Formulas in Solid Mechanics
Tore Dahlberg
Solid 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 and
Ekberg: Failure, Fracture, Fatigue - An Introduction, Studentlitteratur, Lund, Sweden, 2002.
It may be use at examinations in this course.
Contents
1. Definitions and notations
2. Stress, Strain, and Material Relations
3. Geometric Properties of Cross-Sectional Area
4. One-Dimensional Bodies (bars, axles, beams)
5. Bending of Beam Elementary Cases
6. Material Fatigue
7. Multi-Axial Stress States
8. Energy Methods the Castigliano Theorem
9. Stress Concentration
10. Material data
Version 03-09-18
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1. Definitions and notations
Definition of coordinate system and loadings on beam
Mz
Tz
Mx N
My
q ( x)
N Mx
x
Ty
Ty
y
z
A
L
My
Tz
Mz
Loaded beam, length L, cross section A, and load q(x), with coordinate system (origin at the
geometric 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
rad
Moment with respect to axis i
M, Mi
Nm
Normal force
N, P
N (= kg m/s 2)
Shear force in direction i (= y, z)
T, Ti
N
Load
q(x)
N/m
Cross-sectional area
A
m2
Length
L, L0
m
Change of length
δ
m
Displacement in direction x
u, u(x), u(x,y) m
Displacement in direction y
v, v(x), v(x,y) m
Beam deflection
w(x)
m
Second moment of area (i = y, z)
I, Ii
m4
Modulus of elasticity (Young’s modulus)
E
N/m2
Poisson’s ratio
ν
Shear modulus
G
N/m2
Bulk modulus
K
N/m2
Temperature coefficient
α
K− 1
1
2. Stress, Strain, and Material Relations
Normal stress σx
σx =
N
A
∆N = fraction of normal force N
∆A = cross-sectional area element
 ∆N 
or σx = lim 

∆A → 0  ∆A 
Shear stress τxy (mean value over area A in the y direction)
τxy =
Ty
(= τmean)
A
Normal strain εx
Linear, at small deformations (δ << L0)
δ
du(x)
εx =
or εx =
L0
dx
δ = change of length
L0 = original length
u(x) = displacement
Non-linear, at large deformations
L
εx = ln  
 L0 
L = actual length (L = L0 + δ)
Shear strain γxy
γxy =
∂u (x, y) ∂v(x, y)
+
∂y
∂x
Linear elastic material (Hooke’s law)
Tension/compression
σx
εx = + α ∆T
E
∆T = change of temperatur
Lateral strain
εy = − ν εx
Shear strain
τxy
γxy =
G
Relationships between G, K, E and ν
G=
E
2(1+ν)
K=
E
3 ( 1 − 2ν )
2
3. Geometric Properties of Cross-Sectional Area
e
The origin of the coordinate system Oyz is
at the geometric centre of the cross section
O
y
f
dA
z
Cross-sectional area A
A = ⌠ dA
⌡A
Geometric centre (centroid)
dA = area element
e = ζgc = distance from η axis to geometric
centre
f = ηgc distance from ζ axis to geometric
centre
e ⋅ A = ⌠ ζ dA
⌡A
f ⋅ A = ⌠ η dA
⌡A
First moment of area
S y = ⌠ zdA and Sz = ⌠ ydA
⌡A’
⌡A’
Second moment of area
A’ = the “sheared” area (part of area A)
Iy = ⌠ z 2dA
⌡A
Iy = second moment of area with respect to
the y axis
Iz = second moment of area with respect to
the z axis
Iz = ⌠ y 2dA
⌡A
Iyz = ⌠ yzdA
⌡A
Parallel-axis theorems
First moment of area
and
Sη = ⌠ (z + e) dA = eA
⌡A
Second moment of area
Iη = ⌠ (z + e)2 dA = Iy + e 2A ,
⌡A
Iyz = second moment of area with respect to
the y and z axes
Sζ = ⌠ (y + f) dA = fA
⌡A
Iζ = ⌠ (y + f)2 dA = Iz + f 2A ,
⌡A
Iηζ = ⌠ (z + e) (y + f) dA = Iyz + efA
⌡A
3
Rotation of axes
y
Coordinate system Ωηζ has been rotated
the angle α with respect to the coordinate
system Oyz
z
y
dA
z
Iη = ⌠ ζ2 dA = Iy cos2 α + Iz sin2 α − 2Iyz sin α cos α
⌡A
Iζ = ⌠ η2 dA = Iy sin2 α + Iz cos2 α + 2Iyz sin α cos α
⌡A
Iηζ = ⌠ ζη dA = (Iy − Iz ) sin α cos α + Iyz (cos2 α − sin2 α ) =
⌡A
Iy − Iz
sin 2α + Iyz cos 2α
2
Principal moments of area
I1, 2 =
Iy + Iz
±R
2

