MECH 4101 - Mechanics of Deformable Solids Notes
MECH 4101 - Mechanics of Deformable Solids Notes 1.0 Elastic Theory
1
1.1 Stress
1
1.2 Equation of Equilibrium
2
1.3 Stress-Strain Relationship
3
1.4 Vectors and Tensors Notation
4
1.5 Stress at a Point
5
2.0 Elastic Axis-symmetric Deformations
8
2.1 Basic Relation
8
2.2 Pressurized and Thick-Walled Cylinders
9
2.3 Rotating Disks and Long, Rotating Cylinders (T=0)
12
2.4 Thermal Stresses in Thick-Walled Cylinder and Disks
14
2.4.1 Plane stress
15
2.4.2 Plane strain
16
2.4.3 Other End Conditions due to Thermal Loads
19
3.0 Stress Functions
20
3.1 Solution by Polynomial
20
3.2 Stress Concentrations
25
3.2.1 Use of Polar Coordinates
25
3.2.2 Stresses due to Concentration Loads
26
3.2.3 Stress Concentrations around a Small Circular Hole (Kirsh Solution)
29
4.0 Shock and Impact Loading
30
4.1 Stress and Displacement in an Elastic Axial Bar
30
4.2 Effects of Geometry
38
4.3 Stress Wave Propagation under Impact Loading
38
4.4 Changes in Material Properties under Impact Loading
41
5.0 Elasto-Plastic Analysis 5.1 Beam Bending 5.1.2 Elastic-Linear Strain Hardening 5.2 Thick-Walled Cylinders under Internal Pressure
42 44 49 52
5.2.1 Residual Stresses in Internally Pressurized Cylinders
57
5.2.2 The “Shakedown” Condition
59
5.2.3 Using Von Mises
61
1
1.0 Elastic Theory 1.1 Stress Stress is an internal resistance of a material and is defined as an average force exerted over an area. Two types of stresses: normal stress and shear stress arise due to the force acting normal and parallel to the plane, respectively. Mathematically, stress can be written as:
𝐹N 𝐴 𝐹T 𝜏= 𝐴
(1.1)
𝜎=
where: -
(1.2)
𝜎 is normal stress 𝜏 is shear stress 𝐹N is normal force 𝐹T is tangential force 𝐴 is cross-sectional area
A double-suffix notation, 𝜏xy , is used to define the direction of shear stress acting on a plane. The first suffix defines the direction of the outward normal of the plane and the second one represents the direction of the shear stress. It can be shown that: 𝜏xy = 𝜏yx ; 𝜏yz = 𝜏zy ; 𝜏xz = 𝜏zx
(1.3)
In the 3-D case, the stress matrix [𝜎] can be written as: 𝜎xx [𝜎] = [𝜏xy 𝜏xz
𝜏xy 𝜎yy 𝜏yz
𝜏xz 𝜏yz ] 𝜎zz
(1.4)
In solid mechanics, strain is a measure of deformation due to the applied load with reference to the original length. The 3-D normal strain in a solid medium due to an applied load is given by:
𝜕𝑢 𝜕𝑥 𝜕𝑣 𝜀yy = 𝜕𝑦 𝜕𝑤 𝜀zz = 𝜕𝑧 𝜀xx =
(1.5) (1.6) (1.7)
where 𝜀xx , 𝜀yy , and 𝜀zz are normal strains in the x, y and z directions, respectively and u , v , and w are displacements in the x, y, and z directions, respectively.
