Asked by mt
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)=
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=
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)=
ϑ(x)=
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
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
σmax,S= MPa
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)=
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=
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)=
ϑ(x)=
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
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
σmax,S= MPa
Answers
Answered by
Mors
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
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
Answered by
elena
Q2_1_1
TXC=-t_0*L
Q2=1_2
a) 0<=x<L ===>> (2*t_0*L)/(pi*G_0*R4)
b)L<=x<=3L =>>(2*t_0*(2*L-))/pi*G_0*R^4)
c) x=2*L
TXC=-t_0*L
Q2=1_2
a) 0<=x<L ===>> (2*t_0*L)/(pi*G_0*R4)
b)L<=x<=3L =>>(2*t_0*(2*L-))/pi*G_0*R^4)
c) x=2*L
Answered by
elena
The above answers are correct because I got a green chek and not a red X.
Answered by
m
are you shure i put this Q2_1_1
TXC=-t_0*L and its wrong and this one c) x=2*L too
TXC=-t_0*L and its wrong and this one c) x=2*L too
Answered by
superman
yes, I got a green check.
Answered by
ElementarySchoolStudent
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
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
Answered by
ElementarySchoolStudent
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
Answered by
ElementarySchoolStudent
I have only one more chance, please help me and tell me if this answers I put here are ok.
Answered by
ElementarySchoolStudent
elena you are wrong and I lost my 3 chance with your answer.
Answered by
AK
3
τmax= correct
3⋅t0⋅Lπ⋅R3
rτmax= correct
R
xτmax= correct
3⋅L
4
φmax= correct
9⋅t0⋅L28⋅π⋅G0⋅R4
xφmax= correct
3⋅L2
τmax= correct
3⋅t0⋅Lπ⋅R3
rτmax= correct
R
xτmax= correct
3⋅L
4
φmax= correct
9⋅t0⋅L28⋅π⋅G0⋅R4
xφmax= correct
3⋅L2
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