Formulas in Solid Mechanics

Tore DahlbergSolid Mechanics/IKP, Linköping University

Linköping, Sweden

This collection of formulas is intended for use by foreign students in the course TMHL61,Damage Mechanics and Life Analysis, as a complement to the textbook Dahlberg andEkberg: Failure, Fracture, Fatigue - An Introduction, Studentlitteratur, Lund, Sweden, 2002.It may be use at examinations in this course.

Contents Page

1. Definitions and notations 12. Stress, Strain, and Material Relations 23. Geometric Properties of Cross-Sectional Area 34. One-Dimensional Bodies (bars, axles, beams) 55. Bending of Beam Elementary Cases 116. Material Fatigue 147. Multi-Axial Stress States 178. Energy Methods the Castigliano Theorem 209. Stress Concentration 2110. Material data 25

Version 03-09-18

1. Definitions and notations

Definition of coordinate system and loadings on beam

Loaded beam, length L, cross section A, and load q(x), with coordinate system (origin at thegeometric centre of cross section) and positive section forces and moments: normal force N,shear forces Ty and Tz, torque Mx, and bending moments My, Mz

Notations

Quantity Symbol SI Unit

Coordinate directions, with origin at geometric centre of x, y, z m cross-sectional area A

Normal stress in direction i (= x, y, z) σi N/m2

Shear stress in direction j on surface with normal direction i τij N/m2

Normal strain in direction i εi

Shear strain (corresponding to shear stress τij) γij radMoment with respect to axis i M, Mi NmNormal force N, P N (= kg m/s2)Shear force in direction i (= y, z) T, Ti NLoad q(x) N/mCross-sectional area A m2

Length L, L0 mChange of length δ mDisplacement in direction x u, u(x), u(x,y) mDisplacement in direction y v, v(x), v(x,y) mBeam deflection w(x) mSecond moment of area (i = y, z) I, Ii m4

Modulus of elasticity (Young’s modulus) E N/m2

Poisson’s ratio νShear modulus G N/m2

Bulk modulus K N/m2

Temperature coefficient α

xy

zT

T

M M

N

yz

yz

xMTz

MzMy

NxMTy

L

A

q x( )

K− 1

1

2. Stress, Strain, and Material Relations

Normal stress σx

∆N = fraction of normal force N∆A = cross-sectional area element

Shear stress τxy (mean value over area A in the y direction)

Normal strain εx

Linear, at small deformations (δ << L0)δ = change of lengthL0 = original lengthu(x) = displacement

Non-linear, at large deformationsL = actual length (L = L0 + δ)

Shear strain γxy

Linear elastic material (Hooke’s law)

Tension/compression∆T = change of temperatur

Lateral strain

Shear strain

Relationships between G, K, E and ν

σx =NA

or σx = lim∆A → 0

∆N∆A

τxy =Ty

A(= τmean)

εx =δL0

or εx =du (x)

dx

εx = ln

LL0

γxy =∂u (x , y)

∂y+

∂v(x , y)∂x

εx =σx

E+ α ∆T

εy = − ν εx

γxy =τxy

G

G =E

2 ( 1 + ν )K =

E3 ( 1 − 2ν )

2

3. Geometric Properties of Cross-Sectional Area

The origin of the coordinate system Oyz isat the geometric centre of the cross section

Cross-sectional area A

dA = area element

Geometric centre (centroid)

e = ζgc = distance from η axis to geometriccentref = ηgc distance from ζ axis to geometriccentre

First moment of area

A’ = the “sheared” area (part of area A)

