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Mechanics of Materials II.
1 Tensors and their Applications to the Mechanicsof Materials
Basic books:R. Hibbeler, Mechanics of Materials, 7-th edition, Pearson, 2008.George Mase, Continuum Mechanics for Engineers, Second Edition, CRC Press LLC,
1999. Chapter 1. Essential mathematics.
1.1 Vector space.
1. Definitions of vectors.Vector is a directed interval in 3D(2D)-space. Let us chose point O of Euclidian space
as an origin, and the directed interval that connects the origin with point x of this spaceis a vector. All set of such vectors compose a vector space.For every vector a we define the scalar number |a| ≥0 that is the length of the vector
a (the norm of a, the module of a). For every two vectors a and b we can find the angleθ between them.
2. Algebra of vectors.1. Sum of two vectors: c = a+ b
b
a c
1
2. Product of a vector and a scalar
b = αa
3. Scalar product of two vectors
a · b =| a || b | cos θ
Note: a⊥b⇐⇒ a · b = 0.Physical meaning of the scalar product: It is the work A of force F on displacement
vector u :
F
u
A = F · uProperties of the scalar product
(a+ b) · c = (a · c)+ (b · c) ,(a · b) = (b · a).
4. Vector product of two vectors
a× b = |a||b| sin θe.
a
b
θe
Property of vector product1.
(a+ b)× c = a× c+ b× c,2.
a× b = −b× a,3. If b = αa, than a× b = 0.Examples of the vector product applications:
2
1. Moment of a force with respect to the origin.
M = x× F = |x||F| sinθe
M
F
x
h
O
θ
2. Linear velocity of a rotating body
xω
h
θ
v
v = ω × x = |ω| |x| sinθ e,
h = |x| sin θ, |v| = ωh, [ω] =1
sec, [h] = m.
5. Mixed products.5.1. The rule "bac minus cab"
a× (b× c) = b(a · c)− c(a · b).
Note:
(a× b)× c = −c× (a× b) =− a(c · b) + b(c · a)= b(a · c)− a(c · b) 6= a× (b× c).
5.2 Scalar-vector product
V = a · (b× c) = (b× c) · a.
3
Problem:1. Prove that V is the volume of the parallelepiped constructed on the vectors a,b, c.
2. Prove that the equivalence
b× (c× d) + c× (d× b) + d× (b× c) = 0.Here b, c,d are arbitrary vectors.5. Tensor product of two vectors.Tensor product (a⊗ b) is a connected couple of two vectors that don’t interact with
each other.
ab
Properties of the tensor product (definitions!)1.
α(a⊗ b) =(αa)⊗ b = a⊗(αb)2.
(a⊗ b) · c = (b · c)a, c · (a⊗ b) = (c · a)b
3.(a⊗ b)× c = a⊗ (b× c), c× (a⊗ b) = (c× a)⊗ b.
4.(a⊗ b+ c⊗ d) · e = (c⊗ d+ a⊗ b) · e = (b · e)a+ (d · e)c.
4
1.2 Bases and coordinate systems in vector spaces.
1.2.1 Definition of linearly independent vectors.
In order to define a vector we have to introduce a basic or coordinate system in a vectorspace.Definition:Vectors e1,e2, ..., en are called linearly independent if the following equation holds
α1e1 + α2e2 + ...αnen = 0 =⇒ α1 = α2 = ... = αn = 0
Examples:1. 2D- space:
e1
e2
α1e1 + α2e2 = 0 → α1 = α2 = 0.
Note: Any three vectors in a 2D-plane are linearly dependent.2. Two vectors on a line.
e2 e1
α1e1 + α2e2 = 0 → α1 = 1,
e1 −| e1 || e2 |
e2 = 0 α2 = −| e1 || e2 |
3. 3D-space.If three vectors are in one plane, we have
e2
e3
e1
5
α1e1 + α2e2 + α3e3 = 0,=⇒ e3 = −1
α3(α1e1 + α2e2)
If three vectors are not in one plane,
e 2
e 3
e 1
e 4
α1e1 + α2e2 + α3e3 = 0 =⇒ α1 = α2 = α3 = 0
For an arbitrary vector e4 one can fined the coefficiemts α1, α2, α3 such as
e4 = α1e1 + α2e2 + α3e3,
α1e1 + α2e2 + α3e3 − e4 = 0, α1,α2,α3 6= 0.Definitions:1) Maximal number of linearly independent vectors in a vector space is called the
dimension of this space.2) Any n linearly independent vectors in the space of n-dimension compose a basis
of this space.
e 2
e 3
e 1
a
If e1,e2, ....en is the basis, any vector a may be presented in the form.
a = a1e1 + a2e2 + ....+ anen
6
1.2.2 Cartesian bases
e2
e3
e1
Properties of the Cartesian basis:
1. |ei| = 1,
2. ei·ej = δij, δij =
½1, i = j0 i 6= j
- Kronecker’s symbol
3. ei×ej =3X
k=1
εijkek, i, j = 1, 2, 3.
Definition of εijk (Symbol of Levi-Civita)For even permutations of indexes 1, 2, 3 : ε123 = ε312 = ε231 = 1
2
3
1
For odd permutations of 1, 2, 3 : ε132 = ε213 = ε321 = −1
2
3
1
If two indices are the same
εiik = εiki = εiii = 0
Note: We can introduce two types of Cartesian bases the right and left ones:
7
e 2
e 3
e 1
R ig h t S y s te m
e 1
e 3
e 2
L eft S y s te m
For the right system the vector products of the vectorsof the base are defined by theequations:
ei×ej =3X
k=1
εijkek,
e1×e1 = 0, e1×e2= e3, e1×e3= −e2,e2×e1 = −e3, e2×e2= 0, e2×e3= e1,e3×e1 = e2, e3×e2= −e1, e3×e3= 0.
For the left system we have:
ei×ej = −3X
k=1
εijkek.
1.2.3 Presentation of vectors in 3D-space
x =3X
i=1
xiei
x1, x2, x3 are the components of the vector x in the basis ei.If two vectors are equal, their components in any basis are equal
a = b⇔ ai = bi, i = 1, 2, 3.
Einstein rule:Let us not write the sign of sum but imply such a summation with respect to repeating
indices from 1 to 3.
x = xiei,
xj = (x · ej) = |x| cos(dx, ej),(x · ej) = xj(ei·ej) = xiδij = xj.
8
Algebra of vectors in terms of the coordinates.1. Sum of two vectors in terms of coordinates:
a+ b = (aiei) + (biei) = (ai + bi)ei = ciei = c,
ci = ai + bi.
2. Scalar product in terms of coordinates :
a · b = (aiei) · (bjej) = aibj(ei·ej) = aibjδij =
= aibi = a1b1 + a2b2 + a3b3.
3. Vector product:
a× b = (aiei)× (bjej) = aibj(ei×ej) = εijk aibjek = ckek = c,
ck = εijkaibj,
c1 = εij1 aibj = a2b3 − a3b2,
c2 = εij2 aibj = a3b1 − a1b3,
c3 = εij3 aibj = a1b2 − a2b1.
4. Mixed product:
(a× b) · c = (εijk aibjek) · (clel) = εijk aibjck = det
°°°°°°a1, a2, a3b1, b2, b3c1, c2, c3
°°°°°°Problem 1: Prove this equation and the equation
det ||aij|| = εijka1ia2ja3k.
5. Tensor product:
a⊗ b = (aiei)⊗ (bjej) = aibjei⊗ej= a1b1e1⊗e1 + a1b2e1⊗e2 + a1b3e1⊗e3 +
a2b1e2⊗e1 + a2b2e2⊗e2 + a2b3e2⊗e3 +a3b1e3⊗e1 + a3b2e3⊗e2 + a3b3e3⊗e3.
Problem 2: Are these three vectors linearly independent or not? Here e1, e2, e3 arethe vectors of Cartesian basis in 3D-space.a = e1+e2+e3,b = e1−e2 + 2e3,
9
c =5e1 + e2 + 7e3.Probem 3: Prove the equivalences
δii = 3,
δijδji = 3,
δijδjkδki = 3,
δijδjk = δik , amsδsk = amk,
εijkbjbk = 0,
εijkεkpq = δipδjq − δiqδjp.
1.2.4 Changing Cartesian bases in the 2D-vector space
X2` e2` e2
e1 x1
x2x
X1`e1`
ϕ
Let us consider the old basis e1, e2 and the new basis e01, e
02. The same vector x may
be presented in two formsx = xkek = xiei.
The connection between these bases is given by the equation
ei = αjie0j, (1)
In detail: ½e1 = α11e
01 + α21e
02
e2 = α12e01 + α22e
02
,
α11 = (e1·e01) = cosϕ,α21 = (e1·e02) = cos(
π
2+ ϕ) = − sinϕ,
α12 = (e2·e01) = cos(π
2− ϕ) = sinϕ,
α22 = (e2·e02) = α11 = cosϕ.
10
α = kαijk =°°°° cosϕ, sinϕ− sinϕ, cosϕ
°°°° , αT =
°°°° cosϕ, − sinϕsinϕ, cosϕ
°°°° .Note:
ααT = αTα =
°°°° 1, 00, 1
°°°°⇒ αT = α−1.
A matrix that has this property is called orthogonal
αT = α−1.
Using these equation we have
x = xiei = xi αji ej = αji xi| z e0j = x0j|z e0j = x.Thus, we obtain
x0j = αji xi (2)
or:
x01 = cosϕ x1 + sinϕ x2,
x02 = − sinϕ x1 + cosϕ x2
Home task I:1. Prove the equivalence
b× (c× d) + c× (d× b) + d× (b× c) = 0.2. Are these three vectors linearly independent or not? Here e1, e2, e3 are the vector
of a Cartesian basis in 3D-space.a = e1+e2+e3,b = e1−e2 + 2e3,c =5e1 + e2 + 7e3.3. Prove that the following equation is correct
εijk aibjck = det
°°°°°°a1, a2, a3b1, b2, b3c1, c2, c3
°°°°°° .4. Prove the equivalences
11
δii = 3,
δijδji = 3,
δijδjkδki = 3,
δijδjk = δik , amsδsk = amk,
εijkbjbk = 0, det|aij| = εijka1ia2ja3k,
εijkεkpq = δipδjq − δiqδjp.
5. The basis of the Cartesian coordinate system changes as it is shown in figure
e1
e1
e2
e` 2
e3 e3
π/3
x
x = e1 + 2e2 + 3e3 =
= x1e1 + x2e2 + x3e3.
What are the new coordinates xi of the vector x?Note: A vector x in 2D- space may be defined as 2 numbers (x1, x2) that are changed
according to the law (2) if the basis is changed.
12
1.2.5 Vectors in 3D-space
e1
e`1
e3 e`3
x
e2
e`2
Connection between the old and new bases
ei = αjiej,
αji = ei · ej = cos([ei, ej),
x = xiei = xjej,
x = xiei = xi αji ej = αji xiej = xj ej.
xj = αji xi (*)
Properties of the matrix α
α = kαijk , αT = α−1, αTα = I = kδijk .A matrix with such a property is called an orthogonal matrix.Note:
x0 = α x , x = α−1x0 = αTx0 = x0α.
αTij x0j = αT
ij αjk xk = δik xk = xk , αTij = αji.
Inversion of the law (*)
xi = αji xj0
13
1.3 Linear operators in a vector space.
L x
y
L(x) = y
Operator L is the rule of correspondence between vector x and vector y of the samespace.Definition.An operator is called a linear one if it has the following properties1.
L(αx) = αL(x).
2.L(x+ y) = L(x) + L(y).
orL (αx+ βy) = α L(x) + β L(y).
Examples:1. Operator L is defined by the equation
L(x) = a× x,where a is a fixed vector. Let us prove that this operator is lineal.Proof:
L(αx+ βy) = a× (αx+ βy) = a× (αx) + a× (βy) = α a× x| z +β a× y| z == α L(x) + β L(y).
2. Operator L is defined by the equation
L(x) = (a⊗ b) · x = (b · x) a,where a⊗ b is the tensor product of two fixed vectors a and b.Let us prove that this operator is lineal:
L(αx+ βy) = (a⊗ b) · (αx+ βy) = [b · (αx+ βy)]a = [(b·αx) + (b · βy)]a == (b·αx)a+ (b · βy)a = α (b · x)a+ β(b · y)a == α (a⊗ b) · x+ β (a⊗ b) · y = α L(x) + β L(y)
14
1.3.1 Linear operators and bases in a vector space
e2
e3
e1
Let us consider 3D vector space with the Cartesian basis (e1, e2, e3) , and L be a linearoperator in this space. Action of the operator L on vector x
x = xiei
takes the form
y = L(x) = L (xiei) = xi L(ei) = y = yj ej
Action of the operator L on the element of the basis ei has the form
L(ei) = lji ej.
What is the matrix lij in this equation?
L(ei) · ek = lji ej · ek = ljiδjk = lik.
Thus, we have
L (xiei) = lji xi ej = yj ej ⇒⇒ yj = lji xi
Here klijk is a matrix that corresponds to the operator L in the basis (e1, e2, e3).°°°°°°y1y2y3
°°°°°° =°°°°°°l11, l12, l13l21, l22, l23l31, l32, l33
°°°°°°°°°°°°x1x2x3
°°°°°° .We have proved the Theorem:In every fixed basis, any linear operator may be presented as a matrix, and vice verse,
every matrix (3× 3) is a representation of a linear operator
L⇐⇒ klijk
15
Note that components of the matrix klijk are presented in the form
lij = L(ej) · ei.
Home task1. Find the matrix L = klijk of the operators L in a fixed Cartesian basis :
L(x) = a× x.
Here a is a fixed vector with components (a1,a2,a3) in this basis.2. Operator L acts on three vectors x1,x2,x3 of the vector space by the following rules
x1 = e3 =⇒ L(x1) = y1 , y1 = 2e1 + 3e2 + 5e3,
x2 = e2 + e3 ⇒ L(x2) = y2 , y2 = e1,
x3 = e1+e2+e3 ⇒ L(x3) = y3 , y3= e2−e3.
Find the matrix of the operator L in this basis.
1.3.2 Matrices of a linear operator in different Cartesian bases.
e2
e1
e 1 e 2
ϕ
`` X
Y
Let L is a linear operator in 2D-space
L(x) = y⇒½
lijxj = yi in basis eilijxj = yi in basis e0i
The matrix of this operator in the basis (e1,e2,e3) ("old" basis) is klijk, and in the newbasis (e01, e
02, e
03) the matrix of this operator is
°°l0ij°° .What is the connection between thistwo matrixes?
y = yiei = lij xj ei =
= lij xj αki ek = lij αlj x0l αki e
0k = αki αlj lij| z x0l e
0k =
= l0klx0lek = y0ke
0k.
16
Thus, we have the following connection between l0kl and lij :
l0kl = αki αlj lij
Theorem:Every linear operator may be presented in the form
L(x) = L · x, (*)
where L is called the tensor of the second rank. In the basis ei, tensor L has the formL = lij ei⊗ej, where klijk is the matrix of the operator L is the basis eiProof:We know that L(x) = y = yiei, and yi = lijxj.On the other hand:
L · x = (lijei⊗ej) · (xkek) = lij δkj xk ei = lij xj ei = yiei =⇒ yi = lij xj.
That is why the equation (*) holds.
1.3.3 Change of bases (coordinate systems) and presentation of linear oper-ators.
After changing the basis (coordinate system), components of any tensor change similarto the components of a vector. Tensor of the second rank is the following object
L = lijei⊗ej,where ei are basis vectors and lij are component of the tensor in the fixed basis.If the Cartesian basis ei is changed for the basis e0i, and the connection between the
old and new basis has the formei = αjiej,
for the tensor L we have
L = lij αki ek αlj el = αki αlj lij ek ⊗ ei= lkl ek ⊗ el.
It is seen from this equation that the connection of the components of the tensor inthe old and new bases has the form
lkl = αki αlj lij (**)
17
e1
e1`
e2
e2`
e3e3`
Thus, in a fixed basis every tensor is presented as a matrix. But tensor, like vector,is an invariant object that exists without any basis. If we change the coordinate system(the basis), the elements of this matrix change according to the rule (**).
Vector Tensorx is an arrow L(x) = y is a rule
x L(x) = L · xIn a Cartesian bases In a Cartesian basesx = xi ei = xiei L = lijei⊗ej = lij ei ⊗ ej
xi = αijxj lij = αikαjllkl
The total set of the tensors of the second order composes a tensor space.
