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⑧3次元弾性体の応力とひずみとは?「youtube 【初めて学ぶ人のための材料力学の教室】演習問題2/2」をsympyでやってみたい。

Last updated at Posted at 2024-01-28

・演習問題2/2

youtubeです。

途中07:15から (0:00〜10:57)

<再生リスト
https://www.youtube.com/@zairiki/playlists

sympyで

・ver0.1

# ver0.1
from sympy import *
from decimal import Decimal, ROUND_HALF_UP, ROUND_HALF_EVEN
var('E,ν'           ,real=True)
var('σx,σy,σz'    ,real=True)
var('εx,εy,εz,εv',real=True)
var('τxy,τyz,τyz' ,real=True)
var('α ,ΔT'        ,real=True)
rep={E  :200*10**( 3),ν :0.3, \
     α : 10*10**(-6),ΔT:100,  \
     # σx:1  ,σy:1  ,σz :0 
     σy:0  ,σz :0 
     #τxy:0  ,τyz:1,τzx:0,    \
     # τxy:0  ,τyz:1  ,       \
     }
εx    =σx/E-ν*σy/E-ν*σz/E+α*ΔT         ;print("# εx=",εx)
εx_rep=εx.subs(rep)                    ;print("# 0="  ,εx_rep)
σx_ans=solve(εx_rep)[0]                ;print("# σx=",σx_ans);print()
# 
εy    =-ν*σx/E+σy/E-ν*σz/E+α*ΔT        ;print("# εy=",εy)
εy_rep=εy.subs(rep).subs({σx:σx_ans})  ;print("# εy=",εy_rep)
# εx= ΔT*α - ν*σy/E - ν*σz/E + σx/E
# 0= σx/200000 + 0.001
# σx= -200.000000000000

# εy= ΔT*α - ν*σx/E - ν*σz/E + σy/E
# εy= 0.00130000000000000

いつもの? sympyの実行環境と 参考のおすすめです。

(テンプレート)

参考文献

>等分布荷重を受ける単純支持はり
>JSME p85
 
>単純支持ばり 7)
>構造力学公式集 p136

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