EXAMPLES to Lecture no.3 Example No.1:

EXAMPLES to Lecture no.3

Example No.1:

fundamental solution of the second kind.

Solve a doubly-infinite prismatic beam, resting on the Winkler subsoil, loaded by a moment M

o

concentrated at ξ

o

=0; this is the dual case to the one solved for a concentrated force P

in Lecture no.2

Remark: such situations refer to a case of vast foundation slab connected with a (fixed) wall that transmits bending moment Mo or to a beam connected with a (fixed) column; only vertical forces P are analysed in the students’ Project No.1.

I Method – using the general solution for ξ > 0.

1. Let y(ξ) = e

^–ξ

[C

1

cosξ+C

2

sinξ] + e

^+ξ

[C

3

cosξ+C

4

sinξ]. From y(+∞)=0, there is C

3

= C

4

= 0.

2. Antisymmetry y(- ξ ) = -y( ξ ) yields y(-0) = y(0) = -y(0), so y(0) = 0.

Thus C

1

= 0 and y( ξ ) = C

2

e

^–ξ

sin ξ .

3. For the right-hand limit ξ→ 0+ there is M(ξ) → M

o

/2 (as well as M(ξ=0-) = –M

o

/2).

But for ξ > 0: M(ξ) = -EI y” / (L

W

)

²

= -EI (L

W

)

^–2

C

2

[e

^–ξ

sinξ]” = -EI (L

W

)

^–2

C

2

[-2e

^–ξ

cosξ].

4. Result: substituting ξ =0+ we get the solution C

2

= M

o

(L

W

)

²

(4EI)

^–2

, so C

2

= M

o

(BC)

^–1

(L

W

)

^–2

. Finally,

y( ξ ) = M

o

(BC)

^–1

(L

W

)

^–2

e

^–ξ

sin ξ and clearly the function y( ξ ) jest odd; y(- ξ ) = -y( ξ );

in particular

M(ξ) = -EI y”/(L

W

)

²

= (M

o

/2) e

^–ξ

cosξ,

Q(ξ) = -EI y”’/(L

W

)

³

= -M

o

(2L

W

)

^–1

[e

^–ξ

(sinξ+cosξ)].

II Method – using a pair of vertical forces ± V and the solution of the first kind y

1

(x) for P = 1.

1. Vertical concentrated force V is applied at x = 0+dx/2 > 0 and the opposite force (–V) is applied at x = 0–dx/2 < 0. Assume V = M

o

/dx. For every dx, the both forces generate the same moment M

o

, therefore the same value M

o

is kept in the limit passage dx → 0.

2. Using the superposition low y(x) = V [y

1

(x-dx/2) – y

1

(x+dx/2)] = M

o

[y

1

(x-dx/2) – y

1

(x+dx/2)]/dx → M

o

[-dy

1

/dx] = -M

o

dy

1

/d ξ (L

W

)

^–1

= -M

o

(2BC)

^–1

(L

W

)

^–2

[e

^–ξ

(cos ξ +sin ξ )]’,

where y

1

is the known solution for the vertical concentrated force P=1 (first kind).

3. Result: for ξ > 0 there is

y(ξ) = M

o

(BC)

^–1

(L

W

)

^–2

e

^–ξ

sinξ and clearly the function y(ξ) jest odd, y(-ξ) = -y(ξ).

III Method – different versions of the Bleich method

1. In the standard version, the Bleich method uses the fundamental solution of the first kind:

cos ξ e 8 4 ) P ( ξ Q ), cos ξ (sin ξ

e ξ 8 2 ) PL ( ξ M

) ( ξ y C B ) ( ξ r ), sin ξ (cos ξ e ξ

BCL 2 ) P ( ξ y

w ξ w

−

−

=

− −

−

=

=

− +

=

2. It suffices to find the solution y(ξ) on ξ > 0, as this is the odd function (antisymmetry):

y(-ξ) = -y(ξ); therefore in particular y(0)=0, because y(-0) = y(0)= -y(0).

3. Since we „ignore” the left part of the beam, so on this part any loading can be accepted, provided that: y(0+0)=y(0)=0 and M(0+0) = M

o

/2 (limit from the right side).

