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 1. Definitions and notations
Definition of coordinate system and loadings on beam
Mz qx( ) My Tz N Mx Ty x M N y Ty x T z A z M y M L z
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 σ 2 Normal stress in direction i (= x, y, z) i N/m τ 2 Shear stress in direction j on surface with normal direction i ij N/m ε Normal strain in direction i i τ γ Shear strain (corresponding to shear stress ij) ij rad Moment with respect to axis iM, Mi Nm Normal force N, P N (= kg m/s2)
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 xu, u(x), u(x,y)m Displacement in direction yv, v(x), v(x,y)m Beam deflection w(x)m 4 Second moment of area (i = y, z) I, Ii m 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
σ Normal stress x ∆ ∆N = fraction of normal force N σ = N σ = N x or x lim ∆ A ∆A → 0 ∆A A = cross-sectional area element
τ Shear stress xy (mean value over area A in the y direction) T τ = y (= τ ) xy A mean
ε Normal strain x δ Linear, at small deformations ( << L0) δ ( ) δ = change of length ε = ε = du x x or x L = original length L0 dx 0 u(x) = displacement Non-linear, at large deformations δ L L = actual length (L = L0 + ) ε = x ln L0
γ Shear strain xy ∂ ( , ) ∂ ( , ) γ = u x y + v x y xy ∂y ∂x
Linear elastic material (Hooke’s law) Tension/compression σ ∆T = change of temperatur ε = x +α∆T x E
Lateral strain ε =−νε y x Shear strain τ γ = xy xy G
Relationships between G, K, E and ν E E G = K = 2 ( 1 +ν) 3 ( 1 − 2ν)
2 3. Geometric Properties of Cross-Sectional Area
e The origin of the coordinate system Oyz is O y at the geometric centre of the cross section f
dA z Cross-sectional area A ⌠ dA = area element A = dA ⌡A Geometric centre (centroid) ⌠ e = ζ = distance from η axis to geometric e ⋅ A = ζ dA gc ⌡A centre f = η distance from ζ axis to geometric ⌠ gc f ⋅ A = η dA centre ⌡A First moment of area = ⌠ = ⌠ A’ = the “sheared” area (part of area A) Sy zdA and Sz ydA ⌡A’ ⌡A’ Second moment of area
= ⌠ 2 Iy = second moment of area with respect to Iy z dA ⌡A the y axis
= ⌠ 2 Iz = second moment of area with respect to Iz y dA ⌡A the z axis
= ⌠ Iyz = second moment of area with respect to Iyz yzdA ⌡A the y and z axes Parallel-axis theorems First moment of area ⌠ ⌠ Sη = (z + e) dA = eA and Sζ = (y + f) dA = fA ⌡A ⌡A Second moment of area = ⌠ ( + )2 = + 2 , = ⌠ ( + )2 = + 2 , Iη z e dA Iy e A Iζ y f dA Iz f A ⌡A ⌡A
= ⌠ ( + )( + ) = + Iηζ z e y f dA Iyz efA ⌡A
3 Rotation of axes
y Coordinate system Ωηζ has been rotated α z the angle with respect to the coordinate system Oyz y d A z = ⌠ ζ2 = 2 α+ 2 α− α α Iη dA Iy cos Iz sin 2Iyz sin cos ⌡A
= ⌠ η2 = 2 α+ 2 α+ α α Iζ dA Iy sin Iz cos 2Iyz sin cos ⌡A
I − I = ⌠ ζη =( − ) α α+ ( 2 α− 2 α)= y z α+ α Iηζ dA Iy Iz sin cos Iyz cos sin sin 2 Iyz cos2 ⌡A 2
Principal moments of area I + I I − I 2 I = y z ± R where R = y z + I 2 1,2 2 √ 2 yz
I1 + I2 = Iy + Iz
