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Calculation of an elastic hyperstatic plane gantry
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Organization (S): EDF/IMA/MN
Handbook of Validation
V3.90 booklet: Theoretical references of tests in linear statics
Document: V3.90.001
Calculation of an elastic hyperstatic plane gantry
Summary:
The goal of this note is to expose the method of calculation used to determine the reference solution of
case-test SSL 14, entitled: “Plane Gantry articulated in foot”.
One uses the method of the forces (hyperstaticity 1), by taking account only of the energy of inflection: assumption
slim beams.
One considers four loading cases separately.
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Contents
1 isostatic Stresses under real load distributed p on C C
1
....................................................... 4
1.1 Isostatic reactions of supports ....................................................................................................... 4
1.2 Stresses ..................................................................................................................................... 5
1.3 Diagrams ..................................................................................................................................... 6
2 Stresses under concentrated force F1 (downwards) ............................................................................. 7
2.1 Reactions of support ............................................................................................................................. 7
2.2 Stresses ..................................................................................................................................... 7
2.3 Diagrams (F1 downwards) .......................................................................................................... 8
3 Stresses under the concentrated force F2 (towards the left) ................................................................... 8
3.1 Reactions of support ............................................................................................................................. 8
3.2 Stresses ..................................................................................................................................... 9
3.3 Diagrams ..................................................................................................................................... 9
4 Stresses under the couple concentrated (positive) ............................................................................... 10
4.1 Reactions of support ........................................................................................................................... 10
4.2 Stresses ................................................................................................................................... 10
4.3 Diagrams (positive) ................................................................................................................. 11
5 Stresses under the hyperstatic moment X ................................................................................... 12
5.1 Reactions of support ........................................................................................................................... 12
5.2 Stresses ................................................................................................................................... 12
5.3 Diagrams ................................................................................................................................... 13
6 Stresses under specific dummy loads out of C .......................................................................... 13
6.1 Reactions of support ........................................................................................................................... 13
6.2 Stresses ................................................................................................................................... 14
6.3 Diagrams ................................................................................................................................... 14
7 Determination of the hyperstatic moment X ........................................................................................ 15
7.1 Charge distributed p on C C
1
......................................................................................................... 16
7.2 Concentrated loading F1 out of C .......................................................................................................... 17
7.3 Concentrated loading F2 in C1 ......................................................................................................... 18
7.4 Specific couple in C1 ............................................................................................................. 19
7.5 Summary ................................................................................................................................... 20
8 Calculation of displacement out of C ............................................................................................................... 20
8.1 Charge distributed p on C C
1
......................................................................................................... 20
8.2 Concentrated loading F1 out of C .......................................................................................................... 21
8.3 Concentrated loading F2 in C1 ......................................................................................................... 21
8.4 Specific couple in C1 ............................................................................................................. 22
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m M
8.5 Calculation of D =
1 ................................................................................................................ 22
I.E.(internal excitation)
8.6 Summary of displacements CPU and vC .................................................................................... 23
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y
C
One considers the gantry opposite,
p
subjected to various loads.

has

C
C
1

2
Hyperstaticity of degree 1.
F
Hyperstatic unknown factor: X:
1
F

2
moment out of C.

Top-load distributed p on

H
C1 C.
Two forces F1, F2, and a couple
With
in C1.
B
X


tg
= 2a = 0,4 cos

() - 1 = 1,16 =1, 077033
(
)



C
tg
=

=
1
2 A
(+ H)
1,2
GOES
VB
B
=

With
B

2cos
; sin
= B has
H
H
With
B
1
Isostatic stresses under real load distributed p on
C C
1
1.1
Isostatic reactions of supports
p "
p 2

H + H
= 0; V + V
=
;
V
=
With
B
With
B
B
2 cos
8 cos
Part CB is articulated and charged only at its ends:
H
B

BC = 0

H = - V

tg
V
B
B

B
From where isostatic reactions:
p "
3 p "
p "
p "
H
=
tg; V
=
;
H
= -
tg; V
=
With
With
B
B
8 cos
8 cos
8 cos
8 cos
Note:
“tg

=
B
.
8 cos
(
8 A + H)
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1.2 Stresses
Beam
In C1
M
NR

