Determine the following: Reaction support at point B. a. 1350 lb (Upward Direction) b. 1380 lb(Downward Direction) c. 1400 lb (Upward Direction) d. 1420 lb (Downward Direction) Reaction support at point D. a. 600 lb (Upward Direction) b. 700 lb (Downward Direction) c. 800 lb (Upward Direction) d. 900 lb (Downward Direction) Centroid of the triangular load. a. 588 lb (Downward Direction) b. 598 lb (Upward Direction) c. 600 lb(Downward Direction) d. (Upward Direction)
Determine the following: Reaction support at point B. a. 1350 lb (Upward Direction) b. 1380 lb(Downward Direction) c. 1400 lb (Upward Direction) d. 1420 lb (Downward Direction) Reaction support at point D. a. 600 lb (Upward Direction) b. 700 lb (Downward Direction) c. 800 lb (Upward Direction) d. 900 lb (Downward Direction) Centroid of the triangular load. a. 588 lb (Downward Direction) b. 598 lb (Upward Direction) c. 600 lb(Downward Direction) d. (Upward Direction)
International Edition---engineering Mechanics: Statics, 4th Edition
4th Edition
ISBN:9781305501607
Author:Andrew Pytel And Jaan Kiusalaas
Publisher:Andrew Pytel And Jaan Kiusalaas
Chapter6: Beams And Cables
Section: Chapter Questions
Problem 6.76P: The 50-ft measuring tape weighs 2.4 lb. Compute the span L of the tape to four significant figures.
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Derive the shear force equation,bending moment equation for the beam shown. Neglect the weight of the beam.
Determine the following:
Reaction support at point B.
a. 1350 lb (Upward Direction)
b. 1380 lb(Downward Direction)
c. 1400 lb (Upward Direction)
d. 1420 lb (Downward Direction)
Reaction support at point D.
a. 600 lb (Upward Direction)
b. 700 lb (Downward Direction)
c. 800 lb (Upward Direction)
d. 900 lb (Downward Direction)
Centroid of the triangular load.
a. 588 lb (Downward Direction)
b. 598 lb (Upward Direction)
c. 600 lb(Downward Direction)
d. (Upward Direction)
Shear force equation for segment AB.
a.-16.65 psf X squared
b. 16.65 psf X squared
c. -16.75 psf X squared
d. 16.75 psf X squared
Bending moment equation for segment AB.
a.-5.40 psf X cube
b. 5.45 psf X cube
c. -5.55 psf X cube
d. 6.55 psf X cube
Shear force equation for segment BC.
a. 2000 lb - 200 lb/ft ( X)
b. 2100 lb - 300 lb/ft ( X)
c. 2150 lb - 300 lb/ft ( X)
d. 3000 lb - 200 lb/ft ( X)
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.
Show complete solutions and diagram
Transcribed Image Text:A
6 ft-
200 lb/ft
B
CA
1200 lb. ft
★4 ft —— 4 ft -
D
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Solve for the Following
Shear force equation for segment AB.
a.-16.65 psf X squared
b. 16.65 psf X squared
c. -16.75 psf X squared
d. 16.75 psf X squared
Bending moment equation for segment AB.
a.-5.40 psf X cube
b. 5.45 psf X cube
c. -5.55 psf X cube
d. 6.55 psf X cube
Shear force equation for segment BC.
a. 2000 lb - 200 lb/ft ( X)
b. 2100 lb - 300 lb/ft ( X)
c. 2150 lb - 300 lb/ft ( X)
d. 3000 lb - 200 lb/ft ( X)
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
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