Respuesta :
Answer:
The maximum mass the bar can support without yielding = 32408.26 kg
Explanation:
Yield stress of the material ([tex]\sigma[/tex]) = 200 M Pa
Diameter of the bar = 4.5 cm = 45 mm
We know that yield stress of the bar is given by the formula
Yield Stress = [tex]\frac{Maximum load}{Area of the bar}[/tex]
⇒ [tex]\sigma[/tex] = [tex]\frac{P_{max} }{A}[/tex] ---------------- (1)
⇒ Area of the bar (A) = [tex]\frac{\pi}{4}[/tex] ×[tex]D^{2}[/tex]
⇒ A = [tex]\frac{\pi}{4}[/tex] × [tex]45^{2}[/tex]
⇒ A = 1589.625 [tex]mm^{2}[/tex]
Put all the values in equation (1) we get
⇒ [tex]P_{max}[/tex] = 200 × 1589.625
⇒ [tex]P_{max}[/tex] = 317925 N
In this bar the [tex]P_{max}[/tex] is equal to the weight of the bar.
⇒ [tex]P_{max}[/tex] = [tex]M_{max}[/tex] × g
Where [tex]M_{max}[/tex] is the maximum mass the bar can support.
⇒ [tex]M_{max}[/tex] = [tex]\frac{P_{max} }{g}[/tex]
Put all the values in the above formula we get
⇒ [tex]M_{max}[/tex] = [tex]\frac{317925}{9.81}[/tex]
⇒ [tex]M_{max}[/tex] = 32408.26 Kg
There fore the maximum mass the bar can support without yielding = 32408.26 kg