√
 Iy − Iz  2 2

 + Iyz
 2 
where R =
I1 + I2 = Iy + Iz
Principal axes
sin 2α =
− Iyz
R
or cos 2α =
Iy − Iz
2R
A line of symmetry is always a principal
axis
Second moment of area with respect to axes through geometric centre for some
symmetric areas (beam cross sections)
B
y
Rectangular area, base B, height H
BH 3
HB 3
and Iz =
Iy =
12
12
H
z
y
Solid circular area, diameter D
πD 4
Iy = Iz =
64
D
z
y
Thick-walled circular tube, diameters D
and d
π
Iy = Iz = ( D 4 − d 4 )
64
d D
z
4
Thin-walled circular tube, radius R and
wall thickness t (t << R)
t
R
y
Iy = Iz = πR 3t
z
Triangular area, base B and height H
BH 3
HB 3
Iy =
and Iz =
36
48
H
y
B/ 2 B/ 2
z
a
a
y
a
Hexagonal area, side length a
5√
3 a 4
Iy = Iz =
16
a
a
z
2a
y
Elliptical area, major axis 2a and minor
axis 2b
πab 3
πba 3
Iy =
and Iz =
4
4
2b
z
Half circle, radius a (geometric centre at e)
π 8 
4a
Iy =  −  a 4 ≅ 0, 110 a 4 and e =
3π
 8 9π 
a
y
e
z
4. One-Dimensional Bodies (bars, axles, beams)
Tension/compression of bar
Change of length
NL
δ=
or
EA
N, E, and A are constant along bar
L = length of bar
L
L
N(x)
δ = ⌠ ε(x)dx = ⌠
dx
⌡0 E(x)A(x)
⌡0
Torsion of axle
Maximum shear stress
Mv
τmax =
Wv
N(x), E(x), and A(x) may vary along bar
Mv = torque = Mx
Wv = section modulus in torsion (given
below)
Torsion (deformation) angle
Mv L
Θ=
GKv
Mv = torque = Mx
Kv = section factor of torsional stiffness
(given below)
5
Section modulus Wv and section factor Kv for some cross sections (at torsion)
Torsion of thin-walled circular tube, radius
R, thickness t, where t << R,
t
R
y
Wv = 2πR 2t
z
Thin-walled tube of arbitrary cross section
A = area enclosed by the tube
t(s) = wall thickness
s = coordinate around the tube
4A 2
Wv = 2Atmin
Kv =
⌠ [t(s)] − 1 ds
⌡s
(s)
s
t(s)
Area A
y
Thick-walled circular tube, diameters D
and d,
π D4 − d4
π
Kv = (D 4 − d 4)
Wv =
16 D
32
d D
z
y
Solid axle with circular cross section,
diameter D,
πD 3
πD 4
Kv =
Wv =
16
32
D
z
Solid axle with triangular cross section,
side length a
a3
a4 √
3
Kv =
Wv =
20
80
a
y
a/ 2 a/ 2
z
2a
y
Solid axle with elliptical cross section,
major axle 2a and minor axle 2b
π
πa 3b 3
Wv = a b 2
Kv = 2
2
a + b2
2b
z
Solid axle with rectangular cross section b
by a, where b ≥ a
Wv = kWv a 2b
Kv = kKv a 3b
b
y
Kv = 2πR 3t
a
z
for kWv and kKv, see table below
6
Factors kWv and kKv for some values of ratio b / a (solid rectangular cross section)
b/a
kWv
kKv
1.0
1.2
1.5
2.0
2.5
3.0
4.0
5.0
10.0
∞
0.208
0.219
0.231
0.246
0.258
0.267
0.282
0.291
0.312
0.333
0.1406
0.1661
0.1958
0.229
0.249
0.263
0.281
0.291
0.312
0.333
Bending of beam
Relationships between bending moment My = M(x), shear force Tz = T(x), and load q(x) on
beam
dT(x)
dM(x)
d2M(x)
= − q(x) ,
= T(x) , and
= − q(x)
dx
dx
dx 2
Normal stress
N Mz
σ= +
A
I
Maximum bending stress
| M|
I
| σ | max =
where Wb =
Wb
| z | max
I (here Iy) = second moment of area (see
Section 12.2)
Wb = section modulus (in bending)
Shear stress
TSA’
τ=
Ib
SA’ = first moment of area A’ (see Section
12.2)
b = length of line limiting area A’
T
τgc = shear stress at geometric centre
τgc = µ
µ
= the Jouravski factor
A
The Jouravski factor µ for some cross sections
rectangular
triangular
circular
thin-walled circular
elliptical
ideal I profile
1.5
1.33
1.33
2.0
1.33
A / Aweb
7
Skew bending
Axes y and z are not principal axes:
N My (zIz − yIyz ) − Mz (yIy − zIyz )
σ= +
A
Iy Iz − Iyz2
Iy, Iz, Iyz = second moment of area
Axes y’ and z’ are principal axes:
N M1 z’ M2 y’
σ= +
−
A
I1
I2
I1, I2 = principal second moment of area
Beam deflection w(x)
Differential equations