Second moment of area

Iy = second moment of area with respect tothe y axis

Iz = second moment of area with respect tothe z axis

Iyz = second moment of area with respect tothe y and z axes

Parallel-axis theorems

First moment of area

Second moment of area

Oy

z

e

f

dA

A = ⌠⌡A

dA

e ⋅ A = ⌠⌡A

ζ dA

f ⋅ A = ⌠⌡A

η dA

Sy = ⌠⌡A’

zdA and Sz = ⌠⌡A’

ydA

Iy = ⌠⌡A

z 2dA

Iz = ⌠⌡A

y 2dA

Iyz = ⌠⌡A

yzdA

Sη = ⌠⌡A

(z + e) dA = eA and Sζ = ⌠⌡A

(y + f) dA = fA

Iη = ⌠⌡A

(z + e)2 dA = Iy + e 2A , Iζ = ⌠⌡A

(y + f)2 dA = Iz + f2A ,

Iηζ = ⌠⌡A

(z + e) (y + f) dA = Iyz + efA

3

Rotation of axes

Coordinate system Ωηζ has been rotatedthe angle α with respect to the coordinatesystem Oyz

Principal moments of area

I1 + I2 = Iy + Iz

Principal axes

A line of symmetry is always a principalaxis

Second moment of area with respect to axes through geometric centre for somesymmetric areas (beam cross sections)

Rectangular area, base B, height H

Solid circular area, diameter D

Thick-walled circular tube, diameters Dand d

y

z

z

yd A

Iη = ⌠⌡A

ζ2 dA = Iy cos2 α + Iz sin2 α − 2Iyz sin α cos α

Iζ = ⌠⌡A

η2 dA = Iy sin2 α + Iz cos2 α + 2Iyz sin α cos α

Iηζ = ⌠⌡A

ζη dA = (Iy − Iz) sin α cos α + Iyz (cos2 α − sin2 α ) =Iy − Iz

2sin 2α + Iyz cos 2α

I1,2 =Iy + Iz

2± R where R =√

Iy − Iz

2

2

+ Iyz2

sin 2α =− Iyz

Ror cos 2α =

Iy − Iz

2R

y

z

HB

Iy =BH 3

12and Iz =

HB3

12

y

z

DIy = Iz =

πD 4

64

y

z

Dd

Iy = Iz =π64

( D 4 − d 4 )

4

Thin-walled circular tube, radius R andwall thickness t (t << R)

Triangular area, base B and height H

Hexagonal area, side length a

Elliptical area, major axis 2a and minoraxis 2b

Half circle, radius a (geometric centre at e)

4. One-Dimensional Bodies (bars, axles, beams)

Tension/compression of bar

Change of lengthN, E, and A are constant along barL = length of bar

N(x), E(x), and A(x) may vary along bar

Torsion of axle

Maximum shear stressMv = torque = Mx

Wv = section modulus in torsion (givenbelow)

Torsion (deformation) angleMv = torque = Mx

Kv = section factor of torsional stiffness(given below)

y

z

R t

Iy = Iz = πR3t

y

z

H

2B/ 2B/Iy =

BH 3

36and Iz =

HB3

48

y

z

a

a

a

a

a

Iy = Iz =5√ 316

a 4

y

z

2a

b2

Iy =πab 3

4and Iz =

πba 3

4

y

z

a

e Iy =

π8

−8

9π

a 4 ≅ 0,110 a 4 and e =4a3π

δ =NLEA

or

δ = ⌠⌡0

L

ε(x)dx = ⌠⌡0

L N(x)E(x)A(x)

dx

τmax =Mv

Wv

Θ =Mv L

GKv

5

Section modulus Wv and section factor Kv for some cross sections (at torsion)

Torsion of thin-walled circular tube, radiusR, thickness t, where t << R,

Thin-walled tube of arbitrary cross sectionA = area enclosed by the tubet(s) = wall thicknesss = coordinate around the tube

Thick-walled circular tube, diameters Dand d,

Solid axle with circular cross section,diameter D,

Solid axle with triangular cross section,side length a

Solid axle with elliptical cross section,major axle 2a and minor axle 2b

Solid axle with rectangular cross section bby a, where b ≥ a

for kWv and kKv, see table below

y

z

R t

Wv = 2πR2t Kv = 2πR3t

t(s)

(s)

