STRUCTURAL ENGINEERING ROOM
Department of Architecture
91
W2
1 P L L 1 1 P L L 2 2 2 8 E I 4 3 2 4 E Ii 4 3
P L2
2
192 E Ii
48 E Ii 2
The calculation of loads using trapezoidal rules
2
10 P L
P L
192 E Ii
Subsequent relationship is a simplified linearized differential equation crease lines, which
In general, lists any moment of inertia Ic a comparable moment of inertia and further express
solves a double integration. The first integration determines the rotation of the crease lines
the ideal shape for loads
and second crease lines. 2
d y 2
W0
Ic
P L
W1
192 E Ic Ia Ic
We put Ic Ii ,
2
Ia
W3
Ic
,
Ic 2 Ic Ii 96 E Ia P l
1,
Ii
2
2 P L
W1
192 E Ic
2
10 P L
W2
Ic
dz
192 E Ic Ii
8 P L
W3
W2
192 E Ic
2
M ( z) EI ( z )
If we extend the right side of the equation, then we get
Thus, the load get ideal
2
Wo
2
2
10 P L
2
d y
192 E Ic
dz
2
Rotation of the cross section in the support beam are equal. The term will write directly to rotate depending on the force in the form of
M ( z) E I( z)
Ic 2 M ( z) I E I c ( z)
Then to calculate the ideal force we get loads we will continue to work with
a
W0 W1
W2
2
2
1.25
192 E Ic
4
L
2
W1
2
L
P L
4
E Ic
8 L 2 L 15 L 192 2 192 2 192 4
2
9 P L
384 E Ic
Mm lava
Ic Im lava
Mm prava
Ic Im prava Ic
3
1.125
P L
48 E Ic
inertia Ic Ii 2 Ia in span 2 L , we see that the calculation result of the deflection is greater 4
I
triangles load W:
Comparing the equations- moment of inertia constant, and equation – where moment of 1
Ic
, but with diagrams M , called reduced moment diagram.
Decomposition trapezoid area, which is reduced moment diagrams Mz
Deflection at the center of the beam
W0
and
can write
that the result of the equation of rotation of the support is more about 25%.
2
1 EIc
moments or moments of inertia, respectively. both. Generally, for example, to the point we
1
L
. Now leave out constant
On the end between the sectors may occur due to sudden changes in the discontinuous
constant, and the equation, where the moment of inertia Ic Ii 2 Ia in length 2 L , indicates
a
Ic I( z)
16 E Ic
Comparison between the equation in the case of the beam, where the moment of inertia is
ym
M ( z)
P L
2
15 P L
b
M E I
1 E Ic
Wm
1 E Ic
Ic Ic 1 1 1 2 Mm 1 prava S Mm lava S 3 2 3 2 I I m 1 prava m lava
Ic Ic 1 2 1 1 Mm prava S Mm 1 lava S 2 3 2 3 Im prava Im 1 lava
about 12.5%.
Deformation Behaviour of Reinforced Concrete Beams
I
to get perfect