2 In addition to normal strain, shear strain, 𝛾, due to shear force is given by: 𝜕𝑣 𝜕𝑢 + = 2𝜀xy 𝜕𝑥 𝜕𝑦 𝜕𝑤 𝜕𝑣 𝛾zy = 𝛾yz + = 2𝜀yz 𝜕𝑦 𝜕𝑤 𝜕𝑤 𝜕𝑢 𝛾xz = 𝛾zx + = 2𝜀xz 𝜕𝑥 𝜕𝑧 𝛾xy = 𝛾yx
(1.8) (1.9) (1.10)
Six independent stress components 𝜎xx, 𝜎yy, 𝜎zz, 𝜏xy , 𝜏yz , and 𝜏zx are needed to describe the load carried by a 3-D solid at any point. The six independent strain components 𝜀xy , 𝜀yy , 𝜀zz , 𝛾xy ,𝛾yz , 𝛾zx , which are produced by the load, are used to describe the deformation at any point.
1.2 Equation of Equilibrium Consider an infinitesimal element ∆𝑥 × ∆𝑦 × ∆𝑧 onto which non-uniform stresses are applied at different faces, as shown in Figure 1. 𝑦
𝑥
𝑧
Figure 1: An infinitesimal element showing stress distribution.
∑ 𝐹x = 0, gives (𝜎xx + ∆𝜎xx )∆𝑦∆𝑧 − 𝜎xx ∆𝑦∆𝑧 + (𝜏yx + ∆𝜏yx )∆𝑥∆𝑧 − ∆𝜏yx ∆𝑥∆𝑧 + (𝜏zx + ∆𝜏zx )∆𝑥∆𝑦 − ∆𝜏zx ∆𝑥∆𝑦 =0
(1.11)
Dividing both sides by ∆𝑥∆𝑦∆𝑧 and taking the limit as ∆𝑥, ∆𝑦, and ∆𝑧 goes to zero, (1.11) can be rewritten as:
3 𝜕𝜎xx 𝜕𝜏yx 𝜕𝜏zx + + =0 𝜕𝑥 𝜕𝑦 𝜕𝑧
(1.12)
In the presence of body forces, 𝐹x , (1.12) can be rewritten as: 𝜕𝜎xx 𝜕𝜏yx 𝜕𝜏zx + + + 𝐹x = 0 𝜕𝑥 𝜕𝑦 𝜕𝑧
(1.13)
𝜕𝜏xy 𝜕𝜎yy 𝜕𝜏yz + + + 𝐹y = 0 𝜕𝑥 𝜕𝑦 𝜕𝑧
(1.14)
𝜕𝜏xz 𝜕𝜏yz 𝜕𝜎zz + + + 𝐹z = 0 𝜕𝑥 𝜕𝑦 𝜕𝑧
(1.15)
Similarly for ∑ 𝐹y = 0:
And for ∑ 𝐹z = 0:
1.3 Stress-Strain Relationship In general for a 3-D isotropic solid, the stress-strain relationship can be written as: 1 [𝜎 − ν(𝜎yy + 𝜎zz )] E xx 1 𝜀yy = [𝜎yy − ν(𝜎xx + 𝜎zz )] E 1 𝜀zz = [𝜎zz − ν(𝜎xx + 𝜎yy )] E
(1.16)
𝜀xx =
(1.17) (1.18)
E is Young’s modulus and ν is Poisson’s ratio Shear stress is related to the shear strain by: 𝜏xy ; G where shear stress is given by: 𝛾xy =
𝛾yz =
G=
𝜏yz ; G
E 2(1 + ν)
𝛾xz =
𝜏xz ; G
(1.19)
(1.20)
4
1.4 Vectors and Tensors Notation In order to represent stresses, strains using tensor notation, it is important to set up the index notation using Einstein’s summation convention. This notation is used to simplify expressions involving vectors and tensors operations. The direction x, y, and z is referred to as 1, 2, and 3, respectively. A vector 𝑣⃗ in the index notation can be written as; 3
