Irodov 1.304: The Load That Grows with Radius – Amit & Deepali Sinha Foundation

Problem Irodov 1.304

A steel plate of thickness $h$ is a square of side $l$ , with $h \ll l$ . One whole edge of the plate is held rigidly by a vertical axle $OO$ and lies along that axle, so the plate stands in a plane containing the axle, reaches a distance $l$ out from it, and has its free edge parallel to it. The axle is turned with a constant angular acceleration $\beta$ . Taking the density $\rho$ and Young’s modulus $E$ of steel as known, find the deflection $\lambda$ of the plate’s free edge, the sagging being small.

The plate seen from above, looking down the axle at O: it reaches l outward, its equal side running into the page. Hatching marks the held edge, the curved arrow the spin-up β, the dashed curve the deflected surface, sagging by λ. — The plate seen from above, looking down the axle at $O$ : it reaches $l$ outward, its equal side running into the page. Hatching marks the held edge, the curved arrow the spin-up $\beta$ , the dashed curve the deflected surface, sagging by $\lambda$ .

The bending load, on the same plan view. Its intensity vanishes at the axle and grows linearly with distance from it, reaching w_0 at the free edge, so the arrow tails lie on a straight line. Each arrow is a force per unit length. — The bending load, on the same plan view. Its intensity vanishes at the axle and grows linearly with distance from it, reaching $w_0$ at the free edge, so the arrow tails lie on a straight line. Each arrow is a force per unit length.

Given

$l$
$h\ll l$
$\beta$
steel: $\rho$ , $E$ known

Find

Find the deflection

\lambda

of the free edge

Solution

Step 1The plate is a cantilever under a triangular load

Work in the frame turning with the plate. There every element is at rest under two inertial forces: the centrifugal one, which is radial and lies in the plate’s own plane, and the Euler one, which at the distance $x$ from the axle has the magnitude $\beta x$ per unit mass, points backwards against the sense of the spin-up, and is tangential and therefore perpendicular to the plate. The second is the one that bends it. Measure $x$ outward from the axle and take the slice of plate between $x$ and $x+dx$ , which spans the full height $l$ and the full thickness $h$ .

\begin{align*}dm &= \rho h l\,dx \\ df &= (\beta x)\,dm \\ &= \rho h l\beta x\,dx\end{align*}

Dividing that elementary force by the width $dx$ of the slice it acts on gives the load per unit length.

w(x) = \rho h l\beta x

(1)

Its largest value sits at the free edge and is worth a name of its own.

w_0 = w(l) = \rho h\beta l^{2} \tag{2}

The plate's cross-section: a rectangle of height l measured along the axle and thickness h. The load w pushes across the thickness, and the dashed center line parallel to the side l is the neutral line. — The plate’s cross-section: a rectangle of height $l$ measured along the axle and thickness $h$ . The load $w$ pushes across the thickness, and the dashed center line parallel to the side $l$ is the neutral line.

Step 2The bending stiffness of the section

The section that resists this bending is a rectangle whose long side $l$ lies along the neutral line and whose short side $h$ lies along the direction the load pushes, so it is the thickness that enters cubed.

\begin{aligned} I &= \frac{l h^{3}}{12} \\ EI &= \frac{E l h^{3}}{12} \end{aligned} \tag{3}

The square’s upper and lower edges are free, so the section is treated as a beam; a plate prevented from contracting sideways would be stiffer by the factor $1/(1-\nu^{2})$ , and no Poisson ratio appears in a beam.

The lever arm. The dashed line is the section at x; the whole plate is drawn, but the moment counts only the elements beyond it. The element dm of width dξ sits at ξ, and its arm about the section is ξ-x. — The lever arm. The dashed line is the section at $x$ ; the whole plate is drawn, but the moment counts only the elements beyond it. The element $dm$ of width $d\xi$ sits at $\xi$ , and its arm about the section is $\xi-x$ .

