# Bending moment on multi beam sign post

1. Oct 29, 2015

### CompactDisc

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.

2. Oct 29, 2015

### SteamKing

Staff Emeritus
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.

3. Oct 29, 2015

### CompactDisc

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?)?

4. Oct 29, 2015

### SteamKing

Staff Emeritus
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.

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.

5. Oct 30, 2015

### CompactDisc

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

6. Oct 30, 2015

### SteamKing

Staff Emeritus
Unless this is a rather heavy sign, I would expect the magnitudes of Mz1 and Mz2 to be very small.

7. Oct 30, 2015

### CompactDisc

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