Additional Example 2 - maae2202 PDF

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Question 1 An aluminum rod shown in Figure 1(a) has a circular cross section and is subjected to an axial load of 10 kN. A portion of the stress - strain diagram for the material is shown in Figure 1(b). Determine the approximate elongation of the rod when the load is applied. If the load is removed, what is the permanent elongation of the rod? The Young’s modulus of the material is 70 GPa.

(a)

(b) Figure 1 Solution:

 AB =

 BC =

10  1000  20     2 

2

10  1000  15    2 

2

= 31.83 MPa

= 56.59 MPa

1

From the stress - strain diagram, the material in region AB is strained elastically since the yielding stress  Y = 40 MPa >  AB = 31.83 MPa.

 AB =

 AB E

=

31.83 4 = 4.55  10− 70 1000

The material within region BC is strained plastically, since  BC = 56.59 MPa > the yielding stress  Y = 40 MPa. From the stress - strain diagram, for  BC = 56.59 MPa, the corresponding strain  BC = 0.045 The total deformation or elongation of the rod is 600  4.55  10−4 + 400  0.045 = 18.3 mm

When the 10 kN load is removed, segment AB of the rod will be restored to its original length due to elastic deformation. On the other hand, the material in segment BC will only recover elastically along line FG, which is parallel to the linear portion of the stress - strain curve.

 rec =

 BC E

=

56.59 = 8.08  10− 4 70 1000

The remaining plastic strain in segment BC is then

 OG = 0.045 − 8.08  10−4 = 0.0442 Therefore, when the load is removed, the rod remains elongated by an amount

 = OG LBC = 0.0442  400 = 17.7 mm

2

Question 2 Figure 2 shows a steel tube, fixed to a rigid wall; it has an internal diameter D = 37.5 mm and wall thickness t = 1.25 mm. The tube is coaxially jointed to a steel shaft of diameter d = 25 mm via a polymer insert that is perfectly bonded to each member as shown. The shaft is subjected to either an axial load P or a torque T at different stages of the service cycle, and it can displace through a circular opening in the wall. Case (a): Consider the situation when a torque T = 35 N·m is applied on the shaft as shown in Figure 2. (1) What is the mean shear stress on the cylindrical surface of the polymer connector where it is bonded to the tube? (2) What is the mean shear stress on the cross-section of the steel tube at the station C between the rigid wall and the end of the polymer connector? Case (b): Consider the situation when an axial load P = 2 kN is applied on the shaft as shown in Figure 2. (1) What is the mean shear stress on the cylindrical surface of the polymer connector where it is bonded to the tube? (2) What is the mean normal stress on the cross-section of the steel tube at the station C? (3) If the shaft displaced 0.55 mm along its axis relative to the steel tube, what is the shear strain in the polymer, assuming that the steel shaft and tube are rigid compared to the polymer connector? (4) Assuming linear elastic behaviour, determine the approximate shear modulus of the polymer.

Steel tube t Polymer insert D

P

d T

Shaft

C

Figure 2 Solution: Case (a): T = 35 N·m

3

(1) T =

D 37.5 D  L   p , 35  10 3 =   37.5  75   p ,  p = 0.211 MPa 2 2

(2) T =

D 37.5 D  t   t , 35 103 =   37.5 1.25  t ,  t = 12.68 MPa 2 2

Case (b): P = 2 kN (1) P = D  L  p , 2  10 3 =   37.5  75  p ,  p = 0.227 MPa (2) P = D  t   t , 2  10 3 =   37.5 1.25   t ,  t = 13.59 MPa

(3)

(37.5-25)/2 mm

 0.55 mm

tan  =

(4) G =

0.55 = 0.088   37.5 − 25 2

 0.227 = = 2.58 MPa  0.088

Question 3 Rigid bar BCD in Figure 3 is supported by a pin at C and by aluminum rod (1). A concentrated load P is applied to the lower end of aluminum rod (2), which is attached to the rigid bar at D. The cross-sectional area of each rod is A = 0.2 in2. The elastic modulus and thermal expansion of the aluminum material are 10,000 ksi and 23 × 10-6 /C, respectively. With the applied load P, when the temperature of the whole system is increased from room temperature (20C) to 100C, the normal strain in rod (1) is measured as 4  10-3 (tension). Points A and C are pin-fixed. (1) Determine the magnitude of load P. (2) Compute the total deflection of point E.

4

B B

Figure 3 Solution:

f1 Ry

Rx

P

5

(1) Consider the force equilibrium of the whole system 30 P = 20 f1 Since f1 = σ1 A, then 30 P = 20 σ1 A = 20 E ε1e A

 1 =  1e +  1T

1T = T = 23 10−6  (100 − 20 )= 1.84 10−3  1e =  1 −  1T = 4 10− 3 − 1.84  10− 3 = 2.16 10 − 3 −

Therefore P =

20 E 1e A 20 107  2.16 10 3  0.2 = 2880 lb = 30 30

(2) Consider the deformation of aluminum rod (1) 



 l1 = l1 1 = 50 4 10− 3 = 0.2 in, then the rotation angle of rigid bar BCD is calculated as

=

 l 1 0.2 = = 0.01 l BC 20

The deflection of D is lCD = 30  0.01 = 0.3 in The deformation of aluminum rod (2) is computed as

l2 = l2  2

 2 =  2 e +  2T

 2 T = T = 23 10− 6 (100 − 20 )= 1.84 10− 3

6

2 e =

2 E

=

f2 P 2880 −3 = = 7 =1.44 10 EA EA 10  0.2

(

)

 l2 = l2 2 = l2 (2 e + 2 T )= 100 1.44 10− 3 + 1.83 10− 3 = 0.327 in

The total deflection of E is the sum of the deflection of D and the deformation of aluminum rod (2), that is, 0.3 + 0.327 = 0.627 in

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