Respuesta :
Answer:
The correct answer to the following question will be "1.23 mm".
Explanation:
The given values are:
Average normal stress,
[tex]\sigma=200 \ MPa[/tex]
Elastic module,
[tex]E = 77 \ GPa[/tex]
Length,
[tex]L = 570 \ mm[/tex]
To find the deformation, firstly we have to find the equation:
⇒  [tex]\delta=\Sigma\frac{N_{i}L_{i}}{E \ A_{i}}[/tex]
⇒   [tex]=\frac{P(\frac{L}{H})}{E(bt)} +\frac{P(\frac{L}{2})}{E (bt)(\frac{2}{3})}+\frac{P(\frac{L}{H})}{Ebt}[/tex]
On taking "[tex]\frac{PL}{Ebt}[/tex]" as common, we get
⇒   [tex]=\frac{\frac{PL}{Ebt}}{[\frac{1}{4}+\frac{3}{4}+\frac{1}{4}]}[/tex]
⇒   [tex]=\frac{5PL}{HEbt}[/tex]
Now,
The stress at the middle will be:
⇒  [tex]\sigma=\frac{P}{A}[/tex]
⇒   [tex]=\frac{P}{(\frac{2}{3})bt}[/tex]
⇒   [tex]=\frac{3P}{2bt}[/tex]
⇒  [tex]\frac{P}{bt} =\frac{2 \sigma}{3}[/tex]
Hence,
⇒  [tex]\delta=\frac{5 \sigma \ L}{6E}[/tex]
On putting the estimated values, we get
⇒   [tex]=\frac{5\times 200\times 570}{6\times 77\times 10^3}[/tex]
⇒   [tex]=\frac{570000}{462000}[/tex]
⇒   [tex]=1.23 \ mm[/tex] Â
