Week 6 MECH3361/5361

Recap Chapter 4 Modelling and Solution

Deformation occurs in x-y plane, All independent stresses, strains and displacements are the functions of coordinate x and y only.

Plane Stress (thin plate) Plane Strain (thick/long block)

Structure

th,b

h

b

mg

F F

x

y

z

x

y

z x

y

z

σ zz 0 σ zz=ν (σ xx+σ yy )≠0

ε zz ε zz=− νE

(σ xx+σ yy )≠0 0

z-disp w0 w=0

xx Eε xx=σ xx−νσ yy ( E1−ν2 )ε xx=σ xx−( ν

1−ν )σ yy

yy Eε yy=σ yy−νσ xx ( E1−ν2 )ε yy=σ yy−( ν

1−ν )σ xx

xy Eε xy=(1+ν )σ xy ( E1−ν2 )ε xy=[1+( ν

1−ν )]σ xy

E* E∗¿ E E∗¿ E1−ν2

ν∗¿ν ν∗¿ ν1−ν

UnifiedHookes’

Law {E∗ε xx=σ xx−ν∗σ yy

E∗ε yy=σ yy−ν∗σ xx

E∗ε xy=(1+ν∗)σxy

Replacing Solution Methods

From plane stress to plane strain solution: Replace E and by:

E1−ν2

and

ν1−ν .

From plane strain to plane stress solution: Replacing the E and by:

E(1+2ν )(1+ν )2

and

ν1+ν

Superposition Principle

x

y

x

y

1

Week 6 MECH3361/5361

Solution to Cylinder under Internal and External Pressurepo

pi

r

zr

Ri

Ro

Fig. 4.8 Pressurised cylinder

Displacements: {u=c1r+c2

1r=

(1+ν )(1−2ν )E ( pi Ri

2−po Ro2

Ro2−Ri

2 )r+(1+ν )E ( ( pi−po) Ri

2Ro2

Ro2−Ri

2 )1r

¿¿¿¿

Strains:

{εrr=c1−c21r2 =

(1+ν )(1−2ν )E ( pi Ri

2−po Ro2

Ro2−Ri

2 )−(1+ν )E (( pi−po )R i

2 Ro2

Ro2−Ri

2 )1r2 ¿ {εθθ=c1r+c2

1r2 =

(1+ν )(1−2 ν )E ( pi Ri

2− po Ro2

Ro2−R i

2 )+(1+ν )E ( ( pi−po) Ri

2 Ro2

Ro2−Ri

2 )1r2 ¿ ¿¿¿

Stresses:

{σrr (r )= A−Br2 =

pi Ri2−po Ro

2

Ro2−R i

2 −( pi−po )R i

2 Ro2

Ro2−Ri

2 (1r2 ) ¿ {σθθ(r )=A +Br2 =

pi Ri2−po Ro

2

Ro2−R i

2 +( p i− po) Ri

2 Ro2

Ro2−Ri

2 (1r2 ) ¿{σ zz (r )=2ν (pi Ri

2−po Ro2

Ro2−R i

2 ) ¿¿¿ ¿

Saint Venant Principle:Force acting on a small portion of the surface of an elastic body can be:

(1) replaced by another statically equivalent system of forces acting on the same portion of the surface, (2) such redistribution of loading produces substantial change in stress locally but has a negligible

effect on the stress at a distances (or namely, far field) which are large in comparison with a linear dimensions of the surface on which the force are changed”.

-1000

-1000

-2000-2000N/m

2

Week 6 MECH3361/5361

Example 4.4 Application of St Venant Principle: A solid circular shaft of a machine as shown in the figure is subject to tension by a concentrated force P, torsion by torque Q and bending by a bending moment M in the X-Z plane (M=Myy). The radius of the shaft is R and the length is L (L>>R). Describe the boundary conditions by using the Saint Venant (S-V) principle.

Soln:Although we know the resultant force P, M and T on the ends of shaft, we do not know the exact stress distributions that make these resultant forces. However since L>>R, Saint-Venant principle allows us to use the statistically equivalent conditions by ignoring the detailed stress distribution. Therefore, B.C. can be expressed as

At z=0, z=L, 0≤r≤R , 0≤θ≤2 π (flat left and right end faces): ∵dA=(rd θ )(dr )

P=∫0

2 π

∫0

R

σ zz(rd θ)( dr )=∫0

2 π

∫0

R

σ zz rd θ dr

M yy=M=∫0

2 π

∫0

R

rσ zz (rd θ )(dr )cosθ

M xx=∫0

2 π

∫0

R

rσ zz (rd θ )(dr )sin θ=0

−Q=∫0

2π

∫0

R

σ zθr2 dθ dr

σ zr=0

At r=R, −L≤z≤0 , 0≤θ≤2 π (cylindrical surface): σ rr=σ rθ=σrz=0

Chapter 5 Applications5.1 Torsion of Circular Shaft

Step 1 Boundary ConditionAt the fixed end (left): z=0 , 0≤r≤R :u=v=w=0Outer cylindrical surface in-between: 0<z<L , r=R

