Maximum shear stress

• Recall

• In principal system

• Assume • Maximum shear

stress then

( ) ( ) ( )σ σ σ σ σ σ m σXY 1 2 xx 1 2 yy 1 2 zz 1 2 2 1 yz 1 2 2 1 xz 1 2 2 1 xyl l m m n n m n m n l n l n l m l= + + + + + + + +

σ

10

σ σ σ

11 2 1 2

XY 1 2 1 1 2 2 1 2 32 2 2 2 2

1 22

1 1 1 2 2 2

l

l l

l m m n n

l m n l m nm m n n+ + =

= + +

+ + = + + =

σ σ σ1 2 3≥ ≥

max

σ (σ σ )1 (σ σ )2

XY 1 2 1 3 1 2 1 2

1 3

l l l l n n= − = −

τ = −

Examples = 1 1 0

1 1 10 1 1

>> [l,sig]=eig(s)

sig =-0.4142 0 00 1.0000 00 0 2.4142

Maximum shear stress is 1.4142.Why is this reasonable?

Equilibrium when stresses vary from point to point

• Differential element

Differential equilibrium equations• From equilibrium of infinitesimal element

• Why do we have derivatives of stresses but not of body forces?

σσ σ 0 x y zσ σ σ

0x y z

σσ σ 0x y z, , : body force per unit volume in x,y,z directions

xyxx xzx

xy yy yzy

yzxz zzz

x y z

b

b

b

b b b

∂∂ ∂+ + + =

∂ ∂ ∂∂ ∂ ∂

+ + + =∂ ∂ ∂

∂∂ ∂+ + + =

∂ ∂ ∂

Equations of equilibrium repair damage done by kinematic assumptions

• Most commonly used theories start with assumptions on displacements

• Stresses calculated from displacements often do not satisfy equilibrium

• Repair via equilibrium equations improves accuracy

• Stress post-processing in finite element software is possible and desirable

• Check equations in spherical and cylindrical coordinate systems in textbook

Beam theory

• Euler-Bernoulli beam assumptions mean no shear strains

• Equations of equilbrium require shear stresses!

• Cantilever beam under end load P

σ

σ σσ 0 x y y

xx

xy xyxx

My PxyI I

PyI

= =

∂ ∂∂+ = ⇒ = −

∂ ∂ ∂

Deformation of a deformable body

• Deformation will cause a particle P to move from (x,y,z) to (x*,y*,z*)

Strain of a line element• Strain measures are typically based on what

happens to an infinitesimal line element or to angle between to such elements

Displacements• Components of motion

• What is the difference between displacements and deformations?

*

*

*

displacement in x-direction

displacement in y-direction

d

u

isplacement in z-direct

x x

v y y

w nz oz i

=

⇒−

⇒

−

=

⇒

= −

*

*

*

x x u

y y v

z z w

= +

= +

= +

Strain measures

• Length of infinitesimal element before and after deformation

• Engineering strain

• Maginfication factor of a line with direction cosines (l,m,n)

• What does this remind you of?

( ) [ ] ( ) ( )( ) ( ) ( ) ( )

22 2 2

2 2 2 2* * * *

ds dx dy dz

ds dx dy dz

= + +

= + +

[ ]

22

xx xy xz

xy yy yz

xz yz zz

* 10.5 12E E

dsMds

ll m n m

n

ε ε

ε ε εε ε εε ε ε

⎡ ⎤⎛ ⎞= − = +⎢ ⎥⎜ ⎟⎝ ⎠⎢ ⎥⎣ ⎦⎡ ⎤ ⎧ ⎫⎢ ⎥ ⎪ ⎪= ⎨ ⎬⎢ ⎥

⎪ ⎪⎢ ⎥ ⎩ ⎭⎣ ⎦

*E

ds dsds

ε −=

Strain components

.

Reading assignmentSections 2.7-8: Question: Why do we have two kinds of shear

strain?

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