Dr. C. W. Ong

Mechanics AP200 Dr. C. W. Ong

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Textbook : Engineering Mechanics, Dynamics, by Meriam, J.L., Kraige, L.G., John Wiley 1997 4th Ed. Mechanics of Engineering Materials , by P.R. Benham and R.J. Crawford, ELBS, 1988 2 Jump to first page

Assessment method : ■ ■

Course Work 30% (Exercises and tests) Examination 70%

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Vector A vector consists of two things : 1. Magnitude, represented by a number A 2. Direction, represented by a unit vector eˆ A

A

eˆ

  A =Ae A

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In Cartesian coordination system  A = Aeˆ A = A(cos α xˆ + sin α yˆ ) = Ax xˆ + Ay yˆ Y

AY

 A

α AX

X

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Addition of vectors    A = AX i + AY j    B = BX i + BY j     A + B = ( AX + BX )i + ( AY + BY ) j Y

 BY

 AY

 B

   A+ B = C

 Bx

 A  AX

X

 B'

 A

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3-D Cartesian coordination system Z

 A = A eˆ A

   = A cos α i + A cos β j + A cos γ k    = AX i + AY j + AZ k

X

eˆ  A

γ

β Y

α

Direction cosine : cosα , cosβ ,  cosγ A = A 2 +A 2 +A 2 =A X 2

Y

Z 2

=( A cos 2α+A cos 2 β+A 2 cos =A (cos 2α+cos 2 β+cos 2 γ)1 / 2 cos 2α+cos 2 β+cos 2 γ =1

2

γ)1/ 2

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Dot product : Result of dot product is a scalar

 A⋅ B = C

 Definition is : A ⋅ B ≡ AB cosθ

 A θ

A cos θ

 B

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Cross product    Result is a new vector A ×B =C   Definition : A × B = AB sin θ , and  C is ⊥ to the two vectors, with direction determined by right hand rule.

 C

F sinθ

 B

θ

θ

O  A

 F

d

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   ( A × B) ⋅ C

Exercise : Show that is the volume of the parallelepiped formed by the three vectors

    A = AX i + AY j + AZ k   C =C Z k    B = BY j +BZ k

 A  B

   C ( A × B) ⋅ C

Z

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Presentation of vector product in rectangular coordination system     A = AX i + AY j + Az k

    B = BX i + BY j + Bz k  Dot product : ( A ⋅ B) = AX BX + AY BY + AZ BZ

Cross product :

 i  A × B = AX BX

 j AY BY

 k AZ BZ

      − B A k − A B j − B A i = AY BZ i + AX BY k + B X AZ j X Y X Z Y Z 11 Jump to first page

Fundamental of Mechanics of Materials Stress and strain : A

 F

 F

 F

 F  F

 F

Assume F is uniformly distributed over the cross-sectional area A.

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Normal stress σ≡

[

F N / m2 A

]

pascal ( Pa) δF dF = δA → 0 δA dA

Normal stress at a point : σ = lim

δF

P

 F

δA

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Total force F acting on a cross-sectional area is : F = ∫ σ dA A

 F

 F Tensile stress (+)

 F

 F Compressive stress (-)

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Shear stress : Force F is tangential to the area. A A Fixed area Define shear stress :

 F

 F

F τ≡ A

dF At a local point : τ = dA

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Normal strain : A body deformed along the direction of stress. Normal strain is defined to describe the deformation :

x ε≡ 

 F l

x

 F

Shear strain :

x Deformation under shear stress is defined as γ ≡  x F

 16

γ

fixed

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Elastic modulus Stress-strain curve

Work hardening

Stress

Elastic deformation o

Yield pt.

break

Strain

In the elastic region, σ α ε (Hooke's law) Young’s modulus (modulus of elasticity) E = σ / ε Shear modulus : (modulus of rigidity) G = τ /γ 17 Jump to first page

Examp le

dx o

Find the total extension of ✦ the bar. ✦

✦

15mm

5mm W 0.6m

1.2m X

The width of a cross-sectional element at x :

x x (5 × 10 −3 m) = ( m) The stress in this element : 0.6m 120 W=

3 7 2 × 10 N 2 . 88 × 10 The strain σ =of this element: = Pa 2 2 2 ( x / 120) m x

σ 2.88 × 107 / x 2 1.92 × 10−4 ε= = = 9 E 150× 10 x2 18 Jump to first page

2kN

The extension of this element : 1.92 ×10 −4 de = εdx = dx 2 x

∴ The total extension for the whole bar becomes : e = ∫ de = ∫

1 .8

0 .6

1.92 ×10 −4 dx 2 x

= 2.13 x 10-4 m

#

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Bulk modulus :

K=

δp − (δV / V )

δp V + δV

At a local

pointlim: δ V → 0

then

dp K = −V dV

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Poisson's ratio : For homogeneous isotropic materials d + ∆d

d F

F

x

 ■

■ ■ ■

normal strain :

x ε= 

lateral strain : εL Poisson's ratio : value of ν : 0.2 - 0.5

∆d = d

ν ≡ −ε L / ε

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Analysis of stress and strain in Isotropic Materials. σ x ν σy ν σz εx = − − ......(*) E E E σ y ν σx ν σz εy = − − E E E σ z ν σx ν σy εz = − − E E E