d2 
d2
 EI(x)
 = q(x)
w(x)

dx 2 
dx 2
EI
when EI(x) is function of x
d4
w(x) = q(x)
dx 4
when EI is constant
Homogeneous boundary conditions
Clamped beam end
d
w(*) = 0 and
w(*) = 0
dx
where * is the coordinate of beam end
(to be entered after differentiation)
Simply supported beam end
d2
w(*) = 0 and − EI 2 w(*) = 0
dx
x
x=L
x
x=L
Sliding beam end
d
d3
w(*) = 0 and − EI 3 w(*) = 0
dx
dx
x
Free beam end
d2
− EI 2 w(*) = 0
dx
x
d3
and − EI 3 w(*) = 0
dx
8
x=L
x=L
Non-homogeneous boundary conditions
(a) Displacement δ prescribed
w(*) = δ
x
x=L
(a)
(b) Slope Θ prescribed
d
w(*) = Θ
dx
O
(b)
z
M0
(c) Moment M0 prescribed
d2
− EI 2 w(*) = M0
dx
O
x=L M0
x
(c)
P
(d) Force P prescribed
d3
− EI 3 w(*) = P
dx
x=L
x
P
(d)
Beam on elastic bed
Differential equation
d4
EI 4 w(x) + kw(x) = q(x)
dx
EI = constant bending stiffness
k = bed modulus (N/m2)
Solution
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
Boundary conditions as given above
Beam vibration
Differential equation
∂2
∂4
EI 4 w(x, t) + m 2 w(x, t) = q(x, t)
∂x
∂t
EI = constant bending stiffness
m = beam mass per metre (kg/m)
t = time
Assume solution w(x,t) = X(x)⋅T(t). Then the standing wave solution is
T(t) = e iωt and X(x) = C1 cosh (µx) + C2 cos (µx) + C3 sinh (µx) + C4 sin (µx)
where µ4 = ω2m /EI
Boundary conditions (as given above) give an eigenvalue problem that provides the
eigenfrequencies and eigenmodes (eigenforms) of the vibrating beam
9
Axially loaded beam, stability, the Euler cases
Beam axially loaded in tension
Differential equation
d4
d2
EI 4 w(x) − N 2 w(x) = q(x)
N = normal force in tension (N > 0)
dx
dx
Solution
w(x) = wpart(x) + whom(x) where

