lecture notes on mechanics

M. Vable Intermediate Mechanics of Materials: Chapter 1

Stress and Strain

The learning objectives in this chapter are:

• Understanding the concept of stress and the use of double subscripts in determining the direction of stress components on a surface.

• Understanding the concept of strain and the use of small strain and finite difference approximation.

• Understanding the stress and strain transformation in three dimension.

1-1


Internally Distributed Force System

• The intensity of internal distributed forces on an imaginary cut surface of a body is called the stress on a surface.

• The intensity of internal distributed force that is normal to the surface of an imaginary cut is called the normal stress on a surface.

• The intensity of internal distributed force that is parallel to the surface of an imaginary cut surface is called the shear stress on the surface.

• Relating stresses to external forces and moments is a two step process.

Internal Forces EquilibriumStatic Equivalencyand Moments

External Forcesand Moments

Stresses

x

y

z

x

y

z

x

y

z

x

y

z

Mz

T

My

Normal stresslinear in y

Normal stresslinear in z

Uniformshear stressin tangentialdirection.Uniform Normal

Stress σavg

N σavgA= V τavgA=

Uniform ShearStress τavg

Static equivalency

Axial Bending Bending Torsion

1-2


Stress at a Pointi

ΔFj

ΔAi

Outward normal Internal Force

direction ofoutward normal to theimaginary cut surface.

direction of the internal force component.

σijΔFjΔAi---------

⎝ ⎠⎜ ⎟⎛ ⎞

ΔAi 0→lim=

• ΔAi will be considered positive if the outward normal to the surface is in the positive i direction.

• A stress component is positive if numerator and denominator have the same sign. Thus σij is positive if: (1) ΔFj and ΔAi are both positive. (2) ΔFj and ΔAi are both negative.

• Stress Matrix in 3-D:

σxx τxy τxzτyx σyy τyzτzx τzy σzz

Table 1.1. Comparison of number of components

Quantity 1-D 2-D 3-D

Scaler 1=10 1=20 1=30

Vector 1=11 2=21 3=31

Stress 1=12 4=22 9=32

1-3


Stress Element

• Stress element is an imaginary object that helps us visualize stress at a point by constructing surfaces that have outward normal in the coordinate directions.


Stress cube showing all positive stress components

Plane Stress: All stress components on a plane are zero.

σxx τxy 0

τyx σyy 0

0 0 0

Symmetric Shear Stresses: τxy τyx= τyz τzy= τzx τxz=• A pair of symmetric shear stress points towards the corner or away from

the corner.

1-4


1.3 Show the non-zero stress components on the A,B, and C faces of the cube shown in Figure P1.3 and Figure P1.4.

σxx 0= τxy 15– ksi= τxz 0=

τyx 15– ksi= σyy 10ksi C( )= τyz 25ksi=

τzx 0= τzy 25ksi= σzz 20ksi T( )=

Fig. P1.3 1.4

1-5


Stress transformation in two dimension

z

x

y

nt

Horizontal Plane

Verti

calP

lane

Inclined Plane

θ

Outward normalto the inclined plane.

θ

nt

λ

y

x

(a) (c)

σnn σxx θ2cos σyy θ2sin 2τxy θ θcossin+ +=

τnt σxx θ θsincos– σyy θ θcossin τxy θ θ2sin–2cos⎝ ⎠⎛ ⎞+ +=

σtt σxx θ2sin σyy θ2cos 2τxy θ θsincos–+=

Matrix Notation

nx θcos= ny θsin= tx λcos= ty λsin=

True only in 2D: λ 90 θ+= tx ny–= ty nx=

n{ }nxny⎩ ⎭

⎪ ⎪⎨ ⎬⎪ ⎪⎧ ⎫

= t{ }txty⎩ ⎭

⎪ ⎪⎨ ⎬⎪ ⎪⎧ ⎫

= σ[ ]σxx τxyτyx σyy

=

The symmetry of shear stresses σ[ ]T σ[ ]=

σnn n{ }T σ[ ] n{ }=

τnt t{ }T σ[ ] n{ }=

σtt t{ }T σ[ ] t{ }=

1-6


Traction or Stress vector

Mathematically the stress vector {S} is defined as: S{ } σ[ ] n{ }=

Sx σxxnx τxyny+=

Sy τyxnx σyyny+=• pressure is a scaler quantity.• traction is a vector quantity.,• stress is a second order tensor.

σxx

σnn

τnt

σyy

τyx

τxy (nx dA)

(dA)

(dA)

(ny dA)(ny dA)

( nx dA) σxx

σyy

τyx

τxy (nx dA)

(ny dA)(ny dA)

( nx dA) Sx (dA)

Sy (dA)

Statically equivalent force wedge.

S{ } σnn n{ } τnt t{ }+=

Stress vector in different coordinate systems.

x

y

{S}σnn{n}

τnt{t}Sy

Sx

1-7


Principal Stresses and Directions

S{ } σ[ ] p{ } σp p{ }= =OR

S{ }σxx τxyτyx σyy

pxpy⎩ ⎭

⎪ ⎪⎨ ⎬⎪ ⎪⎧ ⎫

σp 0

0 σp

pxpy⎩ ⎭

⎪ ⎪⎨ ⎬⎪ ⎪⎧ ⎫

= =

OR

σxx σp–( ) τxyτyx σyy σp–( )

pxpy⎩ ⎭

⎪ ⎪⎨ ⎬⎪ ⎪⎧ ⎫

0=

Characteristic equation

σp2 σp– σxx σyy+( ) σxxσyy τxy

2–( )+ 0=

Roots: σ1 2, σxx σyy+( ) σxx σyy+( )2 4 σxxσyy τxy2–( )–± 2⁄=

OR

σ1 2,

σxx σyy+2------------------------⎝ ⎠

⎛ ⎞ σxx σyy–2-----------------------⎝ ⎠

⎛ ⎞2

τxy2+±=

• The eigenvalues of the stress matrix are the principal stresses.

• The eigenvectors of the stress matrix are the principal directions.

1-8


Stress Transformation in 3-D

n{ }

nxnynz⎩ ⎭

⎪ ⎪⎪ ⎪⎨ ⎬⎪ ⎪⎪ ⎪⎧ ⎫

= S{ }

SxSySz⎩ ⎭

⎪ ⎪⎪ ⎪⎨ ⎬⎪ ⎪⎪ ⎪⎧ ⎫

= σ[ ]


=

σnn n{ }T σ[ ] n{ }=

τnt t{ }T σ[ ] n{ }=

σtt t{ }T σ[ ] t{ }=

S{ } σ[ ] n{ }=Equilibrium condition: S{ } σnn n{ } τnt tE{ }+=

1-9


Principal Stresses and Directions

S{ } σ[ ] p{ } σp p{ }= =OR


pxpypz⎩ ⎭

⎪ ⎪⎪ ⎪⎨ ⎬⎪ ⎪⎪ ⎪⎧ ⎫

σp 0 0

0 σp 0

0 0 σp

pxpypz⎩ ⎭

⎪ ⎪⎪ ⎪⎨ ⎬⎪ ⎪⎪ ⎪⎧ ⎫

=

OR

σxx σp–( ) τxy τxzτyx σyy σp–( ) τyzτzx τzy σzz σp–( )

pxpypz⎩ ⎭

⎪ ⎪⎪ ⎪⎨ ⎬⎪ ⎪⎪ ⎪⎧ ⎫

0=

• The eigenvalues of the stress matrix are the principal stresses.

• The eigenvectors of the stress matrix are the principal directions.

px2 py

2 pz2+ + 1=

Principal stress convention

Ordered principal stresses in 3-D: σ1 σ2 σ3> >

Ordered principal stresses in 2-D: σ1 σ2>

Principal Angles 0o θx θy θz, , 180o≤ ≤

1-10



σp3 I1σp

2– I2σp I– 3+ 0=

Stress Invariants

I1 σxx σyy σzz+ +=

I2σxx τxyτyx σyy

σyy τyzτzy σzz

σxx τxzτzx σzz

+ +=

I3


=

Roots:

x3 I1– x2 I2x I3–+ 0=

x1 2A αcos I1 3⁄+= x2 3, 2– A α 60o±( )cos I1 3⁄+=

A I1 3⁄( )2 I2 3⁄–=

3αcos 2 I1 3⁄( )3 I1 3⁄( )I2– I3+[ ] 2A3( )⁄=

Principal Stress Matrix σ[ ]

σ1σ2

σ3

=

I1 σ1 σ2 σ3+ +=

I2 σ1σ2 σ2σ3 σ3σ1+ +=

I3 σ1σ2σ3=

1-11


Maximum Shear Stress

τmax maxσ1 σ2–

2------------------σ2 σ3–

2------------------σ3 σ1–

2------------------, ,⎝ ⎠⎛ ⎞=

1-12


Octahedral stresses

• A plane that makes equal angles with the principal planes is called an octa-hedral plane.

σnn σ1n12 σ2n2

2 σ3n32+ +=

τnt σ12n1

2 σ22n2

2 σ32n3

2+ +( ) σnn2–=

n1 n2 n3 1 3⁄= = =

σoct σ1 σ2 σ3+ +( ) 3⁄=

τoct13--- σ1 σ2–( )2 σ2 σ3–( )2 σ3 σ1–( )2+ +=

1-13


1.35 The stress at a point is given by the stress matrix shown. Determine: (a) the normal and shear stress on a plane that has an outward normal at 37o, 120o, and 70.43o, to x, y, and z direction respectively. (b) the principal stresses (c) the second principal direction and (d) the magni-tude of the octahedral shear stress. (e) maximum shear stress

18 12 912 12 6–9 6– 6

ksi

1-14


Strain

• The total movement of a point with respect to a fixed reference coordinates is called displacement.

• The relative movement of a point with respect to another point on the body is called deformation.

• Lagrangian strain is computed from deformation by using the original undeformed geometry as the reference geometry.

• Eulerian strain is computed from deformation by using the final deformed geometry as the reference geometry.

• Relating strains to displacements is a problem in geometry.

KinematicsStrainsDisplacements

Average normal strain

εavLf Lo–

Lo----------------- δ

Lo------= =

• Elongations (Lf > Lo) result in positive normal strains. Contractions (Lf < Lo) result in negative normal strains.

1-15


Average shear strain

π/2

A

B C

Undeformed grid Deformed grid

Wooden Bar with Masking Tape


α

γA1

CB

A



γavπ2--- α–=

• Decreases in the angle (α < π / 2) result in positive shear strain. Increase in the angle (α > π / 2) result in negative shear strain

Units of average strain• To differentiate average strain from strain at a point.

• in/in, or cm/cm, or m/m (for normal strains)

• rads (for shear strains)

• percentage. 0.5% is equal to a strain of 0.005

• prefix: μ = 10-6. 1000 μ in / in is equal to a strain 0.001 in /

1-16


Small Strain Approximation

• Small-strain approximation may be used for strains less than 0.01

• Small normal strains are calculated by using the deformation component in the original direction of the line element regardless of the orientation of the deformed line element.

• In small shear strain (γ) calculations the following approximation may be used for the trigonometric functions: γtan γ≈ γsin γ≈ γcos 1≈

• Small-strain calculations result in linear deformation analysis. • Drawing approximate deformed shape is very important in analysis of

small strains. 1.41 Determine the deformation in bars AP and BP in Fig. P1.41 using small strain approximation.

Fig. P1.41

1-17


Strain at a point

Engineering Strain

εxxΔuΔx-------⎝ ⎠

⎛ ⎞Δx 0→lim x∂

∂u= = εyyΔvΔy------⎝ ⎠

⎛ ⎞Δy 0→lim y∂

∂v= = εzzΔwΔz--------⎝ ⎠

⎛ ⎞Δz 0→lim z∂

∂w= =

γxy γyxΔuΔy------- Δv

Δx------+⎝ ⎠⎛ ⎞

Δx 0→Δy 0→

lim y∂∂u

x∂∂v+== = γyz γzy

ΔvΔz------ Δw

Δy--------+⎝ ⎠⎛ ⎞

Δy 0→Δz 0→

lim z∂∂v

y∂∂w+== =

γzx γxzΔwΔx-------- Δu

Δz-------+⎝ ⎠⎛ ⎞

Δx 0→Δz 0→

lim x∂∂w

z∂∂u+== =

• The partial derivative with respect to a coordinate implies that during the process of differentiation the other coordinates are held constant.