1.3.4 Algebra of second rank tensors
1. Sum of two tensors.L+M = H
L = lij ei⊗ej, M = mij ei⊗ej, H = hij ei⊗ej.
hij = lij +mij
2. Scalar product of tensor and vector
L · a = lij aj ei.
3. Unit tensor I:I = δij ei⊗ej, I · a = a.
4. Scalar product of two tensors:
L ·M = H,
18
L ·M = (lij ei⊗ej) · (mkl ek⊗el) = lij mkl δjk ei⊗el == lik mkl ei⊗el = hil ei⊗el,
Components of the tensor H that is the product of two tensors is the product of twocorresponding matrixes
hil = lik mkl.
5. Inverse tensor L−1 is defined by the equation:
L−1·L = I.
Here I is a unit tensor
L−1 = mijei⊗ej, mik lkl = δil ⇒ kmijk = klijk−1
6. Transpose tensorL = lij ei⊗ej, LT = lji ei⊗ej
If LT= L ⇒ lij = lji , L is called a symmetric tensor.
1.4 Applications of second rank tensors in the mechanics ofsolids.
1.4.1 Stress tensor.
n -n
t(n)
t (-n)
Density of forces t(n)
Here t is the stress vector that acts at point x in the intersection with normal n. Itis clear that vector t depends on orientation of the intersection n. Thus, we can writet = T(n), where T is an operator. What is this propertie of this operator?From the third law of Newton we have
T(−n) = −T(n).
The theorem of Cauchy: T(n) is a linear operator.
19
Let us consider 2D-case:
-e1
e1
e2
-e2
T(n)
T(-e2)
T(-e1)
dx1
dx2
ds n
x2
x1ϕπ/2−
ϕ
π/2−ϕ
The condition of equilibrium of this triangle may be written in the form:
T(−e1) dx2 +T(−e2) dx1 +T(n) ds = 0,Vector n orthogonal to the side ds of the treeangle has the components n1, n2 : n =
n1e1 + n2e2. From the figure we obtain the equations:
ds =qdx21 + dx22,
dx1ds
= cos(π
2− ϕ) = sin(ϕ) ,
dx2ds
= sin(π
2− ϕ) = cos(ϕ),
n1 = cos(ϕ) =dx2ds
,
n2 = sin(ϕ) =dx1ds
.
Thus, the equilibrium equation takes the form:
−T(e1)n1 −T(e2)n2 +T(n) = 0or
T(n) = T(n1e1 + n2e2) = n1 T(e1) + n2 T(e2).
This equation shows us that T(n) is a linear operator.Consequence:The action of this operator on vector n may be presented in the form of scalar product
of this vector with a tensor of the second rank σ
t = T(n) = σ · n
20
σ is called the stress tensor, and in a fixed Cartesian basis this tensor takes the form
σ = σij ei⊗ej , i, j = 1, 2.
1.5 Physical meaning of the components σij of the stress tensorin a fixed Cartesian basish
σ 22
σ 12
σ 11e1
e2
σ 12
σ22
σ11
σ 21
σ 21
T(e1) = σ · e1 = (σijei⊗ej) · e1 == σi1ei = σ11e1 + σ21e2,
T(e2) = σi2ei = σ12e1 + σ22e2,
T(−e1) = −σ11e1 − σ21e2,
T(−e2) = −σ12e1 − σ22e2.
Equilibrium with respect to the center of the square which sides have length dl takesthe form
M = σ21dl2 − σ12dl
2 = (σ12 − σ21)dl2 = 0 ⇒
=⇒ σ12 = σ21.
Thus, σ is a symmetric tensor:
kσijk =°°°° σ11, σ12σ12, σ22
°°°°Note:In the 3D case we have the same properties of the operatorT: T(n) is a linear operator
and it may be presented in the form:
T(n) = σ · n, σ = σijei⊗ej.Here σ is the tensor of the second rank which components have physical meaning that
is indicated in the figure
21
σ 11
σ 21
σ 31
σ 12
σ 22
σ 32
σ 33
σ 23
σ 13
As in the 2D-case σ is a symmetric tensor
σij = σji.
1.5.1 Tensor of the moments of inertia.
Let 0 is the center of weight of the area SZs
x1ds =
Zs
x2ds = 0 ⇒Zs
xids = 0
x 1
e1
x 2
e2
0
x(x1, x2) is a vector of point x in 2D-space.Tensor of the moments of inertia is defined by the equation
J = Jijei⊗ej =Zs
x⊗ xds =
=
Zs
x21dse1⊗e1 +Zs
x1x2ds (e1⊗e2+e2⊗e1) +
Zs
x22 dse2⊗e2,
J11 =
Zs
x21ds , J12 = J21 =
Zs
x1x2ds, J22 =
Zs
x22ds.
This is a symmetric tensor of the second rank.
22
1.5.2 Electric properties of materials.
Electric field inside a cube of electro conductive material.
E =ϕ1 − ϕ0
le2
x2
x1
ϕ 0
ϕ 1
l
E J
Ohm’s law may be written in an invariant form:
J = C ·E, C = cijei⊗ej,Here C is two rank symmetric tensor of the electro conductive properties of the ma-
terial.
1.6 Properties of two rank symmetric tensors
1.6.1 Eigenvectors and eigenvalues of symmetric 2- rank tensors
Let us start with 2D-case
σ = σijei⊗ej, kσijk =°°°° σ11, σ12σ12, σ22
°°°° (1)
σ is a symmetric tensor¡σ= σT
¢for which the following property holds
σ·a=a·σ, σijaj = ajσji.
Definition.Vectors g is called the eigenvector of tensorsσ if it satisfies the equation
σ · g = λg, (2)
where g 6= 0, and λ is a scalar constant.Note :If g is eigenvector, αg is also eigenvector. Thus we can demand that |g| = 1, α = 1
|g|(the normalization condition).
23
Let us try to find the solutions of Eq.(2). This equation in a fixed Cartesian basis isequivalent to the following two equations
σijgj − λgi = 0 , i = 1, 2.
The detailed forms of these equations are
(σ11 − λ)g1 + σ12g2 = 0, (i = 1),
σ12g1 + (σ22 − λ)g2 = 0, (i = 2).
The condition of the existence of non-trivial solution of this system (g1, g2 6= 0) takesthe form
det
°°°° σ11 − λ, σ12σ12, σ22 − λ
°°°° = 0 ⇒ (σ11 − λ)(σ22 − λ)− σ212 = 0
The last equation is called the characteristic equation of tensor σ. It may be rearrengedand written in the form:
λ2 − (σ11 + σ22)λ− σ212 + σ11σ22 = 0 or
λ2 − Iσλ+ IIIσ = 0. (3)
HereIσ = σ11 + σ22, IIIσ = σ11σ21 − σ212 = det kσk
Note:Iσ, IIIσ are invariants of tensor σ (Their values don’t depend on the coordinate
system we chose).
Iσ = σ11 + σ22 = σ11 + σ22,
IIIσ = det kσijk = det kσijk
Home task: To prove these properties of Iσ and IIIσ.The solution of the characteristic equation (3) has the form:
λ1,2 =1
2
∙σ11 + σ22 ±
q(σ11 + σ22)2 − 4(σ11σ22 − σ212)
¸=
=1
2
∙σ11 + σ22 ±
q(σ11 − σ22)2 + 4(σ212)
¸.
Note:
24
It follows from this equation that λ1,2 are real numbers for any symmetric real tensorσ!Let λ = λ1 . In this case
(σ11 − λ1)g(1)1 + σ12 g
(1)2 = 0,
σ12g(1)1 + (σ22 − λ1)g
(1)2 = 0.
Because the determinant of this system is equal to zero the second equation is theconsequence of the first one.From the first equation we obtain
g(1)1 = − σ12
σ11 − λ1g(1)2 .
Taking into account the normalization condition
(g(1)1 )
2 + (g(1)2 )
2 = 1,
we can write ∙µσ12
σ11 − λ1
¶+ 1
¸(g(1)2 )
2 = 1
From here we obtain the component of the first eigenvector g(1)
g(1)1 = − σ12p
(σ11−λ1)2 + σ212, g
(1)2 =
σ11 − λ1p(σ11−λ1)2 + σ212
.
In the same way, for the second eigenvector we have:
g(2)1 = − σ12p
(σ11 − λ2)2 + σ221, g(2)2 =
σ11 − λ2p(σ11 − λ2)2 + σ222
.
,Theorem:Eigenvectors that corresponds to different eigenvalues of tensor σ are orthogonal.Let g(1) is a eigenvector
σ·g(1) = λ1g(1).
If we multiply both parts of this equation with the eigenvector g(2), we obtain
g(2)·σ · g(1) =¡g(2)·σ
¢·g(1) =
¡σ · g(2)
¢·g(1) = λ2(g
(2)·g(1)) = λ1(g(2)·g(1)).
(λ1 − λ2)(g(2)·g(1)) = 0 ⇒ (g(2)·g(1)) = 0 =⇒ g(1)⊥ g(2).
25
1.6.2 The matrix of a tensor in the basis of eigenvectors.
g(2)
g(1)σ1 σ1
σ2
σ2
σ = σijg(i)⊗g(j)
σ · g(1) = λ1g(1) = σi1 g
(i) = σ11g(1) + σ21g
(2) ⇒σ11 = λ1 = σ1, σ21 = σ12 = 0.
σ · g(2) = σ22g(2) = λ2g
(2) ⇒ σ22 = λ2 = σ2
Thus, the matrix of tensor σ in the basis of eigenvalues has the simplest form: it is adiagonal matrix.
kσijk =
°°°° σ1, 00, σ2
°°°°⇒σ = σ1g
(1)⊗g(1) + σ2g(2)⊗g(2).
1.6.3 Normal and tangent stresses in the area with normal n.
This is typical problems that appear in applications:Tensor of stresses at point x is known. What are normal and shear stresses that acts
in a small area with normal n that goes through this point?
g(2)
g(1)
σ n
nϕ
τ n
26
Stress vector acting in the area with normal n
t = σ · n =τnτ + σnn,
n = cos(ϕ)g(1) + sin(ϕ)g(2), τ = − sin(ϕ)g(1) + cos(ϕ)g(2)
Normal and tangent components of vector t take the form :
σn = n · σ · n = σ1n21 + σ2n
22 = σ1 cos
2 ϕ+ σ2 sin2 ϕ,
τn = τ · (σ · n) = σ1τ 1n1 + σ2τ 2n2 = −σ11
2sin 2ϕ+ σ2
1
2sin 2ϕ = −σ1 − σ2
2sin 2ϕ
Final equations for σn and τn are
σn = σ1 cos2 ϕ+ σ2 sin
2 ϕ , τn = −σ1 − σ22
sin 2ϕ.
Let is find extrem values of the function σn(ϕ)
dσndϕ
= −σ1 sin 2ϕ+ σ2 sin 2ϕ = (σ2 − σ1) sin 2ϕ = 0
The solutions of this equation are
2ϕ = kπ, k = 0, 1,=⇒ ϕ = 0, ϕ =π
2.
If ϕ = 0 ⇒ σn = σ1, τ = 0;
If ϕ =π
2⇒ σn = σ2, τ = 0.
Thus σ1 and σ2 are maximal and minimal values of σn.
g ( 1 )
g ( 2 )
If σ1 > σ2,
½σ1 = maxσnσ2 = minσn
¾Let us find maximal value of τn.The function τn(ϕ) has maximum at the point where
the derivative dτndϕis equal to zero
dτndϕ
= −σ1 − σ22
2 cos (2ϕ) = −(σ1 − σ2) cos (2ϕ) = 0.
The solution of the last equation is (σ1 6= σ2)
27
2ϕ =π
2⇒ ϕ =
π
4, sin
³2π
4
´= 1
As a result we obtain
max τ 1 =
¯σ1 − σ22
¯Note 1:
τn =σ1 − σ22
when ϕ =π
4.
Note 2:Let us present σn in the form
σn = σ1 cos2 ϕ+ σ2 sin
2 ϕ = σ11
2(1 + cos 2ϕ) + σ2
1
2(1− cos 2ϕ) =
=σ1 + σ22
+σ1 − σ22
cos 2ϕ,
τn = −σ1 − σ22
sin 2ϕ.
Let us consider equations for σn, τn
σn −σ1 + σ22
=σ1 − σ22
cos 2ϕ,
τn = −σ1 − σ22
sin 2ϕ.
As a consequence of these equations we have
(σn −σ1 + σ22
)2 + τ 2n = (σ1 − σ22
)2.
This is the equation of a circle in coordinates σn, τn that is called Mohr’s circle.
σ1+σ22
2ϕσ2 σ1 σn
σ1-σ22
τ n
28
1.6.4 Geometrical interpretation of a two rank symmetric tensor
Let us consider the line in plane (x1, x2) that is defined by the equation.
x · σ · x = 1. (1)
Here σ is two rank symmetric tensor. If axes x1, x2 coincide with eigenvectors of tensorσ takes the form
σ = σ1e1 ⊗ e1 + σ2e2 ⊗ e2,
where σ1, σ2 are principal values of σ, and σ1, σ2 > 0. As a consequence the Cartesiancoordinates on the line (1) satisfy the equation
σ1x21 + σ2x
22 =
x21¡1/√σ1¢2 + x22¡
1/√σ2¢2 = 1
This is the equation of an ellipse with semiaxes 1/√σ1, 1/
√σ2.
x 2
x1
2σ
1σ1/
1/
If σ1 > 0, σ < 0, the equation for r2(ϕ) takes the form
σ1x21 − |σ2|x22 = 1.
It follows from this equation thatµx1
1/√σ1
¶2−Ã
x2
1/p|σ2|
!2= 1.
29
This is equation for two hyperbolas
X1
X2
Conclusion: An ellipse reflects many properties of a tensor
e1
gg
e2
(1)
(2)
11 σ
=a
2 2
σ = a
1
1
Home task II:1. Find the matrix L = klijk of the operators L in a fixed Cartesian basis :
L(x) = a× x,
where a is a fixed vector with components (a1,a2,a3) in this basis.2. Operator L acts on three vectors of the vector space by the following rules
x1 = e3 =⇒ L(x1) = y1 , y1 = 2e1 + 3e2 + 5e3,
x2 = e2 + e3 ⇒ L(x2) = y2 , y2 = e1,
x3 = e1+e2+e3 ⇒ L(x3) = y3 , y3= e2−e3.
30
Find the matrix of the operator L in this basis.
3. Find eigenvalues and eigenvectors of the tensor σ = σij ei⊗ej
kσijk =°°°° 1, 22, 3
°°°° .
Construct the lines in the plane (x1, x2) that is defined by the equation
1. r(ϕ) =√n · σ · n,
µn =
x
|x|
¶; 2. x · σ · x = 1
2) Tensor σ in a Cartesian basis has the form
kσijk =°°°° 2, 11, σn
°°°° .Find the component σn if it is known that there is an area with normal n that is free
of stresses (σ · n = 0) .Find the components of the normal n to the plane where the stresses don’t act.
1.6.5 Stress tensor in 3D-case
Eigenvectors of a tensor are non-trivial solutions of the following equations
σ · g = λg , |g| = 1.
The detailed form of this equation is⎧⎨⎩ (σ11 − λ)g1 + σ12 g2 + σ13 g3 = 0σ12 g1 + (σ11 − λ)g2 + σ13 g3 = 0σ13 g1 + σ23 g2 + (σ33 − λ)g3 = 0
For the existence of non-trivial solution the determinant of the matrix of this systemshould be equal to zero
det kσij − λδijk = 0 ⇒ λ3 − Iσλ2 + IIσλ
3 − IIIσ = 0
Iσ, IIσ, IIIσ are invariants of the tensor σ
Iσ = σii, IIσ =1
2(σijσji − σiiσjj), IIIσ = det kσijk .
Cubic equation (*) has three real roots λ1, λ2, λ3. These are eigenvalues of tensor σ.For every eigenvalue one can construct a eigenvector.
λ1 −→ g(1), λ2 −→ g(2), λ3 −→ g(3).
31
These three vectors are mutually orthogonal
g(i)·g(j) = δij.
If eigenvectors are chosen as a basis in 3D-space, the matrix of the tensor is diagonal
σ = σijg(i)⊗g(j) ⇒ kσijk =
°°°°°°σ1, 0, 00, σ2, 00, 0 σ3
°°°°°° .Here σ1, σ2, σ3 are principal stresses.Geometric interpretation of the second rank tensor:
r2(n) = n · σ · n = σ1n21 + σ2n
22 + σ23 =
= x21 + x22 + x23
x1 =√σ1n1, x2 =
√σ2n2, x3 =
√σ3n3.