√

√

√


N
N 
N 
x + C3 sinh 
x + C4 cosh 
x
EI
 EI 
 EI 
New boundary condition on shear force (other boundary conditions as given above)
d3
d
T(*) = − EI 3 w(*) + N w(*)
dx
dx
whom(x) = C1 + C2
Beam axially loaded in compression
Differential equation
d2
d4
EI 4 w(x) + P 2 w(x) = q(x)
dx
dx
P = normal force in compression (P > 0)
Solution
w(x) = wpart(x) + whom(x) where

√

√

√


P
P 
P 
x + C3 sin 
x + C4 cos 
x
EI
 EI 
 EI 
New boundary condition on shear force (other boundary conditions as given above)
d3
d
T(*) = − EI 3 w(*) − P w(*)
dx
dx
whom(x) = C1 + C2
Elementary cases: the Euler cases (Pc is critical load)
Case 1
P
P
L, EI
Pc =
Case 2a
Pc =
Case 3
P
P
L, EI
π2EI
4L 2
Case 2b
L, EI
L, EI
π2EI
L2
Pc =
10
π2EI
L2
Pc =
2.05 π2EI
L2
Case 4
P
L, EI
Pc =
4π2EI
L2
5. Bending of Beam
Elementary Cases
Cantilever beam
P
w(x) =
L, EI
x
w(x)
z
PL 3
w(L) =
3EI
M
L, EI
q = Q/L
L, EI
z
ML 2  x 2 
 
2EI  L 2 
w(L) =
ML 2
2EI
w(x) =
qL 4  x 4
x3
x2
 4 − 4 3 + 6 2
24EI  L
L
L 
w(L) =
qL 4
8EI
x
w(x)
q0
L, EI
z
q0 L 4
w(x) =
120EI
x
w(x)
d
ML
w(L) =
dx
EI
d
qL 3
w(L) =
dx
6EI
 x5
x3
x2
 5 − 10 3 + 20 2 
L
L
L 
11 q0 L 4
w(L) =
120EI
q0 L 3
d
w(L) =
dx
8EI
q0 L 4  x 5
x4
x3
x2
w(x) =
 − + 5 4 − 10 3 + 10 2 
120EI  L 5
L
L
L 
q0
L, EI
z
d
PL 2
w(L) =
dx
2EI
w(x) =
x
w(x)
z
PL 3  x 2 x 3 
3 − 
6EI  L 2 L 3 
x
w(x)
q0 L 4
w(L) =
30EI
11
q0 L 3
d
w(L) =
dx
24EI
Simply supported beam
Load applied at x = αL (α < 1), β = 1 − α
L
P
L
w(x) =
+ =1
w(x)
for
x
≤α
L
PL 3 2 2
α β . When α > β one obtains
3EI

1+β
1+β
1 − β2 
wmax = w L
 = w(αL)
3β
3α

3 
x
z
PL 3 
x x3
β (1 − β2) − 3 
6EI 
L L 
w(αL) =
L, EI

√

√
PL 2
d
PL 2
d
w(0) =
α β (1 + β)
w(L) = −
α β (1 + α)
dx
6EI
dx
6EI
MA
MB
w(x) =
MA L MB L
d
w(0) =
+
dx
3 EI 6 EI
x
w(x)
z
L
M
L, EI
w(x) =
L
z
L, EI
x
w(x)
w(L/2) =
L, EI
w(x) =
Q
w(x)
L, EI
w(x) =
Q
z
w(x)
5 QL 3
384 EI
L, EI
12
x
≤α
L
ML
d
w(L) =
(1 − 3α2)
dx
6EI
d
d
QL 2
w(0) = − w(L) =
dx
dx
24EI
QL 3  x 5
x3
x
3 5 − 10 3 + 7 
180EI  L
L
L
d
8 QL 2
w(L) = −
dx
180EI
QL 3  x 5
x4
x3
x
− 3 5 + 15 4 − 20 3 + 8 
180EI  L
L
L
L
d
8 QL 2
w(0) =
dx
180EI
x
for
 x4
x3 x 
 4−2 3+ 
L
L L
d
7 QL 2
w(0) =
dx
180EI
x
z
ML 2 
x x3
(1 − 3β2) − 3 
6EI 
L L 
QL 3
w(x) =
24EI
Q
z
MA L MB L
d
w(L) = −
−
dx
6 EI 3 EI
ML
d
w(0) =
(1 − 3β2)
dx
6EI
x
w(x)
 x x3 
x2 x3
L2   x
MA 2 − 3

