Deflection

A beam that satisfies every stress check can still be unusable. A floor that visibly sags, a lintel that lets a door bind, a gantry that whips under a moving load: none of those is a failure of strength, and all of them are failures. This lesson computes how much a beam moves, which turns out to use a result already derived and to supply, as a bonus, the missing equations for indeterminate beams.

The equation of the deflected shape

The flexure formula of two lessons ago ended in a relation that has not yet been used:

\frac{1}{R} = \frac{M}{E I}

The curvature of the beam at any point is its bending moment divided by its flexural rigidity. From calculus, the curvature of a curve $v (x)$ is

\frac{1}{R} = \frac{d^{2} v / d x^{2}}{{[1 + (d v / d x)^{2}]}^{3 / 2}}

and for a structural beam the slope $d v / d x$ is a small number: a beam deflecting by span over 300 has a maximum slope of order $0.01$ , whose square is $10^{- 4}$ . Dropping it costs a hundredth of a per cent and leaves the Euler-Bernoulli beam equation

E I \frac{d^{2} v}{d x^{2}} = M (x)

with $v$ measured upwards. Integrating once gives the slope and a constant; integrating again gives the deflection and a second constant. The two constants come from boundary conditions: zero deflection at a pin or a roller, zero deflection and zero slope at a built-in end, and continuity of both across any interior point.

Everything about deflection follows from that. There are quicker routes for complicated cases, the moment-area method, Macaulay's step functions, virtual work, but they are all bookkeeping devices over the same integration.

The cantilever, done in full

Take a cantilever of length $L$ , built in at $x = 0$ , carrying a point load $P$ downwards at the free end.

Cutting at $x$ and using the right-hand free body, the moment is $M (x) = - P (L - x)$ , hogging throughout. So

E I v'' = - P (L - x)

Integrating, $E I v' = - P L x + P x^{2} / 2 + C_{1}$ . The built-in end has zero slope, so $v' (0) = 0$ gives $C_{1} = 0$ . Integrating again, $E I v = - P L x^{2} / 2 + P x^{3} / 6 + C_{2}$ , and $v (0) = 0$ gives $C_{2} = 0$ . At the tip,

v (L) = \frac{1}{E I} (- \frac{P L^{3}}{2} + \frac{P L^{3}}{6}) = - \frac{P L^{3}}{3 E I}

The minus sign says downwards, and the magnitude $P L^{3} / 3 E I$ is the standard result. The cube on the length is the thing to carry away: doubling a cantilever's projection multiplies its tip deflection by eight at the same load, and by sixteen if the load is a uniform one that also doubles in total.

Example. A steel cantilever projects $2.5$ m and carries $15$ kN at its tip. Its second moment of area is $8 \times 10^{7}$ mm⁴ and $E = 210$ GPa. Find the tip deflection.

Working in newtons and millimetres, $v = 15 000 \times 2500^{3} / (3 \times 210 000 \times 8 \times 10^{7}) = 4.65$ mm. Against a projection of $2500$ mm that is span over $538$ , which for a cantilever is usually assessed against the projection doubled, and is comfortable.

Now you. A steel cantilever projects $3$ m with $10$ kN at the tip and $I = 1.2 \times 10^{8}$ mm⁴. Find the tip deflection.

Answer

$v = 10 000 \times 3000^{3} / (3 \times 210 000 \times 1.2 \times 10^{8}) = 3.57$ mm.

Repeating the integration with a uniform load $w$ over the whole cantilever, where $M (x) = - w (L - x)^{2} / 2$ , gives a tip deflection of $w L^{4} / 8 E I$ .

The simply supported beam

For a simply supported span $L$ under a uniform load $w$ , the moment was found in the third lesson as $M (x) = w L x / 2 - w x^{2} / 2$ . So

E I v'' = \frac{w L x}{2} - \frac{w x^{2}}{2}

Integrate: $E I v' = w L x^{2} / 4 - w x^{3} / 6 + C_{1}$ . Integrate again: $E I v = w L x^{3} / 12 - w x^{4} / 24 + C_{1} x + C_{2}$ . Now $v (0) = 0$ gives $C_{2} = 0$ , and $v (L) = 0$ gives $w L^{4} / 12 - w L^{4} / 24 + C_{1} L = 0$ , so $C_{1} = - w L^{3} / 24$ . By symmetry the maximum is at midspan, and substituting $x = L / 2$ ,

v_{\max} = - \frac{5 w L^{4}}{384 E I}

The fourth power of the span is why doubling a span is so much more expensive than it looks: the same beam under the same load per metre deflects sixteen times as far, and it also carries four times the moment.

Example. A steel beam spans $8$ m and carries $12$ kN m⁻¹, with $I = 2.5 \times 10^{8}$ mm⁴ and $E = 210$ GPa. Find the maximum deflection and compare it with the usual limits.

Convert: $w = 12$ N mm⁻¹ and $L = 8000$ mm. Then

v = \frac{5 \times 12 \times 8000^{4}}{384 \times 210 000 \times 2.5 \times 10^{8}} = 12.2 mm

The common limits are span over $360$ for the deflection caused by imposed load on a floor and span over $250$ for the total, which here are $22.2$ mm and $32$ mm. The beam passes.

Now you. A steel beam spans $10$ m under $8$ kN m⁻¹ with $I = 3.5 \times 10^{8}$ mm⁴. Find the maximum deflection and check it against span over $360$ .

Answer

$v = 5 \times 8 \times 10 000^{4} / (384 \times 210 000 \times 3.5 \times 10^{8}) = 14.2$ mm, against a limit of $10 000 / 360 = 27.8$ mm. It passes comfortably.

