Question

For the material in the previous question that yields at 200 MPa, what is the maximum mass, in kg, that a cylindrical bar with diameter of 4.5 cm can support without yielding ? Recall that a mass of 1 kg has a weight of 9.81 N at sea-level. Answer Format: X (no decimal) Unit: kg Example: of you calculate 41321.78.... enter 41322

Answer 1

Answer:

The maximum mass the bar can support without yielding = 32408.26 kg

Explanation:

Yield stress of the material ($\sigma$) = 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 = $\frac{Maximum load}{Area of the bar}$

⇒                                $\sigma$ = $\frac{P_{max} }{A}$  ---------------- (1)

⇒ Area of the bar (A) = $\frac{\pi}{4}$ ×$D^{2}$

⇒                            A  = $\frac{\pi}{4}$ × $45^{2}$

⇒                            A = 1589.625 $mm^{2}$

Put all the values in equation (1) we get

⇒ $P_{max}$ = 200 × 1589.625

⇒ $P_{max}$ = 317925 N

In this bar the $P_{max}$ is equal to the weight of the bar.

⇒ $P_{max}$ = $M_{max}$ × g

Where $M_{max}$ is the maximum mass the bar can support.

⇒ $M_{max}$ = $\frac{P_{max} }{g}$

Put all the values in the above formula we get

⇒ $M_{max}$ = $\frac{317925}{9.81}$

⇒ $M_{max}$ = 32408.26 Kg

There fore the maximum mass the bar can support without yielding = 32408.26 kg