µx1√σ1
¶2+
µx2√σ2
¶2+
µx3√σ3
¶2= 1.
x3
x2
g(1)
g(2) σ2
σ1x1
g(3)σ3
x1
g(3)σ31/
1/
1/
1.7 Particular stress states
1.7.1 Pure shear
The pure shear is the stress state that can be schematically represented as follows
32
e1
e2
τ
τ
τ
τ
The corresponding stress tensor has the form
σ = σijei⊗ej, kσijk =°°°° 0, ττ , 0
°°°° .Eigenvectors and eigenvalues of this tensor
σ · g =λg ⇒ (σ − λ1) · g = 0.The detailed form of this equation is½
−λ g1 + τ g2 = 0τ g1 − λ g2 = 0
.
The condition of the existence of non-trivial solutions of this system is.
det
°°°° −λ, ττ ,−λ
°°°° = λ2 − τ 2 = 0.
The roots of characteristic equation are
λ1 = τ , λ2 = −τ .
1. Let us consider the first root λ = τ½−τ g1 + τ g2 = 0τ g1 − τ g2 = 0
.
⇒ g1 = g2, g21 + g22 = 1⇒ g1 = g2 =1√2.
The first eigenvector takes the form
g(1) =1√2(e1+e2).
33
2. The second root of the characteristic equation is λ = −τ ,and for this root thesystem of equation for the eigenvectors takes the form
τg1 + τg2 = 0 ⇒ g1 = −g2 g1 = −g2 =1√2.
The second eigenvector is defined by the equation
g(2) =1√2(−e1+e2).
In the basis of the eigenvectors the stress tensor takes the form
σ = τ(g(1)⊗g(1)−g(2)⊗g(2)).Geometrical interpretation of this tensor is in the picture
g(2)
τ
−τ
g(1)
-τ
τ
Thus, the compression in one direction and tension in another direction applied to asolid element provoke the pure stress state in this element
τ
-τ
τ
-τ
34
1.7.2 Hydrostatic state
This stress tensor of this state is defined the equation
σ = −p1, 1 = δijei⊗ej .Note that if we change the coordinate system, tensor σ does not change
σij = −pαjkαjsδks = −pαikαjk = −pδij ⇒ ααT = I.
Because for any vector g we have σ · g = − pg every vector g is eigenvector inthis case.
1.7.3 Deviator of stress tensor
Let us consider an arbitrary stress tensor
σ = σijei⊗ej.The first invariant of this tensor is
Iσ = σ11 + σ22 + σ33 = σii.
Deviator S of the tensor σ is defined by the equation
S = Devσ = σ − 13Iσ1, σ =
1
3Iσ1+ S.
The first invariant of S is equel to zero
IS = σ11 −1
3Iσ + σ22 −
1
3Iσ + σ33 −
1
3Iσ = Iσ − Iσ = 0.
On the other hand, the first invariant is the sum of diagonal components, and we have
IS = s11 + s22 + s33 = 0.
Let s1, s2,s3 are principal value of S, and g(1),g(2),g(3) are its eigenvectors
S = s1g(1)⊗g(1) + s2g
(2)⊗g(2) + s3g(3)⊗g(3).
IS = s1 + s2 + s3 = 0 , s3 = −s1 − s2.
Thus, we have for the tensor S the following presentation
S = s1(g(1)⊗g(1)−g(3)⊗g(3)) + s2(g
(2)⊗g(2)−g(3)⊗g(3)).
35
S1
S1
-S1
-S 1
S 2 S 2
-S2
-S 2
+
Conclusion: Deviator S is presented the stress state that is a combination of pureshear stress states.Problem for home task III:Hibbeler, Mechanics of Materials, Chapter 9 (Stress transformation)Problems: 9-95,9-96,9-97,9-101,9-103,9-104,9-105,9-112,9-114.
2 Strain tensor (tensor of deformations).
We consider a solid body that changes its position in space because a system of externalforces is applied to this body.
x1
x3
x2
x
ddx u
Vector of displacement of arbitrary point x of the body is
u(x) = u1e1 + u2e2 + u3e3
Components of the vector u are the functions of the position of point x before defor-mation.
ui = ui(x1, x2, x3), i = 1, 2, 3.
36
Vector of the position of point x after deformation (ζ) is
ζ = x+ u, u = ζ − x.
In components this equation has the form
ζi = ζi(x1, x2, x3) = xi + ui(x1, x2, x3), i = 1, 2, 3.
The length of the material element dx in the initial state
|dx| =dS0 =qdx21 + dx22 + dx23, dS20 = dxidxi = δikxixk.
The length of the material element dζ in the final state (after deformation) (ζi =ζi(x1, x2, x3), i = 1, 2, 3)
|dζ| =dS2 = dζ idζ i =∂ζi∂xj
∂ζi∂xk
dxjdxk.
dx
d
The change of the length of the element dx after deformation is
dS2 − dS20 =
∙∂ζi∂xj
∂ζi∂xk− δjk
¸dxjdxk.
If we ihtroduce the vector of displacements u, we can write
ζi = ui + xi,
∂ζi∂xj
=∂(ui + xi)
∂xj=
∂ui∂xj
+ δij,
dS2 − dS20 =
∙µ∂ui∂xj
+ δij
¶µ∂ui∂xk
+ δik
¶− δjk
¸dxjdxn =
=
∙∂uj∂xk
+∂uk∂xj
+∂ui∂xj
∂ui∂xk
¸dxjdxn = 2εjkdxjdxk.
The object
εjk =1
2
∙∂uj∂xk
+∂uk∂xj
¸is called the tensor of deformation (strain tensor). Thus, we have
37
dS2 = dS20 + 2εjkdxjdxk.
Note. Let us calculate the relative change of the length of a small element whichdirection is defined by a unit vector n.Components of the vector n are:
ni =dxidS0
.
thus, we have:
dS = dS0
r1 + 2εjk
dxjdxkdS0
= dS0p1 + 2εjknjnk ' dS0(1 + εjknjnk).
Finally,dS − dS0
dS0= εjknjnk.
Let us consider the displacements of a small vicinity of a point x
x
u
x+u
dx
x
u
x+u
dx
Because the vector dx is small we can write
u(x+ dx) = u(x)+ (u⊗∇) ·dx+ 0(|dx|2) =
(u⊗∇) ·dx =
µ∂ui∂xk
dxk
¶ei = ε · dx+Ω · dx,
∂ui∂xk
dxk =1
2
µ∂ui∂xk
+∂uk∂xi
¶dxk +
1
2
µ∂ui∂xk− ∂uk
∂xi
¶dxk
Finally, we have the equation
u(x+ dx) = u(x) + ε · dx+Ω · dx+ 0(|dx|2). (*)
Geometrical meaning of this equation is the division of the total displacements of asmall element (small vecinity of a point x) on three parts: translation as a rigid body,
38
rotation as a rigid body, and deformation of this slement:
u(x) Ω(x) ε(x)
Let us consider the antisymmetric tensor Ωij
Ωij =1
2
µ∂ui∂xj− ∂uj
∂xi
¶.
This tensor has the following properties Ωij = −Ωji, and we can introduce vector wthat is unically defined by the tensor Ωij
Ωij = −Ωji, wk = −1
2εkijΩij, Ωij = −εkijwk.
Note:
Ωij =1
2εijkεkemΩem =
1
2(δieδjm − δimδje)Ωem =
1
2(Ωij − Ωji) = Ωij.
On the other hand we can write
Ω · dx = Ωijdxj = −εkijwkdxj = εikjwkdxj = w× dxBut w× dx corresponds to displacements of a rigid body by the rotation on angle |w|
around the vector e = w|w| .
d x
w
Thus, the term Ω · dx in the equation (*) corresponds to the rotation of a smallvicinity of point x as a rigid body.
39
2.1 Physical meaning of the components of strain tensor.
2.1.1 The meaning of diagonal components of ε.
dxdxdx
x1
x2
x3
Let us take a material interval along the axis x1
dx = dx1e1, dS20 = dx21
After the deformation this element will have another length dS
dS2 = dS20 + 2εijdxidxj = dx21 + 2ε11dx21 = (1 + 2ε11)dx
21
dS =√1 + 2ε11dx1 ≈ (1 + ε11)dx1 = (1 + ε11)dS0.
From this equation we have
dS − dS0dS0
=dS − dx1
dx1= ε11.
Thus, the component ε11 is the relative change of the length of the element dx1 afterdeformation. By the same way one can show that
dS2 − dx2dx2
= ε22,dS3 − dx3
dx3= ε33.
Note.Let we have an arbitrary small material interval dx with the length |dx| = dS0 directed
along the normal n. After deformation its length will be
dS2 =
µ1 +
2εijdxidxjdS20
¶dS20 , dS =
µ1 + εij
dxidxjdS20
¶dS0.
The deformation εn of the element is defined as follows
εn =dS − dS0
dS0= εij
dxidxjdS20
= εijninj, ni =dxidS0
.
40
Rosettes of extensometers are used for the measuring of the components of deformationtensor on surfaces of solid bodies.Home task 1.For the rosettes in the figure
/4/4
1
2
3
x1
x2
we have: εn1 = 0.1, εn2 = 0.2, εn3 = 0.3.To find components of the strain tensor ε11, ε22, ε12 in the considered point.Home task 2.To show that relative change of the volume of a cube has the form
dV − dV0dV0
= ε11 + ε22 + ε33
.....
A hint:dV0 = (dx1 × dx2) · dx3, dV = (dζ1×dζ2) ·dζ3
41
2.1.2 Physical meaning of non-diagonal components of strain tensor.
Let material elements dx1,dx2 that coincide with the axes x1, x2 before the deformationtake the position dζ(1),dζ(2) after the deformation
d(1)
d(2)dx(1) dx(2)
x1
x2
x3
ζi = ui + xi,∂ζ i∂xi
=∂ui∂xi
+ δij
Scalar product of the vectors dζ(1),dζ(2) is
dζ(1)·dζ(2) = dS(1) · dS(2) cos (ϕ) = (1 + ε11) (1 + ε22) cosϕdx1dx2 =
=∂ζi∂x1
· ∂ζ i∂x2
dx1dx2 =
µ∂ui∂x1
+ δi1
¶µ∂ui∂x2
+ δi2
¶dx1dx2 ≈
≈µ∂u2∂x1
+∂u1∂x2
¶dx1dx2 = 2ε12dx1dx2.
Because ε11, ε22 << 1, we have
ε12 =(1 + ε11) (1 + ε22)
2cosϕ ≈ cosϕ
2=
=1
2cos³π2− γ
´=1
2sin γ ≈ γ
2.
γ =π
2− ϕ.
Here γ is the change of the angle between elements dx1,dx2 after the deformation.The components ε13 and ε23 have similar physical meaning.Note.Because tensor ε = εijei⊗ej is a symmetric tensor of the second rank it has the same
properties as the stress tensor. There are three eigenvalues of this tensor ε1, ε2, ε3 thatare called principal strains and three eigenvectors (g(1),g(2)g(3)) that correspond to every
42
eigenvalue. These vesctors are mutially orthogonal, and in the basis of this vectors thematrix of tensor ε is diagonal
ε = ε1g(1)⊗g(1) + ε2g
(2)⊗g(2) + ε3g(3)⊗g(3).
43
3 Elastic materials. Hooke‘s Law
(Robert Hooke: 1635 - 1703)
One of the most general properties of solid bodies is elasticity. The law of elasticity iscalled Hooke’s law and may be written as a linear connection between stress and straintensors at any point of a solid body
σ = C · ε, ε = S · σ. (1)
Here C is the elastic moduli tensor, S is the compliance tensor. These are four ranktensors symmetric over pair of indexes.
C = Cijklei ⊗ ej ⊗ ek ⊗ el, S =Sijklei ⊗ ej ⊗ ek ⊗ el .Cijkl = Cjikl = Cijlk = Cklij, Sijkl = Sjikl = Sijlk = Sklij,
Hook’s law (1) in the components may be written as follows
σij = Cijklεkl, εij = Sijklσkl.
3.1 Matrix presentation of the elastic moduli tensor.
In the applications, the matrix presentation of four rank tensors is often used. Firstly, letus introduce vector presentation of the second rank tensors
σ ⇒ σ, ε⇒ ε
Here vectors σ and ε are defined by the equations
σ = kσ11, σ22, σ33, σ23, σ13, σ12k ,ε = kε11, ε22, ε33, 2ε23, 2ε13, 2ε12k ,
The Hooke law may be presented now as a linear connection between two vectors σand ε :
σ = cε, c = kcijk , cij = cji.
Here kcijk is a symmetric matrix of the elastic moduli that has the dimension 6× 6.In general case this matrix has 21 independent components
c =
°°°°°°°°°°°°
c11 c12 c13 c14 c15 c16c12 c22 c23 c24 c25 c26c13 c23 c33 c34 c35 c36c14 c24 c34 c44 c45 c46c15 c25 c35 c45 c55 c56c16 c26 c36 c46 c56 c66
°°°°°°°°°°°°44
Let us consider some particular cases of anisotropic materialsOrthotropic material (9 constants)
c =
°°°°°°°°°°°°
c11 c12 c13 0 0 0c12 c22 c23 0 0 0c13 c23 c33 0 0 00 0 0 c44 0 00 0 0 0 c55 00 0 0 0 0 c66
°°°°°°°°°°°°Transversally isotropic material (5 constants)
c =
°°°°°°°°°°°°
c11 c12 c13 0 0 0c12 c11 c13 0 0 0c13 c13 c33 0 0 00 0 0 c44 0 00 0 0 0 c44 0
0 0 0 0 0 (c11−c12)2
°°°°°°°°°°°°Isotropic body (2 constants)
c =
°°°°°°°°°°°°°
c11 c12 c12 0 0 0c12 c11 c12 0 0 0c12 c12 c11 0 0 0
0 0 0 (c11−c12)2
0 0
0 0 0 0 (c11−c12)2
0
0 0 0 0 0 (c11−c12)2
°°°°°°°°°°°°°3.2 Isotropic elatic material.
In the isotropic case the Hook law may be written in the form
σij = λIεδij + 2μεij, Iε = ε11 + ε22 + ε33,
where λ, μ are called Lame’s constants.Another form of the Hooke law for an isotropic material may be written in the form
using the bulk modulus k and shear modulus μ :
σij = kIεδij + 2μ
µεij −
1
3Iεδij
¶=
= (k − 23μ)Iεδij + 2μεij.
45
From this two equations we have
λ = (k − 23μ)
1. Let us consider a hydrostatic case. For an isotropic body we have
ε = ε01.
The Hooke law takes the form:
σij = 3kε0δij ⇒ σ = p1 = 3kε01
Thus, we have.p = 3kε0.
Note:The change of the volume of a small elements is expressed through the first invariant
of the stress tensor ∆vv= 3ε0 = Iε, and using the Hook law we can write
∆v
v= 3ε0 =
p
k= Iε, or
p = kIε.
Thus the bulk modulus k connects the hydrostatic pressure p and the relative volumchange Iε.2. In the case of "pure shear", we have
kσijk =
°°°°°°0 σ12 0σ12 0 00 0 0
°°°°°° , kεijk =
°°°°°°0 ε12 0ε12 0 00 0 0
°°°°°° ,σ12 = 2με12.
3.3 "Technical" form of the Hooke law.
ε11 =1
E[σ11 − ν (σ22 + σ33)] ,
ε22 =1
E[σ22 − ν (σ11 + σ33)] ,
ε33 =1
E[σ33 − ν (σ11 + σ22)] ,
ε12 =1
2μσ12, ε13 =
1
2μσ13, ε23 =
1
2μσ23, μ =
E
2(1 + ν).
Here E is called the Young modulus and ν is the Poisson ratio.
46
If σ22 = σ33 = 0, we have one dimensional stress state, and the components of thestrain tensor take the form
ε11 =1
Eσ11, ε22 = −
ν
Eσ11, ε33 = −
ν
Eσ11.
If a cylindrical specimen is loaded by the force F, the picture of deformation is shownin the figure
Dd
L0
LF
Deformation and stresses may be calculated as follows
ε11 =L− L0L0
, ε22 = ε33 =d−D
D, σ11 =
4F
πD2.
Thus, we can calculate elastic constant E and ν from the equations
E =σ11ε11
, ν = −ε22ε11
.
