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Introduction to Advanced Structural

Prepared by: Antonio Palermo, PhD Student Revised by: Prof. Alessandro Marzani and Prof. Christian Carloni Aim of the Introductory Course The aim of these slides is to help students review some basic concepts of structural mechanics that will be exploited during the course of Advanced Structural Mechanics.

Outline of the Introductory Course • PART I: Cross-sectional Properties. • PART II: Solid Mechanics: Displacements and Strains, and Equilibrium, Constitutive Equations. • PART III: Internal in Beams: Axial , , Shear Force, and . The Euler- Bernoulli model. • PART IV Analysis of Statically determinate and indeterminate Structures.

Links and resources

Contacts: [email protected] [email protected] [email protected]

Suggested reading: • Beer, Johnston, DeWolf, Mechanics of Materials. • Gere and Timoshenko, Mechanics of Materials.

PART I

Cross-sectional Properties Outline of PART I

• Beam: Geometric Model.

• Cross-Sectional Properties: • • First Moments of Area • Centroid • Moments of Area • Transfer of Axis Theorem and of Axes • Principal Axes and Central Ellipse of .

Beam: Geometric Model

A beam is a structural element generated by a planar figure Ω (i.e. cross section) that moves in the remaining normal to the trajectory described by its centroid.

Ω Beam: Geometric Model

Geometric Requirements: Ω=const YES! • Ω(s) constant or can vary continuously:

BEAM YES! Ω(z)

NO! ℎ ≪ 푙 • 푏 ≪ 푙

h 푙 b Cross-sectional Properties

For any cross-section Ω, it is possible to define some quantities that are related only to the cross-section geometry.

Ω • Area A

• Static Moment of Area

• Centroid C C • Moments and Product of Inertia Cross-sectional Properties Area:

퐴 = 푑퐴 Ω

퐴 = [푚2]

푑퐴: Cross-sectional Properties

First Moment of Area: Static Moments

푆푥 = 푦푑퐴 Ω

푆푦 = 푥푑퐴 Ω

푆 = [퐿3]

Cross-sectional Properties Centroid:

The centroid C of a plane figure or two- dimensional shape is the arithmetic mean of all the points in the shape.

The centroid C of an area is the point of intersection of all the straight lines that subdivide the plane figure in equal parts Cross-sectional Properties Centroid:

푆푦 푥퐶 = 퐴

푆푥 푦 = 퐶 퐴

푥퐶, 푦퐶 = [퐿]

Cross-sectional Properties

Static Moment & Centroid: Properties

• 푆푥 푎푛푑 푆푦 can be ⋛ 0 • The static moment 푆 is zero if calculated with respect to a centroidal axis (i.e. the centroid lies on the axis) • The Static Moment calculated with respect to an axis of symmetry = 0. • If an area has an axis of symmetry, the centroid C lies on the axis. • If an area has two axes of symmetry, the centroid C is located at the intersection of the axes .

• 푆 = 푖 푆푖 (domain of integration can be added: geometric decomposition)

Cross-sectional Properties

Why is it called first moment?

Let’s assume to apply a vector at C whose magnitude is the area A of the region W (the of this vector are [L2]). The moment of this vector with respect to y is 퐴푥퐶 , which is the static moment 푆푦 Cross-sectional Properties

Second Moment of Area: :

2 퐼푥 = 푦 푑퐴 Ω

2 퐼푦 = 푥 푑퐴 Ω

퐼 = [퐿4]

Cross-sectional Properties : Product of Inertia and Polar Moment of Area

퐼푥푦 = Ω 푥푦푑퐴 Product o Inertia

2 퐼푂 = Ω 푟 푑퐴 Polar Moment of Area

4 퐼푥푦, 퐼푂 = [퐿 ]

Cross-sectional Properties

Second Moments of Area: Properties

• 퐼푥 푎푛푑 퐼푦 > 0

• 퐼푂 > 0

• 퐼푂 = 퐼푥 + 퐼푦 when O is the origin of the x and y axes.

