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May 25, 2016
Posted by **MAN** on Thursday, July 25, 2013 at 12:39pm.

The shaft ABC is a solid circular cylinder of constant outer diameter 2R and length 3L. The shaft is fixed between walls at A and C and it is composed of two segments made of different materials. The left third of the shaft (AB) is composed of a linear isotropic elastic material of shear modulus G0, while the right two-thirds of the shaft (BC) is composed of a different linear elastic material of shear modulus 2G0. The right segment, BC, is subjected to a uniform distributed torque per unit length t0[N⋅m/m].

Obtain symbolic expressions in terms of R, G0, L, t0, and x for the quantities below. In your answers, leave rationals as fractions and enter G0, t0, and π as G_0, t_0 and pi, respectively.

Q2_1_1 : 100.0 POINTS

The x-component of the reaction torque at C:

TCx= unanswered

You have used 0 of 4 submissions

Q2_1_2 : 60.0 POINTS

The twist rate dφdx(x), and the position x0 along the shaft where the twist rate goes to zero (dφdx(x0)=0):

for0≤x<L,dφdx(x)= unanswered

forL<x≤3L,dφdx(x)= unanswered

dφdx(x0)=0atx0= unanswered

You have used 0 of 4 submissions

Q2_1_3 : 60.0 POINTS

The maximum absolute value of the shear stress in the shaft (τmax) and its location (rτmax, xτmax):

τmax= unanswered

rτmax= unanswered

xτmax= unanswered

You have used 0 of 4 submissions

Q2_1_4 : 100.0 POINTS

The maximum value of the rotation field φ(x) along the shaft (φmax), and the position along the shaft where the maximum rotation occurs (xφmax):

φmax= unanswered

xφmax= unanswered

Q2_2: QUIZ 2, PROBLEM #2

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)=q0xL,withq0=2.76kN/m.

The material moduli are:

For the core, EC=70GPa=E0

For the sleeve, ES=210GPa=3E0

Q2_2_1 : 60.0 POINTS

Obtain a symbolic expression for the internal bending moment resultant in terms of L, q0 (enter as q_0), and x:

M(x)= unanswered

You have used 0 of 4 submissions

Q2_2_2 : 60.0 POINTS

Obtain a symbolic expression for the effective section stiffness of the beam (EI)eff in terms of R0 and E0 (enter these as R_0 and E_0, leave rationals as fractions, and enter π as pi):

(EI)eff= unanswered

You have used 0 of 4 submissions

Q2_2_3 : 60.0 POINTS

Obtain symbolic expressions for the curvature at the neutral axis 1ρ(x) and the slope ϑ(x) of the beam in terms of L, q0, R0, E0, and x (again, leave rationals as fractions and enter π as pi):

1ρ(x)= unanswered

ϑ(x)= unanswered

You have used 0 of 4 submissions

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 unanswered

You have used 0 of 4 submissions

Q2_2_5 : 70.0 POINTS

Obtain the numerical values in MPa for the maximum tensile stresses in the core (σmax,C) and in the sleeve (σmax,S):

σmax,C= MPa unanswered

σmax,S= MPa unanswered

You have used 0 of 4 submissions

No one has answered this question yet.

- MIT 2.01x -
**superman**, Friday, July 26, 2013 at 9:56amQ2_1_1

TCx=-q0L/3 - Physics -
**superman**, Friday, July 26, 2013 at 9:59amQ2_1_1

TCx=-t0L/3 - MIT 2.01x -
**superman**, Friday, July 26, 2013 at 10:16amQ2_1_2

for0¡Üx<L,d¦Õdx(x)=(TBX+2Lt0)/(G0I)

forL<x¡Ü3L,d¦Õdx(x)=(t0(3L-x))/(2G0I)

d¦Õdx(x0)=0atx0 in x=3L - MIT 2.01x -
**superman**, Friday, July 26, 2013 at 10:21amQ2_1_2

1) (TCX+2Lt0)/(G0I) where TCx=-t0L/3

2) t0(3L-x))/(2G0I)

3) x=3L - Physics -
**MAN**, Friday, July 26, 2013 at 4:08pmCan you fix the coding?

- Physics -
**superman**, Friday, July 26, 2013 at 9:23pmQ2_1_2

a) T(x) in 0<=x<=L

T(X)=t_0*L

d*phi/dx=T(x)/(GI) where I=(pi*R^4)/2

so d*phi/dx=(2*t_0*L)/(pi*G_0*R^4)

b) T(x) in l<=x<=3L

T(x)=t_0*(2*L-x)

so

d*Phi/dx=(2*t_0+(2*L-x))/(pi*G_0*R^4)

c)(d*phi/dx)(x)=0 when

0=(2*t_0+(2*L-x))/(pi*G_0*R^4)

so (2*L-x)=0 >>>>>> x=2*L

answer x=2*L - ElementsofStructures -
**ElementarySchoolStudent**, Friday, July 26, 2013 at 11:14pmI got in Q2_1_1

TCX= -2*t_0*L - Physics -
**ElementarySchoolStudent**, Saturday, July 27, 2013 at 12:44amQ2_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 - Physics -
**ElementarySchoolStudent**, Saturday, July 27, 2013 at 12:45amQ2_1_4

B) X PHI MAX=3*L/2 - Physics -
**sick yoda**, Saturday, July 27, 2013 at 11:20amQ2_2_2

(EI)eff=(46*E_0*pi*R_0^4)/4 - Physics -
**MAN**, Saturday, July 27, 2013 at 11:53am2-1-4, 2-2-1, 2-2-3, 2-2-4, 2-2-5

- Physics -
**Insane**, Sunday, July 28, 2013 at 8:56pm2-2-1, 2-2-3, 2-2-4