41573_1_4Lecture 1.4.pdf

LECTURE 1.4

Stress and Strain – Axial Loading (con.) 1

Contents Saint-Venant’s Principle Poisson’s Ratio Stress Concentration: Hole Generalized Hooke’s Law Stress Concentration: Fillet Dilatation: Bulk Modulus Example 2.12 Shearing Strain Elastoplastic Materials Example 2.10 Plastic Deformations Relation Among E, v, and G Residual Stresses Sample Problem 2.5 Example 2.14, 2.15, 2.16 Composite Materials Sample Problems 1-11 Reference: Ferdinand P. Beer, E. Russell Johnston, Jr. & John T. DeWolf, "Mechanics of materials" 5th edition, McGraw-Hill, 2005. (Textbook).

10/26/2011

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Poisson’s Ratio • For a slender bar subjected to axial loading:

x 

x E

y z  0

• The elongation in the x-direction is accompanied by a contraction in the other directions. Assuming that the material is isotropic (no directional dependence),

y  z  0 3

Poisson’s Ratio (con.)

Poisson’s ratio is defined as

y z lateral strain    axial strain x x 4

Generalized Hooke’s Law • For an element subjected to multi-axial loading, the normal strain components resulting from the stress components may be determined from the principle of superposition. This requires: 1) strain is linearly related to stress 2) deformations are small

• With these restrictions:

 x  y  z

x  

E

y   z  



 x E



E



 y  z E

 x  y E



E

E





E

z E

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Dilatation: Bulk Modulus Relative to the unstressed state, the change in volume is











e  1  1   x  1   y 1   z   1  1   x   y   z



 x  y z 

1  2  x  y  z E





 dilatation (change in volume per unit volum e)

For element subjected to uniform hydrostatic pressure, e  p k

31  2  p  E k

E  bulk modulus 31  2 

Subjected to uniform pressure, dilatation must be negative, therefore 0    12 6

Shearing Strain • A cubic element subjected to a shear stress will deform into a rhomboid. The corresponding shear strain is quantified in terms of the change in angle between the sides,

 xy  f  xy 

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Shearing Strain (con.) • A plot of shear stress vs. shear strain is similar to the previous plots of normal stress vs. normal strain except that the strength values are approximately half. For small strains,

 xy  G  xy  yz  G  yz  zx  G  zx

where G is the modulus of rigidity or shear modulus. 8

Example 2.10

SOLUTION: • Determine the average angular deformation or shearing strain of the block.

A rectangular block of material with modulus of rigidity G = 630 MPa is bonded to two rigid horizontal plates. The lower plate is fixed, while the upper plate is subjected to a horizontal force P. Knowing that the upper plate moves through 1.0 mm. under the action of the force, determine a) the average shearing strain in the material, and b) the force P exerted on the plate.

• Apply Hooke’s law for shearing stress and strain to find the corresponding shearing stress.

• Use the definition of shearing stress to find the force P.

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• Determine the average angular deformation or shearing strain of the block.  xy  tan  xy 

1 mm 50 mm

 xy  0.020 rad

• Apply Hooke’s law for shearing stress and strain to find the corresponding shearing stress.  xy  G xy  630MPa0.020 rad   12.6MPa

• Use the definition of shearing stress to find the force P.





P   xy A  12.6 106 Pa 0.2m0.062m  156.2 103 N

P  156.2 kN 10

Relation Among E, v, and G • An axially loaded slender bar will elongate in the axial direction and contract in the transverse directions. • An initially cubic element oriented as in top figure will deform into a rectangular parallelepiped. The axial load produces a normal strain.

• If the cubic element is oriented as in the bottom figure, it will deform into a rhombus. Axial load also results in a shear strain. • Components of normal and shear strain are related,

E  1    2G 11

Sample Problem 2.5 A circle of diameter d = 225 mm is scribed on an unstressed aluminum plate of thickness t = 18 mm. Forces acting in the plane of the plate later cause normal stresses σx = 84 MPa and σz = 140 MPa. For E = 70 GPa and v = 1/3, determine the change in: a) the length of diameter AB, b) the length of diameter CD, c) the thickness of the plate, and d) the volume of the plate.

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SOLUTION: • Apply the generalized Hooke’s Law to • Evaluate the deformation components. find the three components of normal  B A   x d   0.533 103 mm/mm 225mm strain. x   

x E



 y  z E



E

1 1       84 M Pa  0  140 M Pa 3  70 10 M Pa  3

 0.533 10 3 mm/mm

y  

 x



 y  z 

E E E  1.067 10 3 mm/mm

z  

 x  y 



z

E E E  1.600 10 3 mm/mm

 B A  0.12mm

 C D   z d   1.600 103 mm/mm 225mm  C D  0.36mm

 t   y t   1.067 103 mm/mm 18mm 

 t  0.0192mm

• Find the change in volume e   x   y   z  1.067 10 3 mm 3 /mm 3 V  eV  1.067 103 380  380 18mm 3 V  2733mm3 13

Composite Materials

• Fiber-reinforced composite materials are formed from lamina of fibers of graphite, glass, or polymers embedded in a resin matrix.

