(b) The formula for the elasto-plastic tangent is on the formula sheet [{fo }{f.o) [De [Dep] = [De] | [1] [1-[88]] {f.o}T [De] {fo } we have [De] from part (a) so the students just need to calculate the derivative of the yield function with respect to stress, {f} {f}={10-1}. Looking at different components of [Dep] (all in GPa) 1 0 -1 270 135 135 135 0 - 135

Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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can someone explain how we get to this answer 

(b) The formula for the elasto-plastic tangent is on the formula sheet
[Dep] = [de]
109] [11-|
{f₁0 }{f₁0}T [De]
{f₁0}T [De] {f₁0 }
we have [De] from part (a) so the students just need to calculate the derivative of the yield function
with respect to stress, {f, }
{f₁o} = {1 0 -1}.
Looking at different components of [Dep] (all in GPa)
1
0 1
270 135 135
135 0 - 135
{f₁0}{f₁0}T [De]
=
ооо
E
135 270 135 =
0 0 0
1 0 1 ||
135 135 270
J
-135 0 135
Transcribed Image Text:(b) The formula for the elasto-plastic tangent is on the formula sheet [Dep] = [de] 109] [11-| {f₁0 }{f₁0}T [De] {f₁0}T [De] {f₁0 } we have [De] from part (a) so the students just need to calculate the derivative of the yield function with respect to stress, {f, } {f₁o} = {1 0 -1}. Looking at different components of [Dep] (all in GPa) 1 0 1 270 135 135 135 0 - 135 {f₁0}{f₁0}T [De] = ооо E 135 270 135 = 0 0 0 1 0 1 || 135 135 270 J -135 0 135
[De] =
270 135
135
135 270 135
135 135
270
GPa
Transcribed Image Text:[De] = 270 135 135 135 270 135 135 135 270 GPa
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Follow-up Question

hi thank you for your reply when you have put the matrix 135 0 -135,000,-135 0 135 why are the top left and bottom right corner 135 and not 270 ?

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Follow-up Question

sorry also why have you put 1/270 x the matric i dont see that in the formula ?

how is the final result obtained 202.5 135 202.5 etc

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