• If a displacement is only a function of one coordinate, then the partial derivative with respect to that coordinate will be same as ordinary deriva-tive.

εxx xddu x( )=

1-18


Finite Difference Approximation

• Forward difference approximates the slope of the tangent using the point ahead of point i as:

εxx( )i

ui 1+ ui–xi 1+ xi–-----------------------=

• Backward difference approximates the slope of the tangent using the point behind i as:

εxx( )i

ui ui 1––xi xi 1––-----------------------=

• Central difference takes the average value using the point ahead and behind as:

εxx( )i12---

ui 1+ ui–xi 1+ xi–-----------------------

ui ui 1––xi xi 1––-----------------------+=

1-19


1.58 The displacements u and v in the x and y directions respec-tively were measured by Moire' interferometry. Displacements of 16 points on the body and are as given below.

Point u(μmm)

v(μmm) Point u

(μmm)v

(μmm)

1 0.000 0.000 9 0.128 0.3842 -0.112 0.144 10 -0.048 0.3363 -0.128 0.256 11 -0.128 0.2564 -0.048 0.336 12 -0.112 0.1445 0.112 0.176 13 0.048 0.6246 -0.032 0.224 14 -0.160 0.4807 -0.080 0.240 15 -0.272 0.3048 -0.032 0.224 16 -0.288 0.096

Determine the strains εxx εyy, and, γxy at points 1 and 4 shown in Fig-

ure P1.58.

1-20


Strain Transformation

Strain transformation equations in 2-D

εnn εxx θ2cos εyy θ2sin γxy θsin θcos+ +=

εtt εxx θ2sin εyy θ2cos γ– xy θsin θcos+=

γnt 2εxx θsin θcos– 2εyy θsin θcos γxy θ2cos θ2sin–( )+ +=

Stress transformation equations in 2-D

σnn σxx θ2cos σyy θ2sin 2τxy θ θcossin+ +=

σtt σxx θ2sin σyy θ2cos 2τxy θ θsincos–+=

τnt σxx θ θsincos– σyy θ θcossin τxy θ θ2sin–2cos⎝ ⎠⎛ ⎞+ +=

• tensor normal strains = engineering normal strains• tensor shear strains = (engineering shear strains)/ 2

Tensor strain matrix from engineering strains

ε[ ]

εxx εxy γxy 2⁄= εxz γxz 2⁄=

εyx γyx 2⁄= εyy εyz γyz 2⁄=

εzx γzx 2⁄= εzy γzy 2⁄= εzz

=

εnn n{ }T ε[ ] n{ }=

εnt t{ }T ε[ ] n{ }= γnt 2εnt=

εtt t{ }T ε[ ] t{ }=

1-21



εp3 I1εp

2– I2εp I– 3+ 0=

Strain invariants

I1 εxx εyy εzz+ + ε1 ε2 ε3+ += =

I2εxx εxyεyx εyy

εyy εyzεzy εzz

εxx εxzεzx εzz

+ + ε1ε2 ε2ε3 ε3ε1+ += =

I3

εxx εxy εxzεyx εyy εyzεzx εzy εzz

ε1ε2ε3= =

Maximum shear strain

γmax2

------------ maxε1 ε2–

2----------------

ε2 ε3–2

----------------ε3 ε1–

2----------------, ,⎝ ⎠

⎛ ⎞=

1-22


Material Description


• Understand the definition and differences of linear material models.

• Understand the statements and the applications of failure theories.

• Understand the concepts and applications of stress concentration fac-

tor and stress intensity factor in analysis and design.

1-1


Linear Material Models

εxx

εyy

εzz

γyz

γzx

γxy⎩ ⎭⎪ ⎪⎪ ⎪⎪ ⎪⎪ ⎪⎨ ⎬⎪ ⎪⎪ ⎪⎪ ⎪⎪ ⎪⎧ ⎫ C11

C12

C13

C41

C51

C61

C12

C22

C23

C42

C52

C62

C13

C23

C33

C43

C53

C63

C14

C24

C34

C44

C54

C64

C15

C25

C35

C45

C55

C65

C16

C26

C36

C46

C56

C66

σxx

σyy

σzz

τyz

τzx

τxy⎩ ⎭⎪ ⎪⎪ ⎪⎪ ⎪⎪ ⎪⎪ ⎪⎨ ⎬⎪ ⎪⎪ ⎪⎪ ⎪⎪ ⎪⎪ ⎪⎧ ⎫

=

Cij Cji=

• The most general linear anisotropic material requires 21 independent constants.

Monoclinic material• Monoclinic material requires 13 independent material constants. • The z-plane is the plane of symmetry.

εxx

εyy

εzz

γyz

γzx

γxy⎩ ⎭⎪ ⎪⎪ ⎪⎪ ⎪⎪ ⎪⎨ ⎬⎪ ⎪⎪ ⎪⎪ ⎪⎪ ⎪⎧ ⎫ C11

C12

C13

0

0

C16

C12

C22

C23

0

0

C26

C13

C23

C33

0

0

C36

0

0

0

C44

C45

0

0

0

0

C45

C55

0

C16

C26

C36

0

C66

σxx

σyy

σzz

τyz

τzx

τxy⎩ ⎭⎪ ⎪⎪ ⎪⎪ ⎪⎪ ⎪⎪ ⎪⎨ ⎬⎪ ⎪⎪ ⎪⎪ ⎪⎪ ⎪⎪ ⎪⎧ ⎫

=

Orthotropic material • Orthotropic material requires 9 independent constants. • Orthotropic materials have two orthogonal planes of symmetry.

1-2


εxx

εyy

εzz

γyz

γzx

γxy⎩ ⎭⎪ ⎪⎪ ⎪⎪ ⎪⎪ ⎪⎨ ⎬⎪ ⎪⎪ ⎪⎪ ⎪⎪ ⎪⎧ ⎫ C11

C12

C13

0

0

0

C12

C22

C23

0

0

0

C13

C23

C33

0

0

0

0

0

0

C44

0

0

0

0

0

0

C55

0

0

0

0

0

C66

σxx

σyy

σzz

τyz

τzx

τxy⎩ ⎭⎪ ⎪⎪ ⎪⎪ ⎪⎪ ⎪⎪ ⎪⎨ ⎬⎪ ⎪⎪ ⎪⎪ ⎪⎪ ⎪⎪ ⎪⎧ ⎫

=

For plane stress problems

εxxσxxEx--------

νyxEy-------σyy–= εyy

σyyEy--------

νxyEx-------σxx–= γxy

τxyGxy--------= νyx

Ey--------

νxyEx--------=

• The form of equations relating stresses and strain is valid for a specific coordinate system which is called the material coordinate system.

Long Fiber Composite

Transversely isotropic material • Transversely isotropic material requires 5 independent material con-

stants. • Transversely isotropic material are isotropic in a plane.

1-3


εxx

εyy

εzz

γyz

γzx

γxy⎩ ⎭⎪ ⎪⎪ ⎪⎪ ⎪⎪ ⎪⎨ ⎬⎪ ⎪⎪ ⎪⎪ ⎪⎪ ⎪⎧ ⎫ C11

C12

C13

0

0

0

C12

C11

C13

0

0

0

C13

C13

C33

0

0

0

0

0

0

C44

0

0

0

0

0

0

C44

0

0

0

0

0

2 C11 C12–( )

σxx

σyy

σzz

τyz

τzx

τxy⎩ ⎭⎪ ⎪⎪ ⎪⎪ ⎪⎪ ⎪⎪ ⎪⎨ ⎬⎪ ⎪⎪ ⎪⎪ ⎪⎪ ⎪⎪ ⎪⎧ ⎫

=

Isotropic Material• An isotropic material has a stress-strain relationships that are indepen-

dent of the orientation of the coordinate system at a point. • An isotropic body requires only two independent material constants

εxx

εyy

εzz

γyz

γzx

γxy⎩ ⎭⎪ ⎪⎪ ⎪⎪ ⎪⎪ ⎪⎨ ⎬⎪ ⎪⎪ ⎪⎪ ⎪⎪ ⎪⎧ ⎫ C11

C12

C12

0

0

0

C12

C11

C12

0

0

0

C12

C12

C11

0

0

0

0

0

0

2 C11 C12–( )

0

0

0

0

0

0

2 C11 C12–( )

0

0

0

0

0

2 C11 C12–( )

σxx

σyy

σzz

τyz

τzx

τxy⎩ ⎭⎪ ⎪⎪ ⎪⎪ ⎪⎪ ⎪⎪ ⎪⎨ ⎬⎪ ⎪⎪ ⎪⎪ ⎪⎪ ⎪⎪ ⎪⎧ ⎫

=

εxx σxx ν σyy σzz+( )–[ ] E⁄= εyy σyy ν σzz σxx+( )–[ ] E⁄=

εzz σzz ν σxx σyy+( )–[ ] E⁄=

γxy τxy G⁄= γyz τyz G⁄= γzx τzx G⁄=

Comparing C11 1 E⁄= , C12 ν– E⁄= , and 2 C11 C12–( ) 1 G⁄=

G E2 1 ν+( )--------------------=

1-4


εxxεyyεzz⎩ ⎭

⎪ ⎪⎪ ⎪⎨ ⎬⎪ ⎪⎪ ⎪⎧ ⎫

1E---

1 ν– ν–ν– 1 ν–ν– ν– 1

σxxσyyσzz⎩ ⎭

⎪ ⎪⎪ ⎪⎨ ⎬⎪ ⎪⎪ ⎪⎧ ⎫

=

• A material is said to be homogenous if the material properties are the same at all points on the body. Alternatively, if the material constants Cij are functions of the coordinates x, y, or z, then the material is called non-homogenous.

1-5


Plane Stress and Plane Strain

.

Plane Stress

Plane Strain

GeneralizedHooke’s Law

GeneralizedHooke’s Law

σxx τxy 0

τyx σyy 0

0 0 0

εxx γxy 0

γyx εyy 0

0 0 εzzνE--- σxx σyy+( )–=

εxx γxy 0

γyx εyy 0

0 0 0

σxx τxy 0

τyx σyy 0

0 0 σzz ν σxx σyy+( )=

εzz 0≠Free surface

Rigid Surface

Reaction Force σzz 0≠( )

Plane Stress Plane Strain

σxxσyy σyyσxx

σzz 0=( )

Free surface σzz 0=( ) εzz 0=( )

Rigid Surface εzz 0=( ) Reaction Force σzz 0≠( )

2.15 An orthotropic material has the following properties Ex=7,500ksi, Ey= 2,500 ksi, Gxy = 1,250 ksi and νxy= 0.25. Determine the principal direction 1 for stresses and strains at a point on a free sur-face where the following strains were measured

εxx 400– μ= εyy 600 μ= γxy 500– μ=

1-6


Failure Theories

• A failure theory is a statement on relationship of the stress components to material failure characteristics values.

Ductile Material Brittle Material

Characteristic fail-ure stress

Yield stress Ultimate stress

Theories 1. Maximum shear stress2. Maximum octahedral shear stress

1. Maximum normal stress2. Modified Mohr

Maximum shear stress theoryFor ductile materials the theory predicts

A material will fail when the maximum shear stress exceeds the shear stress at yield that is obtained from uni-axial tensile test.

The failure criterion isτmax τyield≤

max σ1 σ2– σ2 σ3– σ3 σ1–, ,( ) σyield≤

Maximum octahedral shear stress theory (Maximum distortion strain energy)For ductile materials the theory predicts

A material will fail when the maximum octahedral shear stress exceeds the octahedral shear stress at yield that is obtained from uni-axial tensile test.

The failure criterion isτoct τyield≤

1-7


12

------- σ1 σ2–( )2 σ2 σ3–( )2 σ3 σ1–( )2+ + σyield≤

Equivalent von-Mises Stress

σvon12

------- σ1 σ2–( )2 σ2 σ3–( )2 σ3 σ1–( )2+ +=

σvon σyield≤

Failure Envelopes for ductile materials in plane stress

σ1

σ2

σyield

σyield

−σyield

−σyield

Maximum octahedral shear stress

Maximum Shear Stress

Maximum normal stress theoryFor brittle materials the theory predicts

A material will fail when the maximum normal stress at a point exceed the ultimate normal stress (σult) obtained from uni-axial tension test.

max σ1 σ2 σ3, ,( ) σult≤

• can be used if principal stress one is tensile and the dominant princi-pal stress.