AArea

s

Wv = 2Atmin Kv =4A2

⌠⌡s

[t(s)] − 1 ds

y

z

Dd

Wv =π16

D 4 − d 4

DKv =

π32

(D 4 − d 4)

y

z

D

Wv =πD 3

16Kv =

πD 4

32

y

z

a

/ 2a / 2a Wv =a 3

20Kv =

a 4 √ 380

y

z

2a

b2

Wv =π2

a b 2 Kv =πa 3b 3

a 2 + b 2

y

z

a

b

Wv = kWv a 2b Kv = kKv a 3b

6

Factors kWv and kKv for some values of ratio b / a (solid rectangular cross section)

b / a kWv kKv

1.0 0.208 0.14061.2 0.219 0.16611.5 0.231 0.19582.0 0.246 0.2292.5 0.258 0.2493.0 0.267 0.2634.0 0.282 0.2815.0 0.291 0.29110.0 0.312 0.312

0.333 0.333

Bending of beam

Relationships between bending moment My = M(x), shear force Tz = T(x), and load q(x) onbeam

Normal stressI (here Iy) = second moment of area (seeSection 12.2)

Maximum bending stress

Wb = section modulus (in bending)

Shear stressSA’ = first moment of area A’ (see Section12.2)b = length of line limiting area A’τgc = shear stress at geometric centreµ = the Jouravski factor

The Jouravski factor µ for some cross sections

rectangular 1.5triangular 1.33circular 1.33thin-walled circular 2.0elliptical 1.33ideal I profile A / Aweb

∞

dT(x)dx

= −q (x) ,dM(x)

dx= T(x) , and

d2M(x)dx 2

= −q (x)

σ =NA

+MzI

| σ |max =|M |Wb

where Wb =I

| z | max

τ =TSA’

Ib

τgc = µTA

7

Skew bendingAxes y and z are not principal axes:

Iy, Iz, Iyz = second moment of area

Axes y’ and z’ are principal axes:

I1, I2 = principal second moment of area

Beam deflection w(x)Differential equations

when EI(x) is function of x

when EI is constant

Homogeneous boundary conditionsClamped beam end

where * is the coordinate of beam end(to be entered after differentiation)

Simply supported beam end

Sliding beam end

Free beam end

σ =NA

+My(zIz − yIyz) − Mz(yIy − zIyz)

Iy Iz − Iyz2

σ =NA

+M1 z’

I1

−M2 y’

I2

d2

dx 2

EI(x)d2

dx 2w(x)

= q (x)

EId4

dx 4w(x) = q (x)

x x=L

w(*) = 0 andd

dxw(*) = 0

x x=L

w(*) = 0 and − EId2

dx 2w(*) = 0

x x=L

ddx

w(*) = 0 and − EId3

dx 3w(*) = 0

x x=L

− EId2

dx 2w(*) = 0 and − EI

d3

dx 3w(*) = 0

8

Non-homogeneous boundary conditions(a) Displacement δ prescribed

(b) Slope Θ prescribed

(c) Moment M0 prescribed

(d) Force P prescribed

Beam on elastic bed

Differential equation

EI = constant bending stiffnessk = bed modulus (N/m2)

Solution

Boundary conditions as given above

Beam vibration

Differential equationEI = constant bending stiffnessm = beam mass per metre (kg/m)t = time

Assume solution w(x,t) = X(x)⋅T(t). Then the standing wave solution is

where µ4 = ω2m /EI

Boundary conditions (as given above) give an eigenvalue problem that provides theeigenfrequencies and eigenmodes (eigenforms) of the vibrating beam

w(*) = δ

x x=L

x x=L

O

z O

0 0MM

x x=LP P

(a)

(b)

(c)

(d)

ddx

w(*) = Θ

− EId2

dx 2w(*) = M0

− EId3

dx 3w(*) = P

EId4

dx 4w(x) + kw(x) = q (x)

w(x) = wpart(x) + whom(x) where

whom(x) = C1 cos (λx) + C2 sin (λx) eλx + C3 cos (λx) + C2 sin (λx) e − λx ; λ4 =k

4EI

EI∂4

∂x 4w(x , t) + m

∂2

∂t2w(x , t) = q (x , t)

T(t) = e iωt and X(x) = C1 cosh (µx) + C2 cos (µx) + C3 sinh (µx) + C4 sin (µx)

