Pravougaona ploč a

Zadatu ploč u i optere ć enje sra č unati inacrtati dijagrame :1. Ugibaw ;2. Prese č nih silaM x , M y , M xy ,T x i T y ;3. Reakcijeoslonaca A x i A y .Geometrijske karakteristike plo č e :Sx=1.9mα=1.25β=0.5γ=0.2h=0.20mυ=0.20Opterećenje ploče:p=12kN /m2

q=16kN /mF=24 kN

α=S y

Sx=¿1.25=

S y

1.9=¿ Sy=1.9∗1.25=2.375m

lx=4∗Sx=4∗1.9=7.6ml y=4∗S y=4∗2.375=9.5m2∗β∗lx=2∗0.5∗7.6=7.6m2∗γ∗l y=2∗0.2∗9.5=3.8m

Konturni uslovi

w18=w4

w17=w3

w16=w2

w15=w2

w14=w1

w19=−w4

w20=−w5

w7=w8=w9=w10=w11=w12=w13=0Proračun elemenata alfa š eme

α 2=1.56251

α2=0.6400

4∗(1+α 2)=10.2500

4∗(1+ 1α 2 )=6.56006∗(α2+ 1α 2 )+8=21.215

α 2∗Sx4=20.3627 0.64002 -6.5600 21.5625 -10.2500 21.2150 -10.2500 1.56252 -6.5600 20.6400

Proračun slobodnihčlanova

k= E∗h3

12∗(1−υ2 )=3.0∗10

7∗0.203

12∗(1−0.202 )=20833,3333

kw0 i=Zi∗α2∗Sx

4

Z1=12∗1,9∗2,375+16∗1,9

1,9∗2,375=18,7368 kN /m2

kw01=18,7368∗20,3627=381,5318

kw05=kw01=381,5318

Z2=12∗1,9∗0,71251,9∗2,375

=3,6000 kN /m2

kw02=3,6000∗20,3627=73,3057

kw04=kw02=73,3057

Z3=24+12∗1,9∗0,7125

1,9∗2,375=8,9186kN /m2

kw03=8,9186∗20,3627=181,6068

Z6=16∗1,9+12∗1,9∗2,375

1,9∗2,375=18,7368kN /m2

kw06=18,7368∗20,3627=381,5318

W 1 W 2 W 3 W 4 W 5 W 6 KW o22,7775 -13,1200 4,0000 0 1,5625 -10,2500 381,5318-6,5600 24,0575 -10,2500 1,5625 0 2,0000 73,30572,0000 -10,2500 22,4950 -10,2500 2,0000 -6,5600 181,60680 1,5625 -10,2500 20,9325 -6,5600 2,0000 73,30571,5625 0 4,0000 -13,1200 19,6525 -10,2500 381,531810,2500 4,0000 -13,1200 4,0000 -10,2500 21,2150 381,5318UGIBI

W 1 W 2 W 3 W 4 W 5 W 684,6938 44,4709 79,9967 60,5498 110,9860 142,1980

kw i [m ]

Stvarniugibi( kw i

k )[m]

W 1 W 2 W 3 W 4 W 5 W 60.004065 0.002135 0.003840 0.002906 0.005327 0.006826

Sileu presecima (Mx)

