Calculate Stress & Strain of Marble Column - 1.5m^2, 2.5x10^4kg

In summary, to calculate the stress of a marble column, you need to know the force acting on the column and the cross-sectional area of the column. The formula for stress is stress = force/area. The stress unit is typically measured in N/m^2 or Pa. To calculate the strain, you need to know the change in length of the column and its original length. The formula for strain is strain = change in length/original length. However, the strain cannot be calculated without the original length. The unit of strain is a dimensionless quantity, often expressed as a percentage or in decimal form. The stress and strain are indicators of the strength of a marble column, and factors such as material properties, dimensions, forces, and environmental
  • #1
zyphriss2
18
0
A marble column of cross-sectional area 1.5 m^2 supports a mass of 2.5×10^4 kg.



What is the stress within the column?
What is the strain?

I have already found the stress in the column by taking (25000kgx9.8 m/s^2)/1.5m^2
The problem is i have no clue how to find the strain given this information. I know strain is delta Y/ original length, or strain/youngs modulus.
 
Physics news on Phys.org
  • #2
Without any information regarding its length or Young's modulus, you can't find the strain.
 
  • #3
This was all that was given to me in the problem...does anyone have any clue.
 

1. How do you calculate the stress of a marble column?

To calculate the stress of a marble column, you need to know the force acting on the column and the cross-sectional area of the column. The formula for stress is stress = force/area. In this case, the cross-sectional area is 1.5m^2 and the force is 2.5x10^4kg. So, the stress of the marble column would be 2.5x10^4kg/1.5m^2 = 1.67x10^4kg/m^2.

2. How do you calculate the strain of a marble column?

To calculate the strain of a marble column, you need to know the change in length of the column and its original length. The formula for strain is strain = change in length/original length. In this case, the original length of the column is not given. So, it is not possible to calculate the strain without this information.

3. What is the unit of stress and strain?

The unit of stress is force per unit area, which is typically measured in newtons per square meter (N/m^2) or pascals (Pa). The unit of strain is a dimensionless quantity, as it is a ratio of two lengths. However, it is often expressed in terms of percentage or in decimal form.

4. How does the stress and strain affect the strength of a marble column?

The stress and strain of a marble column are indicators of its strength. A higher stress value means that the column is under more pressure, which can lead to deformation or failure. Similarly, a higher strain value means that the column has undergone more elongation or compression, which can also lead to failure. Therefore, it is important to keep the stress and strain within a safe limit to maintain the strength of the marble column.

5. What factors can affect the stress and strain of a marble column?

The stress and strain of a marble column can be affected by various factors such as the material properties of the marble, the dimensions and geometry of the column, the magnitude and direction of the force acting on the column, and the environmental conditions such as temperature and humidity. These factors can influence the behavior of the marble column and should be taken into consideration when calculating its stress and strain.

Similar threads

  • Introductory Physics Homework Help
Replies
19
Views
918
  • Engineering and Comp Sci Homework Help
Replies
3
Views
625
  • Introductory Physics Homework Help
Replies
6
Views
2K
  • Introductory Physics Homework Help
Replies
17
Views
2K
  • Introductory Physics Homework Help
Replies
2
Views
1K
  • Introductory Physics Homework Help
Replies
29
Views
5K
  • Introductory Physics Homework Help
Replies
1
Views
1K
  • Introductory Physics Homework Help
Replies
6
Views
2K
  • Introductory Physics Homework Help
Replies
2
Views
1K
  • Introductory Physics Homework Help
Replies
21
Views
6K
Back
Top