Analysis of Elastic Deformable Support in Euler-Bernoulli Beam Theory - Prof. Martinelli, Appunti di Teoria E Progetto Di Strutture

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STRUCTURAL DESIGN

Prof. Paolo Martinelli

Ing. Daniele Andreola

Architectural Engineering

A.Y. 2023 - 2024

Aggregaon between unitsAggregaon between unitsAggregaon between unitsAggregaon between units

Some structural elements must be combined with others to create a structure that encloses a volume (ex. vercal support systems + horizontal spanning systems). A primary structural unit is a discrete structural element. Its dimensions directly related to the funconal dimensions of the building.

Example: roof deck – beams – trusses – columns – foundaons (they all transfer the load to the element on their right ll the ground).

Types of failures and structural responsesTypes of failures and structural responsesTypes of failures and structural responsesTypes of failures and structural responses

Types of collapseTypes of collapseTypes of collapseTypes of collapse

• Axial forces: o Tension: detachment (members pull apart) o Compression:  Short: Crushing  Long: Buckling

Concrete is much stronger in compression than in tension, that’s why we add steel.

• Bending moments: the stress acts perpendicular to the beam cross secon and vary from compression on the top to tension on the bo&om.
• Shear forces: the stress acts parallel to faces and vary from compression on the top and tension on the bo&om.
• Torsion: twisng acon.
• Bearing: localized crushing (important in mber beams).
• Deflecons

BEAMS: GENERAL PRINCIPLES AND ELASTIC BEAM THEORYBEAMS: GENERAL PRINCIPLES AND ELASTIC BEAM THEORYBEAMS: GENERAL PRINCIPLES AND ELASTIC BEAM THEORYBEAMS: GENERAL PRINCIPLES AND ELASTIC BEAM THEORY

Loads applied transverse to the axis of a member cause the member to bow. Φ is called curvature and is equal to 0 if there is no deflecon.

The amount of curvature (1/R) developed in a beam ar a cross secon depends on the magnitude of the external moment present at the secon. The deformaons in the beam associated with this curvature vary linearly from a maximum tensile elongaon on one face of the beam to a maximum compressive deformaon on the other face.

Where there are no deformaons is the neutral axis.

Elasc beam theoryElasc beam theoryElasc beam theoryElasc beam theory

A beam is a solid generated from a plane area (“cross-secon”) that moves remaining normal to the trajectory described by its centre of gravity, said “geometric axis”.

Hypothesis of the theory:

• It’s a slender beam (L >> b,h).
• Small displacements.
• Plane stresses in the xy plane, bidimensional problem
    0. Thanks to this approach we can refer to the sole beam axis, so it’s a mono-dimensional body (we study a line).

We consider these variables:

• Displacement 
  ,  

• Strain 
  ,   

• Stress 
  ,   

Kinemac aspectKinemac aspectKinemac aspectKinemac aspect


,  ^

This is derived using trigonometry. u(x) and v(x) are the displacements of the barycentre point (since it’s solid, the thickness doesn’t change so it stays v(x)). U(x) is the generalized displacement that collects the displacement of the baricenter.

Since the actual shear sffness is too much, we need to use the shear correcon factor (>1):

G  6 5

 F

%FAF^

B

G = B

∗

For a rectangular cross-secon:

 = % H

2 −^ I

2 H

2 +^ I =

2 J

ℎF

4 −^

FL

G =

4 AF^

5 J

ℎF

4 −^

FL 6 = 6

4 AF^

5 J

ℎF

4 −^

FL

F %

M F NMF

So the sffness is reduced. Combining the definions of the variables and the equaons of equilibrium condion, we have:

= = −@A ′

> = B6∗(R^ − )

(@6 R)R^ = −

(@A R)R^ + B6∗(R^ − ) = −

SB6∗(R^ − )TR^ = −

EulerEulerEulerEuler----Bernoulli beam theoryBernoulli beam theoryBernoulli beam theoryBernoulli beam theory

It’s been developed around 1750. It’s a special case but is more used because is simpler. It’s been broadly applied a=er the construcon of the Eiffel Tower (end of 19th century).

It redefines the constuve relaonship as follows:

U

= = @A

$ = 0 (B6∗^ → ∞)

So, there is infinite shear sffness and no angular slip.

