16.20 Techniques of Structural Analysis and Design...

16.20 Techniques of Structural Analysis and DesignSpring 2013

Instructor: Raul RadovitzkyAeronautics & Astronautics

M.I.T

February 15, 2013

Contents

1 Stress and equilibrium 51.1 Internal forces and equilibrium . . . . . . . . . . . . . . . . . . . . . . . . . . 51.2 Stress tensor . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 61.3 Transformation of stress components . . . . . . . . . . . . . . . . . . . . . . 121.4 Principal stresses and directions . . . . . . . . . . . . . . . . . . . . . . . . . 161.5 Mohr’s circle . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 201.6 Linear and angular momentum balance . . . . . . . . . . . . . . . . . . . . . 24

1.6.1 Angular momentum balance . . . . . . . . . . . . . . . . . . . . . . . 27

3

4 CONTENTS

Module 1

Stress and equilibrium

Learning Objectives

• Understand stress as a description of internal forces transmitted in a material.

• Understand the nature of the mathematical description of stress as a second-ordertensor.

• Mathematically describe the implications of local equilibrium at a material point.

• Understand the nature of the symmetry of the stress tensor.

• Quantify stress components in arbitrary orthonormal bases and in particular principalstresses and directions.

1.1 Internal forces and equilibrium

We are going to consider the relation between the external and internal forces exerted on amaterial. External forces come in two flavors: body forces (given per unit mass or volume)and surface forces (given per unit area). If we cut a body of material in equilibrium undera set of external forces along a plane, as shown in Fig.1.1, and consider one side of it, wedraw two conclusions: 1) the equilibrium provided by the loads from the side taken out isprovided by a set of forces that are distributed among the material particles adjacent to thecut plane and that should provide an equivalent set of forces to the ones loading the parttaken out, 2) these forces can now be considered as external surface forces acting on the partof material under consideration.

The stress vector at a point on ∆S is defined as:

t = lim∆S→0

f

∆S(1.1)

If the cut had gone through the same point under consideration but along a plane with adifferent normal, the stress vector t would have been different. Let’s consider the three stressvectors t(i) acting on the planes normal to the coordinate axes. Let’s also decompose each

5

6 MODULE 1. STRESS AND EQUILIBRIUM

n

surface forces

body forces

n

∆S

Figure 1.1: Surface force f on area ∆S of the cross section by plane whose normal is n.

t(i) in its three components in the coordinate system ei (this can be done for any vector) as(see Fig.1.2):

t(i) = σijej (1.2)

e2

e3

e1

x1

t(1)

x2

t(2)

x3

t(3)

σ11

σ12

σ13σ21

σ22

σ23σ31

σ32

σ33

Figure 1.2: Stress components.

σij is the component of the stress vector t(i) along the ej direction.

1.2 Stress tensor

We could keep analyzing different planes passing through the point with different normalsand, therefore, different stress vectors t(n) and one might wonder if there is any relationamong them or if they are all independent. The answer to this question is given by invokingequilibrium on the (shrinking) tetrahedron of material of Fig.1.3. The area of the faces of thetetrahedron are ∆S1, ∆S2, ∆S3 and ∆S. The stress vectors on planes with reversed normals

1.2. STRESS TENSOR 7

t(−ei) have been replaced with −t(i) using Newton’s third law of action and reaction (whichis in fact derivable from equilibrium): t(−n) = −t(n). Enforcing equilibrium we obtain:

t(n)∆S − t(1)∆S1 − t(2)∆S2 − t(3)∆S3 = 0 (1.3)

n

e1

e2

e3−t

(1)

t(n)

−t(2)

−t(3)

Figure 1.3: Cauchy’s tetrahedron representing the equilibrium of a tetrahedron shrinking toa point.

The following relation: ∆Sni = ∆Si derived in the following mathematical aside:

By virtue of Green’s Theorem: ∫V∇φdV =

∫SnφdS

applied to the function φ = 1, we get

0 =

∫SndS

which applied to our tetrahedron gives:

0 = ∆Sn−∆S1e1 −∆S2e2 −∆S3e3

If we take the scalar product of this equation with ei, we obtain:

∆S(n · ei) = ∆Si

or∆Si = ∆Sni

can then be replaced in equation 1.3 to obtain:

∆S(t(n) − (n · e1)t(1) − (n · e2)t(2) − (n · e3)t(3)

)= 0


or

t(n) = n ·(e1t

(1) + e2t(2) + e3t

(3))

(1.4)

The factor in parenthesis is the definition of the Cauchy stress tensor σ:

σ = e1t(1) + e2t

(2) + e3t(3) = eit

(i) (1.5)

Note it is a tensorial expression (independent of the vector and tensor components in aparticular coordinate system). To obtain the tensorial components in our rectangular systemwe replace the expressions of t(i) from Eqn.1.2

σ = eiσijej (1.6)

with:

σij =

σ11 σ12 σ13

σ21 σ22 σ23

σ31 σ32 σ33

Replacing in Eqn.1.4:

t(n) = n · σ (1.7)

or:

t(n) = n · σijeiej = σij(n · ei

)ej =

(σijni

)ej (1.8)

tj = σijni (1.9)

Concept Question 1.2.1. Compute the normal and tangential components of the tractionvector as a function of α.