Connections between different elastic constsnts are presented in this table
λ, μ k, μ E, ν μ, νλ λ k − 2
3μ νE
(1+ν)(1−2ν)2νμ1−2ν
μ μ μ E2(1+ν)
μ
k λ+ 23μ k E
3(1−2ν)2μ(1+ν)3(1−2ν)
E μ(3λ+2μ)λ+μ
9kμ3k+μ
E 2μ(1 + ν)
ν λ2(λ+μ)
3k−2μ6k+2μ
ν ν
3.4 Energy of elastic deformation
1. The energy of tension.Let us consider a cube with the size of the site l at the action of a unidimensional
47
stress field along axis x1 :
11
Total force that acts on the site of the cube is
σ11S = F, l2 = S, V0 = Sl = l3,
Displacement under action of the force F takes the form
ε11 =∂u1∂x1
= const,
u1 = ε11x1 = ε11l.
The work of the force that is equal to the elastic energy of the specimen is calculatedas follows
A =W =1
2Fu1 =
1
2σ11ε11V0.
F
u1
F
u1
The density of the elastic energy w is
W
V0= w =
1
2σ11ε11.
48
2. Shear energy
d 12
γ
The shear force T
T = σ12l2, W =
1
2Tγl = ε12σ12l
3,
d = γl, γ = 2ε12.
The work of the force T or the energy
A =W =1
2Td =
1
2Tγl = ε12σ12l
3
We take into account thatε12 =
γ
2
Finally, we have
W = ε12σ12l3, w =
W
l3= ε12σ12
In general case the density of the energy takes the form
w =1
2[σ11ε11 + σ22ε22 + σ33ε33 + 2σ12ε12 + 2σ13ε13 + 2σ23ε13] =
1
2εijσij.
3. Elastic energy in terms of stresses"Physical" form of the Hooke law for deformation is
σij = kIεδij + 2μ
µεij −
1
3Iεδij
¶.
The inverse law
εij =1
9kIσδij +
1
2μ
µσij −
1
3Iσδij
¶.
49
w(σ) =1
2εijσij =
1
2
∙1
9kIσδij +
1
2μ
µσij −
1
3Iσδij
¶¸σij =
=1
18kI2σ +
1
4μ(σijσij −
1
3I2σ).
In the basis of eigenvectors of tensor σ we have (σ11 = σ1, σ22 = σ2, σ33 = σ3, σ12 =σ13 = σ23 = 0)
w =1
18kI2σ +
1
4μ
µσ21 + σ22 + σ23 −
1
3(σ1 + σ2 + σ3)
2
¶=
=1
18kI2σ +
1
12μ
¡2σ21 + 2σ
22 + 2σ
23 − 2σ1σ2 − 2σ1σ3 − 2σ2σ3
¢=
=1
18kI2σ +
1
12μ
¡(σ1 − σ2)
2 + (σ1 − σ3)2 + (σ2 − σ3)
2¢ ==
1
18kI2σ +
1
2μT 2.
where
Iσ = σ1 + σ2 + σ3, T 2 =1
6
£(σ1 − σ2)
2 + (σ1 − σ3)2 + (σ2 − σ3)
2¤ .w(σ) =
1
18k(σ1 + σ2 + σ3)
2 +1
12μ
£(σ1 − σ2)
2 + (σ1 − σ3)2 + (σ2 − σ3)
2¤ .Note 1. Because the energy is only positive we have
k =E
3(1− 2ν) =2
3
μ(1 + ν)
(1− 2ν) > 0, μ > 0
For the Poisson ratio ν we obtain the following inequalities
1 + ν
1− 2ν ≥ 0 or − 1 ≤ ν ≤ 12.
Note 2. Let us consider a hydrostatic case (σ1 = σ2 = σ3 = p). The elastic energy is
wv =1
2
p2
k=1
2k
µσ1 + σ2 + σ3
3
¶2,
and it is hydrostatic part of the elastic energy.The other part of the total energy wT
wT = w − wv =1
12μ
£(σ1 − σ2)
2 + (σ2 − σ3)2 + (σ3 − σ1)
2¤50
is the energy of shear deformation. If p = σ1 + σ2 + σ3 = 0,
w = wT
An arbitrary stress state we can present as a sum of hydrostatic stress state and deviator(shear) stress state
= +
P
P
P
PS3
S3
S1
S2
P
PS2
S1
σ1
σ1
σ2σ2
σ3
σ3
Home task:A cylindrical specimen is loaded as it is shown in the figure
D
F
M
Lp
h
Parameters of the cylinder are D = 0.2m, h = 0.01m, L = 1m.The values of applied loading:Internal pressure p = 5MPa, axial force P = 105N, moment M = 104NmElastic properties of the material: Young modulus E = 100GPa, Poisson ratio ν = 0.3.Find:
51
1. Maximal normal and shear stresses Max σn, Max τ2. The changes of the diameter and the length of the cylinder ∆L,∆D,3. The angle of rotation of the down top of the cylinder θ.
4 Temperature stresses
Temperature deformationεTij = αijT
αij is the tensor of coefficielts of thermo expansion. For isotropic materials
αij = αδij.
Total deformation is the sum of the temperature and elastic deformation
ε = εe + εT , σ = Cεe.
In the case of the beam between two rigid walls, we have:
ε11 = 0 = εe11 + εT11, εe11 = −εT11 = −αT,σ22 = σ33 = σ12 = σ13 = σ23 = 0,
For the Hooke law in the "technical" form we have:
ε11 =1
Eσ11 = −αT, ε22 = −
ν
Eσ11, ε33 = −
ν
Eσ11.
σ11 = −αET, ε22 = ε33 = ναT.
5 The theory of elasticity
Let us consider an elastic body subjected to an arbitrary system of forces
S1
S2
52
In the application we have to calculate stress and strains tensors in such a body. Thistensors are defined by the equations
σ = σijei⊗ej ε = εijei⊗ej,
εke =1
2
µ∂uk∂ke
+∂w
∂ke
¶, u = uiei.
It is possible to show that the components of the stress tensor should satisfy thefollowing equation
∂σij∂xi
= −qj(x), j = 1, 2, 3.
This equation may be obtained from the condition of equilibrium of a small cubicelement of our body.Differential equations of the elasticity theory may be written in the form
Equilibrium eq. Hooke’s law Strain tensor∂σij∂xi
= −qj(x), σij = cijkeεke, εke =12
³∂uk∂ke+ ∂w
∂ke
´.
We have to add to this system the boundary conditions on the surface of the body
u|s1= u0 σ · n |s1= f0.The sistem of differential equations and boundary conditions may be solved and give
us a unique solution of the problems of the theory of elasticity. The main unknowns ofthe theory are: σ = σ(x), ε = ε(x), u = u(x).
5.1 Example. Deformation of a thick-wall cylinder.
Consider a thick-wall cylinder with internal radius a and external radius b subjected tothe internal presser p. From the symmetry of the problem follows that every differentialelement of the cylinder has the stress state indicated in the figure, and the followingcomponents of the stress tensor are equal to zero:
σrϕ = σrz = σϕz = 0.
53
p a
b
σr+dσr
σϕ σϕ σr
dϕ
r
r+dr
1. Equation of equilibrium of stresses. From the equilibrium of the differential element(projection of all the applied forces on a vertical axis) we have
(σr + dσr)(r + dr)dϕ− σrrdϕ− 2σϕdr sinµdϕ
2
¶= 0.
After simplification that are consequences of the smallness of the differentials dϕ, dr,and dσr
sin
µdϕ
2
¶' dϕ
2, dσrdr ' 0,
we obtain the equation of equilibrium for stresses:
rdσrdr
= σϕ − σr.
2. Deformations. From the symmetry follows that the only essential displacement ofthe cylinder points is the radial displacement u(r)
The radial and tangential deformations are connected with the function u(r) by the equa-tions:
εr =du
dr, εϕ =
u
r, εrϕ = εϕz = εrz = 0.
54
3. Hooke’s law.
σr = λ(εr + εϕ + εz) + 2μεr,
σϕ = λ(εr + εϕ + εz) + 2μεϕ,
σz = λ(εr + εϕ + εz) + 2μεz.
4. Axial stress σzCase a) (cylinder with taps)
σz =pa2
b2 − a2, εz =
pa2
(λ+ 2μ) (b2 − a2)− λ
(λ+ 2μ)(εr + εϕ).
Case b) (cylinder with pistons)
σz = 0, εz = −λ
(λ+ 2μ)(εr + εϕ).
Thus, in general case, one can write:
εz =C
(λ+ 2μ)− λ
(λ+ 2μ)(εr + εϕ),
where
C =pa2
(b2 − a2)or C = 0.
5. Stresses σr, σϕ can be rewritten in the form:
σr =2μλ
λ+ 2μ(εr + εϕ) + 2μεr +
λ
λ+ 2μC,
σϕ =2μλ
λ+ 2μ(εr + εϕ) + 2μεϕ +
λ
λ+ 2μC.
6. From the equilibrium equation for the stresses follows the differential equation forthe function u(r) in the form
r2du2
dr2+ r
du
dr− u
r= 0,
This is Euler equation, and its general solution is:
u = C1r + C2r−1.
As the result, equations for stresses take forms:
σr =4μλ
λ+ 2μC1 + 2μ
µC1 −
C2r2
¶+
λ
λ+ 2μC,
σϕ =4μλ
λ+ 2μC1 + 2μ
µC1 +
C2r2
¶+
λ
λ+ 2μC.
55
7. Boundary conditions at the internal and external radii are:
r = a, σr = −p;r = b, σr = 0.
From these conditions we find
C1 =λ+ 2μ
2μ(3λ+ 2μ)
∙pa2
(b2 − a2)− λ
λ+ 2μC
¸,
C2 =1
2μ
pa2b2
(b2 − a2).
As the result, the equations for stresses take final forms:
σr =pa2
(b2 − a2)
µ1− b2
r2
¶,
σϕ =pa2
(b2 − a2)
µ1 +
b2
r2
¶.
Note: For a thin-wall cylinder we have (h is the thickness of the wall):
b = a+ h ' a, h << a; b+ a ' 2a; r ' a;
σr =pa2
(b+ a)h
µ1− (a+ h)2
r2
¶' pa
2h
µ1− (a+ h)2
r2
¶' −p
2,
σϕ =pa2
(b+ a)h
µ1 +
(a+ h)2
r2
¶' pa
2h
µ1 +
a2
r2
¶' pa
h,
σz =pa2
(b+ a)h' pa
2h.
6 Phenomenon of plasticity
Lüder´s Bands on the surface of a cylindrical specimen
56
ε
y
εp
τ
τy
γ
Yelding sterss is defined from the σ − ε or τ − γ diagrams
σy ≈ 2τ yThe physical mechanism of plasticity is the movement of dislocations:
⊥ ⊥ ⊥
In polycristalls the boarders of grains defined the macroscopic value of σy:
d
σy = σ0 +A√d
7 Criteria of plasticity
We show that the energy of elastic deformation may be presented in the form
w =1
18kI2σ +
1
4μSijSij =
=1
18kI2σ +
1
12μ
£(σ1 − σ2)
2 + (σ1 − σ3)2 + (σ2 − σ3)
2¤ ==
σ2
2k+1
2μT 2.
57
Here the first term depends on hydrastatic part of the etress tensor, and the secondterms depends on the shear stresses
T 2 =1
6
£(σ1 − σ2)
2 + (σ1 − σ3)2 + (σ2 − σ3)
2¤ ,σ =
1
3(σ1 + σ2 + σ3)
Criterion of plasticity may be formulated as follows. The shear part of the elasticenergy is a function of its hydrostatic part:
T = f(σ)
T
7.1 Treska criterion
1. Volume deformation is pure elastic
σ1 = σ2 = σ3 = P =⇒ ε1 = ε2 = ε3 =P
3k
2. If shear stresses exist, plastic deformation also exists.Let
σ1 6= σ2, σ3 = 0, max τ =1
2(σ1 − σ2)
τ σσ
σ1
σ1
58
σ1 6= σ3, σ2 = 0, max τ =1
2(σ1 − σ3)
σ3 6= σ2, σ1 = 0, max τ =1
2(σ2 − σ3).
In general case
|τmax| = max∙1
2|σ1 − σ2| ,
1
2|σ1 − σ3| ,
1
2|σ2 − σ3|
¸Plastic deformations appear if
|τmax| > τ y
7.2 Criterion of Von Mises
Division of the elastic energy on the volume energy and shear energy
w = wv + wT ,
wv =1
18kI2σ, wT =
1
2μT 2
T 2 =1
6
£(σ1 − σ2)
2 + (σ1 − σ3)2 + (σ2 − σ3)
2¤Let us introduce the effective stress
σe =1√2
p(σ1 − σ2)2 + (σ2 − σ3)2 + (σ3 − σ1)2 =
=1√2
q(σ11 − σ22)2 + (σ22 − σ33)2 + (σ33 − σ11)2 + 6(σ212 + σ213 + σ223)
If σ11 6= 0, σ22 = σ33 = σ12 = σ13 = σ23 = 0,
σe = σ11.
The Von Mises criterion.Plastic deformations appear when
wT ≥ wy
(wy is a critical value of the shear elastic energy). In the other words, we have only elasticdeformations if
59
(σ1 − σ2)2 + (σ1 − σ3)
2 + (σ2 − σ3)2 ≤ C2
y = 2σ2ey,
C2y = 2σ
2ey.
What is the value of Cy?1. Basic experiment of torsion of a thin wall cylinder
T
τ = τ y
T ≈2πZ0
τr2hdϕ = 2πτr2h, τ =T
2πr2h
Every element of the cylinder is in the same stress state
h
r
τ
τ
-τ
-τ
σ1 = τ , σ2 = −τ , σ3 = 0.
(2τ y)2 + τ 2y + τ 2y = 6τ
2y = 2(σey)
2 = C2y
60
σey =√3τ y
2. Basic experiment of tension of cylindrical specimen
σ = σy = 2τ y = σ1, σ2 = σ3 = 0
(2τ y)2 + (2τ y)
2 = C2y = 8τ
2y = 2(σey)
2
σey = 2τ y
Thus, we have the following two options1.
C2y = 6τ
2y,
2.C2y = 8τ
2y.
7.3 Graphical presentation of Treska and Von Mises criterion.
Let us introduce the variables
σ1 − σ3 = x, σ2 − σ3 = y, σ1 − σ3 = x− y.
The Treska criterion may be written in the form
max
∙1
2|σ1 − σ3| ,
1
2|σ2 − σ3| ,
1
2|σ1 − σ3|
¸≤ τ y,
|x| ≤ 2τ y, |y| ≤ 2τ y, |x− y| ≤ 2τ y.
61
y
x-2τ y
2τy
2τy
2τy
Cy2 =8 τy
2
x=y
Cy2=6 τ y2
x=-y
2. Von Mires criterion
y = x− 2τ y, x− y = 2τ y, y − x = 2τ y, y = x+ 2τ y
a)x2 + y2 + (x− y)2 ≤ 8τ 2y
This is the equation of an ellipse. What points does this ellipse past
x = 0 ⇒ 2y2 = 8τ 2y ⇒ y = ±2τ y.
y = 0 ⇒ 2x2 = 8τ 2y ⇒ x = ±2τ y
x = y ⇒ 2x2 = 8τ 2y ⇒ x = y = ±2τ yb)
x2 + y2 + (x− y)2 ≤ 6τ 2y
x = 0 ⇒ 2y2 = 6τ 2y y = ±√3τ y ≈ ±1.7τ y
Note: Results of experiments are between these borders. The experiments are per-formed with thin walled cylinders by application of the external moment, tension andinternal pressure.Home task
62
D
F
M
Lp
h
E = 100Gpa, v = 0.3, σy = 300Mpa.
D = 0.2m, h = 0.01m, L = 1m.
1.p = 5Mpa, M = 104Nm,
What is py?(For Treska criterion, for Von Mires Criterion)2.
p = 5Mpa, F = 105N,
What is My?3.
F = 105N, M = 104Nm.
What is Fy?
63
7.4 Ideally elasto-plastic beams
Idealizations in plastic behavior of materials:
ε
σy
σ
ε
σy
σ
ε
σy
σ
Ideally elasto-plastic beam
P
x
z
σ
σy
x
z
σyP=Py
x
z
σyP >Py
σy
x
z
σy
σy
zh
-hb
Moment that acts inside the beam
M = −Zs
σyzds = −b0Z−h
σyzdz + b
hZ0
σyzdz = bσy
∙+h2
2+
h2
2
¸= bh2σy.
ML = bh2σy
Note.