σ rr=σ rθ=σrz=0

At the loading end (right): z=L , 0≤r≤Rσ zz=σzr=0

According to Saint-Venant principle: ∫0

2 π

∫0

R

r2 σ zθ dθ dr=Q (Q=torque)

Step 2: Displacement analysis: For small torsional deformation, we can assume u=w=0 approximately (i.e. no length and diameter change or deformation)Observe the right end angle of twist: we can have: v (z=L)=rφ

3

L

FixEnd

TwistedEnd

zQ

B A

A’

r

z

y

x

drrdd

P

QMM

P

Q

y

x

Week 6 MECH3361/5361

On the cylindrical surface: we can have: v (r=R )=R (φz / L )

At any point in the shaft:v=φ zr

L

Thus displacement functions can be:u=0 , v=φ zr

L, w=0

Step 3: Strain

{εrr=∂u∂r

=0 ¿ {εθθ=1r

∂ v∂θ

+ur=0¿ ¿¿¿

{εrθ=12 (∂ v

∂r+1

r∂u∂θ

−vr )=1

2 (φzL

+0−φzL )=0 ¿ {εθz=

12 (1

r∂w∂ θ

+∂ v∂ z )=φ

2 Lr ¿ ¿¿¿

Step 4: Stress (Hooke’s law): {σ zθ=2 Gεθz=

φGL

r ¿ ¿¿¿

Stress solution satisfies most of the B.C., but it still needs to satisfy

∫0

2 π

∫0

R

r2 σ zθdθ dr=∫0

2 π

∫0

R

r 2( φGL

r )dθ dr=(φGL )∫

0

2 π

dθ ∫0

R

r3 dr=( φGL )2 π 1

4R4=Q

which leads to φ= 2LQ

π GR4= LQ

G ( πR4 /2 )= LQ

GJ (which is the same as Mech of Solids I result!)J=πR4 /2 is the polar moment of inertia and G is the shear modulus.

5.2 Bending BeamStep 1: Boundary Conditions

Stress B.C. At (front and back faces)

x=± b2

, −h2≤ y≤h

2, 0≤z≤L

: σ xx=σ xy=σxz=0

At (top and bottom faces)−b

2≤x≤ b

2, y=±h

2, 0≤z≤L

: σ yy=σ yx=σ yz=0

At (side ends) −b

2≤x≤b

2, −h

2≤ y≤ h

2, z=0 or L

: σ zy=σ zx=0 ,

Moments: M xx= ∫

−h/2

h/2

∫−b/2

b/2

yσ zz dxdy=−M,

M yy= ∫−h/2

h/2

∫−b/2

b/2

xσ zz dxdy=0,

Resultant force: P= ∫

−h/2

h/2

∫−b /2

b /2

σzz dxdy=0

Displacement B.C. To eliminate the rigid body motion, we assume to fix the right-end of the beam. Thus

At x= y=z=0 : u=v=w=0 (assume the central point in the right face is fixed)

4

MM

x

y

z

Lb

h

Symmetric line

Week 6 MECH3361/5361

At z=0 , −b

2≤x≤ b

2, −h

2≤ y≤h

2: w=0 (axial elongation starts from here)

Consider the symmetrical plane, i.e. the yz plane: x=0 , −h

2≤ y≤h

2, 0≤z≤L :

u=0

Step 2: Strain: For a “pure bending”, there is no shear force: ε xy=ε xz=ε yz=0 ,

From the plane assumption in beam: ε zz=cy

(note that as per Poisson’s effect: ε xx= ε yy=−ν cy , ν =Poisson’s ratio) Step 3 Stress: For pure bending, shear strain, shear stress, transverse normal stress are zero:

σ xx=σ yy=σ xy=σ xz=σ yz=0

Hooke’s law: σ zz=Eε zz=E(cy ) . Plug σ zz=cEy into B.C. M xx= ∫

−h/2

h/2

∫−b/2

b/2

yσ zz dxdy=−M,

we have: ∫

−h/2

h/2

∫−b/2

b/2

y( cEy )dxdy= cE ∫−h/2

h/2

∫−b/2

b/2

y2dxdy=cEb 13 ( h3−(−h )3

8 )=cE ( bh3

12 )=−M

Thus c=− M

EI (where I=bh3

12 ) and σ zz=− M

EIEy=−M

Iy

(same as Mech Solids I!)

