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 inne
r 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.
1 answer:
Answer:
The answer is d = 29.43 [mm]
Explanation:
See attachment to follow steps:
- We know the bolt stress, and Stress = Force per unit area (σ =F/A)
- A bolt´s cross section is a circle, we can calculate its area
- We know the bolt´s diameter to be 12[mm], we calculate its area to be 113.10 [mm]
- 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]
- 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]
- Again, we know that σ = F/A, this time we are using the area of the washer
- 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
- 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)
- We solve for d and write down the equation
- Substitute all known values into the equation for the outer diameter of the washer
- 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]
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The question is incomplete. The complete question is :
The Rocket Club is planning to launch a pair of model rockets. To build the rocket, the club needs a rocket body paired with an engine. The table lists the mass of three possible rocket bodies and the force generated by three possible engines.
A 4-column table with 3 rows. The first column labeled Body has entries 1, 2, 3. The second column labeled Mass (grams) has entries 500, 1500, 750. The third column labeled Engine has entries 1, 2, 3. The fourth column labeled Force (Newtons) has entries 25, 20, 30.
Based on Newton’s laws of motion, which combination of rocket bodies and engines will result in the acceleration of 40 m/s2 at the start of the launch?
Body 3 + Engine 1
Body 2 + Engine 2
Body 1 + Engine 2
Body 1 + Engine 1
Solution :
Given :
Body Mass (gram) Engine Force (newtons)
1 500 1 25
2 1500 2 20
3 750 3 30
The body 1 has a mass of 500 gram which is equal to 0.5 kg
And engine 2 has a force of 20 newtons.
We know that according to Newton's laws of motion,
Force = mass x acceleration
20 = 0.5 x acceleration
Acceleration
Therefore, based on laws of motion of Newton, the Body 1 + Engine 2 combination of the rocket bodies and engines will result in an acceleration of at the start of the launch.