Principal Stress and Maximum Shear Stress

In summary, the beam has a steel material with a modulus of elasticity of 210 GN/m^2. The beams supports are made out of steel with a weight of 90833.33 N and a height of 94166.67 N. The maximum shear stress is 94.17 kN and the maximum bending stress is 243.33 kNm.
  • #1
steevee
6
0

Homework Statement



Hi all,

For my CW I have a question on a simple beam, ABCD and its cross-section. Please see attachment for figures

The material of the beam is steel, where modulus of elasticity, E = 210 GN/m^2

I have been asked to calculate the principle stresses and the maximum shear stress at the top of the beam for the loaded system and 2 m from the left of the beam.

Homework Equations



x-bar = (A1x1 + A2x2 + A3x3)/(A1 + A2 + A3)

y-bar = (A1y1 + A2y2 + A3y3)/(A1 + A2 + A3)

Ixx = bd^3/12 + Ah^2

τ = (VAy-bar)/bI

M/I = σ/y = E/R

The Attempt at a Solution



I have worked out the roller support (A) = 94166.67 N and hinge support (D) = 90833.33 N

Next I drew a shear force diagram and a bending moment diagram. From the diagrams, max shear force = 94.17 kN and max bending = 243.33 kNm

Following on from this I concentrated on the z-bar cross-section. I divided the section into three sub-sections and calculated the area of each part and their centroids. From this I calculated x-bar and y-bar and started to calculate Ixx using Ixx = bd^3/12 + Ah^2

I want to ask whether I am on the right lines because I don't know how to carry on from here.
 

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  • #2
The next step would be to start calculating bending and shear stresses at the requested locations.

Once you have calculated those stresses, you may use Mohr's circle to calculate the principal stresses.
 
  • #3
Thanks for your reply SteamKing

Do you mean σx, σy and τxy? If so, how do I do this?
 
  • #4
You've put down the formulas. Don't you know how to use them?
 
  • #5
I think I understand how to get τ but I'm not so sure about σx and σy
The equation is σ = (M/I)×y right? Is this σx or σy?

Also I'm not sure if the method I used to calculate Ixx was correct. Is it as simple as splitting it into three smaller sections because the examples we have covered in class involved integration.
 
  • #6
Bump. Can anyone help please?
 

1. What is the difference between principal stress and maximum shear stress?

Principal stress refers to the maximum or minimum normal stress acting on a plane, while maximum shear stress refers to the maximum shear force acting on a plane. In simpler terms, principal stress is a measure of the force applied perpendicular to the plane, while maximum shear stress is a measure of the force applied parallel to the plane.

2. How are principal stresses and maximum shear stress calculated?

Principal stresses can be calculated using Mohr's circle, which is a graphical method that takes into account the normal and shear stresses acting on a plane. Maximum shear stress is calculated by taking half of the difference between the two principal stresses.

3. What is the significance of principal stress and maximum shear stress in materials testing?

Principal stress and maximum shear stress are important factors in determining the strength and failure of materials under different loading conditions. Understanding these stresses can help engineers design structures and materials that can withstand different types of stress and prevent failure.

4. How do changes in loading conditions affect principal stress and maximum shear stress?

Changes in loading conditions, such as increasing or decreasing the force applied, can result in changes in both the magnitude and direction of principal stress and maximum shear stress. This can significantly impact the strength and stability of a material or structure.

5. What are some real-world applications of principal stress and maximum shear stress?

Principal stress and maximum shear stress are important in a variety of fields, such as civil engineering, mechanical engineering, and materials science. They are used in the design and analysis of structures, machines, and materials to ensure their safety and reliability under different loading conditions.

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