4. In the standard version we use two virtual forces T

1

, T

2

on the left from ξ=0.

There are de facto 4 unknowns – two values of the forces and two positions of the forces, but there are only two boundary conditions at ξ=0; therefore, the positions (for example) can be taken arbitrary, so as to simplify calculations.

Mo

C

y(ξ)

ξ 0

where ξ > 0 denotes the

distance from the load P.

5. If the boundary conditions at ξ=0 are formulated in Q and M, the it is recommended to assume the forces at the dimensionless distances π/4 i π/2 on the left from, because of Q(π/2)=0 and M(π/4)=0 in the fundamental solution (of the first kind). This means a simplification, because

^both forces do not interfere with each other (two separated equations for T1, T2 each instead one coupled system 2x2). Most students would follow this standard way which is correct and good (but … not very good). In details, the governing equations are as follows, for the assumed

π/2 and π/4:

2 )) M 2 / cos( π ) 2 / (sin( π ) 2 / ( π e 8 2

L )) T

4 / cos( π ) 4 / (sin( π ) 4 / ( π e 8 2

L ) T

0 0 ( M

0 )) 2 / sin( π ) 2 / (cos( π 2 / e π BCL 2 )) T 4 / sin( π ) 4 / (cos( π 4 / e π BCL 2 ) T 0 ( y

o w

2 w

1

w 2 w

1

=

− −

− ⋅

− −

− ⋅

= +

=

− + +

− +

=

Solution:

4 / e π L 2 M 4 / e π 2 T T 1

2 , / e π L 2 M T

W 2 o

1 W

2

= − ⋅

o

⋅ + = − ⋅ ⋅ − = ⋅ ⋅ + .

hence for ξ > 0 the function is:

)) 2 / π sin( ξ ) 2 / π (cos( ξ ) 2 / π ( ξ BCL e 2

T

)) 4 / π sin( ξ ) 4 / π (cos( ξ ) 4 / π ( ξ BCL e 2 ) T ( ξ y

w 2 w 1

+ +

+ + ⋅

+ −

+ +

+ +

+ ⋅

= −

We can reduce this a little bit basing on trigonometric tables:

y(ξ) = M

o

·(BC)

^-1

·(L

W

)

^-2

·e

^-ξ

·sin(ξ) and clearly the function y(ξ) jest odd, y(-ξ) = -y(ξ).

6. Coming back to the “very good” version mentioned above, the boundary conditions at ξ

o

=0 are formulated in y and M, therefore it is better to put the virtual forces T

1

, T

2

at π/4 and 3π/4 instead π/4 and π/2, because M(π/4)=0, y(3π/4)=0; this way, both equations are separated.

Indeed, the force T

2

at 3π/4 causes at ξ=0 zero settlement (as it should be) but it is free in moment shaping; this way, the second force T

1

= 0 at π/4 is … unnecessary (T

1

=0). Therefore:

-T

2

L

W

e

^-3π/4

[sin(3π/4)-cos(3π/4)]/4 = M

o

/2, so T

2

= -√2·M

o

·e

^+3π/4

·(L

W

)

^-1.

The solution for dla ξ > 0 is simply:

y(ξ) = -√2·M

o

·e

^+3π/4

·(L

W

)

^-1

·(2BCL

W

)

^-1

·e

^-(3π/4+ξ)

·[cos(3π/4+ξ)+sin(3π/4+ξ)], so

y(ξ) = M

o

·(BC)

^-1

·(L

W

)

^-2

·e

^-ξ

·sin(ξ) and clearly the function y(ξ) jest odd, y(-ξ) = -y(ξ).

Testing question (about finite beams, in general):

would the "new Bleich method" be correct for finite beams using 4 fictional moment loads M

i

based on the above solution, instead of 4 vertical forces T

i

?

Reply:

Certainly YES. Virtual loads on the fictitious part of the beam can be quite arbitrary, e.g. three concentrated forces and one concentrated moment load, etc. The final solution over the entire length of the finite beam will be identical as in the traditional Bleich method, because it meets the same E-B differential equation and the same boundary conditions; the theorem on the uniqueness of the solution of the ordinary differential equation says that there is only one such solution, no matter how derived.

