What is the relationship between normal and shear forces in structural analysis?

In summary, the conversation discussed the use of previous methods to solve a problem involving normal and shear forces. It was determined that the principle of double shear applies and that the normal force and shear force have a relationship of V=P/2. The relationship between normal and shear forces was further clarified in a question involving pin-joints and the correct answer was obtained by assuming the arrow acting on the member is a normal force.
  • #1
savva
39
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Homework Statement


Given in uploaded file View attachment Q8.37.pdf


Homework Equations


σallow=[itex]\frac{P}{A}[/itex]
τallow=[itex]\frac{V}{A}[/itex]

The Attempt at a Solution


Given in uploaded file View attachment Q8.37 WORKING.pdf

Answers are d=16mm, h=10mm
I tried using methods used in previous questions but it didn't work out for me here, if anyone could give a hand, would be greatly appreciated.

I would also like to clarify a couple of other things related to the theory that I'm not sure about.
- I'm not sure what the relationship is between normal force (P) and shear force (V), particularly if given a normal force (P) for example and the question asks to find shear stress. Would Shear force (V) = P/2. I have basically used this principle and it has been generally successful but I am not sure if I'm getting answers correct because double shear applies.

-Secondly, I was doing a question that involved a couple of beams that had pin-joints on each side. Subsequently, I found the external force applied on each pin joint. I have attached a file and drawn an arrow on the diagram. I would like to verify if the arrow acting on the member is a normal force. Because that's what I assumed and when solving the normal stress it gave me the correct answer. The attachment of the problem is below
View attachment Q8.44.pdf
 
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  • #2


savva said:

Homework Statement


Given in uploaded file View attachment 47970


Homework Equations


σallow=[itex]\frac{P}{A}[/itex]
τallow=[itex]\frac{V}{A}[/itex]

The Attempt at a Solution


Given in uploaded file View attachment 47971

Answers are d=16mm, h=10mm
I tried using methods used in previous questions but it didn't work out for me here, if anyone could give a hand, would be greatly appreciated.
The area of a circle is pi d^2/4. You forgot the 4. Otherwise your calcs look good for tensile stress in the rod and 'punching' shear stress in the supporting material.
I would also like to clarify a couple of other things related to the theory that I'm not sure about.
- I'm not sure what the relationship is between normal force (P) and shear force (V), particularly if given a normal force (P) for example and the question asks to find shear stress. Would Shear force (V) = P/2. I have basically used this principle and it has been generally successful but I am not sure if I'm getting answers correct because double shear applies.
Well, as in your example at the eye, 25 kN vert load is 12.5 kN shear load in the curved rod either side of the load (double shear).
-Secondly, I was doing a question that involved a couple of beams that had pin-joints on each side. Subsequently, I found the external force applied on each pin joint. I have attached a file and drawn an arrow on the diagram. I would like to verify if the arrow acting on the member is a normal force. Because that's what I assumed and when solving the normal stress it gave me the correct answer. The attachment of the problem is below
View attachment 47972
Yes it is the normal axial force in the member, causing shear stress in the bolt (double shear). At the left support, however, you have both axial and shear forces in the member. Their resultant is the shear force on the pin...single shear.
 

What is the purpose of finding the diameter and thickness using allowable normal and shear stress equations?

The purpose of finding the diameter and thickness using allowable normal and shear stress equations is to determine the appropriate dimensions for a structural component, such as a beam or column, that can withstand the expected loads and stresses without failing.

What are the allowable normal and shear stress equations?

The allowable normal stress equation is typically represented as σallow = F/A, where σallow is the maximum stress that a material can withstand, F is the applied force, and A is the cross-sectional area of the material. The allowable shear stress equation is usually represented as τallow = V/A, where τallow is the maximum shear stress that a material can withstand, V is the applied shear force, and A is the cross-sectional area of the material.

How do you determine the diameter and thickness using allowable normal and shear stress equations?

To find the appropriate diameter and thickness, you will need to know the expected loads and stresses on the structural component, as well as the material properties. Then, you can rearrange the allowable stress equations to solve for the desired dimension. For example, to find the diameter, you would use the equation d = 2(F/σallow), where d is the diameter, F is the applied force, and σallow is the allowable normal stress for the material.

What factors can affect the determination of diameter and thickness using allowable normal and shear stress equations?

The determination of diameter and thickness using allowable normal and shear stress equations can be affected by several factors, including the type and properties of the material, the geometry of the structural component, and the expected loads and stresses. Additionally, factors such as temperature, corrosion, and fatigue can also impact the maximum allowable stresses for a material.

Are there any limitations to using allowable normal and shear stress equations for determining diameter and thickness?

While allowable normal and shear stress equations are commonly used for determining diameter and thickness, there are some limitations to consider. These equations assume that the material is homogeneous and isotropic, and that the loads and stresses are applied uniformly. In reality, these conditions may not always be met, and other factors such as stress concentrations and material defects may need to be taken into account for more accurate results.

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