Answers
Q2_1_1
TXC=-t_0*L
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 ¦Ð as G_0, t_0 and pi, respectively.
Q2_1_1 : 100.0 POINTS
The x-component of the reaction torque at C:
Q2_1_2 : 60.0 POINTS
The twist rate d¦Õdx(x), and the position x0 along the shaft where the twist rate goes to zero (d¦Õdx(x0)=0):
for 0¡Üx<L,d¦Õdx(x)=
for L<x¡Ü3L,d¦Õdx(x)=
d¦Õdx(x0)=0atx0=
Q2_1_3 : 60.0 POINTS
The maximum absolute value of the shear stress in the shaft (¦Ómax) and its location (r¦Ómax, x¦Ómax):
¦Ómax=
r¦Ómax=
x¦Ómax=
Q2_1_4 : 100.0 POINTS
The maximum value of the rotation field ¦Õ(x) along the shaft (¦Õmax), and the position along the shaft where the maximum rotation occurs (x¦Õmax):
¦Õmax=
x¦Õmax=
11 answers
Answers
Q2_1_2
a) T(x) in 0<=x<=L
T(X)=t_0*L
d*phi/dx=T(x)/(GI) where I=(pi*R^4)/2
so d*phi/dx=(2*t_0*L)/(pi*G_0*R^4)
b) T(x) in l<=x<=3L
T(x)=t_0*(2*L-x)
so
d*Phi/dx=(2*t_0+(2*L-x))/(pi*G_0*R^4)
c)(d*phi/dx)(x)=0 when
0=(2*t_0+(2*L-x))/(pi*G_0*R^4)
so (2*L-x)=0 >>>>>> x=2*L
answer x=2*L
Q2_1_2
a) T(x) in 0<=x<=L
T(X)=t_0*L
d*phi/dx=T(x)/(GI) where I=(pi*R^4)/2
so d*phi/dx=(2*t_0*L)/(pi*G_0*R^4)
b) T(x) in l<=x<=3L
T(x)=t_0*(2*L-x)
so
d*Phi/dx=(2*t_0+(2*L-x))/(pi*G_0*R^4)
c)(d*phi/dx)(x)=0 when
0=(2*t_0+(2*L-x))/(pi*G_0*R^4)
so (2*L-x)=0 >>>>>> x=2*L
answer x=2*L
Q2_1_2 answer is not right
also Q2_1_1 answers are not right either.
fuubo check this
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_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
2) x phi max=3*L/2
2) x phi max=3*L/2
Q2_1_4
1) phi=(9*t_0*L^2)/(2*pi*G_0*R^4)
1) phi=(9*t_0*L^2)/(2*pi*G_0*R^4)
@ElementarySchoolStudent
Thanks, except Q2_1_3 tau max's answer was wrong, others are right.
Thanks, except Q2_1_3 tau max's answer was wrong, others are right.
@@ElementarySchoolStudent
Q2_1_4 1)'s answer was also not rigth.
Q2_1_4 1)'s answer was also not rigth.
fuubo
You are right I think I add wrong and instead of 4 is 3.
If you have more chances try this
tau max=(3*t_0*L)/(pi*R^3)
You are right I think I add wrong and instead of 4 is 3.
If you have more chances try this
tau max=(3*t_0*L)/(pi*R^3)
yes, tau max is right now.
thanks
thanks