Concept Question 1.2.2. Stress Components.Let’s consider the following stress tensor in the (e1, e1, e3) system of coordinates:

σ =

σ11 = 80 MPa (T) σ12 = 30 MPa 0σ21 = 30 MPa σ22 = 40 MPa (C) 0

0 0 0


e2

e1

e3

AB

C

D

E

F

(a) Cube I

e2

e3e1 AB

C

D

E

F

(b) Cube II

Figure 1.4: Cubes in different coordinate systems.

Let’s consider two cubes in different coordinate systems in Figures 1.4(a) and 1.4(b). Thesecubes are comprised of faces A, B, C, D, E, F with the following outward normal for eachof the cubes:

Cube I: A→ e2, B → −e2, C → −e1, D → e1, E → e3, F → −e3.Cube II: A→ −e2, B → e2, C → e3, D → −e3, E → −e1, F → e1.

1. Show the non-zero components on the surfaces of the two cubes in different coordinatesystems. Solution: Cube I : The first subscript of σ11 ans σ12 showsthat the outward normal is in the e1 direction, hence these components will be shownon surfaces C and D in Figure 1.5(a).The outward normal on surface C is in the negative e1 direction, hence the denominatorin the relation 1.1 (see notes, chapter 1) is negative. Therefore on Figure 1.5(a):

• The internal force has to be in the negative e1 direction to produce a positive(tensile) σ11.

• The internal force has to be in the negative e2 direction to produce a positive σ12

The outward normal on surface D is in the positive e1 direction, hence the denominatorin the relation 1.1 is positive. Therefore on Figure 1.5(a):

• The internal force has to be in the positive e1 direction to produce a positive(tensile) σ11.

• The internal force has to be in the positive e2 direction to produce a positive σ12

The first subscript of σ21 and σ22 shows that the outward normal is in the e2 direction,hence these components will be shown on surfaces A and B. The outward normal onsurface A is in the positive e2 direction, hence the denominator in the relation 1.1 ispositive. Therefore on Figure 1.5(a):

• The internal force has to be in the positive e1 direction to produce a positive σ21.

• The internal force has to be in the negative e2 direction to produce a negative(compressive) σ22


The outward normal on surface B is in the negative e2 direction, hence the denominatorin the relation 1.1 is negative. Therefore on Figure 1.5(a):

• The internal force has to be in the negative e1 direction to produce a positive σ21.

• The internal force has to be in the positive e2 direction to produce a negative(compressive) σ22

Cube II : The first subscript of σ11 ans σ12 shows that the outward normal is in the e1

direction, hence these components will be shown on surfaces E and F in Figure 1.5(b).The outward normal on surface E is in the negative e1 direction, hence the denominatorin the relation 1.1 is negative. Therefore on Figure 1.5(b):

• The internal force has to be in the negative e1 direction to produce a positive(tensile) σ11.

• The internal force has to be in the negative e2 direction to produce a positive σ12

The outward normal on surface F is in the positive e1 direction, hence the denominatorin the relation 1.1 is positive. Therefore on Figure 1.5(b):

• The internal force has to be in the positive e1 direction to produce a positive(tensile) σ11.

• The internal force has to be in the positive e2 direction to produce a positive σ12

The first subscript of σ21 and σ22 shows that the outward normal is in the e2 direction,hence these components will be shown on surfaces A and B. The outward normal onsurface A is in the negative e2 direction, hence the denominator in the relation 1.1 isnegative. Therefore on Figure 1.5(b):


• The internal force has to be in the positive e2 direction to produce a negative(compressive) σ22

The outward normal on surface B is in the positive e2 direction, hence the denominatorin the relation 1.1 is positive. Therefore on Figure 1.5(b):


• The internal force has to be in the negative e2 direction to produce a negative(compressive) σ22

Concept Question 1.2.3. Stresses on an inclined face.Consider the tetrahedron shown in Figure 1.7. A set of three mutually orthogonal unit

vectors will be defined: l is a unit vector parallel to vector AB, n is the normal to faceABC, and m is such that m = n× l. The stress vector acting on face ABC is then givenby: tn = σnll + σnmm + σnn.


e2

e1

e3

AB

C

D

E

F

40

30

80

30

30

40

80

30

(a) Cube I

e2

e3e1

AB

C

D

E

F

403030

40

8030

30

80

(b) Cube II

Figure 1.5: Solutions of stress components in cubes in different coordinate systems.

n

e1

e2

e3

A

B

C

O

Figure 1.6: Differential tetrahedron element.


