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 1. Definitions and notations 1 2. Stress, Strain, and Material Relations 2 3. Geometric Properties of Cross-Sectional Area 3 4. One-Dimensional Bodies (bars, axles, beams) 5 5. Bending of Beam Elementary Cases 11 6. Material Fatigue 14 7. Multi-Axial Stress States 17 8. Energy Methods the Castigliano Theorem 20 9. Stress Concentration 21 10. Material data 25 Version 03-09-18
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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)
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)
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.
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
1 Data in this table has been collected from B Sundström (editor): Handbok och Formelsamling iHållfasthetslära, Institutionen för hållfasthetslära, KTH, Stockholm, 1998.
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
1 Data in this table has been collected from B Sundström (editor): Handbok och Formelsamling iHållfasthetslära, Institutionen för hållfasthetslära, KTH, Stockholm, 1998.