Principal axes − I I − I A line of symmetry is always a principal sin 2α= yz or cos 2α= y z R 2R axis
Second moment of area with respect to axes through geometric centre for some symmetric areas (beam cross sections)
B Rectangular area, base B, height H y H BH3 HB3 I = and I = y 12 z 12 z Solid circular area, diameter D y D πD 4 I = I = y z 64 z Thick-walled circular tube, diameters D y d D and d π I = I = ( D 4 − d 4 ) z y z 64
4 Thin-walled circular tube, radius R and R t y wall thickness t (t << R) = =π 3 Iy Iz R t z Triangular area, base B and height H H BH3 HB3 y I = and I = B/ 2 B/ 2 y 36 z 48 z
a a a Hexagonal area, side length a 5√3 y I = I = a 4 a a y z 16 z 2a Elliptical area, major axis 2a and minor axis 2b y 2 b πab3 πba3 I = and I = z y 4 z 4
Half circle, radius a (geometric centre at e) a π y 8 4 4 4a e I = − a ≅ 0, 110 a and e = y 8 9π 3π z
4. One-Dimensional Bodies (bars, axles, beams)
Tension/compression of bar Change of length NL N, E, and A are constant along bar δ= or EA L = length of bar
⌠ L ⌠ L N(x) N(x), E(x), and A(x) may vary along bar δ= ε(x)dx = dx ⌡0 ⌡0 E(x)A(x) Torsion of axle Maximum shear stress
M Mv = torque = Mx τ = v max Wv = section modulus in torsion (given Wv below) Torsion (deformation) angle
M L Mv = torque = Mx Θ= v Kv = section factor of torsional stiffness GKv (given below)
5 Section modulus Wv and section factor Kv for some cross sections (at torsion)
Torsion of thin-walled circular tube, radius R t y R, thickness t, where t << R, = π 2 = π 3 Wv 2 R tKv 2 R t z
(s) Thin-walled tube of arbitrary cross section A = area enclosed by the tube t(s) s t(s) = wall thickness Area A s = coordinate around the tube 2 = = 4A Wv 2Atmin Kv ⌠ − [t(s)] 1 ds ⌡s
Thick-walled circular tube, diameters D y d D and d, π D 4 − d 4 π W = K = (D 4 − d 4) z v 16 D v 32
Solid axle with circular cross section, y D diameter D, πD 3 πD 4 W = K = z v 16 v 32
Solid axle with triangular cross section, a side length a y a 3 a 4 √3 a/2 a/2 W = K = z v 20 v 80
2a Solid axle with elliptical cross section, major axle 2a and minor axle 2b y 2 b π π 3 3 = 2 = a b z Wv ab Kv 2 a 2 + b 2
b Solid axle with rectangular cross section b by a, where b ≥ a y a = 2 = 3 Wv kWv a bKv kKv a b 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 / akWv kKv
1.0 0.208 0.1406 1.2 0.219 0.1661 1.5 0.231 0.1958 2.0 0.246 0.229 2.5 0.258 0.249 3.0 0.267 0.263 4.0 0.282 0.281 5.0 0.291 0.291 10.0 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)on beam dT(x) dM(x) d2M(x) =−q(x), = T(x), and =−q(x) dx dx dx 2
Normal stress I (here I ) = second moment of area (see σ=N + Mz y A I Section 12.2) Maximum bending stress | M| I |σ| = where W = max b | | W = section modulus (in bending) Wb z max b
Shear stress
TS SA’ = first moment of area A’ (see Section τ= A’ Ib 12.2) b = length of line limiting area A’ τ τ =µT gc = shear stress at geometric centre gc A µ = the Jouravski factor The Jouravski factor µ for some cross sections
