Load at the end of a pier.

[Derbyboy]

I plan to build a dock on a rock.

The dock will be 4 m long, with 2 meters on land (rock) and 2 meters over the water. The dock is secured with two fastenings on the rock. One at 10 cm and one at 2 meters.

Can someone help me calculate the force at the outermost part of the dock with 100 kg at the very end?

We can disregard the dock's own weight.

 

[Derbyboy]

What I'm considering is the fastening furthest inland, i.e., at 10 cm.

The idea is to use an L-anchor shoe anchored with anchor adhesive at 30 cm.

At 10 cm, I'll drill through the beam (75x200) and secure an acid-resistant threaded rod M20 into the rock. Then I'll tighten a washer and nut on both sides of the beam.

According to the anchor adhesive specifications, it can withstand 3000 kg (M20) in tensile force before it releases.

The lever arm is 2 meters; how much load is required to reach 3000 kg?

 

[Violina]

You cannot ignore the bridge's own weight, it will be significant

 

[Violina]

With two meters of free hanging, you have a substantial lever arm

 

[Byurn]

You cannot ignore the self-weight of the bridge, it will be significant

The self-weight does not contribute to the tensile force in the innermost attachment.

Set up a force balance and take a look.

 

[Derbyboy]

What does the formula look like?

The self-weight is probably around 200-250 kg without snow and with a moisture content of 16 percent.

 

[Byurn]

What does the formula look like?

The dead weight is perhaps 200-250 kg without snow and with a moisture content of 16 percent.

The formula?
Draw the loads, and you'll see the solution quite easily. See for example this link
https://fysik.ugglansno.se/havstanger/

 

[Plupp 618974]

What does the formula look like?

The dead weight is probably about 200-250 kg without snow and with a moisture content of 16 percent.

You have force and moment equilibrium everywhere; otherwise, the structure is in motion. The simplest way here is to set this around the support at 10 cm, in my opinion.

Google force and moment equilibrium, and it will most certainly explain this very pedagogically.

 

[Derbyboy]

3000/2 = 1500 kg

So if I place 1500 kg 2 meters from the equilibrium position, I get a pulling force of 3000 kg at the innermost point.

 

[Plupp 618974]

3000/2 = 1500 kg

So if I place 1500 kg 2 meters out from the equilibrium point, I get a pulling force of 3000 kg at the innermost position.

Try again!

 

[Derbyboy]

It is roughly, I ignore that the attachment is 10 cm in on the side where the force goes upward.

 

[Byurn]

It's approximate, I'm disregarding the fact that the fastening is 10 cm in on the side where the force is going upward.

You have two meters on both sides of the pivot point.

Your big concern will probably be the deflection in the construction.

 

[Derbyboy]

Ah, since the distance is the same length, 2 meters, it requires 3000 kg down for the traction force at the attachment to be 3000 kg.

 

[Byurn]

Ah since the distance is the same length, 2 meters, it requires 3000 kg down for the pulling force at the attachment point to be 3000 kg.

But then you probably have a dock that has broken due to other reasons

 

[Derbyboy]

Yes, there are different opinions about what my concern is, some say the attachment others say the bending.