Mohr’s theoryFor brittle materials the theory predicts

A material will fail if a stress state is on the envelope that is tangent to the three Mohr’s circles corresponding to: uni-axial ultimate stress in tension, to uni-axial ultimate stress in compression, and to pure shear.

1-8


σT

σC

τS

σ (T)(C)

(CW)

(CCW)

τ

σC σT

τS

Tangent pointsFailure Envelope

Tangent points

Modified Mohr’s Theory

σTσC

σ1

σ2

σT

σC

(-τS,τS)

(τS,-τS)

d Mohr’s Theory

σ2σC------

σ1σT------– 1=

σ1σT------

σ2σC------– 1=

Mohr’s Theory

• If both principal stresses are tensile than the maximum normal stress has to be less than the ultimate tensile strength.

• If both principal stresses are negative than the maximum normal stress must be less than the ultimate compressive strength.

• If the principal stresses are of different signs then for the Modified Mohr’s Theory the failure is governed by

σ2σC------

σ1σT------– 1≤

1-9


2.23 The strains shown by the strain gages were recorded on a free surface of aluminum (E=10,000 ksi, ν = 0.35, σyield = 24 ksi). By how much can the loads be scaled without exceeding the yield stress of alumi-num at the point. Use maximum octahedral shear stress theory.

ba

c

600450

x

yεa 600– μ in in§=

εb 500 μ in in§=

εc 400 μ in in§=

Figure P2.23

Saint-Venant’s Principle

• Two statically equivalent loads systems produce nearly the same stress in regions at a distance that is at least equal to the largest dimension in the loaded region.

W

tP

W

A

PA

t

σ PWt-------≈ at x W>

x x

1-10


Stress Concentration• Large stress gradients in a small region is called stress concentration. • The stress predicted by theoretical models away from the regions of

stress concentration is called the Nominal Stress.

σnom

σnom

σnom

σnom

KconcMaximum StressNominal Stress

---------------------------------------------=

2.28 The stress concentration factor for a flat tension bar with U-shaped notches shown in Figure P2.28 was determined as:

Kconc 3.857 5.066 4rH-----⎝ ⎠

⎛ ⎞– 2.469 4rH-----⎝ ⎠

⎛ ⎞ 2+ 0.258 4r

H-----⎝ ⎠⎛ ⎞ 3

–=

The nominal stress is P/(Ht). Make a chart for the stress concentration factor vs. (r/d) for the following values of (H/d): 1.25, 1.50, 1.75, 2.0. Use of spread sheet is recommended.

2r

2r

2r

t

dHP P

Figure P2.28 2.30 A steel tension bar with U-shaped notches of the type shown in Figure P2.28, is to carry a load P = 12 kips. The yield stress of steel is 30 ksi. The bar has H = 9 in, d = 6 in, t =0.25 in. For a factor of safety of 1.4, determine the value of r if yielding is to be avoided.

1-11


Stress Intensity Factor

σA

A2a

2b

σ

σ

σ

σ

σ

σ

σ

Kconc 1 2a b⁄+( )=

An elliptical hole in an infinite plate.

σij Kintenfij=

Modulus of Toughness.

σ

ε

Stronger MaterialTougher material

σ

ε

Modulus ofToughness

UltimateStress

(a) (b)

• Stress intensity factor depends upon the stress level and the length of the crack.

• Critical stress intensity factor is a material property that is independent of the stress level or crack length.

• A crack becomes unstable (material breaks) when stress intensity fac-tor exceeds the critical stress intensity factor.

1-12


Three modes of relative crack surface movement

Mode 1 Mode 2 Mode 3Crack tip

Crack tipCrack tip

.

2aσnom

σnom

2aτnom

KI σnom πa=

KII τnom πa=τnom

Kequiv KI2 KII

2+=• Microcracks will be assumed to grow in Mode I due to principal stress

one if it is in tension.

2.32 The stresses at a point in plane stress were found to be:σxx 27 ksi T( )= σyy 10 ksi C( )= τxy 15 ksi=

The critical stress intensity factor of the material is 20 ksi in . Determine the critical crack length (a) asuming a small crack exist on a plane -25o from the x-axis. (b) there is no pre-existing crack.

1-13


Basic Structural Analysis


• Understand the limitations of basic theory and how complexities may be added to the basic theories of axial members, torsion of circular shafts, and symmetric bending of beams.

• Understand the concept and use of discontinuity functions in analysis of structural members subjected to discontinuous loads.

Logic in structural analysis

Displacements

Strains

Stresses

External Forcesand

Moments

Equilibrium

Mat

eria

l Mod

els

Static Equivalency

2

1

3

4

Kinematics

Internal Forcesand

Moments

3-1


Preliminaries

Limitations• The length of the member is significantly greater (approximately 10

times) then the greatest dimension in the cross-section. Approximation across the cross-section are now possible as the region of approxima-tion is small.

• We are away from regions of stress concentration, where displace-ments and stresses can be three-dimensional.

• The variation of external loads or changes in the cross-sectional area is gradual except in regions of stress concentration.

• The external loads are such that the axial, torsion and bending prob-lems can be studied individually.

Convention

x (u)

y (v)

z (w)

ρ θ (φ)

• The displacements u, v, and w will be considered positive in the posi-tive x, y, and z -direction, respectively.

• The rotation φ of the cross-section will be considered positive counter-clockwise with respect to the x-axis.

• The external distributed torque per unit length t(x) is positive counter-clockwise with respect to the x-axis.

• The external distributed force per unit length px(x) and py(x) are con-sidered positive in the positive x and y direction, respectively.

3-2


Deformations

Axial Bending Torsion

Assumption 1 Deformations are not function of time.

Assumptions

2-A: Plane sections remain plane and parallel.

2a-B: Squashing deformation is significantly smaller than deformation due to bending.

2b-B: Plane sections before deformation remain plane after deformation.

2c-B: Plane perpendicular to the beam axis remain nearly perpendicular after deformation

2a-T: Plane sections perpen-dicular to the axis remain plane during deformation.

2b-T: All radials lines rotate by equal angle during deformation on a cross-section.

2c-T: Radials lines remain straight during deforma-tion.

u uo x( )= v v x( )=

u y xddv–=

φ φ x( )=

Original Grid

Deformed Grid

xy

z

Original Grid

Deformed Grid

Original Grid

Deformed Grid

Axial

Torsion

Bending(a)

(c)

(b)

Original Grid

(3.1-A) (3.1a-B)

(3.1b-B)

(3.1-T)

3-3


StrainsAxial Bending Torsion

Assumption 3 The strains are small.

εxx xdduo x( )= εxx y

x2

2

dd v x( )–= γxθ ρ

xddφ x( )=

StressesAxial Bending Torsion

Assumption 4 Material is isotropic.Assumption 5 There are no inelastic strains.Assumption 6 Material is elastic. Assumption 7 Stress and strains are linearly related.

Using Hooke’s law σxx E

xdduo x( )= σxx Ey

x2

2

dd v x( )–= τxθ Gρ

xddφ x( )=


Static equivalency

N σxx AdA∫=

Mz yσxx AdA∫– 0= =

My zσxx AdA∫– 0= =

N σxx AdA∫ 0= =

Mz yσxx AdA∫–=

Vy τxy AdA∫=

T ρτxθ AdA∫=

Sign convention +N

+σxx+τxy

+Vy

+Mz

σxxDistribution

Compressivepositive y face

Outward normal

+T

+τxθ

Internal Forces and Moments

(3.2-A) (3.2-B) (3.2-T)

(3.3-A) (3.3-B) (3.3-T)

(3.4a-A)

(3.4b-A)

(3.4c-A)

(3.4a-B)

(3.4b-B)

(3.4c-B)

(3.4-T)

3-4


FormulasSubstituting stresses into equations of internal forces and moments and

Noting xd

duo , x2

2

dd v , and

xddφ are functions of x only while the integration is with

respect to y and z.


Origin Location yE AdA∫ 0= yE Ad

A∫ 0=

Nxd

duo E AdA∫= Mz

x2

2

dd v Ey2 Ad

A∫= T xd

dφ Gρ2 AdA∫=

Assumption 8 Material is homogenous across the cross-section.Origin is at the cen-troid of the cross-section

y AdA∫ 0= y Ad

A∫ 0=

xdduo N

EA-------=x2

2

dd v Mz

EIzz----------= xd

dφ TGJ-------=

A = Area of cross-sectionEA = Axial Rigidity

Izz = Second area moment of inertiaEIzz = Bending rigidity

J= Polar moment of the area.GJ = Torsional rigidity

Stress formulasSubstituting Equations (3.8-A), (3.8-B), and (3.8-T) into Equations (3.3-A), (3.3-B), and (3.3-T)


σxxNA----= σxx

MzyIzz

----------⎝ ⎠⎛ ⎞–=

See section... for shear stresses in bending.

τxθTρJ------=


Assumption 9 Material is homogenous between x1 and x2. Assumption 10 The structural member is not tapered between x1 and x2.Assumption 11 The external loads do not change with x between x1 and x2.Integrating Equations (3.8-A) and (3.8-T)

u2 u1–N x2 x1–( )

EA--------------------------=See Section 3.2.4 for beam deflection. φ2 φ1–

T x2 x1–( )GJ-------------------------=

Deformation formulas

(3.5-A) (3.5-B)

(3.6-A) (3.6-B) (3.6-T)

(3.7-A) (3.7-B)

(3.8-A) (3.8-B) (3.8-T)

(3.9-A) (3.9-B) (3.9-T)

(3.10-A) (3.10-T)

3-5



dx

N N+dNpx(x) dx

Mz Mz + dMz

Vy Vy +dVydx

py(x)dx T t(x) dx

dx

xddN px x( )–=

xddVy py x( )–=

xddMz V– y=

xddT t x( )–=

Equilibrium Equations

Differential Equations

Substituting Equations (3.8-A), (3.8-B), and (3.8-T) into Equations (3.11-A), (3.11a-B), (3.11b-B), and (3.11-T)

.

xdd EA

xdduo

⎝ ⎠⎛ ⎞ px x( )–=

x2

2

dd EIzz

x2

2

dd v

⎝ ⎠⎜ ⎟⎛ ⎞

py x( )= xdd GJ xd

dφ⎝ ⎠⎛ ⎞ t x( )–=

(3.11-A) (3.11a-B)

(3.11b-B)

(3.11-T)

(3.12-A) (3.12-B) (3.12-T)

3-6


3.1 Draw the shear stress due to torsion on the stress cubes at points A and B shown . Is the shear stress positive or negative τxy?

x

y

A

A

B

BT

x

x

3-7


3.7 Determine the contraction of a column shown in Figure P3.7 due to its own weight. The specific weight is γ 0.28 lb in3

⁄= , the modulus of elasticity is E 3 600 ksi,= , the length is L 120 in= , and the radius varies as R 240 x–= , where, R and x are in inches.

xL

R(x)

Figure P3.7

3-8


3.10 A thin cylindrical tube with an outer diameter of 5 inches is fabricated by butt-welding 1/16 inch thick plate along a spiral seam as shown. A through crack of 0.07 inch was observed in the seam. If the critical stress intensity factor for the material is 22 ksi in , determine the maximum torque T that the tube can transmit.

T

T

35o

E

P =4 kips

P =4 kips

Figure P3.10

3-9


Shear Stress in Thin Symmetric Beams

Relative SlidingNo RelativeSliding

Separate Beams Glued Beams

• Assumption of plane section perpendicular to the axis remain perpen-dicular during bending requires the following limitation.

Maximum bending shear stress must be an order of magnitude less than maximum bending normal stress.

Shear stress direction

(a)

P

H 3

1 G

F

I

5

4

2

DC

z

x

A

E

y

B

Normal stress distribution

(b)

�xx � d�xx

(d)

�xx

5

�yx

(c)

�xx � d�xx�xx

3

�zx

(f)

N5

V5 � �yx(tz) �x

N5 � dN55

�xt z

(e)

N3 � dN3

V3 � �zx(ty) �x

N3 3

�x

ty

1

2

3

4

5

3-10


(a)

P

y

x

s

Free surface

1

2

3456

7

8

9z

Free surface

(b)

N

V � �xs(t)�xt

�x

Free surface

1

23

N � dN

Shear Flow: q τxst=• The units of shear flow ‘q’ are force per unit length.

