# Elements of Structures MIT 2.02

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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 ??

• Elements of Structures MIT 2.02 -

Huh?

• Elements of Structures MIT 2.02 -

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 ?

• Elements of Structures MIT 2.02 -

"
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?

• Elements of Structures MIT 2.02 -

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?

• Elements of Structures MIT 2.02 -

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.

• Elements of Structures MIT 2.02 -

no x=0 at the free end

• Elements of Structures MIT 2.02 -

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

• Elements of Structures MIT 2.02 -

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

• Elements of Structures MIT 2.02 -

(EI)eff=350*pi

• Elements of Structures MIT 2.02 -

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

• Elements of Structures MIT 2.02 -

Did you use
Ix=Iy=πd^4/64
?

• Elements of Structures MIT 2.02 -

Ok (EI)eff= 1080*pi is correct

• Elements of Structures MIT 2.02 -

I=pi*R^4/2

• Elements of Structures MIT 2.02 -

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

• Elements of Structures MIT 2.02 -

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

• Elements of Structures MIT 2.02 -

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.

• Elements of Structures MIT 2.02 -

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

• Elements of Structures MIT 2.02 -

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

• Elements of Structures MIT 2.02 -

Good luck!

• Elements of Structures MIT 2.02 -

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

• Elements of Structures MIT 2.02 -

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.

• Elements of Structures MIT 2.02 -

@11YearsOldMITStudent

They're not right.

• Elements of Structures MIT 2.02 -

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.

• Elements of Structures MIT 2.02 -

Can someone update the answers for the other questions?

• Elements of Structures MIT 2.02 -

Please clarify what are the "other" questions.

• Elements of Structures MIT 2.02 -

2_1_1
2_1_2
2_1_3
2_1_4

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