# 1D problem solving with FEM by jovaneli

VIEWS: 53 PAGES: 8

• pg 1
```									                      Ax
A
1

t            4a

C                    2

3P
3a

B

Bx
.

1.
–                                                        2) –
L1=4.a=5.6m                                   L1=3.a=4.2m

F1=14.5*10-4m                                    F2=23.2*10-4m
-               1            2                                   -          2   3

–       – N;         -m                     – Pa

-                                         -2 /              –                         3

1
.

-

-

*

t

t=const)
f1x=f2x=R/2   R=t.L       f1x= f2x=t.L/2,

-
f1x=t.L1/2.
2                                -
f =t.L1/2                           .
-
1        + f1x

(1)
x         2 f2x -
(2)
3    x

2
*

u1=0 ; u3=0

–

*

+ f1x

(          shift+k
[ 1,1

3
*

-

-

(Ni)           (Nj)
Nx1)
fx1=R/2)

(Nx1)               (fx1)          .

-              .

-

4
.

Ax
A             -                         -160                        -103
2
F1=7.4 cm
F1=11 cm2 -160
t
-140
2                   F1=14.5 cm2
C                      F1=14.5 cm
-                                                          -160
3P

B
160                               160
Bx

Ax
Ax
A                         1
L1=a=1.4m
2
i            j
t        4a      3 L2=a=1.4m
(1)        1            2
L3=a=1.4m
4
(2)        2            3
C             2           5 L4=a=1.4m
(3)        3            4
3P                                          (4)        4            5
3a           L5=3*a=4.2m
(5)        5            6
B                        6

Bx               Bx

5
*

-

6
r*Ur=Fr

-1
r

u5= -3.353mm

*

:

<0>

7
*

Ax
A            -
-90
t            -110
-130
-149
C        -
3P

B
160
Bx

8

```
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