9

Axially loaded beam, stability, the Euler cases

Beam axially loaded in tensionDifferential equation

N = normal force in tension (N > 0)

Solution

New boundary condition on shear force (other boundary conditions as given above)

Beam axially loaded in compressionDifferential equation

P = normal force in compression (P > 0)

Solution

New boundary condition on shear force (other boundary conditions as given above)

Elementary cases: the Euler cases (Pc is critical load)

Case 1 Case 2a Case 2b Case 3 Case 4

EId4

dx 4w(x) − N

d2

dx 2w(x) = q (x)

w(x) = wpart(x) + whom(x) where

whom(x) = C1 + C2√ NEI

x + C3 sinh√ N

EIx

+ C4 cosh√ N

EIx

T(*) = − EId3

dx 3w(*) + N

ddx

w(*)

EId4

dx 4w(x) + P

d2

dx 2w(x) = q (x)

w(x) = wpart(x) + whom(x) where

whom(x) = C1 + C2√ PEI

x + C3 sin√ P

EIx

+ C4 cos√ P

EIx

T(*) = − EId3

dx 3w(*) − P

ddx

w(*)

P

L, EI L, EI

P

L, EI

P

L, EI

P

L, EI

P

Pc =π2EI

4L2Pc =

π2EI

L2Pc =

π2EI

L2Pc =

2.05 π2EI

L2Pc =

4π2EI

L2

10

5. Bending of Beam Elementary Cases

Cantilever beam

xz w(x)

L, EIP

w(x) =PL3

6EI

3

x 2

L2−

x 3

L3

w(L) =PL3

3EId

dxw(L) =

PL2

2EI

xz w(x)

L, EIM w(x) =

ML2

2EI

x 2

L2

w(L) =ML2

2EId

dxw(L) =

MLEI

xz w(x)L, EI

q = Q/Lw(x) =

qL4

24EI

x 4

L4− 4

x 3

L3+ 6

x 2

L2

w(L) =qL4

8EId

dxw(L) =

qL3

6EI

xz w(x)L, EI

q0 w(x) =

q0 L4

120EI

x 5

L5− 10

x 3

L3+ 20

x 2

L2

w(L) =11 q0 L4

120EId

dxw(L) =

q0 L3

8EI

xz w(x)L, EI

q0 w(x) =

q0 L4

120EI

−x 5

L5+ 5

x 4

L4− 10

x 3

L3+ 10

x 2

L2

w(L) =q0 L4

30EId

dxw(L) =

q0 L3

24EI

11

Simply supported beam

Load applied at x = αL (α < 1), β = 1 α−

w(x) =PL3

6EIβ

(1 − β2)

xL

−x 3

L3

forxL

≤ α

xz w(x) L, EI

L LP + = 1

w(αL) =PL3

3EIα2 β2 . When α > β one obtains

wmax = wL√1 − β2

3

= w(αL)1 + β3β √ 1 + β

3α

ddx

w(0) =PL2

6EIα β (1 + β)

ddx

w(L) = −PL2

6EIα β (1 + α)

xz w(x) L, EI

MA MB w(x) =L2

6EI

MA

2

xL

− 3x 2

L2+

x 3

L3

+ MB

xL

−x 3

L3

ddx

w(0) =MA L

3 EI+

MB L

6 EId

dxw(L) = −

MA L

6 EI−

MB L

3 EI

w(x) =ML2

6EI

(1 − 3β2)

xL

−x 3

L3

forxL

≤ α

xz w(x) L, EI

L LM

ddx

w(0) =ML6EI

(1 − 3β2)d

dxw(L) =

ML6EI

(1 − 3α2)

w(x) =QL3

24EI

x 4

L4− 2

x 3

L3+

xL

xz w(x) L, EI

Q

w(L /2) =5 QL3

384 EId

dxw(0) = −

ddx

w(L) =QL2

24EI

w(x) =QL3

180EI

3

x 5

L5− 10

x 3

L3+ 7

xL

ddx

w(0) =7 QL2

180EId

dxw(L) = −

8 QL2

180EIx

z w(x) L, EI

Q

w(x) =QL3

180EI

− 3

x 5

L5+ 15

x 4

L4− 20

x 3

L3+ 8

xL

ddx

w(0) =8 QL2

180EId

dxw(L) = −

7 QL2

180EIx

z w(x) L, EI

Q

12

Clamped simply supported beam and clamped clamped beamLoad applied at x = αL (α < 1), β = 1 αOnly redundant reactions are given. For deflections, use superposition of solutionsfor simply supported beams.

