Question

Three wooden planks are fastened together by a series of bolts to form a column. The diameter of each bolt is 12 mm and the inner diameter of each washer is 16 mm, which is slightly larger than the diameter of the holes in the planks. Determine the smallest allowable outer diameter d of the washers, knowing that the average normal stress in the bolts is 36 MPa and that the bearing stress between the washers and the planks must not exceed 8.5 MPa.

Answer 1

Answer:

The answer is d = 29.43 [mm]

Explanation:

See attachment to follow steps:

1. We know the bolt stress, and Stress = Force per unit area (σ =F/A)
2. A bolt´s cross section is a circle, we can calculate its area
3. We know the bolt´s diameter to be 12[mm], we calculate its area to be 113.10 [mm]
4. Going back to step one, we can calculate the force that is tensioning the bolt and compressing the wooden planks. This force is F = 4071.60 [N]
5. Now we want to calculate the minimum outer diameter of the washer and not exceed the bearing strength on the planks which is 8.5[Mpa]
6. Again, we know that σ = F/A, this time we are using the area of the washer
7. We know the ID of the washer, but not the OD (d), writing the equation of the area in terms of d and ID is helpful
8. Now that we know the equation of the area and that it is equal to F/σmax we can solve for d (outer diameter of washer)
9. We solve for d and write down the equation
10. Substitute all known values into the equation for the outer diameter of the washer
11. Compute and get the outer diameter, d = 29.43 [mm]

Any higher value of d, will decrease the Stress when the known Force is acting on the washer. Any lower value of d will increase the Stress and make it exceed the allowable of 8.5 [Mpa]

Answer 2

Stress of an object is the force applied on it per unit area.

The smallest allowable outer diameter of the washers is 29.93 mm.

What is stress?

Stress of an object is the force applied on it per unit area. It can be given as,

$\sigma=\dfrac{F}{A}$

Here, $F$ is the force and $A$ is the area.

Given information-

The diameter of each bolt is 12 mm.

The inner diameter of each washer is 16 mm.

The average normal stress in the bolts is 36 MPa.

The bearing stress between the washers and the planks must not exceed 8.5 MPa.

As the average normal stress in the bolts is 36 MPa. Thus,

$\sigma_{avg}=\dfrac{F}{A_B} \\36=\dfrac{F}{\dfrac{12^2\pi}{4}} \\F=4071.60\rm N$

Suppose the outer diameter pf the washer is $d$. The area of the wooden plank is,

$A=\dfrac{d^2-10^2}{4} \times \pi$

The maximum stress with the above formula can be given as,

$\sigma_{max}=\dfrac{F}{A}$

Put the value of area in the above equation as,

$\sigma_{max}=\dfrac{F}{\dfrac{d^2-10^2}{4} \times \pi}\\8.5=\dfrac{4071.60}{\dfrac{d^2-10^2}{4} \times \pi}\\d=29.93 \rm mm$

Thus the smallest allowable outer diameter of the washers is 29.93 mm.

Learn more about the stress here;

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