1. Determine the stress components, σnl, σnm, and σn, in terms of the stress componentsacting on the faces normal to e1, e2, and e3.

Solution: The stress vectors acting on the faces perpendicular to the axes e1, e2,and e3 are defined by

t1 = σ11e1 + σ12e2 + σ13e3,

t2 = σ21e1 + σ22e2 + σ23e3,

t3 = σ31e1 + σ32e2 + σ33e3,

where σij are the component of the stress tensor.

The normal to the face ABC is defined as n = n1e1 + n2e2 + n3e3, the unit vectorparallel to AB as l = l1e1 + l2e2, and thus m = m1e1 +m2e2 +m3e3.

The traction normal to the surface ABC is commonly written as t(n) = t1n1 + t2n2 +t3n3. Projecting the normal traction respectively onto n, l and m and using the aboveexpressions for t1, t2 and t3 allow us to calculate:

• the component σn:

σn = t(n) · n = (σ11n1 + σ12n2 + σ13n3)n1

+ (σ21n1 + σ22n2 + σ23n3)n2 + (σ31n1 + σ32n2 + σ33n3)n3,

which can rewritten as

σn = σ11n21 + σ22n

22 + σ33n

23 + 2σ12n1n2 + 2σ13n1n3 + 2σ23n2n3.

• the component σnl = t(n) · l:

σnl = σ11n1l1 + σ22n2l2 + σ33n3l3 + σ12(n1l2 + n2l1) + σ13n3l1 + σ23n3l2,

• the component σnm = t(n) · m:

σnm = σ11n1m1 + σ22n2m2 + σ33n3m3

+ σ12(n1m2 + n2m1) + σ13(n1m3 + n3m1) + σ23(n2m3 + n3m2).

1.3 Transformation of stress components

Readings: BC 1.2.1, 1.2.3 (Full 3D)

Readings: BC 1.3.2, 1.3.4 (2D)

1.3. TRANSFORMATION OF STRESS COMPONENTS 13

n

e1

e2

e3

A

B

C

O

Figure 1.7: Differential tetrahedron element.

Consider a different system of cartesian coordinates e′i. We can express our tensor ineither one:

σ = σklekel = σ′mne′me′n (1.10)

We would like to relate the stress components in the two systems. To this end, we take thescalar product of (1.10) with e′i and e′j:

e′i · σ · e′j = σkl(e′i · ek

)(el · e′j

)= σ′mn

(e′i · e′m

)(e′n · e′j

)= σ′mnδimδnj = σ′ij

or

σ′ij = σkl(e′i · ek

)(el · e′j

)(1.11)

The factors in parenthesis are the cosine directors of the angles between the original andprimed coordinate axes.

e1

e2

e∗

1

e∗

2

θ

Figure 1.8: Coordinate systems E = (e1, e2) and E∗ = (e∗1, e∗2), where θ = 25◦.

Concept Question 1.3.1. Stress states on two sets of faces.


The plane stress state at a point is known and characterized by the following stress tensor:

σ =

[σ11 σ12

σ21 σ22

]=

[250 00 250

]in a coordinate system E = (e1, e2), as illustrated in Figure 1.9.

1. Determine the stress components σ∗11, σ∗22, σ∗12 in a coordinate system E∗ = (e∗1, e∗2),

where e∗1 is oriented at an angle of 25 degree with respect to e1. Solution: Thestress tensor σ is given by

σ =

[σ11 σ12

σ21 σ22

]=

[250 00 250

].

Herein σ11 = σ22 and σ12 = 0, hence this tensor corresponds to an hydrostatic stateof stress meaning that the stresses acting on a face with any arbitrary orientation aregiven by σ12 = σ21 = σ∗12 = σ∗21 = 0, and σ11 = σ22 = σ∗11 = σ∗22 = p, where p is theso-called hydrostatic pressure. NB: the result is the same for any value of the angle θ.

This problem can be also solved with the approach performed in the Concept Question1.2.2 on “Stresses on an inclined face”. We know that the stress vectors acting on thefaces perpendicular to the axes e1, and e2 are defined by

t1 = σ11e1 + σ12e2,

t2 = σ21e1 + σ22e2,

where σij are the component of the stress tensor.

