Question

To whoever answers and helps me with these, I would appreciate if you could someone explain how you got the answers and easy ways to notice or find or understand the hmwk. I don't want to just copy it, I want to understand it and learn how I can do it. Thanks!

PROBLEM 1 A. CALCULATE THE AXIAL LOADS (TENSION OR COMPRESSION) IN THE COLUMN BETVWEEN LEVEL 1 AND LEVEL 2. SHOW RESULTS FOR ALL 7 LOAD COMBINATIONS INCLUDED IN THE HANDOUT B. BASED ON THE RESULTS FROM A, WHAT ARE THE MAXIMUM TENSION AND COMPRESSION IN THE COLUMN BETWEEN LEVEL 1 AND LEVEL 2? IS THE COLUMN EVER IN TENSION? P2300 KIPS Pw600 KIPS Po =-450 KPS Pr100 KIPS Ps500 KIPS LEVEL 5 Po450 KIPS R =-300 KIPS LEVEL 4 Po =-450 KIPS Rs-300 KIPS LEVEL 3 Po =-600 KIPS R =-800 KIPS LEVEL 2 LOCATION OF LOAD FOR A AND B LEVEL 1

Answer 1

Dead load D = 1950kip [= 3* 450 + 600] floor live load L = 1400kip[= 2* 300 + 800] prof live load L_r = 100kip snow load S = 500kip rain load R = 0kip wind load W = 0kip earthquake load E = 2300kip(both + end) A) load combination - LRFD 1.4D = 1.4 times 1950 = 2730KIP 1.20 + 1.6l + 0.5 = 1.2(1950) + 1.6(1400) + 0.5(500) = 4830kip 1.215 + 1.6 + (L or 0.5w) = 1.2(1950) + 1.6(500) + 1.0(1400) = 4540kip \r\n 1 + 2D + 1.6(L or S or R) + = 1.2(1950) + 1.6(500) + 1.0 (L or -0.5W) (1400) = 4540kip 12D + 1.0W + L + 0.5 = 1.2(1950) + 10(600) + 1.0(1400) + 0.5(500) = 4590kip 1.2D + 1.0W + L + 0.5 = 1.2(1950) + 1.0(600) + 1.0(1400) + 0.5(500) = 6140kip 1.2D + 1DE + L + 0.25 = 1.2(1950) + 1.0(2300) + 1.0(1400) + 0.2(500) = 1540kip 0.90 + 1.000 = 0.9(1950) + 1.0(600) = 2355kips