• 퐼푥푦 ⋚ 0

• 퐼푥푦 = 0 if either x or y is an axis of symmetry

• 퐼 = 푖 퐼푖 (valid for all the Second Moments of Area)

Cross-sectional Properties Change of coordinates

(퐶) 푥′ = 푥 + 푑푦′푦(퐶) (퐶) 푦′ = 푦 + 푑푥′푥(퐶)

Note that 푑푦′푦(퐶) and 푑푥′푥(퐶) are the coordinates of the centroid C with respect to the Cartesian system ′ Ox’y’. Thus 푑푦′푦(퐶) = 푥퐶 ′ and 푑푥′푥(퐶) = 푦퐶 Cross-sectional Properties Parallel Axes: Static Moment

푆 = 푆 (퐶) + 퐴푑 ′ (퐶) = 퐴푑 ′ (퐶) 푥′ 푥 푥 푥 푥 푥 푆 = 푆 (퐶) + 퐴푑 ′ (퐶) = 퐴푑 ′ (퐶) 푦′ 푦 푦 푦 푦 푦

Note that 푆푥(퐶) = 푆푦(퐶) = 0

′ (퐶) 푆푥′ = 푦 푑퐴 = 푦 + 푑푥′푥(퐶) 푑퐴 = 푆푥(퐶) + 퐴푑푥′푥(퐶) = 퐴푑푥′푥(퐶) Ω Ω Cross-sectional Properties

Transfer-of-axis Theorem: Second Moments of Area

Transfer-of-axis Theorem 2 퐼푥′ = 퐼푥(퐶) + 퐴 푑푥′푥(퐶) 2 퐼푦′ = 퐼푦(퐶) + 퐴 푑푦′푦(퐶)

퐼푥′푦′ = 퐼푥(퐶)푦(퐶) + 퐴푑푥′푥(퐶)푑푦′푦(퐶) Cross-sectional Properties Rotation of the Axes:

푥′ = 푥푐표푠훼 + 푦푠𝑖푛훼 푦′ = 푦푐표푠훼 − 푥푠𝑖푛훼

Cross-sectional Properties

Rotation of the axes: Second Moment of Area

퐼푥 + 퐼푦 퐼푥 − 퐼푦 퐼푥′ = + cos 2훼 − 퐼푥푦푠𝑖푛2훼 2 2 퐼푥 + 퐼푦 퐼푥 − 퐼푦 퐼 = − cos 2훼 + 퐼 푠𝑖푛2훼 푦′ 2 2 푥푦 퐼푥 − 퐼푦 퐼 ′ ′ = 푠𝑖푛2훼 + 퐼 푐표푠2훼 푥 푦 2 푥푦 Cross-sectional Properties Principal axes (1/3) Goal: determine the value of 훼0 for which 퐼푥′ and 퐼푦′ are the maximum and minimum moments of inertia for the cross section (or viceversa) 푑퐼푦′ 푑퐼푥′ = 0 = 0 푑훼 푑훼

2퐼푥푦 푡푔2훼0 = − 퐼푥−퐼푦 Note that if we enforce 퐼푥′푦′ = 0 we obtain the expression above. Thus, when 훼 = 훼0 퐼푥′ and 퐼푦′ attain the maximum and minimum values (or vice versa) and simultaneously 퐼푥′푦′ = 0 Cross-sectional Properties Principal axes (2/3) With 퐶 ≡ 푂 we define 휉, 휂 as the centroidal principal axes:

• 퐼휉, 퐼휂 principal moments of inertia (minimum/maximum moment of inertia or viceversa ) with: • 퐼휉휂 = 0

and:

퐼 +퐼 퐼 −퐼 2 푥(퐶) 푦(퐶) 푥(퐶) 푦(퐶) 2 • 퐼 , 퐼 = ± + 퐼 (퐶) (퐶) 휉 휂 2 2 푥 푦

Cross-sectional Properties Principal axes (3/3): Properties

• If a figure has an axis of symmetry, one of the principal axis is the axis of symmetry. • Any other axis perpendicular to the first one (of symmetry) is the second principal axis.

