Question

The rigid beam is supported by a pin at C and an A992 steel guy wire AB of length 6 ft. If the wire has a diameter of 0.2 in., determine how much it stretches when a distributed load of w=200lb/ft acts on the beam. The wire remains elastic

Answer 1

Answer:

change in length = 0.0913 in

Explanation:

given data

length = 6 ft

diameter = 0.2 in

load of w = 200 lb/ft

solution

we apply here first equilibrium moment about C that is express as

∑M(c)  = 0    .............1

so it can express as to get force in AB

10× 200 × ( 5) - (T cos(30)) × 10  =  0

solve it we get

Tension is wire T(AB)  = 1154.7 lb

and we know modulus of elasticity will be here for A992

E = 29000 ksi

and area will be

Area = $\frac{\pi }{4}\times 0.2^2$  

so change in length will be express as

change in length = $\frac{PL}{AE}$      ................2

put here value and we get

change in length = $\frac{1154.7 \times 6 \times 12}{\frac{\pi }{4}\times 0.2^2 \times 29000 \times 1000}$

change in length = 0.0913 in