A continuous beam with constant EI is shown in the figure. Collapse load for this beam will be equal to

\frac{10 M_{P}}{L}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

\frac{12 M_{P}}{L}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

\frac{8 M_{P}}{L}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

\frac{6 M_{P}}{L}

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

Open in App

Solution

The correct option is D $\frac{6 M_{P}}{L}$
For left span,
$\frac{W L}{2} θ = 2 M_{P} (θ + θ) + M_{P} θ$
$\Rightarrow W = \frac{10 M_{P}}{L}$
For right span,
$W \times \frac{L}{3} \times 2 θ = M_{P} (θ + 2 θ) + M_{P} θ$
$\Rightarrow W = \frac{6 M_{P}}{L}$
Hence $\frac{6 M_{P}}{L}$ will be the collapse load.

Suggest Corrections

Similar questions

\frac{1}{3} r d

L

P

L / 3

M_{P}

E I

P

P L / 2

The rotation of the overhanging end 'A' for the beam loaded as shown in figure is \frac{K M L}{E I} .

Join BYJU'S Learning Program