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P i

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q(x)

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Oznaczamy podpory literami A, B i C

QDWRPLDVWSRáF]HQLDSU]HJXERZHF\IUDPL

1 i 2.

8ZDOQLDP\XNáDGRGZL ]yZL]DVW SXMHP\SRGSRU\UHDNFMDPL

W utwierdzeniu w punkcie A

G]LDáDM QDVW SXMFH UHDNFMH PRPHQW

M

A

RUD] VNáDGRZD

pionowa V

A

i pozioma H

A

NWyUHVRGVLHELHQLH]DOH*QH:SXQNFLH

B

Z\VW SXMHUHDNFMD

R

B

o

P = 3 ql

6 q

.

= 30

o

l

l

3 l

2 l

2 l



= 60

o

6 q

P = 3 ql

.

= 30

o

l

l

3 l

2 l

2 l



= 60

o

A

B

C

2

1

C

R

C

90

o

.

= 30

o



= 60

o



= 90

o

í. 

60

o

II

6 q

P = 3 ql

l

l

3 l

2 l

2 l



= 60

o

A

B

C

2

1



= 60

o

H

A

M

A

V

A

R

B

R

C

I

x

y

2

SLRQRZHM OLQLL G]LDáDQLD QDWRPLDVW Z SXQNFLH

C reakcja R

C

 G]LDáDMFD Z]GáX* SURVWHM

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= 60

o

.

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1

GOD F] FL 

I

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M

A

i V

A

, natomiast

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V

1

, H

1

i H

A

 PDM OLQLH G]LDáDQLD SU]HFKRG]FH SU]H] SXQNW

1

 D ZL F PRPHQW\

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1

VUyZQH]HUX

∑

=

i

I

i1

M

0 :

í

M

A

í

V

A

Â

l = 0 ( * )

=DSLV]HP\ QDVW SQH UyZQDQLH PRPHQWyZ Z]JO GHP SXQNWX

2

 GOD F] FL 

II

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NWyU\P Z\VWSL WH VDPH QLHZLDGRPH : SXQNFLH

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wynosi 4 q.

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1 i 2 jest

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W  F] ü XNáDGX ]JRGQLH ] SRQL*V]\P U\VXQNLHP :\]QDF]DP\ ZDUWRFL Z\SDGNRZ\FK MDNR

SRODILJXUSRGZ\NUHVDPLUR]NáDGXQDW *HQLDREFL*HQLDFLJáHJR

q

W

II

p

4

=

Â

2 l = 8 ql,

2

1

=

II

t

W

Â

q

Â

l = 2 ql

I

l

A

1

H

A

M

A

V

A

H

1

V

1

II

t

W

II

p

W

1

2

1

2

l

l

l

3

4

3

2

l

4 q

2 q

2 l

II

p

W

II

t

W

II

2 l

l

A

1

H

A

M

A

V

A

H

2

V

2

2

l

l

l

3

4

3

2

l

4 q

2

6 q

l + 3 l = 4 l

2 l

1

3

5yZQDQLHPRPHQWyZZ]JO GHPSXQNWX

2

GODF] FL

II

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∑

=

i

II

i2

M

0 :

II

p

W

Â

l +

II

t

W

Â

3

4

l

í

M

A

í

V

A

Â

l = 0 ( ** )

: UyZQDQLX W\P QLH Z\VW SXM QLHZLDGRPH

H

A

, V

2

i H

2

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II

p

W i

II

t

W

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í

M

A

í

V

A

Â

l = 0 ( * )

í

M

A

í

V

A

Â

l +

3

32

Â

ql

2

= 0 ( ** )

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M

A

=

í

3

16

ql

2

, V

A

=

3

16

ql.

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C

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A

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C

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C

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0

=

∑

i

iC

M

:

í

M

A

í

V

A

Â

l

í

R

B

Â

l

í

P

y

Â

l + W

Â

l + l ) = 0

⇒

R

B

=

3

37

ql

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P = 3 ql

l

l

3 l

2 l

2 l



= 60

o

A

B

C

2

1



= 60

o

H

A

M

A

=

í

3

16

ql

2

V

A

=

3

16

ql

R

B

R

C

P

x

= 3 ql

ÂFRV

=

2

3

ql

l

l

3 l

2 l

2 l

A

B

C

2

1



= 60

o

H

A

M

A

=

í

3

16

ql

2

V

A

=

3

16

ql

R

B

R

C

P

y

= 3 ql

ÂVLQ

=

2

3

ql

W =

2

1

Â

q

Â

l + l + 3 l ) = 18 ql

4

∑

=

i

iy

P

0 : V

A

+ R

B

+ R

C

ÂVLQ

í

W

í

P

y

= 0

⇒

R

C

=

9

3

11

ql

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niewiadomej H

A

.

0

=

∑

i

ix

P

: H

A

í

R

C

ÂFRV

í

P

x

= 0

⇒

H

A

=

9

3

10

ql.

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niewykorzystane.

