introduction to advanced structural mechanics
TRANSCRIPT
Introduction to Advanced Structural Mechanics
Prepared by: Antonio Palermo, PhD Student
Revised by: Prof. Alessandro Marzani and Prof. Christian Carloni
Outline of the Introductory Course β’ PART I: Cross-sectional Properties.
β’ PART II: Solid Mechanics: Displacements and Strains, Stress and Equilibrium, Constitutive Equations.
β’ PART III: Internal Forces in Beams: Axial Force, Bending Moment, Shear Force, and Torque. The Euler- Bernoulli beam model.
β’ PART IV Analysis of Statically determinate and indeterminate Structures.
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.
Contacts:
Suggested reading:
β’ Beer, Johnston, DeWolf, Mechanics of Materials.
β’ Gere and Timoshenko, Mechanics of Materials.
Links and resources
Outline of PART I
β’ Beam: Geometric Model.
β’ Cross-Sectional Properties: β’ Area
β’ First Moments of Area
β’ Centroid
β’ Second Moments of Area
β’ Transfer of Axis Theorem and Rotation of Axes
β’ Principal Axes and Central Ellipse of Inertia.
Beam: Geometric Model
A beam is a structural element generated by a planar figure Ξ© (i.e. cross section) that moves in the space remaining normal to the trajectory described by its centroid.
Ξ©
Geometric Requirements:
β’ Ξ©(s) constant or can vary continuously:
β’
BEAM
YES!
h
b
π
β βͺ π π βͺ π
Beam: Geometric Model
YES! Ξ©(z)
NO!
Ξ©=const
For any cross-section Ξ©, it is possible to define some quantities that are related only to the cross-section geometry.
Ξ© β’ Area A
β’ Static Moment of Area
β’ Centroid C
β’ Moments and Product of Inertia
Cross-sectional Properties
C
First Moment of Area: Static Moments
ππ₯ = π¦ππ΄
Ξ©
ππ¦ = π₯ππ΄
Ξ©
π = [πΏ3]
Cross-sectional Properties
Centroid:
The centroid C of a plane figure or two-dimensional shape is the arithmetic mean position 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:
π₯πΆ =ππ¦
π΄
π¦πΆ =ππ₯π΄
π₯πΆ , π¦πΆ = [πΏ]
Cross-sectional Properties
Static Moment & Centroid: Properties
β’ ππ₯ πππ ππ¦ can be β 0
β’ The static moment π is zero if calculated with respect to a centroidal axis (i.e. the centroid lies on the axis)
β’ The Static Moment calculated with respect to an axis of symmetry = 0.
β’ If an area has an axis of symmetry, the centroid C lies on the axis.
β’ If an area has two axes of symmetry, the centroid C is located at the intersection of the axes .
β’ π = πππ (domain of integration can be added: geometric decomposition)
Cross-sectional Properties
Why is it called first moment?
Cross-sectional Properties
Letβs assume to apply a vector at C whose magnitude is the area A of the region W (the dimensions of this vector are [L2]). The moment of this vector with respect to y is π΄π₯πΆ , which is the static moment ππ¦
Second Moment of Area: Moment of Inertia:
πΌπ₯ = π¦2ππ΄
Ξ©
πΌπ¦ = π₯2ππ΄
Ξ©
πΌ = [πΏ4]
Cross-sectional Properties
Second Moment of Area: Product of Inertia and Polar Moment of Area
πΌπ₯π¦ = π₯π¦ππ΄Ξ©
Product o Inertia
πΌπ = π2ππ΄Ξ©
Polar Moment of Area
πΌπ₯π¦ , πΌπ = [πΏ
4]
Cross-sectional Properties
Second Moments of Area: Properties
β’ πΌπ₯ πππ πΌπ¦ > 0
β’ πΌπ > 0
β’ πΌπ = πΌπ₯ + πΌπ¦ when O is the origin of the x and y axes.