• Normal stresses and strains are related by Hooke’s Law but with directionally dependent moduli of elasticity, 14

Composite Materials (con.) x Ex  x

Ey 

y y

z Ez  z

• Transverse contractions are related by directionally dependent values of Poisson’s ratio, e.g.,

 xy

y z   xz   x x

• Materials with directionally dependent mechanical properties are anisotropic. 15

Saint-Venant’s Principle • Loads transmitted through rigid plates result in uniform distribution of stress and strain. • Concentrated loads result in large stresses in the vicinity of the load application point.

• Stress and strain distributions become uniform at a relatively short distance from the load application points.

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Saint-Venant’s Principle (con.)

• Saint-Venant’s Principle: Stress distribution may be assumed independent of the mode of load application except in the immediate vicinity of load application points. 17

Stress Concentration: Hole

Discontinuities of cross section may result in high localized or concentrated stresses.

 max K  ave

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Stress Concentration: Fillet

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Stress Concentration: Hole and Fillet

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Example 2.12 SOLUTION: • Determine the geometric ratios and find the stress concentration factor from Fig. 2.64b. Determine the largest axial load P that can be safely supported by a flat steel bar consisting of two portions, both 10 mm thick, and respectively 40 and 60 mm wide, connected by fillets of radius r = 8 mm. Assume an allowable normal stress of 165 MPa.

• Find the allowable average normal stress using the material allowable normal stress and the stress concentration factor. • Apply the definition of normal stress to find the allowable load.

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• Determine the geometric ratios and find the stress concentration factor from Fig. 2.64b. D 60 mm   1.50 d 40 mm

r 8 mm   0.20 d 40 mm

K  1.82

• Find the allowable average normal stress using the material allowable normal stress and the stress concentration factor.  ave  (b) Flat bars with fillets

 max K



165MPa  90.7 MPa 1.82

• Apply the definition of normal stress to find the allowable load. P  A ave  40 mm 10 mm 90.7 MPa   36.3 103 N

P  36.3 kN 22

Elastoplastic Materials • Previous analyses based on assumption of linear stress-strain relationship, i.e., stresses below the yield stress • Assumption is good for brittle material which rupture without yielding • If the yield stress of ductile materials is exceeded, then plastic deformations occur • Analysis of plastic deformations is simplified by assuming an idealized elastoplastic material • Deformations of an elastoplastic material are divided into elastic and plastic ranges • Permanent deformations result from loading beyond the yield stress 23

Plastic Deformations  A • Elastic deformation while maximum P   ave A  max stress is less than yield stress K

PY 

Y A K

• Maximum stress is equal to the yield stress at the maximum elastic loading • At loadings above the maximum elastic load, a region of plastic deformations develop near the hole

• As the loading increases, the plastic PU   Y A region expands until the section is at a uniform stress equal to the yield  K PY stress

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Residual Stresses • When a single structural element is loaded uniformly beyond its yield stress and then unloaded, it is permanently deformed but all stresses disappear. This is not the general result. • Residual stresses will remain in a structure after loading and unloading if - only part of the structure undergoes plastic deformation - different parts of the structure undergo different plastic deformations • Residual stresses also result from the uneven heating or cooling of structures or structural elements

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Example 2.14, 2.15, 2.16 A cylindrical rod is placed inside a tube of the same length. The ends of the rod and tube are attached to a rigid support on one side and a rigid plate on the other. The load on the rod-tube assembly is increased from zero to 25 kN and decreased back to zero. a) draw a load-deflection diagram for the rod-tube assembly b) determine the maximum elongation c) determine the permanent set d) calculate the residual stresses in the rod and tube.

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Example 2.14, 2.15, 2.16 a) Draw a load-deflection diagram for the rod-tube assembly

Pr Y   r Y Ar  250 106 Pa 48 106 m 2   12 kN  r Y 250 106 Pa δr Y   r Y L  750 mm  L 9 Er

210 10 Pa

 0.89mm

Pt Y   t Y At  310 106 Pa 62 10-6 m 2   19.2 kN  t Y 310 106 Pa δt Y   t Y L   750 mm  L 9 Et

105 10 Pa

 2.21mm

P  Pr  Pt

   r  t 27

Example 2.14, 2.15, 2.16 b,c) determine the maximum elongation and permanent set At a load of P = 25 kN, the rod has reached the plastic range while the tube is still in the elastic range Pr  Pr Y  12 kN

Pt  P  Pr  25  12 kN  13 kN

t 

Pt 13 kN   210 M Pa At 62 10-6 m 2

t

210 106 Pa t  t L  L  750 mm Et 105 109 Pa

 max   t  1.5mm

The rod-tube assembly unloads along a line parallel to 0Yr 19.2 kN  21.6 kN mm  slope 0.89mm P 25 kN     max    1.16mm m 21.6 kN mm  p  0.34mm  p   max     1.5  1.16mm m

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Example 2.14, 2.15, 2.16 Calculate the residual stresses in the rod and tube. Calculate the reverse stresses in the rod and tube caused by unloading and add them to the maximum stresses.