−

xz L, EI

MAL L

P + = 1MA =

PL2

β (1 − β2 )

xz L, EI

MA MB MA =MB

2

xz L, EI

MA ML L

+ = 1 MA =M2

(1 − 3β2 )

xz L, EI

MAQ

MA =QL8

xz L, EI

MAQ MA =

2 QL15

x

zL, EI

MAL L

+ = 1

MBP MA = PL α β2 MB = PL α2 β

x

zL, EI

MA ML L

+ = 1

MB MA = −M β (1 − 3α ) MB = M α (1 − 3β )

xz L, EI

MA

QMB MA = MB =

QL12

xz L, EI

MAQ MB MA =

QL10

MB =QL15

13

6. Material Fatigue

Fatigue limits (notations)

Load Alternating Pulsating

Tension/compression

Bending

Torsion

The Haigh diagram

σa = stress amplitudeσm = mean stressσY = yield limitσU = ultimate strenghtσu, σup = fatigue limitsλ, δ, κ = factors reducing fatigue limits(similar diagrams for σub, σubp and τuv, τuvp)

Factors reducing fatigue limits

Surface finish κ

Factor κ reducing the fatigue limit due tosurface irregularities

(a) polished surface ( κ = 1)(b) ground(c) machined(d) standard notch(e) rolling skin(f) corrosion in sweet water(g) corrosion in salt water

± σu σup ± σup

± σub σubp ± σubp

± τuv τuvp ± τuvp

u

a

m

up

up

upu

Y U

Y

300 600 900 1200 MPa

(a)

(b)

(c)

(d)

(e)(f)(g)

U

1.0

0.8

0.6

0.4

0.2

14

Volume factor λ (due to process)

Factor λ reducing the fatigue limit due tosize of raw material

(a) diameter at circular cross section(b) thickness at rectangular cross section

Volume factor δ (due to geometry)

Factor δ reducing the fatigue limits σub andτuv due to loaded volume.Steel with ultimate strength σU =(a) 1500 MPa(b) 1000 MPa(c) 600 MPa(d) 400 MPa(e) aluminiumFactor δ = 1 when fatigue notch factor Kf >1 is used.

Fatigue notch factor Kf (at stress concentration)

Kt = stress concentration factor (see Section12.8)q = fatigue notch sensitivity factor

Fatigue notch sensitivity factor q

Fatigue notch sensitivity factor q for steelwith ultimate strength σU =(a) 1600 MPa(b) 1300 MPa(c) 1000 MPa(d) 700 MPa(e) 400 MPa

20 40 60 8010 20 30 40

(a)(b)

100mm50mm

1.0

0.8

40 80 1200

(a)(b)

(c)

(d)(e)

Diameter or thickness in mm

1.0

0.9

0.8

Kf = 1 + q (Kt − 1)

1 102 5

(a)

(b)

(c)(d)

(e)

q

r

1.0

0.8

0.6

0.4

0.2

0.1 0.5Fillet radius in mm

15

Wöhler diagram

σai = stress amplitudeNi = fatigue life (in cycles) at stressamplitude σai

Damage accumulation D

ni = number of loading cycles at stressamplitude σai

Ni = fatigue life at stress amplitude σai

Palmgren-Miner’s rule

Failure when ni = number of loading cycles at stressamplitude σai

Ni = fatigue life at stress amplitude σai

I = number of loading stress levels

Fatigue data (cyclic, constant-amplitude loading)

The following fatigue limits may be used only when solving exercises. For a realdesign, data should be taken from latest official standard and not from this table.1

Material Tension Bending Torsionalternating pulsating alternating pulsating alternating pulsatingMPa MPa MPa MPa MPa MPa

Carbon steel141312-00 110 110 110 170 150 150 100 100 100

141450-1 140 130 130 190 170 170 120 120 120

141510-00 230

141550-01 180 160 160 240 210 210 140 140 140

141650-01 200 180 180 270 240 240 150 150 150

141650 460

Stainless steel 2337-02,

Aluminium SS 4120-02, ; SS 4425-06,

1 Data in this table has been collected from B Sundström (editor): Handbok och Formelsamling iHållfasthetslära, Institutionen för hållfasthetslära, KTH, Stockholm, 1998.