β’ πΌπ₯π¦ β 0
β’ πΌπ₯π¦ = 0 if either x or y is an axis of symmetry
β’ πΌ = πΌππ (valid for all the Second Moments of Area)
Cross-sectional Properties
Change of coordinates
π₯β² = π₯(πΆ) + ππ¦β²π¦(πΆ)
π¦β² = π¦(πΆ) + ππ₯β²π₯(πΆ)
Cross-sectional Properties
Note that ππ¦β²π¦(πΆ) and
ππ₯β²π₯(πΆ) are the
coordinates of the centroid C with respect to the Cartesian system Oxβyβ. Thus ππ¦β²π¦(πΆ) = π₯πΆ
β²
and ππ₯β²π₯(πΆ) = π¦πΆβ²
Parallel Axes: Static Moment
ππ₯β² = ππ₯(πΆ) + π΄ππ₯β²π₯(πΆ) = π΄ππ₯β²π₯(πΆ) ππ¦β² = ππ¦(πΆ) + π΄ππ¦β²π¦(πΆ) = π΄ππ¦β²π¦(πΆ)
Cross-sectional Properties
Note that ππ₯(πΆ) = ππ¦(πΆ) = 0
ππ₯β² = π¦β²ππ΄
Ξ©
= π¦(πΆ) + ππ₯β²π₯(πΆ) ππ΄Ξ©
= ππ₯(πΆ) + π΄ππ₯β²π₯(πΆ) = π΄ππ₯β²π₯(πΆ)
Transfer-of-axis Theorem: Second Moments of Area
πΌπ₯β² = πΌπ₯(πΆ) + π΄ ππ₯β²π₯(πΆ)2
πΌπ¦β² = πΌπ¦(πΆ) + π΄ ππ¦β²π¦(πΆ)2
πΌπ₯β²π¦β² = πΌπ₯(πΆ)π¦(πΆ) + π΄ππ₯β²π₯(πΆ)ππ¦β²π¦(πΆ)
Cross-sectional Properties
Transfer-of-axis Theorem
Rotation of the Axes:
π₯β² = π₯πππ πΌ + π¦π πππΌ π¦β² = π¦πππ πΌ β π₯π πππΌ
Cross-sectional Properties
Rotation of the axes: Second Moment of Area
πΌπ₯β² =
πΌπ₯ + πΌπ¦
2+πΌπ₯ β πΌπ¦
2cos 2πΌ β πΌπ₯π¦π ππ2πΌ
πΌπ¦β² =πΌπ₯ + πΌπ¦
2βπΌπ₯ β πΌπ¦
2cos 2πΌ + πΌπ₯π¦π ππ2πΌ
πΌπ₯β²π¦β² =πΌπ₯ β πΌπ¦
2π ππ2πΌ + πΌπ₯π¦πππ 2πΌ
Cross-sectional Properties
Principal axes (1/3) Goal: determine the value of πΌ0 for which πΌπ₯β² and πΌπ¦β² are
the maximum and minimum moments of inertia for the cross section (or viceversa)
ππΌπ¦β²
ππΌ= 0
ππΌπ₯β²ππΌ
= 0
π‘π2πΌ0 = β2πΌπ₯π¦
πΌπ₯βπΌπ¦
Note that if we enforce πΌπ₯β²π¦β² = 0 we obtain the expression
above. Thus, when πΌ = πΌ0 πΌπ₯β² and πΌπ¦β² attain the maximum and
minimum values (or vice versa) and simultaneously πΌπ₯β²π¦β² = 0
Cross-sectional Properties
Principal axes (2/3) With πΆ β‘ π we define π, π as the centroidal principal axes: β’ πΌπ , πΌπ principal moments of inertia (minimum/maximum
moment of inertia or viceversa ) with: β’ πΌππ = 0
and:
β’ πΌπ , πΌπ =πΌπ₯(πΆ)
+πΌπ¦(πΆ)
2Β±
πΌπ₯(πΆ)
βπΌπ¦(πΆ)
2
2
+ πΌπ₯(πΆ)π¦(πΆ)2
Cross-sectional Properties
Principal axes (3/3): Properties
β’ If a figure has an axis of symmetry, one of the principal axis is the axis of symmetry.
β’ 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 π, π, with πΌπ > πΌπ
πΌπ₯ =πΌπ + πΌπ
2+πΌπ β πΌπ
2πππ 2πΌ
πΌπ¦ =πΌπ + πΌπ
2βπΌπ β πΌπ
2πππ 2πΌ
πΌπ₯π¦ =πΌπ β πΌπ
2π ππ2πΌ
Parametric equations of a circle on the plane πΌπ₯, πΌπ₯π¦ .
Note that the expressions above are written as though x and y are two axes that rotate of an angle πΌ with respect to the princial axes
Cross-sectional Properties
Mohr circle: Parametric equations of a circle in the plane πΌπ₯, πΌπ¦.
