Solve for the Following Bending moment equation for segment BC. a. 2000 lb(X) - 100 lb/ft (X squared) - 9600 lb-ft b. 2100 lb(X) - 120 lb/ft (X squared) - 9700 lb-ft c. 21500 lb(X) - 200 lb/ft (X squared) - 9800 lb-ft d. 3000 lb(X) - 200 lb/ft (X squared) - 9850 lb-ft The distance between point A and the shear force VAB=0 a. 6 feet b. 8 feet c. 10 feet d. 12 feet The degree of bending moment equations for segments AB and BC. a. 1st degree and 2rd degree b. 3rd degree and 2nd degree c. 2nd degree and 3rd degree d. None of the above.
Solve for the Following Bending moment equation for segment BC. a. 2000 lb(X) - 100 lb/ft (X squared) - 9600 lb-ft b. 2100 lb(X) - 120 lb/ft (X squared) - 9700 lb-ft c. 21500 lb(X) - 200 lb/ft (X squared) - 9800 lb-ft d. 3000 lb(X) - 200 lb/ft (X squared) - 9850 lb-ft The distance between point A and the shear force VAB=0 a. 6 feet b. 8 feet c. 10 feet d. 12 feet The degree of bending moment equations for segments AB and BC. a. 1st degree and 2rd degree b. 3rd degree and 2nd degree c. 2nd degree and 3rd degree d. None of the above.
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
Related questions
Question
Solve for the Following
Bending moment equation for segment BC.
a. 2000 lb(X) - 100 lb/ft (X squared) - 9600 lb-ft
b. 2100 lb(X) - 120 lb/ft (X squared) - 9700 lb-ft
c. 21500 lb(X) - 200 lb/ft (X squared) - 9800 lb-ft
d. 3000 lb(X) - 200 lb/ft (X squared) - 9850 lb-ft
The distance between point A and the shear force VAB=0
a. 6 feet
b. 8 feet
c. 10 feet
d. 12 feet
The degree of bending moment equations for segments AB and BC.
a. 1st degree and 2rd degree
b. 3rd degree and 2nd degree
c. 2nd degree and 3rd degree
d. None of the above.
Transcribed Image Text:A
6 ft-
200 lb/ft
B
CA
1200 lb. ft
★4 ft —— 4 ft -
D
![per part AB
2001blit
B
200
6
ㅏ
W = 200 X
200.
6FF
K-
6
·X
=
- [ 12 W x] x x 6. - 1/2 1/² - -/ We ²
3
3
MAB = -1/6 · (200) x²
- ½6 (200) x³ =
-5.5555 x ²
3
. - MAB
=
MAB = -5.55 x
-5.55x³;
option (c) is
correct..
Shear Force equcation for
Seyment "BE
X..
J
момент ерженои
Bending Moment concesion
t
WI
=> MAB
;
A
BA
1140016
}
"BC".](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F0ef7dced-e510-451b-b91a-a720e02ad23a%2F8c355a45-3101-4161-a927-ab0e4b1dffa3%2Ffks2llo_processed.png&w=3840&q=75)
Transcribed Image Text:per part AB
2001blit
B
200
6
ㅏ
W = 200 X
200.
6FF
K-
6
·X
=
- [ 12 W x] x x 6. - 1/2 1/² - -/ We ²
3
3
MAB = -1/6 · (200) x²
- ½6 (200) x³ =
-5.5555 x ²
3
. - MAB
=
MAB = -5.55 x
-5.55x³;
option (c) is
correct..
Shear Force equcation for
Seyment "BE
X..
J
момент ерженои
Bending Moment concesion
t
WI
=> MAB
;
A
BA
1140016
}
"BC".
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