𝑣⃗ = 𝑣i ∑ 𝑣i ⃗⃗⃗ 𝑒i
(1.21)
i=1
where {𝑒⃗⃗⃗⃗, 𝑒2 , ⃗⃗⃗⃗⃗} 𝑒3 are the orthonormal unit vectors in 3-D space 1 ⃗⃗⃗⃗⃗ Similarly, a tensor in the index notation can be written as: 3
3
𝑇 𝑇ij = ∑ ∑ 𝑇ij ⃗⃗⃗𝑒 𝑒i ⃗⃗⃗j
(1.22)
i=1 j=1
where {𝑒⃗⃗⃗,i ⃗⃗⃗⃗} 𝑒j are unit dyads that represent an ordered pair of coordinate directions In Einstein’s summation convention, if the index appears more than once, then the summation is repeated for each index over the range of the index. The number of free indices determine whether the index notation is a scalar, a vector, or a tensor. In this method, no free index or rank 0, represents a scalar, one free index or rank 1, means it is a vector, two free indices or rank 2 represents a tensor, and so on. For the Cartesian co-ordinate system where the summation is performed in three mutually- orthogonal space, the rank of the tensor n have 3n elements. For example: 𝜕𝑢i 𝜕2 𝑓 , and are scalars (no free index, rank 0 = 1 element) 𝜕𝑥i 𝜕𝑥i 𝜕𝑥i 𝜕𝑓 ∈ijk , 𝑢i 𝑣j , , and 𝑇ii 𝑣i are vectors (one free index, rank 1 = 3 elements) 𝜕𝑥i 𝑢i 𝑣i , 𝑇ii,
+1 when any two indices i, j, k are equal ∈ijk { −1 when ijk = 1 2 3 or 3 2 1 or 2 3 1 0 otherwise
𝜕𝑢j 𝜕2 𝑓 𝑢i 𝑣j , and , and are tensors (two free indices, rank 2 = 9 element) 𝜕𝑥i 𝜕𝑥i 𝜕𝑥i 𝑐ijk is a tensor (four free indices, rank 4 = 81 elements)
Based on the aforementioned conventions, the strain tensor can be written as: 1 𝜕𝑢i 𝜕𝑢j 𝜀ij = ( + ) 2 𝜕𝑥j 𝜕𝑥i
(1.23)
5 The stress tensor is given by: 𝜎xx 𝑇ij = [𝜏xy 𝜏xz
𝜏xy 𝜎yy 𝜏yz
𝜏xz 𝜏yz ] 𝜎zz
(1.24)
The relationship of strain as a function of stress can be written as:
𝜀ij = where, 𝛿ij = {
1+ν ν 𝑇ij − (𝑇11 + 𝑇22 + 𝑇33 )𝛿ij E E
(1.25)
1 if i = j also known as Kronecker delta 0 if i ≠ j
and the stress in terms of strain is given by: 𝑇ij = λ(𝜀11 + 𝜀22 + 𝜀33 )𝛿ij + 2G𝜀ij
