How to calculate deflection in beam at orientation?

[Uncia_Uncia]

Lets say a hollow rectangle beam of 3" x 5"x 1/4" thk., 48" long is fixed at two ends. This beam is part of attachment of lifting device. If attachment rotates at 20, 30, 45, 60 degree, this beam will also make angle respective angle and Uniformity distributed load will always be right angle to beam.
How should I calculate the deflection of beam?

 

[BAretired]

The angle of the HSS wall makes no difference to the solution if the load is always acting normal to the angle and both ends are fixed. The deflection is calculated exactly as it would be if the angle was zero degrees.

BA

 

[Lomarandil]

If instead, you mean that the load is always vertical and perpendicular to the lifting beam, deflection can be calculated by geometric superposition of the strong axis deflection and weak axis deflection values.

(e.g. use geometry (sin and cos) to separate the resultant load into components acting in the beam strong and weak axes, compute the deflection due to each, and combine).

With a hollow rectangular shape, torsional deflection should be negligible, so no second order effects to speak of (presuming you don't need accuracy to less than 1/10").

 

[rb1957]

mohr's circle for MoI ?

another day in paradise, or is paradise one day closer ?

 

[Uncia_Uncia]

Thank you all for replying. I am still trying to work out the method what @Lomarandil suggested. It is getting complex.

 

[BAretired]

Now you have me confused. Your sketch shows the load normal to the 5" wall of the HSS. If the load W is vertical and the tube is angled at theta from the horizontal, then the load causing minor axis bending is W*cos(theta) while the load causing major axis bending is W*sin(theta).

Major and minor axis deflection can be calculated separately using the above loads but the resultant deflection will be the vector sum of those two deflections.

BA

 

[Uncia_Uncia]

Hello BA,
Even I am confuse honestly. But what you referred as deflection would be the same even if the hollow rectangle pipe would be orientated and in the condition of load normal to pipe, make sense to me,too.

 

[rb1957]

I think the point is to resolve the applied load (and bending) into the principal axes of the section (however rotated), and superimpose.

another day in paradise, or is paradise one day closer ?

 

[JeanLucPicard]

Something like these ? I presumed that the angle α is same along the length of the beam right?

Live long and prosper!

 

[rb1957]

for a unit load at midspan ...
the SE equivalent of the physist's "spherical chicken in a vacuum"

another day in paradise, or is paradise one day closer ?

 

[JeanLucPicard]

Sorry I've totally missed the picture you attached. In addition to the equation that would be the self weight in my attachment you should add the additional u[sub]z,2[/sub] Hence: sqrt( (u[sub]z2[/sub]+u[sub]z[/sub])[sup]2[/sup] + uy[sup]2[/sup][sup][/sup] ). The loading point from the perpendicular loading is going trough the shear center as well, therefore, there is no torsion and restrained torsion.

Live long and prosper!

 

[BAretired]

JeanLucPicard,

The factor 5/384 applies when the supports are hinge and roller. If the ends are fixed as stated in the OP, that should be 1/384.

BA