A cantilever beam is loaded by a force F at its free end.…

A cantilever beam is loaded by a force $F$ at its free end. Find the equation of the elastic curve and the deflection of the free end. Length $l = 6\text{ m}$, $F = 2\text{ t} = 2000\text{ kg}$, modulus of elasticity $E = 2{,}100{,}000\text{ kg/cm}^2$, moment of inertia $I = 30{,}000\text{ cm}^4$.

Step 1: Coordinate system and bending moment

Place the origin at the fixed end (wall), with $x$ toward the free end, $0\le x\le l$. At a section a distance $x$ from the wall, the load $F$ at the tip produces a hogging moment (negative under the sagging-positive convention):

\[ M(x) = -F(l-x). \tag{1} \]

Step 2: Differential equation of the elastic curve

The Euler–Bernoulli beam equation $EI\,y''=M(x)$, with $y(x)$ the downward deflection, gives: \[EI\,y'' = -F(l-x). \tag{2}\]

Step 3: First integration — slope

\[ EI\,y' = -F\!\left(lx-\frac{x^2}{2}\right) + C_1. \]

The wall is rigidly clamped, so the slope vanishes there: $y'(0)=0 \Rightarrow C_1=0$.

Step 4: Second integration — deflection

\[ EI\,y = -F\!\left(\frac{lx^2}{2}-\frac{x^3}{6}\right) + C_2. \]

The wall has no deflection: $y(0)=0 \Rightarrow C_2=0$. Hence:

$y(x) = -\frac{Fx^2$.

The negative sign records that the deflection is in the direction of $F$ (downward); its magnitude is $|y(x)|=\dfrac{Fx^2}{6EI}(3l-x)$.

Step 5: Deflection at the free end

Setting $x=l$: $|y(l)| = \dfrac{Fl^2}{6EI}(3l-l) = \dfrac{Fl^2}{6EI}\cdot2l$:

$\delta = \frac{Fl^3$.

Step 6: Numerical evaluation

Convert to a consistent kg–cm system: $F=2000\text{ kg}$, $l=600\text{ cm}$, $E=2{,}100{,}000\text{ kg/cm}^2$, $I=30{,}000\text{ cm}^4$, $EI = 2{,}100{,}000\times30{,}000 = 6.3\times10^{10}\ \text{kg}\cdot\text{cm}^2$.

\[ \delta = \frac{Fl^3}{3EI} = \frac{2000\times600^3}{3\times6.3\times10^{10}} = \frac{2000\times2.16\times10^{8}}{1.89\times10^{11}} = \frac{4.32\times10^{11}}{1.89\times10^{11}}. \] $\delta = \frac{4.32$.

Step 7: Verification

Differentiate the boxed elastic curve twice and multiply by $EI$ to recover the bending moment:

\[ y' = -\frac{F}{6EI}\bigl(6lx-3x^2\bigr) = -\frac{F}{EI}\!\left(lx-\frac{x^2}{2}\right), \qquad y'' = -\frac{F}{EI}(l-x). \] So $EI\,y'' = -F(l-x) = M(x)$, matching (1). $\checkmark$ Also $y(0)=0$, $y'(0)=0$ as imposed, and $\delta=Fl^3/(3EI)$ has units $\text{kg}\cdot\text{cm}^3/(\text{kg/cm}^2\cdot\text{cm}^4)=\text{cm}$, as required.