I must admit that this unlimited possible selection of virtual loads on the fictitious beam extension is

not intuitive and is surprising - but it's better to believe in mathematics than in intuition

^☺.

Example No.2: Consider a vertical pile in the Winkler subsoil.

Now the beam is rotated, the „springs”

are horizontal, but there is no difference from mathematical point of view.

Assume a constant subsoil coefficient C = const

¹

. If the pile is „very long”, say L >3÷4⋅L

W

, then the beam is (almost) half-infinite on 0 ≤ ξ ≤ +∞,

there are 2 boundary conditions at ξ = 0.

To solve the beam means that y(ξ) is to be found, ξ = x/L

W

≥ 0, because

r(x) = B⋅C⋅y(x) = B⋅C⋅y(ξ) …….. subsoil reaction [kN/m], ϕ(x) = dy(x)/dx = dy(ξ)/dξ /L

W

…….. rotation angle [rad],

M(x) = -EI⋅d

²

y(x)/dx

²

= -EI⋅d

²

y(ξ)/dξ

²

/ (L

W

)

²

…….. bending moment kNm], Q(x) = -EI⋅d

³

y(x)/dx

³

= -EI⋅d

³

y(ξ)/dξ

³

/ (L

W

)

³

…….. shearing force [kN].

Solving:

Assume the boundary conditions at ξ

o

=0 : M(0+0) = 0, Q(0+0) = -H and all above functions vanishing if ξ → ∞.

I Method – using the general solution for ξ > 0.

1. As always, y(ξ) = C

1

⋅e

^-ξ

⋅cosξ + C

2

⋅e

^-ξ

⋅sinξ + C

3

⋅e

^+ξ

⋅cosξ + C

4

⋅e

^+ξ

⋅sinξ

. 2. C

3

= 0 and C

4

= 0 because of vanishing behaviour for ξ → ∞.

3. M(0+0) = 0 so: d

²

y(ξ)/dξ

²

= C

1

⋅2⋅e

^-ξ

⋅sinξ - C

2

⋅2⋅e

^-ξ

⋅cosξ = 0 for ξ=0, thus C

2

= 0 4. Q(0+0) = -H so: d

³

y(ξ)/dξ

³

= C

1

⋅(-2)⋅e

^-ξ

⋅sinξ + C

1

⋅2⋅e

^-ξ

⋅cosξ = H⋅(L

W

)

³

/EI for ξ=0,

thus C

1

= H⋅(L

W

)

³

/(2⋅EI) = 2⋅H/(B⋅C⋅L

W

) 5. Result:

y(ξ) = 2⋅H/(B⋅C⋅L

W

)⋅e

^-ξ

⋅cosξ.

6. In particular, horizontal displacement of the pile-head equals y(0) = C

1

= 2⋅H/(B⋅C⋅L

W

).

IIa Method – the standard Bleich method:

1. The half-infinite beam requires 2 Bleich forces:

T

1

at ξ

1

= π/4 on the left from the load H; this is insensitive to M(0), can correct Q(0), T

2

at ξ

2

= π/2 on the left from the load H; this is insensitive to Q(0), can correct M(0).

This is the standard position of virtual forces, because boundary conditions are formulated in M,Q.

The governing equations for T

1

, T

2

on doubly-infinite beam:

M(0) = -H⋅L

W

/4⋅e

^-0

(sin0-cos0) – T

1

⋅L

W

/4⋅e

^-π/4

(sinπ/4-cosπ/4) – T

2

⋅L

W

/4⋅e

^-π/2

(sinπ/2-cosπ/2) = 0, so T

2

= H⋅e

^+π/2

Q(0+0) = -H/2⋅e

^-0

cos0 – T

1

/2⋅e

^-π/4

cosπ/4 – T

2

/2⋅e

^-π/2

cosπ/2 = -H, so T

1

= H⋅√2⋅e

^+π/4

.

