Irodov 1.301: A Cantilever Under a Couple, Not a Force

Problem Irodov 1.301

At small bendings, an elastic rod’s curve obeys the relation

N(x) = EI\,\frac{d^2y}{dx^2},

where $x$ is measured along the rod from a fixed origin, $y$ is the transverse deflection of the rod’s centreline at that point, $N(x)$ is the internal bending moment at the cross-section at $x$ , $E$ is Young’s modulus, and $I=\int z^2\,dS$ is the moment of inertia of the cross-section, $z$ being the distance of an area element $dS$ from the neutral layer, the rod’s unbent middle surface. A steel rod of square cross-section, side $a$ , is built into a wall at one end (taken as the origin $x=0$ ), with a protruding section of length $l$ and negligible mass. Find the shape of the elastic curve $y(x)$ and the deflection $\lambda$ of the free end, for each of the following loadings applied at the free end:

(a)
a pure bending couple of moment $N_0$ ;
(b)
a transverse force $F$ , perpendicular to the rod’s original direction.

The rod, embedded at O, protrudes a length l and sags to the free end A. Axes x, y (down) originate at O; λ is the deflection of A. — The rod, embedded at $O$ , protrudes a length $l$ and sags to the free end $A$ . Axes $x$ , $y$ (down) originate at $O$ ; $\lambda$ is the deflection of $A$ .

Given

square rod, side $a$ , protruding length $l$

Find

(a) the curve and

\lambda

for an end couple

N_0

; (b) the curve and

\lambda

for an end force

F

Solution

Step 1The moment of inertia of the cross-section

For the square cross-section, $I=\int z^2\,dS$ over a square of side $a$ about an axis through its center is the standard result

I = \int_{-a/2}^{a/2} z^2\,(a\,dz) = a\left[\frac{z^3}{3}\right]_{-a/2}^{a/2} = \frac{a^4}{12}.

This value of $I$ is used in both parts below.

Step 2Part (a): a pure couple

N_0

at the end

Cut the rod anywhere between the wall and $A$ . With no transverse force applied anywhere on the rod, moment balance on the cut piece requires the internal bending moment to equal $N_0$ at every cross-section:

N(x) = N_0 \qquad (0\le x\le l).

Substituting into the governing equation and integrating twice, using the built-in conditions at the wall, zero slope and zero deflection at $x=0$ :

\begin{aligned} EI\,y'' &= N_0\\ EI\,y' &= N_0 x + C_1, \qquad y'(0)=0 \implies C_1 = 0\\ EI\,y &= \frac{N_0 x^2}{2} + C_2, \qquad y(0)=0 \implies C_2 = 0 \end{aligned}

y(x) = \frac{N_0}{2EI}x^2

(1)

The elastic curve is a parabola. The deflection of the free end is $y(l)$ :

\lambda_a = \frac{N_0 l^2}{2EI}.

Step 3Part (b): a transverse force

F

at the end

Now cut the rod at $x$ ; the piece from $x$ to $l$ carries the load $F$ at its far end, a distance $(l-x)$ from the cut, so

N(x) = F(l-x).

Integrating twice with the same built-in conditions:

\begin{aligned} EI\,y'' &= F(l-x)\\ EI\,y' &= F\left(lx-\frac{x^2}{2}\right)+C_1, \qquad y'(0)=0\implies C_1=0\\ EI\,y &= F\left(\frac{lx^2}{2}-\frac{x^3}{6}\right)+C_2, \qquad y(0)=0\implies C_2=0 \end{aligned}

y(x) = \frac{F}{2EI}\left(l-\frac{x}{3}\right)x^2

(2)

The deflection of the free end:

\lambda_b = y(l) = \frac{F}{2EI}\left(l-\frac{l}{3}\right)l^2 = \frac{Fl^3}{3EI}.

Step 4Both results together

With $I=a^4/12$ in each case,

\lambda = \begin{cases} \dfrac{N_0 l^2}{2EI}, & \text{end couple } N_0\\[8pt] \dfrac{Fl^3}{3EI}, & \text{end force } F\end{cases}

(3)

Check

Dimensions: $[N_0 l^2/(EI)] = (\text{N}\cdot\text{m})(\text{m}^2)/(\text{Pa}\cdot\text{m}^4) = \text{m}$ , and $[Fl^3/(EI)]=(\text{N})(\text{m}^3)/(\text{Pa}\cdot\text{m}^4)=\text{m}$ , both lengths as they must be. Also, $\lambda_b$ is exactly what a cantilever’s standard end-load formula gives, and setting $F=0$ or $N_0=0$ in either case correctly returns $\lambda = 0$ : an unloaded rod stays straight.

Remark

The extra power of $l$ in $\lambda_b$ over $\lambda_a$ traces back to the bending moment itself: a couple applies the same $N_0$ at every cross-section, so integrating $y''=N/EI$ twice picks up two factors of $x$ from the integration alone. A transverse force instead makes $N(x)=F(l-x)$ , where the single extra power of $l$ comes from the load being a force rather than a couple, since the moment scale is $Fl$ in place of $N_0$ ; the two integrations of $y''=N/EI$ contribute the same two powers of $x$ in both the couple and force cases, leaving the deflection one power of $l$ steeper overall. The same distinction, a constant moment against one that grows linearly along the beam, is what separates a pure-bending problem from a cantilevered load problem generally.