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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 Page
- Definitions and notations 1
- Stress, Strain, and Material Relations 2
- Geometric Properties of Cross-Sectional Area 3
- One-Dimensional Bodies (bars, axles, beams) 5
- Bending of Beam Elementary Cases 11
- Material Fatigue 14
- Multi-Axial Stress States 17
- Energy Methods the Castigliano Theorem 20
- Stress Concentration 21
- Material data 25
Version 03-09-
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 the geometric centre of cross section) and positive section forces and moments: normal force N , shear forces T (^) y and T (^) z , torque M (^) x , and bending moments M (^) y , 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 ) σ i N/m^2 Shear stress in direction j on surface with normal direction i τ ij N/m^2 Normal strain in direction i ε i Shear strain (corresponding to shear stress τ ij ) γ ij 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 , T (^) i N Load q ( x ) N/m Cross-sectional area A m^2 Length L , L 0 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 m^4 Modulus of elasticity (Young’s modulus) E N/m^2 Poisson’s ratio ν Shear modulus G N/m^2 Bulk modulus K N/m^2
Temperature coefficient α
x y z
T
T
M M
N
y z
y z
Mx
Tz
M (^) z M (^) y
Mx N
Ty
L
A
q x ( )
K−^1
3. Geometric Properties of Cross-Sectional Area
The origin of the coordinate system O yz is at the geometric centre of the cross section
Cross-sectional area A
d A = area element
Geometric centre (centroid)
e = ζgc = distance from η axis to geometric centre f = ηgc distance from ζ axis to geometric centre
First moment of area
A ’ = the “sheared” area (part of area A )
Second moment of area
Iy = second moment of area with respect to the y axis Iz = second moment of area with respect to the z axis Iyz = second moment of area with respect to the y and z axes
Parallel-axis theorems
First moment of area
Second moment of area
O y
z
e
f
d A
A = ⌠
⌡ A
d A
e ⋅ A = ⌠ ⌡ A ζ d A
f ⋅ A = ⌠ ⌡ A η d A
S (^) y = ⌠ ⌡ A ’
z d A and Sz = ⌠ ⌡ A ’
y d A
I (^) y = ⌠ ⌡ A z^2 d A
Iz = ⌠ ⌡ A
y^2 d A
I (^) yz = ⌠ ⌡ A yz d A
S η = ⌠ ⌡ A
( z + e ) d A = eA and S ζ = ⌠ ⌡ A
( y + f ) d A = fA
I η = ⌠ ⌡ A
( z + e )^2 d A = I (^) y + e^2 A , I ζ = ⌠ ⌡ A
( y + f )^2 d A = Iz + f^2 A ,
I ηζ = ⌠ ⌡ A ( z + e ) ( y + f ) d A = I (^) yz + efA
Rotation of axes
Coordinate system Ωηζ has been rotated the angle α with respect to the coordinate system O yz
Principal moments of area
I 1 + I 2 = I (^) y + I (^) z
Principal axes
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)
Rectangular area, base B , height H
Solid circular area, diameter D
Thick-walled circular tube, diameters D and d
y
z
z
y
d A
I η = ⌠ ⌡ A
ζ^2 d A = I (^) y cos^2 α + Iz sin^2 α − 2 I (^) yz sin α cos α
I ζ = ⌠ ⌡ A η^2 d A = I (^) y sin^2 α + Iz cos^2 α + 2 I (^) yz sin α cos α
I ηζ = ⌠ ⌡ A