v

α2= 0,21,5625

=0,128

2∗(1+ v

α 2 )=2,256

M x , 1=k

Sx2∗(2,256∗w1+2∗(−0,128 )∗w2+(−1)∗w6 )=10,3841kNm/m

M x , 2=k

Sx2∗(2,256∗w2+ (−0,128 )∗w1+(−1)∗w3 )=2,6285kNm /m

M x , 3=k

Sx2∗(2,256∗w3+(−1 )∗w4+ (−1 )∗w2+ (−0,128 )∗w6 )=15,8589kNm /m

M x , 4=k

S x2∗(2,256∗w4+(−1)∗w3+(−0,128 )∗w5 )=11,7444 kNm /m

M x , 5=k

Sx2∗(2,256∗w5+(−1 )∗w6+2∗(−0,128 )∗w4 )=25,6747kNm /m

M x , 6=k

Sx2∗(2.625∗w6+(−1 )∗w5+(−1 )∗w1+2∗(−0,128 )∗w3)=28,9861kNm /m

M x , 7=k

Sx2∗( (−1 )∗w5−(−1 )∗w5 )=0kNm /m

-0,128-1 2,256 -1 ¿ { kSx2=M x ,k

-0,128

M x , 8=k

Sx2∗( (−1 )∗w4− (−1 )∗w4 )=0kNm /m

M x , 9=k

Sx2∗(2∗(−0,128 )∗w4 )=−4,2938kNm /m

M x , 10=k

Sx2∗(2∗(−0,128 )∗w3 )=−5,6729kNm /m

M x , 11=k

Sx2∗(2∗(−0,128 )¿w2 )=−3,1536 kNm/m

M x , 12=k

Sx2∗(2∗(−1 )¿w2 )=−24,6376 kNm/m

M x , 13=k

Sx2∗(2∗(−1 )¿w1 )=−46,9218kNm /m

Sileu presecima (My )−v∗α 2=−0,31252∗(1+v∗α2 )=2,625

-1-0,3125 2,625 -0,3125 ¿ { kS y

2 =M y , k

-1

M y ,1=k

Sy2∗(2∗(−1 )∗w2+2,625∗w1+ (−0,3125 )∗w6 )=15,7682kNm /m

M y ,2=k

S y2∗(2,625∗w2+(−1)∗w1+(−0,3125)∗w3)=1,2487kNm /m

M y ,3=k

S y2∗(2,625∗w3+(−1 )∗w6+ (−0,3125 )∗w4+ (−0,3125 )∗w2 )=6,1949 kNm /m

M y , 4=k

S y2∗(2,625∗w4+(−1)∗w5+ (−0,3125 )∗w3 )=4,0702kNm /m

M y ,5=k

S y2∗(2∗(−1 )∗w4+2,625∗w5+(−0,3125)∗w6 )=22,3028kNm /m

M y ,6=k

S y2∗(2,625∗w6+2∗(−1.00 )∗w3+(−0,3125 )∗w5+(−0,3125 )∗w1 )=26,9698kNm /m

M y ,7=k

S y2∗( (−0,3125 )∗w5+ (−0,3125 )∗(−w5 ))=0kNm/m

M y ,8=k

S y2∗( (−0,3125 )∗w4+(−0,3125 )∗(−w4 ))=0kNm/m

M y ,9=k

S y2∗( (−1 )∗w4+ (−1 )∗w4 )=−21,4692kNm /m

M y ,10=k

Sy2∗( (−1 )∗w3+(−1 )∗w3 )=−28,3645kNm /m

M y ,11=k

S y2∗( (−1 )∗w2+(−1 )∗w2)=−15,7681kNm /m

M y ,12=k

S y2∗( (−0,3125 )∗w2+(−0,3125 )∗w2 )=−4,9275kNm /m

M y ,13=k

Sy2∗( (−0,3125 )∗w1+(−0,3125 )∗w1 )=−9,3844 kNm /m

Sileu presecima (Mxy)