Cross-sectional Properties

Mohr circle:

Given the centroidal principal axes 휉, 휂, with 퐼휉 > 퐼휂

퐼휉 + 퐼휂 퐼휉 − 퐼휂 퐼 = + 푐표푠2훼 푥 2 2 퐼휉 + 퐼휂 퐼휉 − 퐼휂 퐼 = − 푐표푠2훼 푦 2 2 퐼휉 − 퐼휂 퐼 = 푠𝑖푛2훼 푥푦 2

Parametric equations of a circle on the plane 퐼푥, 퐼푥푦.

Note that the expressions above are written as though x and y are two axes that rotate of an 훼 with respect to the princial axes Cross-sectional Properties Mohr circle:

Parametric equations of a circle in the plane 퐼푥, 퐼푦. 퐼 −퐼 퐼 +퐼 푅 = 휉 휂 ; 퐶 = 휉 휂 , 0 2 2

*A. Di Tommaso. Geometria delle Masse Cross-sectional Properties Radii of Gyration & Ellipse of Inertia

퐼휉 𝜌 = 퐼 = 퐴𝜌2 휉 퐴 휉 휉

퐼휂 𝜌 = 퐼 = 퐴𝜌2 휂 퐴 휂 휂

Analytical expression:

휉2 휂2 2 + 2 = 1 𝜌휂 𝜌휂

Cross-sectional Properties Radii of Gyration & Ellipse of Inertia The Ellipse of Inertia provides a graphical representation of the inertia properties of the cross-section.

퐼푥 2 𝜌푥 = 퐼푥 = 퐴𝜌푥 퐴

퐼푦 𝜌 = 퐼 = 퐴𝜌2 푦 퐴 푦 푦

*A. Di Tommaso. Geometria delle Masse PART II

Solid Mechanics: Displacements and Strains, Strains and Stresses, Stresses and Forces

Outline of PART II

• Displacements and Strains: compatibility equations

• Strains and Stresses: constitutive equations

• Stresses and Forces: equilibrium equations

Displacements and Strains (1D)

:

푢(푥)

• axial strain:

Phillips, Wadee. Pre course Reading Solid Mechanics [1] 퐴′퐵′ − 퐴퐵 푑푢(푥) 휖(푥) = = 퐴퐵 푑푥 different notation Note: the strain is assumed to be positive if the material/solid 푑푢1 휖11 = elongates and negative viceversa 푑푥1 Displacements and Strains (2D) • displacement:

푢1(푥1, 푥2) and 푢2(푥1, 푥2)

• axial strain:

휕푢1 휖11 = 휕푥1 휕푢2 휖22 = 휕푥2

• shear strain: [1]

휋 휕푢2 휕푢1 훾12 = − 훽 = 휃 − 휆 = + with: 훾12 = 2휖12 2 휕푥1 휕푥2

Displacements and Strains (3D) • displacement: 푢1(푥1, 푥2, 푥3), 푢2(푥1, 푥2, 푥3) and 푢3(푥1, 푥2, 푥3)

• axial strain:

휕푢1 휕푢2 휕푢3 휖11 = , 휖22 = , 휖33 = 휕푥1 휕푥2 휕푥3

• shear strain: 휕푢2 휕푢1 훾12 = + with: 훾12 = 2휖12 휕푥1 휕푥2 휕푢3 휕푢1 훾13 = + with: 훾13 = 2휖13 휕푥1 휕푥3 휕푢3 휕푢2 훾23 = + with: 훾23 = 2휖23 휕푥2 휕푥3

Displacements and Strains (3D) The compatibility equations link displacements and strains in 1D: in 2D: in 3D:

0 0 푑푢1 휕 0 휕 휖11 = 휖 휕 휕푥 휕 0 푑푥1 11 휕푥1 휖 1 푢1 11 휕푥 휕 휖22 = 0 휕푥2 0 2 푢 휖22 2 0 0 휕푥3 푢1 2휖12 휕 휕 휖33 = 휕 휕 0 푢2 휕푥2 휕푥1 2휖12 휕푥 휕푥 휕 푢3 2휖13 2 1 2휖 휕 0 휕푥1 23 휕 휕 휕푥3 0 휕푥3 휕푥2

