Question

A simply supported beam of length 2L is subjected to a moment M at the mid-point x=0 as shown in figure. The deflection in the domain $0\le x\le L$ is
$W=\frac{-Mx}{12EIL}(L-x)(x+c)$
where E is the Young's modulus, I is the area moment of inertia and C is a constant (to be determined).
The slope at the center $x=0$ 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 deflection
$W=\frac{-M}{12EIL}[L{x}^2-x^3+Lcx-C{x}^2]$
$\frac{dW}{dx}=\frac{-M}{12EIL}[2Lx-3x^2+LC-2Cx]$
$\frac{d^{2}W}{d{x}^2}=\frac{-M}{12EIL}[2L-6x-2C]$
$\therefore EI\frac{d^{2}W}{d{x}^2}=\frac{-M}{12L}[2L-6x-2C]$
now $EI\frac{d^{2}W}{d{x}^2}=\frac{-M}{12L}[2L-6x-2C]$
at $x=L,M_x=0$
$\therefore$ -4L - 2C = 0
C = -2L
Slope
$\frac{dW}{dx}=\frac{-M}{12EIL}[2Lx-3x^2-2L^2+4Lx]$
$\therefore {\left|\frac{dW}{dx}\right|}_{x=0}=\frac{ML}{6EI}$
Method II:

$Strainenergy\left(U\right)=2{\int }_o^L\frac{{\left(\frac{Mx}{2L}\right)}^{2}dx}{2EI}$
$=\frac{1}{EI}[\frac{M^2}{4L^2}.\frac{L^3}{3}]=\frac{M^{2}L}{12EI}$
By Castigliano's theorem,
Slope = $\frac{dU}{dM}=\frac{ML}{6EI}$