For the propped cantilever supporting two point loads as shown in
For the propped cantilever supporting two point loads as shown in figure.
A. Plastic collapse load in terms of plastic moment capacity is 4M<sub>p</sub>/L
B. Load corresponding to Plastic hinge formed at Point A and Point B is taken as Collapse load.
C. Plastic collapse load in terms of plastic moment 5M<sub>P</sub>/L
D. Number of plastic hinges formed are 2
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Right Answer is:
SOLUTION
Calculation
Given,
DS = (3 + 1) – 3
= 1
Number of plastic hinges = DS + 1
= 1 + 1 = 2
There is possibility of 2 mechanisms.
Mechanism I
Plastic Hinge at Fixed support and Point B
From similarity of triangles AOB and BOD
\(\theta \times \frac{L}{3} = \alpha \times \frac{{2L}}{3}\)
∴ θ = 2α
By virtual work method
External work = Internal work
\(W \times {\delta _B} + W \times {\delta _c} = {M_p} \times \theta + {M_P} \times \left( {\theta + \alpha } \right)\)
\(W \times \frac{L}{3} \times \theta + W \times \frac{L}{3} \times \alpha = {M_p} \times \theta + {M_P} \times \left( {\theta + \alpha } \right)\)
\(W \times \frac{L}{3} \times 2 \times \alpha + W \times \frac{L}{3} \times \alpha = {M_p} \times 2 \times \alpha + {M_P} \times \left( {2 \times \alpha + \alpha } \right)\)
\(W = \frac{{5{M_P}}}{L}\)
Mechanism II
Plastic Hinge at Fixed support and at Point C
From similarity of triangles AOC and DOC
\(\alpha \times \frac{L}{3} = \theta \times \frac{{2L}}{3}\)
∴ α = 2θ
By virtual work method
External work = Internal work
\(W \times {\delta _c} + W \times {\delta _B} = {M_p} \times \theta + {M_P} \times \left( {\theta + \alpha } \right)\)
\(W \times \frac{L}{3}\alpha + W \times \frac{{2L}}{3} \times \theta = {M_p} \times \theta + {M_P} \times \left( {\theta + \alpha } \right)\)
From similarity of triangles
α = 2 × θ
\(W \times \frac{L}{3} \times 2\theta + W \times \frac{{2L}}{3} \times \theta = {M_p} \times \theta + {M_P} \times \left( {\theta + 2\theta } \right)\)
\(W = \frac{{4{M_P}}}{L}\)
∴ Collapse load = minimum of (\(\frac{{4{M_P}}}{L},\;\frac{{5{M_P}}}{L}\))
∴ \(W = \frac{{4{M_P}}}{L}\)