After writing the axes e∗1 and e∗2 in the components of E = (e1, e2), i.e., e∗1 = cos θ e1 +sin θ e2, and e∗2 = − sin θ e1 + cos θ e2, we can calculate the tractions in the directionse∗1 and e∗2 as

te∗1 = t1 cos θ + t2 sin θ = (σ11e1 + σ12e2) cos θ + (σ21e1 + σ22e2) sin θ,

te∗2 = −t1 sin θ + t2 cos θ = − (σ11e1 + σ12e2) sin θ + (σ21e1 + σ22e2) cos θ.

These expressions and the fact that σ11 = σ22 = p and σ12 = σ21 = 0 enable us tocompute

σ∗11 = te∗1 · e∗1 = σ11cos2θ + σ22sin2θ + 2σ12 sin θ cos θ = p,

σ∗22 = te∗2 · e∗2 = σ11sin2θ + σ22cos2θ − 2σ12 sin θ cos θ = p,

σ∗12 = te∗1 · e∗2 = (σ22 − σ11) sin θ cos θ + σ12

(cos2θ − sin2θ

)= 0.

We can achieve the same conclusion through a rotation of stresses (see the statementof the Problem 1.3.2).

1.3. TRANSFORMATION OF STRESS COMPONENTS 15

e1

e2

e∗

1

e∗

2

θ

Figure 1.9: Coordinate systems E = (e1, e2) and E∗ = (e∗1, e∗2), where θ = 25◦.

Concept Question 1.3.2. Stress rotation formulae in matrix form.

Specialize the general expression for the transformation of stress components Equation(1.11) in the class notes to two dimensions and show that they can be expressed in thefollowing two ways:

σ∗11

σ∗22

σ∗12

=

cos2 θ sin2 θ 2 sin θ cos θsin2 θ cos2 θ −2 sin θ cos θ

− sin θ cos θ sin θ cos θ cos2 θ − sin2 θ

σ11

σ22

σ12

1. Show that this formula can be recast in the following compact matrix form[

σ∗11 σ∗12

σ∗12 σ∗22

]=

[cos θ sin θ− sin θ cos θ

] [σ11 σ12

σ12 σ22

] [cos θ − sin θsin θ cos θ

]Solution: In the

solution of the Problem 1.3.1 we explain how to derive the relation between the stresscomponents σ∗11, σ∗22, σ∗12 expressed in a coordinate system E∗ = (e∗1, e

∗2), and the stress

components σ11, σ22, σ12 expressed in a coordinate system E = (e1, e2). That relationis written here in matrix form: {σ∗11 σ

∗22 σ

∗12}

T = A {σ11 σ22 σ12}T .

Although the matrix A seems to represent a rotation transformation between two vec-tors, it actually represents the rotation transformation between two tensors (remember

that the tensor σ∗ =

[σ∗11 σ∗12

σ∗12 σ∗22

]is symmetric, and thus it can be written as a vector

turning to the Voigt notation). It means that the relation between {σ∗11 σ∗22 σ

∗12} and

{σ11 σ22 σ12} can be also expressed as a change of coordinates between the tensors σ∗

and σ, i.e.

σ∗ = RTσR,

where R =

[cos θ − sin θsin θ cos θ

]in this case.

It is clear now why the plane stress rotation formula and the compact matrix form areequivalent.


1.4 Principal stresses and directions

Readings: BC 1.2.2 (Full 3D)

Readings: BC 1.3.3 2D

Given the components of the stress tensor in a given coordinate system, the determinationof the maximum normal and shear stresses is critical for the design of structures. The normaland shear stress components on a plane with normal n are given by:

tN = t(n) · n= σkinkni

tS =√‖t(n)‖2 − t2N

It is obvious from these equations that the normal component achieves its maximum tN =‖t(n)‖ when the shear components are zero. In this case:

t(n) = n · σ = λn = λIn

or in components:

σkink = λni

σkink = λδkink(σki − λδki

)nk = 0

(1.12)

which means that the principal stresses are obtained by solving the previous eigenvalueproblem, the principal directions are the eigenvectors of the problem. The eigenvalues λ areobtained by noticing that the last identity can be satisfied for non-trivial n only if the factoris singular, i.e., if its determinant vanishes:

|σij − λδij| =

∣∣∣∣∣∣σ11 − λ σ12 σ13

σ21 σ22 − λ σ23

σ31 σ32 σ33 − λ

∣∣∣∣∣∣ = 0

which leads to the characteristic equation:

−λ3 + I1λ2 − I2λ+ I3 = 0

where:

I1 = tr[σ]

= σii = σ11 + σ22 + σ33 (1.13)

I2 = tr[σ−1]det[σ]

=1

2

(σiiσjj − σijσji

)= σ11σ22 + σ22σ33 + σ33σ11 −

(σ12σ21 + σ23σ32 + σ31σ13

)(1.14)

I3 = det[σ]

= σ11σ22σ33 + 2σ12σ23σ31 − σ212σ33 − σ2

23σ11 − σ213σ22 (1.15)

1.4. PRINCIPAL STRESSES AND DIRECTIONS 17

are called the stress invariants because they do not depend on the coordinate system ofchoice.

