ElementarySchoolStudent

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Questions

The following questions were asked by visitors named ElementarySchoolStudent.

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 ??

Answers

The following answers were posted by visitors named ElementarySchoolStudent.

I got in Q2_1_1 TCX= -2*t_0*L

access is your answers are ok, then Q2_1_3 a) shear stress mx=(4*t_0*L)/(pi*R^3) Is it ok?

tau max=(4*t_0*L)/(pi*R^3)

Q2_1_1 TXC=-3/2*t_0*L Q2_1_2: d*phi/dx = (t_0*L)/(pi*G_0*R^4) d*phi/dx =t_0*(3*L-2*x))/(2*pi*G_0*R^4) x0 = 3/2*L Q2_1_3: tau max=(4*t_0*L)/(pi*R^3) r tau max = R x tau max =3*L

fuubo check this Q2_1_1 TXC=-3/2*t_0*L Q2_1_2: d*phi/dx = (t_0*L)/(pi*G_0*R^4) d*phi/dx =t_0*(3*L-2*x))/(2*pi*G_0*R^4) x0 = 3/2*L Q2_1_3: tau max=(4*t_0*L)/(pi*R^3) r tau max = R x tau max =3*L

Q2_1_4 2) x phi max=3*L/2

Q2_1_1 TXC=-3/2*t_0*L Q2_1_2: d*phi/dx = (t_0*L)/(pi*G_0*R^4) d*phi/dx =t_0*(3*L-2*x))/(2*pi*G_0*R^4) x0 = 3/2*L Q2_1_3: tau max=(4*t_0*L)/(pi*R^3) r tau max = R x tau max =3*L

Q2_1_4 B) X PHI MAX=3*L/2

Q2_1_4 1) phi=(9*t_0*L^2)/(2*pi*G_0*R^4)

Q2_1_1 TXC=-3/2*t_0*L Q2_1_2: d*phi/dx = (t_0*L)/(pi*G_0*R^4) d*phi/dx =t_0*(3*L-2*x))/(2*pi*G_0*R^4) x0 = 3/2*L Q2_1_3: tau max=(4*t_0*L)/(pi*R^3) r tau max = R x tau max =3*L Q2_1_4 1) phi=(9*t_0*L^2)/(2*pi*G_0*R^4) 2) x phi max=3*L/2

I have only one more chance, please help me and tell me if this answers I put here are ok.

elena you are wrong and I lost my 3 chance with your answer.

Q2_1_1 TXC=-3/2*t_0*L Q2_1_2: d*phi/dx = (t_0*L)/(pi*G_0*R^4) d*phi/dx =t_0*(3*L-2*x))/(2*pi*G_0*R^4) x0 = 3/2*L Q2_1_3: tau max=(4*t_0*L)/(pi*R^3) r tau max = R x tau max =3*L •MITx: 2.01x - ElementarySchoolStudent, Saturday, July 27, 2013 at 2:12am Q2_1...

Q2_1_1 TXC=-3/2*t_0*L Q2_1_2: d*phi/dx = (t_0*L)/(pi*G_0*R^4) d*phi/dx =t_0*(3*L-2*x))/(2*pi*G_0*R^4) x0 = 3/2*L Q2_1_3: tau max=(4*t_0*L)/(pi*R^3) r tau max = R x tau max =3*L •MITx: 2.01x - ElementarySchoolStudent, Saturday, July 27, 2013 at 2:12am Q2_1...

fuubo You are right I think I add wrong and instead of 4 is 3. If you have more chances try this tau max=(3*t_0*L)/(pi*R^3)

There is a error in tau max the correct answer is tau max =(3*t_0*L)/(pi*R^3)

Q2_1_4 A little error when I got out a factor of 2, I put it in the numerator and must be in the denominator so the correct answer is. 1) phi=(9*t_0*L^2)/(8*pi*G_0*R^4) Q2_1_3: This is the correct answer 3 instead of 4. I added wrong. 1) tau max=(3*t_0*L)...

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 down...

Is it correct ? Q2_2_4 vA=-5.82 cm ??

no x=0 at the free end

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

Ok (EI)eff= 1080*pi is correct

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

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

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

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

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