Question

The rigid pipe is supported by a pin at A and an A-36 guy wire BD. If the wire has a diameter of 0.25 in., determine the load P if the end C is displaced 0.075 in. downward.

Answer 1

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=$arctan\frac{3.125*10^-3}{3}$

  =0.06°

Determination of length BD:  

BD=$\sqrt{3^2+4^2}$

    =60 in

Determination of length BD':  

BD'=$\sqrt{3^2+4^2-2*3*4*cos0.06}$

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:

∈$_{BD}$=  BD'-BD/BD

∈$_{BD}$=5*$10^-4$ in/in

now we can determine the normal stress:

σ$_{BD}$=E.∈$_{BD}$

     = 14.5 ksi

surface cross sectional area is:

$A_{BD}= \frac{\pi }{4} *d^{2} \\A_{BD}= 0.0490 in^{2}$

Force $F_{BD}$ is:

$F_{BD}$=σ$_{BD}$*$A_{BD}$

$F_{BD}$=0.711 kip

determination of static effects

we apply the conditions of equilibrium:

∑$F_{x} =0$

∑$F_{y}=0$

∑M  = 0

from pic (c)

∑∑$M_{C}$=0

$.711*sin53.13*3-P*6-F_{BD} *cos53.13*4=0\\P=568.96$

so P=568.96 lb