M xy ,1=k∗(1−υ )4∗α∗Sx

2 (1∗w3+(−1 )∗w3 )=0kNm /m

M xy ,2=k∗(1−υ )4∗α∗Sx

2 ( (−1 )∗w6 )=−6,3024 kNm /m

M xy ,3=k∗(1−υ )4∗α∗Sx

2 ( (−1 )∗w5+1∗w1 )=−1,1653kNm /m

M xy ,4=k∗(1−υ)4∗α∗Sx

2 (1∗w6 )=6,3022kNm /m

M xy ,5=k∗(1−υ )4∗α∗Sx

2 ( (−1 )∗w3+1∗w3 )=0 kNm/m

M xy ,6=k∗(1−υ )4∗α∗Sx

2 (1∗w4+(−1 )∗w4+1∗w2+(−1 )∗w2 )=0kNm /m

M xy ,7=k∗(1−υ )4∗α∗Sx

2 (1∗w4+(−1 )∗w4 )=0kNm/m

M xy ,8=k∗(1−υ )4∗α∗Sx

2 (1∗w5+1∗w5 )=4,9189kNm/m

1 l -1k-1 k k+1-1 i 1

M xy ,9=k∗(1−υ )4∗α∗Sx

2 (w3−w3 )=0kNm /m

M xy ,10=k∗(1−υ )4∗α∗Sx

2 ( (−1 )∗w4+1∗w4+ (−1 )∗w2+1∗w2 )=0kNm/m

M xy ,11=k∗(1−υ )4∗α∗S x

2 (1∗w3+(−1 )∗w3 )=0kNm /m

M xy ,12=k∗(1−υ )4∗α∗Sx

2 (1∗w1+(−1 )∗w1 )=0kNm /m

M xy ,13=k∗(1−υ )4∗α∗Sx

2 (1∗w2+(−1 )∗w2+1∗w2+(−1 )∗w2 )=0kNm /m

Sileu presecima (Tx)1

α2=0,6400

2∗(1+ 1α 2 )=3,280

0,640 -0,6401 -3,280 k 3,280 -1 ¿ { k

2∗S x3=T x ,k

0,640 -0,640

T x ,1=K

2∗S X3∗( (−3,280 )∗W 6+2∗0,640∗W 3+1∗W 5+(−1 )∗W 1 )=−24,6189kN /mT x ,2=

K

2∗S X3∗¿

T x ,3=K

2∗S X3∗(3,280∗W 2+ (−0,640 )∗W 1+(−3,280 )∗W 4+0,640∗W 5)=−2,6179kN /m

T x , 4=K

2∗SX3∗(3,280∗W 3+(−0,640 )∗W 6+ (−1 )∗W 2+(−1 )∗W 4 )=4,8376kN /m

T x ,5=K

2∗S X3∗(2∗(−0,640 )∗W 3+3,280∗W 6+ (−1 )∗W 1+(−1 )∗W 5 )=12,2710kN /m

T x ,6=K

2∗S X3∗( (−3,280 )∗W 5+3,280∗W 1+2∗0,640∗W 4+2∗(−0,640)∗W 2 )=−4,7862kN /m

Z7=16∗0,95+12∗0,95∗2,375

0,95∗2,375=18,7368kN /m2

Z7=Z13=18,7368kN /m2

Z8=12∗0,98∗0,71250,95∗2,375

=3,6000kN /m2

Z8=Z12=3,6000kN /m2

ZamenjujućeT sile (T )2−v

α2=1,152

2∗(1+ 2−v

α2 )=4,304

1,152 -1,1521 -4,304 k 4,304 -1 ¿ { k

2∗S x3=T x ,k

1,152 -1,152

Ugibi tačakaW 21 ,W 22−slobodnooslonjenaivica (dve tačke vankontura)

W k+2=−W k−2+Zk∗Sx

4

k

W 21=−W 6+Z7∗Sx

4

k=−142,198

k+ 18,7368∗1,9

4

k=101,819

W 22=−W 3+Z8∗Sx

4

k=−79,9967

k+ 3,6000∗1,9

4

k=−33,0811

Ugibi tačakaW 23 ,W 24−uklještena ivica(dve tačke vankontura)

