Bending moment on multi beam sign post

AI Thread Summary
The discussion focuses on calculating bending moments for a sign post supported by two beams under wind loading. The user initially grapples with the moment about the z-axis, questioning whether it becomes negligible or is distributed between the two beams. After drawing a free-body diagram, it is concluded that the moments about z (Mz1 and Mz2) act equally and oppositely to maintain static equilibrium, each bearing half the force. The normal forces (Ry1 and Ry2) counteract the dead load, while the reactions (Rz1 and Rz2) oppose the wind force. The weight of the sign is noted to be relatively light, suggesting that the magnitudes of Mz1 and Mz2 are small.
CompactDisc
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Hey guys,

So just completed a question on a sign post subject to wind loading of 2 kPa where you had to determine bending moments across x and z planes (picture below), to determine normal, shear, max stresses and then determine the factor of safety depending on the beam type you chose. That's all good.

The next question is to do with the same problem but now rather than a single beam holding the sign up, there is a beam either side! Moment about x was relatively similar, just slightly larger area and displaced across the two beams but I am stumped with the moment about z. The weight of the sign is now in equilibrium and in the previous problem Mz = weight of the sign * moment arm.

Is Mz now just negligible (equal to 0), or does Mz halve and get spread across the two beams? Any help would be fantastic! And was also unsure where to post this.

nuvsw.jpg
 
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CompactDisc said:
Hey guys,

So just completed a question on a sign post subject to wind loading of 2 kPa where you had to determine bending moments across x and z planes (picture below), to determine normal, shear, max stresses and then determine the factor of safety depending on the beam type you chose. That's all good.

The next question is to do with the same problem but now rather than a single beam holding the sign up, there is a beam either side! Moment about x was relatively similar, just slightly larger area and displaced across the two beams but I am stumped with the moment about z. The weight of the sign is now in equilibrium and in the previous problem Mz = weight of the sign * moment arm.

Is Mz now just negligible (equal to 0), or does Mz halve and get spread across the two beams? Any help would be fantastic! And was also unsure where to post this.

nuvsw.jpg

Why don't you make a free-body diagram of the sign with two supports? You know, figure out the reactions in the two supports, like what was done with a single support.
 
SteamKing said:
Why don't you make a free-body diagram of the sign with two supports? You know, figure out the reactions in the two supports, like what was done with a single support.

156bhps.jpg


Okay, did the free body diagram for the system (I believe that's right?). In the previous diagram, moment about Z was the weight of the sign multiplied by the length of the centroid of the sign to the centroid of the cross section of the pole.

I'm assuming this still applies but pole 2 is equal and opposite to pole 1 (which would mean Mz2 is reversed?)?
 
CompactDisc said:
156bhps.jpg


Okay, did the free body diagram for the system (I believe that's right?). In the previous diagram, moment about Z was the weight of the sign multiplied by the length of the centroid of the sign to the centroid of the cross section of the pole.

I'm assuming this still applies but pole 2 is equal and opposite to pole 1 (which would mean Mz2 is reversed?)?
Why do you think there would be a moment about the z-axis?

In any event, for the sign to remain in equilibrium, the forces and moments shown at the footings of the support beams should be the reactions which keep the sign in static equilibrium.

You should have reactions Ry1 and Ry2 at the footings pointing up.
You should have reactions Rz1 and Rz2 at the footings pointing in the positive z-direction.

The reaction torques Mx1 and Mx2 should be acting in the opposite sense, to keep the sign upright against the wind loading.

Ditto the two "torques", or My1 and My2. These should be shown acting in the opposite sense, to oppose the wind loading.

Mz1 should act in the opposite direction from what is shown.
 
SteamKing said:
Why do you think there would be a moment about the z-axis?

In any event, for the sign to remain in equilibrium, the forces and moments shown at the footings of the support beams should be the reactions which keep the sign in static equilibrium.

You should have reactions Ry1 and Ry2 at the footings pointing up.
You should have reactions Rz1 and Rz2 at the footings pointing in the positive z-direction.

The reaction torques Mx1 and Mx2 should be acting in the opposite sense, to keep the sign upright against the wind loading.

Ditto the two "torques", or My1 and My2. These should be shown acting in the opposite sense, to oppose the wind loading.

Mz1 should act in the opposite direction from what is shown.

Yeah all of that definitely makes sense, seeing as sum of forces should equal zero and sum of moments should equal zero.

So Ry1 and Ry2 will be the normal forces opposing the dead load of the system.
Rz1 and Rz2 oppose the 2kPa force of the wind trying to push the sign.

There is a moment about Z because the weight of the sign is exerting a force (its weight), onto the beam from a radius (moment arm). Mz1 and Mz2 act equally and oppositely to maintain static equilibrium and therefore should be bearing half of the force each.

So it would still be Mzi = (Weight of sign * Distance to centroid + distance to centroid of post)/2
 
CompactDisc said:
Yeah all of that definitely makes sense, seeing as sum of forces should equal zero and sum of moments should equal zero.

So Ry1 and Ry2 will be the normal forces opposing the dead load of the system.
Rz1 and Rz2 oppose the 2kPa force of the wind trying to push the sign.

There is a moment about Z because the weight of the sign is exerting a force (its weight), onto the beam from a radius (moment arm). Mz1 and Mz2 act equally and oppositely to maintain static equilibrium and therefore should be bearing half of the force each.

So it would still be Mzi = (Weight of sign * Distance to centroid + distance to centroid of post)/2
Unless this is a rather heavy sign, I would expect the magnitudes of Mz1 and Mz2 to be very small.
 
SteamKing said:
Unless this is a rather heavy sign, I would expect the magnitudes of Mz1 and Mz2 to be very small.

Yeah the initial question states 1kN so not too heavy at all! Thank you very much for the help
 
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