Plastic Analysis: Upper Bound Theorem

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Homework Statement:

Find the value of the collapsed load factor when Mp = 120 kN/m

Relevant Equations:

Plastic energy dissipated = work done by loads
Hi,

I have a quick question about part 1 of this upper bound theorem question (in the attached image). Answer says that [itex] \lambda_c = 2.25 [/itex].
Screen Shot 2020-02-19 at 12.12.52 PM.png


First, we know that there is 1 redundancy and therefore there will be a maximum of 2 plastic hinges for failure.

I have found that there needs to be swaying to right of the right side of the structure (by an angle [itex] \theta [/itex]), but I don't have any intuition for why that would be the case. Would someone be able to explain to me why there must be swaying in the structure for the mechanism illustrated? If I update my answer below to include this, then I get the correct answer.

(Initial drawing without the swaying)
IMG_8744.jpg


Doing my original incorrect analysis yields:
[tex] M_p \left( \theta \right) + 2 M_p \left( 2 \theta \right) = 100 \lambda \left( 3 \theta \right) + w \lambda \left( \frac{1}{2} 2 \left( 2 \theta \right) \right) [/tex]
Substituting in the values from the question, that will yield [itex] \lambda = \frac{600}{320} = 1.875 [/itex], which isn't the correct answer.
 

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  • #2
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Homework Statement:: Find the value of the collapsed load factor when Mp = 120 kN/m
Relevant Equations:: Plastic energy dissipated = work done by loads

Hi,

I have a quick question about part 1 of this upper bound theorem question (in the attached image). Answer says that [itex] \lambda_c = 2.25 [/itex]. View attachment 257324

First, we know that there is 1 redundancy and therefore there will be a maximum of 2 plastic hinges for failure.

I have found that there needs to be swaying to right of the right side of the structure (by an angle [itex] \theta [/itex]), but I don't have any intuition for why that would be the case. Would someone be able to explain to me why there must be swaying in the structure for the mechanism illustrated? If I update my answer below to include this, then I get the correct answer.

(Initial drawing without the swaying)
View attachment 257326

Doing my original incorrect analysis yields:
[tex] M_p \left( \theta \right) + 2 M_p \left( 2 \theta \right) = 100 \lambda \left( 3 \theta \right) + w \lambda \left( \frac{1}{2} 2 \left( 2 \theta \right) \right) [/tex]
Substituting in the values from the question, that will yield [itex] \lambda = \frac{600}{320} = 1.875 [/itex], which isn't the correct answer.
In your initial sketch, allegedly without sway, you have included some sway of the left hand column, but not of the right hand column. The line AF produced meets the line DC produced at a point 4 m above C. This is the instantaneous centre of rotation of the beam segment FC, say call it point I. When you draw your sway diagram, there are 3 parts, each with their own centre of rotation, namely A B and I. Apply a theta to A and draw the remaining mechanism to scale. Work from there.
 
  • #3
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When you draw your sway diagram, there are 3 parts, each with their own centre of rotation, namely A B and I. Apply a theta to A and draw the remaining mechanism to scale. Work from there.
Many thanks for your response. Can you explain how B is a centre of rotation, I am struggling to see how this is the case.

Otherwise, with regards to the extra angle [itex] \theta [/itex] at D, is this a correct line of reasoning to follow?
So if we have applied some angle [itex] \theta [/itex] to A in the CW direction, then we must have some angle [itex] \theta [/itex] in the CCW direction at I (intersection of AF and DC) due to [itex] 5 \times \theta_{AF} = 5 \times \theta_{IF} [/itex]. Thus, due to the CCW rotation at I, there will be a displacement at C with the size of [itex] = r_{IC} \theta [/itex]. Due to the equivalence argument again, then we must have a CW angle of [itex] \theta [/itex] at D.
 

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