In compact notation: 흐=Du Displacements and Strains In 3D for a given strain 휖11 휖12 휖13 흐 = 휖12 휖22 휖23 휖 휖 휖 13 23 33 the principal strains 휖푎 and principal directions of strain 풂 ≡ [푎1, 푎2, 푎3], i.e. those directions in which there exist only axial strain and no distorsions, can be found solving the eigenvalue problem:

흐 − 휖푎푰 풂 = 0

흐 2nd order strain tensor

휖푎 principal strain 풂 principal directions 푰 identity quantities to be detrmined

Displacements and Strains

In 2D the principal strains and the principal directions, in the plane 푥1 − 푥2, can be determined also as

• principal strains: 2 휖11 + 휖22 휖11 − 휖22 휖 , 휖 = ± + 휖 2 1 2 2 2 12

• the angle 훼 of the principal directions 푎1, 푎2 w.r.t. the 푥1, 푥2 axes:

2휖12 푡푔2훼 = − 휖11 − 휖22

휖 −휖 휖 +휖 Note: Mohr circle of radius R = 1 2 and centre 퐶 = 1 2 2 2

Strains and Stresses The constitutive equations are the relations between (stress, stress-rate) quantities and (strain, strain-rate) quantities for a material. They describe mathematically the actual behavior of a material.

𝜎 𝜎 𝜎

휖 휖 휖

Uniaxial (1D) stress-strain curves Strains and Stresses In 3D the constitutive equations in linear elasticity read

흈 = 푪흐 where 푪 is the Elasticity matrix, 흈 is the vector collecting the stress components and 흐 is the vector of the strain components

𝜎11 퐶11 퐶12 퐶13 퐶14 퐶15 퐶16 휖11 𝜎22 퐶21 퐶22 퐶23 퐶24 퐶25 퐶26 휖22 𝜎 퐶 퐶 퐶 퐶 퐶 퐶 휖33 33 = 31 32 33 34 35 36 𝜎12 퐶41 퐶42 퐶43 퐶44 퐶45 퐶46 2휖12 𝜎 2휖 13 퐶51 퐶52 퐶53 퐶54 퐶55 퐶56 13 𝜎23 2휖 퐶61 퐶62 퐶63 퐶64 퐶65 퐶66 23

Alike 흐 = 푺흈, where 푺 = 푪−1 is the Compliance matrix. In the general case the material has a 푪 matrix characterized by 36 indipendent coefficients.

Strains and Stresses INDEPENDENT MATERIAL NAME NOTE COEFFICIENTS

The material has NO planes of symmetry, i.e. the material TRICLINIC properties differ in all directions. It is possible to prove that the (GENERAL 21 elasticity and compliance tensor are symmetric. The number of ANISOTROPIC) independent coefficients reduces to 21.

The material has 1 plane of symmetry . Number of coefficients MONOCLINIC 13 reduces to 13.

The material has 3 mutually perpendicular planes of symmetry. ORTHOTROPIC 9 This implies no interaction between normal and shear stresses and strains. Number of coefficients reduces to 9.

The material has one plane in which material properties are TRANSVERSELY 5 independent of the orientation. If 푥 − 푥 is the plane, subscripts ISOTROPIC 1 2 1 and 2 are interchangeable. Number of coefficients reduces to 5.

The material has infinite planes of symmetry, i.e. the material ISOTROPIC 2 properties are independent of orientation. All subscripts are interchangeable. Strains and Stresses For an Isotropic material:

푪 =

푺 = 푪−1=

where 퐸 is the Young’s modulus, 휈 is the Poisson’s ratio Stresses and Forces Cauchy Principle states that upon any surface that divides the body, the of one part of the body on the other is equivalent (equipollent) to the system of distributed forces and couples on the surface dividing the body.

Stress Vector:

푻(풏, 푥)

Body force: 풇 Surface force: 푷 Stresses and Forces

The stress vector 푻, in general, has a component 𝜎푛 along the normal 풏 to the surface, and two tangential components, 휏푐 and 휏푏, on the surface.

