STEM Problem Detail

Problem Statement

Prompt

Give me the solution for problem 7-60

Solution Rendering...

\[
\textbf{Given:}
\]
A beam of total length \(L\) carries a uniformly distributed load of intensity \(w_0\) over its entire length.
The beam is supported by a pin at \(A\) and a roller at \(B\), where \(AB = a\). The beam overhangs beyond \(B\).

\[
\textbf{Reactions at the supports:}
\]

Let the reactions at \(A\) and \(B\) be \(R_A\) and \(R_B\) respectively.

From vertical equilibrium,
\[
R_A + R_B = w_0 L
\]

Taking moments about \(A\),
\[
R_B a = w_0 L \left(\frac{L}{2}\right)
\]
\[
R_B = \frac{w_0 L^2}{2a}
\]

Hence,
\[
R_A = w_0 L - \frac{w_0 L^2}{2a}
\]

\[
\textbf{Bending moment at support } B:
\]

The bending moment at \(B\) is
\[
M_B = R_A a - w_0 a \left(\frac{a}{2}\right)
\]
\[
M_B = \left(w_0 L - \frac{w_0 L^2}{2a}\right)a - \frac{w_0 a^2}{2}
\]
\[
M_B = w_0 La - \frac{w_0 L^2}{2} - \frac{w_0 a^2}{2}
\]

Since the beam overhangs beyond \(B\), this moment is hogging. Its magnitude is
\[
|M_B| = \frac{w_0}{2}\left(L^2 - 2La + a^2\right)
\]

\[
\textbf{Maximum bending moment in span } AB:
\]

The maximum bending moment occurs where the shear force is zero.

At a section at distance \(x\) from \(A\),
\[
V(x) = R_A - w_0 x
\]

Setting \(V(x)=0\),
\[
x = \frac{R_A}{w_0} = L - \frac{L^2}{2a}
\]

The bending moment at this section is
\[
M_{\max} = R_A x - \frac{w_0 x^2}{2}
\]

Substituting for \(x\),
\[
M_{\max} = \frac{w_0}{2}\left(L - \frac{L^2}{2a}\right)^2
\]

\[
\textbf{Condition given:}
\]

The maximum bending moment in span \(AB\) is equal in magnitude to the bending moment at support \(B\),
\[
\frac{w_0}{2}\left(L - \frac{L^2}{2a}\right)^2
=
\frac{w_0}{2}\left(L^2 - 2La + a^2\right)
\]

Cancelling \(\frac{w_0}{2}\),
\[
\left(L - \frac{L^2}{2a}\right)^2 = (L - a)^2
\]

Taking the physically meaningful root,
\[
L - \frac{L^2}{2a} = a - L
\]

\[
\frac{L^2}{2a} = 2L - a
\]

\[
L^2 = 4aL - 2a^2
\]

\[
2a^2 - 4aL + L^2 = 0
\]

Solving,
\[
a = \frac{L}{2}(2 - \sqrt{2})
\]

\[
\textbf{Final Answer:}
\]
\[
\boxed{a = \frac{L}{2}(2 - \sqrt{2})}
\]

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Problem #87