41573_1_4Lecture 1.4.pdf

LECTURE 1.4

Stress and Strain –

Axial Loading (con.)



Poisson’s Ratio



Generalized Hooke’s Law



Dilatation: Bulk Modulus



Shearing Strain



Example 2.10



Relation Among E, v, and G



Sample Problem 2.5



Composite Materials

Contents

Reference: Ferdinand P. Beer, E. Russell Johnston, Jr. & John T. DeWolf, "Mechanics of materials" 5th _{edition, McGraw-Hill, 2005. (Textbook).}



Saint-Venant’s Principle



Stress Concentration: Hole



Stress Concentration: Fillet



Example 2.12



Elastoplastic Materials



Plastic Deformations



Residual Stresses



Example 2.14, 2.15, 2.16



Sample Problems 1-11

Poisson’s Ratio

•

For a slender bar subjected to axial

loading:

•

The elongation in the x-direction is

accompanied by a contraction in the

other directions. Assuming that the

material is isotropic (no directional

dependence),

0





_z

y





0







x _{y z} x

E









Poisson’s Ratio (con.)

Poisson’s ratio is defined as

x

z

x

y





















strain

axial

strain

lateral

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:

E

E

E

E

E

E

z y x z z y x y z y x x

















































Dilatation: Bulk Modulus

Relative to the unstressed state, the change in volume is

For element subjected to uniform hydrostatic pressure,

Subjected to uniform pressure, dilatation must be negative, therefore

























e) unit volum per in volume (change dilatation 2 1 1 1 1 1 1 1                   z y x z y x z y x z y x E e             







1 2



bulk modulus 3 2 1 3           E k k p E p e 1 0  

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

xy

f



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,

where G is the modulus of rigidity

or shear modulus.

zx zx yz yz xy xy

G





G





G











Example 2.10

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.

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

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

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

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

rad 020 . 0 mm 50 mm 1 tan    _{xy xy} xy   

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



630MPa



0.020rad



12.6MPa 

 _xy xy G



• Use the definition of shearing stress to find the force P. kN 2 . 156  P



12.6106Pa





0.2m



0.062m



156.2103N   A P _xy

Relation Among E, v, and G

• An axially loaded slender bar will elongate in the axial direction and contract in the transverse directions.

• Components of normal and shear strain are related,

• 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.

• An initially cubic element oriented as in top figure will deform into a rectangular parallelepiped. The axial load produces a normal strain.











1

2G

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 σ_x = 84 MPa and σ_z = 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.

• Apply the generalized Hooke’s Law to find the three components of normal strain.









mm/mm 10 600 . 1 mm/mm 10 067 . 1 mm/mm 10 533 . 0 M Pa 140 3 1 0 M Pa 84 M Pa 10 70 1 3 3 3 3                         _{ }       E E E E E E E E E z y x z z y x y z y x x            

• Evaluate the deformation components.



0.533103mm/mm





225mm



  _xd A B  



1.600103mm/mm





225mm



  _zd D C  



1.067103mm/mm





18mm



  _yt t   mm 12 . 0   A B  mm 36 . 0   D C  mm 0192 . 0   t 

• Find the change in volume





3 3 3 3 3 mm 18 380 380 10 067 . 1 /mm mm 10 067 . 1              eV V e _x _y _z 3 mm 2733   V SOLUTION:

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

Composite Materials (con.)

•

Transverse contractions are related by directionally dependent

values of Poisson’s ratio, e.g.,

x

z

xz

x

y

xy

_



















•

Materials with directionally dependent mechanical properties

are

anisotropic

.

z

z

z

y

y

y

x

x

x

E

E

E



















Saint-Venant’s Principle

• Loads transmitted through rigid plates result in uniform distribution of stress and strain.

• Stress and strain distributions

become uniform at a relatively short distance from the load application points.

• Concentrated loads result in large stresses in the vicinity of the load application point.

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.

Stress Concentration: Hole

Discontinuities of cross

section may result in high

localized or

concentrated

stresses.

ave

max







K

Example 2.12

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.

SOLUTION:

• Determine the geometric ratios and find the stress concentration factor from Fig. 2.64b.

• Apply the definition of normal stress to find the allowable load.

• Find the allowable average normal stress using the material allowable normal stress and the stress

• Determine the geometric ratios and find the stress concentration factor from Fig. 2.64b. 82 . 1 20 . 0 mm 40 mm 8 50 . 1 mm 40 mm 60      K d r d D

• Find the allowable average normal stress using the material allowable normal stress and the stress

concentration factor. MPa 7 . 90 82 . 1 MPa 165 max ave    K  

• Apply the definition of normal stress to find the allowable load.









N 10 3 . 36 MPa 7 . 90 mm 10 mm 40 3     A _ave P  kN 3 . 36  P

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

Plastic Deformations

• Elastic deformation while maximum stress is less than yield stress

K A A

P _ave  max

• 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 region expands until the section is at a uniform stress equal to the yield stress K A P_Y  Y Y Y U P K A P  

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 also result from the uneven heating or

cooling of structures or structural elements

• 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

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.

a) Draw a load-deflection diagram for the rod-tube assembly

   







   

 





mm 89 . 0 mm 750 Pa 10 210 Pa 10 250 kN 12 m 10 48 Pa 10 250 9 6 2 6 6             L E L δ A P r Y r Y r Y r r Y r Y r   

   







   

 





mm 21 . 2 mm 750 Pa 10 105 Pa 10 310 kN 2 . 19 m 10 62 Pa 10 310 9 6 2 -6 6            L E L δ A P t Y t Y t Y t t Y t Y t    t r t r P P P       

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

 





mm 750 Pa 10 105 Pa 10 210 M Pa 210 m 10 62 kN 13 kN 13 kN 12 25 kN 12 9 6 t 2 6 -t                 L E L A P P P P P P t t t t t r t Y r r     mm 5 . 1 max t  

The rod-tube assembly unloads along a line parallel to 0Y_r



1.5 1.16



mm mm 16 . 1 mm kN 6 . 21 kN 25 slope mm kN 6 . 21 0.89mm kN 2 . 19 max p max                    m P m mm 34 . 0  p 

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.























210 162.75



M Pa 47.25M Pa M Pa 5 . 75 M Pa 5 . 325 250 M Pa 75 . 162 Gpa 105 10 55 . 1 M Pa 5 . 325 GPa 10 2 10 55 . 1 mm mm 10 55 . 1 mm 750 mm 16 . 1 , , 3 3 3                                         t t t residual r r r residual t t r r E E L            

Example 2.14, 2.15, 2.16

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.

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.

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 = ky

2

) of the rubber when the brakes are

applied (Fig. 3b), determine the shearing strain in the rubber.

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.

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.

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

Example Problem 7

A short cylindrical rod of ASTM–A36 structural steel,

having an original diameter of d

_o

and length L

_o

is placed

in a compression machine and squeezed until its length

becomes L

_f

. Determine the new diameter of the rod.

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.

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.

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.

Example Problem 11

The rectangular concrete block shown in Fig. 11 is subjected to

loads that have the resultants P

_x

= 100 kN, P

_y

= 150 kN, and P

_z

=

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