Question

A simply supported beam of length 2L is subjected to a moment M at the mid - point x = 0 as shown in the figure. The vertical deflection in the domain $0\le x\le L$ is given by $y=\frac{-Mx}{12EIL}(L-x)(x+C)$ Where E is Young's modulus, I is the area moment of inertia and C is a constant (to be determined).
The slope at the centre x = 0 is

• A. $\frac{ML}{\left(2EI\right)}$
• B. $\frac{ML}{\left(3EI\right)}$
• C. $\frac{ML}{\left(6EI\right)}$
• D. $\frac{ML}{\left(12EI\right)}$

Answer 1

The correct option is C $\frac{ML}{\left(6EI\right)}$

Equation of vertical deflection is modified as,
$y=\frac{-M}{12EIL}[L{x}^2-x^3+LCx-C{x}^2]$
$\frac{dy}{dx}=\frac{-M}{12EIL}[2Lx-3x^2+LC-2Cx]$
$\frac{d^{2}y}{d{x}^2}=\frac{-M}{12EIL}[2L-6x-2C]$
$\therefore EI\frac{d^{2}y}{d{x}^2}=\frac{-M}{12L}[2L-6x-2C]$
Now $EI\frac{d^{2}y}{d{x}^2}=\frac{-M}{12L}[2L-6x-2C]$
At $x=L,M_x=\frac{EI{d}^{2}y}{d{x}^2}=0$
$\therefore C=-2L$
$\therefore$ Slope $=\frac{dy}{dx}=\frac{-M}{12EIL}[2Lx-3x^2-2L^2+4Lx]$
$\therefore {\left|\frac{dy}{dx}\right|}_{x=0}=\frac{ML}{6EI}$