64
The momentMy that corresponds to the beginning of plastic deformations is calculatedas for the elastic beam.
z
σy
Maximal stresses in the beam of a rectangular cross-section (bx2h)
σy =Myh
Jy, Jy =
2
3bh3
As a result we have
My =2
3bh2σy =
2
3ML
ExampleTo find the limit value of the force P that can support the beam in the figure with
rectangular intersection (bx2h, 2h is the vertical size of the cross-section of the beam)
L
L/2
PR1 R2
ML
ML
Solution.Equation of the force equilibrium
PL −R1 −R2 = 0.
Moment equilibrium with respect to the center of the beam
ML −R1L
2= −ML =⇒ R1 =
4ML
L.
65
Moment equilibrium with respect to the end of the beam
ML −R1L+ PLL
2= 0
From this equation we have
ML = −PLL
2+ 4ML =⇒ PL =
6ML
L.
Final equation for the limit force is
PL =6bh2
Lσy.
Home taskTo find limit value of the force P for the following beam of square intersection.
L
L/4P
66
8 The theory of elastic beams
8.1 Tension of direct elastic beamsF
e2
e3
y
z
xe1
External forces applied to the beam:Boundary force F = Fe1,Body forces q(x, y, z) = q(x)e11. Geometrical hypothesis of the theory of beam tensionAny orthogonal cross-section of the beam that is plane before deformation remains
plane after the deformation.
e2
e3
e1
uu
P’
P
W
V
x
zy
e3
e1
P
Q
u
uW
W
x
zFrom this hypothesis follow the equations for the components of the displacement
vector u of the points of the beam
67
u(x, y, z) = ue1 + ve2 + we3,
u(x, y, z) = u(x).
Note: the components v and w may be function of (x, y, z)2. Statical hypothesis of the theory of beam tension
σ11
A
The stress state of the beam is unidimensional:
σ11 = σxx 6= 0, σ22 = σ33 = σ12 = σ13 = σ23 = 0.
Consequences of the main hypotheses1. Deformation of the beam.The component of the strain tensor along x-axis
ε11 = εxx =∂u
∂x= εxx(x)
2. Equilibrium of stresses (σ11(x) = σ(x))
σ(x+dx)
xx+dx
σ(x)
F x x+dx
F+dFq(x)
F x x+dx
F+dFq(x)
The force F (x) that acts in the cross-section of the beam with coordinate x
F (x) =
ZS
σ(x)ds = σ(x)S(x), q(x) = q(x)S(x).
The force in the cross-section F (x+ dx)
68
F (x+ dx) = σ(x+ dx)S(x+ dx) = F (x) + dF (x).
The equilibrium equation (projection of all the forces on x-axis)
F + dF − F + q(x)dx = 0,
dF
dx= −q(x) =⇒ d
dx[σ(x)S(x)] = −q(x).
3. The Hook’s law
ε11 =1
E[σ11 − ν(σ22 + σ33)].
Here E is Young’s module, ν is Poisson’s coefficient. From the static hypothesis of thetheory:
σ22 = σ33 = 0, σ11 = σ,
ε11 = ε, ε =σ
E, σ(x) = Eε(x) = E
du
dx.
Note: The components ε22 and ε33 of the strain tensor are not equal to zero
ε22 = −νε, ε33 = −νε.Thus, we have the following system of equations:
d
dx[σ(x)S(x)] = −q(x), σ(x) = E(x)ε(x), ε(x) =
du
dx
This system is reduced to the following governing equation of the theory of beamtension for the displacement u(x) along the beam
d
dx[E(x)S(x)
du
dx] = −q(x).
For the construction of a unique solution we need boundary conditions.1. If the left end of the beam (x = 0) is fixed,
u(0) = 0.
2. If force F acts on the right end (x = L), we haveµdu
dx
¶x=L
=F
ES(L).
69
3. If the displacement of the right end is known,
u(L) = u0.
Note:In order not to have deflection or torsion, the tension force F should not have moment
in respect center of weight of the intersection of beam.
M =
Zσxy ds =
Zσxz ds = 0, σxy = 0.
Example 1. A cylindrical beam
F
L
ESd2u
dx2= 0 =⇒ d2u
dx2= 0
The general solution of this equation is
u(x) = c1 + c2x.
The boundary condition at x = 0,
u(0) = 0 =⇒ c1 = 0
B.C. at x = L, µ∂u
∂x
¶x=L
=F
SE
Thus, we havedu
dx= c2 =⇒ c2 =
F
SE.
The final solution is
u(x) =F
SEx, ε =
F
SE, σ =
F
S.
70
Note.Some facts from the theory of linear differential equationsTheoremEvery homogeneous differential equation of the order "n"
au(x)dnu
dxn+ an−1(u)
dn−1u
dxn−1+ ...+ a0(x)u(x) = 0 (1)
has “n” linearly independent solutions.
u1(x), u2(x), ... , un(x)
The general solution of this equation has the form:
u(x) = c1u1(x) + c2u2(x) + ...+ cnun(x),
where c1, c2, ...., cn are arbitrary constants. To find c1, c2, ... , cn, we need “n" additionalconditions.Example 2 A conical beam.Radius of the beam r(x) = r0 − αx, the area of the cross-section S(x) = π(r0 − αx)2,Young’s modulus E is constant,
α =r0 − r(x)
x= tan
β
2
Fβ
L
r0
Boundary Conditions
x = 0, u(0) = 0,
x = L,du
dx=
F
ES(L).
To find: u(x), ε(x), σ(x)
71
Solution.The governing equation
d
dx[ES(x)
du
dx] = 0
The first integration gives
ES(x)du
dx= c1,
du
dx=
c1Eπ(r0 − αx)2
.
The second integration
u =c1Eπ
Zdx
(r0 − αx)2+ c2.
Calculation of the indefinite integral
Zdx
(r0 − αx)2=
⎧⎨⎩ t = r0 − αxdt = −αdxdx = − 1
αdt
⎫⎬⎭ = − 1α
Zdt
t2=1
αt=
1
α(r0 − αx).
General solution of the governing equation
u =c1
Eαπ(r0 − αx)+ c2.
The boundary condition
x = 0, u =c1
Eαπr0+ c2 = 0,
x = L,du
dx=
c1Eπ(r0 − ∂x)2
=F
Eπ(r0 − αL)2.
As a result we have
c1 = F, c2 = −F
Eαπr0.
Final result:
u(x) =F
Eαπ
∙1
r0 − αx− 1
r0
¸=
Fx
Eπr0(r0 − αx),
Strains
ε(x) =∂u
∂x=
F
Eπ(r0 − ∂x)2=
F
ES(x)=
σ(x)
E,
Stresses
72
σ(x) =F
S(x)=
F
π(r0 − αx)2.
Example 3 (Home task)The radius of a direct beam depends on x in the following way:
r = r(x) = r0e−βx
Body forces are the weight forces acting in the direction of the axis of the beam
q = ρg, q = S(x)ρg = πr2
0e−2βxρg.
To find: u(x), ε(x), σ(x)Solution:The governing equation
d
dx[ES(x)
du
dx] = − πr
2
0e−2βxρg,
Eπr2
0
d
dx[e−2βx
du
dx] = − πr
2
0e−2βxρg.
The first integration
d
dx[e−2βx
du
dx] = − ρg
Ee−2βx =⇒ e−2βx
du
dx=
ρg
2βEe−2βx + c1
The second integration
du
dx=
ρg
2βE+ e2βxc1 =⇒ u(x) =
ρg
2βEx+
e2βx
2βc1 + c2.
Boundary condition
u(0) = 0 =⇒ c12β+ c2 = 0,µ
∂u
∂x
¶x=L
= 0 =⇒ ρg
2βE+ e2βLc1 = 0.
As a result we have
c1 = −ρg
2βEe−2βL, c2 =
ρg
4β2Ee−2βL.
73
The final solution:
u(x) =ρg
2βEx− ρg
4β2Ee−2β(L−x) +
ρg
4β2Ee−2βL =
=ρg
2βE[x+
e−2βL
2β(1− e2βx)],
ε(x) =∂u
∂x=
ρg
2βE[1− e−2β(L−x)],
σ(x) =ρg
2β[1− e−2β(L−x)].
Note 1.Volume of the beam below point x :
V =
LZx
πr2dx = πr0
LZx
e−2βxdx = − πr202β
e−2βx|Lx
=πr202β[e−2βx − e−2βL].
The weight of the volume below point x
Mg = ρV g =πρr20g
2β[e−2βx − e−2βL] = F (x).
Stress that acts in the intersection x
σ(x) =F (x)
S(x)=
ρg
2β[1− e−2β(L−x)].
Home task:A cylindrical beam has the length 10 m, aqnd the radius 0.5m. The density of the
material of the beam is 104 kgm3 , and elastic modulus is E = 100GPa.What is the difference
in the length of the beam in its horizontal and vertical position?
8.2 The error of the theory of the beam tension
1. Details of the force application
x
F
F/4
74
Saint - Venant principle.The system of forces statically equivalent to zero disturbs the stress state only in a
small neighborhood of the forces application.
F1F2
F0
Fn
x
If F0 + F1 + F2 + ...+ FN = 0, =⇒ σ(x)˜e−γx,
and γ = 0(r0).2. Fast changing of the cross-section size
F
The error of the theory is essential in the indicated region.The condition of the application of the beam theory.
dD
dx<< 1
k is the stress concentration factors (see Gere and Timoshenko pp 137-139).
8.3 Torsion of direct beams
Let us consider a direct beam with a circular cross-section.
θM
One end of the beam is fixed, and torsion momentM is applied to the other end.In this case, the hypothesis of plane intersection is exact statement, and shear defor-
mation is a linear function of the distance from the center of the beam
εrϕ = 2γ.
75
Here γ is the change of the angle between two orthogonal sides of the infinitesimalcube (dr, rdϕ, dx) .
rdr
dϕ
From the geometric consideration we have
γ = rdθ
dx.
Here θ is the angle of rotation of the craoo-section.Hooke’s law give us for the shear stress σrϕ in the beam the equation
σrϕ = τ = μγ, εrϕ =γ
2=
σrϕ2μ
γ dθ
dx
r
The moment in the cross-section with the coordinate x
M(x) =
2πZ0
dϕ
r2Zr1
τ(r)r2dr = μ(x)dθ
dxJp(x), (1)
Jp(x) =
2πZ0
dϕ
r2Zr1
r3dr =π
2(r42 − r41). (2)
Jp is called the polar moment of inertia.
76
Equilibrium of an infinitesimal element of the beam
dx
M+dM
M
m(x)
M + dM +m(x)dx−M = 0
gives us the following equation
dM
dx= −m(x).
Substituting here M from Eq.(1) we obtain the governing equation for the angle ofrotation θ(x) in any cross-section of the beam
d
dx
∙μ(x)Jp(x)
dθ
dx
¸= −m(x).
The boundary conditions:
θ(0) = 0, μJpdθ
dx
¯x=L
=M0.
Example.L
x0 M0
Solution in the interval Iθ1(x) = c1 + c2x.
Solution in the interval II
θ2(x) = c3 + c4(L− x).
77
Boundary conditions:
θ1(0) = 0 =⇒ c1 = 0,
θ2(L) = 0 =⇒ c3 = 0.
Conditions in the border of the intervals x = x0.
θ1(x0) = θ2(x0), μJpdθ1dx
¯x=x0
− μJpdθ2dx
¯x=x0
=M0.
From these equations we have
c2x0 − c4(L− x0) = 0, c2 + c4 =M0
μJp,
Thus, we have
c4 =c2x0
(L− x0), c2 =
L− x0L
M0
μJp.
The final solution
θ(x) =L− x0
L
M0
μJpx, M(x) =
L− x0L
M0, x < x0;
θ(x) =x0L
M0
μJp(L− x), M(x) = −x0
LM0, x > x0;
Home task:
L
x0 M0
M1
Note.The hypothesis of plane cross-sections is not correct for non-cylindrical beams.
8.4 Deflection of direct beams
x y
z
h >> L
e2e3
e1
z
h
z
h
y
78
We will consider direct beams with:1. Slowly changing intersections S(x) ≈ const.2. Cross-section S(x) is symmetric with respect to z-axis.3. The forces are applied in the direction of the z-axis, the vector of the applied
moments are directed along the y-axis.4. For the description of the geometry of deformation we choose the coordinate system
with the axis x that passes through the centers of weight of intersections of the beam. Itis assumed that all such centers are situated in one direct line.Z
S
ydS =
ZS
zdS = 0, dS = dydz.
5. The material of the beam is isotropic with Young modulus E and Poisson ratio ν.
8.4.1 Geometrical hypothesis of the theory (Hypothesis of direct normals).
y S’
x
S
1. An arbitrary cross-section S that is plane and orthogonal to the surface of the beambefore deformation is plane and orthogonal to the surface of the beam after deformation.2. Angle ϕ of the rotation of any intersection S is small: ϕ << 1.Consequences of the geometrical hypothesis
x
S
e3e2
e1 oU
U0
S1
O1
Displacement of an arbitrary point in the cross-section S
U = ue1 + ve2 + we3.
79
U0 is the displacement of the center of weight of the cross-section S.
U = U0+∆U
From the geometrical hypothesis of the theory, we have in the line y = 0
U0 = u0(x)e1 + w0(x)e3,
∆U = ∆u(x, z)e1 +∆w(x, z)e3
ΔuΔw
-z-z
z
x
ϕ
∆u(x, z) = −z sinϕ, ∆w(x, z) = −z(1− cosϕ).If ϕ is small, we have
sinϕ = ϕ− ϕ3
3!+ ...,
cosϕ = 1− ϕ2
2!+ ....
∆U(x, z) ≈ −zϕ, ∆w = 0.
u(x, z) ≈ u0(x)− zϕ, w(x, z) ≈ w0(x).
What is angle ϕ?dw0(x)
dx= tanϕ =
sinϕ
cosϕ≈ ϕ.
The final approximation of the displacement vector of a point (x, 0, z) is
U = [u0(x)− zdw0dx]e1 + w0(x)e3.
80
This equation is the consequence of the hypothesis of direct normals. In the followingwe will write w0(x) = w(x).Deformation of the beam: εxx
εxx =∂u
∂x=
du0dx− z
d2w
dx2= ε0(x)− z
d2w
dx2.
8.4.2 Static hypothesis.
Normal stresses σxx are the most essential in the beam, the stresses σyy and σzz may beneglected.
σzz
σyy
σxx
σzz
σxx >> σyy
σxx >> σzz
Hook’s low and the static hypothesis
εxx =1
E[σxx − ν(σyy + σzz)] =
σxxE=⇒ σxx = Eεxx,
εyy =1
E[σyy − ν(σxx + σzz)] = −
νσxxE
,
εzz =1
E[σzz − V (σyy + σxx)] = −
νσxxE
,
σxx = E
∙ε0(x)− z
d2w
dx2
¸.
Note 1: The stress distribution is lineal in every intersection.Note 2: let us calculate the shear deformation
εxz =1
2
µ∂w
∂x+
∂u
∂z
¶=1
2
µdw
dx− dw
dx
¶= 0.
It follows from the Hooke law that
σyz = 2μεxz = 0.
These equations are a consequence of the hypothesis of direct normals.
81
8.4.3 Forces and moment in the beam
The rules of signs for the moments
yz
x+ My
+q
+V
yz
x+ My
+q
+V+V
σxx > 0 tension,
σxx < 0 compression.
The total force that acts along the beam axis
F =
ZS
σxxdS =
ZS
E
∙ε0(x)− z
d2w
dx2
¸dydz =
= Eε0(x)S(x)−Ed2w
dx2
ZS
zdydz = Eε0(x)S(x).
If F = 0 =⇒ ε0(x) = 0.
As the result, we have for the deformation and stress in the x-direction.
εxx = −zd2w
dx2, σxx = −Ez
d2w
dx2.
The total moment acting in the cross-section
My = −ZS
zσxxdS =
Zs
Ez2d2w
dx2∂s = E
d2w
dx2
Zs
z2dS,
The integral in this equation is called the moment of inertia of the cross-section of thebeam with respect to the y-axis.Z
S
z2dS =
ZS
z2dzdy = Jy
82
Finally, we have
My = EJyd2w
dx2=⇒ d2w
dx2=
My
EJy,
σxx = −zMy
Jy.
8.4.4 Equilibrium of an element of the beam
z
yV+dV
V
M+dMM
q
dx
Equilibrium of forces:−V + (V + dV ) + qdx = 0,
.dV
dx= −q.
Equilibrium of moments with respect to point O.