rectangular 1.5 triangular 1.33 circular 1.33 thin-walled circular 2.0 elliptical 1.33
ideal I profile A / Aweb
7 Skew bending Axes y and z are not principal axes: M (zI − yI )−M (yI − zI ) σ=N + y z yz z y yz − 2 Iy, Iz, Iyz = second moment of area A Iy Iz Iyz
Axes y’ and z’ are principal axes: M z’ M y’ σ=N + 1 − 2 I1, I2 = principal second moment of area A I1 I2
Beam deflection w(x) Differential equations d2 d2 EI(x) w(x) = q(x) dx 2 dx 2 when EI(x) is function of x
d4 EI w(x)=q(x) dx 4 when EI is constant
Homogeneous boundary conditions Clamped beam end x x=L d w(*)=0 and w(*)=0 dx where * is the coordinate of beam end (to be entered after differentiation) Simply supported beam end x x=L d2 w(*)=0 and − EI w(*)=0 dx 2
Sliding beam end x x=L d d3 w(*)=0 and − EI w(*)=0 dx dx 3
Free beam end x x=L d2 d3 − EI w(*)=0 and − EI w(*)=0 dx 2 dx 3
8 Non-homogeneous boundary conditions (a) Displacement δ prescribed w(*)=δ x x=L (a) Θ (b) Slope prescribed O d w(*)=Θ (b) dx z O
M00x x=L M (c) Moment M0 prescribed d2 (c) − EI w(*)=M dx 2 0 P x x=L P (d) Force P prescribed (d) d3 − EI w(*)=P dx 3
Beam on elastic bed Differential equation d4 EI w(x)+kw(x)=q(x) dx 4 EI = constant bending stiffness k = bed modulus (N/m2) Solution ( )= ( )+ ( ) w x wpart x whom x where
λ −λ k w (x)={C cos (λx)+C sin (λx)} e x +{C cos (λx)+C sin (λx)} e x ; λ4 = hom 1 2 3 2 4EI Boundary conditions as given above
Beam vibration Differential equation ∂4 ∂2 EI = constant bending stiffness EI w(x, t)+m w(x, t)=q(x, t) ∂x 4 ∂t 2 m = beam mass per metre (kg/m) t = time Assume solution w(x,t)=X(x)⋅T(t). Then the standing wave solution is ( )= iωt ( )= (µ )+ (µ )+ (µ )+ (µ ) T t e and X x C1 cosh x C2 cos x C3 sinh x C4 sin x µ4 ω2 where = m /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 w(x)−N w(x)=q(x) dx 4 dx 2 N = normal force in tension (N >0) Solution ( )= ( )+ ( ) w x wpart x whom x where
N N N w (x)=C + C x + C sinh x + C cosh x hom 1 2√ EI 3 √ EI 4 √ EI New boundary condition on shear force (other boundary conditions as given above) d3 d T(*)=−EI w(*)+N w(*) dx 3 dx Beam axially loaded in compression Differential equation d4 d2 EI w(x)+P w(x)=q(x) dx 4 dx 2 P = normal force in compression (P >0) Solution ( )= ( )+ ( ) w x wpart x whom x where
P P P w (x)=C + C x + C sin x + C cos x hom 1 2√ EI 3 √ EI 4 √ EI New boundary condition on shear force (other boundary conditions as given above) d3 d T(*)=−EI w(*)−P w(*) dx 3 dx
Elementary cases: the Euler cases (Pc is critical load) Case 1 Case 2a Case 2b Case 3 Case 4 P P P P P
L, EI L, EI L, EI L, EI L, EI
π2EI π2EI π2EI 2.05 π2EI 4π2EI P = P = P = P = P = c 4L 2 c L 2 c L 2 c L 2 c L 2
10 5. Bending of Beam Elementary Cases
Cantilever beam P PL3 x 2 x 3 L, EI w(x)= 3 − 6EI L 2 L 3 x z w(x) PL3 d PL2 w(L)= w(L)= 3EI dx 2EI
M ML2 x 2 L, EI w(x)= 2EI L 2 x z w(x) ML2 d ML w(L)= w(L)= 2EI dx EI