(1.26)
Finally, the equilibrium equations for elasticity can be written as: 𝜕𝑇ij +𝐹 =0 𝜕𝑥i
1.5 Stress at a Point
Transforming the coordinate system by angle θ and knowing 𝜏yx = 𝜏xy: 𝜎n (AC) + [𝜏yx (BC) cos 𝜃] − 𝜎x (AB) cos 𝜃 + 𝜏yx (AB) sin 𝜃 − 𝜎y (BC) sin 𝜃 = 0
Divide every term by area AC: BC AB AB BC 𝜎n + [𝜏yx ( ) cos 𝜃] − 𝜎x ( ) cos 𝜃 + 𝜏yx ( ) sin 𝜃 − 𝜎y ( ) sin 𝜃 = 0 AC AC AC AC where: BC AB = sin 𝜃 and = cos 𝜃 ; areas are constant AC AC
(1.27)
6 It can be further simplified to: 𝜎n + 𝜏yx sin 𝜃 cos 𝜃 − 𝜎x cos2 𝜃 + 𝜏yx cos 𝜃 sin 𝜃 − 𝜎y sin2 𝜃 = 0 𝜎n − 𝜎x cos 2 𝜃 − 𝜎y sin2 𝜃 + 2𝜏yx cos 𝜃 sin 𝜃
For Mohr’s circle, if the shear stress tends a clockwise (C.W.) rotation, the element is positive. Using: sin2 𝜃 =
1 − cos 2𝜃 1 + cos 2𝜃 ; cos 2 𝜃 = ; 2 sin 𝜃 + cos 𝜃 = sin 2𝜃 2 2
1 1 𝜎n = (𝜎x − 𝜎y ) + (𝜎x + 𝜎y ) cos 2𝜃 + 𝜏yx sin 𝜃 2 2 1 𝜏s = (𝜎x + 𝜎y ) sin 2𝜃 −𝜏xy cos 2𝜃 2 Squaring ① and ② and adding them: 1 2 𝜏s = (𝜎x − 𝜎y ) + 𝜏2xy 4
Comparing this to the equation of a circle: (𝑥 − 𝑎)2 + (𝑦 − 𝑏)2 = 𝑟 2
…so: 1 2 𝑟 = (𝜎x − 𝜎y ) + 𝜏2xy 4
① ②
7 Hence the Mohr’s circle:
𝜎x + 𝜎y 𝜎ave = 2 1 1 𝜎n = (𝜎x − 𝜎y ) + (𝜎x + 𝜎y ) cos 2𝜃 + 𝜏yx sin 𝜃 2 2 1 𝜏s = (𝜎x + 𝜎y ) sin 2𝜃 −𝜏xy cos 2𝜃 2 𝜎1,2 =
𝜎x + 𝜎y 𝜎x − 𝜎y 2 2 ± √( ) + 𝜏xy 2 2
tan 2𝛷 =
𝜏xy 1 2 (𝜎x − 𝜎y )
𝜎x − 𝜎y 2 2 |𝜏max | = √( ) + 𝜏xy 2 𝜎x − 𝜎y |𝜏max | = 2 (𝜎x − 𝜎y ) tan 2𝜃s = − 2𝜏xy
Principals of Solutions in Solid Mechanics: i) ii) iii) iv)
condition of state equilibrium stress-strain relationship condition for compatibility boundary conditions (BCs)
Methods of Solution a) engineering (strength of material) approach - simplifying assumptions related to the geometry of deformation, the state of distribution of stress/strain across the cross section - useful in the simple beam bending and torsion problem b) “exact” – mathematical solution to the governing equations c) numerical solution (FEM, BEM) d) experiments
