This is my last chance and I need to see if my calculus is correct.
Is it correct ?
Q2_2_4
vA=-5.82 cm ??
Huh?
Yes I have lost 3 chances and I almos sure this is the correct answer but I hve a doubt with the minus sign.
The problem is this.
The composite beam AB, of length L=2m, is free at A (x=0) and fixed at B (x=L) and is composed of a round cylindrical core of constant radius R0=1cm bonded inside a sleeve of thickness R0 (outer radius 2R0=2cm). The beam is loaded, as indicated, by a downward linearly varying distributed load per unit length of magnitude
q(x)=qxL,with
q0=2.76kN/m.
The material moduli are:
For the core, EC=70GPa=E0
For the sleeve, ES=210GPa=3E0
This is I want to know.
Q2_2_4 : 70.0 POINTS
Obtain the numerical value (in cm) for the displacement at the free end, vA=v(x=0):
vA= ....cm
Is it correct ?
Q2_2_4
vA=-5.82 cm ?
"
q(x)=qxL,with
q0=2.76kN/m.
"
Most of the time, x is measured from the fixed end (of a cantilever). Is this the case?
I guess I did not read that x=0 at the free end (A), and the fixed end (B) is x=L.
Also, do you mean
q(x)=q0*x*L?
What did you get for EI of the composite beam?
Do you get 8050π for the EI of the composite beam? I get 8050π
For some reason, I get δ=-0.1164, which is exactly double your number.
no x=0 at the free end
is q(x)=q0*x
or is
q(x)=q0*x*L (as you had it above?)
No (EI)eff=350ð for the composite beam, remember the radius is in cm, E_0 in GPa.
(EI)eff=350*pi
I have for the core
I0=2.5π*10^-9
and for the sheath
I1=3.75π*10^-8
Multiplied by the corresponding E gives me
EI0=175π (core) and
EI1=7875π (sheath).
Total(effective)=8050π