q = Q/L qL4 x 4 x 3 x 2 w(x)= − 4 + 6 24EI L 4 L 3 L 2 x L, EI qL4 d qL3 z w(x) w(L)= w(L)= 8EI dx 6EI
4 5 3 2 q q0 L x x x 0 w(x)= − 10 + 20 120EI L 5 L 3 L 2 L, EI x w(x) 4 3 z 11 q0 L d q0 L w(L)= w(L)= 120EI dx 8EI
4 5 4 3 2 q q0 L x x x x 0 w(x)= − + 5 − 10 + 10 120EI L 5 L 4 L 3 L 2 L, EI x w(x) 4 3 z q0 L d q0 L w(L)= w(L)= 30EI dx 24EI
11 Simply supported beam Load applied at x = αL (α < 1), β =1− α PL3 x x 3 x w(x)= β (1 −β2) − for ≤α P + = 1 6EI L L 3 L L L x PL3 w(αL)= α2 β2 . When α>β one obtains z w(x) L, EI 3EI 1 −β2 1 +β 1 +β w = wL = w(αL) max √ 3 3β √ 3α
d PL2 d PL2 w(0)= αβ(1 +β) w(L)=− αβ(1 +α) dx 6EI dx 6EI
L 2 x x 2 x 3 x x 3 MA MB w(x)= M 2 − 3 + + M − 6EI A L L 2 L 3 B L L 3
d MA L MB L d MA L MB L x w(0)= + w(L)=− − z w(x) L, EI dx 3 EI 6 EI dx 6 EI 3 EI
ML2 x x 3 x w(x)= (1 − 3β2) − for ≤α M 6EI L L 3 L L L d ML d ML x w(0)= (1 − 3β2) w(L)= (1 − 3α2) z w(x) L, EI dx 6EI dx 6EI
QL3 x 4 x 3 x Q w(x)= − 2 + 24EI L 4 L 3 L
x 5 QL3 d d QL2 w(L/2)= w(0)=− w(L)= z w(x) L, EI 384 EI dx dx 24EI
QL3 x 5 x 3 x w(x)= 3 − 10 + 7 Q 180EI L 5 L 3 L
x d 7 QL2 d 8 QL2 w(0)= w(L)=− z w(x) L, EI dx 180EI dx 180EI
QL3 x 5 x 4 x 3 x w(x)= − 3 + 15 − 20 + 8 Q 180EI L 5 L 4 L 3 L
x d 8 QL2 d 7 QL2 w(0)= w(L)=− z w(x) L, EI dx 180EI dx 180EI
12 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.
P + = 1 PL 2 MA M = β(1 −β ) L L A 2 x z L, EI
M M MB A B M = A 2 x z L, EI M MA M + = 1 M = (1 − 3β2 ) L L A 2 x z L, EI Q QL M M = A A 8 x z L, EI Q 2 QL M M = A A 15 x z L, EI = αβ2 = α2 β MA P MB MA PL MB PL L L x L, EI z + = 1 =− β( − α) = α( − β) MA M MB MA M 1 3 MB M 1 3 L L x L, EI z + = 1 Q QL M M M = M = A B A B 12 x z L, EI Q QL QL M M M = M = A B A 10 B 15 x z L, EI
13 6. Material Fatigue
Fatigue limits (notations)
Load Alternating Pulsating ±σ σ ±σ Tension/compression u up up ±σ σ ±σ Bending ub ubp ubp ±τ τ ±τ Torsion uv uvp uvp
The Haigh diagram a Y σ a = stress amplitude σ m = mean stress σ u Y = yield limit up σ U = ultimate strenght σ σ u, up = fatigue limits u λ, δ, κ = factors reducing fatigue limits up m σ σ τ τ (similar diagrams for ub, ubp and uv, uvp) up Y U
Factors reducing fatigue limits Surface finish κ Factor κ reducing the fatigue limit due to 1.0 (a) surface irregularities (b) 0.8 (a) polished surface ( κ =1) (c) (b) ground 0.6 (c) machined (d) standard notch 0.4 (d) (e) rolling skin (e) (f) corrosion in sweet water 0.2 (f) (g) corrosion in salt water (g)
300 600 900 1200 MPa U
14 Volume factor λ (due to process) Factor λ reducing the fatigue limit due to 1.0 size of raw material
0.8 (a) diameter at circular cross section (b) thickness at rectangular cross section (a) 20 40 60 80 100mm (b) 10 20 30 40 50mm
Volume factor δ (due to geometry) δ σ Factor reducing the fatigue limits ub and τ uv due to loaded volume. 1.0 σ (a) Steel with ultimate strength U = (b) (a) 1500 MPa 0.9 (b) 1000 MPa (c) (c) 600 MPa 0.8 (d) (d) 400 MPa (e) (e) aluminium δ Factor = 1 when fatigue notch factor Kf > 0 40 80 120 1 is used. Diameter or thickness in mm
