My WebLink
|
Help
|
About
|
Sign Out
Home
Browse
Search
922 W Park Ln - Plan
PBA
>
Building
>
ProjectDox
>
P
>
Park Ln
>
922 W Park Ln
>
922 W Park Ln - Plan
Metadata
Thumbnails
Annotations
Entry Properties
Last modified
2/13/2026 5:02:30 AM
Creation date
2/13/2026 5:00:26 AM
Metadata
Fields
Template:
Plan
Permit Number
101124745
Full Address
922 W Park Ln
Street Number
922
Street Direction
W
Street Name
Park
Street Suffix
Ln
Jump to thumbnail
< previous set
next set >
There are no annotations on this page.
Document management portal powered by Laserfiche WebLink 9 © 1998-2015
Laserfiche.
All rights reserved.
/
330
PDF
Print
Pages to print
Enter page numbers and/or page ranges separated by commas. For example, 1,3,5-12.
After downloading, print the document using a PDF reader (e.g. Adobe Reader).
Show annotations
View images
View plain text
Shear wall height H = 8 ft <br /> <br /> <br /> <br />Length (ft) Type H/B Concept <br />5.5 Wall 1.45 Shear Wall <br />4 Opening <br />7.3 Wall 1.1 Shear Wall <br />Shear capacity adjustment factor – cl.4.3.5.6 <br />Sum of perforated shear wall segment lengths ∑bi=5.5+7.3=12.8 ft <br />Total area of wall Awall=16.8×8=134.4 ft 2 <br />Total full height sheathed area A fhs=5.5×8+7.3×8=102.4 ft2 <br />Shear capacity adjustment factor (eqn. 4.3 -6) Co = min(Awall/(3×Ao + Afhs), 1.0)= 0.931 <br />Perforated shear wall capacity <br />Maximum shear force under wind loading V w_max=0.6×WL/1000 =0.206 Kips <br />Shear capacity for wind loading V w=(Vn×Co×∑bi/2000) =5.208 Kips <br />ratio V w-max / Vw =0.04 <br />PASS - Shear capacity for wind load exceeds maximum shear force <br />Maximum shear force under seismic loading V s_max=0.7×Eq/1000 =0.287 Kips <br />Shear capacity for seismic loading V s=(Vn×Co×∑bi/(2.8×1000)) =3.72 Kips <br />ratio V s_max / Vs =0.077 <br />PASS - Shear capacity for seismic load exceeds maximum shear force <br />Chord capacity for chord1 <br />Load Combination : 0.6DL - 0.7EQ <br />Maximum tensile force in chord T=0 lbs <br />Maximum applied tensile stress f t=T/Aen=0 lb/in2 <br />Design tensile stress F t'=Ft×CD×CMt×Ctt×CFt×Ci=1380 lb/in2 <br />Ratio0/1380=0 <br />PASS - Design tensile stress exceeds maximum applied tensile stress <br />Load Combination : DL + 0.7EQ <br />Maximum compressive force in chord P=407.388 lbs <br />Maximum applied compressive stress f c=P/Ae=33.256 lb/in2 <br />Design compressive stress F c'=Fc×CD×CMc×Ctc×CFc×Ci×Cp=596.16 <br />lb/in2Ratio33.26/596.16=0.06 <br />PASS - Design compressive stress exceeds maximum applied compressive stress <br />Page 64 of 290
The URL can be used to link to this page
Your browser does not support the video tag.