𝜎푛 = 푻 풏, 푥 풏

휏푐 = 푻 풏, 푥 풄

휏푏 = 푻 풏, 푥 풃

2 2 흉 = 휏푐 + 휏푏 Stresses and Forces The stress vector 푻 can be related to the the

𝜎11 𝜎12 𝜎13 흈 = 𝜎21 𝜎22 𝜎23 𝜎 𝜎 𝜎 31 32 33 as

𝜎 𝜎 𝜎 푛 11 12 13 1 푻 풏, 푥 = 흈풏 = 𝜎21 𝜎22 𝜎23 푛2

𝜎31 𝜎32 𝜎33 푛3

Note: from the balance of angular moment it can be proved that 𝜎12=𝜎21, 𝜎13=𝜎31 and 𝜎23=𝜎32, i.e. the stress tensor has 6 indipendent components. Stresses and Forces The equilibrium equations link forces and stresses:

푑𝑖푣 흈 + 풇 = ퟎ equations:

휕𝜎11 휕𝜎12 휕𝜎13 + + + 푓1 = 0 휕푥1 휕푥2 휕푥3 휕𝜎12 휕𝜎22 휕𝜎23 + + + 푓2 = 0 휕푥1 휕푥2 휕푥3 휕𝜎13 휕𝜎23 휕𝜎33 + + + 푓3 = 0 휕푥1 휕푥2 휕푥3 on the boundary:

푷 = 흈풏

Stresses and Forces In 3D for a given stress tensor 𝜎11 𝜎12 𝜎13 흈 = 𝜎21 𝜎22 𝜎23 𝜎 𝜎 𝜎 31 32 33 the principal stresses 𝜎푎 and principal directions of stress 풂 ≡ [푎1, 푎2, 푎3], i.e. those directions in which there exist only axial stress and no shear stress, can be found solving the eigenvalue problem:

흈 − 𝜎푎푰 풂 = 0

흈 2nd order strain tensor

𝜎푎 principal strain 풂 principal directions 푰 quantities to be detrmined

PART III

Internal Forces Outline of PART III

• Internal forces: • Area • First Moments of Area • Centroid • Second Moments of Area • Transfer of Axis Theorem and Rotation of Axes • Principal Axes and Central Ellipse of Inertia.

Introduction 1/2 • A beam is in equilibrium under the action of external forces. • Each portion of the beam must be in equilibrium under the action of external forces. • If we cut the beam, the equilibrium of each portion is ensured by a distribution of stresses equivalent to a force S applied at the centroid of the cross-section and a whose moment is M.

Note: The external forces are not shown! Introduction 2/2

The components of force S with respect to the Cartesian axes: • P axial force;

• Vx shear force in the x direction;

• Vy shear force in the y direction; P The components of the moment M of the couple with respect to the Cartesian axes:

• My ;

• Mz bending moment;

• Mx torsion; Axial Loading Equilibrium: • The resultant of the internal forces S for an axially loaded member is normal to a section cut perpendicular to the member axis. S=N.

• 푀 = 0 → 푀푥, 푀푦, 푀푧 =0 ; • The only non-zero internal force is 푁.

• 푁 > 0 tension; 푁 < 0 compression.

Axial Force: 푁 = 푃

Normal stress: 푃 𝜎 = 퐴

1 - 51 Axial Loading Displacement and strains: • The Cross-section translates along the beam axis and remains normal to the axis. • The total elongation is δ

Axial Strain:

δ 휀 = 퐿

Axial Loading Stress-strain Curve: Ductile Material

Constitutive Equation:

Linear Elasticity

Hooke’s law: 𝜎 = 퐸휀 Axial Loading

Deformation under axial loading

• From Hooke’s Law: 𝜎 𝜎 = E휀 휀 = E

• From Equilibrium: 푃 푃 𝜎 = 휀 = 퐴 EA

• From the definition of strain:    L • Equating and solving for the deformation, PL   AE Pure Bending Loading

Equilibrium:

• The beam is subjected to equal and opposite couples, whose moment is M, acting in the same longitudinal plane.