Fatigue notch factor Kf (at stress concentration) = + ( − ) Kf 1 q Kt 1 Kt = stress concentration factor (see Section 12.8) q = fatigue notch sensitivity factor Fatigue notch sensitivity factor q q 1.0 Fatigue notch sensitivity factor q for steel σ (a) with ultimate strength U = (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 2510 Fillet radius r in mm
15 Wöhler diagram
a σ ai = stress amplitude ai Ni = fatigue life (in cycles) at stress σ amplitude ai
0 1234567 log N
Damage accumulation D
ni ni = number of loading cycles at stress D = σ N amplitude ai i σ Ni = fatigue life at stress amplitude ai
Palmgren-Miner’s rule
Failure when ni = number of loading cycles at stress I σ ni amplitude ai ∑ = 1 σ Ni = fatigue life at stress amplitude ai i = 1 Ni 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
σ =± Stainless steel 2337-02, u 270 MPa σ =± σ =± Aluminium SS 4120-02,ub 110 MPa ; SS 4425-06, u 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 σ = circumferential stress σ = R σ = R σ ≈ ) t t p and x p ( z 0 σ t 2t x = longitudinal stress p = internal pressure R = radius of pressure vessel t = wall thickness (t << R) σ Rotational symmetry in structure and load (plane stress, i.e. z =0) Differential equation for rotating circular plate d2 u 1 d u u 1 −ν2 u = u(r) = radial displacement + − =− ρω2r ρ d r 2 r d r r 2 E = density ω = angular rotation (rad/s) Solution 2 B0 1 −ν u(r)=u + u = A r + − ρω2 r 3 hom part 0 r 8E
Stresses +ν + ν B 3 2 2 B 1 3 2 2 σ (r)=A − − ρω r and σφ(r)=A + − ρω r r r 2 8 r 2 8
where EA EB A = 0 and B = 0 1 −ν 1 +ν
Boundary conditions σ r or u must be known on inner and outer boundary of the circular plate Shrink fit δ= ( )− ( ) δ uouter p uinner p = difference of radii p = contact pressure u = radial displacement as function of p Plane stress and plane strain (plane state) σ τ τ Plane stress (in xy-plane) when z =0, xz = 0, and yz =0 τ τ ε Plane strain (in xy-plane) when xz =0, yz = 0, and z = 0 or constant Stresses in direction α (plane state) σ(α) = σ cos2(α) + σ sin2(α) + 2τ cos(α)sin(α) y () x y xy () τ(α) = − (σ −σ ) sin(α)cos(α) + τ (cos2(α) − sin2(α)) x y xy x σ(α) = normal stress in direction α τ(α) = shear stress on surface with normal in direction α
17 σ Principal stresses1,2 and principal directions at plane stress state
σ +σ σ −σ 2 σ =σ ± R = x y ± x y +τ2 1,2 c 2 √ 2 xy τ σ −σ ψ xy x y 1 = angle from x axis (in xy plane) to sin(2ψ )= or cos(2ψ )= σ 1 R 1 2R direction of principal stress 1
Strain in direction α (plane state) ε(α) = ε 2(α) + ε 2(α) + γ (α) (α) x cos y sin xy sin cos y γ(α) = (ε −ε) ( α) + γ ( α) x y x sin 2 xy cos 2 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) ε +ε ε −ε 2 γ 2 ε =ε ± R = x y ± x y + xy 1,2 c 2 √ 2 2
γ ε −ε ψ xy x y 1 = angle from x axis (in xy plane) to sin(2ψ )= or cos(2ψ )= ε 1 2R 1 2R direction of principal strain 1
