Question

The pole is supported by a pin at C and an A-36 steel guy wire AB. If the wire has a diameter of 0.2 in., determine how much it stretches when a horizontal force of 2.5 kip acts on the pole.

Answer 1

We have that the wire stretch  is mathematically given as

Ls=0.304inches

Force

Question Parameters:

• the wire has a diameter of 0.2 in.
• when a horizontal force of 2.5 kip acts on the pole

Generally the equation for the moment at point C   is mathematically given as

FabCos60(7)-2.5*4=0

Fab=2857lb

Normal stress on wire

$S=\frac{Fab}{Aab}\\\\S=2857/(\pi/4)*0.2^2\\\\S=90.95ksi$

Therefore

The length at normal state is

$Ln=\frac{7*12}{sin60}\\\\Ln=96.99inch\\\\Where\\\\Eab=0.003136in\\\\Therefore\\\\\Stretch length is given as\\\\Ls=0.003136in*96.99inch$

Ls=0.304in

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