π =πΌπβπΌπ
2; πΆ =
πΌπ+πΌπ
2, 0
*A. Di Tommaso. Geometria delle Masse
Cross-sectional Properties
Radii of Gyration & Ellipse of Inertia
ππ =πΌπ
π΄ πΌπ= π΄ππ
2
ππ =πΌπ
π΄ πΌπ= π΄ππ
2
Analytical expression:
π2
ππ2 +
π2
ππ2 = 1
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.
ππ₯ =πΌπ₯π΄ πΌπ₯= π΄ππ₯
2
ππ¦ =πΌπ¦
π΄ πΌπ¦= π΄ππ¦
2
*A. Di Tommaso. Geometria delle Masse
Cross-sectional Properties
Outline of PART II
β’ Displacements and Strains: compatibility equations
β’ Strains and Stresses: constitutive equations
β’ Stresses and Forces: equilibrium equations
Displacements and Strains (1D)
β’ displacement:
π’(π₯)
β’ axial strain:
π(π₯) =π΄β²π΅β² β π΄π΅
π΄π΅=ππ’(π₯)
ππ₯
Phillips, Wadee. Pre course Reading Solid Mechanics [1]
Note: the strain is assumed to be positive if the material/solid elongates and negative viceversa π11 =
ππ’1ππ₯1
different notation
β’ displacement: π’1(π₯1, π₯2) and π’2(π₯1, π₯2) β’ axial strain:
π11 =ππ’1ππ₯1
π22 =ππ’2ππ₯2
β’ shear strain:
πΎ12 =π
2β π½ = π β π =
ππ’2
ππ₯1+ππ’1
ππ₯2 with: πΎ12 = 2π12
[1]
Displacements and Strains (2D)
β’ displacement:
π’1(π₯1, π₯2, π₯3), π’2(π₯1, π₯2, π₯3) and π’3(π₯1, π₯2, π₯3)
β’ axial strain:
π11 =ππ’1
ππ₯1, π22 =
ππ’2
ππ₯2, π33 =
ππ’3
ππ₯3
β’ shear strain:
πΎ12 =ππ’2
ππ₯1+ππ’1
ππ₯2 with: πΎ12 = 2π12
πΎ13 =ππ’3
ππ₯1+ππ’1
ππ₯3 with: πΎ13 = 2π13
πΎ23 =ππ’3
ππ₯2+ππ’2
ππ₯3 with: πΎ23 = 2π23
Displacements and Strains (3D)
in 2D:
π11π222π12
=
π
ππ₯10π
ππ₯2
0π
ππ₯2π
ππ₯1
π’1π’2
in 3D:
π11π22π332π122π132π23
=
π
ππ₯100π
ππ₯2π
ππ₯30
0π
ππ₯20π
ππ₯10π
ππ₯3
00π
ππ₯30π
ππ₯1π
ππ₯2
π’1π’2π’3
in 1D:
π11 =ππ’1ππ₯1
π=Du In compact notation:
Displacements and Strains (3D) The compatibility equations link displacements and strains
In 3D for a given strain tensor
the principal strains ππ and principal directions of strain π β‘[π1, π2, π3], i.e. those directions in which there exist only axial strain and no distorsions, can be found solving the eigenvalue problem:
π β πππ° π = 0
π 2nd order strain tensor
ππ principal strain
π principal directions
π° identity matrix
π =
π11 π12 π13π12 π22 π23π13 π23 π33
Displacements and Strains
quantities to be detrmined
In 2D the principal strains and the principal directions, in the plane π₯1 β π₯2, can be determined also as β’ principal strains:
π1, π2 =π11 + π22
2Β±
π11 β π222
2
+ π122
β’ the angle πΌ of the principal directions π1, π2 w.r.t. the π₯1, π₯2 axes:
π‘π2πΌ = β2π12
π11 β π22
Note: Mohr circle of radius R =π1βπ2
2 and centre πΆ =
π1+π2
2
Displacements and Strains
Strains and Stresses The constitutive equations are the relations between kinetics (stress, stress-rate) quantities and kinematics (strain, strain-rate) quantities for a material.
They describe mathematically the actual behavior of a material.
