Formül;TL2 5 için c - cz -;(1– 0,630b/a)M-(lzy.2– ly.y) + M,(1;.z – Izy -y)1.ly – IzyOx + dy, tan28p =Tmaks =Gab?Czab'G"mVQTDItМy2tAOx + Oy x-dy cos20 + Trysin20 , Jort =Ox + 0yTy'y' = -sin20 + Trycos20 , Omaks.min =+r(Ox –+ TyTmaks =A 3 m long steel angle shown in the figure has a section thickness of 12.7 mm and across sectional area of 4350 mm ^ 2. Find the resulting torsion angle in degrees.T = 500 Nm ve G = 77,2 GPa(°)O 5,34O 3,89O 4,87O 7,86O 8.72O O O O

Answered: Image /qna-images/answer/a272ad35-1dfb-4c9c-aa12-d41caf703350.jpg

form TL My .ly - ly , +,- ce20 + ttin20. a , tan20- sin20 + Tycor20, Imaksmin - + )' + rà, 2 Tmak A 3 m long steel angle shown in the figure has a section thickness of 12.7 mm and a cross sectional area of 4350 mm ^ 2. Find the resulting torsion angle in degrees. T= 500 Nm ve G = 77,2 GPa () 5.34 3.89 4,87 7,86 8,72

form TL My .ly - ly , +,- ce20 + ttin20. a , tan20- sin20 + Tycor20, Imaksmin - + )' + rà, 2 Tmak A 3 m long steel angle shown in the figure has a section thickness of 12.7 mm and a cross sectional area of 4350 mm ^ 2. Find the resulting torsion angle in degrees. T= 500 Nm ve G = 77,2 GPa () 5.34 3.89 4,87 7,86 8,72

Elements Of Electromagnetics

7th Edition

ISBN:9780190698614

Author:Sadiku, Matthew N. O.

Publisher:Sadiku, Matthew N. O.

ChapterMA: Math Assessment

Section: Chapter Questions

Problem 1.1MA

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Question

Formül;
TL
2 5 için c - cz -;(1– 0,630b/a)
M-(lzy.2– ly.y) + M,(1;.z – Izy -y)
1.ly – Izy
Ox + dy, tan28p =
Tmaks =
Gab?
Czab'G"
m
VQ
TD
It
Мy
2tA
Ox + Oy x-dy cos20 + Trysin20 , Jort =
Ox + 0y
Ty'y' = -
sin20 + Trycos20 , Omaks.min =
+r
(Ox –
+ Ty
Tmaks =
A 3 m long steel angle shown in the figure has a section thickness of 12.7 mm and a
cross sectional area of 4350 mm ^ 2. Find the resulting torsion angle in degrees.
T = 500 Nm ve G = 77,2 GPa
(°)
O 5,34
O 3,89
O 4,87
O 7,86
O 8.72
O O O O

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