For the given question and solution,Why is taux'y' after the 25 degree rotation = positive Rsin(phi) = 1.05MA instead of negative Rsin(phi) ? how can i determine the sign convention after rotation? i get mixed up with the sign convention of the shear stress after rotation. Problem 7.387.38 Solve Prob. 7.16, using Molr's circle.7.13 through 7.16 For the given state of stress, determine the normaland shearing stresses after the eleiment shown has been rotated (a) 25° clock-wise, (b) 10° connterclockwise,56 MPa35 MPa6x =6,: 56MPA35 MPGGave6xt228M PaPoints:X: (0, -35 Mpa)Y: (56 MPa, 35 MPa)C: (28 MPa, o)ol FX'(MPa)FXtan 20p-FC351.252820p- 51,34*R = FE + Fx* - f2s* + 35*44.8 M PG(a) e :25° )20 - So° )9: 51.34° - 50° = 1.34°6, : 6aveR cus 9- 16.8 MPA -: Rsin 9= {.0S MPa6y: - 5e + Rcos p = 72.8 MPa(b) 0 = lo° )20 = 20°)51.34(MPa): 51.34° + 20°- 71.34°- 13.7 MPaR sin 9423 M PaX'6, : 6. + R cus g =42: 4 MPaCuS- So°.

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Elements Of Electromagnetics

7th Edition

ISBN:9780190698614

Author:Sadiku, Matthew N. O.

Publisher:Sadiku, Matthew N. O.

ChapterMA: Math Assessment

Section: Chapter Questions

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For the given question and solution,

Why is taux'y' after the 25 degree rotation = positive Rsin(phi) = 1.05MA instead of negative Rsin(phi) ? how can i determine the sign convention after rotation?

i get mixed up with the sign convention of the shear stress after rotation.

Problem 7.38
7.38 Solve Prob. 7.16, using Molr's circle.
7.13 through 7.16 For the given state of stress, determine the normal
and shearing stresses after the eleiment shown has been rotated (a) 25° clock-
wise, (b) 10° connterclockwise,
56 MPa
35 MPa
6x =
6,: 56MPA
35 MPG
Gave
6xt
2
28M Pa
Points:
X: (0, -35 Mpa)
Y: (56 MPa, 35 MPa)
C: (28 MPa, o)
ol F
X'
(MPa)
FX
tan 20p-
FC
35
1.25
28
20p- 51,34*
R = FE + Fx* - f2s* + 35*
44.8 M PG
(a) e :
25° )
20 - So° )
9: 51.34° - 50° = 1.34°
6, : 6ave
R cus 9
- 16.8 MPA -
: Rsin 9
= {.0S MPa
6y: - 5e + Rcos p = 72.8 MPa
(b) 0 = lo° )
20 = 20°)
51.34
(MPa)
: 51.34° + 20°
- 71.34°
- 13.7 MPa
R sin 9
423 M Pa
X'
6, : 6. + R cus g =
42: 4 MPa
CuS
- So°.

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