Step 4. Displacements:

{ε xx=∂ u∂ x

=−ν cy ¿ {ε yy=∂ v∂ y

=−νcy ¿ ¿¿¿ Integrate:

{u=−νcxy+ f 1 ( y , z ) ¿{v=−12

ν cy 2+ f 2( x , z )¿ ¿¿¿

We need to determine functions f1, f2, and f3 by using the displacement B.C:

At z=0 , −b

2≤x≤b

2, −h

2≤ y≤ h

2: w=0 f 3( x , y )=0

At x=0 , −h

2≤ y≤h

2, 0≤z≤L :

u=0 f 1( y , z )=0Substitution of u,v,w into the shear strain leads to:

ε xy=∂ u∂ y

+∂ v∂ x

=0 →∂ f 2 ( x , z )

∂ x=νcx

ε yz=∂ w∂ y

+∂ v∂ z

=0 →∂ f 2 ( x , z )

∂ z=−cx

We can have:f 2( x , z )=1

2ν cx2+g1 (z ), f 2 ( x , z )=−1

2cz2+g2( x )

Since f2 must be unique, we can obtain: g1 (z )=−1

2cz 2+d1

and g2 (x )= 1

2ν cx 2+d2

Thus:f 2( x , z )=1

2c (νx2−z2 )+d

where d is a constant. Thus:

v=−12

ν cy 2+ f 2( x , z )=−12

νcy 2+[ 12

c (νx2−z2)+d ]=12

c [ν ( x2− y2)− z2 ]+d

5

Week 6 MECH3361/5361

At x= y=z=0 : u=v=w=0 , therefore d=0, we have: v=1

2c [ ν ( x2− y2 )−z2 ]

Finally:

{u=−ν cxy ¿{v=12

c [ ν ( x2− y2 )−z2 ]¿ ¿¿¿

Tute Q3 (week 5). Show that stress equilibrium equations can be written in displacements asSoln:First check x-direction equilibrium:∂σ xx

∂ x+

∂σ xy

∂ y+

∂ σ xz

∂ z+ρf x=0

Step 1: From strain-displacement relation:

{ε xx=∂ u∂ x

, ε yy=∂ v∂ y

, ε zz=∂ w∂ z ¿ ¿¿¿

Step 2: Apply the generalized Hooke’s law: {σ xx=2 με xx+ λI1

ε

σxy=2 με xy

σ xz=2 με xz

where I 1

ε=ε xx+ε yy+ε zz=∂u∂ x

+ ∂ v∂ y

+∂ w∂ z . Thus

{σ xx=2 μ(∂ u∂ x )+λ (∂u

∂ x+ ∂ v

∂ y+ ∂ w

∂ z )σ xy=2 μ 1

2 ( ∂u∂ y +

∂ v∂ x )

σ xz=2 μ 12 ( ∂ w

∂ x +∂ u∂ z )

Step 3: Apply the Equilibrium Equation

{∂ σ xx

∂ x= ∂

∂ x [2 μ(∂u∂ x )]+ ∂

∂ x [ λ(∂ u∂ x

+∂ v∂ y

+∂ w∂ z ) ]=2 μ ∂2 u

∂ x2 +λ ∂2 u∂ x2 +λ [ ∂2 v

∂ y∂ x+

∂2 w∂ z ∂ x ]

∂ σxy

∂ y= ∂

∂ yμ( ∂u

∂ y+ ∂v∂ x )=μ ∂2 u

∂ y2 +μ ∂2 v∂ x∂ y

∂σ xz

∂ z= ∂

∂ zμ ( ∂w

∂ x+ ∂u

∂ z )=μ ∂2w∂ x∂ z

+μ ∂2 u∂ z2

6

{ ¿ ( λ + μ ) ∂ I1

ε

∂ x + μ∇ 2u + ρ f x = 0 ,

( λ + μ ) ∂ I1

ε

∂ y + μ ∇2 v + ρ f y = 0 ,

( λ + μ) ∂ I1

ε

∂ z + μ ∇2 w + ρ f z = 0.

