Determine the resultant normal force across the cross section at point B. Express your answer to three significant figures and i
nclude appropriate units.
1 answer:
Complete Question
A force of F =460 lb. acts on the curved member as shown. Determine the resultant normal force across the cross section at point B. Express your answer to three significant figures and include appropriate units.(Figures Attached)
Answer:
Resultant Normal Force = 55. 02 lb.
Explanation:
In Order to find the resultant normal force we must first resolve the figure to find out the x and y components of the force across the cross-section at point B. Hence after resolving the figure we can find the resultant normal force by finding the vertical component of the force: ∑ F_y
The normal force acting at point B will be:
∑ F_y :-
N_F + 460(4/5)sin〖30〗^° - 460(3/5) sin〖60〗^° = 0
N_F + 184 - 239.02 = 0
N_F = 55. 02 lb.
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Answer:
a) Ef = 0.755
b) length of specimen( Lf )= 72.26mm
diameter at fracture = 9.598 mm
c) max load ( Fmax ) = 52223.24 N
d) Ft = 51874.67 N
Explanation:
a) Determine the true strain at maximum load and true strain at fracture
True strain at maximum load
Df = 9.598 mm
True strain at fracture
Ef = 0.755
b) determine the length of specimen at maximum load and diameter at fracture
Length of specimen at max load
Lf = 72.26 mm
Diameter at fracture
= 9.598 mm
c) Determine max load force
Fmax = 52223.24 N
d) Determine Load ( F ) on the specimen when a true strain et = 0.25 is applied during tension test
F = 51874.67 N
attached below is a detailed solution of the question above
