Physics
posted by helpless .
Composite bar with end load (two segments)
The composite bar BCD in the figure is composed of an inner aluminum cylindrical core of length 2L=2 m and radius R=1 cm and a sleeve of steel of length L=1 m and thickness R=1 cm surrounding the aluminum core in the CD section of the bar. The Youngâ€™s moduli are: E(Al) = 70 GPa, and E(steel)=210 GPa. The bar is fixed at B and a concentrated tensile load F is applied at the free end D. The resulting total elongation of bar BCD, is δ=1 cm. Determine the normal stress in the steel sleeve. (in MPa)
i[dot]imgur[dot]com/McHq9LJ.png

Helpless it is, 184 MPA
Do you have figured out Problem 1 and 2? 
Anyone?

Problem 1 and 2?

Problem 1 and 2 please!

Determine the normal stress in the steel sleeve.
Obtain the numerical value of the displacement of section.
pllllllllllllllllllllz help? 
86.8 MPa

problem3_1 second part,
Obtain the numerical value of the displacement of section c....!
plz help? 
@jojo
0.4mm 
3_2(a)
3_2(b)
plz help. 
3_3 problem
plz help. 
HW3_2A
(2*p_0*L)/27p_0*(L/3x+(3*x^2)/(4*L))
(p_0*2*L/27)
HW3_2B
(2*p_0*L*x)/(27*E*A)p_0/(E*A)*((L*x)/3x^2/2+x^3/(4*L))
p_0*2*L/(27*E*A)*(xL) 
HW3_3A
u1 = a*L*x1;
u2 = 2*a*(x2.^2L*x2);
HW3_3B
N1 = 3*a*L*A*E+0*x1;
N2 = 6*a*A*E*(2*x2L);
HW3_3C
fx1 = 0 + 0*x1;
fx2 = 12*a*A*E + 0*x2;
HW3_3D1
Fx=3*a*L*A*E
x=L/2
HW3_3D2
R(B)=3*a*A*E*L
R(C)= 6*a*A*E*L 
E2_2:
E2_3:
PLZ help 
E2_2
1)epsilon_0*x/L*(Lx/2)
2)epsilon_0*L/2
3)epsilon_0/(2*L)*(Lx)^2
4)epsilon_0*L/2
E2_3
1)112
2)0.001
3)0.004
4)0.004 
E3_1?
HW3_4? 
HW3_3 Inverse PROBLEM ON STATICALLY INDETERMINATE COMPOSITE BAR (PART 4)

HW3_4B: 0.5

E3_1
1)50
2)50,100
3)12*a*L*E_1*A_2*(x/L1),30
4)0
5)12*E_1*A_2*a
HW3_3D1
Fx=3*a*L*A*E
x=L/2
HW3_3D2
Rx(B)=3*a*A*E*L
Rx(C)= 6*a*A*E*L 
HW3_4A
2.66
HW3_4B
0.49 
problem hw3_1b
second part?
anyone gave the answer? 
thanks alot regz.

thanks regz.

thanks anonymous.

problem 4_1?
SD truss problem with the method of joints...............! 
hw3_1(b)
0.4mm
@heyyo 
plz help
problem 4_1? 
HW4_1
1)
1.17
1.17
1.4142
0.936
2.35
2)
1.414
0.702
0.702
3)
4.05
4.05
2.59
6.49