The hypothesis of this theory are the same as before:

• It’s a slender beam (L >> b,h).
• Small displacements.
• Plane stresses in the xy plane, bidimensional problem ( = 0). But also:
• Rigid roto-translaon of the cross-secon.
• Plane secon.
• Cross-secon always orthogonal to the beam axis.

Since $ = 0 → −() + 'W' = 0 →  = 'W'.

Being B6∗^ → ∞, the shear stress is indeterminate from the constuve equaons > = B6∗$ so it has to be derived by equilibrium from the indeterminate equilibrium equaons. The kinemac is described by the field of displacements only as a funcon of the deformed configuraon assumed by the axis line ( elasc line ). The relave elasc energy is: D  =

F +^1

2 @A

F =^1

:F

=F

@A

This is more approximated than the Timoshenko model but is sufficiently accurate for us. We have:

⎧= = −@A

F>

F^

;/<7 1<$+$$+. ,*X

F=

F^ + 2 = 0^ ;/<7 .&+,+%/+

→ −YZ

[]

[^^ + _(^) = `

This is the Euler-Bernoulli equaon , that describes the relaonship between the beam’s deflecon and the applied load.

MOHR ANALOGYMOHR ANALOGYMOHR ANALOGYMOHR ANALOGY

It’s a calcula on approach to define deflec ons and rota ons of the beam.

Mohr theorem : the elas c line of a beam coincides with the fic ous moment diagram M* due to the fic ous load q*=M/EI equal to the curvature of the real beam.

The elas c line is the expression of the deforma on of the beam (small deforma ons). The curvature in small displacements is the second deriva ve of the deforma on v (while the first deriva ve is the slope
):

 = ^ = −

Where I is the iner a to the sec on (related to the shape), E is the young modulus (related to the material), EI combined is the s ffness of the sec on. The deforma on v depends on x.

In a real beam:

 ^ = −^

 

In a fic ous beam:

∗ ^ = −

We change the variables but they’re the same expression (if I know the solu on for one, I can find the other).

1. You have to know the moment so you know the curvature (real-second).
2. You can find the fic ous load (fic ous-second).
3. You can find its rela ve moment M* (fic ous-first).
4. You can now find the deflec on for the real beam (real-first).

∗

  

First corollary of the theorem:  = ∗^!

"# !"^ = 

∗

Second corollary of the theorem: !#! =
and !

∗ ! = 

We also need to change boundaries condi ons between the two worlds (real/fic ous):

REAL FICTITIOUS External hinge v=0
≠ 0 External hinge M=0 V≠ 0 Fixed end v=0
= 0 Free end M=0 V= 0 Free end v≠ 0
≠ 0 Fixed end M≠ 0 V≠ 0 External slider v≠ 0
= 0 External slider M≠ 0 V= 0 External middle hinge vL=vR=0
& =
' ≠ 0 Internal hinge ML=MR= 0 VL=VR≠ 0 Internal hinge vL=vR≠ 0
& ≠
' External middle hinge ML=MR ≠ 0 VL≠VR Internal slider vL≠vR
& =
' External middle slider ML≠MR VL=VR External middle slider vL=vR
& =
' = 0 Internal slider ML=MR VL=VR= 0

* ∗/^0

2

3

 = 0 → @ * ? A

B

2

3

 = 0 → * ? A

B 

2

3

+ * A

B

2

3

Where we call the first integral
?? and the second one
?3. So it becomes:

?? +
?3 = 0 → = −

This is called compa bility equa on.

In our case, from the tables:
?? = (^) C^2 and
?3 = − (^) ED2 → = D

If we consider the shear deformability, we obtain:

=

F

1 − 6J 3

1 + 3J 3

Where J 3 = (^) MN2L" → < D^.

If we consider the case of imposed rota on/displacement at the fixed end, we just need to change
?3 using the tables:

?? = P → =

P

Q

Q

In sta cally determined structures, temperature varia ons or imposed movements have no consequences in internal ac ons. We now solve a generic case:

It has 3 degrees of freedom and 6 degrees of constraint.

?? =
=

Q

Q

So it’s symmetric.

The compa bility equa ons of the end rota ons are: R

???^ +
?^ +
?3^ = 0 ?? +
^ +
^3 = 0 The flexibility matrix is: S = T

(^)?
U^ which is symmetric and the main diagonal is posi ve.