+
M
− 
+
B
6EI   L
 L L 3 
L 2 L 3
d
7 QL 2
w(L) = −
dx
180EI
Clamped simply supported beam and clamped clamped beam
Load applied at x = αL (α < 1), β = 1 − α
Only redundant reactions are given. For deflections, use superposition of solutions
for simply supported beams.
PL
P
+ =1
MA
MA =
β (1 − β2 )
L
L
2
x
z
L, EI
MA
MB
MA =
MB
2
MA =
M
(1 − 3β2 )
2
MA =
QL
8
MA =
2 QL
15
x
z
MA
L, EI
M
L
L
+ =1
x
z
L, EI
Q
MA
x
L, EI
z
Q
MA
MA
P
L
z
MA
L
L, EI
MB
MA = PL α β2
x
L, EI
L
MB
MA = − M β (1 − 3α )
x
+ =1
Q
MA = MB =
MB
MA
z
z
L, EI
QL
12
x
L, EI
Q
MA
MB = PL α2 β
+ =1
M
L
z
x
L, EI
z
MA =
MB
x
13
QL
10
MB =
QL
15
MB = M α (1 − 3β )
6. Material Fatigue
Fatigue limits (notations)
Load
Alternating
Pulsating
Tension/compression
± σu
σup ± σup
Bending
± σub
σubp ± σubp
Torsion
± τuv
τuvp ± τuvp
The Haigh diagram
a
σa = stress amplitude
σm = mean stress
σY = yield limit
σU = ultimate strenght
σu, σup = fatigue limits
λ, δ, κ = factors reducing fatigue limits
(similar diagrams for σub, σubp and τuv, τuvp)
Y
u
up
u
m
up
up
Y
U
Factors reducing fatigue limits
Surface finish κ
1.0
0.8
0.6
0.4
0.2
Factor κ reducing the fatigue limit due to
surface irregularities
(a)
(b)
(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
(c)
(d)
(e)
(f)
(g)
300 600 900 1200 MPa
U
14
Volume factor λ (due to process)
Factor λ reducing the fatigue limit due to
size of raw material
1.0
(a) diameter at circular cross section
(b) thickness at rectangular cross section
0.8
(a) 20
(b) 10
40
20
60
30
80 100mm
40 50mm
Volume factor δ (due to geometry)
1.0
Factor δ reducing the fatigue limits σub and
τuv due to loaded volume.
Steel with ultimate strength σU =
(a) 1500 MPa
(b) 1000 MPa
(c) 600 MPa
(d) 400 MPa
(e) aluminium
Factor δ = 1 when fatigue notch factor Kf >
1 is used.
(a)
(b)
0.9
(c)
(d)
(e)
0.8
0
40 80 120
Diameter or thickness in mm
Fatigue notch factor Kf (at stress concentration)
Kt = stress concentration factor (see Section
Kf = 1 + q (Kt − 1)
12.8)
q = fatigue notch sensitivity factor
Fatigue notch sensitivity factor q
q
1.0
Fatigue notch sensitivity factor q for steel
(a)
with ultimate strength σU =
(a) 1600 MPa
0.8 (b)
(b) 1300 MPa
0.6 (c)
(c) 1000 MPa
(d)
(d) 700 MPa
0.4 (e)
(e) 400 MPa
0.2
0.1
0.5 1 2 5 10
Fillet radius r in mm
15
Wöhler diagram
a
σai = stress amplitude
Ni = fatigue life (in cycles) at stress
amplitude σai
ai
0 1 2 3 4 5 6 7
log N
Damage accumulation D
D=
ni = number of loading cycles at stress
amplitude σai
Ni = fatigue life at stress amplitude σai
ni
Ni
Palmgren-Miner’s rule
Failure when
I n
i
∑ =1
i = 1 Ni
ni = number of loading cycles at stress
amplitude 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 real
design, data should be taken from latest official standard and not from this table.1
Material
Carbon steel
141312-00
141450-1
141510-00
141550-01
141650-01
141650
Tension
alternating pulsating
MPa
MPa
± 110
± 140
± 230
± 180
± 200
Bending
alternating pulsating
MPa
MPa
Torsion
alternating pulsating
MPa
MPa
110 ± 110
130 ± 130
± 170
± 190
150 ± 150
170 ± 170
± 100
± 120
100 ± 100
120 ± 120
160 ± 160
180 ± 180
± 240
± 270
± 460
210 ± 210
240 ± 240
± 140
± 150
140 ± 140
150 ± 150
Stainless steel 2337-02, σu = ± 270 MPa
Aluminium SS 4120-02, σub = ± 110 MPa; SS 4425-06, σu = ± 120 MPa
1
Data in this table has been collected from B Sundström (editor): Handbok och Formelsamling i
Hållfasthetslära, Institutionen för hållfasthetslära, KTH, Stockholm, 1998.
16
7. Multi-Axial Stress States
Stresses in thin-walled circular pressure vessel
σt = circumferential stress
R
R
σt = p
and σx = p
(σz ≈ 0) σ = longitudinal stress
x
t
2t
p = internal pressure
R = radius of pressure vessel
t = wall thickness (t << R)
Rotational symmetry in structure and load (plane stress, i.e. σz = 0)
Differential equation for rotating circular plate
u = u(r) = radial displacement
d2 u 1 d u u
1 − ν2 2
−
ρω
+
=
−
r
ρ = density
E
d r2 r d r r2
ω = angular rotation (rad/s)
Solution
B0 1 − ν2
u (r) = uhom + upart = A0 r + −
ρ ω2 r 3
r
8E
Stresses
σr (r) = A −
where
B 3+ν
−
ρ ω2 r 2
2
8
r
A=
E A0
1−ν
and
σφ(r) = A +
and
B=
B 1 + 3ν
−
ρ ω2 r 2
2
8
r
E B0
1+ν
Boundary conditions
σr or u must be known on inner and outer boundary of the circular plate
Shrink fit
δ = difference of radii
p = contact pressure
u = radial displacement as function of p
δ = uouter(p) − uinner(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) = σx cos2(α) + σ y sin2(α) + 2τxy cos(α)sin(α)
y
τ(α) = − (σx − σ y ) sin(α)cos(α) + τxy (cos (α) − sin (α))
2
( )
2
σ(α) = normal stress in direction α
τ(α) = shear stress on surface with normal in direction α
17
( )
x
Principal stresses σ1, 2 and principal directions at plane stress state
σ1, 2 = σc ± R =
sin(2ψ1) =
τxy
R
σx + σ y
±
2