My + dMy −My + V dx−x+dxZx
qξdξ = 0.
−dMy − V dx− q(dx)2
2= 0,
dMy
dx= −V, =⇒ d3w
dx3= − V
EJy.
After differentiation of this equation we obtain
d2My
dx2= −dV
dx= q.
83
Substituting here the equation for the moment My we obtain the governing equationof the beam deflection theory.
d2
dx2[EJy
d2w
dx2] = q(x).
Here w(x) is the vertical displacement of the neutral line of the beam (the line of theweight centers)
wx
q
If this equation is solved,
σxx = −Ezd2w
dx2
Note1.E and Jy may be functions of x (E = E(x), Jy = Jy(x))!
8.4.5 Boundary conditions of the beam theory
x
y
1)
w=0 , dx
Cantilever enddw = 0.
2) Simple support
w=0, My=0 w(0)=0, w´´(0)=0
84
xF r e e e n d
3 )
M y = 0 w ´ ´ ( 0 ) = 0V = 0 w ´ ´ ´ ( 0 ) = 0
4)
M
V V
M
W´´=M/EJ,
W´´´=V/EJ
W´´=M/EJ,
W´´´= - V/EJ
w ´(0 )= 0 ,w ´´´(0 )= 0 .5 )
8.5 Shear stresses in the beam theory
M M+dM
V+dV
V
M M+dM
V+dV
V
σF*
τ
F*+dF*σ + dσ
z
85
F ∗ =
ZS∗
σds = −My
Jy
ZS∗
zds =My
JyS(z), S(z) = −
ZS∗
zds,
F ∗ + dF ∗ =My + dMy
JyS(z)
Equation of equilibrium of an element of the beam
F ∗ + dF ∗ − F ∗ + τ(z)b(z)dx = 0,
dF ∗ =dMy
JyS(z) = −τ(z)b(z)dx.
τ
dx b(z)
F*+dF*F*
τ
dx b(z)
F*+dF*F*
dMy
dx= −τ(z)b(z)Jy
S(z)= −V =⇒ τ =
V S(z)
Jyb(z)
S(z) = −ZS
zdS = −zZ
−max z
zdz
b(z)/2Z−b(z)/2
dy
Example:(Rectangular cross-section)
2h
b
z
-h
z
86
S(z) = −zZ
−h
zdz
b/2Z−b/2
dy = −bz2
2
¯z−h=
b
2(h2 − z2)
τ(z) =3V
2b2h3b
2(h2 − z2∗) =
3V
4bh3(h2 − z2)
-h
h
τ
z
Jy =2βh3
3, max τ =
3V
4βh
Note 1:The presence of shear stresses is a controversy with the hypothesis of direct normals.
2γ =1
2
µ∂u
∂z+
∂w
∂x
¶,
∂u
∂z= −dw
dx,
u = −zdwdx
, w = w(x) =⇒ 2γ =1
2(−dw
dx+
dw
dx) = 0,
τ =γ
μ= 0.
Note 2:For long beams σxx >> τ .Let us consider an example
L
PP/2 P/2
This is statically determined beam, and an internal moment has the form
87
My = −P2x, x <
L
2
My = −P2x+ P (x− L
2) =
P
2(x− L), x > L/2.
Maximal values of the moment and the shear force are:
maxM = −PL4, maxV =
P
2Normal and shear stresses in the beam
σ = −zMy
Jy= z
PL
4· 3
2bh3=3
8
PLz
bh3, τ =
3V
4bh3(h2 − z2).
max |σ| = 3
8
PL
bh2, max |τ | = 3P
8bh,
max |τ |max |σ| =
h
L<< 1.
Thus, we obtain finallymax |σ| >> max |τ | .
8.6 The main restrictions of the theory of beams
1. Let us consider a cantilever beam subjected to the moment at the free end.
M
Internal moment and shear force in this case are
V = 0, M =M.
Stress state in this beam is defined by the equation: σ = −zMJy
Let us consider the beam with the intersection in the figure subjected by the systemof forces with the equivalent force and moment equal to zero. The Saint Venant principletells that the stresses are localized in a small vicinity of the point of application.
88
But if the central part of the beam is very narrow, the stresses will exist along all thebeam, and all the beam will be involved in deformation.
Thus, for thin walled beams the Saint-Venant principle does not serve.2. When does the hypothesis of direct normals serve?
MM
The role of shear stresses in the process of deformation
PP-h
h
P-h
h
M = PL, maxσII =hM
Jy=3M
2bh2=3PL
2bh2,
Jy =bh3
12=2bh3
3, maxσII =
3PL
2bh2,
P =⇒ P
n, h =⇒ h
n, maxσI =
3PLn2
2n bh2=3PL
2bh2n,
89
maxσI = maxσII n
d2w
dx2=1
R=
M
EJ,
1
RII=3PL
2Ebh3,
1
RI=3
2
PL
Ebh3n2
Home taskTo develop the differential equation of the deflection of the beam laying on an elastic
support. Elastic characteristic of the support is g(x) = kw(x).
8.7 Methods of the solution of the equation of the beam theory.The method of initial parameters.
Governing equation of the beam deflection theory
d2
dx2
∙EJ(x)
d2w
dx2
¸= q(x).
x
If EJ = const, we haved4w
dx4=
q(x)
EJ. (*)
Homogeneous equation:d4w
dx4= 0.
General solution of Eq.(*):
w(x) = c1 + c2x+ c3x2
2+ c4
x3
3!+ w∗(x).
Here w∗(x) is any partial solution of Eq.(*).Theorem:Partial solution w∗(x) of Eq.(*) is:
w∗ (x) =
xZ0
(x− ξ)3
3!
q(ξ)
EJdξ
Proof.
90
w0∗(x) =
xZ0
(x− ξ)2
2
q(ξ)
EJdξ, w00∗(x) =
xZ0
(x− ξ)q(ξ)
EJdξ,
w000∗ (x) =
xZ0
q(ξ)
EJdξ, wIV
∗ (x) =q(ξ)
EJ.
Thus, the general solution of the governing equation (*) has the form:
w(x) = c0 + c1x+ c2x2
2+ c3
x3
3!+
1
EJ
xZ0
(x− ξ)3
3!q(ξ)dξ
Note:
w(0) = w0 = c0, w0(0) = w1 = c1, w00(0) = w2 = c2, w000(0) = w3 = c3.
w(x) = w0 + w1x+ w2x2
2+ w3
x3
3!+
1
EJ
xZ0
(x− ξ)3
3!q(ξ)dξ
This is the basic equation of theMethod of Initial Parameters.Example 1:
z
M
Because q = 0 we have
w(x) = w0 + w1x+ w2x2
2+ w3
x3
3!.
Boundary conditions at x = 0
w(0) = w0 = 0, w00(0) = w2 =
M0
EJ.
Boundary conditions at x = L
w(L) = 0, w0(L) = 0,
91
Thus, we have
w(x) = w1x+M0
EJ
x2
2+ w3
x3
3!and the boundary conditions take the forms
w(L) = w1L+M0
EJ
L2
2+ w3
L3
3!= 0,
w0(L) = w1 +M0
EJL+ w3
L2
2= 0.
Canonical form of this system is
w1 + w3L2
6= −M0L
2EJ,
w1 + w3L2
2= −M0
EJL .
Solution of this system is
w1 = −M0
4EJL, w3 = −
3M0
2EJL.
Final solution w(x) is
w(x) = − M0
4EJLx+
M0
EJ
x2
2− 3M0
2EJL
x3
6= −M0L
4EJx³1− x
L
´2.
Example 2.
L
z
x
q
General solution
w(x) = w0 + w1x+ w2x2
2+ w3
x3
3!+
x
1
EJ
Z0
(x− ξ)3
3!q(ξ)dξ =
= w0 + w1x+ w2x2
2+ w3
x3
3!− q
EJ
(x− ξ)4
24
x
|0
=
= w0 + w1x+ w2x2
2+ w3
x3
3!+
q
24x4.
92
Boundary conditions at x = 0
w0 = w1 = 0.
Boundary condition at x = L
w(L) = 0, w00(L) = 0.
w(L) = w2L2
2+ w3
L3
6+
q
24EJL4 = 0,
w00(L) = w2 + w3L+q
2L2 = 0
The systems for the constants w2, w3 is
w2 + w3L
3= − q
12EJL2,
w2 + w3L = − q
2EJL2.
Solution of this system
w2 =q
8EJL2, w3 = −
5
8EJqL.
Final solution:
w(x) =q
8EJL2 · x
2
2− 5
8EJqL
x3
6+
q
24x4 =
qx2L2
48EJ
∙3− 5x
L+ 2
³xL
´2¸.
Home task.
q0
q = q0(1−x
L)
93
Solution
I(x) =q0x
4
120EJ
³5− x
L
´;
w(x) =q0EJ
x2(L− x)(x− 2L)2120L
=q0
120EJL4³xL
´2 ³1− x
L
´³2− x
L
´2.
Constants
w0 = w1 = 0, w2 =L2q015EJ
, w3 = −2Lq05EJ
.
Example 3.
x
Solution in the interval 0 < x < L1
w(1)(x) = w2x2
2+ w3
x3
6Solution in the interval L1 < x < L
w(2)(x) = w(2)0 + w
(2)1 (x− L1) + w
(2)2
(x− L1)2
2!+ w
(2)3
(x− L1)3
3!Conditions at x = L1
w(1)(L1) = w(2)(L1) ⇒ w(2)0 = w
(1)2
L212+ w
(1)3
L316,£
w(1)¤0(L1) =
£w(2)
¤0(L1) ⇒ w
(2)1 = w
(1)2 L1 + w
(1)3
L212,
EJ1£w(1)
¤00(L1) = EJ2
£w(2)
¤00(L1)⇒ w
(2)2 =
hw(1)2 + w
(1)3 L1
i J1J2,
EJ1£w(1)
¤000(L1) = EJ2
£w(2)
¤000(L1)⇒ w
(2)3 = w
(1)3
J1J2.
Conditions at x = L
w(2)(L) = 0,£w(2)
¤00(L) =
M
EJ2,
This is two equation for two unknown constants w(1)2 , w(1)3 .
Home taskTo finish this problem.
94
8.8 Concentrated force.
x0
x
P
x0
x
P
xxx0 x0+h
qh(x,x0)P/h
Total force that correspond to this distribution
hZ0
qh(x, x0)dx =p
hh = P.
The limit value of this distribution is
limh→0
qh(x, x0) = Pδ(x− x0).
Here δ(x− x0) is called Dirac’s delta-function.Definition of δ-function.1. (Intuitive)
δ(x− x0) =
½0, x 6= x0∞ , x = x0
,
Zδ(x− x0)dx = 1.
2.(Mathematical).Let f(x) belong to a class of continuos functions. δ(x−x0) is a functional on this class
of functions defined by the equation:
(δ, f) =
Zf(x)δ(x− x0)dx = f(x0).
a b
f(x)
x0 x0+h
1/h
Let us prove the last equation if
δ(x− x0) = limh→0
δh(x, x0).
95
Here
δh(x, x0) =
½1h, x0 < x < x0 + h0, x < x0 or x > x0 + h
,
Proof:
bZa
f(x)δh(x, x0)dx =
x0+hZx0
f(x)δh(x, x0)dx ≈
≈ f(x0)
x0+hZx0
δh(x, x0)dx = f(x0).
limh→0
bZa
f(ξ)qh(ξ)dξ = limh→0
P
h
x0+hZx0
f(ξ)dξ ≈ limh→0
P
hf(x0)h = Pf(x0).
If this formula applied to the integral term in the equation of the method of initialparameters, we have
limh→0
xZ0
(x− 3)33!
qh(ξ)dξ =
½0, x < x0
(x−3)33!
P, x > x0.
Let us introduce function Ψ(x) :
Ψ(x) =
½0, x < 0x3
6, x > 0
.
x0
Ψ(x) Ψ(x- x0)
The general solution of the equation of the beam deflection in the presence of concen-trated forces takes the form
96
w(x) = w0 + w1x+ w2x2
2+ w3
x3
3!+
xZ0
(x− ξ)3
3!
q(ξ)dξ
EJ+
+MXk=0
Pk
EJΨ(x− xk).
x
P0P1
x1x0
Derivatives of Ψ - function:
Ψ0(x) =
½0, x < 0x2
2, x > 0
, Ψ00(x) =
½0, x < 0x, x > 0
, Ψ000(x) =
½0, x < 01, x > 0
.
Example.
Taking into account the boundary conditions at x = 0 we have general solution forthe vertical deflection in the form
w(x) = w2x2
2+ w3
x3
6+
P
EJΨ(x− L
2).
Boundary conditions at x = L
w(L) = w00(L) = 0,
These conditions give us the system for the constants equation w2, w3
w2L2
2+ w3
L3
6= − P
6EJ(L− L
2)3 = − P
48EJL3,
w2 + w3L = − PL
2EJ.
97
Solution of this system:
w2 =3PL
16EJ, w3 = −
11P
16EJ.
Final solution
w(x) =PL
32EJx2µ3− 11
3
x
L
¶+
P
EJΨ(x− L
2).
Home task
L/2 L/4
w(x) = w2x2
2+ w3
x3
3+ I(x) +
P
EJΨ(x− 3L
4),
I(x) =
½q
24EJx4, x < L
2q
24EJ
£x4 − (x− L
2)4¤, x > L
2
.
Boundary conditions at x = L
w(L) = w0(L) = 0.
Constantsw2 =
L
192EJ(9P + 11Lq) , w3 = −
5P + 13Lq
32EJ.
Final solution
w(x) =
⎧⎨⎩x2
384EJ[9LP + 11L2q − 10Px− 26Lqx+ 16qx2] , x < L
2;
− x2
384EJ[L4q − 8L3qx+ L(−9P + 13Lq)x2 + 2(5P − 3Lq)x3] , L
2< x < 3L
4;
− 1384EJ
(L− x)2 [22LP + L2q − 54Px− 6Lqx] , x > 3L4.
.
98
8.8.1 Concentrated moment applied in the middle part of a direct beam
x0
M
x0
M
z
x
-M/h M/h
hx0 x0 + h
The displacement that correspond to two forces applied to the beam is
limh→0
∙− M
hEJΨ(x− x0) +
M
hEJΨ(x− x0 − h)
¸=
−M
EJlimh→0
∙Ψ(x− x0)−Ψ(x− x0 − h)
h
¸= −M
EJΨ0(x− x0).
Let us introduce function Ψ1(x)
Ψ1(x− x0) = Ψ0(x− x0) =
½0, x < x0
(x−x0)22
, x > x0.
Derivatives of this function:
Ψ01(x− x0) =
½0, x < x0
(x− x0), x > x0, Ψ00
1(x− x0) =
½0, x < x01, x > x0
.
General equation of the method of initial parameters
w(x) = w0 + w1x+ w2x2
2!+ w3
x3
3!+
xZ0
(x− ξ)3
3!
q(ξ)
EJ)dξ +
+Xi
Pi
EJΨ (x− xi)−
Xj
Mj
EJΨ1 (x− xj) .
Example
99
M0
x
z
w(x) = w1x+ w3x3
6− M0
EJΨ1
µx− L
2
¶.
Boundary conditions
w(L) = w1L+ w3L3
6− M0
EJ
L
8= 0,
w0(L) = w1 + w3L2
2− M0
EJ
L
2= 0.
Final system for the constants w1, w3:
w1 + w3L2
6=
M0
EJ
L
8
w1 + w3L2
2=
M0
EJ
L
2
Final solution
w1 = − 116
M0
EJL, w3 =
9
8
M0
EJL,
w(x) = −M0Lx
16EJ
∙1− 3
³xL
´2¸− M0
EJΨ1
µx− L
2
¶.
100
8.8.2 Continuous beam with several supports.
x
z
y
qP
L1
L2
L3
R1 R2 R3
P q
Let us change the supports for the force of reaction R1, R2, R3. This are new unknownsof the problemGeneral solution take the form:
w(x) = w0 + w1x+ w2x2
2+ w3
x3
3+
xZ0
(x− ξ)3
3!q(ξ)dξ +
+P
EJΨ(x− xi)−
3Xj=1
Rj
EJΨ(x− xj).
Boundary conditions:
w00(0) = 0, w000(0) = 0,
w(L) = 0, w0(L) = 0.
and the conditions at the supports
w(L1) = 0, w(L2) = 0, ..., w(Ln) = 0
give us all number of necessary equations for the unknowns of the problem.