Case with an external load:

? =^ = −

9Q

Case with an imposed rota on:

?3 = −P
3 = 0

→ (^)? = W 2 P → = −  2 P

Case with an imposed displacement:

?3 = − X 2
3 = X 2

→ (^)? =  2 YCX 2 Z → = −  2 YCX 2 Z

A compa bility equa on is:

[

C?
C

W?
W

?C
?W

C
W

CC
CW

WC
WW

\ ]

?

C W

^ + ]

C

W

^ = ]

^

Where the green line is always posi ve, the yellow lines are symmetric and the blue lines are equal to zero.

?? = C^2
? = E^2

Q

3 +^

Q

3 =^

2Q

9Q

R

?? +
+
3 = 0 →^?^ =

9Q = −

9Q

Now it’s a determined system so we can solve it with equilibrium:

=C =

Elasc deformable support (ROTATIONAL FLEXIBILITElasc deformable support (ROTATIONAL FLEXIBILITElasc deformable support (ROTATIONAL FLEXIBILITElasc deformable support (ROTATIONAL FLEXIBILITY)Y)Y)Y)

 = J

Where
is the rota on (unitless) and k is the s ffness in N/m (∞ for clamping and 0 for hinge). Se = ?f (0 for clamping and ∞ for hinge) in m/N.

We consider an example:

Case 1:
?? = (^) C^2 + gh
(^)? = (^) E^2

Case 2:
? = (^) E^2
= (^) C^2

Case 3:
3 = i

" ?E

Assuming Se = (^) C^2 : R

???^ +
?^ +
?3^ = 0 ?? +
^ +
^3 = 0

→ ? = W3? 9Q = − ?3? 9Q

Elasc deforElasc deforElasc deforElasc deformable support (TRANSLATIONAL FLEmable support (TRANSLATIONAL FLEXIBILITY)mable support (TRANSLATIONAL FLEmable support (TRANSLATIONAL FLEXIBILITY)XIBILITY)XIBILITY)

F = Jk @ Where @ is the deflec on and Jk is the s ffness (∞ for no support and 0 for rigid support). Sk =

Jk

We consider an example:

The kinema c is: From equilibrium:

??^ =? 2 YSk? 2 Z = l 2 m" where Sk? 2 = @

Case 1:
?? =
??^ + (^) C^2 = l 2 m" + (^) C^2
(^)? = (^) E^2 − 2 l 2 "m

Case 2:
? = (^) E^2 − 2 l 2 m"
= (^) C^2 + 2 2 l"m

Case 3:
?3 = i

" ?E +^

i lm 2
^3 =^

i2" ?E − 2^

i lm 2

Assuming Sk = 2

n C:^?^ = −^

? o 9Q^ = −^

? o 9Q

3 + ^

! - = 3^

Because we called (^)  = &*.

6 + ^

! = −^

! - = −3&5*^ ,1 − 6

! - = −3&5*^ ,

Because we called (^) 6 = &5* and 768^9 = / 5.

When k 0 becomes smaller, the contribu on of shear disappears.

Different cross seconsDifferent cross seconsDifferent cross seconsDifferent cross secons

E and G depend on the material; I and A* depend on the geometry.

Poisson coefficient: : = 0,15 ÷ 0,3 where 0,15 is usually concrete and 0,3 usually steel. This coefficient is related to the expansion on the orthogonal direc on. 
= 2.1 + :0 ≅ 2,3 ÷ 2,6^ we can take 2,5 (5/2).  is tabled for each cross-sec on geometry.

For a rectangular secon :

 =

 -^ =^

4B

Where λ is the slenderness and is equal to B = (^) C.

For a double T secon : !∗^ ≅ .ℎ − D0E ≅ 0,33!

 ≅ 2?D ,

ℎ − D

2?Dℎ

Where 2?D is the horizontal area and C7G^ is the distance from the barycentre.

/% =

 =^

4B

From k to ki we change the asymptote. For the rectangular sec on (thin line): If B > 4 → slender beam If B < 4 → squat beam

For the double T sec on (thick line): If B > 9 → slender beam If B < 9 → squat beam

DISPLACEMENT METHODDISPLACEMENT METHODDISPLACEMENT METHODDISPLACEMENT METHOD

It’s a method to solve stacally indeterminate structures, based on the Principle of Virtual Work, specifically on its sufficient condion: while the force method had the unknown of forces and moments, now the unknown are displacements and rotaons, so instead of downgrading the structure, now we add constraints.