The shear flow along the center-line of the cross-section is drawn in such a direction as to satisfy the following rules:

• the resultant force in the y-direction is in the same direction as Vy.• the resultant force in the z-direction is zero.• it is symmetric about the y-axis. This requires shear flow will change

direction as one crosses the y-axis on the center-line.

3.15 Assuming a positive shear force Vy, (a) sketch the direction of the shear flow along the center-line on the thin cross-sections shown. (b) At points A, B, C, and D, determine if the stress component is τxy or τxz and if it is positive or negative.

y

z

A

B

C

D

3-11


Bending Shear Stress Formula

(b)

(c)

V � �sx t dx

t

dx1

23

Ns � dNs

Ns � � �xx dAAs

s*

Free surfaceFree surface

4567

8

9dxNs* � dNs*

Ns* � � �xx dAAs*

V* � �s*x t dx

(a)

y

x

s

s

s*

Free surface

1

2

3456

7

8

9 z

Free surface

dxAs*As

Ns dNs+( ) Ns– τsxt xd+ 0= τsxtxd

dNs–=

τsxt –xd

d σxx AdAs

∫ –xd

d MzyIzz

---------–⎝ ⎠⎛ ⎞ Ad

As

∫ xdd Mz

Izz----- y Ad

As

∫= = =

• As is the area between the free surface and the point where shear stress is being evaluated.

Define: Qz y AdAs

∫= τsxt xdd MzQz

Izz-----------=

Assumption 1 The beam is not tapered.

q tτsxQzIzz------⎝ ⎠⎜ ⎟⎛ ⎞

xddMz QzVy

Izz-----------⎝ ⎠⎜ ⎟⎛ ⎞

–= = = τsx τxsVyQzIzzt-----------

⎝ ⎠⎜ ⎟⎛ ⎞

–= =

3-12


Calculation of Qz y Ad

As

∫=

• As is the area between the free surface and the point where shear stress is being evaluated.

• Qz is zero at the top surface as the enclosed area As is zero. • Qz is zero at the bottom surface (As=A) by definition of centroid.

y

z

ys

As

Line along which Shear stress isbeing found.

Qz Asys=

Neutral Axis

Centroid of As

y2 Qz A2y2=Centroid of A2

A2

• Qz is maximum at the neutral axis.Bending shear stress at a section is maximum at the neutral axis.

.

Table 3.2. Stresses and Strains

AxialSymmetric Bending

TorsionAbout z-axis About y-axis

StrainsStresses Strains Stresses Stresses Stresses Strains

σxxNA----= εxx

σxxE--------= σxx

MzyIzz

----------⎝ ⎠⎛ ⎞–= σxx

MyzIyy

----------⎝ ⎠⎛ ⎞–= εxx

σxxE--------= σxx 0= εxx 0=

σyy 0= εyyνσxx

E-----------⎝ ⎠⎛ ⎞–= σyy 0= σyy 0= εyy

νσxxE-----------⎝ ⎠

⎛ ⎞–= σyy 0= εyy 0=

σzz 0= εzzνσxx

E-----------⎝ ⎠⎛ ⎞–= σzz 0= σzz 0= εzz

νσxxE-----------⎝ ⎠

⎛ ⎞–= σzz 0= εzz =

τxy 0=

τxz 0=

τyz 0=

γxy 0=

γxz 0=

γyz 0=

τxsVyQzIzzt

-------------⎝ ⎠⎛ ⎞–=

τyz 0=

τxsVzQyIyyt-------------⎝ ⎠

⎛ ⎞–=

τyz 0=

γxsτxsG-------=

γyz 0=

τxθTρJ-------=

τyz 0=

γxθτxθG-------=

γyz 0=

3-13


3.18 A positive shear force Vy= 10 kN acts on the thin cross-sec-tion shown in Figure P3.18 (not drawn to scale). The cross-section has a uniform thickness of 10 mm. Determine the equation of shear flow along the center lines and sketch it.

y

z 100 mm

100 mm

25 mm 25 mm

Figure P3.18

3-14


3.21 A highway sign uses a 16 inch hollow pipe as a vertical post and 12 inch hollow pipe for horizontal arms. The pipes are one inch thick. Assume a uniform wind pressure of p acts on the sign boards and the pipes. Note the pressure on the pipes acts on the projected area of Ld, where L is the length of pipe and d is the diameter of the pipe. The yield stress pipes is 40 ksi. For a factor of safety of 2 determine the maximum wind pressure. Use maximum octahedral shear stress theory.

3 ft

17 ft

6 ft 6 ft1 ft 8 ft

BA

5ft

3-15


3.23 A beam resting on an elastic foundation has a distributed spring force that depends upon the deflections a point acting as shown in Figure P3.23. Show the differential equation governing the deflection of the beam is as given below:

x2

2

dd EIzz

x2

2

dd v

⎝ ⎠⎜ ⎟⎛ ⎞

kv+ p=

Mz Mz + ΔMz

Vy Vy +ΔVy

dx(kdx)v

p

Figure P3.23

3-16


Discontinuity Functions

a

∞

x

x a–⟨ ⟩ 1–

ε2--- ε

2---

ap

ε2---

ε2--- x

P = p ε

a

P pε( )ε 0→lim

p ∞→lim= or x a–⟨ ⟩ 1– 0 x a≠

∞ x a→⎩ ⎭⎨ ⎬⎧ ⎫= x a–⟨ ⟩ 1– xd

a ε–( )

a ε+( )

∫ 1=

Delta Function: x a–⟨ ⟩ 1–

x a–⟨ ⟩ 1– xd∞–

x

∫ x a–⟨ ⟩ 1– xd∞–

a ε–( )

∫ x a–⟨ ⟩ 1– xda ε–( )

a ε+( )

∫ x a–⟨ ⟩ 1– xda ε+( )

x

∫+ + 1= =

x a–⟨ ⟩0 x a–⟨ ⟩ 1– xd∞–

x

∫0 x a<1 x a>⎩ ⎭

⎨ ⎬⎧ ⎫

= =

x

x a–⟨ ⟩0

x

x a–⟨ ⟩1

x

x a–⟨ ⟩2

a a a

x a–⟨ ⟩n 0 x a≤

x a–( )n x a>⎩ ⎭⎨ ⎬⎧ ⎫

=

x a–⟨ ⟩n xd∞–

x

∫x a–⟨ ⟩n 1+

n 1+( )--------------------------= n 0≥

3-17


Doublet Function: x a–⟨ ⟩ 2– 0 x a≠∞ x a→⎩ ⎭

⎨ ⎬⎧ ⎫= x a–⟨ ⟩ 2– xd

∞–

x

∫ x a–⟨ ⟩ 1–=

xdd x a–⟨ ⟩ 1–

x a–⟨ ⟩ 2–= xdd x a–⟨ ⟩0

x a–⟨ ⟩ 1–=

xdd x a–⟨ ⟩n

n x a–⟨ ⟩n 1–= n 1≥

• The function delta function x a–⟨ ⟩ 1– and the doublet function x a–⟨ ⟩ 2– become infinite at x = a. Alternatively stated these functions are singu-lar at x = a. and are referred to as singularity functions.

• The entire class of functions x a–⟨ ⟩n for positive and negative ‘n’ are called the discontinuity functions.

3-18


Axial Displacement

xddu N

EA-------= xddN px x( )–=

Differential Equation: xdd EA xd

du⎝ ⎠⎛ ⎞ px x( )–=

Boundary Conditions u or N

ax

NFN F x a–⟨ ⟩

0–=

px F x a–⟨ ⟩1–=

Template equations

Example 3.7

2 kips

4 kips4 kips

2 kips25 in

60 in

20 in

B

D

C

A

x

px 10 4000 x 25–⟨ ⟩ 1– 8000 x 85–⟨ ⟩ 1–+ +( ) lb in⁄= (E1)

• Differential equation

xdd EA xd

du⎝ ⎠⎛ ⎞ 10 4000 x 25–⟨ ⟩ 1– 8000 x 85–⟨ ⟩ 1–+ +[ ]–= (E2)

• Boundary Conditionsu 0( ) 0= (E3)

u 105( ) 0= (E4)

3-19


Torsional Rotation

xddφ T

GJ-------= xddT t x( )–=

Differential Equation: xdd GJ xd

dφ⎝ ⎠⎛ ⎞ t x( )–=

Boundary Conditions φ or T

ax

TT

T T x a–⟨ ⟩0–=

t T x a–⟨ ⟩1–=

Template equations

3-20


3.36 The external torque on a drill bit varies as a quadratic function to a maximum intensity of q in-lb/in as shown Figure P3.36. If the drill bit diameter is d, its length L, and modulus of rigidity G, determine (a) the maximum shear stress on the drill bit. (b) the relative rotation of the end of the drill bit with respect to the chuck. L

q x2

L2------⎝ ⎠⎜ ⎟⎛ ⎞

in lb ⁄ in–

x

Figure P3.36

3-21


3.38 An aluminum alloy (G = 28 GPa) hollow shaft has a critical stress intensity factor of 22 ksi in . The shaft has a thickness of 1/4 in and an outer diameter of 2 in and is loaded as shown in Figure P3.38. What is the critical crack length at which the shaft be taken out of service?

BA2 ft. 3.5 ft.

C

T = 165 in-kips

Figure P3.38

3-22


Beam Deflection

2nd order differential equation:x2

2

dd v Mz

EIzz----------=

xddVy py x( )–=

xddMz V– y=

4th order differential Equation: x2

2

dd EIzz

x2

2

dd v

⎝ ⎠⎜ ⎟⎛ ⎞

py x( )=

Boundary Conditions• Group 1 v or Vy

and

• Group 2 xddv or Mz

ax

Mz

Vy

Mz0

M–x a<x a>⎩ ⎭

⎨ ⎬⎧ ⎫

=

Mz M x a–⟨ ⟩0–=

py M x a–⟨ ⟩ 2––=

a

P

x

Mz

Vy

Mz0

P x a–( )–x a<x a>⎩ ⎭

⎨ ⎬⎧ ⎫

=

Mz P x a–⟨ ⟩1–=

py P x a–⟨ ⟩ 1––=

ax

Mz

Vy

Mz

0

w x a–( )2

2-----------------------–

x a<x a>

⎩ ⎭⎪ ⎪⎨ ⎬⎪ ⎪⎧ ⎫

=

Mz w x a–⟨ ⟩2

2-------------------–=

py w x a–⟨ ⟩0–=

M w

Template equations Template equations Template equations

x

y

x

y

x

y

3-23


3.43 The displacement of the beam in the y-direction, in section AB of the beam shown in Figure P3.43 is given by

v1 5 x3 20x2–( ) 10 6–( ) in= and in section BC is given

v2 5 x3 800x– 8000+( ) 10 6–( ) in= . If the bending rigidity (EI) is

135 (106) lbs-in2, determine the moment MB and the reaction force at B.

Some complex Loadingy

20 in 20 in 40in

MB

BxA C D

Figure P3.43

3-24


3.44 In terms of w, L, E, and I, determine the deflection and slope at x = L of the beam shown in Figure P3.44.

L

x

yw x2

L2------⎝ ⎠⎜ ⎟⎛ ⎞

A B

wL2

wL Figure P3.44

3-25


3.53 Determine the deflection of the beam at point C in terms of E, I, w, and L for the beam shown in Figure P3.53.

Figure P3.53

3-26


Composite Structural Members

• Assumption 8 on material homogenity across the cross-section is not valid.

The learning objective of this chapter is:Understand the incorporation and implications of material in-homoge-neity across the cross-section in the theories for axial members, circu-lar shafts in torsion, and symmetric bending of beams.

Composite Axial Members

1

x

y

z

x

y

z

σxx E xddu x( )=


N σxx AdA∫= Mz yσxx Ad

A∫– 0= = or

N xddu x( ) E Ad

A∫=

Location of origin: yE AdA∫ 0=

4-1


Formulas for composite axial rods

N xddu E Ad

A∫ xd

du E1 AdA1

∫ E2 AdA2

∫ ⋅ ⋅ En AdAn

∫+ + + += =

N xddu EjAj

j 1=

n

∑=

σxx( )iNEi

EjAj

j 1=

n

∑

--------------------= u2 u1–N x2 x1–( )

EjAj

j 1=

n

∑

-------------------------=

Location of axial force application (origin)

ηc

ηiEiAi

i 1=

n

∑

EiAi

i 1=

n

∑

-------------------------=

4.3 A wooden rod (EW = 2000 ksi) and steel strip (Es = 30,000 ksi) are fastened securely to each other and to the rigid plates as shown in Figure P4.3. Determine (a) the location h of the line along which the external forces must act to produce no bending. (b) the maximum axial stress in steel and wood.