√
 σx − σ y  2 2

 + τxy
 2 
or cos(2ψ1) =
ψ1 = angle from x axis (in xy plane) to
direction of principal stress σ1
σx − σ y
2R
Strain in direction α (plane state)
ε(α) = εx cos2(α) + εy sin2(α) + γxy sin(α)cos(α)
γ(α) = (εy − εx ) sin(2α) + γxy cos(2α)
ε(α) = normal strain in direction α
γ(α) = shear strain of element with normal in direction α
y
x
Principal strains and principal directions (plane state)
ε1, 2 = εc ± R =
sin(2ψ1) =
εx + εy
±
2
γxy
2R

√
 εx − εy  2  γxy  2

 + 
 2   2
or cos(2ψ1) =
ψ1 = angle from x axis (in xy plane) to
direction of principal strain ε1
εx − εy
2R
Principal stresses and principal directions at three-dimensional stress state
The determinant
 σx τxy τxz


| S−σI| =0
Stress matrix S = τyx σ y τyz


gives three roots (the principal stresses)
 τzx τzy σz
(contains the nine stress components σij)
1 0 0
Unit matrix I = 0 1 0
0 0 1
Direction of principal stress σi (i = 1, 2, 3) is given by
nix, niy and niz are the elements of the unit
(S − σi I) ⋅ ni = 0
and
vector ni in the direction of σi
T
( T means transpose)
ni ⋅ ni = 1
18
Principal strains and principal directions at three-dimensional stress state
Use shear strain εij = γij / 2 for i ≠ j
The determinant
| E−εI| =0
gives three roots (the principal strains)
 εx

