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Plane stress state in
rectangular coordinatesIndirect solving method
8/16/2019 s1.1 SPTcc Gsr Fdu Updated
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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
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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
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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
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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
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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
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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
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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
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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
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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
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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
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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
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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
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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
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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
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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]
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100 kN/m2
17
SAP2000 – Plane finite elements (Area sections)
a)
b)
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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]
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“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
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20
COSMOS – plane finite elements PLANE2D (Ux, Uy)
PLANE SAP
ETh
vmv
umu
max
7
000326,0
1033,3
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stresses
21
x
PLANE SAPmkN max2
max 7,3346
[kN/m2 ]
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stresses
22
y
[kN/m2 ]
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23
xy
[kN/m2 ]
stresses