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August 5, 2015

August 5, 2015

Posted by **ElementarySchoolStudent** on Saturday, July 27, 2013 at 2:32pm.

Is it correct ?

Q2_2_4

vA=-5.82 cm ??

- Elements of Structures MIT 2.02 -
**bobpursley**, Saturday, July 27, 2013 at 2:50pmHuh?

- Elements of Structures MIT 2.02 -
**11YearsOldMITStudent**, Saturday, July 27, 2013 at 3:00pmYes 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 -
**MathMate**, Saturday, July 27, 2013 at 4:33pm"

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 -
**MathMate**, Saturday, July 27, 2013 at 4:38pmI 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 -
**MathMate**, Saturday, July 27, 2013 at 5:26pmDo 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 -
**ElementarySchoolStudent**, Saturday, July 27, 2013 at 5:41pmno x=0 at the free end

- Elements of Structures MIT 2.02 -
**MathMate**, Saturday, July 27, 2013 at 5:47pmis q(x)=q0*x

or is

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

- Elements of Structures MIT 2.02 -
**ElementarySchoolStudent**, Saturday, July 27, 2013 at 5:48pmNo (EI)eff=350ð for the composite beam, remember the radius is in cm, E_0 in GPa.

- Elements of Structures MIT 2.02 -
**ElementarySchoolStudent**, Saturday, July 27, 2013 at 5:49pm(EI)eff=350*pi

- Elements of Structures MIT 2.02 -
**MathMate**, Saturday, July 27, 2013 at 5:58pmI 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 -
**MathMate**, Saturday, July 27, 2013 at 6:00pmDid you use

Ix=Iy=πd^4/64

?

- Elements of Structures MIT 2.02 -
**ElementarySchoolStudent**, Saturday, July 27, 2013 at 6:07pmOk (EI)eff= 1080*pi is correct

- Elements of Structures MIT 2.02 -
**ElementarySchoolStudent**, Saturday, July 27, 2013 at 6:22pmI=pi*R^4/2

- Elements of Structures MIT 2.02 -
**ElementarySchoolStudent**, Saturday, July 27, 2013 at 6:33pmI got a new delta=-43.38 cm but I'm not sure, I see it to high.

- Elements of Structures MIT 2.02 -
**ElementarySchoolStudent**, Saturday, July 27, 2013 at 6:42pmmy 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 -
**MathMate**, Saturday, July 27, 2013 at 7:05pm1. 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 -
**ElementarySchoolStudent**, Saturday, July 27, 2013 at 7:09pmmy 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 -
**ElementarySchoolStudent**, Saturday, July 27, 2013 at 7:12pmOk, thanks a lot MathMate. I'm sure the answer is -5,82cm

- Elements of Structures MIT 2.02 -
**MathMate**, Saturday, July 27, 2013 at 7:17pmGood luck!

- Elements of Structures MIT 2.02 -
**ElementarySchoolStudent**, Saturday, July 27, 2013 at 8:45pmsigma 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 -
**11YearsOldMITStudent**, Saturday, July 27, 2013 at 9:16pmIn 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 -
**fuubo**, Saturday, July 27, 2013 at 10:06pm@11YearsOldMITStudent

They're not right.

- Elements of Structures MIT 2.02 -
**MathMate**, Saturday, July 27, 2013 at 10:22pmI 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 -
**anonymous**, Sunday, July 28, 2013 at 1:41amCan someone update the answers for the other questions?

- Elements of Structures MIT 2.02 -
**MathMate**, Sunday, July 28, 2013 at 6:50amPlease clarify what are the "other" questions.

- Elements of Structures MIT 2.02 -
**Ash**, Monday, July 29, 2013 at 1:08am2_1_1

2_1_2

2_1_3

2_1_4

Please?