Strain matrix E = εyx

 εzx
εxy
εy
εzy
εxz
εyz

εz 
I = unit matrix
Direction of principal strain εi (i = 1, 2, 3) is given by
nix, niy and niz are the elements of the unit
(E − εi I) ⋅ ni = 0
and
vector ni in the direction of εi
T
T
(
means transpose)
ni ⋅ ni = 1
Hooke’s law, including temperature term (three-dimensional stress state)
εx =
1
[σ − ν(σy + σz )] + α ∆T
E x
εy =
1
[σ − ν(σz + σx )] + α ∆T
E y
εz =
1
[σ − ν(σx + σ y )] + α ∆T
E z
γxy =
τxy
G
γyz =
τyz
G
γzx =
α = temperature coefficient
∆T = change of temperature (relative to
temperature giving no stress)
τzx
G
Effective stress
The Huber-von Mises effective stress (the deviatoric stress hypothesis)
σvM
=√
σ2x + σ2y + σ2z − σx σ y − σ y σz − σz σx + 3τ2xy + 3τ2yz + 3τ2zx

e
=

√
1
{(σ1 − σ2)2 + (σ2 − σ3)2 + (σ3 − σ1)2 }
2
The Tresca effective stress (the shear stress hypothesis)
σTe = max [ | σ1 − σ2 | , | σ2 − σ3 | , | σ3 − σ1 | ] = σprmax − σprmin
19
(pr = principal stress)
8. Energy Methods
the Castigliano Theorem
Strain energy u per unit of volume
Linear elastic material and uni-axial stress
σε
u=
2
Total strain energy U in beam loaded in tension/compression, torsion, bending, and
shear
L
Mt(x)2 Mbend(x)2
N(x)2
T(x)2 
 dx
Utot = ⌠ 
+
+
+β
⌡0  2EA(x) 2GKv(x) 2EI(x)
2GA(x)
Mt = torque = Mx
Mbend = bending moment = My
Cross section
β
Kv = section factor of torsional stiffness
β = shear factor, see below
µ
Shear factor β
6/5
3/2
2
A ⌠  SA’ 
β = 2   dA
I ⌡A  b 
10/9
4/3
2
2
A/A web
β is given for some cross sections in the
table (µ is the Jouravski factor, see Section
12.3 One-Dimensional Bodies)
A/A web
Elementary case: pure bending
M1
Only bending momentet Mbend is present.
The moment varies linearly along the beam
with moments M1 and M2 at the beam ends.
One has
Mbend(x) = M1 + (M2 − M1)x /L, which gives
L
{M12 + M1 M2 + M22 }
Utot =
6EI
The second term is negative if M1 and M2
have different signs
M2
L, EI
x
M1
M
M2
The Castigliano theorem
δ=
∂U
∂P
and Θ =
δ = displacement in the direction of force P
of the point where force P is applied
Θ = rotation (change of angle) at moment
M
∂U
∂M
20
9. Stress Concentration
Tension/compression
Maximum normal stress at a stress concentration is σmax = Kt σnom, where Kt and σnom
are given in the diagrams
Kt
3.0
r
P
P
B
thickness h
P
=
nom
bh
B/b
2.0
1.5
1.2
1.1
1.05
1.01
2.0
1.5
0.1
0.2
r/b
Tension of flat bar with shoulder fillet
P
P
P
b
2.5
thickness h
=
nom
2.0
P
bh
B/b
2.0
1.2
1.1
1.05
1.01
1.5
1.0
0
0.1
0.2
0.3
0.4 0.5 0.6 r/b
r
Kt
3.0
P
P
d
D
P
B
Tension of flat bar with notch
r
Kt
3.0
r
b
2.5
1.0
0
Kt
3.0
D
d
2.5
2.5
nom
2.0
=
4P
d2
nom
D/d
2.0
1.5
1.2
1.1
1.05
1.01
1.5
1.0
0
0.1
0.2
2.0
D/d
1.5
1.2
1.1
1.05
1.01
1.0
0.3 r/d
