Answer:
Explanation:
(1) first part of solution: we know stress= Force/Area ; Force= 400lb and Area as given
(2) Second part of the solution its the standard formula for Hoop stress and the formula explains all the terms
(3) Third part of the solution we know Strain= (Change in length)/original length and also the strain are substituted in terms of strain and youngs modulus
Further explanation are given in the attach document.
The normal stress in the band is 1600 psi. The pressure exerted on the cylinder is 25 psi. The distance elongation of the half band stretch is 0.0140 in
From the given information, the stress in the steel band can be determined by using the formula:
[tex]\mathbf{\sigma _1 = \dfrac{F}{A}}[/tex]
where;
∴
[tex]\mathbf{\sigma _1 = \dfrac{400 }{ 2 \times \dfrac{1}{8} }}[/tex]
[tex]\mathbf{\sigma _1 = \dfrac{400 \times 8 }{ 2 }}[/tex]
[tex]\mathbf{\sigma _1 = 1600 \ psi}[/tex]
The hoop stress as a result of the pressure applied while tightening the bolts is estimated by using the expression:
[tex]\mathbf{\sigma = \dfrac{pr}{t}}[/tex]
[tex]\mathbf{1600= \dfrac{p\times 8}{1/8}}[/tex]
p = 25 psi
The distance elongation of the half band stretch is computed by using the formula:
[tex]\mathbf{\delta = \varepsilon _1L}[/tex]
[tex]\mathbf{\delta = \dfrac{\sigma}{E}\times \pi \times r_{mean}}[/tex]
[tex]\mathbf{\delta = \dfrac{1600}{29*10^{5}}\times \pi \times (8+ \dfrac{1}{16})}[/tex]
[tex]\mathbf{\delta =0.01397 \ in}[/tex]
[tex]\mathbf{\delta \simeq0.0140 \ in}[/tex]
Learn more about normal stress here:
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