CE $537-$ TIMBER AND MASONRY DESIGN $(4$ UNITS$)$ REV.#$15/4/2024$

TAKE HOME DESIGN PROJECT $/$ DATE: $05/04/2024/$ DUE: $5/13/2024$ at $6$:$00$ PM

ROOF FRAMING PLAN

FIGURE #A

CE $537-$ TIMBER AND MASONRY DESIGN $(4$ UNITS$)$ REV.#$15/4/2024$ TAKE HOME DESIGN PROJECT $/$ DATE: $05/04/2024/$ DUE: $5/13/2024$ at $6$:$00$ PM PROBLEM #$3$:

Given:

$(18$ points total$)$S$\_($s$)=1\mathrm{.}745$g$,$ and S$\_(1)=0\mathrm{.}602$g$.$Site $($Soil$)$ Class $=$ B $($Rock$)$ and Risk Category II should be assumed.T$\_($L$)=8\mathrm{.}0$sRedundancy Factor, rho$=1\mathrm{.}0$Assume Seismic diaphragm force coefficient $=$C$\_($s$)$Specified: ASCE $7-16($no Addendums to ASCE $7-16$ are considered$)$:Chapters $11$ & $12.$Excerpts from ASCE $7-16($some tables & some formulas $-$ see next $2$ pages$)$ are provided just for your convenience.Find:For the longitudinal $($East $-$ West$)$ direction:A$.$ Natural Period, T~~T$\_($a$)($s$)$; seismic base shear coefficient, C$\_($s$)$; Seismic Design Category $($SDC$)$; Tributary Height $($Seismic$),$ h$\_("$T; $")$ Effective Seismic Weight, W $($kips$)$ for Seismic Base Shear Force and Base Shear, V $($kips$)$B$.$ Roof Diaphragm $-$ Strength level: $(10$ points$)$The uniform seismic force w$\_($L$)$ distributed to the roof diaphragm in PLF$.$The unit diaphragm seismic shear v $($PLF$)$ adjacent to the longitudinal walls.C$.$ Roof Diaphragm $-$ ASD level:$(6$ points$)$The uniform seismic force w$\_($L$)$ distributed to the roof diaphragm in PLF$,$ adjusted to an ASD level.The unit diaphragm seismic shear v $($PLF$)$ adjacent to the transverse longitudinal walls adjusted to an ASD level.$(2$ points$)$Page $7$ of $10$