Answer:
P=568.96 lb
Explanation:
The steel wire BD is made of material A — 36 and has a diameter d = 0.25 in.It is necessary to determine the force P if the point C moves by 0.075 in downwards.
Ф=[tex]arctan\frac{3.125*10^-3}{3}[/tex]
=0.06°
Determination of length BD:
BD=[tex]\sqrt{3^2+4^2}[/tex]
=60 in
Determination of length BD':
BD'=[tex]\sqrt{3^2+4^2-2*3*4*cos0.06}[/tex]
BD'=5.0025 ft
Normal strain is the ratio of the length of the element before and after deformation.
We have all the necessary data to be able to calculate normal strain.
For BD
formula is:
∈[tex]_{BD}[/tex]= BD'-BD/BD
∈[tex]_{BD}[/tex]=5*[tex]10^-4[/tex] in/in
now we can determine the normal stress:
σ[tex]_{BD}[/tex]=E.∈[tex]_{BD}[/tex]
= 14.5 ksi
surface cross sectional area is:
[tex]A_{BD}= \frac{\pi }{4} *d^{2} \\A_{BD}= 0.0490 in^{2}[/tex]
Force [tex]F_{BD}[/tex] is:
[tex]F_{BD}[/tex]=σ[tex]_{BD}[/tex]*[tex]A_{BD}[/tex]
[tex]F_{BD}[/tex]=0.711 kip
determination of static effects
we apply the conditions of equilibrium:
∑[tex]F_{x} =0[/tex]
∑[tex]F_{y}=0[/tex]
∑M = 0
from pic (c)
∑∑[tex]M_{C}[/tex]=0
[tex].711*sin53.13*3-P*6-F_{BD} *cos53.13*4=0\\P=568.96[/tex]
so P=568.96 lb