• The internal forces in the generic cross-section must satisfy the condition.

Fx   x dA  0 M y   z x dA  0

M z    y x dA  M Pure Bending Loading Displacement and Strains 1/2

2D Beam in pure bending:

• member remains symmetric

• bends uniformly to form a circular arc

• Cross-sections remain plane and perpendicular to the axis of the beam

• The outermost fibers will shorten and the bottom outermost fibers will elongate

• a neutral surface must exist that is parallel to the upper and lower surfaces and for which the length does not change

Pure Bending Loading Displacement and Strains 2/2 Consider a beam segment of length L. After deformation, the length of the neutral surface remains L. At other sections:

L    y    L  L    y     y  y y       (strain varies linearly) x L   c c  m  or ρ    m y     x c m Pure Bending Loading Navier’s Formula: Constitutive Equation: • For a linearly elastic material: y   E   E x x c m y    (stress varies linearly) c m • 1) Static equilibrium: • 2) Static equilibrium: y  y  Fx  0   x dA    m dA M   y x dA   y  m  dA   c    c     I 0   m y dA M  m y 2 dA  m c  c  c Mc First moment with respect to   m I neutral plane is zero y y dA  S  0 Substituting  x    m  z c M The neutral surface must pass    y x Navier’s Formula through the centroid of the cross- I section. Pure Bending Loading A Maximum normal stress & Section Modulus:

• The maximum normal stress due to bending,

M M   c  m I S I  section moment of inertia I W   section modulus c • For a rectangular beam cross section,

b I 1 bh 3 w   12  1 bh 2  1 Ah c h 2 6 6

• Structural steel beams (I beams and H beams) are designed to have a large section modulus. Pure Bending Loading

Deformation under pure bending

• Deformation due to bending moment M is quantified by the curvature of the neutral surface

1   1 Mc    m  m   c Ec Ec I M   EI Eccentric Axial Loading in a Plane of Symmetry • The eccentric loading determines an axial force F and a couple Pd at cross section C. F  P

M  Pd • Principle of superposition: stress distribution due to eccentric loading is determined by superposing the uniform stress due to a centric load and the linear stress distribution due to pure bending

 x   x centric   x bending P My   A I Unsymmetric bending • Analysis of pure bending is limited to members subjected to bending couples acting in a plane of symmetry.

• Members remain symmetric and bend in the plane of symmetry and the neutral axis of the cross-section coincides with the axis of the couple.

• For situations in which the bending couples do not act in a plane of symmetry, the neutral axis of the cross-section will not coincide with the axis of the couple (x) and the beam will not bend in the plane of the couple (y-z).

Unsymmetric bending Principle of superposition :

• Resolve the couple vector into components along the centroidal principal axes. M z  M cos M y  M sin • Superpose the stress distributions

M z y M y z  x    Iz I y • The neutral axis is found by enforcing:

M z y M y z M cos y M sin z  x  0       Iz I y Iz I y y I tan   z tan z I y General Case of Eccentric Axial Loading • Consider a straight member subject to equal and opposite eccentric forces.

• The eccentric force is equivalent to the system of a centric force and two couples. P  centric force

M y  Pa M z  Pb

• By the principle of superposition, the combined stress distribution is

P M z y M y z  x    A Iz I y

• If the neutral axis lies on the cross-section, it may be found from M M P z y  y z  Iz I y A Shear Flow on the Horizontal Face • Transverse loading applied to a beam results in normal and shearing stresses in transverse sections.

• Distribution of normal and shearing stresses satisfies

Fx   xdA  0 M x  y xz  z xy dA  0 Fy   xydA  V M y   z xdA  0 Fz   xzdA  0 M z   y x   0

• By reciprocity of shear stress, when shearing stresses are exerted on the vertical faces of an element, equal stresses must be exerted on the horizontal faces • Longitudinal shearing stresses must exist in any member subjected to transverse loading. Shear flow 1/2

• For equilibrium of beam element F  0  H    dA  x   C D  A M  M H  D C  y dA I A • Note: S  y dA z  A dM M  M  x  V x D C dx • Substituting:

VQ H  x I H VS q   z  shear flow x I Shear flow 2/2 • Shear flow,

H VS q   z  shear flow x I • where

S  y dA z  A

 first moment of area above y1 I   y 2dA AA'  second moment of full cross section • Same result found for lower area

H VS' q    q x I S  S' 0  first moment with respect to neutral axis H  H Shear stress

• The average shearing stress on the horizontal face of the element is obtained by dividing the shearing force on the element by the area of the face.

