[ ] 2. Grafički rad

2. grafički rad

a=4.0m

b=4.8m

β=1.2

F=40 kN /m'

ν=0.2

0≤ ρ≤1W=A+B ρ2

1≤ρ≤ βW ¿=C1+C2 ρ2+C3 ln ρ+C4 ρ

2 ln ρ

1

[ ] 2. Grafički rad

Granični i prelazni uslovi

Prelazni uslovi za ρ=1.0

1 )W=0

2 )W ¿=0

3 ) dWdr

=dW¿

dr

4 )M r=M r¿

Granični uslovi za ρ=β=1.2

5 )M r¿=0

6 )T r¿=P

dWdr

=2Bρ

d2Wdr2

=2B

d3Wdr3

=0

dW ¿

dr=1a

(2C2ρ+C3 ρ−1+2C4 ρ ln ρ+C4 ρ )

d2W ¿

dr 2= 1a2

(2C2−C3ρ−2+2C4 ln ρ+3C4 )

d3W ¿

dr 3= 1a3

(2C3 ρ−3+2C4 ρ−1)

M r=−K [ d2Wdr2 + νrdWdr ]=−K [2B+ 0.2

ρ∙a2B ρ]=−K ∙2.1 ∙B

M φ=−K [ 1r dWdr +ν d2Wdr2 ]=−K [ 1ρ ∙a 2 Bρ+0.2∙2B]=−K ∙0.9 ∙B

T r=−K [ d3Wdr3 + 1rd2Wdr2

− 1r2dWdr ]=−K [ 1ρ ∙a 2B− 1

( ρ∙a )22Bρ ]=−K B ∙0.375

ρ

2

[ ] 2. Grafički rad

M r¿=−K [ d2W ¿

dr 2+ νrdW ¿

dr ]=−K [ 142 (2C2−C3 ρ−2+2C4 ln ρ+3C4 )+ 0.2ρ∙ 414

(2C2ρ+C3 ρ−1+2C4 ρ ln ρ+C4 ρ ) ]=−K [0.15 ∙C2−0.05 ∙C3 ρ−2+0.15 ∙C4 ln ρ+0.2 ∙C4 ]

M φ¿=−K [1r dW ¿

dr+ν d

2W ¿

dr 2 ]=−K [1r 1a (2C2 ρ+C3 ρ−1+2C4 ρ ln ρ+C4 ρ )+ν 1a2

(2C2−C3ρ−2+2C4 ln ρ+3C4 )]=−K [ 1ρ∙4 14 (2C2ρ+C3 ρ−1+2C4 ρ ln ρ+C4 ρ )+0.2 142

(2C2−C3 ρ−2+2C4 ln ρ+3C4 )]=−K [0.15 ∙C2+0.05 ∙C3 ρ−2+0.15C4 ln ρ+0.1∙C4 ]

T r¿=−K [ d3W ¿

dr3+ 1rd2W ¿

dr2− 1r2dW ¿

dr ]=−K [ 1a3 (2C3 ρ−3+2C4 ρ−1 )+ 1ρ ∙a

1a2

(2C2−C3ρ−2+2C4 ln ρ+3C4 )− 1( ρ ∙a )2

1a

(2C2 ρ+C3ρ−1+2C4 ρ ln ρ+C4 ρ )]=−K ∙0.0625 ∙C4 ρ−1

ρ=1.0{1 °W=0→A+B=02 °W ¿=0→C1+C2=0

3 °dWdr

=dW¿

dr→0.5C2+C30.25+0.25C4−2B=0

4 °M r=M r¿→0.15 ∙C2−0.05 ∙C3+0.2∙C4−2.1 ∙B=0

ρ=1.2{5 °M r¿=0→0.15 ∙C2−0.035∙C3+0.227 ∙C4=06 ° T r

¿=P→0.052∙C4−40=0

A=−16.827/K

B=16.827 /K

C1=−937.500 /K

C2=937.500 /K

C3=−971.154 /K

C4=−769.231 /K

Ugibi

KW=A+B ρ2

KW ¿=C1+C2ρ2+C3 ln ρ+C4 ρ

2 ln ρ

ρ0=0→KW=−16.827

ρ1=0.5→KW=−16.827+16.827 ∙0.52=−12.620

ρ2=1→KW=0

ρ3=1.2→KW=−937.500+937.500 ∙1.22−971.154 ∙ ln 1.2−769.231 ∙1.22 ∙ ln1.2=33.48

3

[ ] 2. Grafički rad

16.82712.620

33.480

12.620

33.480

M r

M r=−K ∙2.1 ∙B

M r¿=−K [0.15 ∙C2−0.05 ∙C3 ρ−2+0.15 ∙C4 ln ρ+0.2∙C4 ]

ρ0=0→M r , 0=−K ∙2.1∙ 16.827K

=−35.337kNm /m

ρ1=0.5→M r , 1=−K ∙2.1 ∙ 16.827K

=−35.337 kNm/m

ρ2=1→M r ,2=−K ∙2.1∙ 16.827K

=−35.337kNm /m

ρ3=1.2→M r ,3=0

35.337

4

[ ] 2. Grafički rad

M φ

M φ=−K ∙0.9∙ B

M φ¿=−K [0.15 ∙C2+0.05 ∙C3ρ−2+0.15C4 ln ρ+0.1 ∙C4 ]

ρ0=0→M φ ,0=−K ∙0.9 ∙ 16.827K

=−15.144kNm /m

ρ1=0.5→M φ ,1=−K ∙0.9 ∙ 16.827K

=−15.144 kNm/m

ρ2=1→M φ, 2=−K ∙0.9 ∙ 16.827K

=−15.144kNm /m

ρ3=1.2→M φ, 3=−K [0.15 ∙ 937.500K−0.05 ∙ 971.154

K1.2−2+0.15−769.231

Kln 1.2+0.1∙−769.231

K ]=−8.944 kNm/m

15.1448.944 8.944

T r

T r=−K B ∙0.375ρ

T r¿=−K ∙0.0625 ∙C4 ρ

−1

ρ0=0→T r , 0=−K

16.827K

∙0.375

0=0

ρ1=0.5→T r ,1=−K

16.827K

∙0.375

0.5=−12.620kN /m

ρ2=1→T r ,2=−K

16.827K

∙0.375

1=−6.310kN /m

5

[ ] 2. Grafički rad

ρ2=1→T r ,2¿=−K ∙0.0625 ∙−769.231

K1−1=48.077 kN /m

ρ3=1.2→Tr ,3=−K ∙0.0625 ∙−769.231K

1.2−1=40.064 kN /m

6