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- To: <seaint(--nospam--at)seaint.org>
- Subject: flexible diaphragm question
- From: "The Buckinghams" <srichmnd(--nospam--at)medford.net>
- Date: Mon, 28 Oct 2002 19:07:08 -0800
I have a question on flexible horizontal diaphragms. A quick sketch of the problem will help explain it. Draw a plan view with north-up (in the Y-direction) and east-right (in the X-direction).
Draw a horizontal (east-west) gridline named "A". Offset this line 10 feet north and call it "B". Offset line "B" 30 feet north and call it "C". Offset line "C" 10 feet north and call it "D".
Now draw a vertical (north-south) gridline named "1". Offset this line 50 feet east and call it line "2". Now offset line "2" 10 feet east and call it "3".
The perimeter of the building (all shear walls) is A1-D1-D2-C2-C3-B3-B2-A2-A1.
Draw an "element" from C2 to B2.
It should be a 50’ by 60’ building with two 10’ re-entrant corners on the east side.
For simplicity, say the roof dead load is 20 psf, the walls don’t weigh anything and the seismic coefficient is 0.2.
For a north-south quake, wall A1-D1 gets 5000# shear (100 plf), wall line A2-D2 gets 5600# shear (280 plf for wall A2-B2 and wall C2-D2) and 600# shear for wall B3-C3 (20 plf). No real question there.
Now consider an east-west quake. How is the 11,200# of total roof shear distributed? Is it 5000# to wall D1-D2, 5000# to wall A1-A2, 600# to C2-C3 and 600# to B2-B3 ? Is that a valid analysis?
It seems to me that points C2 and B2 are fixed to both the shear walls (C2-C3 and B2-B3) and to the chord on the 50’x50’ diaphragm, so the assumption that the 50’x50’ flexible diaphragm behaves like a simply supported beam isn’t valid, and the shear distributed to C2-C3 and B2-B3 will be more than 600#.
Do I add two 50’ long collector struts (C1-C2 and B1-B2) and assume that the shear force of the 50’x50’ diaphragm is distributed based on tributary area to lines A,B,C & D? This would mean only a narrow 5’ strip is tributary to the 50’ long walls, giving shear forces of 1000# (a tiny 20 plf) to walls D1-D2 and A1-A2 and 4600# (a whopping 460 plf) to walls C2-C3 and B2-B3 ? This doesn’t seem right to me.
Do I have "two diaphragms" or "one diaphragm split into two sub diaphragms"?
What if I designed the 50’ walls for 5000#, designed the 10’ walls for 4600# (worst case for each wall) and did not put in collector struts C1-C2 and B1-B2 ? Is there a possible failure mode that I’m not seeing ? This also seems somewhat over designed.
Does anyone have a good reference book showing lots of complicated diaphragms split into sub-diaphragms? It seems like all my reference books show how to analyze a simple rectangle and leave it at that.
Thank you in advance for any insight into this problem.
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