q(x)=q0xL ?
Is it q(x)=q_0*x*L (linear) or q(x)=q_0*L (linear, but constant)?
What is the relation between x and L?
Write this precisely, please
Q2_1: QUIZ 2, PROBLEM #1
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
8 answers
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_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.
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_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.
@ElementarySchoolStudent
Q2_1_3 taumax and Q2_1_4 phi are not right.
Q2_1_3 taumax and Q2_1_4 phi are not right.
I didn't see the drawing (so I don't know the direction of t0), but:
Q2_1_3:
tau max=|tau min|=(3*t_0*L)/(pi*R^3)
r tau max = R (it's obvious)
x tau max =3*L
Q2_1_4:
phi max=(9*t_0*L^2)/(8*pi*G_0*R^4)
x phi max=3*L/2
dphi/dx is a derivative of phi and it's zero is at 3*L/2 (refer to Q2_1_2), so the maxima of function phi is at the same point x=3*L/2
Q2_1_3:
tau max=|tau min|=(3*t_0*L)/(pi*R^3)
r tau max = R (it's obvious)
x tau max =3*L
Q2_1_4:
phi max=(9*t_0*L^2)/(8*pi*G_0*R^4)
x phi max=3*L/2
dphi/dx is a derivative of phi and it's zero is at 3*L/2 (refer to Q2_1_2), so the maxima of function phi is at the same point x=3*L/2
the answers Q2_1_3 and Q2_1_4 are correct!!, tnks!
What about the Q2_2? Provide me a drawing or a better (and EXACT) description of this problem
Q2_2: I've found it by myself
Here, the solutions
Q2_2_1 -(q_0)/(6*L)*x^3
Q2_2_2 46*E_0*(pi*R_0^4)/4
Q2_2_3 a) -(q_0*x^3)/(69*L*E_0*pi*R_0^4)
b) -(q_0)/(276*L*E_0*pi*R_0^4)*(x^4-L^4)
Q2_2_4 -5.8
Q2_2_5 a)50.9 b)305.6
Now, go and f# yourselves, lazy sob's ;)
Here, the solutions
Q2_2_1 -(q_0)/(6*L)*x^3
Q2_2_2 46*E_0*(pi*R_0^4)/4
Q2_2_3 a) -(q_0*x^3)/(69*L*E_0*pi*R_0^4)
b) -(q_0)/(276*L*E_0*pi*R_0^4)*(x^4-L^4)
Q2_2_4 -5.8
Q2_2_5 a)50.9 b)305.6
Now, go and f# yourselves, lazy sob's ;)
Thanks Mors
Your answers are all right and help me a lot.
Your answers are all right and help me a lot.
7 answers