0.2

30 in 60 in

20kips

h

30 in

Figure P4.3

4-2


4.8 A for use in a building is modeled as shown in Figure P4.8. The column is constructed by reinforcing concrete with nine steel circu-lar bars of diameter 1 inch. The modulus of elasticity for concrete and iron are Ec= 4,500 ksi and Ei = 25,000 ksi. Determine the maximum axial stress in concrete and steel.

12 in

9 ft150 kips150 k

9 ft

9 ft

200 kips200 k

Figure P4.8

4-3


Composite Shafts

τxθ Gρ xddφ x( )=


T ρτxθ AdA∫= or T xd

dφ Gρ2 AdA∫=

Outward normal

+T+τxθ

Formulas for composite shafts

T xddφ G1ρ

2 AdA1

∫ G2ρ2 Ad

A2

∫ ⋅ ⋅ Gnρ2 Ad

An

∫+ + + +=

T xddφ GjJj

j 1=

n

∑=

τxθ( )iGiρT

GjJj

j 1=

n

∑

-------------------------= φ2 φ1–T x2 x1–( )

GjJj

j 1=

n

∑

-------------------------=

4-4


4.12 A solid steel (G = 80 GPa) shaft 3 m long is securely fastened to a hollow bronze (G = 40 GPa) shaft that is 2 m long as shown Figure P4.12. Determine (a) the magnitude of maximum shear stress in the shaft. (b) the rotation of section at 1 m from the left wall.

TextTT

2 m1 m

=10kN-m

Figure P4.12

4-5


Composite BeamsEn

y

Ei

E2

E1

�c

σxx Eyx2

2

dd v x( )–=


N σxx AdA∫ 0= = Mz yσxx Ad

A∫–= or

Mzx2

2

dd v Ey2 Ad

A∫=

Location of origin: yE AdA∫ 0=

Formulas for composite beams

Mz x2

2

dd v E1y2 Ad

A1

∫ E2y2 AdA2

∫ ⋅ ⋅ Eny2 AdAn

∫+ + + +=

Mzx2

2

dd v Ej Izz( )j

j 1=

n

∑= σxx( )iEiyMz

Ej Izz( )jj 1=

n

∑

---------------------------–=

Location of neutral axis (origin) : ηc

ηiEiAi

i 1=

n

∑

EiAi

i 1=

n

∑

-------------------------=

Bending shear stress in composite beamsEquilibrium equation

4-6


τsxt xddNs– – xd

d σxx AdAs

∫= =

τsxt – xdd E– yMz

Ej Izz( )jj 1=

n

∑

--------------------------- AdAs

∫ xdd Mz

Ej Izz( )jj 1=

n

∑

--------------------------- Ey AdAs

∫ xdd MzQcomp

Ej Izz( )jj 1=

n

∑

---------------------------= = =

Qcomp Ey AdAs

∫=

τsx τxsQcompVy

Ej Izz( )jj 1=

n

∑ t

----------------------------------–= =

Qcomp Ej Qz( )jj 1=

ns

∑=

4-7


4.16 The cross-section of a composite beam with a coordinate sys-tem that has an origin at C is shown . The normal strain at point A due to bending about the z-axis is εxx 200– μ= , and the modulus of elasticity of the materials are E1 8000 ksi= and E2 2000 ksi= (a) Plot the stress distri-bution across the cross-section. (b) Determine the maximum bending normal stress in each material. (c) Determine the equivalent internal bending moment Mz.

z

y

1 in

1 in

4 in

4 in

4-8


4.20 A wooden rod (EW = 2000 ksi) and steel strip (Es = 30,000 ksi) are fastened securely to rigid plates as shown. Deter-mine (a) the maximum intensity of the load w, if the allowable bending normal stresses in steel and wood are 20 ksi, and 4 ksi.respectively. (b) the magnitude of the maximum shear stress in the beam corresponding to the load in part (a).

Steel

Wood

0.25 in

2 in

4 in

6 ft6 ft

y w

x

4-9


Inelastic Structural Behavior

The following assumptions will be dropped.

Assumption 5: There are no inelastic strains.Assumption 6: Material is elastic. Assumption 7: Stress and strains are linearly related.

The learning objectives of this chapter are:

• Understand the incorporation of thermal and initial strains in the theory and analysis of axial members.

• Understand the analysis techniques for incorporating elastic-perfectly plas-tic material behavior in axial members, circular shafts, and symmetric beams.

• Understand the incorporation of non-linear material models into the basic simplified theories on structural members.

5-1


Effects of Temperature

Nor

mal

str

ess

�

Normal strain �

Test performed at T0 � �T

Test performed at T0

�

� �T

O O1

E�

ε σE--- α ΔT+=

εxx σxx ν σyy σzz+( )–[ ] E⁄ α ΔT+=

εyy σyy ν σzz σxx+( )–[ ] E⁄ α ΔT+=

εzz σzz ν σxx σyy+( )–[ ] E⁄ α ΔT+=

γxy τxy G⁄=

γyz τyz G⁄=

γzx τzx G⁄=

Mechanical Strain Thermal Strain

5-2


Temperature effects in axial members

Kinematics: εxx xddu x( )=

Assume an initial strain of εo

Stresses: εxxσxxE

-------- εo+xd

du= = or σxx Exd

du εo–⎝ ⎠⎛ ⎞=

Internal Axial Force: N σxx AdA∫=

Assume material homogenity across the cross-section and εo is uniform.

N E xddu Eεo–⎝ ⎠

⎛ ⎞ AdA∫ xd

du E AdA∫ Eεo Ad

A∫– xd

duEA EAεo–= = =

xddu N

EA------- εo+=

σxxNA----=

Assumptions 9 through 11 are assumed valid. N, E and A are constant between x1 and x2 Assume εo also does not change with x

u2 u1–N x2 x1–( )

EA--------------------------= εo x2 x1–( )+ or δ NLEA-------- εoL+=

• δ is the deformation of the bar in the undeformed direction. • If N is a tensile force then δ is elongation.• If N is a compressive force then δ is contraction. • Deformation of a member shown in the drawing of approximate deformed

geometry must be consistent with the internal force in the member that is shown on the free body diagram.

• The sign of εoL due to temperature changes must be consistent with the force N shown on the free body diagram.

5-3


• Pre-strains (stresses) can be analyzed by using εo as negative to the actual initial strain.

5-4


5.5 The stress at a point, material properties, and change in tem-perature are as given below. Calculate εxx, εyy, γxy, εzz, and σzz (a) assum-ing plane stress, and (b) assuming plane strain.

σxx 300 MPa C( )= σyy 300 MPa T( )= τxy 150 MPa=

G 15 GPa= ν 0.2= α 26.0μ Co⁄= ΔT 75oC=

5-5


5.16 The rigid bar in Figure P5.16 is horizontal when the unit is put together by finger-tightening the nut. The pitch of the threads is 0.125 inch. Develop a table in steps of quarter turn of the nut that can be used for prescribing the pre-tension in bar B. Maximum number of quar-ter turns is limited by the yield stress.

Bar A Bar B

Modulus of Elasticity 10,000 ksi 30,000 ksi

Yield Stress 24 ksi 30 ksi

Area of cross-section 0.5 in2 0.75 in2

A B

Rigid

5 in 15 in

50 in

Figure P5.16

5-6


5.22 Determine the axial stress in bar A of problem 5.16 assuming that the nut is turned 1 full turn and the temperature of bar A is decreased by 50oF. The coefficients of thermal expansion for bar A is αst = 22.5 μ /

oF.

5-7


Non-linear material models• Elastic-perfectly plastic in which the non-linearity is approximated by a

constant. • Linear strain hardening model in which the non-linearity is approximated

by a linear function.• Power law model in which the non-linearity is approximated by one term

non-linear function.

Elastic-perfectly plastic�

�

�yield

�yield

�yield

�yield

� � ��yield

� � �yield

� �

E�

�

�

�yield

�yield

�yield

�yield

� � ��yield

� � �yield

� �

G�

σ

σyield

Eεσ– yield⎩

⎪⎨⎪⎧

=

ε εyield≥

εyield– ε εyield≤ ≤

ε ε– yield≤

τ

τyield

Gγτ– yield⎩

⎪⎨⎪⎧

=

γ γyield≥

γyield– γ γyield≤ ≤

γ γ– yield≤

• The set of points forming the boundary between the elastic and plastic region on a body, is called the elastic-plastic boundary.

1. On the elastic-plastic boundary the strain must be equal to the yield strain, and stress equal to yield stress.

2. Deformations and strains are continuous at all points including points at the elastic plastic boundary.

5-8


Linear strain hardening material model� � �yield � E2(� � �yield)

� � ��yield � E2(� � �yield) � � ��yield � G2(� � �yield)

�

�

�yield

�yield

�yield

�yield

� �

E1�

�

�

�yield

�yield

�yield

�yield

� �

G1�

� � �yield � G2(� � �yield)

σ

σyield E2 ε εyield–( )+

E1ε

σ– yield E2 ε εyield+( )+⎩⎪⎨⎪⎧

=

ε εyield≥


ε εyield≥

Power Law

�

�

Strain hardening0 � n � 1

Strain softeningn � 1

n � 1

� � �E(��)n

� � E�n

0 � n � 1

�

�

Strain hardening0 � n � 1

Strain softeningn � 1

n � 1

� � �G(��)n

� � G�n

0 � n � 1

σEεn

E ε–( )n–⎩⎨⎧

=ε 0≥ε 0<

5-9


Elastic-perfectly plastic axial members• The plot of the applied force vs. the deflection at that point in the direction

of the applied force is called the load deflection curve.• The load at which the structure exhibits unbounded deformation is called

the collapse load.�

�

�yield

�yield

�yield

�yield

� � ��yield

� � �yield�

� E

�

�

�

�yield

�yield

�yield

�yield

� � ��yield

� � �yield

� �

G�

Elastic: δ NLEA--------=

5-10


5.25 A force F is applied to the roller that slides inside a slot as shown in Figure P5.25. Both bars have an area of cross-section of A = 100 mm2, modulus of elasticity E = 200 GPa, and a yield stress of 250 MPa. Bar AP and BP have lengths of LAP= 200 mm and LBP= 250 mm respectively. Draw the load deflection curve and determine the collapse load.

F

110°

P

A

B

Figure P5.25

5-11


5.29 Three steel (E = 200 GPa, σyield = 200 MPa) bars shown in Figure P5.29 have lengths of LA=4 m, LB=3 m and LC= 2 m respectively.

All bars have the same cross-sectional area of 500 mm 2. Draw the load deflection curve for the structure and determine the collapse load.

A

50�

5 m

4 m 3 m

0.0009 mD E

C

B

F

ORigid

5

Figure P5.29

5-12


Elastic-perfectly plastic circular shafts�

�

�yield��

�yield��

�yield��

�yield��

� � ��yield��

� � �yield��

� �

G�

Unl

oadi

ng

A B

OC

• Before yield stress the material stress-strain relationship is represented by Hooke’s Law and after yield stress the stress is assumed to be constant.

• To determine the strain (deformation) in the horizontal portion AB of the curve we have to use the requirement that deformation must be continuous.

• Unloading (elastic recovery) from a point in the plastic region is along line BC which is parallel to the linear portion of the stress strain curve OA.

Kinematic: γxθ ρ xddφ x( )=

Elastic: τxθTρJ------= φ2 φ1–

T x2 x1–( )GJ-------------------------=

At the elastic-plastic boundary: γyield ρy xddφ=

If ρy is not a function of x: φ2 φ1–γyield

ρy------------ x2 x1–( )=

Stress distribution: τ

τyieldρρy

---------------- ρ ρy≤

τyield ρ ρy≥⎩⎪⎨⎪⎧

=

Internal equivalent torque: T ρτxθ AdA∫=

5-13


5.43 The shaft shown in Figure P5.43 made from elastic - perfectly plastic material has a shear yield stress of 200 MPa and a shear modulus of G = 80 GPa. The plastic zone in section AB is 25 mm deep. Deter-mine: (a) the torque Text (b) the rotation of section at B. (c) the residual stress in AB when the external torque Text is removed.

