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
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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
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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
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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
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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
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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
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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
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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
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unanswered
You have used 0 of 4 submissions
2-1-4, 2-2-1, 2-2-3, 2-2-4, 2-2-5
To solve problem Q2_1, we need to find the x-component of the reaction torque at C (TCx).
First, let's analyze the forces and torques acting on the shaft.
Since the left segment AB is composed of a linear isotropic elastic material, it does not experience any torque. So, we only need to consider the right segment BC.
The torque per unit length on segment BC is given as t0. The torque per unit length can be written as the product of the shear modulus and the twist rate:
t0 = G(x) * dφdx(x)
For 0 ≤ x ≤ L, G(x) = G0 and for L < x ≤ 3L, G(x) = 2G0.
Let's solve for the twist rate in each segment:
For 0 ≤ x ≤ L:
t0 = G0 * dφdx(x)
dφdx(x) = t0 / G0
For L < x ≤ 3L:
t0 = 2G0 * dφdx(x)
dφdx(x) = t0 / (2G0)
Now, to find TCx, we integrate the twist rate expression from 0 to 3L:
TCx = ∫(dφdx(x))dx from 0 to 3L
TCx = ∫[t0 / G0 * ((0 ≤ x ≤ L) + 2G0)^-1 * (L < x ≤ 3L)]dx from 0 to 3L
TCx = ∫[t0 / G0 * [1/(0 ≤ x ≤ L) + 1/(2G0)(L < x ≤ 3L)]]dx from 0 to 3L
Solving the integral will give us the expression for TCx in terms of R, G0, L, t0, and x.
Now, for the twist rate dφdx(x) and the position x0 where the twist rate goes to zero (dφdx(x0) = 0), we need to solve the twist rate expression in each segment separately.
For 0 ≤ x ≤ L:
dφdx(x) = t0 / G0
For L < x ≤ 3L:
dφdx(x) = t0 / (2G0)
We leave the rational solutions as fractions.
For the maximum absolute value of the shear stress in the shaft (τmax) and its location (rτmax, xτmax), we need to consider both segments of the shaft.
In segment AB (0 ≤ x ≤ L):
τmax(AB) = G0 * (dφdx(x) / R)
In segment BC (L < x ≤ 3L):
τmax(BC) = 2G0 * (dφdx(x) / R)
To find the maximum shear stress, we need to determine which segment has a higher value and take the maximum.
For the position xτmax, we need to find the x-value at which the maximum shear stress occurs.
Finally, for 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), we need to find the maximum rotation in each segment and compare them.
In segment AB (0 ≤ x ≤ L):
φmax(AB) = ∫[(t0 / G0)dx] from 0 to L
In segment BC (L < x ≤ 3L):
φmax(BC) = ∫[(t0 / (2G0))dx] from L to 3L
We leave the rational solutions as fractions.
By following these steps and performing the integrals, we can obtain the symbolic expressions for TCx, dφdx(x), x0, τmax, rτmax, xτmax, φmax, and xφmax in terms of R, G0, L, t0, and x.