• Dell
In summary, the conversation discusses the behavior of steel under load and the calculation of displacement and stress in different steel types. The P(delta) graphs were used to determine the displacement of point C and the maximum stress in each steel portion. The speaker found the correct answer for the mild steel, but the answer for the high strength steel was incorrect. The conversation also questions the change of gradient in the graph for point C and identifies the elastic and plastic portions of the graph.
Dell
in the following question, i am dealing with steel which is assumed to be have elasto-plactic behaviour,- its stress strain graph will be linear to the yield stress with a slope of "E" and then linear with a slope of zero.

i calculated and drew the P(delta) graphs for the 2 steel types and added them to get the displacement of point "c"

knowing that P=1080KN and looking for the maximum displacement of C, i went to the 3rd graph and said

(1080-950)/(x-2.5)=(1140-950)/(3.5-2.5)

==> x=3.1842
===> x=0.31842mm

which is correct,

now looking for the permanent deflection of "c" i know that at x=0.31842mm mild steel will have plastic deformations but high strenght steel will not yet, so i went to the 1st graph and said
(475)/(2.5)=(475)/(3.1842-x)
==> x=0.6842
===> x=0.06842mm

but the correct answer is 0.0382mm which is half of the answer i got
what am i doing wrong??

then for the second part i need to find the maximum stress in each of the steel portions,
for the mild steel i reached the plastic region- the maximum strenght, so as far as i can see the maximum stress felt in the mild steel is the stress corresponding to the 0.31842 displacement
250MPA,
which according to the question is the right answer

for tyhe high strength steel i think the maximum stress wil be at the maximum deflection, but since this is before the yield stress,
665/3.5=x/3.1842
x=605KN
x/A=318MPa

but the correct answer is 818Mpa

now i know p=1080KN
P/A=568Mpa

and using the correct answers 250-818=-568MPa

so i know that i am in the right direction but don't know what to do,
also how is it that the stress they found in the high strength steel is more than the yield stress?

Can you explain the change of gradient in the graph for point C? Which parts are elastic, and which plastic?

its not a stress strain curve- its a force displacement curve,
the first part is totally elastic, the second part elastic(one steel is plastic one elastic) and the 3rd is totally plastic.

## 1. What is elasto-plastic steel?

Elasto-plastic steel is a type of steel that exhibits both elastic and plastic behavior when subjected to axial loading. This means that it can be stretched to a certain point without permanent deformation, but will eventually yield and deform when the load exceeds its yield strength.

Axial loading refers to a type of structural loading where the force is applied along the longitudinal axis of a member. In the case of elasto-plastic steel, this means that the force is applied along the length of the steel, causing it to stretch or compress.

Under axial loading, elasto-plastic steel will initially behave elastically, meaning it will deform proportionally to the applied load. However, once the load exceeds the yield strength of the steel, it will begin to exhibit plastic behavior and permanently deform. This behavior is known as the yield point.

## 4. What is the difference between elastic and plastic behavior?

Elastic behavior refers to the ability of a material to return to its original shape after being deformed under a certain load. In contrast, plastic behavior refers to the permanent deformation of a material when subjected to a load that exceeds its yield strength. Elasto-plastic steel exhibits both elastic and plastic behavior under axial loading.

## 5. How is elasto-plastic steel used in structural engineering?

Elasto-plastic steel is commonly used in structural engineering for its ability to withstand high loads and deform without failure. It is often used in the construction of buildings, bridges, and other structures that require high strength and ductility. It is also used in the design of earthquake-resistant structures, as it can absorb and dissipate energy during seismic events.

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