101
Example:
L/4 L/4 L/4 L/4
P0 M0P0 M0
R
x
General solution
w(x) = w2x2
2+ w3
x3
6+
P0EJ
Ψ(x− L
4)− M0
EJΨ1(x−
3L
4)− R
EJΨ(x− L
2).
Boundary conditions at x = L and x = L/2 :
w2L2
2+ w3
L3
6− R
EJ(L
2)21
6=
M0L2
32EJ− 2P0L
3
128EJ,
w2L+ w3L2
2− RL2
8EJ=
M0L
4EJ− 9P032EJ
,
w2L2
8+ w3
L3
48= − P0
EJ
L3
384
Home task.
q
L/2 L/2
102
w(x) = w2x2
2+ w3
x3
6+ I(x)− R
EJΨ
µx− L
2
¶,
W2 = − L2q
16EJ, W2 =
3Lq
8EJ, R =
7Lq
8.
I(x) =
½0, x < L
2;
q24EJ
¡x− L
2
¢4, x > L
2
.
Home task.
L/4 L/4 L/4 L/4
P0
M0
w(x) = w2x2
2+ w3
x3
6− M0
EJΨ1(x−
L
2)− R1
EJΨ(x− L
4)− R2
EJΨ(x− 3L
4),
w2 =M0 + LP022EJ
, w3 = −6(LP0 +M0)
11EJL,
R1 = −3(9LP0 + 20M0)
22L, R2 =
48M0 + 37LP022L
.
9 Deflection of the beams with non-symmetric areasof intersections
x
z
y
103
Vector of displacement of an arbitrary point of the beam
U = ue1 + ve2 + we3
The component u(x, y, z) takes the form
u(x, y, z) ∼= u0(x)− ϕz − ψy,
ϕ =dw
dx, ψ =
dv
dx.
Deformation along the x−axis:
ε11 =∂u
∂x=
du0dx− z
d2w
dx2− y
d2v
dx2.
Stress distribution in every intersection
σ11 = Eε11 = E
∙ε0(x)− z
d2w
dx2− y
d2v
dx2
¸.
1. Total force in the intersection
F =
ZS
σ11dx = Eε0(x)S(x); if F = 0, ⇒ ε0 = 0
2. Total moments in the intersection
My = −ZS
zσ11ds = Ed2w
dx2
ZS
z2ds+Ed2v
dx2
ZS
yzds =
= Ed2w
dx2Jyy +E
d2v
dx2Jyz,
Mz = −ZS
yσ11ds = Ed2w
dx2Jyz +E
d2v
dx2Jzz.
Thus, we have the following equations connected moments and derivatives
My = Ed2w
dx2Jyy +E
d2v
dx2Jyz, Mz = E
d2w
dx2Jyz +E
d2v
dx2Jzz
d2w
dx2=
MyJzz −MzJyz
E£JzzJyy − J2yz
¤ , d2u
dx2=
MzJyy −MyJyz
E£JzzJyy − J2yz
¤If Mz = 0,
d2w
dx2=
MyJzz
E£JzzJyy − J2yz
¤ , d2u
dx2=
−MyJyz
E£JzzJyy − J2yz
¤ .104
Let us change the denotation of axes
x =⇒ x3, y =⇒ x1, z =⇒ x2,
and introduce the tensor
J =
ZS
x⊗ xds =ZS
x21dse1⊗e1 +Z
x1x2ds (e2⊗e1+e1⊗e2) +Z
x12dse3⊗e3 =
= J11e1⊗e1 + J12 (e2⊗e1+e1⊗e2) + J22e2⊗e2.
J is the tensor of the moments of inertia:
g(2)
g(1)
e1
e2
x1
x2
ζ1
ζ2
As every two rank symmetric tensor J has two eigenvalues J (1), J (2) and two eigenvec-tors g(1),g(2)
: J (1), J (2) =⇒ g(1),g(2).
J = J (1)g(1)⊗g(1) + J2g(2)⊗g(2), J (12) = 0.
In the basis of eigenvectors the equation of deflection is simplified
d2w
dζ22=
M2
EJ (2),
d2v
dζ21=
M1
EJ (1)
Here ZS
ξ1ξ2ds = 0,
ZS
ξ21ds = J (1),
ZS
ξ22ds = J (2).
105
The general case we have
d4w
dζ41=
q1EJ1
,d4v
dζ42=
q2EJ2
Example 1:
x1
x2
4 5
3
z=x2
y=x1
x=x3
P
3
x1
4
x2
Position of the center of weight of the intersection
3
x1
4
x2
ZS
(x− xc)dS = 0,
ZS
xdS = 6xc.
1
6
⎡⎣Zs
x1dse1 +
Zs
x2dse2
⎤⎦ = 2e1 +4
3e2 = xc1e1 + xc2e2.
xc1 = 2, xc2 =4
3.
106
-2 -1
- 4/3
x1
8/3
x2=4/3(x1+1)
Tensor of the moments of inertia
J =
ZS
x⊗ xds =ZS
x21dse1⊗e1 +ZS
x1x2ds(e1⊗e2+e2⊗e1) +ZS
x22dse2⊗e2
= Jijei⊗ej.
J11 =
ZS
x21ds =
1Z−2
x21dx1
43(x1+1)Z− 43
dx2 =
1Z−2
x21
∙4
3(x1 + 1) +
4
3
¸dx1 = 3,
J12 =
1Z−2
x1
43(x1+1)Z− 43
x22dx2 = 2, J22 =
1Z−2
dx1
43(x1+1)Z− 43
x22dx2 =16
3.
J = 3e1⊗e1 + 2(e1⊗e2+e2⊗e1) +16
3e2⊗e2.
Eigenvalues of tensor J
det kJij − λδijk = det°°°° 3− λ, 2
2, 163− λ
°°°° = λ2 − 253λ+ 12 = 0.
Roots of this equationλ1 = 1.851, λ2 = 6.482
Eigenvectors
λ1 = 1.851 =⇒ g(1) = −0.867e1 − 0.498e2,λ2 = 6.482 =⇒ g(2) = 0.498e1 − 0.867e2
107
J = 1.851g(1)⊗g(1) + 6.482g(2)⊗g(2)
e1
e2g(2)
g(1)y
P
P2e2
P1
e1
g(2)
g(1)
Example 1(continuation)
x
z
y
y
P
P2e2
P1
e1
g(2)
g(1)
z
P
e2
g(1)
e(1)
g(2)
P2 P1
In the basis of eigenvectors the system of equations for the components of deflectiontakes the form
108
d2w
dx2=
P2(L− x)
EJ2,
d2v
dx2=
P1(L− x)
EJ1.
J2 = 6.482cm4, J1 = 1.387cm4.
Solution of these equations
w =P2
6.482E
∙Lx2
2− x2
3
¸=
P26 · 6.482E (3L− x), u =
P1x2
6 · 1.851E (3L− x)
Vector of displacement
U = vg(1) + wg(2) + ue3 = v(−0.867e1 + 0.498e2) + w(0.957e1 + 0.807e2) + ue3 =
= ve1 + we2 + ue3,
v = −0.867v + 0.498w, w = 0.498u+ 0.807w.
P = P1g(1) + P2g
(2) = Pe1,
P1 = P · g(1) = P (e1·g(1)) = −0.867P,P2 = P · g(2) = 0.498P
Home task:
x
zy
P
P
1
3
2
L=20
P=10N
10 Energy of deformation of elastic beam.
1. The energy of deflection
σ11 = −zMy
Jy, ε11 =
1
Eσ11 = −
zM
EJy.
109
w =1
2σ11ε11 =
z2M2y
2EJ2y,
W =
ZV
wdv =
LZ0
dx
ZS
z2M2y (x)
2EJ2y (x)ds =
LZ0
M2y
2EJ2ydx
ZS
z2ds =
LZ0
M2y
2EJ2ydx.
2. The energy of tension
F
w =1
2σ11ε11 =
F 2
2ES2, Wt =
LZ0
F 2
2ESdx
3. The energy of torsion
M
ε12 =τ
2μτ = r
Mm(x)
Jp= σ12 w = σ12ε12 =
r2M2m
2μJ2p
Wm =
LZ0
M2m(x)
2μJp(x)dx
Total elastic energy of a beam
W =
LZ0
M2y
2EJydx+
LZ0
F 2
2ESdx+
LZ0
M2m
2μJpdx.
110
10.1 Castigliano theorem
1.Elastic energy W = W (P1, P2, ...Pn) is a function of he applied forces. If the forcePn increases on dPn , the energy will change as following
Wt =W (Pn) +∂W
∂PndPn (1)
dPndPndPn
Pn
On the other hand, the increment of energy is
dPn dPn
+
Pn
At the first stage the change of the energy is
dW = dA =1
2dPndun.
At the second stage we have
Wt =W (P1, P2, ...Pn) + undPn +1
2dPndun. (2)
111
If we compare now Eqs.(1) and (2), we obtain
∂W
∂Pn= un.
Example 1.To find the displacement δ under the force P applied to the beam
P
δ
W =
LZ0
M2
2EJdx =
LZ0
P 2(L− x)2
2EJdx = − P 2
2EJ
0ZL
z2dz = −P2z3
6EJ
0
|L
=P 2L3
6EJ.
δ =∂W
∂P=
PL3
3EJ
Note:Using the Castigliano theorem to find vertical displacement δ at point x0 if the moment
M is applied to the end of the beam
δ Mx0
M
QRa
Max0
M(x) = −Rax+Ma , x < x0,
M(x) = −Rax+Q(x− x0) +Ma , x > x0.
Ra = Q, Ma =M +Qx0.
M(x) = −Qx+M +Qx0 = Q(x0 − x) +M, x < x0
M(x) =M, x > x0
W =1
2EJ
⎡⎣ x0Z0
[Q(x0 − x) +M ]2 dx+M2
LZx0
dx
⎤⎦ ,112
∂W
∂Q=
2
2EJ
x0Z0
[Q(x0 − x) +M ] (x0 − x)dx |Q=0=
=M
EJ
x0Z0
(x0 − x)dx = −M
EJ
0Zx0
ζdζ =M
EJ
x202.
δ(x0) =M
EJ
x202.
Note.The Castigliano theorem is correct for the generalized forces and the corresponding
generalized displacements. For instance, if moment M acts on a beam we have
∂W
∂M= ω,
where ω is the angle of rotation in the point of application of the moment.Home task.To find horizontal displacement δ at the point of application of the force P to the
beam system shown in the picture.
δ
L1
L2
P
The beams have square intersections: h1 × h1 for the first beam and h2 × h2 for thesecond beam.
113
11 Impact action on beams
Example 1.To find maximal horizontal displacement of the beam after the collision with the body
of mass m that moves with the horizontal velocity v.
m v x
z
P
Let T is the kinetic energy of the body, and W is the elastic energy of the beam inthe moment of collision.
T =mv2
2, W =
LZ0
M2
2EJdx
The equation of the balance of energy has the form
W = T.
Calculation of the elastic energy
M = P (L− x)⇒W =
LZ0
P 2(L− x)2
2EJdx = − P 2
2EJ
0ZL
z2dz =P 2
6EJL3
Balance of energymv2
2=
P 2
6EJL3 ⇒ P =
r3EJmv2
L3,
δ =Pl3
3EJ=
r3EJmv2
L3L3
3EJ=
rmv2L3
3EJ.
Example 2.Let a body of mass m fall down on the beam shown in the picture. What is the
maximal displacement in the center of the beam after the collision?
114
m
PdH
δ
Potential energy of the body
m =F
g, Π = mg(H + δ)
Balance of energy (potential energy of the body is equal to the elastic energy of thebeam)
W = Π
W =
LZ0
M2
2EJdx =
1
EJ
L2Z0
P 2x2
4EJdx =
P 2L3
96EJ=24δ2EJ
L3
∂W
∂P= δ =
PL3
48EJ⇒ P =
48EJ
L3δ
mg(δ +H) =24EJ
L3δ2 ⇒ δ2 − L3mg
24EJδ − L3mg
24EJH = 0
Displacement in the center of the beam is
δ =mgL3
48EJ+
"µmgL3
48EJ
¶2+ (2H)
µmgL3
48EJ
¶# 12
Note that the static displacement δst under the weight of the body (F = mg) is
δst =mgL3
48EJ⇒ δd = δst +
£δ2st + 2Hδst
¤ 12 .
1. If H = 0, we haveH = 0⇒ δd = 2δst
2. If H >> δstH >> δst ⇒ δd =
p2Hδst
115
Note:
P(t)
δx
W=0
W´´=0
W=0
W´´=0
For H = 0 the exact equation of the problem is
EJ∂4W
∂xy− ρ
∂2W
∂t2= P (t)δ(x− L
2), P (t) = PH(t)
Initial Condition
w (x, t) |t=0 = 0,µ∂w
∂t
¶t=0
= 0.
The behavior of the solution is similar to the picture
5 10 15 20 25 30
0.5
1
1.5
2
116
Home task problem.The gate of the American football is constructed from cylindrical beams of diameter
D = 0.1m with the thickness of the walls h = 0.01m and has the following sizes:L1 = 2m,L2 = 4m,L3 = 1.5m. The material of the cylinders is steel with the followingproperties: E = 100GPa, ν = 0.3, σy = 300MPa. The ball with the mass 1kg moveswith the velocity v and clashes with the gate in its angle (see figure). What should bethe velocity v of the ball in order to cause plastic deformation in the gate after collision?
L2
L1
L3
v
117
12 Stability of elastic beams (columns)
Examples of unstable elastic systems.1. External pressure applied to a cylindrical shell
P
Note. For the stable case we have for stresses
σ =PR
h.
2. Moment applied to a thin walled specimen
M
M
M
M
For the stable state
τ =M
2πR2h, τ < τ y.
3. An axial force applied to an elastic beam
118
σ =P
S
12.1 Euler Problem
z
xP
w
z
xP
Equation of deflection:
d2w
dx2=
M
EJ, M = −Pw,
d2w
dx2+
P
EJw = 0 , k2 =
P
EJ,
d2w
dx2+ k2w = 0 ⇒ w = A cos (kx) +B sin (kx)
119
Boundary conditions
w(0) = 0⇒ A = 0, w(L) = B sin kL = 0,
Because B 6= 0, we have
kL = nπ,
rP
EJL = nπ, P =
(nπ)2
L2EJ, n = 1, 2, ...
Final solution
w(x) = B sin
ÃrPn
EJx
!= B sin(
nπ
Lx)
n = 1 ; P1 =π2
L2EJ
n = 2 ; P2 =4π2
L2EJ
n = 3 ; P3 =ϑπL
L2EJ
Questions:1) What is the value of constant B if P = Pn ?2) If P1 < P < P2 ,=⇒ A = B = 0 or not ?A more careful non linear analysis of the problem says:1. If P < P1 , then A = B = 0.2. If P1 < P < P2 −→ The solution A = B = 0 is not stable!3) w(x) cannot be described in elementary functions.
120
12.2 The dependence of Pn on boundary conditions
Example 1.
z
x
P
xδ w
d2w
dx2=
M
EJ, M = P (δ − w)
Differential equation of the deflection
d2w
dx2+ k2w =
P
EJδ = k2δ , k2 =
P
EJ
General solutionw = A cos k x+B sin k x+ δ.
Boundary condition at x = 0 :
w(0) = δ ⇒ A = 0.
Boundary condition at x = L :
w(L) = δ +B sin k L = 0, B = − δ
sin k L,
w(L) = Bk cos kL = 0,
cos kL = 0, kL = π(n− 12), n = 1, 2, 3, ....
rPn
EJL = π(n− 1
2) , Pn =
π2(n− 12)2
L2EJ
n = 1 : P1 =π2
4L2EJ,
n = 2 : P2 =ϑπ2
4L2EJ
Final solution:
w = δ(1− sin kxsin kL
)
121
Note:δ = w(0) is not defined in the solution.Minimal critical force
P1 =π2
4L2EJ.
Note:In the case of Euler beam: P1 = π2
L2EJ.
Example 2:
z
x
M Mw M
P
d2w
dx2=
M
EJ, M(x) = −Pw +M,
d2w
dx2+ k2w =
M
EJ, k2 =
P
EJ.
General solution:
w = A cos kx+B sin kx+M0
P
Boundary conditions at x = 0
w(0) = A+M0
P= 0 ⇒ A = −Mo
P.
w(0) = −Mo
Pk sin kx+Bk cos kx |x=0= 0⇒ B = 0
The solution
w(x) =Mo
P[1− cos (kx)] .
Boundary conditions at x = L
122
w(L) =Mo
P[1− cos (kL)] = 0⇒ kL = 2πn,
w(L) = −Mo
Pk sin (kL) = 0 ⇒ kL = 2πn.