3 p "
NR
= -
Iso

y

8 cos
+


V
p
V

=

With
Viso
tg

8 cos
0

p "
H
M
= -
y tg
With
Iso

8 cos
Beam C B
2
M
NR

p "
NR
= -

+
Iso
8 cos
y



V
p
V
= -
tg
V
Iso
B

8 cos
0

p "
H
= -

B
M
y
Iso
tg

8 cos
Beam C C
1
M

px
= -
cos -
sin +


V
NR
H
V
Iso
With
With
sin

cos
p


X
NR
= -
tg + 3 tg - 8 tg
+
p

y

8



px


V
= H sin - V cos +


Iso
With
With

cos

cos

p "

=
(tg tg - 3+8 X/“)
GOES

8
2
H

px
With
M
= -
+ V X - H y
Iso
With
With
2 cos
X


p
x2 3 “Xy tg

=
-
+
-
(= 0 in C)!

cos
2
8
8


p "
2 A + H

M
= -
Iso
(
2 S
S 2a
H
3
bh
8 A + H)
(
)





B -
+
+
with
S
X
= cos [
0, B]
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Beam C C2
NR
NR
= H cos - V

Iso
B
B sin

p


= -
(tg + tg)
y
M


8
+
V
V
= H sin + V

Iso
B
B

cos

p "
= -

(tg .tg -) 1

8
M

= H y + V (“- X)
V
Iso
B
B
B

p "
H

= -
(y tg - (“- X)) (= 0 C
in
)!
B

8

X
cos

1.3 Diagrams


B =



2 cos
+
+
p

-
- H tg

pH "
tg
8cos
8

cos
p

p
=

p

p
- pbh
- 3
-
tg

- tg

8 (a+h)

8 cos
8cos

8cos

8cos

NR
V
M
Iso
Iso
Iso
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2
Stresses under concentrated force F1 (downwards)
2.1 Reactions
of support
F1
H + H
=
With
B
0;
C
V + V
= F
With
B
1
;
C
C
1
2
H
H
With
B

AC = 0 =


BC
V
V
With
B
V
V

With
B
With
B
H
H
With
B
From where:
1
1
1
1
H
=
F tg
; V

1
=
F
;
H
1
= - F tg
;
V
1
=
F
With
With
B
B
2
2
2
2 1
2.2 Stresses
Beam
In C1:

1
NR
= - F
Iso
1

2

1
tg
V
=
F

Iso

2 1

1
M
= - F y tg
Iso


2 1
Beam C B
2
:

1
NR
= - F
Iso
1

2

1
tg
V
= - F

Iso

2 1

1
M
= - F y tg
Iso


2 1
Beam C C
1
:

1
NR
= - F



1
+
Iso
(tg cos sin)

2

1
V
=
F



1
-
Iso
(tg sin cos)

2

1
M
= - F

1
-
Iso
(y tg X)


2
Beam C C2:

1
NR
= - F



1
+
Iso
(tg cos sin)

2

1
V
= - F



1
-
Iso
(tg sin cos)

2

1
M
= - F

1
- -
Iso
(y tg (X))


2
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2.3 Diagrams
(F1 downwards)
F1
- H tg
2
- F1h tg
= - F lh
1
2
4 (a+h)
- F1/2
- F1/2
NR
V
M
Iso
Iso
Iso
3
Stresses under the concentrated force F2 (towards the left)
3.1 Reactions
of support
· HA + HB = F2;
C
C
+ VB GOES = 0;
1
C
V
2
B + H F2
= 0
F

2
HB

·
BC =
V
V

0
With
V
B
B
With
B
H
H
With
B
From where:

H
H
H
H
H
= F


1 1 -
tg
;
V
F
;
H
F
tg
;
V
F
With
With
2
B
1
B






=
=
= - 2



Note:
H
H

H
2a + H
=
1 -

“tg
(
;
tg
2 A + H)





=
(
2 A + H)
2
2
H "
“- (
4 A + ah)
tg sin - cos
= - 2 (
-
=
B has + H)
;
tg cos
sin
4 (
B has + H)
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3.2 Stresses
Beam A C1:

NR
= - F H/“
Iso

2


H
V
= F

2 1 -
tg
Iso









H
M
= - F y

2
1 -
tg
Iso








Beam C B
2
:

NR
= F H/“
Iso

2

H
V
= F
tg
Iso

2 “H
M
= - F
y tg
Iso


2 “
Beam C C
1
:


H
H

NR
= - F



2
1 -
tg
cos
sin
Iso






+ “





H
H

V
= F



2
1
tg
sin
cos
Iso






- “




H

H

M

= F
X

2
- 1 - tg
y
Iso






Beam C C2:

H
NR
= F



2
+
Iso
(tg cos sin)



H
V
= F



2
-
Iso
(tg sin cos)

H
M
= F

2

- -
Iso
(y tg (X))



3.3 Diagrams
(
)
- H2 has + H
F2
+
2 A
(+ H)
-
- H
H2
F
F
2
2

2 (+ H has)
M
NR
Iso
Iso
Viso
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4
Stresses under the concentrated couple (positive)
4.1 Reactions
of support
· HA + HB = 0;
V
C
WITH + VB
= 0;


C
V

2
B +
= 0
H

B
·
BC =

0
V
V
V
With
B
B
With
B
H
H
With
B
From where:
H
= - tg “; V
= “
With
With
H
= tg “; V
= - “
B
B
Note:
tg
=
1

(
2 A + H)
4.2 Stresses
Beam A C1:
NR
= - “
Iso

V
= - tg “
Iso
M
=
y

tg “
Iso
Beam C B
2
:
NR
= “
Iso

V
= tg “
Iso
M
=
y

tg “
Iso
Beam C C

1
:
NR
=
tg cos -


Iso
(
sin)




V
= -
tg sin +

Iso
(
cos)




M
=
+ tg -


Iso
(X y
)


Beam C C2:


NR
=
tg cos +


Iso
(
sin)




V
=
tg sin -

Iso
(
cos)




M
=
tg - “-
Iso
(y
(X))


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4.3 Diagrams
(positive)
(
)
- H +2a
2 (+ H has)
-
H
H
2 A
(+h)
2 (+ H has)
- tg
-/
/





NR
M
Iso
Viso
Iso
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5
Stresses under moment X hyperstatic
5.1 Reactions
of support
X
X
+
H + H
=
With
B
0;
C
V + V
=
C
C2
With
B
0;
1
V
=
B
0;
V
V
H (+ H has) - X
=
With
B
B
0
With
B
H
HB
With
From where reactions:
X
X
H
= -
;
V
= 0
; H
=
;
V
=
With
0
+ H has
With
B
+ H has
B
5.2 Stresses
Beam A C1:

NR
=
X
0

X
V
= -
X
+ H has

X
M
=
y
X


+ H has
Beam C B
2
:

NR
=
X
0

X
V
=
X
+ H has

X
M
=
y
X


+ H has
Beam C C
1
:

X
NR
=
cos


X
+ H has


X
V
= -
sin
X
+ H has

X
X
M
=
y
=
+

X
(H X tg)


+ H has
+ H has
Beam C C2:

X
NR
=
cos


X
+ H has


X
V
=
sin
X
+ H has

X
M
=
y
X


+ H has
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5.3 Diagrams
X
X
X sin
X cos
+ H has
+
+ H has
X H
X H
has
has
+ H
+ H
O
O
X
+ H has
NR
V
M
X
X
X
6
Stresses under specific dummy loads out of C
In order to calculate displacement out of C, using Principe of work Virtuels (cf.
paragraph [§8]), it is necessary to establish the diagrams of stresses under the action of two
“fictitious” forces F and G applied out of C.
6.1 Reactions
of support
H + H
= - F
G
With
B
;
V + V
= - G
F
With
B
;
H
H
C
With
B

AC = 0 =


BC
C
C
V
V
2
With
B
1
V
V

With
B
With
B
H
H
With
B
From where:
1
1
H
= -
+

;
= -
+ cot G
With
(F G tg)
GOES
(G F
)
2
2
1
1
H
= -
-

;
= -
- cot G
B
(F G tg)
VB
(G F
)
2
2
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6.2 Stresses
Beam A C1:

1
N
=

(g+ F cotg)

2

1
v = - (F + G tg)

2

1
m
=
(F +g tg) y


2
Beam C B
2
:

1
N
=

(G F cotg)