Concept Question 1.4.1. Principal stresses.Let’s consider the following state of stress:

σ =

σ11 σ12 σ13

σ21 σ22 σ23

σ31 σ32 σ33

=

200 50 −8050 300 100−80 100 −100

.1. Determine both, the principal stresses and the principal directions.

Solution: We have a stress tensor σ given by

σ =

200 50 −8050 300 100−80 100 −100

.To determine (1) the principal stresses and (2) the principal directions we have tocompute the eigenvalues and eigenvectors, respectively. After the calculation, we obtaina stress tensor σ∗ with the eigenvalues in the diagonal, and a matrix M∗ with thecorresponding eigenvectors in columns:

σ∗ =

331.64 0 00 215.12 00 0 −146.76

,and

M∗ =

−0.2562 −0.9335 0.2508−0.9510 0.1970 −0.2381−0.1729 0.2995 0.9383

.We can calculate the invariants I1, I2 and I3 of the stress tensor to verify is these resultsare correct. We obtain I1 = I∗1 = 400, I2 = I∗2 = −8900, and I3 = I∗3 = −10470000. Itmeans that is the same tensor but expressed in different coordinate systems. As one ofthe matrices is diagonal, we can conclude that is the matrix of the principal stresses.

Concept Question 1.4.2. Principal stresses and transformation.Let’s consider the following state of stress:

σ =

3 1 11 0 21 2 0

.1. Using the stress invariants, determine both, the principal stresses and directions.


2. Determine the traction vector on a plane with a unit normal n = (0, 1, 1)/√

(2).Solution: 1. First, we calculate the three invariants:

I1 = tr(σ)

= σkk

= σ11 + σ22 + σ33

= 3

I2 = tr(σ−1)det(σ)

=1

2(σiiσjj − σijσji)

= σ11σ22 + σ22σ33 + σ11σ33 − σ212 − σ2

23 − σ213

= −6

I3 = det(σ)

= σ11σ22σ33 + 2σ12σ23σ31 − σ212σ33 − σ2

23σ11 + σ213σ22

= −8

This leads to the following characteristic equation:

det[σij − λδij] = −λ3 + I1λ2 − I2λ+ I3 = 0

= −λ3 + 3λ2 + 6λ− 8 = 0

The roots of this equation are found to be λ = 4, 1 and −2. Back-substituting thefirst root into the fundamental system gives:

−n(1)1 + n

(1)2 + n

(1)3 = 0

n(1)1 − 4n

(1)2 + 2n

(1)3 = 0

n(1)1 + 2n

(1)2 − 4n

(1)3 = 0

Solving this system, the normalized principal direction is found to ben(1) = (2, 1, 1)/

√(6). In similar fashion, the other two principal directions are n(2) =

(−1, 1, 1)/√

(3) and n(3) = (0,−1, 1)/√

(2).

2. The traction vector on the specified plane is calculated by using the relation:

T ni =

3 1 11 0 21 2 0

0

1/√

2

1/√

2

=

2/√

2

2/√

2

2/√

2

.

Concept Question 1.4.3. Stress invariants for plane stress state.Let’s introduce the following two quantities:

I1 = σ11 + σ22,

I2 = σ11σ22 − σ212.

1.4. PRINCIPAL STRESSES AND DIRECTIONS 19

1. In the case of plane stress problems, show that these two quantities are invariant.

2. Prove this invariance by showing that these quantities are identical when computed interms of the principal stresses and in terms of stresses acting on a face at an arbitraryorientation.

Solution: The invariants of the stresstensor are the quantities that remain unchangeable under coordinate transformations.In the three-dimensional case they are defined as

I1 = σ11 + σ22 + σ33,

I2 = σ11σ22 + σ22σ33 + σ11σ33 − σ212 − σ2

23 − σ213,

I3 = σ11σ22σ33 − σ11σ223 − σ22σ

213 − σ33σ

212 + 2σ12σ13σ23.

