s1.1 SPTcc Gsr Fdu Updated

8/16/2019 s1.1 SPTcc Gsr Fdu Updated

http://slidepdf.com/reader/full/s11-sptcc-gsr-fdu-updated 1/23

Plane stress state in

rectangular coordinatesIndirect solving method



Example: Simple supporter beam with full length

uniform distributed load

2

3

3

3

3

3

2

3

2

3

6

2

322

32

4

5

3

2

36

xybh

q x

bh

qb

q y

bh

q y

bh

q

y

bh

q y

bh

q yl

bh

q y x

bh

q

xy

y

x

l/2 l/2

x

y

q

b

hP1

P2

P

2



3

Checking the plane stress relations accordingly with the indirect

solving method

2

3

3

3

3

3

2

3

2

3

6

2

322

32

4

5

3

2

36

xybh

q xbh

qb

q ybh

q ybh

q

ybh

q ybh

q yl

bh

q y x

bh

q

xy

y

x

presumptively knownstress relations

1st set of conditions: condition of equilibrium (the stress relations mustsatisfy the differential equations of equilibrium)

0

0

y x

y x

y xy

yx x

xybh

q

x

x

3

12

xybh

q

y

yx

3

12

yx xy (duality of the shear stresses)

0

y x

yx x

and

bh

q y

bh

q

y

y

2

36 2

3

2

3

6

2

3 y

bh

q

bh

q

x

xy

and

0

y x

y xy



4

2nd set of conditions: condition of compatibility (or continuity) of thedeformations the equation is expressed in terms of stresses

0)(2

2

2

2

2

2

2

22

y x y x

y y x x y x

y

bh

q

x

x

32

212

ybh

q

y

y

32

212

2

3

2

3

2

3

12

5

3

2

36 ybh

q

bh

qlbh

q xbh

q

y

x

0

x

y

xybh

q

x

x

3

12

y

bh

q

y

x

32

2 24

bh

q ybh

q

y

y

2

36 2

3

02

2

x

y

012

02412

)(333

2 ybh

q y

bh

q y

bh

q y x



5

3rd set of condition: limit conditions (expressed in terms of stresses){thecompatibility of the stresses and the loads on the boundary of the domain}

x

σy

σy

σx σx

τ xy

τ

yx

τ xy

τ yx

y

0;; yx xy y x

A B

C D

AB: the upper side of the plane domain

2

h y (equation of the boundary straight line)

b

q

b

qh

bh

qh

bh

qh

y y

222

3

2

2 3

32

02

623

2

32

h x

bhq x

bhqh

y xy l/2 l/2

x

y

q

b

hP1 P2P

q/b

CD: the lower side of the plane domain

2

h y (equation of the boundary straight line)

0222

3

2

2 3

32

b

qh

bh

qh

bh

qh

y y

02

6

2

3 2

32

h xbh

q xbh

qh

y xy



6

A B

C D

q

ql/2 ql/2

AC: the left side of the plane domain

2

l x (equation of the boundary straight line)

3

3

3

3

2

3

2

32

4

5

34

5

3

2

3

2

6 ybh

q ybh

q ybh

q ybh

q yl

bh

q yl

bh

ql

x x

2

3

2

32

3

4

3

2

6

22

3 y

bh

ql

bh

ql yl

bh

ql

bh

ql

x

xy

z

y

y

dy dA=bdy

b

2

2

3

3

3

32

04

5

34

5

3 h

h A A

l x x

bdy ybh

q ybh

qdA y

bh

q ybh

qdA N

2

2

2

2

5

3

2

2

34

3

23

32

0

5

4

35

34

5

34

5

3 h

h

h

h

h

h A A

l x x

y

h

q y

h

qbdy y

bh

q ybh

q ydA y

bh

q ybh

q ydA M

2

2

2

2

3

3

2

2

2

3

2

32 244

3

3

3

4

33

4

33

4

3 h

h

h

h

h

h

A A

l x xy

qlqlql y

h

ql yh

qlbdy y

bh

ql

bh

qldA y

bh

ql

bh

qldAT

BD: the right side of the plane domain

2

l x (equation of the boundary straight line) Similar with AC side



Computing the deformations

E

E GG

E

E

xy xy

xy xy

x y y

y x x

12

12;

1

1

b

q ybh

q ybh

q ybh

q ybh

q ybh

ql y x

bh

q

E x

22

324

5

3

2

361 3

3

3

33

22

3

3

33

22

3

3

3

4

5

3

2

36

22

321 ybh

q ybh

q ybh

ql y xbh

q

b

q ybh

q ybh

q

E y

2

3

2

3

123123 xy

Ebh

q x

Ebh

q xy

Ebh

q x

Ebh

q xy

7



x

v

y

u

y

v

x

u

xy y x

;;