Two more standard cases complete the set most design work needs. A simply supported beam with a central point load $P$ deflects $P L^{3} / 48 E I$ , and one with a moment $M_{0}$ applied at one end deflects a maximum of $M_{0} L^{2} / (9 \sqrt{3} E I)$ .

Superposition, and sizing for stiffness

Because the beam equation is linear in $v$ , the deflection under several loads is the sum of the deflections under each separately. That turns a table of four or five standard cases into a method for almost any real loading: split the load into recognisable pieces, look each one up, and add.

Superposition also inverts cleanly, which is how beams get sized for stiffness rather than strength. Rearranging the uniform load result for $I$ ,

I_{required} = \frac{5 w L^{4}}{384 E v_{limit}}

Example. A floor beam spans $6$ m under $10$ kN m⁻¹ and must not deflect more than span over $360$ . What second moment of area does it need?

The limit is $6000 / 360 = 16.7$ mm, so $I = 5 \times 10 \times 6000^{4} / (384 \times 210 000 \times 16.7) = 4.82 \times 10^{7}$ mm⁴. Compare that with the strength requirement: the moment is $10 \times 36 / 8 = 45$ kN m, so at $275$ MPa the section modulus needed is $164 000$ mm³, which for a beam $300$ mm deep corresponds to $I$ of about $2.5 \times 10^{7}$ mm⁴. Stiffness asks for roughly twice as much section as strength does, which is typical for steel floors and is why deflection, not stress, usually chooses the beam.

Now you. A floor beam spans $5$ m under $12$ kN m⁻¹ with the same limit of span over $360$ . What second moment of area does it need?

Answer

The limit is $5000 / 360 = 13.9$ mm, so $I = 5 \times 12 \times 5000^{4} / (384 \times 210 000 \times 13.9) = 3.35 \times 10^{7}$ mm⁴.

The reason is a scaling one. Stress limits are set by material strength, which for steel has risen over the last century; deflection limits are set by $E$ , which has not moved at all. Higher grade steel buys strength and buys nothing whatever in stiffness, so as grades improve, more and more members end up sized by deflection.

A shortcut used constantly on site is the span-to-depth ratio. Because $I$ scales as $d^{3}$ and $Z$ as $d^{2}$ , the deflection under a given load per metre depends mostly on $L / d$ , and a rule such as "a simply supported steel floor beam at about $L / 20$ , a timber joist at about $L / 16$ , a reinforced concrete slab at about $L / 26$ " gets a first size in one step. It is a way to start, not a way to finish.

The propped cantilever

The first lesson set aside structures with more reactions than equations. Deflection supplies the missing ones.

Example. A beam of span $L$ is built in at $A$ and simply propped at $B$ , carrying a uniform load $w$ . Find the reaction at the prop.

There are four unknowns, two forces and a fixing moment at $A$ , and one force at $B$ , against three equations. Take the prop force $R$ as the redundant, remove the prop, and treat the result as a plain cantilever carrying two things: the uniform load, and an upward point load $R$ at the tip.

Under $w$ alone the tip would drop $w L^{4} / 8 E I$ . Under $R$ alone it would rise $R L^{3} / 3 E I$ . The prop is rigid, so the true deflection there is zero, and that is the compatibility condition:

\frac{w L^{4}}{8 E I} = \frac{R L^{3}}{3 E I} \Rightarrow R = \frac{3 w L}{8}

Everything else follows from statics. The built-in end takes $w L - 3 w L / 8 = 5 w L / 8$ , and its fixing moment is $w L^{2} / 2 - (3 w L / 8) L = w L^{2} / 8$ , hogging. The largest sagging moment works out at $9 w L^{2} / 128$ , about $56$ per cent of the simply supported value, and the maximum deflection is $w L^{4} / 185 E I$ , about two fifths of the simply supported one. Adding a prop buys a great deal.

Note what disappeared: $E$ and $I$ cancelled, so the answer holds for any material and any section, provided the beam is uniform. That cancellation is a general feature of indeterminate structures with one material throughout, and it is why the classical results can be tabulated at all.

Now you. A propped cantilever of $6$ m span carries $20$ kN m⁻¹. Find the prop reaction, the reaction at the built-in end, and the fixing moment.

Answer

$R = 3 \times 20 \times 6 / 8 = 45$ kN at the prop, so the built-in end takes $120 - 45 = 75$ kN, and the fixing moment is $w L^{2} / 8 = 20 \times 36 / 8 = 90$ kN m, hogging.

What the integration leaves out

Three limits are worth stating.

The equation used only bending. Beams also deflect because of shear distortion, which the derivation ignored entirely. For a slender beam that is right to within a per cent or two, but for a span less than about ten times the depth the shear contribution becomes significant, and for a sandwich panel with a soft core it can dominate.

The small slope approximation was made explicitly, so the results describe deflections that are small compared with the span. A fishing rod, a leaf spring or a diving board bends far enough that the exact curvature has to be kept, and the resulting elastica problem has no elementary solution.

And $E I$ was assumed constant and known. For steel that is fine. For reinforced concrete it is not, because the section cracks in tension under service load and its effective $I$ drops to somewhere between a third and a half of the uncracked value, and it goes on deflecting for years afterwards under creep, typically doubling the initial figure. Concrete deflection calculations therefore carry uncertainties of tens of per cent, which is why concrete codes prefer to control deflection with span-to-depth rules rather than by calculating it.

Every result so far, in bending and in deflection alike, has assumed the beam fails when the material gives way. For a member in compression that assumption is simply false, and the next lesson shows how badly.