2

1
v = - (F - G tg)

2

1
m
= - (F - G tg) y


2
Beam C C
1
:

1
1
N
=

(F +g tg) cos + (g+ F cotg) sin

2
2

1
1
v = - (F + G tg) sin + (G + F cot G) cos

2
2

1
1
m
= + (F + G tg) y - (G + F cot G) X


2
2
Beam C C2:

1
1
N
= -

(F - G tg) cos + (G F cotg) sin

2
2

1
1
v = - (F - G tg) sin - (G - F cot G) cos

2
2

1
1
m
= - (F - G tg) y - (G - F cot G) (“- X)


2
2
6.3 Diagrams
Here diagrams of stresses under the action of the two “fictitious” forces F and G. One considers
here: F 0, G F
cotg.
G
+
F
- fh/2
fh
gh
2
+

4

has
(+ H)
+ gh
4 A
(+ H)
N
v
m
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7
Determination of moment X hyperstatic
One places oneself in elasticity; one considers only the energy of inflection, the beams being slim. The state
naturalness is supposed to be virgin (not prestressings nor of displacement of support).
The complementary potential is then:

(
2
2
M
+ M1 X

+ 1
Iso
)
(MR. MR. X
Iso
)
F * (X) =

+
I.E.(internal excitation)

I.E.(internal excitation)

1

2
posts
frames
It is stationary with balance, from where:


M2

2
M
1
1

M1 Miso
M1

M
X
=


+
X = -
Iso
S


I.E.(internal excitation)
pot
1


I.E.(internal excitation)
charp
2


-
I.E.(internal excitation)
pot
1


=
I.E.(internal excitation)
charp
2
.
.
.
.
The coefficient of flexibility is the sum of:

M2
2 H H
2
1
I.E.(internal excitation)
3 I.E.(internal excitation)

has
H


=
+
posts
1
1

M2
2 B
H2 1 has 2
ah

1


I.E.(internal excitation)
I.E.(internal excitation)



has
H + 3

has

+ H
2


=
+

+
2
2
(has

+ H)
frames

that is to say:
3
2
2

2
H
(B h3 +a + a3h)
E
=


(
+
+ H) 2 has 3I
3I
1
2


Numerical application:
In the example considered:
-
I = 2I
4
= 5 0
, 10
m4
1
2

H = 2 A = 8 m; “= 20 m; B =
1 1
, 6
2
From where:
2
=
2
H
19 B
H
E (

+
+ H) 2 has I
3

2
1




2353 45347
3
.
m
One studies one after the other the various loadings to calculate the second members S.
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7.1 Charge
distributed
p on C C
1
The second member S due to p is:
MR. M
2 H
H
3


1
Iso
Pb H
-

2
pH B


=

I.E.(internal excitation)
3 I.E.(internal excitation)



has
2
1
1
+ H (
8 A + H)
= E
24
posts
(+ H has) I1
MR. M
pb2h
2


1
Iso
1 H
1 A

1
phb (H
3 + has)
-




=
I.E.(internal excitation)
(
8 A
2
2
+ H) I.E.(internal excitation) 2

has
2
+ H + 6

+ H has
= E
48
C C
(+ H has) I
2
2


+

-
MR. M
1
p


1
Iso
B
has
H
has
2 2
S
S (2a 3h) bh
H
S
ds



=

I.E.(internal excitation)
I.E.(internal excitation)
(
8 A + H) 2 0



B -
+
+


B
C C
2
2



+
1
1
2
p " B
2
2
=

H - ah - has
2
(2
)
E (+ H has) I
48
2
From where:
3
2
2
2
2
p " B 4h
(
hb H
3 + has)
(bh - ah-a)
S
=


+
+

E (+ H has) 2
96 I
I
I
1
2
2


Numerical application:
I = 2I
;
H = 2 A;
p = 3.000 NR/m
1
2
(downwards)
2
p " bh2
13
S
=

4h +
B
E (+ H has) 2 I
96


2


1



43.946 02189
.
NR m4

From where:
· moment out of C:
X
= 18.672.994
.
NR m
· reaction in a:
B "
X
Pb " 8 - X
H
= p
-
=
With
(
8 A + H)
+ H has
(+ H has)
H
=
With
5.175 37
.
NR
3pb
V
=
-
With
0
4
V
=
With
24.233 24
.
NR
Handbook of Validation
V3.90 booklet: Theoretical references of tests in linear statics
HI-75/96/037 - Ind A