x f dyv

y f dxu

y

x

2

1

y f x f x Ebh

ql x

Ebh

q x

Ebh

q x

Ebh

q123

2

33

3

3 2

3

2

3

5

122

22

3

2224

32

11

4

3

4

3

5

6

2c x x

Ebh

ql x

Ebh

q x

Ebh

q x

Ebh

q x f

c y y f

8



1

3

3

3

33

23

3

22

3

24

5

3

2

32

,

c y x Eb

q xy

Ebh

q

xy Ebh

q

xy Ebh

q

xy Ebh

q

xy Ebh

ql

y x Ebh

q

y xu

22

3

2224

3

4

3

22

3

222

3

24

3

4

3

4

3

5

6

2

10

3

4

33

24

3

2

,

c x x Ebh

ql x

Ebh

q x

Ebh

q x

Ebh

q y

Ebh

q

y

Ebh

q y

Ebh

ql y x

Ebh

q y

Eb

q y

Ebh

q y

Ebh

q y xv

00,2

0,00,2

2

1

v yl

xP

vu yl xP

0

16

3

16

3

10

3

32

4

3

422

3

4

2

1

Ebh

ql

Ebh

ql

Ebh

ql

Ebh

qlc

Eb

qlc

9



Eb

ql x Eb

q xy

Ebh

q

xy Ebh

q xy Ebh

q xy Ebh

q xy Ebh

ql y x Ebh

q y xu

422

3

24

5

3

2

32,

3

3

3

33

23

3

3

422

3

4

2

3

2224

3

4

3

22

3

222

3

24

3

16

3

16

3

10

3

32

4

3

4

3

5

6

2

10

3

4

33

24

3

2,

Ebh

ql

Ebh

ql

Ebh

ql

Ebh

ql

x Ebh

ql x

Ebh

q x

Ebh

q x

Ebh

q y

Ebh

q

y Ebh

q y

Ebh

ql y x

Ebh

q y Eb

q y

Ebh

q y

Ebh

q y xv

10



2

3 12;

12 h

I bh A

bh I

2

2

2

24

max 2,192,11384

5

l

h

l

h

EI

qlv

2

24

max 16,21384

52,0

l

h

EI

qlv

EI

hql

EI

hql

EI

ql

Ebh

ql

Ebh

ql

Ebh

qlv

Eb

qlu

y xP

M R M v

192

3

120

3

384

5

16

3

10

3

32

5

04

0,0 2222422

3

4

max

..max,

11



T h/2

h/2

x

y

max max

dx

T

T

max max

maxT dv

dx

Comparison with the results provided by means of Strength of

Materials

taking into account the influence of shear force in the calculation of

bending deformations

G

maxmax

dx

dvT

max

;0max

T A

T k

Shear coefficient

(depending of the shape

of the cross-section

average

k

max0

5,1

5,12

3

0

maxmax

k

A

T

A

T

bh

T

bI

TS

average

b

h

12



c x EbhqvT 2

213

x

Ebh

q x

Ebh

q x

Gbh

q

dx

dvqxT

T

13125,15,1

cvl

xPP T

0

2, 21

2

2

415,1 xl EbhqvT

Ebh

qlv xP

T 2

1375,00

2

24224

..max, 14,21

3845

121375,0

3845

lh

EI ql

EI hql

EI qlv M S T M

Th E M S

T M vl

h

EI

qlv

.

max2

24..

max, 88,21384

52,0

13

T

T

GA

T k

dx

dv0



Numerical application

1 m 1 m

x

y

q = 10 kN/m

10 cm

30 cm

E = 300000 daN/cm2

= 0,2

P

beamlengthmedium1067,65 h

l

14



2maxmax

22

2

maxmax 33,33532,0;33,3333

68 mkN

bqmkN

bhql

W M SM EThSM

m Eb

qluu ETh

P

SM

P

71033,34

;0

ml

h

EI

qlv

m EI

qlv

SM

T M

SM

M

000328642,088,21384

5

000308642,0384

5

2

24

max,

4

max,

mlh

EI qlv ETh 000323642,016,21

3845

2

24

max

15



16

SAP2000 – frame finite elements (FRAME)

With shear forceinfluence

( A T automatically

calculated)

Without shear

force influence

(A T → )

Finite Element Method solution

ETh

vmax

SM

M vmax,

[m]

[m]



100 kN/m2

17

SAP2000 – Plane finite elements (Area sections)

a)

b)



2/)(98,3345 maxmax

2

max

EThSM mkN

2/)(00032604,0

1033,3

maxmax,

7

EThSM

T M

ETh

vvmv

umua), b)

“PLANE” finite elements for Plane stress

(Ux, Uz)

18

x

[mm]



“SHELL” finite elements with membrane behavior

(Ux, Uz, Ry) a), b)

EThmkN max

2

max 81,3367

19

EThSM

T M

PLANE

ETh

vv

vmv

umu

maxmax,

7

and between

00032586,0

10333,3

[mm]

x



20

COSMOS – plane finite elements PLANE2D (Ux, Uy)

PLANE SAP

ETh

vmv

umu

max

7

000326,0

1033,3



stresses

21

x

PLANE SAPmkN max2

max 7,3346

[kN/m2 ]



stresses

22

y

[kN/m2 ]



23

xy

[kN/m2 ]

stresses

Documents

s1.1 SPTcc Gsr Fdu Updated