2. Therefore: y(ξ) = H/(2BCL

W

)⋅e

^-ξ

⋅(cosξ + sinξ) + H⋅√2⋅e

^+π/4

/(2BCL

W

)⋅e

^-(ξ+π/4)

⋅[cos(ξ+π/4) + sin(ξ+π/4)] + H⋅e

^+π/2

/(2BCL

W

)⋅e

^-(ξ+π/2)

⋅[cos(ξ+π/2) + sin(ξ+π/2)] , or simply

y(ξ)= 2⋅H/(B⋅C⋅L

W

)⋅e

^-ξ

⋅cosξ ,

because cos(ξ+π/4) + sin(ξ+π/4) = √2⋅cosξ oraz cos(ξ+π/2) + sin(ξ+π/2) = -sinξ + cosξ.

IIb Method – another version of the Bleich method (for more experienced users)

1. In the fundamental solution, the force P=H causes a discontinuity of Q=±H/2 at ξ = 0±0. The right-hand value is Q(0+0) = -H/2, so the required boundary condition Q(0+0) = -H will be satisfied if taking the force P=2H instead of H at ξ = 0; the moment is non-zero, however.

1 acceptable for cohesive soils;

for noncohesive ones, an increase of C with depth is more suitable; we will come to this case in the next example.

L C

y(ξ)

H ξ→∞

0

2. Next, it is necessary to reduce the moment M(0) to 0, but keeping the condition for Q which is already correct. For such a purpose, a virtual force T placed on the left from P=2H at ξ = π/2 is usefull, cos(π/2), which does not change Q=0. The condition M(0) = 0 due to the action of both 2H and T gives T = 2⋅H⋅e

^+π/2

.

3. Result:

y(ξ) = 2H/(2BCL

W

)⋅e

^-ξ

⋅(cosξ + sinξ) + 2H⋅e

^+π/2

/(2BCL

W

)⋅e

^-(ξ+π/2)

⋅[cos(ξ+π/2) + sin(ξ+π/2)], so y(ξ) = 2⋅H/(B⋅C⋅L

W

)⋅e

^-ξ

⋅cosξ,

since sin(ξ+π/2) = cosξ, cos(ξ+π/2) = -sinξ .

IIc Method – the standard Bleich metod (for still more experienced users):

1. To tell the truth, the forces H or 2H used in IIa and IIb at ξ = 0 seem to be overcomplicated, because it is possible to solve the problem with no force applied at ξ = 0, like for M

o

/2 in the Example 1. Indeed, use 2 standard Bleich forces at π/4, π/2:

M(0) = -T

1

⋅L

W

/4⋅e

^-π/4

(sinπ/4-cosπ/4) – T

2

⋅L

W

/4⋅e

^-π/2

(sinπ/2-cosπ/2) = 0, so T

2

= 0,

Q(0+0) = -T

1

/2⋅e

^-π/4

cosπ/4 – T

2

/2⋅e

^-π/2

cosπ/2 = -T

1

/2⋅e

^-π/4

cosπ/4 – 0 = -H, so T

1

= 2√2⋅H⋅e

^+π/4

at π/4.

2. As it is seen, there is only 1 virtual force – neither 3 as in IIa, nor 2 as in IIb; therefore this approach should be recommended as the simplest one.

Result:

y(ξ) = H⋅2√2⋅e

^+π/4

/(2BCL

W

)⋅e

^-(ξ+π/4)

⋅[cos(ξ+π/4) + sin(ξ+π/4)], so y(ξ) = 2⋅H/(B⋅C⋅L

W

)⋅e

^-ξ

⋅cosξ ,

since cos(ξ+π/4) + sin(ξ+π/4) = √2⋅cosξ.

Conclusion, in fact the same as in the Example1:

Here, two virtual forces T

i

are different than in IIa , IIb (though even at the same points)- despite this, the solutions are the same on the entire half-line ξ ≥ 0; no wonder, the uniqueness theorem should be recalled:

If a function y( ξ ):

1) satisfies on the interval (a,b) a linear ordinary differential equation with constant coefficients, 2) satisfies all boundary conditions at a and b,

then there exists one and only one such solution on the interval (a,b).

Of course, due to the linearity of the equation, we use constantly the superposition law which:

if y

i

( ξ ) all satisty a linear differential equation

²

, then the sum y( ξ )= Σ y

i

( ξ ) also satisfy this equation.

