How to use this lab
The stage shows the beam, then its shear force V, bending moment M and deflection diagrams. Add a load with the buttons in the controls and watch the safety factor. Drag loads, load ends and supports along the beam. Hover (or touch) anywhere on the diagrams to read V, M, deflection and stress at that section. Ready-made real-world cases are in Scenarios.
Results
1. Static equilibrium and support reactions
Why it matters: floors, bridges, shelves and balconies are all beams. The engineer must check two things: the beam must not break (stress) and must not sag too much (deflection). A beam is a slender member loaded perpendicular to its axis. Before we can find what happens inside it, we find the forces the supports exert on it. For a planar body at rest, the resultant force and resultant moment must both vanish:
Each support provides certain reaction components:
- Roller: one vertical force (free to slide and rotate).
- Pin: vertical and horizontal forces (free to rotate). With only vertical loads the horizontal reaction is zero.
- Fixed (built-in): vertical force, horizontal force and a moment; prevents both translation and rotation.
A pin plus a roller, or a single fixed support, gives exactly the three unknowns the three equilibrium equations can solve: the beam is statically determinate. A propped cantilever (fixed + roller) has one extra unknown and a fixed–fixed beam has two extra in bending; these are statically indeterminate, and we need the beam's deformation (compatibility) to find the reactions.
2. Internal forces and sign convention
Cut the beam at a section $x$. To keep each piece in equilibrium, the cut face must carry an internal shear force $V$ (which tries to slide one piece past the other) and bending moment $M$ (which tries to bend it). This lab uses the most common textbook convention:
Computed from the left part of the beam, with upward forces positive and anticlockwise applied couples $C$ positive:
3. Load, shear and moment relationships
Equilibrium of a tiny slice $dx$ carrying a downward distributed load $w(x)$ gives the two differential relationships that make diagrams quick to sketch:
- No distributed load: $V$ is constant and $M$ is a straight line.
- Uniform load: $V$ is linear and $M$ is a parabola. A triangular load gives a quadratic $V$ and a cubic $M$.
- A point load makes $V$ jump by the load; an applied couple $C$ makes $M$ jump by $C$ (down for an anticlockwise couple in this convention).
- $M$ is a maximum or minimum where $V = 0$ or changes sign. That is where to look for the critical section.
- The change in $M$ between two points equals the area under the shear diagram between them.
4. Bending stress and section properties
When a beam sags, its top fibres are squashed and its bottom fibres stretched. Somewhere between lies the neutral axis, which is neither stretched nor squashed (the centre line for symmetric sections). In pure bending, plane sections remain plane and the strain varies linearly with distance $y$ from the neutral axis. With a linear elastic material this gives the flexure formula:
$I=\int_A y^2\,dA$ is the second moment of area about the neutral axis and $c$ the distance from the neutral axis to the outermost (extreme) fibre, where stress is largest. Material far from the neutral axis does most of the work, which is why I-beams and hollow sections are efficient.
| Section | Area A | Second moment I | c |
|---|---|---|---|
| Rectangle $b\times h$ | $bh$ | $\dfrac{bh^3}{12}$ | $h/2$ |
| Solid circle, diameter $d$ | $\dfrac{\pi d^2}{4}$ | $\dfrac{\pi d^4}{64}$ | $d/2$ |
| Tube $D$, wall $t$, $d_i = D-2t$ | $\dfrac{\pi(D^2-d_i^2)}{4}$ | $\dfrac{\pi (D^4-d_i^4)}{64}$ | $D/2$ |
| Box $b\times h$, wall $t$ | $bh-(b-2t)(h-2t)$ | $\dfrac{bh^3-(b-2t)(h-2t)^3}{12}$ | $h/2$ |
| I-beam: flanges $b\times t_f$, web $t_w$ | $2bt_f+(h-2t_f)t_w$ | $\dfrac{bh^3-(b-t_w)(h-2t_f)^3}{12}$ | $h/2$ |
In this lab the I-beam web thickness is taken as $t_w = 0.6\,t_f$, a typical proportion for rolled sections. The safety factor shown is $SF = f_y/\sigma_{\max}$, using the yield strength (steel, aluminium), the characteristic bending strength (timber) or the mean tensile strength $f_{ctm}$ (plain concrete; a conservative cracking limit in bending).
