En este post haremos un completo ejercicio de vigas, reacciones, esfuerzos y perfiles HEB. El ejercicio que os adjunto un PDF os permitirá entenderlo mejor

1. http://www.elrincondelingeniero.com/
In the structure ABCDEFG shown in the
next figure, C is a pin, E and G are roller
supports, and A is a simply support.
b)
Determine:
0
x
a)
Reaction forces (HA, VA, VE, VG) using
the equations of equilibrium.
2
V
b)
Shear forces and Bending moment
laws in the beam. Analytical
expression as a function of x
(horizontal coordinate from A).
c)
F
A
B
1m
3 kN.m
C
1m
1m
4
E
2m
1m
2
4→
2x
x
4 x
1 →
.
4
x
.
6
q x
q
1
x
l
2
9
V
V x
G
.
3
V
F
6 kN/m
D
0→
d) Knowing
that
the
maximum
allowable normal stress of the cross
section of the beam is 100 MPa and
considering a safety coefficient equal
to 4, determine the HEB profile that
optimizes the structure.
4 kN
0→
x
M
3
1
M
V
Draw the Shear force and Bending
moment diagrams of the beam
(showing maximum values).
0→
Force laws
2x
1
0→H
x
2
6 1
ƺ
dƺ
2
6 1
6ƺ 1
ƺ
4
1m
V x
6 x
4
x
1
4
4
Solution:
a)
Reaction forces
M
0 → 2V
M
0
2.7
4.6
F
4
0→
M
2x
M x
3
3V
1
. 6.2. 1
2
0→
4
3
0
M x
1
9 x
6ƺ 1
6 x
M
ƺ
dƺ
4
6
x
4
1
2
4
4
12
ƺ
2
ƺ
12

2. http://www.elrincondelingeniero.com/
M
→
x
4
0 → ∄ ∈ 4,6
7 kN. m
7
M
V x
M x
M
6
dM x
dx
z kN. m
Considering that z is the horizontal
216
coordinate starting from G (x +z = 7):
c)
.
Force laws diagrams
d)
σ
Selection of the HEB‐profile
M
.y
I
σ
100 MPa
4
W
M
σ
W
7 kN. m
25 MPa
M
I
25 MPa
280 cm
And now entering the chart:
M
W
280 cm
311cm →