 

 L



 1.16mm  1.55 10 3 mm mm 750 mm

 r   Er   1.55 10 3 210 GPa   325.5 M Pa

 t   Et   1.55 10 3 105 Gpa  162.75 M Pa  residual,r   r   r  250  325.5 M Pa  75.5 M Pa

 residual,t   t   t  210  162.75 M Pa  47.25 M Pa

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Example Problem 1

Determine the normal strain in the members AB and CB of the pin-connected plane structure shown in Fig. 1 if point B moves leftward 3 mm, after load P is applied. Assumption: Axial deformation is uniform throughout the length of each member.

Fig. 1 30

Solution of Example Problem 1

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Example Problem 2 When loaded, the 400 mm by 400 mm square plate of Fig. 2 deforms into a shape in which diagonal BD elongates 0.2 mm and diagonal AC contracts 0.4 mm while they remain perpendicular and side AD remains horizontal. Calculate the average strain components in the xy plane.

Fig. 2 32

Solution of Example Problem 2

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Example Problem 3

The handbrakes on a bicycle consist of two blocks of hard rubber attached to the frame of the bike, which press against the wheel during stopping (Fig. 3a). Assuming that a force P causes a parabolic deflection (x = ky2 ) of the rubber when the brakes are applied (Fig. 3b), determine the shearing strain in the rubber.

Fig. 3 34

Solutions of Example Problem 3

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Example Problem 4 The stress-strain curves for a structural steel bar are shown in Fig. 4. Note that, the entire diagram and its initial portion are plotted using a strain scale N and an enlarged strain scale M in the figure, respectively. Determine: (a) The strains at yield point and fracture of the material. (b) The % elongation of the bar for a 50-mm gage length.

Fig. 4 36

Solution of Example Problem 4

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Example Problem 5 A 10 mm by 10 mm square ABCD is drawn on a member prior to loading. After loading, the square becomes the rhombus shown in Fig. 5. Determine: (a) The modulus of elasticity. (b) Poisson's ratio.

Fig. 5

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Solution of Example Problem 5

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Example Problem 6

Figure 6 shows a steel block subjected to an axial compression load of 400 kN. After loading, if dimensions b and L are changed to 40.02 and 199.7 mm, respectively, calculate: (a) Poisson's ratio. (b) The modulus of elasticity. (c) The final value of the dimension a. (d) The shear modulus of elasticity. Given: a = 60 mm, b = 40 mm, L = 200 mm

Figure 6 40

Solution of Example Problem 6

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Example Problem 7

A short cylindrical rod of ASTM–A36 structural steel, having an original diameter of do and length Lo is placed in a compression machine and squeezed until its length becomes Lf . Determine the new diameter of the rod. Given: do = 30 mm, Lo = 50 mm, Lf = 49.7 mm, v = 0.3

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Solution of Example Problem 7

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Example Problem 8

The cast-iron pipe shown in Fig. 8 is under an axial compressive load P. Determine: (a) The change in length ΔL. (b) The change in diameter ΔD . (c) The change in thickness Δt . Given: D = 130 mm, t = 15 mm, L = 0.5 m, P = 200 kN, E = 70 GPa, v = 0.3. Assumption: Buckling does not occur.

Fig. 8 44

Solution of Example Problem 8

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Example Problem 9

The aluminum rod, 50 mm in diameter and 1.2 m in length, of a hydraulic ram is subjected to the maximum axial loads of ±200 kN. What are the largest diameter and the largest volume of the rod during service? Given: E = 70 GPa, v = 0.3.

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Solution of Example Problem 9

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Example Problem 10 A 20-mm-diameter bar is subjected to tensile loading. The increase in length resulting from the load of 50 kN is 0.2 mm for an initial length of 100 mm. Determine: (a) The conventional and true strains. (b) The modulus of elasticity.

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Solution of Example Problem 10

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Example Problem 11 The rectangular concrete block shown in Fig. 11 is subjected to loads that have the resultants Px = 100 kN, Py = 150 kN, and Pz = 50 KN. Calculate: (a) Changes in lengths of the block. (b) The value of a single force system of compressive forces applied only on the y faces that would produce the same deflection as do the initial forces. Given: E = 24 GPa, v = 0.2

Fig. 11

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Solution of Example Problem 11

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Solution of Example of Problem 11 (con.)

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