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

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_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.

@ElementarySchoolStudent

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

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 ;)

Thanks Mors

Your answers are all right and help me a lot.

To solve these problems, we need to use principles of mechanics and solid mechanics. We will derive symbolic expressions for the given quantities in terms of the given variables. Here is how we can approach each problem:

Q2_1_1: The x-component of the reaction torque at C (TCx)
To find TCx, we need to consider the torque equilibrium of the shaft. As the shaft is fixed at both ends, the torque at C will be the equal and opposite of the torque at A. We can express TCx as a function of x using the distributed torque t0. Using the equation for torque equilibrium, we can write:

TCx = -∫(0 to L) t0*(L - x) dx

Q2_1_2: The twist rate dφdx(x) and the position x0 where dφdx(x0) = 0
To find dφdx(x), we need to consider the torsional deformation of the shaft. We can find the twist rate separately for the AB and BC segments of the shaft.

For 0 ≤ x ≤ L, the shaft segment AB:
The twist rate can be expressed as:
dφdx(x) = -G0/((πR^2)/2)*(L - x)

For L < x ≤ 3L, the shaft segment BC:
The twist rate can be expressed as:
dφdx(x) = -2G0/((πR^2)/2)*(3L - x)

To find x0 where dφdx(x0) = 0, we can set each equation equal to zero and solve for x.

Q2_1_3: The maximum absolute value of the shear stress in the shaft (τmax) and its location (rτmax, xτmax)
To find τmax, we need to find the maximum shear stress in the shaft. We can express τmax as a function of x using the given shear modulus G0 and the radius of the shaft R. We can consider the left AB segment and the right BC segment separately.

For 0 ≤ x ≤ L, the shaft segment AB:
τmax = (G0/R)*(L - x)

For L < x ≤ 3L, the shaft segment BC:
τmax = (2G0/R)*(3L - x)

We can find the corresponding positions (rτmax, xτmax) by substituting the expressions for τmax into x.

Q2_1_4: 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)
To find φmax, we need to consider the torsional deformation of the shaft. We can find the maximum rotation separately for the AB and BC segments of the shaft.

For 0 ≤ x ≤ L, the shaft segment AB:
φmax = G0/(2*(πR^2)/2) * (L^2/2)

For L < x ≤ 3L, the shaft segment BC:
φmax = 2G0/(2*(πR^2)/2) * ((3L^2/2) - (2L*x) + (x^2/2))

We can find the corresponding position xφmax by substituting the expressions for φmax into x.

Q2_2_1: The internal bending moment resultant (M(x))
To find M(x), we need to consider the bending deformation of the composite beam. We can express M(x) as a function of x using the given variables L, q0, and the moment M(x) at any point x can be obtained using the equation for bending moment.
M(x) = ∫(0 to x) q(x)*(L - x) dx

Q2_2_2: The effective section stiffness of the beam (EI)eff
To find (EI)eff, we need to consider the effective section properties of the composite beam. We can express (EI)eff as a function of R0 and E0 using the equations for the moment of inertia of a circular section.
(EI)eff = [(π/4)*(R0^4) - (π/4)*[(R0-R)^4]] * E0

Q2_2_3: The curvature at the neutral axis (1ρ(x)) and the slope (ϑ(x)) of the beam
To find 1ρ(x) and ϑ(x), we need to consider the curvature and slope of the beam. We can express 1ρ(x) and ϑ(x) as functions of L, q0, R0, E0, and x using the equations for curvature and slope.
1ρ(x) = [(q0/(2*E0)) * (L^2 - x^2)]
ϑ(x) = [(q0/(2*E0)) * (L*x - (x^2/2))]

Q2_2_4: The displacement at the free end (vA)
To find vA, we need to consider the displacements of the beam. Since the beam is fixed at B and free at A, the displacement at A will be zero.
vA = 0

Q2_2_5: The maximum tensile stresses in the core (σmax,C) and in the sleeve (σmax,S)
To find σmax,C and σmax,S, we need to consider the stresses in the core and sleeve of the composite beam. We can express σmax,C and σmax,S as functions of R0, E0, and the curvature at the neutral axis.
σmax,C = [(R0/2)] * 1ρ(x)
σmax,S = [(3/2)*R0] * 1ρ(x)

By using these approaches, the symbolic expressions and numerical values for the given quantities can be obtained.