ζη d A = ( I (^) y − Iz ) sin α cos α + I (^) yz (cos^2 α − sin^2 α ) =
I (^) y − Iz 2
sin 2α + I (^) yz cos 2α
I 1 , 2 =
I (^) y + Iz 2
± R where R =
I (^) y − Iz 2
2
sin 2α =
− I (^) yz R
or cos 2α =
I (^) y − Iz 2 R
y
z
H
B I (^) y =
BH^3
and Iz =
HB^3
y
z
D I (^) y = Iz =
π D^4 64
y
z
d D I (^) y = Iz = π 64
( D^4 − d^4 )
Section modulus W v and section factor K v for some cross sections (at torsion)
Torsion of thin-walled circular tube, radius R , thickness t , where t << R ,
Thin-walled tube of arbitrary cross section A = area enclosed by the tube t ( s ) = wall thickness s = coordinate around the tube
Thick-walled circular tube, diameters D and 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 2 a and minor axle 2 b
Solid axle with rectangular cross section b by a , where b ≥ a
for k Wv and k Kv, see table below
y
z
R t
W v = 2 π R^2 t K v = 2 π R^3 t
t(s)
(s)
Area A
s
W v = 2 At min K v =
4 A^2
⌡ s [ t ( s )] −^1 d s
y
z
d D
W v =
π 16
D^4 − d^4 D
K v =
π 32
( D^4 − d^4 )
y
z
D
W v =
π D^3 16
K v =
π D^4 32
y
z
a a / 2 a / 2 W v =^
a^3 20
K v =
a^4 √ 3
y
z
2 a
2 b W v =
π 2
a b^2 K v =
π a^3 b^3 a^2 + b^2
y
z
a
b
W v = k Wv a^2 b K v = k Kv a^3 b
Factors k Wv and k Kv for some values of ratio b / a (solid rectangular cross section)
b / a k Wv k Kv
Bending of beam
Relationships between bending moment M (^) y = M ( x ), shear force T (^) z = T ( x ), and load q ( x ) on beam
Normal stress I (here I (^) y ) = second moment of area (see Section 12.2)
Maximum bending stress
W b = section modulus (in bending)
Shear stress SA’ = first moment of area A ’ (see Section 12.2) b = length of line limiting area A ’ τgc = shear stress at geometric centre μ = the Jouravski factor The Jouravski factor μ for some cross sections
rectangular 1. triangular 1. circular 1. thin-walled circular 2. elliptical 1. ideal I profile A / A web
d T ( x ) d x
= − q ( x ) ,
d M ( x ) d x
= T ( x ) , and
d^2 M ( x ) d x^2
= − q ( x )
σ =
N
A
Mz I
| σ | (^) max =
| M |
Wb
where Wb =
I
| z | (^) max
τ =
TSA ’
Ib
τgc = μ
T
A
Non-homogeneous boundary conditions (a) Displacement δ prescribed
(b) Slope Θ prescribed
(c) Moment M 0 prescribed
(d) Force P prescribed
Beam on elastic bed
Differential equation
EI = constant bending stiffness k = bed modulus (N/m 2 ) Solution
Boundary conditions as given above
Beam vibration
Differential equation 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
where μ^4 = ω^2 m / EI Boundary conditions (as given above) give an eigenvalue problem that provides the eigenfrequencies and eigenmodes (eigenforms) of the vibrating beam
w (*) = δ
x x=L
x (^) x=L
O
z O
M 0 M 0
P x^ x=L P
(a)
(b)
(c)
(d)
d d x
w (*) = Θ
− EI
d^2 d x^2
w (*) = M 0
− EI
d^3 d x^3
w (*) = P
EI
d^4 d x^4
w ( x ) + kw ( x ) = q ( x )
w ( x ) = w part( x ) + w hom( x ) where
w hom( x ) = { C 1 cos (λ x ) + C 2 sin (λ x )} eλ x^ + { C 3 cos (λ x ) + C 2 sin (λ x )} e − λ x^ ; λ^4 =
