itself depends on. This is a short, deep beam (L < 10h): shear deflection, neglected by Euler–Bernoulli theory, is no longer small.

M_{max} = \frac{P L}{4} + \frac{w L^2}{8}

Assumes: both load cases put their maximum moment at midspan, so the maxima add legitimately — superposing maxima is NOT generally legal; it works here only because the peaks coincide

\sigma = \frac{M_{max}\,(h/2)}{I}

Assumes: flexure formula at the outer fiber, midspan · Valid while: Bending stress exceeds the yield strength — the outer fiber has yielded and every linear-elastic number here (superposition included) stops being the truth.

Source: Gere, J. M., & Goodno, B. J., Mechanics of Materials, 9th ed., Cengage, 2018 — flexure formula (§5.5), shear-deflection and small-deflection limits, App. E section properties.

\mathrm{SF} = \frac{\sigma_y}{\sigma}

Assumes: margin against first yield at the outer fiber, not against collapse

Source: Gere, J. M., & Goodno, B. J., Mechanics of Materials, 9th ed., Cengage, 2018 — flexure formula (§5.5), shear-deflection and small-deflection limits, App. E section properties.

m = \rho\, b\, h\, L

Assumes: prismatic beam, uniform density

Source: Gere, J. M., & Goodno, B. J., Mechanics of Materials, 9th ed., Cengage, 2018 — flexure formula (§5.5), shear-deflection and small-deflection limits, App. E section properties.

Derivation

Steps marked modeling step are where physics enters by citation — every other line is machine-proven to follow from them, and the cited models are independently re-derived in the test pipeline where possible. See what is and isn't machine-verified.

M_{P} = \frac{P x}{2}

1. The modeling step, load case one: by symmetry each support carries P/2, so a cut at x (left of midspan) sees the internal moment M_P = (P/2)·x — a triangle peaking at midspan with value PL/4. — statics: method of sections, point load modeling step

M_{w} = \frac{w x \left(L - x\right)}{2}

2. Load case two: each support carries wL/2, and the load already passed contributes −wx·(x/2), leaving M_w = wx(L−x)/2 — a parabola, also peaking at midspan, with value wL²/8. — statics: method of sections, distributed load modeling step

M_{max} = \frac{L^{2} w}{8} + \frac{L P}{4}

3. Evaluate both at x = L/2 and add. Adding the MAXIMA is legal here only because both diagrams peak at the same place — superpose two off-center loads and the combined peak sits somewhere new, smaller than the sum of the individual peaks. The fields always add; their maxima only sometimes do. — superpose the moment diagrams at their shared peak

\delta = \delta_{P} + \delta_{w}

4. The headline act: EI·v″ = M_P(x) + M_w(x) is linear, and integration is linear, so the deflection under both loads is exactly the deflection under each, added. δ_P = PL³/48EI and δ_w = 5wL⁴/384EI are the two most-quoted rows of every beam table (Shigley Table A-9); the build cannot integrate, so the test pipeline re-derives both rows from EI·v″ = M(x) with a computer algebra system, constants and all. — superposition: linear ODE ⇒ responses add

\sigma = \frac{6 M_{max}}{b h^{2}}

5. Flexure formula with the rectangular section folded in: σ = 6M/(bh²). Depth enters squared — the size-depth configuration runs this backwards to find the joist that meets a code margin, which is how every floor over your head was actually chosen. — flexure formula, rectangular section

How it fails

The widget’s margin is first yield at the midspan outer fiber, under static loads. Floors and girders find other ways:

Serviceability governs before strength. Most real beams are sized by deflection limits (L/360 live-load for floors and plastered ceilings, L/240 for less crack-sensitive finishes — the stricter number exists because plaster cracks), not by yield — a beam at SF = 3 against yield can still feel like a trampoline. The two-knob widget makes the split visible: stiffness is $EI$ ‘s business, strength is $\sigma_y$ ‘s.
Lateral-torsional buckling. A deep, narrow beam bent about its strong axis can roll sideways and twist long before the outer fiber yields — the compression flange is a column (Euler’s problem in disguise) that buckles out of the loading plane. This is why long unbraced steel beams get lateral bracing and why the flexure formula alone cannot size them.
Shear, the forgotten diagram. Euler–Bernoulli ignores shear deflection (the L > 10h warn), but shear stress also peaks at the supports — short heavy beams and notched wood beams fail there, parallel to the grain in timber’s case, while the bending formula was still smiling.
Local crushing at supports and loads. The reactions enter through small contact areas; web crippling (steel) and bearing crushing (wood) at the supports are checked separately in every design code. The model’s knife-edge supports are a fiction the hardware has to survive.
The superposition trap. Adding load cases is exact only while the structure stays linear. Two classic violations: cables and membranes (geometry changes with load — stiffening), and beam-columns (an axial load multiplies the lateral deflection, so axial and lateral cases cannot be added). Use table rows outside their linear home and the sum is fiction with confident digits.
Vibration. A beam is a spring with distributed mass; footfall near its natural frequency is how a structurally sound footbridge becomes infamous. Stiffness-per-mass, not strength, sets the tune — another reason $E$ and $\rho$ team up in the Ashby charts.

