For a beam supported at both ends with a mass in the middle, the flexural modulus is given by:
$$E_b = \frac{F L^3}{48I d}$$
with I as the area moment
$$I = \frac{1}{12}wh^3$$
How does the expression for modulus change if we have one fixed end as in the diagram below?
ja72 states in this answer that
$$E_b = \frac{F L^3}{3I d}$$
But I'm not clear on the derivation.
The fundamental beam equation is
$$\dfrac{\text{d}^2}{\text{d}x^2}\left(EI\dfrac{\text{d}^2w}{\text{d}x^2}\right) = q$$
Which basically translates to "the fourth derivative of the deflection function is equal to the applied load". In fact
All deflection results are obtained using this equation.
To simplify this answer, I'm going to start off from the second derivative, the bending moment, since it (and the ones after it) is trivial to find by inspection.
For the simply-supported beam with a concentrated load at midspan, we have:
$$\begin{align} M &= \begin{cases} \dfrac{Fx}{2} &\text{ for } x \in [0,\dfrac{L}{2}] \\ \dfrac{F(L-x)}{2} &\text{ for } x \in [\dfrac{L}{2}, L] \\ \end{cases} \\ EI\theta = \int M\text{d}x &= \begin{cases} \dfrac{Fx^2}{4} + C_1 &\text{ for } x \in [0,\dfrac{L}{2}] \\ \dfrac{FLx}{2} - \dfrac{Fx^2}{4} + C_2 &\text{ for } x \in [\dfrac{L}{2}, L] \\ \end{cases} \\ EI\delta = EI\int \theta\text{d}x &= \begin{cases} \dfrac{Fx^3}{12} + C_1x + C_3 &\text{ for } x \in [0,\dfrac{L}{2}] \\ \dfrac{FLx^2}{4} - \dfrac{Fx^3}{12} + C_2x + C_4 &\text{ for } x \in [\dfrac{L}{2}, L] \\ \end{cases} \end{align}$$
We know that $\delta(0) = \delta(L) = 0$ and that $\theta\left(\dfrac{L}{2}^+\right) = \theta\left(\dfrac{L}{2}^-\right)$ and $\delta\left(\dfrac{L}{2}^+\right) = \delta\left(\dfrac{L}{2}^-\right)$ (that is, the deflection and angle is continuous at $\dfrac{L}{2}$).
So you solve it:
$$\begin{gather} \delta(0) = C_3 = 0 \\ \delta(L) = \dfrac{FL^3}{4} - \dfrac{FL^3}{12} + C_2L + C_4 = 0 \\ \therefore C_4 = -\dfrac{FL^3}{6} - C_2L \\ \theta\left(\dfrac{L}{2}^+\right) = \theta\left(\dfrac{L}{2}^-\right) \\ \therefore \dfrac{FL^2}{16} + C_1 = \dfrac{FL^2}{4} - \dfrac{FL^2}{16} + C_2 \\ \therefore C_1 = \dfrac{FL^2}{8} + C_2 \\ \delta\left(\dfrac{L}{2}^+\right) = \delta\left(\dfrac{L}{2}^-\right) \\ \therefore \dfrac{FL^3}{96} + \dfrac{FL^3}{16} + \dfrac{C_2L}{2} = \dfrac{FL^3}{16} - \dfrac{FL^3}{96} + \dfrac{C_2L}{2} - \dfrac{FL^3}{6} - C_2L \\ \therefore C_2 = -\dfrac{9FL^2}{48} \\ \therefore C_1 = -\dfrac{3FL^2}{48} \\ \therefore C_4 = \dfrac{FL^3}{48} \end{gather}$$
Now, by inspection we can easily see the deflection is at the midspan, so let's calculate it (doesn't matter which $\delta$ equation you choose).
$$\delta\left(\dfrac{L}{2}\right) = \dfrac{1}{EI}\left(\dfrac{FL^3}{96} - \dfrac{3FL^3}{96}\right) = -\dfrac{FL^3}{48EI}$$
The same process can be repeated for a cantilevered beam, only it's much simpler:
$$\begin{align} M &= FL - Fx \\ EI\theta = \int M\text{d}x &= FLx - \dfrac{Fx^2}{2} + C_1 \\ EI\delta = EI\int \theta\text{d}x &= \dfrac{FLx^2}{2} - \dfrac{Fx^3}{6} + C_1x + C_2 \\ \theta(0) &= C_1 = 0 \\ \delta(0) &= C_2 = 0 \\ \therefore EI\delta &= \dfrac{FLx^2}{2} - \dfrac{Fx^3}{6} \\ \therefore \delta(L) &= \dfrac{FL^3}{3EI} \end{align}$$
'I' doesn't change for the different situations. 'I' is a property of the cross-section of the beam - a rectangle of width w and depth h. The Eb equations are correct for the situations set out, with the following caveats:
The first situation is not for a beam that is fixed at both ends, it is for a beam that is supported at both ends but not 'fixed'. 'Fixed' implies a moment continuity, in which case the '48' in the equation would be '192'.
The second situation needs the end condition to be fixed - if it is supported but not fixed (which is what is shown in the diagram), it is a mechanism.