Code_Aster ®
Version
4.0
Titrate:
Calculation of an elastic hyperstatic plane gantry
Date:
01/09/99
Author (S):
F. VOLDOIRE
Key:
V3.90.001-A Page:
17/24
7.2 Charge
specific
F1 out of C
The second member is obtained using:
MR. M
2 H
H
3


1
Iso
F H
1

-

2
F H
1


=

I.E.(internal excitation)
3 I.E.(internal excitation)



has
2
1
1
+ H 4 (+ H has) = E
12
posts
(+ H has) I1
MR. M
2b
F
1
Iso
1 H
1
H
1
has

-





=

I.E.(internal excitation)
I.E.(internal excitation)
4 (has
2
2
+ H) 2

+ H has + 6

+ H has
frames
2
F B " (
H H
1
3 + has)
= E (

+ H) 2 has I
24
2
From where:
2
FH2 H2
(
B H
3
1
+ has)
S
=


+

E (+ H has) 2
24
I
I

1
2


Numerical application:
I = 2I
;
H = 2 A; F = 20.000 NR
1
2
1
(downwards)
2
FH2
S
1
=

2 + 7
2
[H B]
E (+ H has) I
24
1





97.485.127 76
,
NR m4

From where
· moment out of C:
X
= 41.422.161
.
NR m
· reaction in a:
1

X
F “4
1
- X
H
=
F1
-
=
With
4
+ H has
+ H has
(+ H has)
H
=
With
4.881 4866
.
NR
1
V
=
F1 -
With
0
2
V
=
With
10.000 0
. NR
Handbook of Validation
V3.90 booklet: Theoretical references of tests in linear statics
HI-75/96/037 - Ind A

Code_Aster ®
Version
4.0
Titrate:
Calculation of an elastic hyperstatic plane gantry
Date:
01/09/99
Author (S):
F. VOLDOIRE
Key:
V3.90.001-A Page:
18/24
7.3
Concentrated loading F2 in C1
The second member is obtained using:
MR. M
H
3
1
Iso
H
F
(
H2 has
2
+ H)
2
F H (2a
2
+ H)
-
=



=

I.E.(internal excitation)
3 I.E.(internal excitation)



has
2
1
1
+ H2 (+ H has)
E
12
WITH C
(+ H has) I1
1
2
MR. M
H
4
1
H
(- F H
2
)
2
-
-
=
F H
Iso
2


=

I.E.(internal excitation)
3 I.E.(internal excitation)



has
2
1
1
+ H (
2 A + H)
E
12
C B
(+ H has) I
2
1
MR. M
B
F H2
1
Iso
2
(a+ H) 1 H
1
has

-





=
I.E.(internal excitation)
I.E.(internal excitation)
(
2 A
2
2
+ H)
2

6
C C


+ H has +

+ H has
1
2
F bh
2
2
2
(h3 +7ah+2a)
= E (

+ H) 2 has I
24
2
MR. M
B
2
2
1
Iso
- F H
2
1 H
1 A

2
- F bh (H
2
3 + has)
-




=
I.E.(internal excitation)
I.E.(internal excitation)
(
2 A
2
2
2
+ H) 2

6
24
C C


+ H has +

+ H has = E (+ H has) I2
2
From where:
2
F ha 2h2
B

S
=
2

+
(H
3 + has)
E (+ H has) 2
12 I
I
1
2

Numerical application:
I = 2I
;
H = 2 A; F = 10.000 NR
1
2
2
(towards the left)
2
F H2 has
S
=
2
(2h + 7b)
E (+ H has) 2 I
12
1





19.497.025 55 NR m
4
.
From where:
· moment out of C:
X
= 8.284 4321
.
NR m
· reaction in a:
2a + H
X
F (has
2
+ H/2) - X
H
= F2
-
=
With
(
2 A + H)
+ H has
(+ H has)
H
=
With
5.976.297
.
NR
V
= F H
With
2
/“
V
=
With
4.000 00
.
NR
Handbook of Validation
V3.90 booklet: Theoretical references of tests in linear statics
HI-75/96/037 - Ind A

