# Finding the Force P from Normal Stress

• Northbysouth
In summary: However, if the pin is too large or the shear area is not sufficient, then the link can fail and the beam will be unable to support its own weight.
Northbysouth

## Homework Statement

In the steel structure shown, a 6‐mm‐diameter pin is used at C and 10‐mm‐diameter pins are
used at B and D. The ultimate shearing stress is 150 MPa at all connections, and the ultimate normal stress is 400 MPa in link BD. Knowing that a factor of safety of 3.0 is desired, determine the largest load P that can be applied at A. Note that link BD is not reinforced around the pin holes.

F.S = σultallow

σ = F/A

## The Attempt at a Solution

I understand the parts concernging the ultimate shearing stress, my confusion is with the normal stress.

When I draw a FBD of beam AB, I assign the force at B, By, to point upwards and the force at C, Cy, to point downwards.

ƩMc = -By(.120) + P(0.280)

By = 2.33P

My question is, does this this result mean that member BD is in compression or tension? Having drawn a FBD of AC I understand that the forces are acting on beam AC, thus while By points in the +y-direction, if I were to draw a FBD of BD, the force, By, would point in the opposite direction thereby indicating that member BD is in tension.

Hence when I use the normal stress equation

σ = F/A

where F is the internal force and A is the cross sectional area, why is the diameter of the pin removed form the width of member BD?

#### Attachments

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1. Link BD is in tension.

2. The diameter of the pin is removed from link BD width because that material is not present and cannot be considered when calculating stress. If you look at the pin at point B or D and the link, draw a FBD of the link and make a cut horizontally thru the diameter of the pin. Only the material outside the pin hole can support any load.

If link BD had been in compression instead of tension, would you still need to remove the diameter of the pin?

If link BD were in compression, then you would have to look at the shear area available to keep pin B from tearing out of the pinhole. In this case, a vertical cut on either side of the pin thru the link would be analyzed to determine how much area is able to resist the tendency of the pin to shear thru the end of the link.

If there is sufficient shear area and the pin is OK, then buckling of the link itself should be checked.

Based on the given information, it appears that the force on member BD is in tension. This is because the force at B is pointing in the opposite direction of the force at C, indicating that BD is being pulled apart.

In terms of the normal stress equation, the diameter of the pin is not taken into account because it is assumed that the load is being distributed evenly across the entire cross sectional area of member BD. This is a simplification, but it is a common assumption in structural engineering calculations. However, if the diameter of the pin is significant in comparison to the cross sectional area of BD, it should be taken into account in the calculation.

## 1. What is the definition of normal stress in relation to finding the force P?

Normal stress is the force per unit area that is applied perpendicular to a surface. In the context of finding the force P, it refers to the stress acting on a surface in a direction perpendicular to the surface.

## 2. How is the force P calculated from normal stress?

The force P can be calculated from normal stress by multiplying the normal stress by the surface area that it is acting on. This can be represented by the equation P = σA, where P is the force, σ is the normal stress, and A is the surface area.

## 3. What are some common units of measurement for normal stress and force P?

Normal stress is typically measured in units of force per unit area, such as Pascals (Pa) or pounds per square inch (psi). Force P is measured in units of force, such as Newtons (N) or pounds (lbs).

## 4. Can normal stress and force P have different magnitudes and directions?

Yes, normal stress and force P can have different magnitudes and directions. Normal stress is a scalar quantity, meaning it only has magnitude, while force P is a vector quantity, meaning it has both magnitude and direction.

## 5. How is normal stress and force P related to each other?

Normal stress and force P are directly related to each other through the surface area that the stress is acting on. The greater the surface area, the greater the force P will be for a given normal stress. This relationship is represented by the equation P = σA.

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