Indeed, if d

⁴

y

1

(ξ)/dξ

⁴

+4y

1

(ξ)=0 and d

⁴

y

2

(ξ)/dξ

⁴

+4y

2

(ξ)=0,

so for y

1

(ξ)+y

2

(ξ) there is also d

⁴

[y

1

(ξ)+y

2

(ξ)]/dξ

⁴

+4[y

1

(ξ)+y

2

(ξ)]=0, because 0+0=0.

2they do, because in the Bleich method we use the fundamental solutions of the Euler-Bernoulli equation, though shifted generally

Let’s check this as an exercise.

1.

Do not take into account H,M

o

at ξ ⁼⁰

Follow the standard example IIc using only 2 virtual forces T

2

, T

1

and the doubly-infinite beam;

boundary conditions at ξ = 0 are taken:

M(0) = -T

1

⋅L

W

/4⋅e

^-π/4

(sinπ/4-cosπ/4) – T

2

⋅L

W

/4⋅e

^-π/2

(sinπ/2-cosπ/2), and must be M(0) = M

o

, so immediately T

2

= -4M

o

/L

W

⋅e

^+π/2

.

Q(0) = -T

1

/2⋅e

^-π/4

cosπ/4 – T

2

/2⋅e

^-π/2

cosπ/2, and must be Q(0) = -H, so T

1

= 2√2⋅H⋅e

^+π/4

. The result is composed of 2 terms of course:

y(ξ) = 2√2⋅H⋅e

^+π/4

/(2BCL

W

)⋅e

^-(ξ+π/4)

⋅[cos(ξ+π/4) + sin(ξ+π/4)] – – 4M

o

/L

W

⋅e

^+π/2

/(2BCL

W

)⋅e

^-(ξ+π/2)

⋅[cos(ξ+π/2) + sin(ξ+π/2)].

or after some reductions

y(ξ) = 4H/(2B⋅C⋅L

W

)⋅e

^-ξ

⋅cosξ + 4M

o

/(2B⋅C⋅(L

W

)

²

)⋅e

^-ξ

⋅(sinξ – cosξ),

because cos(ξ+π/4) + sin(ξ+π/4) = √2⋅cosξ oraz cos(ξ+π/2) + sin(ξ+π/2) = -sinξ + cosξ.

2.

Do take into account H,M

o

at ξ =0:

Follow the example IIa using using 2 virtual forces T

2

, T

1

, 2 real loadings H, M

o

at ξ=0 and the doubly-infinite beam; boundary conditions at ξ=0 are taken:

M(0) = -H⋅L

W

/4⋅e

^-0

(sin0-cos0) + M

o

/2⋅e

^–0

cos0 –

– T

1

⋅L

W

/4⋅e

^-π/4

(sinπ/4-cosπ/4) – T

2

⋅L

W

/4⋅e

^-π/2

(sinπ/2-cosπ/2), and must be M(0) = M

o

, so immediately T

2

= (H-2M

o

/L

W

)⋅e

^+π/2

.

Q(0+0) = -H/2⋅e

^-0

cos0 – M

o

/(2L

W

)⋅[e

^–0

(sin0+cos0)] –T

1

/2⋅e

^-π/4

cosπ/4 – T

2

/2⋅e

^-π/2

cosπ/2, and must be Q(0+0) = -H, so immediately T

1

= √2⋅(H-M

o

/L

W

)⋅e

^+π/4

.

The result is composed of 4 terms of course

y(ξ) = H/(2BCL

W

)⋅e

^–ξ

⋅[cos(ξ) + sin(ξ)] + M

o

(BC)

^–1

(L

W

)

^–2

e

^–ξ

sinξ + + √2⋅(H-M

o

/L

W

)⋅e

^+π/4

/(2BCL

W

)⋅e

^-(ξ+π/4)

⋅[cos(ξ+π/4) + sin(ξ+π/4)] + + (H-2M

o

/L

W

)⋅e

^+π/2

/(2BCL

W

)⋅e

^-(ξ+π/2)

⋅[cos(ξ+π/2) + sin(ξ+π/2)].

or after some reductions

y(ξ) = 4H/(2B⋅C⋅L

W

)⋅e

^-ξ

⋅cosξ + 4M

o

/(2B⋅C⋅(L

W

)

²

)⋅e

^-ξ

⋅(sinξ – cosξ),

because cos(ξ+π/4) + sin(ξ+π/4) = √2⋅cosξ oraz cos(ξ+π/2) + sin(ξ+π/2) = -sinξ + cosξ.