W k+2=Zk∗S x

4

k+8∗(1+ 1α 2 )¿W k−1−

4α2

∗(W i−1+W l−1 )−W k−2

W 23=Z12∗Sx

4

k+8∗(1+ 1α 2 )¿W 2−

4α2

∗(W 1+W 11)−W 3

W 23=3,6000∗1,94

k+8∗(1+0,640 )∗44,4709− 4

1,5625∗(84,6938+0 )−79,9967=333,5610

W 24=Z13∗Sx

4

k+8∗(1+ 1α 2 )¿W 1−

4α2

∗(W 2+W 2 )−W 6

W 23=18,7386∗1,94

k+8∗(1+0,640 )∗84,6938− 4

1,5625∗(44,4709+44,4709 )−142,198=333,5610

T x ,7=k

2∗S x3∗¿

T x ,8=k

2∗S x3∗¿

T x ,12=k

2∗Sx3∗(W 3−W 23)=−18,4841kN /m

T x ,13=k

2∗Sx3∗(W 6−W 24)=−61,4728kN /m

Reakcijeoslonaca ( A7 , A8 , A12 , A13 )−slobodno oslonjena ivica A7 , A8

2∗2−vα2

=2,304

4∗(1+ 2−v

α2 )=8,608

Ax ,7=K2∗SX

3∗(Z7∗Sx4k+8,608∗w20+ (−2 )∗w21+2∗(−2,304 )∗w19)=−46.3725kN /m

Ax ,8=K2∗SX

3∗(Z8∗Sx4k+8,608∗w19+ (−2 )∗w22+ (−2,304 )∗w20)=−11.1112kN /m

-2,304 l-2 8,608 k +Zk∗Sx

4

k∗{ k2∗Sx

3=Ax ,k

-2,304 i

−uklještena ivica A12 , A134

α2=2.560

8∗(1+ 1α 2 )=13,120

Ax ,12=K2∗SX

3 ∗( Z8∗Sx4k+13,120∗w2+ (−2 )∗w3+(−2,560 )∗w1)=18,4841kN /m

Ax ,13=K2∗SX

3 ∗( Z7∗Sx4k+13,120∗w1+2∗(−2,560 )∗w2+(−2 )∗w6)=61,4728kN /m

Sileu presecima (Ty)α 2=1.56252∗(1+α 2)=5.125

T y ,1=K

2∗S y3∗( (−5,125 )∗W 2+1,5625∗W 3+5,125∗W 2+(−1,5625 )∗W 3 )=0kN /m

T y ,2=K

2∗S y3∗( (−1 )∗W 16+ (−5,125 )∗W 1+1,5625∗W 6+1∗W 2)=−7,90776kN /m

T y ,3=K

2∗S y3∗( (−1 )∗W 17+(−5,125 )∗W 6+1,5625∗W 5+1,5625∗W 1+1∗W 3)=−15,7883kN /m

T y ,4=K

2∗Sy3∗(−W 18+ (−5,125 )∗W 5+1,5625∗W 6+W 4 )=−12.9369kN /m

T y ,5=K

2∗S y3∗(5.125∗W 4+ (−5,125 )∗W 4+(−1,5625)∗W 3+1,5625∗W 3 )=0kN /m

T y ,6=K

2∗S y3∗(5.125∗W 3+(−5,125 )∗W 3+(−1,5625 )∗W 4+1,5625∗W 4+(−1,5625 )∗W 2+1,5625∗W 2 )=0kN /m

Z9=Z10=Z11

ZamenjujućeT sile (T )α 2∗(2−v )=2,8125

-2,560 l-2 13,120 k +Zk∗Sx

4

k∗{ k2∗Sx

3=Ax ,k

-2,560 i

-1-1,5625 5,125 -1,5625

k ¿ { k

2∗S y3 =T y, k

1,5625 -5,125 1,56251

2∗(1+α 2∗(2−v ) )=7,625

UgibitačakaW 25 ,W 26 ,W 27−uklještenaivica (dve tačke vankontura)

Wm=Zk∗S y

4

k+8∗(1+α 2 )¿W i−4∗α

2∗(W i−1+W i+1 )−W h

W 25=Z9∗Sy

4

k+8∗(1+α 2)¿W 4−4∗α

2∗(W 8+W 3 )−W 5=630,3055

W 26=Z10∗S y

4

k+8∗(1+α2 )¿W 3−4∗α

2∗(W 4+W 2)−W 6=841,3550

W 27=Z11∗S y

4

k+8∗(1+α 2 )¿W 2−4∗α

2∗(W 3+W 12 )−W 1=326,9803

T y ,9=k

2∗S y3∗¿

T x ,10=k

2∗S y3∗¿

T y ,11=k

2∗S y3∗¿

-1-1,5625 5,125 -1,5625

k ¿ { k

2∗S y3 =T y, k

1,5625 -5,125 1,56251

Reakcijeoslonaca ( A9 , A10 , A11)−uklještena ivica4∗α2=6,2508∗(1+α2)=20,500

A y, 9=K2∗S y

3∗(Z9∗S y4

k+20.500∗w4+ (−6.250 )∗w3+ (−2.0 )∗w5)=19.3827kN /m

A y, 10=K2∗Sy

3∗(Z9∗S y4

k+20.500∗w3+ (−6.250 )∗w4+ (−6.250 )∗w2+(−2.0 )∗w6)=26.0948kN /m

A y, 11=K2∗S y

3∗( Z9∗Sy4k+20.500∗w2+ (−6.250 )∗w3+(−2.0 )∗w1)=9.049kN /m

k-1 k k+1-6,250 20,500 -6,250 +Zk∗S y

4

k∗{ k2∗S y

3 =A y , k

-2,000