Tension of circular bar with shoulder
fillet
= 4 P2
d
0
0.1
0.2
0.3 0.4
0.5 0.6 r/d
Tension of circular bar with U-shaped
groove
21
a
2r
0
0
B
Kt
B/a= 5
3.2
3.0
3
2.8
2.6
2.5
2.4
2.25
B
2.2
nom = B - 2 r
2
0
2.0
0
0.1
0.2
0.3
0.4
r/a
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
d
d
Kt
Kt
Mb
Mb
Mb
Mb
3.0
3.0
D
B
2.5
thickness h
d/h = 0
0.25
0.5
2.0
1.0
2.0
1.5
nom
=
2.5
=
nom
2.0
Mb
D 3 d D2
32
6
1.5
6 Mb
(B-d)h
2
1.0
1.0
0
0.2
0.4
Bending of flat bar with hole
0.6 d/B
0
0.1
0.2
Bending of circular bar with hole
22
0.3 d/D
r
Kt
3.0
Mb
Mb
B
2.0
=
1.0
0.1
hb 2
0.2
r
Mb
3.0
2.5
thickness h
Mb
hb 2
1.0
0
0.1
0.2
0.3
0.4
0.5
0.6 r/b
Bending of flat bar with notch
Kt
r
Mb
Mb
3.0
D
2.5
B/b
1.2
1.1
1.05
1.01 1.02
1.5
d
D
6 Mb
=
nom
2.0
r/b
Bending of flat bar with shoulder fillet
Kt
b
6 Mb
B/b
6.0
2.0
1.2
1.05
1.01 1.02
1.5
Mb
B
thickness h
nom
r
Mb
b
2.5
0
Kt
3.0
d
2.5
=
nom
2.0
32 M b
d3
2.0
D/d
6.0
2.0
1.2
1.05
1.01 1.02
1.5
1.0
0
0.1
0.2
=
nom
d3
D/d
1.5
1.2
1.05
1.01 1.02
1.0
0.3 r/d
Bending of circular bar with shoulder
fillet
32M b
0
0.1
0.2 0.3 0.4
0.5 0.6 r/d
Bending of circular bar with U-shaped
groove
23
Torsion
Maximium shear stress at stress concentration is τmax = Kt τnom, where Kt and τnom are
given in the diagrams
r
Kt
3.0
r
Kt
Mv
Mv
3.0
Mv
Mv
d
D
D
d
2.5
2.5
nom
=
16 M v
d3
nom
2.0
=
2.0
D/d
2.0
1.3
1.2
1.1
1.5
1.0
0
0.1
0.2
1.5
1.2
1.05
1.01
1.0
0
0.1
0.2
0.3
0.4 0.5 0.6 r/d
Torsion of circular bar with notch
r
Kt
4.0
D/d
0.3 r/d
Torsion of circular bar with shoulder
fillet
16 M v
d3
d
Kt
2.0
Mv
Mv
D
d /4
3.5
7d
8
nom
=
1.5
3.0
Mv
D 3 d D2
16
6
d
2.5
nom =
16 M v
d3
2.0
1.0
0
0.05
0.10
r/d
Torsion of bar with longitudinal keyway
0
0.1
0.2
Torsion of circular bar with hole
24
0.3 d/D
10. Material data
The following material properties may be used only when solving exercises. For a real
design, data should be taken from latest official standard and not from this table (two values
for the same material means different qualities).1
Material
Young’s
modulus
E
GPa
ν
Carbon steel
141312-00
206
0.3
141450-1
205
0.3
141510-00
205
0.3
141550-01
205
0.3
141650-01
206
0.3
141650
206
0.3
α106
Ultimate
strength
K-1
MPa
12
360
460
430
510
510
640
490
590
590
690
860
11
Yield limit
tension/
compression
MPa
bending
torsion
MPa
MPa
>240
260
140
>250
290
160
>270
360
190
>310
390
220
>550
610
>320
Offset yield strength Rp0.2 (σ0,2)
Stainless steel
2337-02
Aluminium
SS 4120-02
196
0.29 16.8
70
23
SS 4120-24
70
23
SS 4425-06
70
23
>490
>200
170
215
220
270
>340
>65
1
>170
>270
Data in this table has been collected from B Sundström (editor): Handbok och Formelsamling i
Hållfasthetslära, Institutionen för hållfasthetslära, KTH, Stockholm, 1998.
25