Asked 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 Young’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
Answered by FLu
Helpless it is, 184 MPA

Do you have figured out Problem 1 and 2?
Answered by Any
Anyone?
Answered by My
Problem 1 and 2?
Answered by Neon
Problem 1 and 2 please!
Answered by me
Determine the normal stress in the steel sleeve.
Obtain the numerical value of the displacement of section.
pllllllllllllllllllllz help?
Answered by Pa
-86.8 MPa
Answered by jojo
problem3_1 second part,
Obtain the numerical value of the displacement of section c....!
plz help?
Answered by regz
@jojo

0.4mm
Answered by jojo
3_2(a)
3_2(b)
plz help.
Answered by hony
3_3 problem
plz help.
Answered by regz
HW3_2A

(2*p_0*L)/27-p_0*(L/3-x+(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)/3-x^2/2+x^3/(4*L))

p_0*2*L/(27*E*A)*(x-L)
Answered by regz
HW3_3A
u1 = -a*L*x1;

u2 = 2*a*(x2.^2-L*x2);

HW3_3B
N1 = -3*a*L*A*E+0*x1;

N2 = 6*a*A*E*(2*x2-L);

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
Answered by jojo
E2_2:
E2_3:
PLZ help
Answered by regz
E2_2
1)epsilon_0*x/L*(L-x/2)
2)epsilon_0*L/2
3)-epsilon_0/(2*L)*(L-x)^2
4)epsilon_0*L/2

E2_3
1)112
2)0.001
3)-0.004
4)-0.004
Answered by roni
E3_1?
HW3_4?
Answered by roni
HW3_3 Inverse PROBLEM ON STATICALLY INDETERMINATE COMPOSITE BAR (PART 4)
Answered by Anonymous
HW3_4B: 0.5
Answered by regz
E3_1
1)-50
2)50,100
3)12*a*L*E_1*A_2*(x/L-1),-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
Answered by regz
HW3_4A
2.66

HW3_4B
0.49
Answered by heyyo
problem hw3_1b
second part?
anyone gave the answer?
Answered by roni
thanks alot regz.
Answered by jojo
thanks regz.
Answered by roni
thanks anonymous.
Answered by roni
problem 4_1?
SD truss problem with the method of joints...............!
Answered by roni
hw3_1(b)
0.4mm
@heyyo
Answered by jojo
plz help
problem 4_1?
Answered by regz
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
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