Principal stresses and principal directions at three-dimensional stress state σ τ τ The determinant x xy xz | −σ |= S I 0 Stress matrix S = τ σ τ gives three roots (the principal stresses) yx y yz τ τ σ zx zy z σ (contains the nine stress components ij) 100 Unit matrix I = 010 001
σ Direction of principal stress i (i = 1, 2, 3) is given by
(S −σ I)⋅n = 0 nix, niy and niz are the elements of the unit i i σ and vector ni in the direction of i T ⋅ = ( T means transpose) ni ni 1
18 Principal strains and principal directions at three-dimensional stress state Use shear strain ε =γ / 2 for i ≠ j ε ε ε ij ij x xy xz The determinant = ε ε ε Strain matrix E yx y yz | E −εI |=0 ε ε ε zx zy z gives three roots (the principal strains) I = unit matrix ε Direction of principal strain i (i = 1, 2, 3) is given by
(E −ε I)⋅n = 0 nix, niy and niz are the elements of the unit i i ε and vector ni in the direction of i T ⋅ = ( T means transpose) ni ni 1
Hooke’s law, including temperature term (three-dimensional stress state) 1 ε = [σ −ν(σ +σ)] + α∆T x E x y z α = temperature coefficient ∆T = change of temperature (relative to 1 temperature giving no stress) ε = [σ −ν(σ +σ)] + α∆T y E y z x 1 ε = [σ −ν(σ +σ)] + α∆T z E z x y τ τ τ γ = xy γ = yz γ = zx xy G yz G zx G
Effective stress The Huber-von Mises effective stress (the deviatoric stress hypothesis) σvM = σ2 +σ2 +σ2 −σσ −σ σ −σσ + τ2 + τ2 + τ2 e √x y z x y y z z x 3 xy 3 yz 3 zx
= 1 {(σ −σ)2 +(σ −σ)2 +(σ −σ)2 } √2 1 2 2 3 3 1
The Tresca effective stress (the shear stress hypothesis) σT = [|σ −σ |,|σ −σ |,|σ −σ |]=σpr −σpr e max 1 2 2 3 3 1 max min (pr = principal stress)
19 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 2 ( )2 ( )2 2 ⌠ N(x) Mt x Mbend x T(x) U = + + +β dx tot ⌡ ( ) ( ) ( ) ( ) 0 2EA x 2GKv x 2EI x 2GA x
Mt = torque = Mx Kv = section factor of torsional stiffness β Mbend = bending moment = My = shear factor, see below
Cross sectionβµ Shear factor β 2 A ⌠ SA’ 6/5 3/2 β= dA 2 ⌡ I A b 10/9 4/3 β is given for some cross sections in the table (µ is the Jouravski factor, see Section 2 2 12.3 One-Dimensional Bodies)
A/A web A/Aweb
Elementary case: pure bending
M12M Only bending momentet Mbend is present. L, EI The moment varies linearly along the beam
with moments M1 and M2 at the beam ends. One has M (x)=M +(M − M )x /L, which gives M1 x bend 1 2 1 L U = {M 2 + M M + M 2 } M M2 tot 6EI 1 1 2 2 The second term is negative if M1 and M2 have different signs
The Castigliano theorem ∂U ∂U δ = displacement in the direction of force P δ= and Θ= ∂P ∂M of the point where force P is applied Θ = rotation (change of angle) at moment M
20 9. Stress Concentration Tension/compression σ σ σ Maximum normal stress at a stress concentration is max = Kt nom, where Kt and nom are given in the diagrams r K t K t r 3.0 P P 3.0 P P Bb Bb