8 Compatibility Equation in Terms of Stress ∂ ∂ ( 2 + 2 ) (𝜎x + 𝜎y ) = 0 ∂x ∂y ∇2 (𝜎x + 𝜎y ) = 0 True when: 𝐹x , 𝐹y , and 𝑇 are zero.
Plane strain (𝜀z = 0): ∇2 (𝜎x + 𝜎y + Eα𝑇) = −
∂𝐹x ∂𝐹y + ) ∂y 1 − ν ∂x 1
(
Plane stress (𝜎z = 0): ∇2 (𝜎x + 𝜎y + Eα𝑇) = −(1 + ν) (
∂𝐹x ∂𝐹y + ) ∂x ∂y
2.0 Elastic Axis-symmetric Deformations -
axis-symmetric problems in stress analysis occur in numerous practical-pressure vessels, compound cylinders, turbine/compressor disks, flywheels not only geometry has to be similar, loads applied along an axis has to be similar
2.1 Basic Relation -
general transformation of Cartesian to cylindrical polar 𝑥 = 𝑟 cos 𝜃 𝑦 = 𝑟 sin 𝜃 𝑥2 + 𝑦2 = 𝑟2 tan 𝜃 =
𝑦 𝑥
2𝑟
d𝑟 d𝑟 = 2𝑥 = cos 𝜃 d𝑥 d𝑥
∂ ∂𝑟 ∂ ∂𝜃 ∂ = + ∂x ∂x ∂r ∂x ∂𝜃
2𝑟
d𝑟 d𝑟 = 2𝑦 = sin 𝜃 d𝑦 d𝑦
∂ ∂𝑟 ∂ ∂𝜃 ∂ = + ∂y ∂y ∂r ∂y ∂𝜃
d𝜃 𝑦 sin 𝜃 =− 2 − d𝑥 𝑟 𝑟
∂ ∂ sin 𝜃 ∂ = cos 𝜃 − ∂x ∂r 𝑟 ∂𝜃
d𝜃 𝑥 cos 𝜃 =− 2 d𝑦 𝑟 𝑟
∂ ∂ cos 𝜃 ∂ = sin 𝜃 + ∂y ∂r 𝑟 ∂𝜃
9
In matrix form: ∂ cos 𝜃 ∂x = [ ∂ sin 𝜃 [∂y]
∂ sin 𝜃 𝑟 ] ∙ [ ∂r ] cos 𝜃 ∂ 𝑟 ∂𝜃
−
-
for axis-symmetric, all displacements are 𝑓(𝑟) and independent of 𝜃 𝑟 = 0 (pressurized
-
cylinders and rotating disks) shear in 𝜏rθ = 0 (principle stress) 𝜎r , 𝜎θ are 𝑓(𝑟)
For equilibrium: d𝜎r 𝜎r − 𝜎θ + + 𝐹r = 0 d𝑟 𝑟
Strain-Displacement Relationship
Plane stress (𝝈𝐳 = 𝟎)
d𝑢 d𝑟 } 2.2 𝑢 𝜀θ = 𝑟
(thin rotating disk; open-ended cylinder)
𝜀r =
Hooke’s Law 1 𝜀r = [𝜎r − ν(𝜎θ + 𝜎z )] + α∆𝑇 E 1 𝜀θ = [𝜎θ − ν(𝜎r + 𝜎z )] + α∆𝑇 E 1 𝜀z = [𝜎z − ν(𝜎r + 𝜎θ )] + α∆𝑇 E
E [𝜀 + ν𝜀θ (1 + ν)α𝑇] 1 − ν2 r E [𝜀 + ν𝜀r (1 + ν)α𝑇] 𝜎θ = 1 − ν2 θ 𝜎r =
Plane strain (𝜺𝐳 = 𝟎) (long cylinders with restrained ends) 𝜎z = ν(𝜎r − 𝜎θ ) − Eα𝑇 1 − ν2 ν 𝜎 ] + (1 + ν)α𝑇 [𝜎r − E 1−ν θ 2 1−ν ν 𝜀θ = 𝜎 ] + (1 + ν)α [𝜎θ − E 1−ν r 𝜀r =
2.2 Pressurized and Thick-Walled Cylinders Under pressure loads, the stresses in the cylinder are given by Lamé’s solution: 𝜎r = 𝐴 −
𝐵 𝑟2
𝜎θ = 𝐴 +
𝐵 𝑟2
where A and B are found from the given boundary conditions.