The results are the same but the version No.1 is shorter; so, as such - it could be recommended.

y(ξ)

H,Mo ξ→∞

0

The same in other words and more general.

“To be or not to be” question:

is it necessary to take into account explicit loads H,M

o

applied to the beam end or not (when using the Bleich method)?

Answer:

It doesn’t matter.

Example No.3:

Consider a Winkler vertical beam with horizontal springs and the subsoil coefficient which depends on depth x > 0, like C(x) = m·x or C(x) = Co + m·x + n·x², etc.

Comment:

such rotated situation corresponds to a vertical pile or wall loaded by the horizontal force H if the subsoil stiffness increases with depth; this happens in noncohesive soils.

A much more general analytical tool can be used here – polynomial series expansions

³

.

Main steps are as follows.

1. Every continuous function defined on a finite interval can be approximated with any accuracy by polynomial series (the Weierstrass Theorem). Apply this to the beam deformation line

y(x)= Σ a

i

x

ⁱ

, where the number of terms can be even infinite. Using 3 first derivatives of y(x)= Σ a

i

x

ⁱ

and the Euler-Bernoulli relation, internal forces can be found, like M(x), Q(x).

2. On the unloaded interval (q

o

= 0), the E-B equation says that EI⋅d

⁴

y(x)/dx

⁴

= – B⋅C⋅ y(x), for B = const, EI = const.

3. Since y(x) is a polynomial and C(x) is a polynomial too, so the fourth order derivative of y(x) = Σ a

i

x

ⁱ

can be calculated step-by-step, next y(x) is multiplied by the polynomial C(x) and finally both sides of the E-B equation are to be compared.

4. Let for example C(x) = m·x, where m = const > 0.

Denote L

Z

= [EI/(mB)]

^1/5

and introduce dimensionless variable ξ = x / L

Z

, so y(ξ) = Σ α

i

ξ

ⁱ

=α

o

+ α

1

ξ + α

2

ξ

²

+α

3

ξ

³

+ ... , where α

i

= a

i

(L

Z

)

ⁱ

.

In the next point no.5, the corresponding coefficients at ξ

ⁱ

are compared on both sides of the E-B equation – as mentioned in point no.3 – but using reformulated equation in ξ variable, i.e.

d

⁴

y(ξ)/dξ

⁴

= – ξ·y(ξ).

5. Immediately we see that α

4

= 0 (why?). After easy operations, the following set of recurrent relations can be found: α

i+4

= – α

i–1

i! / (i+4)! for i = 1,2,3,...

6. This means that all expansion coefficients of y(ξ) are known, so y(ξ) itself is known, but with the exception of α

o

, α

1

, α

2

, α

3

.

The 4 missing coefficients can be found by making use of 4 pre-defined boundary conditions at 2 beam ends. From the mathematical point of view, all 4 conditions could be set on one end but physically this is impossible – displacements and forces at the same place cannot be assigned;

therefore, there are 2 of them at each end.

For the linear function C(x) = m·x, this method was used by Zawrijew and so it is sometimes called.

Comment:

similar idea of polynomial expansions has been also applied by Gorbunov-Posadov to beams and slabs resting on the elastic half-space, not on the Winkler subsoil; numerical difficulties are, however, much greater and the accuracy is less certain.

3in XIXc. there was a lot of important problems of physics solved this way (differential equations with functional coefficients); nowadays, 150 years later, the method is still attractive – particularly if supported by symbolic operations (Mathematica^® and others); some solutions used in the old Polish National Code for piles PN-83/B-02482 are also based on C=C(x).

Note that standard solutions for the Winkler beams are also not free of expansions, because without (hidden) expansions the computer can do nothing with functions like exp{±x} or cos(x).

H

EI,B,L

x y

C(x)