A propped cantilever beam shown in the figure has a plastic momen
A propped cantilever beam shown in the figure has a plastic moment capacity of M0
The collapse load is
A. <span class="math-tex">\(\frac{{4{M_0}}}{L}\)</span>
B. <span class="math-tex">\(\frac{{6{M_0}}}{L}\)</span>
C. <span class="math-tex">\(\frac{{8{M_0}}}{L}\)</span>
D. <span class="math-tex">\(\frac{{12{M_0}}}{L}\)</span>
Please scroll down to see the correct answer and solution guide.
Right Answer is: B
SOLUTION
Concept:
To make a collapse mechanism, no of plastic hinges required = Ds +1.
Where
Ds is static indeterminacy
In this Case, Ds = 1
⇒ Two plastic hinges will be required to form collapse mechanism. The possible location of plastic hinge is B and C. There will no plastic hinge at A because at A bending moment is zero due to hinge connection.
Thereafter, use principle of virtual work i.e.
External work done + Internal work done = 0
Calculation:
External work done = \(\omega {\rm{\Delta }} = \omega \times \frac{L}{2}\;Q = \frac{{\omega LQ}}{2}\)
Internal work done = - Mp θ – Mp (θ +θ) = -3 Mp θ
By virtual work
\(\frac{{\omega L\theta }}{2} = 3{M_P}\theta \)
\(\Rightarrow \omega = \frac{{6{M_p}}}{L}\)
