Answer:
δ_AB = 0.0333 in
Explanation:
Given:
- The complete question is as follows:
" The rigid beam is supported by a pin at C and an A−36
steel guy wire AB. If the wire has a diameter of 0.5 in.
determine how much it stretches when a distributed load of
w=140 lb / ft acts on the beam. The material remains elastic."
- Properties for A-36 steel guy wire:
Young's Modulus E = 29,000 ksi
Yield strength σ_y = 250 MPa
- The diameter of the wire d = 0.5 in
- The distributed load w = 140 lb/ft
Find:
Determine how much it stretches under distributed load
Solution:
- Compute the surface cross section area A of wire:
A = π*d^2 / 4
A = π*0.5^2 / 4
A = π / 16 in^2
- Apply equilibrium conditions on the rigid beam ( See Attachment ). Calculate the axial force in the steel guy wire F_AB
Sum of moments about point C = 0
-w*L*L/2 + F_AB*10*sin ( 30 ) = 0
F_AB = w*L*L/10*2*sin(30)
F_AB = 140*10*10/10*2*sin(30)
F_AB = 1400 lb
- The normal stress in wire σ_AB is given by:
σ_AB = F_AB / A
σ_AB = 1400*16 / 1000*π
σ_AB = 7.13014 ksi
- Assuming only elastic deformations the strain in wire ε_AB would be:
ε_AB = σ_AB / E
ε_AB = 7.13014 / (29*10^3)
ε_AB = 0.00024
- The change in length of the wire δ_AB can be determined from extension formula:
δ_AB = ε_AB*L_AB
δ_AB = 0.00024*120 / cos(30)
δ_AB = 0.0333 in