Code_Aster ®
Version
4.0
Titrate:
Calculation of an elastic hyperstatic plane gantry
Date:
01/09/99
Author (S):
F. VOLDOIRE
Key:
V3.90.001-A Page:
19/24
7.4 Couples
specific
in C1
The second member is obtained using:
MR. M
3


1
- 2h
H
H
2
-
-

=

H
Iso


=
I.E.(internal excitation)
3 I.E.(internal excitation)



has
2
1
1
+ H (
2 A + H)
E
6
posts
(+ H has) I1
MR. M
B

1
Iso
(H + 2a) 1 H
1
has

-




=
I.E.(internal excitation)
I.E.(internal excitation)
(
2 A
2
2
+ H) 2

6
C C


+ H has +

+ H has
1
2
(H + 2a) (H
3 + has)
=
B
E (

+ H) 2 has I
24
2
MR. M
B


1
- H 1 H
1 A

2
- hb (H
3 + has
Iso
)
-




=
I.E.(internal excitation)
I.E.(internal excitation)
(
2 A
2
2
2
+ H) 2

6
24
C C


+ H has +

+ H has = E (+ H has) I
2
2
From where:
2
- 2h3
(
ab H
3 + has)
S
=


-

E (+ H has) 2 12 I
I
1
2

Numerical application:
I = 2I
;
H = 2 A; = - 100 000 NR m
1
2
(direction hands clock)
2
-
S
=

3
2
[H - has (B h3+a)]
E (+ H has) I
6
1



11.571.281 93
.
NR m
4
From where
· moment out of C:
X
= 4.916 7243
.
NR m
· reaction in a:
-
X
- 2 - X
H
=
-
=
With
(
2 A + H)
+ H has
(+ H has)
H
=
With
4.576.394
.
NR

V
=
With

V
=
With
5.000.000
.
NR
Handbook of Validation
V3.90 booklet: Theoretical references of tests in linear statics
HI-75/96/037 - Ind A

Code_Aster ®
Version
4.0
Titrate:
Calculation of an elastic hyperstatic plane gantry
Date:
01/09/99
Author (S):
F. VOLDOIRE
Key:
V3.90.001-A Page:
20/24
7.5 Summary
CAS
Moment out of C
Reactions of A (NR)
(N.m)
Ha
GOES
p on C C
18672.994
5175.37
24233.240
1
F1 out of C
41422.161
4881.487
10000.000
F
8284.432
5976.297
4000.000
2 in C1
out of C
4916.724
4576.394
5000.000
1
TOTAL
22033.31
43233.24
Notice
Recall: in post A C1: normal effort = - GOES, shearing action = ha.
8
Calculation of displacement out of C
One considers also only the elastic energy of inflection (slim beams). By applying Principe
of Travaux virtual on the structure subjected to the fictitious forces of the paragraph [§6], working in
sought displacements, one calculates the numbers W and D depending linearly on F and G:

m (M + X M1


+
Iso
)
m (M
X M
Iso
1)
F U + G v
=
W
Xd

+
=
+
, (
F, G
C
C
)
I.E.(internal excitation)
I.E.(internal excitation)

1

pot
charp
2
.
.
8.1 Charge
distributed
p on C C
1

m M
2h
gh "
- pbh "
2
- gpbh3 2

Iso

=



=

I.E.(internal excitation)
3 I.E.(internal excitation)
(
4 A
2
1
1
+ H) (
8 A + H)
E
+
96
posts
(H has) I1

m M
2
- p " hb2
Iso

2

2
(F (a+h) +g) (ha)


=
EI2
E
+
384
C C
(H has) I
1
2

M
B
pbh "
fh
G H


2
- pb2 " H2 “- 2
+
Iso
(G
F (H has))

-




=

I.E.(internal excitation)
I.E.(internal excitation)
3
(
8 A + H) 2
(
4 A + H) =
E
2
+
192
C C
2
2
(H has) I
2
2
From where:
2
- pbh " 4g H2