In the plane stress approach we assume that σ33 = 0, σ13 = 0, and σ23 = 0. Byreplacing these stresses in the above equations we obtain the invariants for the planestress state

I1 = σ11 + σ22,

I2 = σ11σ22 − σ212,

I3 = 0,

which correspond with the values indicated in the statement of the problem.

Given the stress components σ11, σ22 and σ12 on two orthogonal faces, we can writethe principal stresses σp11 and σp22 as

σp11 =σ11 + σ22

2+ ∆,

σp22 =σ11 + σ22

2−∆,

where ∆ =√(

σ11−σ222

)2+ σ2

12.

These expressions make straightforward the following calculations

Ip1 = σp11 + σp22 = σ11 + σ22 ⇒ Ip1 = I1.

Ip2 = σp11σp22 − (σp12)2 = σ11σ22 − (σ12)2 ⇒ Ip2 = I2.

If the stress components σ11, σ22 and σ12 on two orthogonal faces are known, thestresses acting on a face with an arbitrary direction θ can be calculated as

σ∗11 =σ11 + σ22

2+

(σ11 − σ22

2

)cos 2θ + σ12 sin 2θ,

σ∗22 =σ11 + σ22

2−(σ11 − σ22

2

)cos 2θ − σ12 sin 2θ,

σ∗12 = −(σ11 − σ22

2

)sin 2θ + σ12 cos 2θ.


These expressions make straightforward the following calculations

I∗1 = σ∗11 + σ∗22 = σ11 + σ22 ⇒ I∗1 = I1.

I∗2 = σ∗11σ∗22 − (σ∗12)2 = σ11σ22 − (σ12)2 ⇒ I∗2 = I2.

1.5 Mohr’s circle

Readings: BC 1.3.6

The Mohr’s circle, named after Christian Otto Mohr (1835-1918), is a two-dimensional graph-ical representation of the state of stress at a point M . A point M , belonging to the circle,has the coordinates (σM , τM) in the reference axes (σ,τ) corresponding to the normal andshear stresses at this point, respectively. Figure 1.10 shows a point M belonging to a Mohr’scircle corresponding to the following state of stress:

σM =

(σ11 σ12

σ12 σ22

)The principal stresses, herein denoted σI and σII (Figure 1.10), are located on the axis

σ for which the shear component is zero. The center of the circle, giving the value of theaverage normal stress, is located on the horizontal axis a distance 1

2(σI +σII) from the origin

and the radius of the circle is given by:

R =

√[1

2(σ11 − σ22)

]2

+ σ212 (1.16)

The principal stresses are obtained as follows:

σI =1

2(σ11 + σ22) +

√[1

2(σ11 − σ22)

]2

+ σ212

σII =1

2(σ11 + σ22)−

√[1

2(σ11 − σ22)

]2

+ σ212 (1.17)

Also, the maximum shear stress σmax12 is obtained on the Mohr’s circle (Figure 1.10) andis equal to the value of the radius of the circle. The stress components of a point M areacting on a particular plane oriented at an angle θ to the principal directions. In practice,on the Mohr’s circle, the diameter of the circle joining the coordinates (σ22, σ12) and (σ22,-σ12) of the point M makes an angle of 2θ with the horizontal axis, as depicted in Figure1.10. Therefore, the circle describes the stress state of any point located on plane at allorientations, i.e 0 < θ < π.

1.5. MOHR’S CIRCLE 21

σ

τ

b

bM(σ22,−σ12)

b

M (σ11,σ12)

b

σ∗

11

b

σ∗

22

2θ

b

σmax12

R

Figure 1.10: Mohr’s circle for two-dimensional stress

Concept Question 1.5.1. Mohr’s circle derivation.Let’s consider a Mohr’s circle of radius R and principal stresses σ∗11 and σ∗22 which are

points belonging to both, the circle and the σii axis.

1. Derive the expression of the radius R as a function of the stress components.

2. Derive the expressions of the principal stresses.

Solution: 1. In Figure 1.11, from the Pythagore theorem we have:

σ2m + σ2

12 = R2

and

σm =σ11 − σ22

2hence: (

σ11 − σ22

2

)2

+ σ212 = R2

finally we obtain the following relation for the radius of the circle:

R =

√[1

2(σ11 − σ22)

]2

+ σ212

2. The value of principal stresses is equal to the ordinate of the origin (12(σ11 + σ22)) of

the Mohr’s circle + or − the radius of the circle R:

σ∗11 =1

2(σ11 + σ22) +

√[1

2(σ11 − σ22)

]2

+ σ212

σ∗22 =1

2(σ11 + σ22)−

√[1

2(σ11 − σ22)

]2

+ σ212 (1.18)


σi

σij

b

b(σ22,σ12)

b

(σ11,−σ12)

σm

R

b

σ∗

11

b

σ∗

22

2θ∗

Figure 1.11: Mohr circle.