σz

τ zx

τ xz

τ zy

τ yz

τ xy τ yx σx

z

σy

x

y

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Rewrite (*) by x ⇒ xx, y ⇒yy , z ⇒ zz : ε xx = (σ xx − ν σyy − ν σzz ) / E ε yy = (σ yy − ν σzz − ν σxx ) / E ε zz = (σ zz − ν σxx − ν σyy ) / E

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−E Exercise : Verify K = 3(1 − 2ν ) Solution From (*)

1 +ν ν ε xx = σ xx − (σ xx + σ yy + σ zz ) E E (1 +ν ) ν ε yy = σ yy − (σ xx + σ yy + σ zz ) E E (1 +ν ) ν ε zz = σ zz − (σ xx + σ yy + σ zz ) E E

(1 − 2ν ) ε xx + ε yy + ε zz = (σ xx + σ yy + σ zz ) E 24 Jump to first page

For hydrostatic pressure σ xx = σ yy = σ zz = ∆p

ε xx = ε yy = ε zz = ε

4 4 π (r − ∆r )3 − π r 3 ∆V 3 3 = 4 3 V πr 3

∴

(symmetry) ∆r

r

∆r ≅ −3 ( ) = −3ε r (1 − 2ν ) =− (3∆p ) E ∆p −E K ≡− = ∆V / V 3(1 − 2ν )

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Shear stress and strain Take moment about the z axis, total torque = 0, y so τ xy = τ yx τ

τ

∆y z

xy

∆x

yx

x

Similarly, take moment about x-axis, ⇒ τ yz = τ zy Take moment about y-axis, ⇒ τ

xz

=τ

zx

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Shear strain :

γ xy = τ xy / G γ yz = τ yz / G γ zx = τ zx / G

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dy dx Shear deformation is usually defined by : , ,...etc. ∆x ∆y dy dx and However, pure rotation may cause non-zero ∆x ∆y

which cannot be used to represent any deformation One better method is to select axes appropriately, such that dy dx = = ε xy = ε yx = γ/2 ∆x ∆y y y Pure y rotation ∆x

dy

x

∆x

dy

x

γ 2

γ /2

x

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Example : Show that ε

n

=ε

yx

(AC)2(1+ε n)2 = (AD)2 + (AD)2 -2(AD)2 cos(90o+γ ) 2(AD)2 (1+ε n)2 = 2(AD)2+2(AD)2 sin γ 1+2ε n = 1 + sin γ

≈ 1+γ y

2ε n ≈ γ since γ = 2 ε ε

n=

ε

yx

C’ yx

C

γ 2

D’ A

γ /2

D

x

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εn =

Example : Show that σ yx (lW) sin 45o x2 = LW σ direction)

n

σn 2G

(equilibrium along n-

= 2(l cos 45o) W σ

n

Therefore σ yx = σ n σ xy σ n = = 2ε n From definition : γ xy = y

σ

σ

G

σn ∴ε n = 2G

G

yx

σ

l

xy

σ

γ /2

A 30

γ /2

l xy

yx

L σ

n

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E = 2G 1+υ

Example : Show

1 +ν ν ε xx = σ xx − (σ xx + σ yy + σ zz ) (from p.24) E E -σ Set σ σ ε

xx

zz xx

=σ

n

=-σ

,

n

σ

= 0, =ε

σ

n

-σ

n

ε n = (1+ν ) σ example)

∴

yy

n

/E = σ

n

n

/2G (previous

E = 2G 1+υ

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n

Example The cube of isotropic material has side 20mm, E = 60 GPa, ν = 0.3.

✦

(i) Here

ε x = 0, σ y = 0 and

− 12 × 103 N 7 σz = = − 3 × 10 Pa −3 2 (20 ×10 m)

✦

Since :

✦

∴ σ x = -9 x 106 Pa (compressive stress)

1 ε x = (σ x − υ σy − υ σz ) E (i) Find the force exerted by 1 7 the restraining walls upon the ∴ 0 = [ σ − 0 − 0 . 3 × ( − 3 × 10 )] x 9 block. 60 ×10

(ii)Find the strain in y direction z y 12kN

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The force exerted by the restraining wall : ✦ Aσ = (20 x 10-3 m)2 x (-9 x 106 Pa) x ✦

= -3.6 x 103 N (compressive force) ✦ (ii) The strain in y direction : 1 ε y = (σ y − υ σz − υ σx ) E x 1 7 6 = [ 0 − 0 . 3 × ( − 3 × 10 ) − 0 . 3 × ( − 9 × 10 )] 9 60 × 10 Jump to first page = 1.95 × 10− 4 # ✦

Elastic Strain Energy F

F  dW = Fdx

dx

✦

The energy store in the material :

✦

∴ The energy store :

x dU = Fdx = AE ( )dx 

AE U =∫ ( x)dx e=extension 0  1 AEe 2 1 e 2 = = E ( ) ( A) 2  2  1 2 = Eε ⋅ V 2 e

F/A E= x/

✦

∴ Energy density within the material : U 1 2 1 σ2 u ≡ = Eε = V 2 2 E

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Similarly for shear strain : dx

F

U =∫



  F ⋅ dx = ∫ Fdx

F/A τ G= = x/ γ

2 1 1 τ ∴u = G γ 2 = 2 2G

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