G " (
B H
3 - has) - 2 Bfrs (+ H has) 2
W
=


+
2



E (+ H has)
384
I
I
1
2

Numerical application:
I = 2I
;
H = 2 A;
p = 3.000 NR/m
1
2
(downwards)
2

5
9

- p bh2 "
W
=
G " 2h + B
fbh
E (+ H has) 2 I



2 - 2


192
1



- 2154065922
.
NR m
3
Handbook of Validation
V3.90 booklet: Theoretical references of tests in linear statics
HI-75/96/037 - Ind A

Code_Aster ®
Version
4.0
Titrate:
Calculation of an elastic hyperstatic plane gantry
Date:
01/09/99
Author (S):
F. VOLDOIRE
Key:
V3.90.001-A Page:
21/24
8.2 Charge
specific
F1 out of C

m M
H
gh "
- F H
3 2
2


1
2
- F gh
Iso

=

1


=

I.E.(internal excitation)
3 I.E.(internal excitation)
(
4 A
2
1
1
+ H) (
4 A + H)
E
+
48
posts
(H has) I1

m M
B
gh "
- F H
2 2
2


1
2
- F gbh
Iso

=

1


=

I.E.(internal excitation)
3 I.E.(internal excitation)
(
4 A
2
.
2
2
+ H) (
4 A + H)
E
+
48
charp
(H has) I2
From where (it is noted that W does not depend on F for this loading):
2
- F gh2 2
H
B
W
=

1
+
E (+ H has) 2
48
I
I
1
2
Numerical application:
I = 2I
;
H = 2 A; F = 20.000 NR
1
2
1
(downwards)
2g
- F H2 2

W
=

1
(H + 2b)
E (+ H has) 2 I
48
1




315.100.365 0
. NR m
5
8.3 Charge
specific
F2 in C1

m M
H
F H



2
fh
gh

- fh
gh

Iso
- 2a + H
+
H
+


=


I.E.(internal excitation)
3 I.E.(internal excitation)
(
2 A
1
1
+ H)
(
) 2 (4a

+ H) + 2
(
4 A + H)
posts

2
- F
3
2 H
=

2
“2
2
(Ag + F (a+h))
E (+ H has) I
24
1

m M
2
- F bh2
Iso

2
2
“2
2
(Ag + F (a+h))


=
I.E.(internal excitation)
.
2
E
+
24
charp
(H has) I2
From where:
2
- F H2
H
B
W
=

2
2

2
(Ag +2f (a+h)) +
E (+ H has)
24
I
I
1
2
Numerical application:
I = 2I
;
H = 2 A; F = 10.000 NR
1
2
2
(towards the left)
2
-
3
2
+
=

2
(
F H H
2b
G + 9gh)
(
)
W
E (+ H has) I
48
1




- 315100365
.
NR m
4
Handbook of Validation
V3.90 booklet: Theoretical references of tests in linear statics
HI-75/96/037 - Ind A

Code_Aster ®
Version
4.0
Titrate:
Calculation of an elastic hyperstatic plane gantry
Date:
01/09/99
Author (S):
F. VOLDOIRE
Key:
V3.90.001-A Page:
22/24
8.4 Couples
specific
in C1

m M
H
H

fh
G H

fh
G " H
Iso

+


=


I.E.(internal excitation)
3 I.E.(internal excitation)
(
2 A
1
1
+ H) 2
(
4 A

+ H) + -
+

2 (4a+h)
posts

2
h3 " G
=

E (+ H has) 2 I
24
1

m M
B


fh
G " H
fh
G " H
Iso
- 2a + H
+
H


=


I.E.(internal excitation)
3 I.E.(internal excitation)
(
2 A
.
2
2
+ H)
(
)

2
(
4 A + H) +
-
+

2 (4a+h)
charp

- 2
bh
=


2
“2
2
(Ag + F (a+h))
E (+ H has) I
24
2
Numerical application:
I = 2I
;
H = 2 A; = - 100.000 Nm
1
2
2
H2
W
=


2
(G (H-b)-9 fhb)
E (+ H has) I
24
1

- 266.666.667
.
NR m
3
m M
8.5 Calculation
of
D =
1
I.E.(internal excitation)

m M
2h
G " H
H
2
G " h3
1

=



=

I.E.(internal excitation)
3 I.E.(internal excitation)
(
4 A
2
1
1
+ H) has + H
E
+
12
posts
(H has) I1

m M
2b

G "
1
1 H
1 A
G H
2
B (
H H
3 + has)
=



=
I.E.(internal excitation)
I.E.(internal excitation)
2

has
2
2
2

+ H + 6

+ H has (
4 A + H)
E
+
24
frame
(H has) I2
From where (it is noted that D does not depend on F):
2
G H
“2h2
(
B H
3 + has)
D
=


