Irodov 1.302: Two Supports Beat One by Sixteen – Amit & Deepali Sinha Foundation

Problem Irodov 1.302

A steel girder of length $l$ rests freely on two supports, one at each end, a pin at one end and a roller at the other, so neither end resists rotation. The moment of inertia of its cross-section about the neutral axis is $I$ , exactly as defined in Problem 1.301. Neglecting the girder’s own mass and assuming the sag stays slight, find the deflection $\lambda$ at the middle of the girder produced by a force $F$ applied there, perpendicular to the girder’s length.

The girder, pinned at one end and free to roll at the other, sagging under the central force F. The dashed curve is the deflected shape (exaggerated); λ, the deflection the problem asks for, is the sag at the midpoint, directly below F. — The girder, pinned at one end and free to roll at the other, sagging under the central force $F$ . The dashed curve is the deflected shape (exaggerated); $\lambda$ , the deflection the problem asks for, is the sag at the midpoint, directly below $F$ .

Given

length $l$
moment of inertia $I$ (Problem 1.301)
force $F$ at the midpoint
steel, mass of girder neglected

Find

\lambda

, the deflection at the midpoint

Solution

Step 1The support reactions

The whole girder as a free body for this step. The two supports answer the central load F with the upward reactions R_A and R_B, and the sag is left out of the drawing, which is a statement of equilibrium only. — The whole girder as a free body for this step. The two supports answer the central load $F$ with the upward reactions $R_A$ and $R_B$ , and the sag is left out of the drawing, which is a statement of equilibrium only.

By symmetry the two supports share the central load equally; equilibrium of the whole girder confirms it. Balancing the vertical forces, then taking moments about the support $B$ ,

\begin{align*}R_A+R_B &= F \\ R_A\cdot l &= F\cdot\frac{l}{2}\end{align*}

and the second of these gives $R_A$ at once, the first then giving $R_B$ .

R_A = R_B = \frac{F}{2}

Step 2The internal bending moment

Measure $x$ from support $A$ , and let $y(x)$ , positive in the sag direction, obey the governing relation of Problem 1.301, $N(x)=EI\,y''$ , with $N$ carrying the sign of the transverse force that produces it. Cutting the girder at $0\le x\le l/2$ and isolating the segment nearer $A$ , the only force on that segment is the upward reaction $R_A$ , a distance $x$ from the cut.

N(x) = EI\,y'' = -R_A\,x = -\frac{F}{2}x, \qquad 0\le x\le\frac{l}{2}.

Step 3Integrating with the support and symmetry conditions

The girder is simply supported, so $y(0)=0$ ; by the left-right symmetry of the load and the supports, the slope must vanish at the midpoint, the point of greatest sag. Integrating twice and fixing each constant from these two conditions:

\begin{aligned} EI\,y' &= -\frac{F}{4}x^2+C_1, & y'\!\left(\frac{l}{2}\right)=0 \;\Longrightarrow\; C_1=\frac{Fl^2}{16}\\[4pt] EI\,y &= -\frac{F}{12}x^3+\frac{Fl^2}{16}x+C_2, & y(0)=0 \;\Longrightarrow\; C_2=0 \end{aligned}

Step 4The deflection at the midpoint

The maximum sag $\lambda$ is $y(l/2)$ , the same point where the slope above was set to zero.

EI\,y\!\left(\frac{l}{2}\right) = -\frac{F}{12}\left(\frac{l}{2}\right)^3+\frac{Fl^2}{16}\cdot\frac{l}{2} = -\frac{Fl^3}{96}+\frac{Fl^3}{32} = \frac{Fl^3}{48}.

\lambda = \frac{Fl^3}{48EI}

(1)

Check

Dimensions: $[Fl^3/(EI)] = (\text{N})(\text{m}^3)/\bigl[(\text{N/m}^2)(\text{m}^4)\bigr] = \text{m}$ , a length, as $\lambda$ must be. And a route that never integrates reproduces the box: since the slope vanishes at the midpoint, each half of the girder behaves as a cantilever of span $l/2$ built in at that midpoint and carrying its own support reaction $F/2$ at the far tip, so Problem 1.301(b)’s tip formula gives $(F/2)(l/2)^3/(3EI) = Fl^3/(48EI)$ , the same $\lambda$ .

Remark

The same force $F$ produces a tip deflection $Fl^3/(3EI)$ on a cantilever of the same length $l$ , Problem 1.301(b). The simply supported girder here sags only $1/16$ as much for the same $F$ and $l$ : sharing the bending between two supports, one on each side of the load, is sixteen times stiffer than carrying the whole span from a single built-in end. That factor of sixteen is pure geometry. Both deflections carry $E$ and $I$ in exactly the same place, so the two moduli cancel out of the ratio entirely: no choice of material and no reshaping of the cross-section can shift it, and only where the girder is held, and by how many supports, buys the gain.