Upload
ivankafadar
View
224
Download
0
Embed Size (px)
Citation preview
7/23/2019 Programski Drvene IK
http://slidepdf.com/reader/full/programski-drvene-ik 1/5
UNIVERZITET U TRAVNIKUFAKULTET ZA TEHNIČKE STUDIJE
-Programski zadatak-
DRVENE KONSTRUKCIJE
Kandidat: Mentor:
Ivan Kafadar profdr !o"o So#do
7/23/2019 Programski Drvene IK
http://slidepdf.com/reader/full/programski-drvene-ik 2/5
Travnik, e!r"ar #$%&'
7/23/2019 Programski Drvene IK
http://slidepdf.com/reader/full/programski-drvene-ik 3/5
-Savijanje
M max=q∗l
2
8=1,25∗4
2
8=2,5kNm=250kNcm
σ m= M max
W ≤σ md=1000
N
cm2
W pot = M max
σ md=250∗10
3
1000=250cm
W a=a3
6→a=
3
√ 250∗6=11,45 cm
-Smicanje
Qmax=
q∗l
2 =
1,25∗4
2 =2,5kN
τ mII =1,5∗Qmax
A ≤ τ md=90
N
cm2
A pot =
1,5∗Qmax
τ md=1,5∗2,5∗10
3
90=
41,66 cm
2
A=a2→a=
2√ 41,66=6,45cm
Usvojen poprečni presjek a=12 cm
7/23/2019 Programski Drvene IK
http://slidepdf.com/reader/full/programski-drvene-ik 5/5
-Kontrola ugiba
E=10000 MPa=10000∗102 N
cm2
I =a4
12=12
4
12=1728cm
4
f =
5
384∗q∗l
4
EI =
5
384∗1,25∗10
3∗400
4
102∗1000000∗1728
=2,41cm
f dop=l
300=400
300=1,33<2,41 cm
Poprečni presjek ne zadovoljava!
Usvaja se vei poprečni presjek a=1cm
I =a4
12=14
4
12=3201,33 cm
4
f =
5
384∗q∗l
4
EI =
5
384∗1,25∗10
3∗400
4
102∗1000000∗3201,33
=1,30cm<1,33 cm=f dop
Usvojeni poprečni presjek zadovoljava"