k 4 EI
EI
∂^4
∂ x^4
w ( x , t ) + m
∂^2
∂ t^2
w ( x , t ) = q ( x , t )
T ( t ) = e i ω t^ and X ( x ) = C 1 cosh (μ x ) + C 2 cos (μ x ) + C 3 sinh (μ x ) + C 4 sin (μ x )
Axially loaded beam, stability, the Euler cases
Beam axially loaded in tension Differential 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 compression Differential 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 ( P c is critical load)
Case 1 Case 2a Case 2b Case 3 Case 4
EI
d^4 d x^4
w ( x ) − N
d^2 d x^2
w ( x ) = q ( x )
w ( x ) = w part( x ) + w hom( x ) where
w hom( x ) = C 1 + C 2 √
N
EI
x + C 3 sinh
√
N
EI
x
√
N
EI
x
T (*) = − EI
d 3 d x^3
w (*) + N d d x
w (*)
EI
d^4 d x^4
w ( x ) + P d^2 d x^2
w ( x ) = q ( x )
w ( x ) = w part( x ) + w hom( x ) where
w hom( x ) = C 1 + C 2 √
P
EI
x + C 3 sin
√
P
EI
x
√
P
EI
x
T (*) = − EI
d 3 d x^3
w (*) − P
d d x
w (*)
P
L, EI L, EI
P
L, EI
P
L, EI
P
L, EI
P
P c =
π^2 EI 4 L^2
P c =
π^2 EI L^2
P c =
π^2 EI L^2
P c =
2.05 π^2 EI L^2
P c =
4 π^2 EI L^2
Simply supported beam
Load applied at x = α L (α < 1), β = 1 −α
w ( x ) =
PL^3
6 EI
β
( 1 − β^2 )
x L
x^3 L^3
for
x L
≤ α
x z w(x) L, EI
L L
P +^ = 1
w (α L ) =
PL^3 3 EI
α^2 β^2. When α > β one obtains
w max = w
L
√
1 − β^2 3
= w (α L ) 1 + β 3 β √
1 + β 3 α
d d x w (^0 ) =^
PL^2 6 EI α β (^1 + β)^
d d x w ( L ) = −^
PL^2 6 EI α β (^1 + α)
x z w(x)^ L, EI
M A M B (^) w ( x ) = L^2 6 EI
M A
2
x L − 3
x^2 L^2
x^3 L^3
x L −
x^3 L^3
d d x w ( 0 ) =
M A L 3 EI
M B L 6 EI
d d x w ( L ) = −
M A L 6 EI −
M B L 3 EI
w ( x ) =
ML^2
6 EI
( 1 − 3 β^2 )
x L
x^3 L^3
for
x L
≤ α
x z w(x) L, EI
L L
M
d d x w ( 0 ) = ML 6 EI ( 1 − 3 β^2 ) d d x w ( L ) = ML 6 EI ( 1 − 3 α^2 )
w ( x ) =
QL^3
24 EI
x^4 L^4
x^3 L^3
x L
x z w(x)^ L, EI
Q
w ( L / 2 ) =
5 QL^3 384 EI
d d x w ( 0 ) = −
d d x w ( L ) =
QL^2 24 EI
w ( x ) =
QL^3
180 EI
x^5 L^5
x^3 L^3
x L
d d x
w ( 0 ) =
7 QL^2
180 EI
d d x
w ( L ) = −
8 QL^2
180 EI
x z w(x) L, EI
Q
w ( x ) =
QL^3
180 EI
x^5 L^5
x^4 L^4
x^3 L^3
x L
d d x
w ( 0 ) =
8 QL^2
180 EI
d d x
w ( L ) = −
7 QL^2
180 EI
x z w(x) L, EI
Q
Clamped simply supported beam and clamped clamped beam
Load applied at x = α L (α < 1), β = 1 α Only redundant reactions are given. For deflections, use superposition of solutions for simply supported beams.
x z L, EI
M A
L L
P + = 1
M A =
PL
β ( 1 − β^2 )
x z L, EI
M A M B
M A =
M B
x z L, EI
M A M
L L
+ = 1 M
A =^
M
( 1 − 3 β^2 )
x z
L, EI
M A
Q
M A =
QL
x z L, EI
M A
Q M
A =^
2 QL
x
z
L, EI
M A
L L
P M B M A =^ PL^ α β
2 M
B =^ PL^ α
(^2) β
x
z
L, EI
M A M
L L
M B M A = −^ M^ β (^1 −^3 α )^ M B =^ M^ α (^1 −^3 β )
x z L, EI
M A
Q
M B M A =^ M B =^
QL
x z L, EI
M A Q^ M B M A =^
QL
M B =
QL
Volume factor λ (due to process) Factor λ reducing the fatigue limit due to size 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 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 K f > 1 is used.
