Question

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,with
q0=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)= ........
Ï‘(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

Answer 1

Is it correct ?

Q2_2_4

vA=-5.82 cm ??

Answer 2

Q2_2_2
23/2*R_0^4*E_0*pi

Q_2_2$
-5,82 cm

Answer 3

Q2_2_5

STRESS CORE=9 MPa

stress sleve= 73 MPa

Is it correct ?

Answer 4

Q 2_2_1?
Q 2_2_3?
Q 2_2_4?
please reply

Answer 5

Q2_2_5 is incorrect

Answer 6

Q 2_2_1
q_0*x^3/(6*L)

Q 2_2_3
q_0*x^3/(6*L)/(23/2*R_0^4*E_0*pi)

Q 2_2_4
-5.82

Answer 7

2_2_1 & 2_2_3 are incorrect!

Answer 8

2_1_1,
1_2,
1_3,
1_4
please?

Answer 9

2_1_1 -3/4*t*L
2_1_2 (t*L)/(pi*G*R^4)
(t*(3*L-2*x))/(2*pi*G*R^4)
3/4*L
2_1_3 (3/2*t*L)/(pi*R^3)
R
3/2L

Answer 10

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)= ................

Answer 11

2_1_3 is wrong

Answer 12

Anyone have any idea for

q2_2_1
Q2_2_3
Q2_2_5

Appreciate your help. Thanks :)

Answer 13

M(x) = -q_0*x^3/(6*L)

Answer 14

thanks simon76 :)

Answer 15

Q2_2_3

1/rho_x = -(q_0*x^3/(6*L))/(46/4*E_0*pi*R_0^4)

because

Q2_2_2

EI_eff = (46/4*E_0*pi*R_0^4)

Answer 16

Q2_2_3

Just try this for the slope:

(q_0*(L^4-x^4)/(24*L))/(46/4*E_0*pi*R_0^4)

but I'm not sure

Answer 17

Q2_2_3

Just try this for the slope:

(q_0*(L^4-x^4))/(24*L))/(46/4*E_0*pi*R_0^4)

but I'm not sure

Answer 18

any one Q2_2_5

Answer 19

2_2_5

50.9
305.6

Answer 20

please any one
2_1_1
and
2_2_3
v(x)

Thanks!!

Answer 21

2_2_1

-q_0*x^3/(6*L)

Q2_2_2

23/2*R_0^4*E_0*pi

Q2_2_3

-(q_0*x^3/(6*L))/(46/4*E_0*pi*R_0^4)

V(x)????

Q2_2_4

-5.82

Q2_2_5

50.9
305.6

Answer 22

2_2_3

v(x)= (q_0*(L^4-x^4)/(24*L))/(46/4*E_0*pi*R_0^4)

Answer 23

2_1_1,
1_2,
1_3,
1_4
please?