Cantilever Beam (End Load)
A beam fixed at one end, loaded at the other — the fruit-fly of structures. One widget shows why stiffness (E) and strength (σ_y) are independent axes: swap steel for titanium and deflection goes UP while the safety factor also goes up.
- stress
- mass-cost
Fixed-Fixed Beam (UDL)
A beam built rigidly into a wall at BOTH ends under a uniform load. Two equilibrium equations, four unknown reactions — indeterminate to the second degree — so two compatibility conditions (zero slope and zero deflection at a released end) close the system, and the build solves the coupled 4×4 group exactly. The fixing moment at each wall governs, and none of the reactions cares about the material.
- stress
- mass-cost
Propped Cantilever (UDL)
A cantilever with a prop under its free end: one redundant support turns a determinate beam into a statically indeterminate one. Equilibrium alone cannot find the three reactions — compatibility (the prop deflects to zero) supplies the missing equation, and the build solves the coupled 3×3 system exactly.
- stress
- mass-cost
Circular Plate under Uniform Pressure (Clamped vs Simply Supported)
Push uniform pressure on a flat circular plate — a tank head, a porthole, a valve cover — and how hard it deflects and where it cracks depend entirely on the RIM. Bolt it down (clamped) and it is stiff and hottest at the edge; rest it on a ring (simply supported) and it sags four times as far and is hottest at the center. This is the page where Poisson's ratio moves a STRESS: the simply-supported stress carries ν, the clamped-edge stress carries no material property at all.
- stress
Curved Beam in Bending (Winkler — Crane Hook, C-Clamp, Press Frame)
Bend a bar that is already curved and the neutral axis walks off the centroid, toward the inside of the curve — the inner fibers run hotter than the straight-beam Mc/I ever predicts. This is the Winkler formula behind a crane hook, a C-clamp, and a press frame: σ = M·c/(A·e·r), the tiny eccentricity e = r_c − r_n doing all the work, plus the direct P/A a hook's load adds on top.
- stress
Transverse Shear in Beams (τ = VQ/Ib, Shear Flow, Fastener Spacing)
A beam does not only bend — the shear force V drags its layers past one another, and that longitudinal shear is what a built-up beam's nails or bolts actually carry. The stress is a parabola (peak 3V/2A at the neutral axis, zero at the surfaces), and the shear flow q = VQ/I sets the fastener spacing. Statics and geometry only: no stiffness enters at all.
- stress

Chains with

Outputs whose SI dimension and quantity kind match another THING's input — the only wires the planner's connectionLegal accepts (invariant 2, computed at build time, not hand-listed). Wire these on the chaining demo.

Composite Bar (Core + Sleeve)
- delta L
- delta_P L
- delta_w L
- h L
Impact Loading (Falling Mass, Energy Method)
- delta b
- delta d
- delta h
- delta L
- delta_P b
- delta_P d
- delta_P h
- delta_P L
- delta_w b
- delta_w d
- delta_w h
- delta_w L
- h b
- h d
- h h
- h L
- m_beam m
Symmetric Two-Bar Truss
- delta d
- delta L
- delta_P d
- delta_P L
- delta_w d
- delta_w L
- h d
- h L
Thermal Assembly (Two-Segment Bar Between Rigid Walls)
- delta L_1
- delta L_2
- delta_P L_1
- delta_P L_2
- delta_w L_1
- delta_w L_2
- h L_1
- h L_2
Cantilever Beam (End Load)
- delta b
- delta h
- delta L
- delta_P b
- delta_P h
- delta_P L
- delta_w b
- delta_w h
- delta_w L
- h b
- h h
- h L
Circular Plate under Uniform Pressure (Clamped vs Simply Supported)
- delta a
- delta t
- delta_P a
- delta_P t
- delta_w a
- delta_w t
- h a
- h t
- sigma q
- sigma sigma_allow
Curved Beam in Bending (Winkler — Crane Hook, C-Clamp, Press Frame)
- delta r_i
- delta r_o
- delta w
- delta_P r_i
- delta_P r_o
- delta_P w
- delta_w r_i
- delta_w r_o
- delta_w w
- h r_i
- h r_o
- h w
Fixed-Fixed Beam (UDL)
- delta b
- delta h
- delta L
- delta_P b
- delta_P h
- delta_P L
- delta_w b
- delta_w h
- delta_w L
- h b
- h h
- h L

+ 24 more THINGs its outputs can legally feed (showing the first 8 in course order).

Sources

Budynas, R. G., & Nisbett, J. K., Shigley's Mechanical Engineering Design, 10th ed., McGraw-Hill, 2015 — ch. 4 (Deflection and Stiffness): deflections by integration and by superposition, and Table A-9 (beam deflection cases; the simply-supported center-load and uniform-load rows used here).
Gere, J. M., & Goodno, B. J., Mechanics of Materials, 9th ed., Cengage, 2018 — flexure formula (§5.5), shear-deflection and small-deflection limits, App. E section properties.

Simply Supported Beam (Center Load + UDL)

Try it

Governing relations

Derivation

How it fails

Related THINGs

Chains with

Sources