




Study with the several resources on Docsity
Earn points by helping other students or get them with a premium plan
Prepare for your exams
Study with the several resources on Docsity
Earn points to download
Earn points by helping other students or get them with a premium plan
Community
Ask the community for help and clear up your study doubts
Discover the best universities in your country according to Docsity users
Free resources
Download our free guides on studying techniques, anxiety management strategies, and thesis advice from Docsity tutors
A step-by-step calculation of base shear in seismic analysis, following the guidelines of the national structural code of the philippines (nscp) 2015. It demonstrates the application of various factors and coefficients, including seismic zone factor, near source factors, and response modification factor, to determine the base shear value. A practical example of how to calculate base shear for a specific structure.
Typology: Study notes
1 / 8
This page cannot be seen from the preview
Don't miss anything!
Na = 1. Nv = 1.
Assumed distance = greater than 15km from the known active fault Seismic Zone Factor (Zone 4): >10km Z = 0.40 (NSCP 2015 Table 208-3) Near Source Factors: Na = 1.0 (NSCP 2015 Table 208-5) Nv = 1.0 (NSCP 2015 Table 208-6) Seismic Response Coefficient: Ca = 0.44Na = 0.44(1.0) = 0.44 (NSCP 2015 Table 208- 7)
Cv = 0.96Nv = 0.96(1.0) = 0.96 (NSCP 2015 Table 208- 8) Response Modification Factor: R = 8.5 Kn R = 8.5 , for Concrete SMRF (NSCP 2015 Table 208-11A)
HORIZONTAL BASE SHEAR, V: W = 250 kN ( total seismic weight based on tributary area ) V = (Cv I W) / R T (NSCP 2015 ; 208-8) V = 1.536 (1.0) (150) / 8.5 (0.46244) = 58.61 kN Base Shear Need Not to Exceed, V = (2.5Ca I W) / R (NSCP 2015 ; 208-9) V = 2.5(0.528)(1.0)(150) / 8.5 = 23.29 kN Base Shear Shall Not be Less Than, V = (0.1Ca I W) (NSCP 2015 ; 208-10) V = 0.1 (0.528) (1.0) (150) = 7.92 kN Base Shear Shall Not be Less Than,V = (0.8 Z Nv I W) /R (NSCP 2015 ; 208-11) V = 0.8 (0.40) (1.6) (1.0) (150) / 8.5 = 9.04 kN Therefore use, V= 23.29 kN