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I solved the principal stresses but having a hard time on determing the theta-P (orientation of principla plances), could you help me?

I solved the principal stresses but having a hard time on determing the theta-P (orientation of principla plances), could you help me?

Elements Of Electromagnetics
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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I solved the principal stresses but having a hard time on determing the theta-P (orientation of principla plances), could you help me?

Figure 9-42 Full Alternative Text 9-43. Determine the principal stresses in the box beam at point B. Show the results on an element located at this point. Prob. 9-43 2 ft- 10 kip 4 in. A 6 in. B -1.5 ft- 4 in. HA B 4 kip 0.5 ft 13 in. 13 in. -2 ft-
Transcribed Image Text:Figure 9-42 Full Alternative Text 9-43. Determine the principal stresses in the box beam at point B. Show the results on an element located at this point. Prob. 9-43 2 ft- 10 kip 4 in. A 6 in. B -1.5 ft- 4 in. HA B 4 kip 0.5 ft 13 in. 13 in. -2 ft-
Step 4) Solve Principal Stresses and Orientation Gin 6 (6)³ + (6) (6) (0)² 12 - [4 (4)² + (4) (4) (0)² ] 86-6666 in4 Iz= · 5 Q = (1.9) ((6.3) - (4-2)) in ³ ·19 in ³ = T=VQ It = (2 kip) (19 in ³) (86.6666m 4) (2 m) Jx=0 OT Jy=0 Txy=-0.2192 Ksi 01₁/2 = (J₂² +Jy) = √ (Tv=Jy)² + [ny² Jx Exy = 0.2192 15. = -0.2192 1sí + = ( 0 + ₂0 ) + √ ( 0 = 0 ) ² + (-0.2192je 2 O B 02= Op = 1/2 tan " ( _2 +Zxy) Jx-Jy EAS 212 Problem 1 Continued) oz-(9)-J1=j+ (-0.? 3* Т2=-0. 2192 ksi -0.2192 Ksi "error" y = y = 1. a in Homework #8 (1.5.18)-(1.8) 18-8 Silva, Martin C
Transcribed Image Text:Step 4) Solve Principal Stresses and Orientation Gin 6 (6)³ + (6) (6) (0)² 12 - [4 (4)² + (4) (4) (0)² ] 86-6666 in4 Iz= · 5 Q = (1.9) ((6.3) - (4-2)) in ³ ·19 in ³ = T=VQ It = (2 kip) (19 in ³) (86.6666m 4) (2 m) Jx=0 OT Jy=0 Txy=-0.2192 Ksi 01₁/2 = (J₂² +Jy) = √ (Tv=Jy)² + [ny² Jx Exy = 0.2192 15. = -0.2192 1sí + = ( 0 + ₂0 ) + √ ( 0 = 0 ) ² + (-0.2192je 2 O B 02= Op = 1/2 tan " ( _2 +Zxy) Jx-Jy EAS 212 Problem 1 Continued) oz-(9)-J1=j+ (-0.? 3* Т2=-0. 2192 ksi -0.2192 Ksi "error" y = y = 1. a in Homework #8 (1.5.18)-(1.8) 18-8 Silva, Martin C
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