Irodov 1.300: Bowed by Its Own Weight – Amit & Deepali Sinha Foundation

Problem Irodov 1.300

A steel beam of rectangular cross-section is built horizontally into a wall at one end, with a length $l = 6.0$ m protruding freely, and sags very slightly under its own weight. The beam’s cross-section has thickness (height) $h = 10$ cm. Treating the steel’s density and Young’s modulus as known material constants, find the radius of curvature of the beam’s neutral layer at the point where it emerges from the wall.

The cantilevered girder, one end built into the wall at O; the dashed curve traces the sagged neutral layer (exaggerated), l the protruding length and h the cross-sectional thickness. — The cantilevered girder, one end built into the wall at $O$ ; the dashed curve traces the sagged neutral layer (exaggerated), $l$ the protruding length and $h$ the cross-sectional thickness.

Given

$l = 6.0$ m
$h = 10$ cm $= 0.10$ m
steel: $\rho$ , $E$ known

Find

the radius of curvature

R

of the neutral layer at

O

Solution

Step 1The bending moment at the wall

Treat the protruding length as a cantilever loaded by its own weight, a uniformly distributed load $q$ per unit length, where $q = \rho A g$ , $A$ the cross-sectional area and $\rho$ the steel density. Cutting the beam at a distance $x$ from the wall, the overhanging piece beyond the cut has length $l-x$ and weight $q(l-x)$ acting at its own midpoint, so the bending moment there is

M(x) = q(l-x)\cdot\frac{l-x}{2} = \frac{q(l-x)^2}{2}.

At the wall, $x=0$ , this is

M(0) = \frac{q l^2}{2}. \tag{1}

Step 2The cross-section’s second moment of area

For a rectangular cross-section of width $b$ (into the page) and thickness $h$ , bending about the horizontal axis through its centroid,

I = \frac{b h^3}{12}. \tag{2}

The self-weight load per unit length is the density times the cross-sectional area times $g$ .

q = \rho (bh) g \tag{3}

Step 3Curvature from the bending moment

The Euler-Bernoulli relation between curvature and bending moment, $1/R = M/(EI)$ , evaluated at $O$ with (1), (2) and (3), gives

\frac{1}{R} = \frac{M(0)}{EI} = \frac{q l^2/2}{E\,bh^3/12} = \frac{\rho b h g\, l^2/2}{E\,bh^3/12}.

\dfrac{1}{R} = \dfrac{6\rho g l^2}{E h^2}

(4)

Step 4Evaluating

Using (4) and the steel values $\rho = 7.8\times10^3\ \text{kg/m}^3$ , $E = 2.0\times10^{11}$ Pa, $g = 9.8\ \text{m/s}^2$ , $l = 6.0$ m, $h = 0.10$ m,

\frac{1}{R} = \frac{6(7.8\times10^3)(9.8)(6.0)^2}{(2.0\times10^{11})(0.10)^2} = \frac{1.651\times10^7}{2.0\times10^9} = 8.25\times10^{-3}\ \text{m}^{-1}.

Taking the reciprocal gives the radius of curvature itself.

R = \frac{1}{8.25\times10^{-3}\ \text{m}^{-1}} = 1.21\times10^{2}\ \text{m}.

R = \dfrac{E h^2}{6\rho g l^2} = 1.2\times10^{2}\ \text{m} = 0.12\ \text{km}

(5)

Check

Units: $[E]/([\rho][g][l^2]) = (\text{N/m}^2)/\big((\text{kg/m}^3)(\text{m/s}^2)(\text{m}^2)\big) = \big(\text{kg}/(\text{m}\,\text{s}^2)\big)/(\text{kg/s}^2) = \text{m}^{-1}$ , so $h^2E/(6\rho g l^2)$ carries one extra $\text{m}^2$ against that $\text{m}^{-1}$ and comes out a length, as $R$ must be. Also $R\propto 1/l^2$ : doubling the protruding length quarters the radius of curvature at the wall, since both the load and its lever arm at $O$ grow with $l$ .

Remark

A concentrated tip load would give a bending moment linear in the protruding length, $M=Fl$ , since only its lever arm grows with $l$ . Here the load is the beam’s own weight, distributed along the whole span, so both the total weight ( $\propto l$ ) and its lever arm about $O$ ( $\propto l$ ) contribute a factor of $l$ each, giving the quadratic $M(0)=ql^2/2$ found above. Self-weight is therefore doubly punishing for a longer cantilever, compared to carrying a fixed load at the tip.