Question

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 ??

Answer 1

Huh?

Answer 2

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 ?

Answer 3

"
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?

Answer 4

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?

Answer 5

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.

Answer 6

no x=0 at the free end

Answer 7

is q(x)=q0*x
or is
q(x)=q0*x*L (as you had it above?)

Answer 8

No (EI)eff=350ð for the composite beam, remember the radius is in cm, E_0 in GPa.

Answer 9

(EI)eff=350*pi

Answer 10

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

Answer 11

Did you use
Ix=Iy=πd^4/64
?

Answer 12

Ok (EI)eff= 1080*pi is correct

Answer 13

I=pi*R^4/2

Answer 14

I got a new delta=-43.38 cm but I'm not sure, I see it to high.

Answer 15

my delta equation is
delta=-q_o(x-5xL^5+4L^5)/(120LEI)

en x=0 at the free end

delta=(-q_0*L^4)/(30EI)

where EI=1080*pi

thus

delta= (-q_0*L^4)/(32400*pi)

so

delta=-0,4338 m =-43,38 cm

Answer 16

1. I suggest you check your EI.

2. You have not confirmed
q(x)=q0*x*L (as you have written).
I think you mean q(x)=q0*(x/L)
If that's the case, I also get δ=-0.0582 as you did.

I think the large δ comes from the erroneous EI.
If you use EI=8050π, you'd get δ=-0.0582 as I have, and as you had before.

Answer 17

my delta equation is
delta=-q_o(x-5xL^5+4L^5)/(120LEI)

en x=0 at the free end

delta=(-q_0*L^4)/(30EI)

where EI=8050*pi

thus

delta= (-q_0*L^4)/(241500*pi)

so

delta=-0,0582 m =5,82 cm

Answer 18

Ok, thanks a lot MathMate. I'm sure the answer is -5,82cm

Answer 19

Good luck!

Answer 20

sigma max en core and sigma max I sleeve
I got 47 MPa in core and 35 MPa in sleeve are this correct ?

Answer 21

In this problem

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



Now i got

Q2_2_5

max STRESS in CORE=9 MPa

and max stress in sleeve= 73 MPa

Are this values correct ? Ples help me this are the last values to finish and I have only more chance and I will pass the course.

Answer 22

@11YearsOldMITStudent

They're not right.

Answer 23

I have quite different values as you have. It would help if you show your work so we can compare notes.

My approach would be:

Since the beam is composite, there is only one value of 1/r at each cross section x, which is given by
M(x)/EI.

For a cantilever beam, M(x) is evidently at the fixed end, equal to
(q0*L/2)*(L/3)=q0*L^2/6=1840 N-m

EI had been calculated before and is equal to 8050π
Thus 1/r=M(L)/EI=0.22857/π=0.07276 (approx.)

Recall that
σ=Ey/r
where r is the radius of curvature and 1/r approximately equals M/EI for large r.

So for the core,
σ0=E0*y0*(1/r)
where E0=70 Gpa
y0=0.01=distance from neutral axis
=70*10^9*0.01*(1/r)
=50.9 MPa

For the sheath,
σ1=E1*y1*(1/r)
where E1=210 GPa
y1=0.02 = distance from neutral axis
=210*10^9*0.02*(1/r)
= 305.6 MPa (approx. 44 ksi)


Since this is going to be your last life, I would like you to compare my work with yours and be completely convinced of any number before you make your last attempt.

Answer 24

Can someone update the answers for the other questions?

Answer 25

Please clarify what are the "other" questions.

Answer 26

2_1_1
2_1_2
2_1_3
2_1_4
Please?