Hiner Notes Problem 16

[golamycine]

Dear ALL,

I would appreciate any help with the Problem 16 of Hiner Notes as follows:

In E-W Direction:

Part 4: Calculate Shear Force to Walls 1A and 1B

The solution shows that the force is distributed to the wall relative to their respective ridgidities (H/D). The equation used is also in the Masour notes (P-24). My question is that why the relative rigidity is used when the diaphragm is flexible. Should not it be based on the tributory area? I am confused any help is appreciated.

Seismic is just not my area and this is the last part of my PE left. Appreciate any help and Good Luck to all

 

[sac_engineer]

Refer to Section 8.6 in Hiner's manual. Because the problem involves special reinforced masonry shear walls, you need to use relative rigidities, regardless of the diaphragm.

Good luck!

 

[golamycine]

Refer to Section 8.6 in Hiner's manual. Because the problem involves special reinforced masonry shear walls, you need to use relative rigidities, regardless of the diaphragm.
Good luck!

Thank you for your quick reply. The solution also uses the relative rigidities for caculcation of drag force in E-W direction. The relative rigidities for N-S walls are not used in the draf foce calculation.

I understand that the N-S direction is much easier to see since the drag force would be half of the diaphragm shear (wall covers only half of the diaphrapgm length on one side) but if we use the same principal as in E-W direction should not the answer be same?

Thanks in advance

 

[sac_engineer]

Computing the drag force in both NS and EW is consistent; there's no relative rigidity calc for the drag force in the EW direction. Part B is asking for an additional step in calculating the shear force in the EW direction, hence the required relative rigidity calc.

 

[golamycine]

Computing the drag force in both NS and EW is consistent; there's no relative rigidity calc for the drag force in the EW direction. Part B is asking for an additional step in calculating the shear force in the EW direction, hence the required relative rigidity calc.

Actually there is.....Page 2-39 of Hiner notes second line from the bottom of the page drag force of 3,050 lbs is calculated for Wall 1A based on the relative rigidity and also on page 2-40 for Wall 1B the relative rigidity is used to calculate 12,350 lbs of drag force. These values are used for caluclating drag force in Part 5

I also sent this question to Mr. Hiner and pasting his response below which does make some sense. I still think that by this argument the 20 foot wall on the drag force should be calculated based on the rigidity of that wall rather than its length???that would mean the wall is only removing 5,130 lbs of force instead of 9,880 lbs when it is considered as flexible?

I am posting his response below:

"Here is a response to someone asking very similar question … see if this might clear it up for you as well.

<I thought a flexible Diaphragm imparts the load evenly along the length—so the force is distributed by length of member not by rigidity of receiving member.>

That works only for wood shear walls … where the load is assumed to be resisted by multiple shear walls on a wall line based on their respective wall lengths. It is assumed that a shear wall that is twice as long is also twice as stiff (or rigid). But for masonry or concrete shear walls, a wall that is twice as long may be 4 to 6 times more rigid.

A flexible diaphragm imparts the load equally to each end of the diaphragm (i.e., V1 = V2 = wL /2). But if you have multiple masonry or concrete walls on one of those wall lines, you need to distribute the force based on relative rigidity (not based on the length of the shear walls).