1 2 3 4 5 6 70

a

log N

ai

D =ni

Ni

∑i = 1

I ni

Ni

= 1

± ± ± ± ± ±± ± ± ± ± ±±± ± ± ± ± ±± ± ± ± ± ±

±

σu = ± 270 MPa

σub = ± 110 MPa σu = ± 120 MPa

16

7. Multi-Axial Stress States

Stresses in thin-walled circular pressure vessel

σt = circumferential stressσx = longitudinal stressp = internal pressureR = radius of pressure vesselt = wall thickness (t << R)

Rotational symmetry in structure and load (plane stress, i.e. σz = 0)

Differential equation for rotating circular plateu = u(r) = radial displacementρ = densityω = angular rotation (rad/s)

Solution

Stresses

where

Boundary conditionsσr or u must be known on inner and outer boundary of the circular plate

Shrink fit

δ = difference of radiip = contact pressureu = radial displacement as function of p

Plane stress and plane strain (plane state)

Plane stress (in xy-plane) when σz = 0, τxz = 0, and τyz = 0

Plane strain (in xy-plane) when τxz = 0, τyz = 0, and εz = 0 or constant

Stresses in direction α (plane state)

σ(α) = normal stress in direction ατ(α) = shear stress on surface with normal in direction α

σt = pRt

and σx = pR2t

(σz ≈ 0)

d2 u

d r 2+

1r

d ud r

−u

r 2= −

1 − ν2

Eρω2r

u (r) = uhom + upart = A0 r +B0

r−

1 − ν2

8Eρ ω2 r 3

σr(r) = A −B

r 2−

3 + ν8

ρ ω2 r 2 and σφ(r) = A +B

r 2−

1 + 3ν8

ρ ω2 r 2

A =E A0

1 − νand B =

E B0

1 + ν

δ = uouter(p ) − uinner(p )

σ(α) = σxcos2(α) + σysin2(α) + 2τxycos(α)sin(α) y

x

( )( )

τ(α) = − (σx − σy) sin(α)cos(α) + τxy(cos2(α) − sin2(α))

17

Principal stresses and principal directions at plane stress state

ψ1 = angle from x axis (in xy plane) todirection of principal stress σ1

Strain in direction α (plane state)

ε(α) = normal strain in direction αγ(α) = shear strain of element with normal in direction α

Principal strains and principal directions (plane state)

ψ1 = angle from x axis (in xy plane) todirection of principal strain ε1

Principal stresses and principal directions at three-dimensional stress state

The determinant

gives three roots (the principal stresses)

(contains the nine stress components σij)

Direction of principal stress σi (i = 1, 2, 3) is given by

nix, niy and niz are the elements of the unitand vector ni in the direction of σi

( T means transpose)

σ1,2

σ1,2 = σc ± R =σx + σy

2±√

σx − σy

2

2

+ τxy2

sin(2ψ1) =τxy

Ror cos(2ψ1) =

σx − σy

2R

ε(α) = εx cos2(α) + εy sin2(α) + γxy sin(α)cos(α) y

xγ(α) = (εy − εx) sin(2α) + γxy cos(2α)

ε1,2 = εc ± R =εx + εy

2±√

εx − εy

2

2

+

γxy

2

2

sin(2ψ1) =γxy

2Ror cos(2ψ1) =

εx − εy

2R

Stress matrix S =

σx τxy τxz

τyx σy τyz

τzx τzy σz

| S − σ I | = 0

Unit matrix I =

1 0 00 1 00 0 1

(S − σi I) ⋅ ni = 0

niT ⋅ ni = 1

18

Principal strains and principal directions at three-dimensional stress state

Use shear strain

The determinant

gives three roots (the principal strains)I = unit matrix

Direction of principal strain εi (i = 1, 2, 3) is given by

nix, niy and niz are the elements of the unitand vector ni in the direction of εi

