Apply the boundary condition of slope to find c1;
at x=4, dy/dx=0 because at the fixed support the
slope is zero.
=> c1= 16 + 18 - 0.666= 33.33
Substituting the value of c1 into Eq. 2 we get the
** ** EI (dy/dx) = - 2[x^{2}/2]
- 4[x-1]^{3}/6
+ 4[x-3]^{3}/6 +33.33
Finally the equation of slope is written as
**EI (dy/dx) = - 2[x**^{2}/2]
- 4[x-1]^{3}/6
+ 4[x-3]^{3}/6 +33.33
Eq. 3
and integrating again Eq. 3, we get;
EI y = -2[x^{3}/6] - 4 [x-1]^{4}/24
+33.33x +4[x-3]^{4}/24 + c2
Eq.4
To get the value of c2 we apply the boundary
condition for deflection; because the support at D is fixed and
rigid (unyielding) Therefore at x=4, y=0.
Substituting these values
of x and y in Eq. 4 we get the value of c2= - 98.5 and the __equation for deflection__ can be written as
**EI y = -2[x**^{3}/6] - 4 [x-1]^{4}/24
+33.33x +4[x-3]^{4}/24 - 98.5
.Eq. 5
Slope at 2m from
free end; we put x=2 in Eq.3 (neglect the terms which become
negative)
** dy/dx = (28.66)/EI**
Deflection at 2m
from free end; put x=2 in Eq. 5 (neglect the terms which
become negative)
** y = (-
34.66)/EI**
Use
moment
of inertia calculator for the value of Moment of inertia for
T-beam section
I_{xx}
= 363.76 cm^{4}
EI= (200x10^{9}
N/m^{2})x(363.76
cm^{4} ) = 727.52 kN m^{2}
Substituting this
value of EI in the expressions of slope and deflection we get;
**dy/dx = 28.66/727.52 = 0.0394 rad**
**
y = - 34.66/727.52 = - 0.047 m **(negative sign
indiactes that the deflection is downward)
You can also use our
Slope
deflection calculator for different combinations of load on
cantilever.
For other problems please visit our
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