+

E (+ H has) 2 24 I
I
1
2

Numerical application:
I = 2I
;
H = 2 A
1
2
2
H2
D
=
G
(2h + 7b)
E (+ H has) 2 I
24
1




4.874 2564
.
m4
Handbook of Validation
V3.90 booklet: Theoretical references of tests in linear statics
HI-75/96/037 - Ind A

Code_Aster ®
Version
4.0
Titrate:
Calculation of an elastic hyperstatic plane gantry
Date:
01/09/99
Author (S):
F. VOLDOIRE
Key:
V3.90.001-A Page:
23/24
8.6
Summary of displacements CPU and vC
-
I
4
= 5 0
, 10
m4
1
E
= 210.000 MPa
CAS
X
X D
wV
pressure on C C
18672.994
91016960.3
­ 184930109.4
1
F1 out of C
41422.161
201902233.4
­ 315100365.0
F
8284.432
40380445.6
­ 63020073.0
2 in C1
out of C
4916.724
23965373.4
14775091.25
1
CAS
W
U
(m)
v
(m)
H
C
C
pressure on C C
83519999.94
0.0110476
­ 0.012422374
1
F1 out of C
0.00
0.00
­ 0.01497330
F
­ 226872262.8
­ 0.03000956
­ 0.00299466
2 in C1
out of C
206790328.5
0.0273532
­ 0.001215646
1
Note:
2
D
=
G D
4
, with: D
= 4.874 2564
.
m
E (+ H has) 2 I1
2
W
=

2
(G W + F W
V
H) to see higher
E (+ H has) I1
2
2
U
=
W
;
v
=
2
2
(W + Xd
C
H
C
V
)
E (+ H has) I
E (has
1
+ H) I1
2
10
-
1
-
- 4
=.

NR m
E (+ H has)
1 32275132 10
2
1
I
Handbook of Validation
V3.90 booklet: Theoretical references of tests in linear statics
HI-75/96/037 - Ind A

Code_Aster ®
Version
4.0
Titrate:
Calculation of an elastic hyperstatic plane gantry
Date:
01/09/99
Author (S):
F. VOLDOIRE
Key:
V3.90.001-A Page:
24/24
Aster comparison - reference analytical (R.)
CAS
Moment out of C
Reaction
Reaction V
Displacement
With
Displacement vC
(N.m)
HA (NR)
(NR)
CPU (m)
(m)
p on
R:
18672.994
5175.37
24233.24
0.0110476
­ 0.012422374
C C
Aster:
18673.20
5175.36
24233.2
0.0110472
­ 0.0124233
1
F
41422.161
4881.487
10000.00
0.00000
­ 0.01497330
1 out of C
R:
Aster:
41422.40
4881.47
10000.0
0.0000
­ 0.0
F
R:
8284.432
5976.297
4000.00
­ 0.03000956
­ 0.00299466
2 in C1
Aster:
8284.34
5976.31
4000.0
­ 0.0300098
­ 0.00299450
out of C
R:
4916.724
4576.394
5000.00
0.0273532
­ 0.001215646
1
Aster:
4916.62
4576.38
5000.0
0.0273536
­ 0.00121583
Note:
Aster calculation was carried out by taking very slim elements, so that: S " 2 << I. Thus,
the energy of inflection is prevalent. The values of Aster calculation result from case-test VPCS called SSLL14,
with the following data:
-
-
I
4
=
.
m4
;
I
4
=
.
m4
5 0 10
2 5 10
;
E
= 210.000 MPa
1
2
,

H = 2 A = 8 m; “= 20 m; B =
11
. 6,
2
p = 3.000 NR/m (downwards),
F =
NR
1
20 000
(downwards),
F =
NR
2
10 000
(towards the left),
= - 100.000 Nm.
Handbook of Validation
V3.90 booklet: Theoretical references of tests in linear statics
HI-75/96/037 - Ind A