∑

=

i

iA

M

0 :

í

M

A

í

W

Â

l

í

P

y

Â

l + R

B

Â

l + R

C

ÂVLQ

Â

l =

=

í

(

í

3

16

Â

ql

2

)

í

ql

Â

l

í

2

3

ql

Â

l +

3

37

ql

Â

l +

9

3

11

ql

Â

2

3

Â

l =

= ql

2

(

3

16

íí

2

27

+

3

148

+

6

77

)

= 0

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M

A

=

í

3

16

ql

2

=

í

ql

2

, V

A

=

3

16

ql = 5,33 ql, H

A

=

9

3

10

ql = 1,92 ql

R

B

=

3

37

ql = 12,33 ql, R

C

=

9

3

11

ql = 2,12 ql

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A

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∑

=

i

iA

M

0 :

í

M

A

í

W

Â

l

í

P

y

Â

l + R

B

Â

l + R

C

ÂVLQ

Â

l =

=

í

(

í

ql

2

)

í

ql

Â

l

í

ql

Â

l + 12,33 ql

Â

l + 2,12 ql

Â

0,866

Â

l =

= 0,0014 ql

2

§

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]DRNUJOHQLDZDUWRFLUHDNFML

W tym samym zadaniu wyznaczymy reakcje

EH] NRQLHF]QRFL UR]ZL]\ZDQLD XNáDGX GZX

UyZQD ] GZLHPD QLHZLDGRP\PL 5R]ZD*DQ\ XNáDG EHONRZ\ SRG]LHOLP\ QD SRMHG\QF]H

EHONL]ZDQHSRGXNáDGDPL

l

l

3 l

2 l

2 l

P

x

=

2

3

ql

A

B

C

2

1



= 60

o

H

A

=

9

3

10

ql

M

A

=

í

3

16

ql

2

V

A

=

3

16

ql

R

B

=

3

37

ql

R

C

=

9

3

11

ql

P

y

=

2

3

ql

W =

2

1

Â

q

Â

l + l + 3 l ) = 18 ql

5

3RQL*V]\U\VXQHNSU]HGVWDZLDVFKHPDWSUDF\XNáDGXEHONRZHJR

5R]ZL]\ZDQLH]DGDQLDUR]SRF]\QDP\RG F] FL 

II ), która w punkcie 1 i 2

 SRáF]RQD MHVW

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W przegubach 1 i 2

QD F] ü 

II

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H

1

, V

1

, H

2

i V

2

.

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V

1

i V

2

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F] FL 

II

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1

 GOD F] FL 

II

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niewiadoma V

2

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H

1

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1

i H

2

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1

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1.

∑

=

i

II

i1

M

0 : V

2

Â

l

í

II

p

W

Â

l

í

II

t

W

Â

3

2

l = 0

⇒

V

2

=

3

14

ql

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V

1

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2

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II

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0

=

∑

i

II

iy

P

: V

1

+ V

2

í

II

p

W

í

II

t

W = 0

⇒

V

1

=

3

16

ql

1

2

H

1

H

2

V

2

V

1

II

p

W

l

l

II

t

W

l

3

4

3

2

l

2 l

II

II

6 q

4 q

1

2

H

1

H

2

V

2

V

1

I

4 q



= 60

o

l

l

3 l

2 l

2 l

P = 3 ql



= 60

o

1

A

H

A

M

A

V

A

B

C

2

R

B

R

C

H

1

H

2

V

2

V

1

III

6

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1

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I

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I

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V

A

.

∑

=

i

I

iy

P

0 : V

A

í

V

1

= 0

⇒

V

A

=

3

16

ql

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I

XNáDGX

∑

=

i

I

iA

M

0 :

í

M

A

í

V

1

Â

l = 0

⇒

M

A

=

í

3

16

ql

2

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B

i R

C

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2

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kolorem czarnym.

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III

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III

t

W

,

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P

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1

A

H

A

M

A

V

A

H

1

V

1

=

3

16

ql

I

l

l

3 l

2 l

B

C

2



= 60

o

R

B

R

C

H

2

P

x

= 3 ql

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=

2

3

ql

P

y

= 3 ql

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=

2

3

ql

V

2

=

3

14

ql

3

8

l

3

4

l

III

t

W

=

2

1

Â

q

Â

l = 8 ql

III

l

3 l

2 l

B

C

2

4 q



= 60

o

P = 3 ql



= 60

o

R

B

R

C

H

2

V

2

=

3

14

ql

III

7

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C

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III

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R

B

.

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2

i R

C

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C

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0

=

∑

i

III

iC

M

:

III

t

W

Â

3

8

l + V

2

Â

l

í

P

y

Â

l

í

R

B

Â

l = 0

⇒

R

B

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3

37

ql

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R

C

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i

III

iy

P

: R

B

í

V

2

í

P

y

+ R

C

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í

III

t

W

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⇒

R

C

=

9

3

11

ql

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H

A

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=

∑

i

ix

P

: H

A

í

R

C

ÂFRV

í

P

x

= 0

⇒

H

A

=

9

3

10

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V

1

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2

w przegubach 1 i 2

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sprawdzenia.