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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