We consider an example:

The structure has 3 degrees of freedom and 5 degrees of constraints, so it makes 2 degrees of redundancy. We now add constraints in rigid block, which is a parcular rotaonal spring with infinite sffness ( = ∞).

This respects compability (in the force method respected equilibrium).

It’s a superposion of 3 cases:

Case 1

Case 2

Case 3

We apply the Principle of Virtual Work (opposite of the force method):

• Kinemac quanty → Auxiliar
• Stac quanty → Real

^ 

Where the le3 side is the internal work, and the right side is the external. S represents the structure, 

represents any possible variaon of ,   is the curvature, P is the real load (force),  is the displacement

and this is mulplied by  because otherwise it would be unitary.

BEAM ON ELASTIC SOILBEAM ON ELASTIC SOILBEAM ON ELASTIC SOILBEAM ON ELASTIC SOIL

It’s a mechanical model to study beams supported by soil. The equaon of the beam of 4th^ order, from Euler- Bernoulli, is:

−

^ +^ ^ = 0

Where EI is the sffness, v is the deformaon and P is the load.

We need to include the contribuon of the soil, so we consider a poron of the beam:

The soil is represented by translaonal springs.

We can isolate a small poron of the beam:

r is the reacon, connuous because the soil is everywhere. It can also be wri%en as:

Because the force is proporonal to displacement  =  .

The unit of measurement of k(x) is N/m^3 , because usually is N/m but soils is in 3 direcons, b and v are both in meters, and r(x) is in N/m.

The total load is the difference between the load and the reacon, so that we include the contribuon of the soil:  =   −  

For foundaons, our field of applicaon, k(x) is constant because the type of soil is generally constant, even though in some cases the soil can change. The spring can’t transfer shear stress, so in this model we can’t reproduce cohesive soil (the one that transfers shear).

Typically, we use bilateral springs, that work both in compression and in tension (while unilateral work only in one direcon). We’ll verify that v > 0 , so the springs work in compression.

We now put the new P(x) in the Euler-Bernoulli equaon:

^

^

Where 4^ =  →  = ^  (^)  , so it’s basically a rao between soil sffness and beam sffness.

This soil model is called WINKLER MODEL.

We obtain:

 =  + 

Where vg is the v general (general integral of the associated homogeneous), and vp is the v parcular (parcular integral that depends on the load applied).

The soluon is:  = !"#$%& '()  + * +,' - + !#$%. '()  + / +,' -

 =

Where A, B, C and D are integraon constants, that can be found with 4 boundary condions.

We can define the wavelength λ , used to disnguish long and short beams, as:

λ = 22 → λ =

 =^



Boundary condi ons (B.C.)Boundary condi ons (B.C.)Boundary condi ons (B.C.)Boundary condi ons (B.C.)

We need to find the boundary condions at the extremies of the beam. We look for these informaons:

• Displacement ( v )
• Rotaon ( v’ )
• Moment ( v’’ ) → ^44 = − (^) ^5
• Shear ( v’’’ ) → ^444 = − (^) ^6

(Non sono da sapere a memoria tu i casi ma bisogna capire il meccanismo per riprodurlo.)

()  = 0 → 78 = 0 → 

9 = 0 → ^44 = 0 !! !)

9 = 0 → ^44 = 0 !! !)

i )  = 0 → ;8 = −^

< = → 

= ^4 = 0

,>?>(,) (' @!, A, 'BCC!>B

9 = 0 → ^44 = 0

It’s a symmetric problem, so we consider x in the axis of symmetry and just half of shear contribuon (then we’ll sum le7 and right).

i )  = 0 → ;8 = −^ → 

 9 = 0 → ^44 = 0

()  = 2: → 78 = 0 → 

9 = 0 → ^44 = 0 !! !)

i )  = 0 → ;

 = 0 ?)>( − 'BCC!>(+ D,:!C

9 = − E= → ^44 = =E

9 = 0 → ^44 = 0

i )  = 0 → ;

8 = 0 → ^444 = 0

9 = − → ^44 = E

9 = 0 → ^44 = 0 !! !)