Pn =4πn2
L2EJ , n = 1, 2, ...
minPn =4π2
L2EJ.
Note that this force is four times more than Euler force.Example 3.
P PR
W´
z
W´´
z
x
z
Differential equation of deflection
d2w
dx2=
M
EJ.
Equation for the bending moment
M = −Pw +Rx.
Final equation
w + k2w =Rx
EJ, k =
rP
EJ.
General solution of this equation
w = A cos kx+B sin kx+R
Px.
123
Boundary conditions:
w(0) = 0⇒ A = 0,
w(L) = B sin kL+R
PL = 0,
w(L) = Bk cos kL+R
P= 0.
Condition of the existence of a non-trivial solution of the system of last two equationsfor B and R is
det
¯sin kL, RL
P
k cos kL, RP
¯=
R
Psin kL− kL cos kL
R
P= 0.
A short form of this equation is
tan (kL) = kL.
If a new variable z is introduced, we have
kL = z ⇒ tan (z) = z.
-10 -5 5 10
-10
-7.5
-5
-2.5
2.5
5
7.5
10
The roots of this equation
z1 = 4.493, z2 = 7.725, z3 = 10.904, .......
Thus, minimal value of z = kL is
kL = 4.49, P1 =(4.49)2
L2EJ,
P1 =π2
(μL)2EJ, μ =
π
4.49= 0.7
124
Example 4.
R
z
xP
3L/4δ
w =M
EI
M(x) = −Ph(x) = −P (δ − |w(x)|), x <L
4;
w(x) < 0, δ = −w(0), h(x) = −w(0) + w(x),
M(x) = −P (w − w(0)) = −P (w + δ), x <L
4
M(x) = −P (w + δ) +R(x− L
4), x >
L
4
Differential equation of deflection (k2 = PEJ)
d2w
dx2+ k2w = −k2δ, x <
L
4,
d2w
dx2+ k2w = −k2δ + k2
R
P(x− L
4), x >
L
4.
General solution
w = A1 cos kx+B1 sin kx− δ , x <L
4,
w = A2 cos kx+B2 sin kx− δ +R
P(x− L
4), x >
L
4.
Boundary conditions
w(0) = −δ ⇒ A1 = 0,
w(L
4− 0) = w(
L
4+ 0) = 0,
w(L
4− 0) = w(
L
4+ 0),
w(L) = 0, w(L) = 0.
125
As a result we have the following system of equation for the unknowns: δ,B1, A2, B2, R.
−δ +B1 sin(kL
4) = 0,
−δ +A2 cos(kL
4) +B2 sin(k
L
4) = 0,
B1k cos(kL
4) +A2k sin k
L
4−B2k cos k
L
4− R
P= 0,
−δ +A2 cos(kL) +B2 sin kL+R
P
3L
4= 0,
−A2k sin(kL) +B2k cos(kL) +R
P= 0.
The condition of the existence of non-trivial solution of this system is
det
°°°°°°°°°°−1, sin
¡kL4
¢, 0, 0, 0
−1, 0, cos¡kL4
¢, sin
¡kL4
¢, 0
0, kL cos¡kL4
¢, kL sin
¡kL4
¢, −kL cos
¡kL4
¢, −L
P
−1, 0, cos (kL) , sin (kL) 3L4P
0, 0, −kL sin (kL) , kL cos (kL) , LP
°°°°°°°°°°= 0
After expansion of the determinant we obtain the equation
F (kL) = −LkL4P
∙3kL cos(kL) + 8 sin
µkL
4
¶+ 2 sin
µkL
2
¶− 6 sin (kL)
¸= 0
Let us consider the function (kL = x)
f(x) = 3 cos(x) +8
xsin³x4
´+2
xsin³x2
´− 6
xsin (x) ,
which roots coincide with the roots of function F (x).The graph of function f(x) is
10 20 30 40 50
-2
2
4
126
The first roots of the equation f(x) = 0 are:
x1 = 3.448, x2 = 7.853, x3 = 13.9606, x4 = 17.278, x5 = 20.1764, ....
Thus,we have
Pn =x2nL2
EJ.
minPn = P1 =3.4482
L2EJ.
Home task:
L
60º 45º
Pcr =?
E = 100GPa, L = 7m, D = 0.1m, h = .006m.
12.3 Compression of non-ideal beams.
Pv0
w
z
x
Differential equation of deflection:
d2w
dx2=
M
EI, M = −P [ν0(x) + w(x)] ,
d2w
dx2+ k2w = −k2v0(x) , k2 =
P
EJ. (1)
127
Let the initial deflection be:v0(x) = δ sin
πx
l.
General solution of Eq.(1) is:
w = A cos kx+B sin kx+D sinπx
l.
The constant
D
∙−π
2
l2+ k2
¸sin
π
lx = −k2δ sin π
lx,
D = − k2δ
k2 − (πl)2=
k2δ
(πl)2 − k2
,
Boundary conditions
w(0) = 0⇒ A = 0,
w(L) = B sin kL = 0⇒ B = 0.
Final solution:
w(x) =k2δ
( πL)2 − k2
sinπx
L, k2 =
P
EJ,
Pc =π2
L2EJ,
P
Pc=
k2¡πL
¢2 ,w(x) =
P
Pc − Pδ sin
πx
l, w(
L
2) =
P
Pc − Pδ.
w(L/2)
δ 3
δ 2
δ 1
Pc P
δ3 > δ2 > δ1
128
12.4 Complex deflection of the beam
z
x
Differential equation of deflection of statically determined beam
d2w
dx2=
M
EJ, M = −Pw +My,
d2w
dx2+ k2w =
My
EJBoundary conditions
w(0) = w(L) = 0
12.4.1 Method of initial parameters of statically determined beam
General solution of the differential equation of the problem has the form
w(x) = c1 cos kx+ c2sin kx
k+
Z x
0
1
ksin k(x− z) m(z) dz.
x = 0, c1 = w(0), w(0) = c2.
Proof.
w0(x) =
Z x
0
cos k(x− z)m(z)dz,
w0(x) = m(x)− k
Z x
0
sin k(x− z)m(z)dz,
w0(x) + k2w0(x) = m(x)− k
Z x
0
sin k(x− z)m(z)dz + k
Z x
0
sin k(x− z)m(z)dz =
= m(x).
Note: If the force P has another direction thhe equation of the problem takes the form
d2w
dx2− k2w =
My
EI,
129
and its general solution is
w(x) = c1 cosh kx+ c2sinh kx
k+
Z x
0
1
ksinh k(x− z) m(z) dz.
x = 0, c1 = w(0), w(0) = c2.
12.4.2 Method of initial parameters of statically non-determined beam
The equation of deflection in the case of compression may be written in the form
d4w
dx2+ k2
d2w
dx2=
q(x)
EJ,
General solution of this equation is
w(x) = c0 + c1x+ c2(1− cos(kx))
k2+ c3
1
k2
µx− sin(kx)
k
¶+
+1
k2EJ
xZ0
∙(x− z)− sin k(x− z)
k
¸q(z)dz.
Proof.
w0(x) = c1 + c2sin(kx)
k+ c3
(1− cos(kx))k2
+1
k2EJ
xZ0
[1− cos(k(x− z))] q(z)dz,
w00(x) = c2 cos(kx) + c3sin(kx)
k+
1
kEJ
xZ0
sin(k(x− z))q(z)dz,
w000(x) = −c2k sin(kx) + c3 cos(kx) +1
EJ
xZ0
cos(k(x− z))q(z)dz,
wIV (x) = −c2k2 cos(kx)− c3k sin(kx)−k
EJ
xZ0
sin(k(x− z))q(z)dz +q(x)
EJ.
As a result we have
wIV + k2w00 =q(x)
EJ.
Note:w(0) = c0, w0(0) = c1, w00(0) = c2, w000(0) = c3.
In the case of the other direction of the field P the differential equation of the deflectiontakes the form
130
d4w
dx2− k2
d2w
dx2=
q(x)
EI,
and its general solution may be taken in the form
w(x) = c0 + c1x− c2(1− cosh(kx))
k2− c3
1
k2
µx− sinh(kx)
k
¶+
− 1
k2EJ
xZ0
∙(x− z)− sinh k(x− z)
k
¸q(z)dz.
Example 1
qP
General solution
w = c2(1− cos(kx))
k2+ c3
1
k2
µx− sin(kx)
k
¶+
q
k2EJ
xZ0
∙(x− z)− sin k(x− z)
k
¸dz =
= c2(1− cos(kx))
k2+ c3
1
k2
µx− sin(kx)
k
¶− q
2k2EJ
£2− k2x2 + 2 cos(kx)
¤.
Boundary conditions
w(L) = c2(1− cos(kL))
k2+ c3
1
k2
µL− sin(kL)
k
¶− q
2k2EJ
£2− k2L2 + 2 cos(kL)
¤= 0,
w(L) = c2 cos (kL) + c3sin (kL)
k+
q
k2EJ[1− cos(kL)] = 0.
Solution of the system:
c2 =qL(−2 + 2 cos(kL) + kL sin(kL))
2EJk(kL cos(kL)− sin(kL)) , c3 =q(2− (2 + k2L2) cos(kL))
2EJk(kL cos(kL)− sin(kL)) .
Example 2
131
z
xL
PB
PA
P M
Equilibrium equations of forces and moments
PA + PB = 0, PAL =M0, PA =M0
L.
Differential equation of the deflection
w00 + k2w =My(x)
EJ, My(x) = PAx−M0 =
M0
L(x− L).
General solution
w(x) = A sin(kx) +B cos(kx) +M0
k2L(x− L).
Boundary conditions
w(0) = B − M0
k2= 0, B =
M0
k2,
w(L) = A sin(kL) +M0
k2cos(kL) = 0,
w(x) = −Mo
k2cos(kL)
sin(kL)sin(kx) +
Mo
k2cos(kx) +
M0
k2L(x− L), k =
rP
EJ.
Note:
If sin kL = 0, kL = π, P = Pc =π2
L2EJ, w(x) =⇒∞.
Example 3
QRMo
P
132
General solution for the given boundary conditions
w(x) =1
k
Z x
0
sin k(x− ζ)m(ζ)dζ, m(x) =My(x)
EJ
Equations of equilibrium of the beam
Mo = QL
2, R = Q,
My(z) = Mo −Rx = Q(L
2− x), x <
L
2,
My(z) = Mo −Rx+Q(x− L
2) = 0, x >
L
2.
Thus, we have the general solution for given boundary conditions
w(x) =1
k
xZ0
sin k(x− z)Q
EI(L
2− z)dz, x <
L
2,
w(x) =1
k
L/2Z0
sin k(x− z)Q
EI(L
2− z)dz, x >
L
2.
After the calculation of integrals we have
w(x) =Q
2k2EI
∙L(1− cos kx)− 2x+ 2sin kx
k
¸, x <
L
2,
w(x) =Q
2k2EI
∙−L cos kx− 2
k
µsin k(x− L
2)− sin kx
¶¸, x >
L
2.
Example 4
k21 =P
EJ1, k22 =
P
EJ2, J2 = 4J1.
w + k21w = 0, x <L
2, w + k22w = 0, x >
L
2,
w1(x) = A1 cos k1x+A2 sin kx , w2(x) = A3 cos k2x+A4 sin k2x
133
Boundary conditions
w1(0) = 0⇒ A1 = 0,
w1(L
2) = w2(
L
2), w1(
L
2) = w2 = (
L
2), w2(L) = 0,
A2 sink1L
2−A3 cos
k2L
2−A4 sin
k2L
2= 0,
A2 sink1L
2−A3 cos
k2L
2−A4 sin
k2L
2= 0,
A2k1 cosk1L
2+A3k2 sin
k2L
2−A4k2 cos
k2L
2= 0,
A3 cos k2L+A4 sin k2L = 0.
Matrix form of the system°°°°°°sin k1L
2,− cos k2L
2,− sin k2L
2
k1 cosk1L2, k2 sin
k2L2,−k2 cos k2L2
0, cos k2L, sin k2L
°°°°°°°°°°°°A2A3A4
°°°°°° =°°°°°°000
°°°°°°Condition of the existence of non-trivial solution
det k....k = 0
sink1L
2
∙k2 sin
k2L
2sin k2L+ k2
cos k2L
2cos(k2L)
¸− k1 cos
k1L
2=∙
− cos k2L2sin k2L+ sin
k2L
2cos k2L
¸= 0.
This equation may be rewritten in the form
tank1L
2+
k1k2tan
k2L
2= 0.
Because k1 = 2k2, we have
tank1L
2+ 2 tan
k1L
4= 0, tan
z
2+ 2 tan
z
4= 0
The first roots of this equation are
k1L|1 = 3.82, k1L|2 = 8.74
k21 =(3.82)2
L2=
P
EJ1, Pc =
(3.82)2
L2EJ1 =
14.6EJ1L2
.
134
Table of critical forces
Pc =π2EJ
(μL)2
µ=1 µ=2 µ=0.5 µ=0.7 µ=1 µ=0.5
135
The home task problem.1. To find the critical value of the force F when buckling of the beam L1 occures.
Young modulus of the material of the beams is E, the upper beam has the moment ofinertia J , amd the low beam - J1.
F
L/2 L/2
L1
2. The body of mass m=20 kg falles down on the gorisontal beam (see figure). Whatshoud be vertical distance H (see figure) in order the beam L1 will lose its direct form.
L/2 L/2
L1
H
m
Parameters of the beams are: L1 = 1m, L2 = 2m, E = 100GPa; beam L1 has squareintersection 0.05m× 0.05m, beam L is cylinder of the radius 0.06m.
136
12.5 Stability and the potential energy of elastic systems
1.- Identification of stability
Potential energy of the system is minimal in the stable state!What is the potential energy Π of an elastic system?
Pλ
Π =W −A =1
2
Z L
0
EJ(w00)2dx− Pλ, λ =
Z L
0
dλ.
dx
dx
dλ
α
137
dλ = (1− cosα)dx ≈ (12α2)dx ≈
1
2(w)2dx.
cosα = 1− α2
2+
α4
4!− ......., α = w,
Π =1
2
Z L
0
EJ(w00)2dx− P
2
Z L
0
(w0)2dx.
Example:
w = δ sinπ
Lx, w0 =
δπ
Lcos
π
Lx, w00 = −δ(π
L)2 sin
π
Lx
Π(δ) =EJ
2
Z L
0
δ2(π
L)4 sin2
³πLx´dx− P
2δ2(
π
L)2Z L
0
cos2³πLx´dx =
=π2δ2
4L
∙π2
L2EJ − P
¸1. If P < π2
L2EJ, Π(δ) > 0, and δ = 0 corresponds to a stable state.
Π(δ)
δ
2. If P > π2
L2EJ, Π(δ) < 0, and δ = 0 corresponds to a non-stable state.
Π (δ)
δ
138
3. If P = π2
L2EJ, Π(δ) = 0, and δ = 0 corresponds to a neutral state.
Π(δ)
δ
Note 1: In the neutral state Π(δ) = 0, and we have
Pcr =
R L0EJ(w00)2dxR L0(w0)2dx
For approximate calculation of Pcr let us choose w in the form:
w = δx(L− x), w0 = δ(L− 2x), w00 = −2δ.The integrals on the equation for Pcr take the following formsZ L
0
EJ(w00)2dx = 4EJLδ2,
Z L
0
(w0)2dx = δ2Z L
0
(L− 2x)2dx = δ2L3
3.
Thus, the value of critical force is
Pcr =4EJδ2Lh
δ2L3
3
i = 12EJ
L2,
Note:Exact value of the critical force is
Pce =π2EJ
L2=9.87EJ
L2,
Pcr − Pce
Pce= 0.22 = 22%
Let us choose the firm of deflection as follows
w00(x) = αx(L− x).
After integration we get
w0 = α
∙Lx2
2− x3
3− L3
12
¸,
w = α
∙L
6x3 − 1
12x4 − 1
12L3x
¸=
αL4
12
³xL
´³1− x
L
´hxL
³xL− 1´− 1i.
139
This form satisfies the following conditions
w(0) = 0, w(L) = 0,
w00(0) = 0, w00(L) = 0.
In this case we haveZ L
0
EJ(w00)2dx =EJL5α2
30,
Z L
0
(w0)2dx =17α2L2
5040
Pcr =168EJ
17L2≈ 9.88
EJ
L2,
Pcr − Pce
Pce=9.88− π2
π2≈ 0.1%
140