¿

Week 6 MECH3361/5361

So

LHS=∂ σ xx

∂ x+∂ σ xy

∂ x+∂ σ xz

∂ x+ρf x

¿ [2 μ∂2u∂ x2 + λ∂2 u

∂ x2 +λ (∂2 v∂ y∂ x

+∂2 w∂ z ∂ x )]+[ μ∂2 u

∂ y2 +μ∂2 v∂ x∂ y ]+[μ∂2 w

∂ x∂ z+μ∂2 u

∂ z2 ]+ ρf x

¿ ( λ+μ )∂∂ x

[∂ u∂ x

+∂ v∂ y

+∂ w∂ z⏟

I1ε

]+μ [∂2 u∂ x2 +

∂2 u∂ y2 +

∂2 u∂ z2⏟

∇2u

]+ρf x

= ( λ+μ )∂ I 1

ε

∂ x +μ ∇2 u+ρf x

Similarly we can work out the rest two equilibrium equations in y- and z-directions.Example (Previous quiz question): Aluminium (E=69GPa; =0.33, Al=23.6×10-6) and Mild steel (E=207GPa; =0.30; st=11.3×10-6) bars are assembled to bear a F=60MN load as shown. The increase in the temperature is 10C. It is assumed that the rigid body remains horizontal all the time. Determine the stress and strain states in the vertical bars.SolnStep 1: Extract the compatibility conditions. When the rigid bar remains horizontal: ε Al=ε st

No external force is applied in x- and z-directions: ∴ σxx=σ zz=0

Step 2: F.B.D. This is an indeterminate system, two equations are required to solve for two

unknowns σ Al , yy and σ st , yy ,

Step 3: Force Equilibrium Equation of the rigid bar: ∑ F yy=0 :

∑ F yy=0=2F Al+ Fst−F=(2 A Al ) (−σ Al , yy )+ A st (−σ st , yy )−60×106=0

∴ (2 A Al) σ Al , yy+ Ast σst , yy=−60×106

7

StAl Al

F=60MN

AAl=1m2 ASt=2m2 AAl=1m2

Rigid Bar

x

y

zx

y

z

StAl Al

F=60MN

AAl=1m2 ASt=2m2 AAl=1m2

Rigid Bar

StAl Al

F=60MN

Rigid Bar

FAl Fst FAl

F.B.D.

x

y

zx

y

zx

y

zx

y

z

Week 6 MECH3361/5361

Step 4: Deformation compatibility condition: ε Al , yy=εst , yy

{ε Al , yy=1EAl

[ σ Al , yy−ν (σ Al , xx+σ Al , zz )]+α Al ( ΔT )=σ Al , yy

E Al+αAl ( ΔT ) ¿ ¿¿¿

LHS=RHS:

σ Al , yy

EAl+α Al ( ΔT )=

σ st , yy

E st+α st ( ΔT )

Step 5: Solve for the following two equations {1EAl

σ Al , yy−1Est

σst , yy=α st ( ΔT )−αAl ( ΔT ) ¿ ¿¿¿

{169×109 σ Al , yy−1207×109 σst , yy=11. 3×10−6× (10 )−23 . 6×10−6×(10 ) ¿ ¿¿¿

From (2) ∴ σst , yy=−30×106−σ Al , yy

Sub into (1) 169×109

σ Al , yy−1207×109 (−30×106−σ Al , yy )=11.3×10−6×(10 )−23. 6×10−6×(10 )

(169×109 +1207×109 )σ Al , yy=(11. 3−23. 6 )×10−6×10+−30×106

207×109

σ Al , yy=(11.3−23 . 6 )×10−6×10+−30×106

207×109

( 169×109 + 1

207×109 )=−13 .865 MPa

σ st , yy=−30×106−σ Al , yy=−30×106−(−13 . 865×106 )=−16 .135 MPaStep 6: Stress status

[σ Al ]=[0 0 00 −13 .865 00 0 0 ] [σ St ]=[0 0 0

0 −16 .135 00 0 0 ]

Step 7: Strain statusAluminium:

{ε xx=1E [ σ xx−ν ( σ yy+σ zz) ]+α ( ΔT )=−ν

Eσ yy+α ( ΔT ) ¿ {ε yy=

1E [ σ yy−ν ( σ zz+σ xx) ]+α ( ΔT )=1

Eσ yy+α ( ΔT ) ¿ ¿¿¿

8

Week 6 MECH3361/5361

{ε xx=−νE σ yy+α ( ΔT )=−0 .3

69×109 ×(−13 .865 )×106+23 .6×10−6×10=2 .963×10−4¿ {ε yy=

1E σ yy+α ( ΔT )=1

69×109 ×(−13 .865)×106+23 .6×10−6×10=0 .3506×10−4¿ ¿¿¿

[ εAl ]=[2 .963 0 00 −0 . 3506 00 0 2 .963 ]×10−4

Mild Steel:

{ε xx=−νE σ yy+α ( ΔT )=−0 .3

207×109 ×(−16 .135×106 )+11.3×10−6×10=1 .364×10−4¿ {ε yy=

1E σ yy+α ( ΔT )=1

207×109 ×(−16 . 135×106 )+11.3×10−6×10=0. 3505×10−4¿ ¿¿ ¿

[ εSt ]=[1. 364 0 00 0.3505 00 0 1 .364 ]×10−4

9