Fatigue notch factor K (^) f (at stress concentration)
K t = stress concentration factor (see Section 12.8) q = fatigue notch sensitivity factor
Fatigue notch sensitivity factor q
Fatigue notch sensitivity factor q for steel with ultimate strength σU = (a) 1600 MPa (b) 1300 MPa (c) 1000 MPa (d) 700 MPa (e) 400 MPa
(a)
(b)
100mm
50mm
(a)
(b)
(c)
(d)
(e)
Diameter or thickness in mm
K f = 1 + q ( K t − 1 )
(a)
(b)
(c)
(d)
(e)
q
r
Fillet radius in mm
Wöhler diagram
σa i = stress amplitude Ni = fatigue life (in cycles) at stress amplitude σa i
Damage accumulation D
ni = number of loading cycles at stress amplitude σa i Ni = fatigue life at stress amplitude σa i
Palmgren-Miner’s rule
Failure when ni = number of loading cycles at stress amplitude σa i Ni = fatigue life at stress amplitude σa i 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 Tension Bending Torsion alternating pulsating alternating pulsating alternating pulsating MPa MPa MPa MPa MPa MPa
Carbon steel 141312-00 110 110 110 170 150 150 100 100 100 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 i Hållfasthetslära, Institutionen för hållfasthetslära, KTH, Stockholm, 1998.
a
log N
a i
D =
ni Ni
i = 1
I (^) ni Ni
σu = ± 270 MPa
σub = ± 110 MPa σu = ± 120 MPa
Principal stresses and principal directions at plane stress state
ψ 1 = angle from x axis (in xy plane) to direction of principal stress σ 1
Strain in direction α (plane state)
ε(α) = normal strain in direction α γ(α) = shear strain of element with normal in direction α
Principal strains and principal directions (plane state)
ψ 1 = angle from x axis (in xy plane) to direction of principal strain ε 1
Principal stresses and principal directions at three-dimensional stress state
The determinant
gives three roots (the principal stresses)
(contains the nine stress components σ ij )
Direction of principal stress σ i ( i = 1, 2, 3) is given by n (^) ix , niy and niz are the elements of the unit and vector n i in the direction of σ i ( T^ means transpose)
σ 1 , 2
σ 1 , 2 = σc ± R =
σ x + σ y 2
σ x − σ y 2
2
sin( 2 ψ 1 ) =
τ xy R
or cos( 2 ψ 1 ) =
σ x − σ y 2 R
ε(α) = ε x cos^2 (α) + ε y sin^2 (α) + γ xy sin(α)cos(α) y
γ(α) = (ε x y − ε x )^ sin(^2 α) + γ xy cos(^2 α)
ε 1 , 2 = εc ± R =
ε x + ε y 2 ±
ε x − ε y 2
2
γ xy 2
2
sin( 2 ψ 1 ) =
γ xy 2 R
or cos( 2 ψ 1 ) =
ε x − ε y 2 R
Stress matrix S =
σ x τ xy τ xz τ yx σ y τ yz τ zx τ zy σ z
| S − σ I | = 0
Unit matrix I =
( S − σ i I ) ⋅ n i = 0
n i^ T^ ⋅ n i = 1
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 ε i ( i = 1, 2, 3) is given by nix , niy and niz are the elements of the unit and vector n i in the direction of ε i ( T^ means transpose)
Hooke’s law, including temperature term (three-dimensional stress state)
α = temperature coefficient ∆ T = change of temperature (relative to temperature giving no stress)
Effective stress
The Huber-von Mises effective stress (the deviatoric stress hypothesis)
The Tresca effective stress (the shear stress hypothesis)
ε ij = γ ij / 2 for i ≠ j Strain matrix E =
ε x ε xy ε xz ε yx ε y ε yz ε zx ε zy ε z
| E − ε I | = (^0)
( E − ε i I ) ⋅ n i = 0
n i^ T^ ⋅ n i = 1
εx =
E
[σx − ν(σy + σ z )] + α ∆ T
εy =
E
[σy − ν(σz + σ x )] + α ∆ T
εz =
E
[σz − ν(σx + σ y )] + α ∆ T
γ xy =
τ xy G
γ yz =
τ yz G
γ zx =
τ zx G
σevM^ = (^) √σ x^2 + σ y^2 + σ z^2 − σ x σ y − σ y σ z − σ z σ x + 3 τ xy^2 + 3 τ yz^2 + 3 τ zx^2
√
{(σ 1 − σ 2 )^2 + (σ 2 − σ 3 )^2 + (σ 3 − σ 1 )^2 }
σeT^ = max [ | σ 1 − σ 2 | , | σ 2 − σ 3 | , | σ 3 − σ 1 | ] = σmaxpr^ − σminpr^ (pr = principal stress)
