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?
Mechanics of Materials (MindTap Course List)
9th Edition
ISBN:9781337093347
Author:Barry J. Goodno, James M. Gere
Publisher:Barry J. Goodno, James M. Gere
Chapter7: Analysis Of Stress And Strain
Section: Chapter Questions
Problem 7.2.14P: Repeat the previous problem using ? = 50° and stresses on the rotated element: sy1= 70 MPa, ??y1=-82...
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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?
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](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F652e5396-12b6-4e5e-865e-eecd6b9e3fee%2Fc54cc0e4-f936-4472-a127-6d77afb45b66%2Fjgryj2i_processed.png&w=3840&q=75)
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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