5. Deflection: the Euler–Bernoulli equation
A beam bends more where the moment is larger and less where it is stiffer. For small slopes, the curvature of the deflected axis $y(x)$ is proportional to the bending moment and inversely proportional to the stiffness $EI$ ($E$ = Young's modulus of the material, $I$ = shape stiffness):
Integrating twice gives slope $\theta = dy/dx$ and deflection $y$; the two constants of integration come from the support conditions ($y=0$ at a pin or roller, and $y=0,\ \theta=0$ at a fixed end). $E$ is Young's modulus and $EI$ is the flexural rigidity. Doubling the depth of a rectangular beam multiplies $I$ by 8 and divides the deflection by 8.
Each example below can be loaded into the lab so you can check every number against the live readouts.
Example 1: simply supported beam with a point load and a UDL
A 6 m steel beam (pin A at 0, roller B at 6 m) carries a 20 kN point load at $x = 2$ m and a uniform load of 5 kN/m over the whole span. Section: I-beam $h=300$, $b=150$, $t_f=12$ mm ($t_w = 7.2$ mm). Self-weight ignored.
- Reactions. Total UDL $= 5\times6 = 30$ kN at $x=3$ m. Moments about A: $6R_B = 20(2) + 30(3) = 130 \Rightarrow R_B = 21.67$ kN. Vertical equilibrium: $R_A = 50 - 21.67 = 28.33$ kN.
- Shear. $V = 28.33 - 5x$ for $0<x<2$, so $V(2^-) = 18.33$ kN; the point load drops it to $V(2^+) = -1.67$ kN; it then falls linearly to $-21.67$ kN at B.
- Maximum moment. $V$ changes sign at $x=2$ m, so $M_{\max} = 28.33(2) - 5(2)^2/2 = 46.67$ kN·m.
- Section. $I = \frac{0.15(0.3)^3 - (0.15-0.0072)(0.276)^3}{12} = 8.730\times10^{-5}$ m$^4$, $Z = I/0.15 = 5.820\times10^{-4}$ m$^3$ (582 cm³).
- Stress. $\sigma_{\max} = M/Z = 46.67\times10^3 / 5.820\times10^{-4} = 80.2$ MPa, so $SF = 275/80.2 = 3.43$.
- Deflection at mid-span by superposition ($EI = 200\times10^6 \times 8.730\times10^{-5} = 17\,460$ kN·m²): $\delta_{\text{UDL}} = \frac{5wL^4}{384EI} = \frac{5(5)(6^4)}{384(17460)} = 4.83$ mm, $\delta_P = \frac{Pa(3L^2-4a^2)}{48EI} = \frac{20(2)(108-16)}{48(17460)} = 4.39$ mm, total $9.22$ mm. The true maximum (9.24 mm at about $x = 2.9$ m) sits slightly towards the point load.
Example 2: timber cantilever with a triangular load
A 3 m timber cantilever (fixed at A, $x=0$) carries a triangular load rising from 0 at the wall to 4 kN/m at the free end, plus a 2 kN point load at the tip. Section: rectangle $b=100$, $h=250$ mm, C24 timber ($E = 11$ GPa).
- Resultants. Triangle: $\tfrac12(4)(3)=6$ kN acting $\tfrac23(3)=2$ m from the wall. Tip load 2 kN at 3 m.
- Reactions. $R_A = 6+2 = 8$ kN upward. Fixed-end moment $M_A = 6(2)+2(3) = 18$ kN·m (anticlockwise on the beam), i.e. a hogging moment $M(0)=-18$ kN·m.
- Section. $I = bh^3/12 = 0.1(0.25)^3/12 = 1.302\times10^{-4}$ m$^4$, $Z = bh^2/6 = 1.042\times10^{-3}$ m$^3$, so $\sigma_{\max} = 18/1.042\times10^{-3} = 17.3$ MPa and $SF = 24/17.3 = 1.39$. Adequate in the elastic sense but below the 2.0 you might want.
- Tip deflection (superposition, $EI = 11\times10^6\times1.302\times10^{-4} = 1432$ kN·m²): point load $\frac{PL^3}{3EI} = \frac{2(27)}{3(1432)} = 12.57$ mm; triangle increasing to $w_0$ at the free end $\frac{11 w_0 L^4}{120 EI} = \frac{11(4)(81)}{120(1432)} = 20.74$ mm. Total 33.3 mm, which is $L/90$: too flexible for a floor.