Concept Question 1.5.2. Mohr’s circle.Let’s consider the following state of stress:

σ =

[80 4040 −20

].

1. Draw the Mohr’s circle of this state of stress

2. Using the Mohr’s circle, determine the principal stresses and the corresponding direc-tions.

3. Using the Mohr’s circle, calculate the stresses on axes rotated 60 degrees counterclock-wise from the reference axes.

4. Compare these results with the ones obtained analytically? Solution: The stresstensor σ is given by

σ =

[σ11 σ12

σ21 σ22

]=

[80 4040 −20

].

1. We obtain the Mohr’s circle (Figure 1.12) by plotting the two points with the co-ordinates (σ22,−σ12) and (σ11,σ12), respectively. The line between these two pointsintersects the σii−axis at the center of the circle.

2. The principal stresses are the stresses for which the shear components are zero,i.e when the stress matrix is diagonal. Hence, σ∗11 and σ∗22 are the two intersectionpoints between the circle and the σii−axis. These stresses are (σ∗11)plot ≈ 94 MPa and(σ∗22)plot ≈ −34 MPa, respectively. The direction is given by the angle 2θ between the

1.5. MOHR’S CIRCLE 23

line joining (σ22,−σ12) and (σ11,σ12) and the σii−axis. Herein, (θ∗)plot ≈ 19o.

3. For an angle 2θ equals to 60o from the reference axis (or σii−axis), we obtain thestresses σ11 ' 39.6 MPa, σ22 ' 20.3 MPa and σ12 ' −63.3 MPa.

4. The principal stresses and corresponding directions are calculated analytically withthe following equations

σ∗11 =σ11 + σ22

2+ ∆,

σ∗22 =σ11 + σ22

2−∆,

tan 2θp =2σ12

σ11 − σ22

where ∆ =√(

σ11−σ222

)2+ σ2

12 is the shear stress component, around 0 when the stress

tensor is expressed in the principal directions system of coordinates.

∆ =

√(80−−20

2

)2

+ (40)2

= 64.0312 MPa

so

σ∗11 =80− 20

2+ 64.0312

= 94.0312 MPa

and

σ∗22 =80− 20

2− 64.0312

= −34.0312 MPa

and

tan 2θp =2× 40

80 + 20= 0.8

Hence, we obtain σp11 = 94.0312, σp22 = −34.0312, and θp = 19.33, and these results arein agreement with our previous results.


σi

σij

b

b

(σ22,−σ12)

b(σ11,σ12)

b

σ∗

11 = 94 MPab

σ∗

22 = −34 MPa 2θ∗

b(σ11, σ12)

b

(σ22, −σ12)

2θ

Figure 1.12: Mohr circle.

1.6 Linear and angular momentum balance

Readings: BC 1.1.2 (derivation by local equilibrium of differential volume element)

We are going to derive the equations of momentum balance in integral form, since this isthe formulation that is more aligned with our “integral” approach in this course. We startfrom the definition of linear and angular momentum. For an element of material at positionx of volume dV , density ρ, mass ρdV which remains constant, moving at a velocity v, thelinear momentum is ρvdV and the angular momentum x× (ρvdV ). The total momenta ofthe body are obtained by integration over the volume as:∫

V

ρvdV and

∫V

x× ρvdV

respectively. The principle of conservation of linear momentum states that the rate of changeof linear momentum is equal to the sum of all the external forces acting on the body:

D

Dt

∫V

ρvdV =

∫V

fdV +

∫S

tdS (1.19)

where DDt

is the total derivative. The lhs can be expanded as:

D

Dt

∫V

ρvdV =

∫V

D

Dt(ρdV )v +

∫V

ρ∂v

∂tdV

but DDt

(ρdV ) = 0 from conservation of mass, so the principle reads:∫V

ρ∂v

∂tdV =

∫V

fdV +

∫S

tdS (1.20)

Now, using what we’ve learned about the tractions and their relation to the stress tensor:

1.6. LINEAR AND ANGULAR MOMENTUM BALANCE 25

∫V

ρ∂v

∂tdV =

∫V

fdV +

∫S

n · σdS (1.21)

This is the linear momentum balance equation in integral form. We can replace the surfaceintegral with a volume integral with the aid of the divergence theorem:∫

S

n · σdS =

∫V

∇ · σdV

and then (1.21) becomes: ∫V

(ρ∂v

∂t− f −∇ · σ

)dV = 0

Since this principle applies to an arbitrary volume of material, the integrand must vanish:

ρ∂v

∂t− f −∇ · σ = 0 (1.22)

This is the linear momentum balance equation in differential form. In components:

σji,j + fi = ρ∂vi∂t

Concept Question 1.6.1. Stress and equilibrium.Let’s consider an elastic structural member for which the stress field is expressed as

follows:

σ =

−x31 + x2

2 5x3 + 2x22 x1x

33 + x2

1x2

5x3 + 2x22 2x3

1 + 12x2

2 0x1x

33 + x2

1x2 0 4x22 − x3

3

.1. Determine the body force distribution for equilibrium in static. Solution: We

apply the following relation of equilibrium:

∇ · σ + F = 0

∇ · σ =

∂σ11

∂x1

+∂σ12

∂x2

+∂σ13

∂x3∂σ21

∂x1

+∂σ22

∂x2

+∂σ23

∂x3∂σ31

∂x1

+∂σ32

∂x2

+∂σ33

∂x3

Which leads to:

−3x21 + 4x2 + 3x1x

23 + F1 = 0

x2 + F2 = 0

−3x23 + x3

3 + 2x1x2 + F3 = 0


The body force distribution, as obtained from these expressions, is therefore:

F1 = 3x21 − 4x2 − 3x1x

23

F2 = −x2

F3 = −2x1x2 − x33 +−3x2

3

The state of stress and body force at any specific point within the member may beobtained by substituting the specific values of x1, x2 and x3 into the previous equations.

Concept Question 1.6.2. Stress fields in static equilibrium.

Let’s consider a structure in equilibrium and free of body forces. Are the following stressfields possible?

1. σ =

c1x1 + c2x2 + c3x1x2 −c3x2

2

2− c1x2

−c3x2

2

2− c1x2 c4x1 + c1x2

.

2. σ =

[3x1 + 5x2 4x1 − 3x2

4x1 − 3x2 2x1 − 4x2

].

3. σ =

x21 − 2x1x2 + cx3 −x1x2 + x2

2 −x1x3

−x1x2 + x22 x2

2 −x2x3

−x1x3 −x2x3 (x1 + x2)x3

. Solution: To solve this

problem we turn to the momentum equation

∂σji∂xj

+ ρfi = ρ∂2ui∂t2

.

As the structure is in equilibrium (steady state) and free of body forces, the terms

ρ∂2ui∂t2

and ρfi are null. Then, the equilibrium equations become

∂σji∂xj

= 0.

For a 2D stress field we have:

∂σji∂xj

=

∂σ11

∂x1

+∂σ12

∂x2

∂σ21

∂x1

+∂σ22

∂x2

(1.23)

1.6. LINEAR AND ANGULAR MOMENTUM BALANCE 27

For a 3D stress field we have:

∂σji∂xj

=

∂σ11

∂x1

+∂σ12

∂x2

+∂σ13

∂x3

∂σ21

∂x1

+∂σ22

∂x2

+∂σ23

∂x3

∂σ31

∂x1

+∂σ32

∂x2

+∂σ33

∂x3

(1.24)

1. For the first stress field we obtained:

∂σji∂xj

=

[c1 + c3x2 +−c3x2 − c1

c1

]=

[0c1

](1.25)

This stress field does not satisfy the equilibrium equation.

2. For the second stress field we obtained:

∂σji∂xj

=

[3− 34− 4

]=

[00

](1.26)

This stress field does satisfy the equilibrium equation.

3. For the third stress field we obtained:

∂σji∂xj

=

2x1 − 2x2 − x1 + 2x2 − x1

−x2 + 2x2 − x2

−x3 − x3 + x1 + x2

=

00

−2x3 + x1 + x2

(1.27)

This stress field does not satisfy the equilibrium equation

1.6.1 Angular momentum balance and the symmetry of the stresstensor

The principle of conservation of angular momentum states that the rate of change of angularmomentum is equal to the sum of the moment of all the external forces acting on the body:

D

Dt

∫V

ρx× vdV =

∫V

x× fdV +

∫S

x× tdS (1.28)

It can be conveniently written as∫S

(xitj − xjti

)dS +

∫V

(xifj − xjfi

)dV =

∫V

ρ(xi∂vj∂t− xj

∂vi∂t

)dV


Using ti = σkink, the divergence theorem and (1.22), this expression leads to (see homeworkproblem): ∫

V

(σij − σji)dV = 0

which applies to an arbitrary volume V , and therefore, can only be satisfied if the integrandvanishes. This implies:

σij = σji (1.29)

16.20 Techniques of Structural Analysis and Design...

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