191k views
1 vote
(b) A steel column is 3 m long and 0.4 m diameter. It carries a load of 50MN.Given that the Modulus of elasticity is 200GPa, calculate the Compressive stress and determine how much the Column is compressed.

1 Answer

4 votes

Answer:

Step-by-step explanation:

To calculate the compressive stress and the amount of compression in the steel column, we can use the following formulas:

1. Compressive Stress:

Compressive stress (σ) is calculated by dividing the force (load) applied to the column by the cross-sectional area of the column.

σ = F / A

where:

σ = Compressive stress

F = Load applied to the column

A = Cross-sectional area of the column

2. Compression:

The amount of compression (∆L) can be calculated using Hooke's Law, which states that the deformation of a material is directly proportional to the applied force.

∆L = (F * L) / (A * E)

where:

∆L = Amount of compression

F = Load applied to the column

L = Original length of the column

A = Cross-sectional area of the column

E = Modulus of elasticity of the material

Given the following information:

- Load (F) = 50 MN (50 * 10^6 N)

- Length (L) = 3 m

- Diameter (d) = 0.4 m

- Modulus of elasticity (E) = 200 GPa (200 * 10^9 Pa)

First, let's calculate the cross-sectional area (A) of the column using the diameter:

A = π * (d/2)^2

= π * (0.4/2)^2

≈ 0.1257 m^2

Now, we can calculate the compressive stress (σ):

σ = F / A

= (50 * 10^6 N) / 0.1257 m^2

≈ 398.408 MPa

Finally, we can calculate the amount of compression (∆L):

∆L = (F * L) / (A * E)

= ((50 * 10^6 N) * 3 m) / (0.1257 m^2 * 200 * 10^9 Pa)

≈ 0.5962 m (or 596.2 mm)

Therefore, the compressive stress in the steel column is approximately 398.408 MPa, and the column is compressed by approximately 0.5962 m (or 596.2 mm).

User Logan H
by
7.6k points