10 Example 1: Thick-walled open-ended cylinder subject to internal pressure: 𝑝 = 100 MPa. Find the maximum stress in the cylinder given: 𝑟i = 25 mm 𝑟o = 50 mm Boundary conditions: @ 𝑟 = 𝑟i
@ 𝑟 = 𝑟o
𝐵 𝜎r = 𝐴 − 2 𝑟i
𝜎r = 𝐴 −
𝐵 ① (25 mm)2 (−𝑝 due to compression) −𝑝 = 𝐴 −
0=𝐴−
𝐵 𝑟o2
𝐵 ② (50 mm)2
②→① −100 MPa =
𝐵 𝐵 𝐵 𝐵 − − 100 MPa = − −3 2 −3 2 2 (50 × 10 m) (25 × 10 m) 0.0025 m 6.25 × 10−4 m2 (6.25 × 10−4 m2 )𝐵 − (0.0025 m2 )𝐵 1.56 × 10−6 m4 2 − 156.25 N ∙ m = (−1.88 × 10−4 m2 )𝐵 𝑩 = 𝟖. 𝟑𝟑 × 𝟏𝟎𝟒 𝐍 − 100 MPa =
𝐵 8.33 × 10−4 N 0=𝐴− 𝐴= (50 mm)2 25 × 10−3 m2 𝑨 = 𝟑. 𝟑𝟑 × 𝟏𝟎𝟕 𝐏𝐚 8.33 × 104 N 𝜎r = 3.33 × 10 Pa − 𝑟2 8.33 × 104 N 𝜎θ = 3.33 × 107 Pa + 𝑟2 7
(𝝈𝐫 )𝐦𝐚𝐱 = −𝟏𝟎𝟎 𝐌𝐏𝐚 𝝈𝛉 @ 𝒓 = 𝒓𝐢 𝟏𝟔𝟔. 𝟔 𝐌𝐏𝐚 𝝈𝛉 @ 𝒓 = 𝒓𝐨 𝟔𝟔. 𝟔 𝐌𝐏𝐚
In general, the boundary conditions are: 𝐵 𝑝i − 𝑝o 𝑘 2 𝐴 = 2 𝑟o 𝑟i 𝑘2 − 1 ; where 𝑘 = 2 𝐵 (𝑝 − 𝑝o )𝑟o 𝑟i 𝑟 = 𝑟o 𝜎ro = −𝑝o − 𝑝o = 𝐴 − 2 𝐵 = i 2 𝑟o } 𝑘 −1 𝑟 = 𝑟i 𝜎ri = −𝑝i − 𝑝i = 𝐴 −
For internal pressure ONLY (𝑝o = 0): 𝑝i 𝑟o2 𝑝i 𝑟o2 𝜎r = 2 (1 − 2 ) ; 𝜎θ = 2 (1 + 2 ) 𝑘 −1 𝑟 𝑘 −1 𝑟
For external loading ONLY (𝑝i = 0): 𝑝i 𝑘 2 𝑟i2 −𝑝o 𝑟o2 𝜎r = 2 ( 2 − 1) ; 𝜎θ = 2 (1 − 2 ) 𝑘 −1 𝑟 𝑘 −1 𝑟
11
𝜎θ > 𝜎z > 𝜎r
Example: A long, thick-walled cylinder with ends rigidly restrained is embedded in a rigid wall and subjected to an internal pressure of 𝑝i . What is the maximum p allowed to avoid yielding according to Tresca’s criteria? (Tresca:
𝜎θ + 𝜎r 𝜎y = ) 2 2 𝑟=ro
1 [𝜎 − ν(𝜎r − 𝜎θ )] E z 𝜎z = ν(𝜎θ + 𝜎r ) 𝑢 (𝜀θ )ro = 0 = 0; 𝑢 is the displacement 𝑟o 1 𝜀θ = [𝜎θ − ν(𝜎r + 𝜎z )]𝑟o 0 E
𝑟i = 20 × 10−3 m 𝑟o = 50 × 10−3 m ν = 0.33 E = 60 GPa 𝜎y = 80 MPa 1 1 = 400; 2 = 2500 2 𝑟o 𝑟i
𝜀z = 0
let: 𝑎 = (𝜎θ )ri & 𝑏 = (𝜎θ )ro @ 𝑟i : 𝑎 + 𝑃i 80 = 2 2 𝑎 + 𝑃i = 80 MPa ① @ 𝑟o : 𝜎θ = 𝑏, 𝜎r = −𝑃o 𝑢r 2 = 0
𝑎 + 𝑃i 𝑃i − 𝑃o = −2 + 400 m 2500 m−2 + 400 m−2 𝑏 + 𝑃o 400 m−2 + 400 m−2 2500 m−2
