B Force exerted at the end of a beam

[Tone L]

Hi all.
Please see image I created as reference to this question.I have a uniform rod which is fixed to a structure, at the end of the rod there is a point mass attached to it.

I am trying to calculate the downward force (torque) made at the end of the rod where the mass resides. I want to know how much stress is being put on the fixed point.

mass of Rod = 2kg
mass of Mass = .5kg

length of Rod = 4.5 m

I am not sure if this is a torque problem.

 

[russ_watters]

I am trying to calculate the downward force (torque) made at the end of the rod where the mass resides. I want to know how much stress is being put on the fixed point...

I am not sure if this is a torque problem.

"Torque", "stress", "force"? You'll have to tell us what you want, but I would certainly hope the force and torque are easy to calculate! [caveat about the construction...]

Is this homework or real-world?

 

[berkeman]

@Tone L -- in your other thread last week, you said that you understood why I mentioned I-Beams and using Triangles in construction of mechanical structures. But now you've posted another structure that uses neither of those. How come?

Ah ! Obvious. Thanks :D

 

[Tone L]

@Tone L -- in your other thread last week, you said that you understood why I mentioned I-Beams and using Triangles in construction of mechanical structures. But now you've posted another structure that uses neither of those. How come?

For simplicity. The force exerted at the end of the beam should only be a function of beam length, mass and mass of object at the end, correct? The structure of the beam should have negligible, if any effect on the force exerted at the end of the beam?

 

[Tone L]

"Torque", "stress", "force"? You'll have to tell us what you want, but I would certainly hope the force and torque are easy to calculate! [caveat about the construction...]

Is this homework or real-world?

Real-world, I guess its not easy enough, hence me posting it..?

 

[jbriggs444]

Real-world, I guess its not easy enough, hence me posting it..?

How are you affixing the beam to the top of the post? i.e. where is the downforce applied?

Nails will pull out. Screws into end grain are not a whole lot better. My choice would likely be a metal strap fastened to the left side of the post (screwed into side grain and under transverse load = strong), bent over the edge and then fastened to the top of the beam (screwed into side grain and under transverse load = strong).

Now you have a simple lever. Strap, hard point and load.

 

[berkeman]

For simplicity. The force exerted at the end of the beam should only be a function of beam length, mass and mass of object at the end, correct? The structure of the beam should have negligible, if any effect on the force exerted at the end of the beam?

To a first order, but the deflection and deformation of the beam will be affected by its rigidity (which comes from its construction). It will not take much to plastically deform the beam, given what you have shown.

 

[Tone L]

How are you affixing the beam to the top of the post? i.e. where is the downforce applied?

Nails will pull out. Screws into end grain are not a whole lot better. My choice would likely be a metal strap fastened to the left side of the post (screwed into side grain and under transverse load = strong), bent over the edge and then fastened to the top of the beam (screwed into side grain and under transverse load = strong).

Now you have a simple lever. Strap, hard point and load.

The beam will be attaching to a metal post, not wood sorry for confusion. The post has 1/4-20 tapped holes, it will definitely not pull out.

 

[jbriggs444]

The beam will be attaching to a metal post, not wood sorry for confusion. The post has 1/4-20 tapped holes, it will definitely not pull out.

Good. So now where are those holes located? (or where can we count the force centers of the bolt heads that are doing the hold-down work).

 

[russ_watters]

Real-world, I guess its not easy enough, hence me posting it..?

Well, you know the weight of the weight and the length of the beam and you know the weight of the beam. Do you know the equation for torque? Can't you just plug in the numbers? That will get you halfway to what you need. We're trying to get you to try a little bit here - you'll learn more and become more self sufficient. This is why for homework we require an attempt at a solution and this is similar to homework.

As said, exactly what is going on at the connection can be complicated. If you look closely at bridges you will see simplified connections like pins and rollers. This isn't to make the math easier, it's required for the bridge to not self destruct. Yours probably won't, but you can still use or assume simplified connections. This really is a simple matter of calculating torques from linear forces and vice versa. There's 4 linear forces and 4 torques.

Step 1: Calculate the applied torque at the wall.
Step 2: Calculate the downward force at the back connection to counter the torque at the wall.
Step 3: Calculate the upward force at the front connection.

 

[Tone L]

Sure I guess I was confused when calcualting torque. The equation seems to require an angle relative to the axis of rotation, of course. However isn't the mass parelell to the axis of rotation, theta = 180/0?
Torque=l*F*sin(theta).

 

[jbriggs444]

Sure I guess I was confused when calcualting torque. The equation seems to require an angle relative to the axis of rotation, of course. However isn't the mass parelell to the axis of rotation, theta = 180/0?
Torque=l*F*sin(theta).

Ummm, how are you measuring theta. It is the angle between what and what?

Angles are normally measured between two [directed] lines that intersect in a point. So what two lines are being considered here?

 

[Tone L]

Ummm, how are you measuring theta. It is the angle between what and what?

Angles are normally measured between two [directed] lines that intersect in a point. So what two lines are being considered here?

The fixed point and mass, which i assumed to be a point mass.

 

[jbriggs444]

The fixed point and mass, which i assumed to be a point mass.

Those are two points. Angles are measured between lines, not points. Torque, in particular is determined based on the angle made between two lines:

From the reference point to the point where a force is applied.
From the point where the force is applied in the direction of the force.
The sine of this angle is relevant.

 

[Tone L]

Those are two points. Angles are measured between lines, not points. Torque, in particular is determined based on the angle made between two lines:

From the reference point to the point where a force is applied.
From the point where the force is applied in the direction of the force.
The sine of this angle is relevant.

Hm.. So one line would be the structure/post and the other line would be the beam/rod, making a perpendicular angle?

 

[jbriggs444]

Hm.. So one line would be the structure/post and the other line would be the beam/rod, making a perpendicular angle?

No.

Where are you placing the reference axis about which we are computing torque? I would recommend placing it at the right hand edge of the post where the beam rests.

 

[russ_watters]

@Tone L Hint: the two relevant lines are drawn correctly in your diagram!

 

[Tone L]

No.

Where are you placing the reference axis about which we are computing torque? I would recommend placing it at the right hand edge of the post where the beam rests.

Yes that it is where I located it, thanks for the help.