perform analysis of lateral load resisting systems

Question 1

seismic load effects - additive - max

Answer 1

E = Eh + Ev or E = ρ Qe + .2 Sds D

Question 2

seismic load effects - counteractive - min

Answer 2

E = Eh - Ev or E = -ρ Qe - 0.2 Sds D

Question 3

horizontal seismic load effect, Eh

Answer 3

Eh = ± ρ Qe where Qe is horizontal seismic forces ρ = 1.3 for SDC = D, E, or F

Question 4

verticle seismic load effect, Ev

Answer 4

Ev = ± 0.2 Sds D, exceptions it is permited to use Ev = 0 for either where Sds ≤ 0.125

Question 5

seismic load effect including over strength factor, Ω0

Answer 5

Em = Emh + Ev, Em = Ω Qe + 0.2 Sds D

Question 6

seismic load effect including over strength factor, Ω0

Answer 6

Em = - Emh - Ev, Em = - Ω Qe - 0.2 Sds D

Question 7

horizontal seismic load effect w overstrength factor, Emh

Answer 7

Emh = ± Ω0 Qe

Question 8

strength design or load & reistance factor design

Answer 8

see pg. 1-67

Question 9

alternative basic (ASD) load combinations

Answer 9

see pg 1-68

Question 10

load combinations with over strength factor

Answer 10

see pg 1-68

Question 11

calculated deflection of a level, δx

Answer 11

δx = Cd δxe / I

Question 12

calculated story drift, ∆x

Answer 12

∆x = δx - δx-1 given δx is amplified top & amplified at bottom

Question 13

calculated (max inelastic) 2nd story drift

Answer 13

calculated ∆2 = δ2 - δ1

Question 14

story drift limit, ∆ax

Answer 14

see pg. 1-54

Question 15

diaphragm design force, Fpx

Answer 15

Fpx = (∑ Fi / ∑Wi) x Wpx with ≤ 0.4 Sds I Wpx max and ≥ 0.2 Sds I Wpx min.

Question 16

allowable stress design (ASD) - max. unit diaphragm shear

Answer 16

Vmax = (.7(max. shear) / diaphragm depth = .7 V1 / d (units of plf)

Question 17

strength design (SD or LRFD)

Answer 17

Vmax = max. shear / diaphragm depth = V1/d

Question 18

chord force

Answer 18

CF = moment / depth = M/d where uniform load, Ws: M = Ws L^2/8 therefore CF = Ws L^2/8 d