( T means transpose)

Hooke’s law, including temperature term (three-dimensional stress state)

α = temperature coefficient∆T = change of temperature (relative totemperature giving no stress)

Effective stress

The Huber-von Mises effective stress (the deviatoric stress hypothesis)

The Tresca effective stress (the shear stress hypothesis)

εij = γij / 2 for i ≠ j

Strain matrix E =

εx εxy εxz

εyx εy εyz

εzx εzy εz

| E − ε I | = 0

(E − εi I) ⋅ ni = 0

niT ⋅ ni = 1

εx =1E

[σx − ν(σy + σz)] + α ∆T

εy =1E

[σy − ν(σz + σx)] + α ∆T

εz =1E

[σz − ν(σx + σy)] + α ∆T

γxy =τxy

Gγyz =

τyz

Gγzx =

τzx

G

σevM = √σx

2 + σy2 + σz

2 − σxσy − σyσz − σzσx + 3τxy2 + 3τyz

2 + 3τzx2

=√ 12

(σ 1 − σ2)2 + (σ2 − σ3)2 + (σ3 − σ1)2

σeT = max [ | σ1 − σ2 | , | σ2 − σ3 | , | σ3 − σ1 | ] = σmax

pr − σminpr (pr = principal stress)

19

8. Energy Methods the Castigliano Theorem

Strain energy u per unit of volume

Linear elastic material and uni-axial stress

Total strain energy U in beam loaded in tension/compression, torsion, bending, andshear

Mt = torque = Mx Kv = section factor of torsional stiffnessMbend = bending moment = My β = shear factor, see below

Cross section Shear factor β

β is given for some cross sections in thetable (µ is the Jouravski factor, see Section12.3 One-Dimensional Bodies)

Elementary case: pure bending

Only bending momentet Mbend is present.The moment varies linearly along the beamwith moments M1 and M2 at the beam ends.One hasMbend(x) = M1 + (M2 M1)x /L, which gives

The second term is negative if M1 and M2

have different signs

The Castigliano theorem

δ = displacement in the direction of force Pof the point where force P is appliedΘ = rotation (change of angle) at momentM

u =σ ε2

Utot = ⌠⌡0

L

N(x)2

2EA(x)+

Mt(x)2

2GKv(x)+

Mbend(x)2

2EI(x)+ β

T(x)2

2GA(x)

dx

β µ

6/5 3/2

10/9 4/3

2 2

A/A A/Awebweb

β =A

I2⌠⌡A

SA’

b

2

dA

L, EIMM1 2

M1

M M2

x −

Utot =L

6EI M1

2 + M1 M2 + M22

δ =∂U∂P

and Θ =∂U∂M

20

9. Stress ConcentrationTension/compression

Maximum normal stress at a stress concentration is σmax = Kt σnom, where Kt and σnom

are given in the diagrams

Tension of flat bar with shoulder fillet Tension of flat bar with notch

Tension of circular bar with shoulder Tension of circular bar with U-shapedfillet groove

K t

0

B bP

nom =bhP

P

r

r/b

B/b

thickness h

3.0

2.5

2.0

1.5

1.0

2.01.51.21.11.051.01

0.1 0.2

K t

0

nom =bhP

r

r/b

B/b

PPB b

thickness h

3.0

2.5

2.0

1.5

1.0

2.01.21.1

1.051.01

0.1 0.2 0.3 0.4 0.5 0.6

K t

0

nom =

r

PPD d

r/d

d2

D/d

4 P

3.0

2.5

2.0

1.5

1.0

2.01.51.21.1

1.051.01

0.1 0.2 0.3

K t

0

nom =

r

PPD d

r/d

d2

D/d

4 P

3.0

2.5

2.0

1.5

1.0

1.21.1

1.051.01

0.1 0.2 0.3 0.4 0.5 0.6

21

Tension of flat bar with hole

Bending

Maximum normal stress at a stress concentration is σmax = Kt σnom, where Kt and σnom

