Engineering, 03.03.2020 04:30 estermartinez

If the wire has a diameter of 0.5 inin., determine how much it stretches when a distributed load of 140 lb/ftlb/ft acts on the beam. The material remains elastic. Express your answer to three significant figures and include appropriate units.

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Answer from: alissa3329

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

If the wire has a diameter of 0.5 inin., determine how much it stretches when a distributed load of

Answer from: Quest

this is true

explanation:

i looked it up

Answer from: Quest

What? im so confused

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