2.5 thickness h 2.5 thickness h P = P nom bh = nom 2.0 2.0 bh B/b B/b 2.0 1.5 2.0 1.5 1.2 1.5 1.2 1.1 1.1 1.05 1.05 1.01 1.01 1.0 1.0 0 0.1 0.2 r/b 0 0.1 0.2 0.3 0.4 0.5 0.6 r/b Tension of flat bar with shoulder fillet Tension of flat bar with notch
r r K t K t 3.0 P P 3.0 P P D d Dd
2.5 2.5 4 P = 4 P nom 2 = d nom d2 2.0 2.0 D/d 2.0 D/d 1.5 1.5 1.2 1.5 1.2 1.1 1.1 1.05 1.05 1.01 1.01 1.0 1.0 0 0.1 0.2 0.3 r/d 0 0.1 0.2 0.3 0.4 0.5 0.6 r/d Tension of circular bar with shoulder Tension of circular bar with U-shaped fillet groove
21 a 2r 0 B 0 K t B/a= 5 3.2 3.0 3 2.8
2.6 2.5 2.4 2.25 B 2.2 = 2 nom B - 2r 0 2.0 0 0.1 0.2 0.3 0.4 r/a Tension of flat bar with hole
Bending σ σ σ Maximum normal stress at a stress concentration is max = Kt nom, where Kt and nom are given in the diagrams d K t K t d MbbM Mb Mb 3.0 3.0 D B
2.5 2.5 Mb thickness h = nom 3 2 d/h = 0 DDd 32 6 2.0 0.25 2.0 0.5 1.0 1.5 2.0 1.5 6 M = b nom (B-d)h 2 1.0 1.0 0 0.2 0.4 0.6 d/B 0 0.1 0.2 0.3 d/D Bending of flat bar with hole Bending of circular bar with hole
22 r K t K t Mb r Mb M M 3.0 b b 3.0 Bb Bb
2.5 thickness h 2.5 thickness h 6Mb = 6Mb nom hb2 = 2.0 2.0 nom 2 B/b hb B/b 6.0 2.0 1.2 1.5 1.2 1.5 1.1 1.05 1.05 1.02 1.01 1.01 1.02 1.0 1.0 0 0.1 0.2 r/b 0 0.1 0.2 0.3 0.4 0.5 0.6 r/b Bending of flat bar with shoulder fillet Bending of flat bar with notch
r r K t K t M M Mb Mb b b 3.0 3.0 D d Dd
2.5 2.5 32 M = b 32Mb nom = d3 nom 3 2.0 2.0 d D/d 6.0 D/d 2.0 1.5 1.5 1.2 1.2 1.05 1.05 1.01 1.02 1.02 1.0 1.0 1.01 0 0.1 0.2 0.3 r/d 0 0.1 0.2 0.3 0.4 0.5 0.6 r/d Bending of circular bar with shoulder Bending of circular bar with U-shaped fillet groove
23 Torsion τ τ τ Maximium shear stress at stress concentration is max = Kt nom, where Kt and nom are given in the diagrams r r K t K t 3.0 Mv Mv 3.0 Mv Mv D d Dd
2.5 2.5 16 M = v 16 Mv nom 3 = d nom d 3 2.0 2.0
D/d D/d 1.5 2.0 1.5 1.3 1.2 1.2 1.05 1.1 1.0 1.0 1.01 0 0.1 0.2 0.3 r/d 0 0.1 0.2 0.3 0.4 0.5 0.6 r/d Torsion of circular bar with shoulder Torsion of circular bar with notch fillet
K t r K t d 4.0 2.0 M vvM D
d /4 3.5 7d Mv = 8 nom D 3 d D 2 16 6 3.0 1.5 d 16M 2.5 = v nom d 3
2.0 1.0 0 0.05 0.10 r/d 0 0.1 0.2 0.3 d/D Torsion of bar with longitudinal keyway Torsion of circular bar with hole
24 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 να106 Ultimate Yield limit modulus strength tension/ bending torsion E compression GPa K-1 MPa MPa MPa MPa
Carbon steel 141312-00 206 0.3 12 360 >240 260 140 460 141450-1 205 0.3 430 >250 290 160 510 141510-00 205 0.3 510 >320 640 141550-01 205 0.3 490 >270 360 190 590 141650-01 206 0.3 11 590 >310 390 220 690 141650 206 0.3 860 >550 610
σ Offset yield strength Rp0.2 ( 0,2) Stainless steel 2337-02 196 0.29 16.8 >490 >200 Aluminium SS 4120-02 70 23 170 >65 215 SS 4120-24 70 23 220 >170 270 SS 4425-06 70 23 >340 >270
1 Data in this table has been collected from B Sundström (editor): Handbok och Formelsamling i Hållfasthetslära, Institutionen för hållfasthetslära, KTH, Stockholm, 1998.
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