are given in the diagrams

Bending of flat bar with hole Bending of circular bar with hole

2

3

5B/a=

nomB=

0

0 0

K t

r/a

a

B

0

2 r

2 r

B -

2.5

2.25

0.1 0.2 0.3 0.42.0

2.2

2.4

2.6

2.8

3.0

3.2

K t

0

nom=

MMd

M

b b

b

B

(B-d)h 2

6

d/B

d/hthickness h

3.0

2.5

2.0

1.5

1.0

0.25

0.5

1.0

2.0

= 0

0.2 0.4 0.6

K t

0

nom =d

MM

3

D

d

d/D

M

D D6

2

b b

b

32

3.0

2.5

2.0

1.5

1.00.1 0.2 0.3

22

Bending of flat bar with shoulder fillet Bending of flat bar with notch

Bending of circular bar with shoulder Bending of circular bar with U-shapedfillet groove

K t

0

B b

nom =

r

r/b

B/b

Mb Mb

Mb6

hb2

thickness h

3.0

2.5

2.0

1.5

1.0

6.02.01.2

1.051.021.01

0.1 0.2

K t

0

nom =

r

r/b

B/b

B b

Mb Mb

Mb6

hb2

thickness h

3.0

2.5

2.0

1.5

1.0

1.21.1

1.051.021.01

0.1 0.2 0.3 0.4 0.5 0.6

K t

0

nom =

r

D d

r/d

d

D/d

3

Mb Mb

Mb32

3.0

2.5

2.0

1.5

1.0

6.02.01.2

1.051.021.01

0.1 0.2 0.3

K t

0

nom =

r

D d

r/d

d

D/d

Mb Mb

Mb323

3.0

2.5

2.0

1.5

1.0

1.21.051.021.01

0.1 0.2 0.3 0.4 0.5 0.6

23

Torsion

Maximium shear stress at stress concentration is τmax = Kt τnom, where Kt and τnom aregiven in the diagrams

Torsion of circular bar with shoulder Torsion of circular bar with notchfillet

Torsion of bar with longitudinal keyway Torsion of circular bar with hole

K t

0

nom =

r

D d

r/d

d

D/d

MM

3

vv

Mv16

3.0

2.5

2.0

1.5

1.0

2.01.31.21.1

0.1 0.2 0.3

K t

0

nom =

r

D d

r/d

d

M M

3

D/d

v v

vM16

3.0

2.5

2.0

1.5

1.0

1.21.051.01

0.1 0.2 0.3 0.4 0.5 0.6

K t

0 r/d

r

d

8

d 3nom = v

/4 7d d

16M

4.0

3.5

3.0

2.5

2.00.05 0.10

K t

0

nom=

d

MM

3

D

d

d/D

M

v v

v

D16

D6

2

2.0

1.5

1.00.1 0.2 0.3

24

10. Material data

The following material properties may be used only when solving exercises. For a realdesign, data should be taken from latest official standard and not from this table (two valuesfor the same material means different qualities).1

Material Young’s ν α106 Ultimate Yield limitmodulus strength tension/ bending torsionE compressionGPa K-1 MPa MPa MPa MPa

Carbon steel141312-00 206 0.3 12 360 >240 260 140

460141450-1 205 0.3 430 >250 290 160

510141510-00 205 0.3 510 >320

640141550-01 205 0.3 490 >270 360 190

590141650-01 206 0.3 11 590 >310 390 220

690141650 206 0.3 860 >550 610

Offset yield strength Rp0.2 (σ0,2)

Stainless steel2337-02 196 0.29 16.8 >490 >200AluminiumSS 4120-02 70 23 170 >65

215SS 4120-24 70 23 220 >170

270SS 4425-06 70 23 >340 >270

1 Data in this table has been collected from B Sundström (editor): Handbok och Formelsamling iHållfasthetslära, Institutionen för hållfasthetslära, KTH, Stockholm, 1998.

25