ejercicio estructuras articuladas
TRANSCRIPT
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The truss shown in the figure is subjected to one vertical force of 4 kN at node 6 and
one horizontal load of 4 kN at node 4.
Data:
Diameter of each bar: ,
Young’s Modulus: .
Length of every length of the structure except in f) is equal to L m.
Determine:
a) DSI of the truss and its implications.
b) Reaction forces.
c) Using the method of the joints, the axial forces in all members of the structure
indicating if they are tensile or compressive.
d) Using the method of the sections, the axial forces in bars 2, 5 and 9, indicating
if they are tensile or compressive.
e) Internal forces diagram.
f) If the length of every member in the truss is equal to one meter, determine the
maximum elongation that a member of the structure will experiment due to the
set of loads applied in the structure.
:
2
4
5
31
37
5
6
8 94
1
2
6
4 kN
4 kN
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1
1. DSI
4 3 1
∑2. 2. 1 2. 2. 2 13. 2. 3 1 2. 5 1 24
3. 1 3. 9 1 24
0
1
Alternatively
2 9 4 2.6 1
Where:
m: number of bars
r: number of reaction forces
j: number of joints
Thus, the structure is entirely linked and
its reactions forces cannot be determined
directly with the equations of equilibrium.
However, it is possible to calculate them
using the method of the nodes.
2. Reaction forces
0 → . 2 4. 0
0 → 4 2 →
0 → 4 0 1
3. Calculation of internal forces by the
mehod of the nodes
Due to the geometry of the truss, all the
angles are equal to 45° or 90°, therefore all
oblique forces are multiplied by √ .
Node 4
0 →
0 →
Node 6
0 →
0 →
Node 5
0 →√22
√22
4 0
0 → 0
√
√
2
4
5
31
37
5
6
8 94
1
2
6
4 kN
4 kN
4 kN
N4
5
N5
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2
Node 1
0 →√22. 2. √2 0 2
Node 2
0 → 0 3
0 →
Node 3
0 →√22. 2. √2 0 4
Now, with equations (1), (2), (3) and (4) we
obtain the next results:
4. Internal forces by the method of sections
0 → 0
0 → 2 .√22
0
√
0 → 2 .√22
4 0
Obtaining the same results by both
methods.
5. Internal forces diagram
6. Calculation of the maximum increment of
length the structure will experiment due
to the system of loads of the exercise.
. .∆→ ∆
∆4. 10 . 1
210. 10 . . 0,020,015
N1H1
N4
2 kN
1
N2
3
N5
2 kN
4 kN
H3
2
4
5
1
3
9
5
6
84
1
2
4 kN
N2
N5
N9