100 mm

Figure P5.43

5-14


Elastic-perfectly plastic beamsB

OC

A

Unloa

ding

� � E

�

�

�

�yield

�yield

�yield

�yield� � ��yield

� � �yield

• Before yield stress, the material stress-strain relationship is represented by Hooke’s Law and after yield stress, the stress is assumed to be constant.

• To determine the strain (deformation) in the horizontal portion AB of the curve we have to use the requirement that deformation must be continuous.

• Unloading (elastic recovery) from a point in the plastic region is along line BC which is parallel to the linear portion of the stress-strain curve OA.

Stress distribution and changes in neutral axis location

y y

�a

a

�yield

�yield

A

B

z

A

B

C

y

z

y

�a

a

�yield

�A

(a) (b)

Symmetric Stress Distribution Asymmetric Stress Distribution

Mz yσxxdAA∫–=

5-15


• The moment at which the maximum bending normal stress just reaches the yield stress is called the elastic moment and will be designated by Me.

• The internal moment for which the entire cross-section becomes fully plas-tic is called the plastic moment and will be designated by Mp

• The ratio of the plastic moment to the elastic moment is called the shape factor for the cross-section and will be designated by f.

fMpMe------=

5-16


5.55 An elastic - perfectly plastic material has a yield stress of σyield = 40 ksi. Point A in Figure P5.56 is at yield stress due to bending of the beam. (a) Determine the location of the neutral axis assuming it is in the web.(b) The applied moment that produced the state of stress.

5.56 For the cross-section shown in Figure P5.56 determine the shape factor

0.5 in

0.5 in

0.5 in

4 in

8 in

7 in

A

Figure P5.56

5-17


Non-linear models in structural membersKinematics


εxx xddu x( )= εxx y

x2

2

dd v x( )–= γxθ ρ xd

dφ x( )=

σEεn

E ε–( )n–⎩⎨⎧

=ε 0≥ε 0<

τGγn

G γ–( )n–⎩⎨⎧

=γ 0≥γ 0<

Static equivalency (Internal Forces and Moments)

Axial Bending TorsionN σxx Ad

A∫= N σxx Ad

A∫ 0= =

Mz yσxx AdA∫–=

Vy τxy AdA∫=

T ρτxθ AdA∫=

5-18


5.59 A circular solid shaft of radius R is made from a non-linear material that has a shear stress-shear strain relationship given by τ = Κ γ0.4. Show

τmax 0.5411 TR3------= φ2 φ1– 0.2154 x2 x1–( ) T

KR3.4-------------⎝ ⎠⎛ ⎞ 2.5

=

where τmax is the maximum shear stress at a section, T is the internal torque at the section, φ2 , and φ1 are the rotation of section at x1 and x2.

5-19


5.62 The stress-strain curve in tension, for a material is given by

σ Kε0.5= . For a rectangular cross-section show that the bending normal

stress is given by:

σxx

5 2–bh2------------- y

h---⎝ ⎠⎛ ⎞ 0.5

Mz

5 2bh2---------- y

h---⎝ ⎠⎛ ⎞ 0.5

Mz⎩⎪⎪⎨⎪⎪⎧

=y 0>

y 0<h

b

y

z

5-20


Unsymmetric Bending of Beams

• Drop the limitation that the beam has a plane of symmetry and the loading is in the plane of symmetry.

• Assume loading is such that there is no twisting of the cross-section.The learning objectives of this chapter are:

• Understand the theory, its limitations, and its application in design and analysis of unsymmetric bending of beam.

• Understand the concept of shear center and how to determine its loca-tion.

1-1


Theory

x

y

zA

BCD

Pz

(Mz)ext

Py

pz(x)

(My)extpy(x)

Theory objective is:• Relate the internal shear forces Vy, Vz and internal moment My, Mz to

displacements v and w and obtain the stresses in unsymmetric bend-ing.

Deformation BehaviorAssumption 1 The loads are such that there is no axial or torsional defor-

mation.

χ

ψ

uo

A

B

C

D

A1

B1

C1

D1

OO1

No twist implies: γyz z∂∂v

y∂∂w+ 0= =

v x y z, ,( ) v x y,( )= w x y z, ,( ) w x z,( )=

Assumption 2 Squashing action is significantly smaller then bending action.

εyy y∂∂v 0≈= εzz z∂

∂w 0≈=

v v x( )= w w x( )=

1-2


Assumption 3 Plane sections before deformation remain plane after deformation.

u uo ψ– y χ– z=

ψ ψtan≈xd

dv=

ψ

y

v(x)

y

x

y

y ψsin yψ–≈–

χ χtan≈xd

dw=

χ

z

w(x)

z

x

z

z χsin zχ–≈–

Assumption 4 Plane perpendicular to the axis remain nearly perpendicu-lar after deformation.

u y xddv– z xd

dw–=

Strain DistributionAssumption 5 Strains are small.

εxx xddu y

x2

2

dd v– z

x2

2

dd w–= =

Material ModelAssumption 6 Material is isotropicAssumption 7 Material is elastic.Assumption 8 Stress and strains are linearly relatedAssumption 9 There are no inelastic strain.Hooke’s Law: σxx Eεxx=

σxx Eyx2

2

dd v– Ez

x2

2

dd w–=

1-3


Internal forces and moments

x

y

z

y

z

x

y

zMz

dVy = τxy dA

dN = σxx dA

Vy

dVz = τxz dAMy

Vz

O O

Shear center

ey

ez

(a) (b)

N σxx AdA∫ 0= =

Mz yσxx AdA∫–= My zσxx Ad

A∫–=

Vy τxy AdA∫= Vz τxz Ad

A∫=

T y ey–( )τxz z ez–( )τxy–[ ] AdA∫ 0= =

• The maximum normal stress σxx in the beam should be nearly an order of magnitude (factor of 10) greater than the maximum shear stress τxy and τxz.

Sign Convention

+Mz

+My

xz

+τxy +Vy

xy

+τxz +Vz

σxx

σxx

1-4


Bending Formulas

Substituting σxx Eyx2

2

dd v– Ez

x2

2

dd w–= into internal moment expression.

Mzx2

2

dd v Ey2 Ad

A∫ x2

2

dd w Eyz Ad

A∫+= My

x2

2

dd v Eyz Ad

A∫ x2

2

dd w Ez2 Ad

A∫+=

Assumption 10 Material is homogenous across the cross-section.

Mz EIzzx2

2

dd v EIyz

x2

2

dd w+= My EIyz

x2

2

dd v EIyy

x2

2

dd w+=

Area moment of inertia

Izz y2 AdA∫= Izz y2 Ad

A∫= Iyz yz Ad

A∫=

Moment Curvature Relationship

x2

2

dd v 1

E---IyyMz IyzMy–

IyyIzz Iyz2–

----------------------------------⎝ ⎠⎜ ⎟⎛ ⎞

= x2

2

dd w 1

E---IzzMy IyzMz–

IyyIzz Iyz2–

----------------------------------⎝ ⎠⎜ ⎟⎛ ⎞

=

Stress Formula

σxx

IyyMz IyzMy–

IyyIzz Iyz2–

----------------------------------⎝ ⎠⎜ ⎟⎛ ⎞

y–IzzMy IyzMz–

IyyIzz Iyz2–

----------------------------------⎝ ⎠⎜ ⎟⎛ ⎞

z–=

Location of origin

Centroid: y AdA∫ 0= z Ad

A∫ 0=

• The origin of the coordinate system must be the centroid of a homoge-nous cross-section

• Normal stress σxx in bending varies linearly with y and z on a homog-enous cross-section.

1-5


Neutral Axis (σxx = 0)

N.A. equation: y βtan( )z= βtanIzz Iyz Mz My⁄( )–

Iyz Iyy Mz My⁄( )–-------------------------------------------=

• The orientation of the neutral axis depends upon the shape of cross-section as well as the external loading.

• Bending normal stress σxx is maximum at the point which is the far-thest from the neutral axis.

• The displacement of the beam is always perpendicular to the neutral axis.

Equilibrium equations

My My + dVz

Vz+dVzdx

Mz Mz + dMzVy

Vy +dVydx

py dx

o o

pz dx.

xddVy py–= xd

dVz pz–=

xddMz V– y= xd

dMy V– y=

1-6


Area Moment of InertiasParallel axis theorem

y

z

cd dz

dy

yc

zc

Iyy IycycAdz

2+=

Izz IzczcAdy

2+=

Iyz IyczcAdydz+=

• Iyy and Izz are always positive and minimum about the axis passing through the centroid of the body.

• Iyz can be positive or negative. • If either y or z axis is an axis of symmetry then Iyz will be zero.Coordinate TransformationDefinition 1 The coordinate system in which the cross moment of iner-

tia is zero is called the principal coordinate system.Definition 2 The moment of inertias in the principal coordinate system

are called principal moment of inertias.

yn

t

z

yθ

z

(y, z)(n, t)

P y cosθ

y sinθ

θ

z cosθz sinθ

n y θ z θsin+cos=t y θsin– z θcos+=

Inn t2 AdA∫ Iyy θ2cos Izz θ2sin 2Iyz– θ θsincos+= =

Itt n2 AdA∫ Iyy θ2sin Izz θ2cos 2Iyz θ θsincos+ += =

Int nt AdA∫ Iyy Izz–( ) θ θsincos Iyz θ2cos θ2sin–( )+= =

2θptan2Iyz–

Iyy Izz–( )--------------------= I1 2,

Iyy Izz+( )2-----------------------

Iyy Izz–2------------------⎝ ⎠

⎛ ⎞2

Iyz2+±=

• Area moment of intertias are second order tensors.

1-7


6.5 The internal bending moments on the cross-section shown in Figure P6.5 were determined to be My 10– in kips–= and Mz 12– in kips–= . Determine (a) the orientation of the neutral axis. (b) the maximum bending normal stress.

3 in1 in

6 in

1 in

1 in

1.5 in

y

z

1.72 in

0.25 in

C

Figure P6.5

1-8


6.12 A cantilever beam is loaded such that there is no twist. The distributed load acts in the y-z plane at an angle of 240 from the x-y plane as shown in Figure P6.12. On a section at x = 60 in, determine: (a) the orientation of the neutral axis. (b) the bending normal stress at points A and B.

y

z

24op = 50 lb/in

x

96 in24 in

3 in1 in

6 in

1 in

1 in

1.5 in

y

z

1.72 in

0.25 in

A

B

Figure P6.12

6.13 The modulus of elasticity for the beam in problem 6.12 is E = 30,000ksi. Determine the deflection of the beam at x = 60 inch and show that it is perpendicular to the neutral axis.

1-9


Shear stress in thin open sections

A differential element of a thin open section.

ns

x

zy

Free Surface

Free Surface

Free Edge

t

Free Edge

dxA

BC

D

NS

NS+dNS τsx t dx

dx

tA

BC

D

NS dNS+( ) NS– τsxt xd+ 0= or

Equilbrium Equations: τsxt – xddNs=

Axial Force: Ns σxx AdAs

∫=

τsxt xdd σxx Ad

As

∫–=

Definition 3 The direction of the s-coordinate is from the free surface towards the point where shear stress is being calculated.

Definition 4 The area As is the area between free edge and the point at which the shear stress is being evaluated.

τsxt xdd IyyMz IyzMy–

IyyIzz Iyz2–

----------------------------------⎝ ⎠⎜ ⎟⎛ ⎞

y–IzzMy IyzMz–

IyyIzz Iyz2–

----------------------------------⎝ ⎠⎜ ⎟⎛ ⎞

z– AdAs

∫–=

τsxt xdd IyyMz IyzMy–

IyyIzz Iyz2–

----------------------------------⎝ ⎠⎜ ⎟⎛ ⎞

y AdAs

∫IzzMy IyzMz–

IyyIzz Iyz2–

----------------------------------⎝ ⎠⎜ ⎟⎛ ⎞

z AdAs

∫+=

We define the first moment of the area As as:

1-10


Qz y AdAs

∫= Qy z AdAs

∫=

Assumption 11 The beam is not tapered.