Example 3: fixed–fixed versus simply supported
An 8 m beam carries a UDL of 10 kN/m. Compare simple supports with both ends built in (steel I-beam $h=300$, $b=150$, $t_f=12$ mm, $EI = 17\,460$ kN·m²).
| Quantity | Simply supported | Fixed–fixed |
|---|---|---|
| Reactions | 40 kN each | 40 kN, end moments 53.3 kN·m |
| Mid-span moment | $wL^2/8 = 80.0$ kN·m | $wL^2/24 = 26.7$ kN·m |
| Support moment | 0 | $-wL^2/12 = -53.3$ kN·m |
| Max deflection | $\frac{5wL^4}{384EI} = 30.5$ mm | $\frac{wL^4}{384EI} = 6.11$ mm |
| σ max | 137.5 MPa (SF 2.00) | 91.6 MPa (SF 3.00) |
Fixing the ends cuts the peak moment by a third and the deflection by a factor of five, but the critical section moves to the supports, which must now resist moment. The fixed-end moment cannot be found from statics alone: it comes from the condition that the slope is zero at both ends.
Each challenge sets up the lab and then checks your design live. Change the section, supports or loads in the lab until the badge turns green.
Open experiments
- With one point load on a simply supported beam, drag it from end to end. Where is the bending moment largest, and what is the maximum value as a function of $P$ and $L$?
- Keep the load and section fixed and double the span. By what factor does the maximum deflection change? (Predict first: for a central point load, $\delta \propto L^3$.)
- Switch a rectangular section from "flat" ($b > h$) to "upright" ($h > b$) with the same area. Compare $I$, stress and deflection.
- Apply an anticlockwise moment at mid-span of a simply supported beam with no other load. Explain why the reactions form a couple and why the moment diagram jumps.
- Choose plain concrete and watch the safety factor. Concrete is strong in compression but cracks at a few MPa of tension, which is why real concrete beams carry steel reinforcement in the tension zone.
Standard cases (span $L$, flexural rigidity $EI$)
| Case | Max moment | Max deflection |
|---|---|---|
| Simply supported, central point load $P$ | $PL/4$ | $\dfrac{PL^3}{48EI}$ |
| Simply supported, point load $P$ at $a$ ($b=L-a$) | $Pab/L$ | at mid-span: $\dfrac{Pa(3L^2-4a^2)}{48EI}$ ($a\le L/2$) |
| Simply supported, UDL $w$ | $wL^2/8$ | $\dfrac{5wL^4}{384EI}$ |
| Cantilever, tip load $P$ | $PL$ (at wall) | $\dfrac{PL^3}{3EI}$ |
| Cantilever, UDL $w$ | $wL^2/2$ | $\dfrac{wL^4}{8EI}$ |
| Cantilever, triangular load 0 at wall to $w_0$ at tip | $w_0L^2/3$ | $\dfrac{11w_0L^4}{120EI}$ |
| Cantilever, triangular load $w_0$ at wall to 0 at tip | $w_0L^2/6$ | $\dfrac{w_0L^4}{30EI}$ |
| Propped cantilever, UDL $w$ | $wL^2/8$ (at wall) | $\dfrac{wL^4}{185EI}$ |
| Fixed–fixed, UDL $w$ | $wL^2/12$ (ends), $wL^2/24$ (mid) | $\dfrac{wL^4}{384EI}$ |
| Fixed–fixed, central point load $P$ | $PL/8$ | $\dfrac{PL^3}{192EI}$ |
Material data used in the lab
| Material | E (GPa) | Strength used (MPa) | Density (kg/m³) |
|---|---|---|---|
| Structural steel S275 | 200 | 275 (yield) | 7850 |
| Aluminium 6061-T6 | 69 | 240 (0.2% proof) | 2700 |
| Softwood timber C24 | 11 | 24 (bending) | 420 |
| Plain concrete C30/37 | 33 | 2.9 (mean tensile, fctm) | 2400 |
Typical serviceability limits: floor beams $L/360$, roofs $L/250$, cantilevers $L/180$ (span measured as the cantilever length). Design codes also apply partial safety factors to loads and materials; this lab shows the plain elastic safety factor $f/\sigma$.