A cantilever beam of length $2a$ is loaded at the tip with force $F$. The beam is supported in the middle by a roller (pin). Find the reaction moment at the built-in end of the beam as $\alpha Fa$, where $\alpha =$ ....................... (round off to one decimal place).
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Use deflection compatibility when beams have intermediate supports. Treat each load separately and enforce zero displacement at the redundant support.
Step 1: Beam layout.
- Cantilever of length $2a$, built at left.
- Midpoint at $a$ supported by roller.
- Load $F$ applied at free end.
Step 2: Redundancy.
Support at $a$ makes beam statically indeterminate. Use compatibility.
Let reaction at roller = $R$.
Step 3: Deflection approach.
Deflection at $a$ (due to $F$ and $R$) must be zero.
Deflection at $a$ due to $F$:
For cantilever length $2a$, load at tip:
\[
y_F(x) = \frac{F}{6EI}x^2(3L - x), L = 2a
\]
At $x=a$:
\[
\delta_F = \frac{F}{6EI}a^2(3(2a) - a) = \frac{F}{6EI}a^2(5a) = \frac{5Fa^3}{6EI}
\]
Deflection at $a$ due to $R$ (upward):
$R$ at $a$ acts like downward load on cantilever at $a$.
Deflection at $a$ = $\frac{Ra^3}{3EI}$.
Step 4: Compatibility.
Net deflection = 0:
\[
\frac{5Fa^3}{6EI} - \frac{Ra^3}{3EI} = 0
\]
\[
R = \frac{5F}{2}
\]
Step 5: Reaction moment at built end.
Moment at built-in = $M = Ra - F(2a)$.
\[
M = \frac{5F}{2}a - 2Fa = \frac{F}{2}a
\]
So,
\[
M = 0.5 Fa
\]
Wait — check carefully: sign convention.
Correct derivation shows final $\alpha = 0.5$.
\[
\boxed{\alpha = 0.5}
\]
