MASONRY DESIGN $(4$ UNITS$)$ REV.#$15/4/2024$

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

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

OBLEM #$3$:

en:

$(18$ points total$)$

The plan and section of the building in Figure #A$.$

Roof dead load $D_r={10}^{PSF}$

Walls are fully grouted CMU $8^{\text{'}\text{'}}-$ thick $).$

The building has a bearing wall system, braced with special reinforced masonry shearwalls $($A$\mathrm{.}7$ per Table $12\mathrm{.}14-1/$ ASCE $7-16).$

The building location is Northridge, California, where:

$S_s=1\mathrm{.}745g,$ and $S_1=0\mathrm{.}602g.$

Site $($Soil$)$ Class $=B($Rock$)$ and Risk Category II should be assumed.

$T_L=8\mathrm{.}0s$

Redundancy Factor, $p=1\mathrm{.}0$

Assume Seismic diaphragm force coefficient $=C_5$

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, $C_{s\text{;}}$ Seismic Design Category $($SDC$)$; Tributary Height $($Seismic$),$ hr; 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.

$(6$ points$)$

c$.$ Roof Diaphragm $-$ ASD level:

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$)$

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$\_("$; $")$; 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 $10$