Step 3The bending moment carried by a section

Cut the plate at $x$ and keep the outboard piece. The moment the section transmits is the moment about that section of every elementary force beyond it, each one weighted by its own arm $\xi-x$ , with the load intensity taken from (1).

\begin{align*}M(x) &= \int_x^l w(\xi)\,(\xi-x)\,d\xi \\ &= \rho h l\beta\int_x^l \bigl(\xi^{2}-x\xi\bigr)\,d\xi \\ &= \rho h l\beta\left[\frac{\xi^{3}}{3}-\frac{x\xi^{2}}{2}\right]_{x}^{l}\end{align*}

M(x) = \rho h l\beta\left(\frac{l^{3}}{3}-\frac{l^{2}x}{2}+\frac{x^{3}}{6}\right)

(4)

Step 4The deflected shape from the moment

The Euler-Bernoulli relation makes the curvature proportional to the moment (4). The held edge neither moves nor turns, which supplies both constants of integration,

y(0) = 0, \qquad y'(0) = 0,

so integrating once from the held edge gives the slope and twice gives the deflection.

\begin{align*}EI\,y''(x) &= M(x) \\ EI\,y'(x) &= \int_0^x M(t)\,dt \\ &= \rho h l\beta\left(\frac{l^{3}x}{3}-\frac{l^{2}x^{2}}{4}+\frac{x^{4}}{24}\right) \\ EI\,y(x) &= \int_0^x EI\,y'(t)\,dt \\ &= \rho h l\beta\left(\frac{l^{3}x^{2}}{6}-\frac{l^{2}x^{3}}{12}+\frac{x^{5}}{120}\right)\end{align*}

The deflection asked for is this at the free edge, where the three fractions collapse into one.

\begin{align*}EI\lambda &= \rho h l\beta\, l^{5}\left(\frac{1}{6}-\frac{1}{12}+\frac{1}{120}\right) \\ &= \rho h l\beta\, l^{5}\cdot\frac{20-10+1}{120} \\ &= \frac{11}{120}\,\rho h\beta l^{2}\cdot l^{4}\end{align*}

\lambda = \frac{11}{120}\,\frac{w_0\,l^{4}}{EI}

(5)

Step 5The deflection in the plate’s own dimensions

Substituting the section’s $I$ from (3) and the peak load $w_0$ from (2) into (5) clears the geometry out of the answer.

\begin{align*}\lambda &= \frac{11}{120}\,\frac{(\rho h\beta l^{2})\,l^{4}}{E\,(l h^{3}/12)} \\ &= \frac{11}{120}\cdot\frac{12\,\rho\beta l^{5}}{E h^{2}}\end{align*}

\lambda = \frac{11\,\rho\beta l^{5}}{10\,E h^{2}} = 1.1\,\frac{\rho\beta l^{5}}{E h^{2}}

(6)

Check

Dimensions first: $\rho\beta l^{5}/(Eh^{2})$ carries $(\text{kg}/\text{m}^{3})(1/\text{s}^{2})(\text{m}^{5})$ over $\bigl(\text{kg}/(\text{m}\,\text{s}^{2})\bigr)(\text{m}^{2})$ , which is $(\text{kg}\,\text{m}^{2}/\text{s}^{2})/(\text{kg}\,\text{m}/\text{s}^{2}) = \text{m}$ , a length. Second, a bound the answer must respect: replace the true load by the uniform load $w_0$ over the whole span, which is nowhere smaller than the true one, and the standard uniformly loaded cantilever gives $w_0l^{4}/(8EI)$ , that is $1.5\,\rho\beta l^{5}/(Eh^{2})$ . The boxed $1.1$ sits below that ceiling, as a load that thins out toward the held edge must.

Remark

The thing worth keeping is where the lever arm starts. The material beyond a section presses on it with a shear force as well as with a couple, and only the couple bends the fibers, so the arm in (4) runs from the cut. Measuring it from the axle instead adds the transmitted shear times its own distance out, and it changes the beam coefficient from $11/120$ to $3/20$ , a factor $18/11$ that would inflate the boxed $1.1$ to $1.8$ . The slip is an easy one to make on a spinning body, where every other question asked of it is naturally answered about the rotation axis.