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

The maximum mass a cylindrical bar with a diameter of 4.5 cm can support without yielding, given the material yields at 200 MPa, is 32416 kg.

Step-by-step explanation:

To calculate the maximum mass that a cylindrical bar with a diameter of 4.5 cm can support without yielding, given that the material yields at 200 MPa, we start by converting the yield stress into an appropriate unit and calculating the cross-sectional area of the bar.

The yield stress is 200 MPa, which is equivalent to 200 x 106 N/m2. The cross-sectional area, A, of the cylindrical bar is calculated using the formula for the area of a circle, A = πr2, where r is the radius in meters. The diameter of 4.5 cm is equivalent to 0.045 m, making the radius 0.0225 m. Therefore, A = π(0.0225 m)2 ≈ 0.00159 m2.

The maximum force, F_max, that this cross-sectional area can support without yielding is calculated by multiplying the yield stress by the area: F_max = 200 x 106 N/m2 x 0.00159 m2 = 318000 N.

To calculate the maximum mass, m, we use the relationship between force and weight: F = m x g, where g is the acceleration due to gravity (9.81 m/s2). Rearranging for mass gives us m = F_max / g. So the maximum mass the bar can support without yielding is m = 318000 N / 9.81 m/s2 ≈ 32416 kg. The answer, rounded to the nearest kilogram as per the instructions, is 32416 kg.

Answer 2

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

Step-by-step 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 =
`(Maximum load)/(Area of the bar)`

⇒
`\sigma` =
`(P_(max) )/(A)` ---------------- (1)

⇒ Area of the bar (A) =
`(\pi)/(4)` ×
`D^(2)`

⇒ A =
`(\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)` =
`(P_(max) )/(g)`

Put all the values in the above formula we get

⇒
`M_(max)` =
`(317925)/(9.81)`

⇒
`M_(max)` = 32408.26 Kg

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