H q x VSz x  ave    A A Iz b x VS  z Iz b

• On the upper and lower surfaces of the beam,

tyx= 0. It follows that txy= 0 on the upper and lower edges of the transverse sections. Shear stress: Example

• For a narrow rectangular beam: VS 3 V  y 2    z  1  xy  2  Iz b 2 A  c  3 V   max 2 A Shearing Stresses in Thin-Walled Members • Consider a segment of a wide-flange beam subjected to the vertical shear V. • The longitudinal shear force on the element is VS H  x I

• The corresponding shear stress is H VS      zx xz t x It

• Previously found a similar expression for the shearing stress in the web VS   xy It • NOTE:  xy  0 in the flanges  xz  0 in the web Shearing Stresses in Thin-Walled Members

• The variation of shear flow across the section depends only on the variation of the first moment. VS q   t  I

• For a box beam, q grows smoothly from zero at A to a maximum at C and C’ and then decreases back to zero at E. Torque Equilibrium: • Net of the internal shearing stresses is an internal torque, equal and opposite to the applied torque

T    dF    dA • Although the net torque due to the shearing stresses is known, the distribution of the stresses is not

• Distribution of shearing stresses is statically indeterminate – must consider beam deformations

• Unlike the normal stress due to axial loads, the distribution of shearing stresses due to torsional loads can not be assumed uniform. Torque Displacement

• The angle of twist is proportional to the applied torque and to the element length.

  T   L • When subjected to torsion, every cross-section of a circular beam remains plane and undistorted. • Cross-sections for hollow and solid circular beam remain plain and undistorted because a circular beam is axisymmetric.

• Cross-sections of noncircular (non- axisymmetric) beam are distorted when subjected to torsion (warping displacement). Torque Shearing Strain • Consider an interior section of the beam. As a torsional load is applied, an element on the interior cylinder deforms into a rhombus.

• Since the ends of the element remain planar, the shear strain is equal to angle of twist.

• It follows that  L   or   L

• Shear strain is proportional to twist and radius c    and    max L c max Torque Constitutive Equation: • Multiplying the previous equation by the shear modulus  G  G c max From Hooke’s Law,   G , so     c max The shearing stress varies linearly with the J  1 c4 2 radial position in the section. • Recall that the sum of the moments from the internal stress distribution is equal to the torque on the beam at the section,  2  T    dA  max   dA  max J c c • The results are known as the elastic torsion formulas, J  1 c4  c4 Tc T 2  2 1    and   max J J Deformation under Torque • Recall that the angle of twist and maximum shearing strain are related, c   max L • In the elastic range, the shearing strain and shear are related by Hooke’s Law,  Tc   max  max G JG • Equating the expressions for shearing strain and solving for the angle of twist, TL   JG • If the torsional loading or beam cross-section changes along the length, the angle of rotation is found as the sum of segment T L    i i i JiGi Thin-Walled Hollow section • Summing forces in the x-direction on AB,  Fx  0   At Ax B t Bx  t   t  t  q  shear flow A A B B shear stress varies inversely with thickness

• Compute the beam torque from the integral of the moments due to shear stress

dM0  p dF  p t ds  qpds  2q dA

T   dM0   2q dA  2qA T   2tA • Angle of twist: TL ds    4A2G t Torsion of Noncircular Members • Previous torsion formulas are valid for circular section

• Planar cross-sections of noncircular element do not remain planar and stress and strain distribution do not vary linearly

• For uniform rectangular cross-sections, T TL     max 2 3 c1ab c2ab G

• At large values of a/b, the maximum shear stress and angle of twist for other open sections are the same as a rectangular bar.