The ultimate collapse load (P) in terms of plastic moment M_{p} by kinematic approach for a propped cantilever of length L with P acting at its mid-span as shown in the figure, would be

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GATE CE 2014 Official Paper: Set 1

Option 3 : \(P = \frac{{6{M_p}}}{L}\)

CT 3: Building Materials

2962

10 Questions
20 Marks
12 Mins

The following procedure to be adopted to find the collapse load.

1. Find the all possible locations of plastic Hinge. In this case plastic hinges will be at A, B only. There will no plastic hinge at C because at C bending moment is zero due to hinge connection.

2. No of plastic hinge required to form collapse mechanism, N = Degree of static indeterminacy + 1

Degree of static indeterminacy for the given beam = 1

**⇒**** N = 2 **i.e. two plastic hinge will be needed to form mechanism and that will be point A and B.

3. Use principle of virtual work i.e.

External work done + Internal work done = 0

__Calculation:__

Possible hinge 0 elastic hinge.

External work done \(= P{\rm{\Delta }} = \left( {\frac{{PL\theta }}{2}} \right)\)

Internal work done: \(\begin{array}{*{20}{c}} { - \left( {{M_P}\theta } \right)}& + &{{M_P}\left( { - \theta - \theta } \right)}& = &{ - 3{M_P}\theta }\\ \downarrow &{}& \downarrow &{}&{}\\ {at\;A}&{}&{at\;B}&{}&{} \end{array}\)

**By virtual work:**

W_{ext} + W_{int} = 0

\(\Rightarrow \frac{{PL\theta }}{2} + \left( { - 3{M_P}\theta } \right) = 0 \Rightarrow P = \frac{{6{M_P}}}{L}\)