Docsity
Docsity

Prepare for your exams
Prepare for your exams

Study with the several resources on Docsity


Earn points to download
Earn points to download

Earn points by helping other students or get them with a premium plan


Guidelines and tips
Guidelines and tips

Elasticity: Equilibrium, Strain-Displacement, and Stress-Strain Relations, Exercises of Structures and Materials

An in-depth analysis of elasticity, focusing on equilibrium, strain-displacement, and stress-strain relations. It covers the concepts of strain compatibility equations, hooke's law, and the elasticity tensor. The document also discusses the importance of static determinacy and the implications for calculating stresses.

Typology: Exercises

2011/2012

Uploaded on 07/26/2012

raam
raam 🇮🇳

4.4

(14)

90 documents

1 / 14

Toggle sidebar

This page cannot be seen from the preview

Don't miss anything!

bg1
Let’s first review a bit
from Unified, saw that there are 3 basic considerations in elasticity:
1. Equilibrium
2. Strain - Displacement
3. Stress - Strain Relations (Constitutive Relations)
Consider each:
1. Equilibrium (3)
Σ Fi = 0, Σ Mi = 0
Free body diagrams
Applying these to an infinitesimal element
yields 3 equilibrium equations
Figure 4.1 Representation of general infinitesimal
element
pf3
pf4
pf5
pf8
pf9
pfa
pfd
pfe

Partial preview of the text

Download Elasticity: Equilibrium, Strain-Displacement, and Stress-Strain Relations and more Exercises Structures and Materials in PDF only on Docsity!

from Unified, saw that there are 3 basic considerations in elasticity:Let’s first review a bit …

Equilibrium

Strain - Displacement

Stress - Strain Relations (

Constitutive

Relations)

Consider each:

Equilibrium

F

i = 0,

M

i = 0

Free body diagrams

yieldsApplying these to an infinitesimal element

equilibrium equations

Figure 4.

elementRepresentation of general infinitesimal

∂σ

11

∂σ

21

∂σ

31

  • f

1

=

∂ y 1 ∂ y 2 ∂ y 3

∂σ

12

∂σ

22

∂σ

32

  • f

2

=

∂ y 1 ∂ y 2 ∂ y 3

∂σ

13

∂σ

23

∂σ

33

  • f

3

=

∂ y 1 ∂ y 2 ∂ y 3 ∂ ∂ +

m mn

n

y

f

Strain - Displacement

Based on geometric considerations

Linear considerations

I.e.,

small

strains only --

we will talk about

large strains later

(and

infinitesimal

displacements only)

Stress - Strain

mn

E

mnpq

pq

Static Determinance Let’s review the “4th important concept”:^ we’ll come back to this …

There are there possibilities

(as noted in U.E.)

a. A structure is not sufficiently restrained

(fewer reactions than d.o.f.)

freedom degrees of

DYNAMICS

b. Structure is exactly (or “simply”) restrained

(# of reactions = # of d.o.f.)

determinate)STATICS (statically

Implication

: can calculate stresses via

equilibrium (

as done in Unified

c. Structure is overrestrained

(# reactions > # of d.o.f.)

INDETERMINATESTATICALLY

simultaneously …must solve for reactions

with stresses, strains, etc.

in this case, you must employ the stress-strain equations

--> Overall, this yields for elasticity:

15 unknowns

and

15 equations

6 strains =

ε mn

3 equilibrium (

σ )

6 stresses =

σ mn

6 strain-displacements (

ε )

3 displacements = u

m

6 stress-strain (

σ

ε )

IMPORTANT POINT:

equilibrium (equilibrium).material) as they depend on geometry (strain-displacement) and The first two sets of equations are “universal” (independent of the

Only

the stress-strain equations are

dependent on the material.

Note that deformations (u

m ) must be continuous

doesn’t makesingle-valued functions for continuity. (or it

physical sense

Step 2

: Now consider the case where there are gradients in the strain field

∂ε

12

∂ε

12

∂ y 1 ∂ y 2

∂ Take derivatives on both sides:structure This is the most general case and most likely in a general 2 ε 12

1  ∂ 3 u 1 ∂ 3 u 2 

2

2

y

y 2

y y 2 y y 2

1

1

1

Step 3

: rearrange slightly and recall other strain-displacement equations

∂ u 1 = ε 1

u 2

ε 2

∂ y 1 ∂ y 2

2 ε 12

∂^

2 ε 11

2 ε 22

(^) 

2

y

y 2

y 2

y (^12)



1

they are all related to the 3 displacements. So, the gradients in strain are related in certain ways since

Same for other 5 cases …

Let’s now go back and spend time with the …

Stress-Strain Relations and the Elasticity

Tensor

Law: The basic relation between force and displacement (recall 8.01) is Hooke’sgeneralize it to encompass all cases. In Unified, you saw particular examples of this, but we now want to

F = kx

spring constant (linear case)

2212

(symmetry in switching

last two

indices

  1. From thermodynamic considerations

(1st law of thermo)

E

mnpq

= E

pqmn

(symmetry in switching

pairs of

indices

Also note that:

Since

σ mn

=

σ nm

are only 6!

, the apparent 9 equations for stress

 With these symmetrics, the resulting equations are: σ 11

(^) 

E

1111

E

1122

E

1133

2E

1123

2E

1113

2E

1112

ε 11

(^) 

E

1122

E

2222

E

2233

2E

2223

2E

2213

2E

2212

^

ε 22

^

σ 22

^

σ 33 (^) 

E

1133

E

2233

E

3333

2E

3323

2E

3313

2E

3312

ε 33

(^) 

σ 23

^

E

1123

E

2223

E

3323

2E

2323

2E

1323

2E

1223

^

ε 23

^

σ 13 (^) 

E

1113

E

2213

E

3313

2E

1323

2E

1313

2E

1213

ε 13 (^) 

σ 12 (^) 

E

1112

E

2212

E

3312

2E

1223

2E

1213

2E

1212

ε 12 (^) 

Results in

independent components of the elasticity tensor

Along diagonal (6)

Upper right half of matrix (15)

[don’t worry about 2’s]

Also note

: 2’s come out automatically…don’t put them in

For example:

σ 12

= … E

1212

ε 12

  • E

1221

(^) ε 21 (^) …

= … 2E

1212

ε 12 …

These E

mnpq

can be placed into

3 groups

Extensional strains to extensional stresses

E

1111

E

1122

E

2222

E

1133

E

3333

E

2233

e.g.,

σ 11

= … E

1122

ε 22 …

Shear strains to shear stresses

E

1212

E

1213

E

1313

E

1323

E

2323

E

2312

Isotropic

Cubic

“Transversely Isotropic”*

Tetragonal

Orthotropic

Monoclinic

Anisotropic

# of Independent

Components of E

mnpq

Material Type

Useful

Engineering Materials

LaminatesComposite Basic

Composite

Ply

Metals

(on average)

Good Reference

BMP, Ch. 7

*not in BMP

For

orthotropic material

s (which is as complicated as we usually get),

there are no coupling terms in the

principal axes of the material

When you apply an extensional stress, no shear strains arise

e.g., E

1112

(total of 9 terms are now zero)

When you apply a shear stress, no extensional strains arise

(some terms become zero as for

previous condition)

(stresses) in another planeShear strains (stresses) in one plane do not cause shear strains

( E

1223

, E

1213

, E

1323

With these additional terms zero, we end up with

independent

components:

and the equations are: