Irodov 1.303: Self-Weight Bends a Girder Two Ways – Amit & Deepali Sinha Foundation

Problem Irodov 1.303

A rectangular steel girder has thickness (cross-sectional height) $h$ , and its own weight is now the only load on it, distributed uniformly along its length. Using the same governing relation as Problem 1.301, $N(x)=EI\,d^2y/dx^2$ , find the deflection $\lambda$ this self-weight produces in two separate cases:

the girder is built into a wall at one end, with a protruding section of length $l$ , and $\lambda$ is the drop of the free end below its original, undeflected position
the girder, of total length $2l$ , rests freely on two end supports, a pin at one and a roller at the other, and $\lambda$ is the sag at the midpoint

Case (a): the cantilever, built in at the wall, sagging under its own distributed weight q per unit length. The dashed curve is the deflected shape, drawn exaggerated, and not the undeflected reference direction; λ is the drop of its free end. — Case (a): the cantilever, built in at the wall, sagging under its own distributed weight $q$ per unit length. The dashed curve is the deflected shape, drawn exaggerated, and not the undeflected reference direction; $\lambda$ is the drop of its free end.

Case (b): the same girder doubled to length 2l and simply supported at both ends, sagging under the same distributed self-weight. Again the dashed curve is the exaggerated deflected shape; λ is its sag at the midpoint. — Case (b): the same girder doubled to length $2l$ and simply supported at both ends, sagging under the same distributed self-weight. Again the dashed curve is the exaggerated deflected shape; $\lambda$ is its sag at the midpoint.

Given

rectangular cross-section, thickness $h$
steel: $\rho$ , $E$ known
self-weight only, no other load

Find

(a)

\lambda

for a cantilever of length

l

, (b)

\lambda

for a simply supported girder of length

2l

Solution

Step 1The cross-section and the distributed self-weight

For a rectangular cross-section of width $b$ and thickness $h$ , bending about the horizontal centroidal axis, the moment of inertia and the self-weight per unit length are, exactly as in Problem 1.300,

I = \frac{bh^3}{12}, \qquad q = \rho(bh)g, \tag{1}

where $\rho$ is the density of steel. Both cases below reuse these.

Step 2Part (a): the cantilever’s governing equation

Cut the protruding section at distance $x$ from the wall, $0\le x\le l$ . The moment there is carried by the overhanging weight beyond the cut, exactly as found in Problem 1.300, and the girder is built in at $x=0$ , so both the deflection and the slope vanish there.

EI\,y'' = N(x) = \frac{q}{2}(l-x)^2, \qquad y(0)=0,\ \ y'(0)=0. \tag{2}

Step 3Part (a): integrating to the tip

Integrating twice and fixing the constants from the wall conditions:

\begin{aligned} EI\,y' &= -\frac{q}{6}(l-x)^3+C_1, & y'(0)=0 \;\Longrightarrow\; C_1=\frac{ql^3}{6}\\[4pt] EI\,y &= \frac{ql^3}{6}x+\frac{q}{24}(l-x)^4+C_2, & y(0)=0 \;\Longrightarrow\; C_2=-\frac{ql^4}{24} \end{aligned}

The tip deflection is $y(l)$ , where the $(l-x)^4$ term vanishes.

EI\,y(l) = \frac{ql^3}{6}\cdot l-\frac{ql^4}{24} = \frac{ql^4}{6}-\frac{ql^4}{24} = \frac{ql^4}{8}.

\lambda_a = \frac{ql^4}{8EI}

(3)

Step 4Part (b): the simply supported girder’s governing equation

The girder now has total length $2l$ , and by symmetry each support carries half the total weight $R$ . Measure $x$ from the left support, $0\le x\le l$ (the near half, by symmetry). The segment nearer that support carries the upward reaction $R$ and the downward weight $qx$ of the piece itself, with the sign convention of Problem 1.302.

R = \frac{q(2l)}{2} = ql,

Taking moments about the cut, that reaction and the segment’s own weight give the bending moment there, and the support and the symmetry supply the two conditions the integration will need.

EI\,y'' = N(x) = -R\,x+(qx)\cdot\frac{x}{2} = \frac{q}{2}x^2-qlx, \qquad y(0)=0,\ \ y'(l)=0\ (\text{symmetry}).

Step 5Part (b): integrating to the midpoint

\begin{aligned} EI\,y' &= \frac{q}{6}x^3-\frac{ql}{2}x^2+C_1, & y'(l)=0 \;\Longrightarrow\; C_1=\frac{ql^3}{3}\\[4pt] EI\,y &= \frac{q}{24}x^4-\frac{ql}{6}x^3+\frac{ql^3}{3}x+C_2, & y(0)=0 \;\Longrightarrow\; C_2=0 \end{aligned}

The midpoint deflection is $y(l)$ .

EI\,y(l) = \frac{q}{24}l^4-\frac{ql}{6}l^3+\frac{ql^3}{3}l = ql^4\left(\frac{1}{24}-\frac{1}{6}+\frac{1}{3}\right) = \frac{5}{24}ql^4.

\lambda_b = \frac{5ql^4}{24EI}

(4)

Step 6Eliminating the unknown cross-section

Both boxed results still carry the width $b$ hidden inside $q$ and $I$ . Dividing $q$ by $I$ from (1) removes it.

\frac{q}{I} = \frac{\rho(bh)g}{bh^3/12} = \frac{12\rho g}{h^2}. \tag{5}

Substituting this into (3) and (4) gives both deflections purely in terms of $\rho$ , $g$ , $l$ , $E$ and $h$ .

\lambda_a = \frac{l^4}{8E}\cdot\frac{12\rho g}{h^2}, \qquad \lambda_b = \frac{5l^4}{24E}\cdot\frac{12\rho g}{h^2}.

For (a), the cantilever of length $l$ ,

\lambda_a = \frac{3}{2}\frac{\rho g l^4}{Eh^2}

(6)

For (b), the simply supported girder of length $2l$ ,

\lambda_b = \frac{5}{2}\frac{\rho g l^4}{Eh^2}

(7)

Check

The cantilever’s governing equation (2) at $x=0$ gives the curvature at the wall, $N(0)/(EI)=ql^2/(2EI)$ , which after the substitution (5) is $6\rho g l^2/(Eh^2)$ , exactly Problem 1.300’s boxed curvature for the same cantilever. And $\rho g l^4/(Eh^2)$ carries units of (kg/m $^3$ )(m/s $^2$ )(m $^4$ )/(Pa $\cdot$ m $^2$ ), which reduces to a meter, correct for both boxed deflections.

Remark

The two boxed results are not directly comparable term for term, because the problem deliberately compares a cantilever of length $l$ against a simply supported girder of twice that length, $2l$ . Written instead at equal total span $L$ , a self-loaded cantilever of length $L$ sags $qL^4/(8EI)$ at its tip while a simply supported girder of the same length $L$ sags only $5qL^4/(384EI)$ at its center, a factor of $48/5\approx9.6$ stiffer for the same span and the same weight per unit length: a second support removes far more sag than the doubled span and doubled weight of case (b) give back, which is exactly why case (b) still comes out only $5/3$ times case (a) despite carrying twice the length and twice the total weight.