τsxtIyy xd

dMz Iyz xddMy–

IyyIzz Iyz2–

----------------------------------------

⎝ ⎠⎜ ⎟⎜ ⎟⎜ ⎟⎜ ⎟⎛ ⎞

Qz

Izz xddMy Iyz xd

dMz–

IyyIzz Iyz2–

----------------------------------------

⎝ ⎠⎜ ⎟⎜ ⎟⎜ ⎟⎜ ⎟⎛ ⎞

Qy+=

q τsxtIyyQz IyzQy–

IyyIzz Iyz2–

----------------------------------⎝ ⎠⎜ ⎟⎛ ⎞

– Vy

IzzQy IyzQz–

IyyIzz Iyz2–

----------------------------------⎝ ⎠⎜ ⎟⎛ ⎞

– Vz= =

Shear centerFrom statics we know that any distributed force can be replaced by a force and a moment at any point, or, by a single force (and no moment) at a specific point. The specific point at which the shear stress (shear flow) can be represented by just shear forces Vy and Vz (components of a single force) and no internal torque is called the shear center. .

C

y

z

Shear Center

O

C

y

z

O

ey

ez

VyVz Shear Center

τxs

Definition 5 Shear center is a point in space at which the shear stress due to bending can be replaced by statically equivalent internal shear forces and no internal torque.

orShear center is a point in space such that if the line of action of external forces pass through the point then the cross-section will not twist.

1-11


6.24 A thin cross-section of uniform thickness t is shown in Figure P6.24. If shear stresses were to be found at points A and B what values of Qy and Qz are needed for the calculation. Assume t a« and gap at D is of negligible thickness. Report the values of Qy and Qz in terms of t and a.

y

z

C

a

a

45o

45o

A

B D

Figure P6.24

1-12


6.25 Shear forces on the cross-section shown in Figure P6.25 were calculated as Vy 10 kips= and Vz 5 kips–= . The cross section has a uniform thickness of 1/8 in. Determine the bending shear stresses at points A and B and report your answers as τxy and τxz

y

z

2 in 2 in

1 in

1 in

1 in

1 in

1 / 8 in

C

A

B1 in

1 in

2 / 7 in

.

Figure P6.25

1-13


6.27 A cantilever beam is loaded such that there is no twist. The cross-section has a uniform thickness of 0.5 inch. Calculations show that Iyy 80.25 in4= and Izz 166.00 in4= . Determine the bending normal stress and bending shear stress at point A.

y

z

Figure P6.27

A

2.75 in

8 in

8 inc

2.5 in

y

z

Ax

2 ft.5 ft.

8 kips.

2 kips

1-14


6.33 The cross-sections shown in Figure P6.33 has a uniform thickness t. Assuming t a« determine the location of shear center with respect to point A

Figure P6.33

1-15


6.36 The cross-section shown in Figure P6.36 has a uniform thick-ness t. Assuming t a« determine the location of shear centers with respect to point A.

a

1.5 a

1.5 a

A

Figure P6.36

1-16


6.38 The cross-section shown in Figure P6.38 has a uniform thick-ness t and boundaries made from circular arcs. Assuming t a« determine the location of shear centers with respect to point A in terms of radius a and angle α.

A

a

αα Figure P6.38

1-17


Shear stresses in thin closed sections

Cz

Cz

s

qo

s

yy(a) Thin closed section. (b) An imaginary cut in closed section.

AB

A

B

qc qo q+=qc is the shear flow in the closed section at any point, q is the shear flow of the open section, and qo is the unknown shear flow at the starting point that has to be determined.

Shear strain can be written as:

γxs s∂∂u

x∂∂vs+

τxsG

-------= =

u and vs are displacement in the x and s direction, respectively, and G is the shear modulus of elasticity.

s∂∂u sd

sA

sB

∫τxsG

-------x∂

∂vs– sd∫°= or u sB( ) u sA( )–τxsG

-------x∂

∂vs– sd∫°=

Assumption 1 through Assumption 3 implies: Cross-section shape and dimen-sion undergoes negligible change. This implies that no point on the cross-section moves relative to the other in the s-direction i.e., vs = 0 in pure bending. Noting that u sB( ) u sA( )= we obtain:

qct

-----⎝ ⎠⎛ ⎞ sd∫°

qo q+t

---------------⎝ ⎠⎛ ⎞ sd∫° 0= =

If the thickness is uniform across the cross-section.

qo1S--- q sd∫°–=

where, S is the total path length of the perimeter of the cross-section.

1-18


6.47 The thin cross-section shown in Figure P6.47 is subjected to a shear force Vy V= acting through the shear center. Determine the shear stress at points A and B in terms of V, a, and t.

a/2

a

a

A

B

y

zC

Figure P6.47

6.48 The thin cross-section shown is subjected to a shear force Vz V= acting through the shear center. Starting with point A determine the shear stress at points A and B in terms of V, a, and t.

6.49 Determine the shear center of the cross-section shown in Fig-ure P6.47.

1-19


6.67 A cantilever beam is loaded as shown in Figure P6.67. The cross-section has a uniform thickness of t = 1/4 in. Determine the normal and shear stress at points A and B in cartesian coordinates on a section next to the wall.

y

z10 ft

1 kips/ft

10 kips

C

5 in

a = 10

a = 10 in

A

B

y

zC

6.72 in

Figure P6.67

1-20


Energy Methods

• Minimum-energy principles are an alternative to statement of equilib-rium equations.

Displacements

EquilibriumM

ater

ial M

odel

s

Static Equivalency

2

1

3

4

Kinematics

Strains

StressesInternal Forcesand

Moments

External Forcesand

Moments

Energy Methods


• Understand the perspective and concepts in energy methods.• Learn the use of dummy unit load method and Castigliano’s theorem

for calculating displacements in statically determinate and indetermi-nate structures.

7-1


Strain Energy

• The energy stored in a body due to deformation is called the strain energy.

• The strain energy per unit volume is called the strain energy density and is the area underneath the stress-strain curve up to the point of deformation.

ε

σ

O

Uo = Strain energy density

Uo = Complimentary strain energy den

dε

dσ

dUo = σ dε

AdUo = ε dσ

Strain Energy: U Uo VdV∫= [

Strain Energy Density: Uo σ εd

0

ε

∫=

Units: N-m / m3, Joules / m3, in-lbs / in3, or ft-lb/ft.3

Complimentary Strain Energy Density: Uo ε σd

0

σ

∫=

• The strain energy density at the yield point is called Modulus of Resil-ience.

7-2


• The strain energy density at rupture is called Modulus of Toughness.

YieldPoint

σ

ε

Stronger MaterialTougher material

σ

ε

σ

ε

Modulus ofResilience

Modulus ofToughness

RuptureStress

7-3


Linear Strain Energy Density

Uniaxial tension test: Uo σ εd

0

ε

∫ Eε( ) εd

0

ε

∫Eε2

2--------- 12---σε= = = =

Uo12---τγ=

• Strain energy and strain energy density is a scaler quantity.

Uo12--- σxxεxx σyyεyy σzzεzz τxyγxy τyzγyz τzxγzx+ + + + +[ ]=

x

Ay

z

dx dV=Adx

1-D Structural Elements

Axial strain energy• All stress components except σxx are zero.

σxx Eεxx= εxx xddu x( )=

UA12---Eεxx

2 VdV∫

12---E xd

du⎝ ⎠⎛ ⎞

2Ad

A∫ xd

L∫

12--- xd

du⎝ ⎠⎛ ⎞

2E Ad

A∫ xd

L∫= = =

UA Ua xdL∫= Ua

12---EA xd

du⎝ ⎠⎛ ⎞

2=

• Ua is the strain energy per unit length.

UA Ua xdL∫= Ua

12---

N2

EA-------=

7-4


Torsional strain energy• All stress components except τxθ in polar coordinate are zero

τxθ Gγxθ= γxθ ρ xddφ x( )=

UT12---Gγxθ

2 VdV∫

12---G ρ xd

dφ⎝ ⎠⎛ ⎞

2Ad

A∫ xd

L∫

12--- xd

dφ⎝ ⎠⎛ ⎞

2Gρ2 Ad

A∫ xd

L∫= = =

UT Ut xdL∫= Ut

12---GJ xd

dφ⎝ ⎠⎛ ⎞

2=

• Ut is the strain energy per unit length.

UT Ut xdL∫= Ut

12---

T2

GJ-------=

Strain energy in symmetric bending about z-axisThere are two non-zero stress components, σxx and τxy.

σxx Eεxx= εxx yx2

2

dd v–=

UB12---Eεxx

2 VdV∫

12---E y

x2

2

dd v

⎝ ⎠⎜ ⎟⎛ ⎞

2

AdA∫ xd

L∫

12--- x2

2

dd v

⎝ ⎠⎜ ⎟⎛ ⎞

2

Ey2 AdA∫ xd

L∫= = =

UB Ub xdL∫= Ub

12---EIzz

x2

2

dd v

⎝ ⎠⎜ ⎟⎛ ⎞

2

=

• where Ub is the bending strain energy per unit length.

UB Ub xdL∫= Ub

12---

Mz2

EIzz----------=

The strain energy due to shear in bending is: US12---τxyγxy Vd

V∫

12---

τxy2

E------- Vd

V∫= =

As τmax σmax« Us UB«

7-5


Work• If a force moves through a distance, then work has been done by the

force.dW Fdu=

• Work done by a force is conservative if it is path independent.• Non-linear systems and non-conservative systems are two indepen-

dent description of a system.

uL

P

T

vL

θxd

dv=M

P

δW PδuL=

WorkLoading Mode

δW TδφL=

δW PδvL=

δW MδθL=

u(x)

δW p x( )δu x( ) xd0

L

∫=

p(x)

t (x)

φ x( )

φL

v(x)

p(x)

δW t x( )δφ x( ) xd0

L

∫=

δW p x( )δv x( ) xd0

L

∫=

• Any variable that can be used for describing deformation is called the generalized displacement.

• Any variable that can be used for describing the cause that produces deformation is called the generalized force.

7-6


Virtual Work

• Virtual work methods are applicable to linear and non-linear systems, to conservative as well as non-conservative systems.

The principle of virtual work:The total virtual work done on a body at equilibrium is zero.

δW 0=• Symbol δ will be used to designate a virtual quantity

δWext δWint=

Types of boundary conditions

x

z

y

ε

TPx

M Pyε

TPx

MKinematic variable Statical variable

u or N

v

Mz

Vy

NT

Displacement and rotation specified Internal forces and moment specified

at this end at this end to meet equilibrium

φ

Vy

θ xddv= Mz

or

or

or

T

Py

(Primary variable) (Secondary variable)

Geometric boundary conditions (Kinematic boundary conditions)(Essential boundary conditions):

Condition specified on kinematic (primary) variable at the boundary.

Statical boundary conditions(Natural boundary conditions)

Condition specified on statical (secondary)variable at the boundary.

7-7


Kinematically admissible functions• Functions that are continuous and satisfies all the kinematic boundary

conditions are called kinematically admissible functions. • actual displacement solution is always a kinematically admissible

function• Kinematically admissible functions are not required to correspond to

solutions that satisfy equilibrium equations.

Statically admissible functions• Functions that satisfy satisfies all the static boundary conditions, sat-

isfy equilibrium equations at all points, and are continuous at all points except where a concentrated force or moment is applied are called statically admissible functions.

• Actual internal forces and moments are always statically admissible.• Statically admissible functions are not required to correspond to solu-

tions that satisfy compatibility equations.

7-8


7.3 Determine a class of kinematically admissible displacement functions for the beam shown in Figure P7.3.

L

xA BL

CwL2

w

Figure P7.3

7.4 For the beam and loading shown in Figure P7.3 determine a statically admissible bending moment.

7-9


Virtual displacement method• The virtual displacement is an infinitesimal imaginary kinematically

admissible displacement field imposed on a body.

x

actual displacement

kinematically admissible displacementδv = virtual displacement

• Of all the virtual displacements the one that satisfies the virtual work principle is the actual displacement field.

Virtual Force Method• The virtual force is an infinitesimal imaginary statically admissible

force field imposed on a body.• Of all the virtual force fields the one that satisfies the virtual work

principle is the actual force field.

7-10


7.7 The roller at P shown in Figure P7.7 slides in the slot due to the force F = 20kN. Both bars have a cross-sectional area of A = 100 mm2 and a modulus of elasticity E = 200 GPa. Bar AP and BP have lengths of LAP= 200 mm and LBP= 250 mm respectively. Deter-mine the axial stress in the member AP by virtual displacement method.

P

110oF

B

A

Figure P7.7

7-11


7.8 A force F = 20kN is appled to pin shown in Figure P7.8. Both bars have a cross-sectional area of A = 100 mm2 and a modulus of elasticity E = 200 GPa. Bar AP and BP have lengths of LAP= 200 mm and LBP= 250 mm respectively. Using virtual force method determine the movement of pin in the direction of force F.

