Bi 1 : Kt cu gio b tng lin khi nhp dm bn c l loi ct chng lin tip
s dng kt cu nh hnh ct ng trn MIK c khong cch cc ct theo phng dc v
ngang l 1,5x1,5m. Ti cc im k vn khun, ngi ta xc nh cc lc S1,S2,S3
do cc ti trng gm : ct thp, va b tng, trng lng vn khun, ti trng thi
cng, S1 = 23kN, S2 = 29,5kN, S3 = 12kN. Khi lng ring ca b tng bt =
2500kb/m3. Theo phng ngang, trng lng vn khun y Pvk = 1,5kN/m2, trng
lng dm dc v x m ca gio Pg = 0,8kN/m2, ti trng thi cng do dm v va ri
qvr = 0,5kN/m2. Xc nh s lng cc hng ct chng trn mt ct ngang v xc nh
lc dc bt li nht do ti trng thng ng tc dng ln ct chng ?
Bi 2 : Cu dm thp lin hp bao gm 5 nhp gin n c chiu di bng nhau l
L=33m c thi cng bng bin php lao ko dc trn ng trt con ln c s dng mi
dn. Dm thp c trng lng Pd= 7465kN/m. Mi dn c chiu di La = 10m v c
chiu cao thay i, trng lng ti v tr lin kt vi dm ch l Pa= 2140kN/m, u
dn hng l P1 = 1160kN/m. Cm dm u tin c lp v lao ra nhp gm 2 dm ch,
ngi ta lp ln lt tng nhp v lao v tr nhp b tr 1 b tr 1 tr tm cch m
khong L1=10m. Khong cch t trng tm mi dn n v tr im t K l xa = , vi k
= Kim tra iu kin n nh chng lt ca nhp lao ?