Uniaxial (1D) stress-strain curves
π π π
π π π
In 3D the constitutive equations in linear elasticity read
π = πͺπ
where πͺ is the Elasticity matrix, π is the vector collecting the stress components and π is the vector of the strain components
π11π22π33π12π13π23
=
πΆ11 πΆ12πΆ21 πΆ22
πΆ13 πΆ14πΆ23 πΆ24
πΆ15 πΆ16πΆ25 πΆ26
πΆ31 πΆ32πΆ41 πΆ42
πΆ33 πΆ34πΆ43 πΆ44
πΆ35 πΆ36πΆ45 πΆ46
πΆ51 πΆ52πΆ61 πΆ62
πΆ53 πΆ54πΆ63 πΆ64
πΆ55 πΆ56πΆ65 πΆ66
π11π22π332π122π132π23
Alike π = πΊπ, where πΊ = πͺβ1 is the Compliance matrix. In the general case the material has a πͺ matrix characterized by 36 indipendent coefficients.
Strains and Stresses
MATERIAL NAME INDEPENDENT COEFFICIENTS
NOTE
TRICLINIC (GENERAL
ANISOTROPIC) 21
The material has NO planes of symmetry, i.e. the material properties differ in all directions. It is possible to prove that the elasticity and compliance tensor are symmetric. The number of independent coefficients reduces to 21.
MONOCLINIC 13 The material has 1 plane of symmetry . Number of coefficients reduces to 13.
ORTHOTROPIC 9 The material has 3 mutually perpendicular planes of symmetry. This implies no interaction between normal and shear stresses and strains. Number of coefficients reduces to 9.
TRANSVERSELY ISOTROPIC
5 The material has one plane in which material properties are independent of the orientation. If π₯1 β π₯2 is the plane, subscripts 1 and 2 are interchangeable. Number of coefficients reduces to 5.
ISOTROPIC 2 The material has infinite planes of symmetry, i.e. the material properties are independent of orientation. All subscripts are interchangeable.
Strains and Stresses
For an Isotropic material:
πͺ =
πΊ = πͺβ1=
Strains and Stresses
where πΈ is the Youngβs modulus, π is the Poissonβs ratio
Stresses and Forces Cauchy Principle states that upon any surface that divides the body, the action 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.
π»(π, π₯)
Stress Vector:
π
Body force:
π·
Surface force:
The stress vector π», in general, has a component ππ along the normal π to the surface, and two tangential components, ππ and ππ, on the surface.
ππ = π» π, π₯ π ππ = π» π, π₯ π
ππ = π» π, π₯ π
π = ππ2 + ππ
2
Stresses and Forces
The stress vector π» can be related to the the Cauchy stress tensor
as
π =
π11 π12 π13π21 π22 π23π31 π32 π33
π» π, π₯ = ππ =
π11 π12 π13π21 π22 π23π31 π32 π33
π1π2π3
Stresses and Forces
Note: from the balance of angular moment it can be
proved that π12=π21, π13=π31 and π23=π32, i.e. the stress tensor has 6 indipendent components.
The equilibrium equations link forces and stresses:
πππ£ π + π = π
scalar equations:
ππ11ππ₯1
+ππ12ππ₯2
+ππ13ππ₯3
+ π1 = 0
ππ12ππ₯1
+ππ22ππ₯2
+ππ23ππ₯3
+ π2 = 0
ππ13ππ₯1
+ππ23ππ₯2
+ππ33ππ₯3
+ π3 = 0
on the boundary:
π· = ππ
Stresses and Forces
In 3D for a given stress tensor
the principal stresses ππ and principal directions of stress π β‘[π1, π2, π3], i.e. those directions in which there exist only axial stress and no shear stress, can be found solving the eigenvalue problem:
π β πππ° π = 0
π 2nd order strain tensor
ππ principal strain
π principal directions
π° identity matrix
π =
π11 π12 π13π21 π22 π23π31 π32 π33
quantities to be detrmined
Stresses and Forces
Outline of PART III
β’ Internal forces: β’ Area
β’ First Moments of Area
β’ Centroid
β’ Second Moments of Area
β’ Transfer of Axis Theorem and Rotation of Axes
β’ Principal Axes and Central Ellipse of Inertia.
Introduction 1/2 β’ A beam is in equilibrium under the action of external forces.
β’ Each portion of the beam must be in equilibrium under the action of external forces.
β’ 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 couple whose moment is M.
Note: The external forces are not shown!
The components of force S with respect to the Cartesian axes:
β’ P axial force;
β’ Vx shear force in the x direction;
β’ Vy shear force in the y direction;
The components of the moment M of the couple with respect to the Cartesian axes:
β’ My bending moment;
β’ Mz bending moment;
β’ Mx torsion;
Introduction 2/2
P
Axial Loading
1 - 51
Equilibrium:
β’ 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.