P

110o

F

B

A

40o

Figure P7.8

7-12


Dummy unit load method• This is a virtual force method that is formalized.• Can be used for axial, torsion or bending problems.

Application to beam bendingDisplacement Calculations

Consider two beams.BEAM 1: Actual beam with actual internal moment M1(x) and actual dis-placement v1(x).BEAM 2: A beam with same supports as beam 1 with a unit force placed at point xp at which we want to calculate the displacement. M2(x) be the statically admissible bending moment and v2(x) be the kinematically admissible displacement for beam 2.Note: No relationship between M2 and v2The internal and external virtual work for beam 2:

δWint M2 x( ) θ2d0

L

∫ M2 x( ) xddθ2 xd

0

L

∫ M2 x( ) xdd

xddv2

⎝ ⎠⎛ ⎞ xd

0

L

∫ M2 x( )x2

2

d

d v2 xd0

L

∫= = = =

δWext 1( )v2 xP( )=

By theorem of virtual work: v2 xP( ) M2 x( )x2

2

d

d v2 xd0

L

∫=

v1(x) is a kinematically admissible displacement field, hence can be used for v2(x).

v1 xP( ) M2 x( )x2

2

d

d v1 xd0

L

∫M2 x( )M1 x( )

EI------------------------------- xd0

L

∫= =

7-13


Slope CalculationsBEAM 1: Actual beam with actual internal moment M1(x) and actual dis-placement v1(x).BEAM 2: A beam with same supports as beam 1 with a unit moment placed at point xp at which we want to calculate the slope. M2(x) be the statically admissible bending moment and v2(x) be the kinematically admissible displacement for beam 2.Note: No relationship between M2 and v2The internal and external virtual work for beam 2:

δWint M2 x( ) θ2d0

L

∫ M2 x( ) xddθ2 xd

0

L

∫ M2 x( ) xdd

xddv2

⎝ ⎠⎛ ⎞ xd

0

L

∫ M2 x( )x2

2

d

d v2 xd0

L

∫= = = =

δWext 1( ) xddv2 xP( )=

By theorem of virtual work: xddv2 xP( ) M2 x( )

x2

2

d

d v2 xd0

L

∫=

v1(x) is a kinematically admissible displacement field, hence can be used for v2(x).

xddv1 xP( ) M2 x( )

x2

2

d

d v1 xd0

L

∫M2 x( )M1 x( )

EI------------------------------- xd0

L

∫= =

7-14


7.21 Using dummy unit load method, find the reaction force at A and deflection at B in terms of P, E,I, and L for the beam shown in Figure P7.21.

L 2L

P

A Bx

Figure P7.21

7-15


Castigliano’s theorem• Simple and more elegant way of finding reaction forces and/or

moments for statically indeterminate structures.Instead of a unit force we consider a force F applied at xp in the dummy unit load method. For linear systems the corresponding statically admis-

sible moment (M̃2) would be F multiplied by M2.

M̃2 FM2= or M2 F∂∂M̃2= v1 xP( )

M2 x( )M1 x( )EI------------------------------- xd

0

L

∫1

EI------ F∂∂M̃2M1 x( )⎝ ⎠

⎛ ⎞ xd0

L

∫= =

The actual moment is a statically admissible moment, and hence we can substitute M̃2 M1= we obtain the following:

v1 xP( ) 1EI------ F∂

∂M1M1 x( )⎝ ⎠⎛ ⎞ xd

0

L

∫1

2EI--------- F∂∂M1

2

⎝ ⎠⎜ ⎟⎛ ⎞

xd0

L

∫ F∂∂ M1

2

2EI--------- xd0

L

∫ F∂∂UB= = = =

• The derivative of the complimentary strain energy with respect to a force at xp gives the deflection in the direction of the force at xp.

xddv1 xP( ) M∂

∂UB=

• The derivative of the complimentary strain energy with respect to a moment at xp gives the slope in the direction of the moment at xp.

• Performing the derivative with respect to force and moment before performing integration will generally result in less algebra.

• The integrals obtained after taking the derivative with respect to force and moment result in integrals that are identical to the dummy unit load method for finding reactions.

7-16


7.27 Using Castigliano’s theorem, find the reaction force at A and deflection at B in terms of P, E,I, and L for the beam shown in Figure P7.21.

L 2L

P

A Bx

7-17

yy)---------

θ2sin– )

I 1 3⁄ )I 2 I 3+ ] 2A3( )⁄

γyz z∂∂v

y∂∂w+= γzx x∂

∂wz∂

∂u+=

σxx

E 1 ν–( )εxx νε yy+[ ]1 2ν–( ) 1 ν+( )

---------------------------------------------------=

σ2–

2----------

σ2 σ3–

2------------------

σ3 σ1–

2------------------, ,

ax

M

Vy

w

x

y

Mz wx a–⟨ ⟩ 2

2--------------------–=

py w x a–⟨ ⟩ 0–=

FORMULA SHEET

Axial (Rods) Torsion (Shafts) Bending (Beams)

σij

∆F j

∆Ai----------

∆ Ai 0→

lim= σnn σxx θ2cos σyy θ2

sin 2τ xy θ θcossin+ += τnt σxx θ θsincos– σyy θ θcossin τ xy θ2cos θ2

sin–( )+ += 2θptan2τ xy

σxx σ–(-------------------=

σ1 2,σxx σyy+( )

2----------------------------

σxx σyy–

2-----------------------

2

τ xy2+±= εnn εxx θ2

cos εyy θ2sin γxy θsin θcos+ += γnt 2εxx θsin θcos– 2εyy θsin θcos γxy θ2

cos(+ +=

σnn n{ } T σ[ ] n{ }= τnt t{ } T σ[ ] n{ }= σtt t{ } T σ[ ] t{ }= S{ } σ[ ] n{ }=

σp3 I 1σp

2– I 2σp I– 3+ 0= I 1 σxx σyy σzz+ += I 2σxx τ xy

τ yx σyy

σyy τ yz

τ zy σzz

σxx τ xz

τ zx σzz

+ += I 3

σxx τ xy τ xz

τ yx σyy τ yz

τ zx τ zy σzz

=

x3

I 1– x2

I 2x I 3–+ 0= x1 2A αcos I 1 3⁄+= x2 3, 2– A α 60o±( )cos I 1 3⁄+= A I 1 3⁄( )2

I 2 3⁄–= 3αcos 2 I 1 3⁄( )3 (–[=

σoct σ1 σ2 σ3+ +( ) 3⁄= τoct13--- σ1 σ2–( )2 σ2 σ3–( )2 σ3 σ1–( )2

+ += εxx x∂∂u= εyy y∂

∂v= εzz z∂∂w= γxy y∂

∂ux∂

∂v+=

εxx σxx ν σ yy σzz+( )–[ ] E⁄= α∆T+ γxy τ xy G⁄= G E 2 1 ν+( )[ ]⁄= σxx εxx νε yy+[ ] E

1 ν2–( )

-------------------= εzzν

1 ν–------------

– εxx εyy+( )=

εxx

σxx

Ex--------

ν yx

E y--------σyy–= γxy

τ xy

Gxy---------=

ν yx

E y--------

νxy

Ex--------= σvon

1

2------- σ1 σ2–( )2 σ2 σ3–( )2 σ3 σ1–( )2

+ +=σ2

σC-------

σ1

σT------– 1≤ τmax max

σ1--------=

K I σnom πa= K II τnom πa= Kequiv K I2

K II2

+=

σ

σyield

Eεσ– yield

=

ε εyield≥


ε ε– yield≤

σ

σyield E2 ε εyield–( )+

E1ε

σ– yield E2 ε εyield+( )+

=

ε εyield≥


ε εyield≥

σEεn

E ε–( )n–

=ε 0≥ε 0<

ax

NF

ax

TT

ax

Mz

Vy

Mx

y

a

P

x

Mz

Vy

x

y

N F x a–⟨ ⟩ 0–= px F x a–⟨ ⟩ 1–

= T T x a–⟨ ⟩ 0–= t T x a–⟨ ⟩ 1–

= Mz M x a–⟨ ⟩ 0–=

py M x a–⟨ ⟩ 2––=

Mz P x a–⟨ ⟩ 1–=

py P x a–⟨ ⟩ 1––=

U o12---σε= U o

12--- σxxεxx σyyεyy σzzεzz τ xyγxy τ yzγyz τ zxγzx+ + + + +[ ]=

F 1=( )v1 xP( )M 2 x( )M 1 x( )

EI--------------------------------- xd

0

L

∫= M 1=( )xd

dv1 xP( )M 2 x( )M 1 x( )

EI--------------------------------- xd

0

L

∫= v1 xP( )F∂

∂U B=xd

dv1 xP( )M∂

∂U B=

Unsymmetric Bending

zxd

dw– v v x( )= w w x( )=

x2

2w

Ezx

2

2

d

d w τ xy 0 σxx«≠ τ xz 0 σxx«≠

M z yσxx AdA∫–= M y zσxx Ad

A∫–=

z τ xz AdA∫=

I yzM y

I yz2

–-----------------

yI zzM y I yzMz–

I yyI zz I yz2

–------------------------------------

z–

Qz I yzQy–

yI zz I yz2

–------------------------------

V y

I zzQy I yzQz–

I yyI zz I yz2

–--------------------------------------

– V z

I yzM y

z I yz2

–---------------------

x2

2

d

d w 1E---

I zzM y I yzMz–

I yyI zz I yz2

–-----------------------------------------

=

x

Mz V– y=xd

dV z pz x( )–=xd

dM y V– z=

Axial (Rods) Torsion (Shafts) Symmetric Bending (Beams)Displace-ments

Strains

Stresses

InternalForces &Moments

StrainEnergy

C. StrainEnergy

u x y z, ,( ) u x( )= φ x y z, ,( ) φ x( )=u x y z, ,( ) y

xddv

–= v v x( )= w 0= u x y z, ,( ) yxd

dv–=

εxx xddu= γxθ ρ

xddφ

= εxx yx

2

2

d

d v–= εxx y

x2

2

d

d v– z

d

d–=

σxx Eεxx Exd

du= = τ xθ Gγxθ Gρ

xddφ

= = σxx Eεxx Eyx

2

2

d

d v–= = τ xy 0 σxx«≠ σxx Ey

x2

2

d

d v– –=

N σxx AdA∫= T ρτxθ Ad

A∫= N σxx Ad

A∫ 0= =

M z yσxx AdA∫–= V y τ xy Ad

A∫=

N σxx AdA∫ 0= =

V y τ xy AdA∫= V

σxxNA-----=

τ xθTρJ

-------= σxx

M z y

I zz-----------

–=

q τ xstV yQz

I zz--------------

–= =

σxx

I yyMz –

I yyI zz

--------------------

–=

q τ xstI yy

I y

---------

–= =

xddu N

EA-------= u2 u1–

N x2 x1–( )EA

---------------------------=xd

dφ TGJ-------= φ2 φ1–

T x2 x1–( )GJ

--------------------------=

x2

2

d

d v M z

EI zz-----------= v

M z

EI-------- xd∫ dx C1x C2+ +∫=

x2

2

d

d v 1E---

I yyMz –

I yyI z

---------------------

=

σxx( )i

NEi

E j A jj 1=

n

∑-----------------------= τ xθ( )

i

GiρT

G jJ jj 1=

n

∑-----------------------------=

σxx( )i

Ei yM z

E j I zz( )j

j 1=

n

∑-------------------------------–= q τ xst

QcompV y

E j I zz( )j

j 1=

n

∑-------------------------------------

–= =

u2 u1–N x2 x1–( )

E j A j∑---------------------------= φ2 φ1–

T x2 x1–( )G jJ j∑[ ]

--------------------------= vM z

E j I zz( )j∑

---------------------------- xd∫ dx C1x C2+ +∫=

xddN px x( )–=

xddT t x( )–=

xd

dV y py x( )–=xd

dMz V– y=xd

dV y py x( )–=d

d

xdd

EAxd

duo px x( )–= xd

dGJ

xddφ

t x( )–=

x2

2

d

dEI zz

x2

2

d

d v

py x( )=

U a EAxd

du

22⁄= U t GJ

xddφ

22⁄= U b EI zz

x2

2

d

d v

2

2⁄=

U a N2

2EA( )⁄= U t T2

2GJ( )⁄= U b M z2

2EI zz( )⁄=

lecture notes on mechanics

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