For a fixed beam with a concentrated load w at $\frac{1}{4}$ of

$For a fixed beam with a concentrated load w at $\frac{1}{4}$ of$

| For a fixed beam with a concentrated load w at $\frac{1}{4}$ of span from one end, the ultimate load is

A. $\frac{{16{M_p}}}{{3L}}$

B. $\frac{{4{M_p}}}{L}$

C. $\frac{{32{M_p}}}{{3L}}$

D. $\frac{{6{M_p}}}{L}$

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Given:

Fixed beam with concentrated load L/4 from one end

Calculation:

Number of Plastic hinges = D_s + 1 = (4 - 2) + 1 = 3

i.e. at A, B and C

From geometry,

$\Delta = \frac{L}{4}{\theta _A}$ and ${\theta _A} = 3{\theta _B}$

According to virtual work principle,

External work done = Internal work Done

$W\Delta = {M_P}{\theta _A} + {M_P}{\theta _A} + {M_P}{\theta _B} + {M_P}{\theta _B}$

$W\frac{L}{4}{\theta _A} = 2{M_P}{\theta _A} + \frac{{2{M_P}{\theta _A}}}{3}$

$\Rightarrow W = \frac{{32}}{3}\frac{{{M_P}}}{L}$

For a fixed beam with a concentrated load w at \(\frac{1}{4}\) of