Calculating Angular Deflection in a Welded Steel Bracket

[hatchelhoff]

Homework Statement

The Figure shows a welded steel bracket loaded by a force F = 5 kN.

Homework Equations

[/B]

The Attempt at a Solution

I know that the total deflection is the deflection of the beam due to F. And I also need to take into account the angular deflection at the joint between the beam and the bracket.

Im not sure how to find the angular deflection at the joint between the beam and the bracket.[/B]

 

[Chestermiller]

You need to apply the macroscopic conditions, to determine MB and Rx as a function of F and the geometry.

 

[PhanthomJay]

Im not sure how to find the angular deflection at the joint between the beam and the bracket.

since you are apparently allowed to use tables, look it up in a table that gives the rotation angle at the midpoint of the end plate, using the moment at the end of the cantilever as the applied couple at that point. Otherwise , use the calculus of beam theory.

 

[hatchelhoff]

You need to apply the macroscopic conditions, to determine MB and Rx as a function of F and the geometry.

MB = F*Distance = 5KN*410 mm = 2050 NM
R1 = R2 = MB/L = 2050NM/0.41M = 5000 N

Im not sure how to use these values to find the angular deflection.

 

[hatchelhoff]

since you are apparently allowed to use tables, look it up in a table that gives the rotation angle at the midpoint of the end plate, using the moment at the end of the cantilever as the applied couple at that point. Otherwise , use the calculus of beam theory.

I can't seem to find any such table. Can you point me int the right direction.

 

[CivilSigma]

What do you mean by angular deflection? The rotation in Radians ?
I'm not too sure why you would need this value to compute the maximum deflection of the cantilever beam.

Can you please elaborate.

 

[Chestermiller]

MB = F*Distance = 5KN*410 mm = 2050 NM
R1 = R2 = MB/L = 2050NM/0.41M = 5000 N

Im not sure how to use these values to find the angular deflection.

So you have MB. From the solution to the second problem, you can get the initial slope. You multiply that by the length of the beam to get the additional dip at the far end.

 

[PhanthomJay]

Ask and it shall be given unto thee, seek and ye shall find. Merry Christmas.

http://virtual.cvut.cz/beams/
Table 2 fig 8. Then follow Mr. Miller's last post.

 

[hatchelhoff]

CivilSigma I do mean the rotation in radans. In need this because the force F causes a deflection in the beam, but it also causes a deformation of the upright bracket which causes an additional dip at the far end.

PhanthomJay Merry christmas to you also. Thanks for your link.

Chestermiller I made an error in my previous post . I should have said that R1 = R2 = MB/L = 2050NM/0.3M = 6833.3 N
I have now calculated the slope at the pivot point to be 0.00669 radans

 

[hatchelhoff]

I have found the deflection of the beam due to bending to be 2.66 mm.
I have found the deflection of the beam due to the deformation of the bracket to be 2.74 mm.
this gives me a total deflection of 5.4 mm

 

[Chestermiller]

I have found the deflection of the beam due to bending to be 2.66 mm.
I have found the deflection of the beam due to the deformation of the bracket to be 2.74 mm.
this gives me a total deflection of 5.4 mm

I don't want to have to go through this entire problem and do the detailed calculations. My goal was just to point out the general concept of how to approach this problem. I assume you understood the concept. The rest was up to you. If you have questions about the general concept, I will be glad to address them.

 

[hatchelhoff]

Thanks I understood the concept, and I have no questions.