β’ π = 0 β ππ₯, ππ¦, ππ§ =0 ;
β’ The only non-zero internal force is π.
β’ π > 0 tension; π < 0 compression.
Axial Force: π = π
Normal stress:
π =π
π΄
β’ The Cross-section translates along the beam axis and remains normal to the axis.
β’ The total elongation is Ξ΄
Axial Strain:
π =Ξ΄
πΏ
Displacement and strains:
Axial Loading
Linear Elasticity
Hookeβs law: π = πΈπ
πΈ
Constitutive Equation:
Stress-strain Curve: Ductile Material Axial Loading
β’ From the definition of strain:
L
β’ Equating and solving for the deformation,
AE
PL
β’ From Equilibrium: π =
π
π΄ π =
π
EA
β’ From Hookeβs Law:
π = Eπ π =π
E
Deformation under axial loading
Axial Loading
Pure Bending Loading
β’ The beam is subjected to equal and opposite couples, whose moment is M, acting in the same longitudinal plane.
MdAyM
dAzM
dAF
xz
xy
xx
0
0
β’ The internal forces in the generic cross-section must satisfy the condition.
Equilibrium:
2D Beam in pure bending:
Pure Bending Loading
β’ Cross-sections remain plane and perpendicular to the axis of the beam
β’ a neutral surface must exist that is parallel to the upper and lower surfaces and for which the length does not change
β’ bends uniformly to form a circular arc
β’ member remains symmetric
Displacement and Strains 1/2
β’ The top outermost fibers will shorten and the bottom outermost fibers will elongate
Consider a beam segment of length L.
After deformation, the length of the neutral surface remains L.
At other sections:
mx
m
m
x
c
y
cΟ
c
yy
L
yyLL
yL
or
linearly) varies (strain
Pure Bending Loading Displacement and Strains 2/2
β’ For a linearly elastic material:
linearly) varies (stressm
mxx
c
y
Ec
yE
β’ 1) Static equilibrium:
dAyc
dAc
ydAF
m
mxx
0
0
First moment with respect to neutral plane is zero
The neutral surface must pass through the centroid of the cross-section.
β’ 2) Static equilibrium:
yI
M
c
y
I
Mc
c
IdAy
cM
dAc
yydAyM
x
mx
m
mm
mx
ngSubstituti
2
Constitutive Equation:
0 zSdAy
Navierβs Formula
Pure Bending Loading Navierβs Formula:
β’ The maximum normal stress due to bending,
modulus section
inertia of moment section
c
IW
I
S
Mc
I
Mm
β’ For a rectangular beam cross section,
Ahbhh
bh
c
Iw
612
61
3
121
2
β’ Structural steel beams (I beams and H beams) are designed to have a large section modulus.
b
A Maximum normal stress & Section Modulus:
Pure Bending Loading
β’ Deformation due to bending moment M is quantified by the curvature of the neutral surface
EI
M
I
Mc
EcEcc
mm
11
Deformation under pure bending
Pure Bending Loading
β’ The eccentric loading determines an axial force F and a couple Pd at cross section C.
β’ 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
I
My
A
P
xxx
bendingcentric
PdM
PF
Eccentric Axial Loading in a Plane of Symmetry
β’ Analysis of pure bending is limited to members subjected to bending couples acting in a plane of symmetry.
β’ 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).
β’ 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.
Unsymmetric bending
Principle of superposition :
β’ Resolve the couple vector into components along the centroidal principal axes.
sincos MMMM yz
β’ Superpose the stress distributions
y
y
z
zx
I
zM
I
yM
β’ The neutral axis is found by enforcing:
tantan
sincos
y
z
yzy
y
z
zx
I
I
z
y
I
zM
I
yM
I
zM
I
yM
0
Unsymmetric bending
β’ Consider a straight member subject to equal and opposite eccentric forces.
β’ The eccentric force is equivalent to the system of a centric force and two couples.
PbMPaM
P
zy
force centric
β’ By the principle of superposition, the combined stress distribution is
y
y
z
zx
I
zM
I
yM
A
P
β’ If the neutral axis lies on the cross-section, it may be found from
A
Pz
I
My
I
M
y
y
z
z
General Case of Eccentric Axial Loading
00
0
00
xzxzz
xyxyy
xyxzxxx
yMdAF
dAzMVdAF
dAzyMdAF
β’ Distribution of normal and shearing stresses satisfies
β’ Transverse loading applied to a beam results in normal and shearing stresses in transverse sections.
β’ 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
β’ Longitudinal shearing stresses must exist in any member subjected to transverse loading.
Shear Flow on the Horizontal Face
β’ For equilibrium of beam element
A
CD
A
DCx
dAyI
MMH
dAHF 0
xVxdx
dMMM
dAyS
CD
A
z
β’ Note:
flowshearI
VS
x
Hq
xI
VQH
z
β’ Substituting:
Shear flow 1/2
flowshearI
VS
x
Hq z
β’ Shear flow,
β’ where
section cross full of moment second
above area of moment first
'
AA
A
z
dAyI
y
dAyS
2
1
β’ Same result found for lower area
HH
SS
qI
VS
x
Hq
axis neutral to
respect withmoment first
'
'
0
Shear flow 2/2
β’ 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.
bI
VS
xb
x
I
VS
A
xq
A
H
z
z
z
zave
β’ 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
Shearing Stresses in Thin-Walled Members β’ Consider a segment of a wide-flange
beam subjected to the vertical shear V.
β’ The longitudinal shear force on the element is
xI
VSH
It
VS
xt
Hxzzx
β’ The corresponding shear stress is
β’ NOTE: 0xy
0xz
in the flanges
in the web
β’ Previously found a similar expression for the shearing stress in the web
It
VSxy
β’ The variation of shear flow across the section depends only on the variation of the first moment.
I
VStq
β’ For a box beam, q grows smoothly from zero at A to a maximum at C and Cβ and then decreases back to zero at E.
Shearing Stresses in Thin-Walled Members
Torque
dAdFT
β’ Net of the internal shearing stresses is an internal torque, equal and opposite to the applied torque
β’ Although the net torque due to the shearing stresses is known, the distribution of the stresses is not
β’ Unlike the normal stress due to axial loads, the distribution of shearing stresses due to torsional loads can not be assumed uniform.
β’ Distribution of shearing stresses is statically indeterminate β must consider beam deformations
Equilibrium:
β’ The angle of twist is proportional to the applied torque and to the element length.
L
T
β’ When subjected to torsion, every cross-section of a circular beam remains plane and undistorted.
β’ Cross-sections of noncircular (non-axisymmetric) beam are distorted when subjected to torsion (warping displacement).
β’ Cross-sections for hollow and solid circular beam remain plain and undistorted because a circular beam is axisymmetric.
Torque Displacement
Shearing Strain β’ Consider an interior section of the beam. As
a torsional load is applied, an element on the interior cylinder deforms into a rhombus.
β’ Shear strain is proportional to twist and radius
maxmax and
cL
c
LL
or
β’ It follows that
β’ Since the ends of the element remain planar, the shear strain is equal to angle of twist.
Torque
Jc
dAc
dAT max2max
β’ Recall that the sum of the moments from the internal stress distribution is equal to the torque on the beam at the section,
4
21 cJ
41
422
1 ccJ and max
J
T
J
Tc
β’ The results are known as the elastic torsion formulas,
β’ Multiplying the previous equation by the shear modulus
max
Gc
G
max
c
From Hookeβs Law, G , so
The shearing stress varies linearly with the radial position in the section.
Constitutive Equation: Torque
β’ Recall that the angle of twist and maximum shearing strain are related,
L
c max
β’ In the elastic range, the shearing strain and shear are related by Hookeβs Law,
JG
Tc
G max
max
β’ Equating the expressions for shearing strain and solving for the angle of twist,
JG
TL
β’ If the torsional loading or beam cross-section changes along the length, the angle of rotation is found as the sum of segment rotations
i ii
ii
GJ
LT
Deformation under Torque
β’ Summing forces in the x-direction on AB,
shear stress varies inversely with thickness
flowshear
0
qttt
xtxtF
BBAA
BBAAx
tA
T
qAdAqdMT
dAqpdsqdstpdFpdM
2
22
2
0
0
β’ Compute the beam torque from the integral of the moments due to shear stress
Thin-Walled Hollow section
t
ds
GA
TL24
β’ Angle of twist:
Torsion of Noncircular Members
β’ 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.
Gabc
TL
abc
T3
22
1
max
β’ For uniform rectangular cross-sections,
β’ Previous torsion formulas are valid for circular section
β’ Planar cross-sections of noncircular element do not remain planar and stress and strain distribution do not vary linearly