𝑎 = 0.8125𝑃1 𝑃i = 2.3125𝑃2 𝑃o = 25.6 MPa
12 𝑏 − ν[−𝑃o ν(b − 𝑃o )] = 0 ν 𝑏=− 𝑃 − 0.5𝑃o ② 1−ν o
𝑃i = 59.2 MPa 𝒂 = (𝝈𝛉 )𝐫𝟏 𝟐𝟎. 𝟖 𝐌𝐏𝐚 𝒃 = (𝝈𝛉 )𝐫𝟐 − 𝟏𝟐. 𝟖 𝐌𝐏𝐚
2.3 Rotating Disks and Long, Rotating Cylinders (T=0) -
similar to analysis as in the Lamé’s equation, for pressurized cylinders, except now due to rotation there is a body force, Fr
Plane stress case (i.e. disk): 𝐵 3+ν 2 2 − 𝜌𝑟 𝜔 𝑟2 8 𝐵 1 + 3ν 2 2 𝜎θ = 𝐴 + 2 − 𝜌𝑟 𝜔 𝑟 8
A and B are constants for a given geometry and speed (found using boundary conditions)
𝜎r = 𝐴 −
Example: A thin steel disk with 𝑟i = 80 mm, 𝑟o = 380 mm is shrink-fitted onto a rigid shaft. The radial interference (𝛿 = 0.1 mm). At what speed will the shrink fit loosen? 𝐸 = 200 GPa ν = 0.3 𝜌 = 7 850
kg m3
When the shrink fit loosens, 𝜎r = 0 @ 𝑟 = 𝑟i @ 𝑟 = 𝑟o 𝜎(ro )2 = 0 @ 𝑟 = 𝑟i 𝐵 3 + 0.3 kg 0=𝐴− − ( ) (7 850 ) (0.08 m)2 𝜔2 ① (0.08 m)2 8 m3 𝐴 = 488.3𝜔2 @ 𝑟 = 𝑟o 𝐵 = 2.933𝜔2 𝐵 3 + 0.3 kg 0=𝐴− −( ) (7 850 3 ) (0.38 m)2 𝜔2 ② 2 (0.38 m) } 8 m 𝜀θ =
𝑢 𝑟 (𝜎θ − ν𝜎r ) 𝑟 E
𝛿 = (𝑢)rmo − (𝑢)rmi 0.1 m
13 𝜎θ @ 𝑟 = 𝑟1 (0.1 × 10−3 m)(200 × 109 Pa) 𝐵 3 + 0.3 kg =𝐴+ −( ) (2 830 3 ) (0.08 m)2 𝜔2 (0.08 m)2 (0.08 m)2 8 m 𝝎 = 𝟓𝟏𝟒. 𝟔
𝐫𝐚𝐝 𝝎 = 𝟒 𝟗𝟏𝟒 𝐑𝐏𝐌 𝐬
Plane strain (long rotating cylinders) 𝐵 3 + ν∗ 2 2 − 𝜌𝑟 𝜔 𝑟2 8 𝐵 1 + 3ν∗ 2 2 𝜎θ = 𝐴 + 2 − 𝜌𝑟 𝜔 𝑟 8 𝜎r = 𝐴 −
ν∗ =
ν 1−ν
Example: A long, solid steel shaft of 200 mm radius is rotating @ 300 RPM with its ends restrained from longitudinal displacement. Find the total longitudinal force exerted at the ends. E = 200 GPa, ν = 0.3, 𝜌 = 7 850 ν∗ =
kg m3
0.3 0.42857 0.7
B. Cs. 𝜀z = 0 𝜀z =
1 [𝜎 − ν(𝜎r − 𝜎θ )] + α∆𝑇 𝝈𝐳 = 𝛎(𝝈𝐫 − 𝝈𝛉 ) E z
𝐵 = 0 for a solid steel shaft 𝜎z = ν [(𝐴 − ν [(𝐴 −
𝑟 @ 𝑟o = 200 mm,
𝐵 3 + ν∗ 2 2 𝐵 1 + 3ν∗ 2 2 − 𝜌𝑟 𝜔 ) + (𝐴 + − 𝜌𝑟 𝜔 )] 𝑟2 8 𝑟2 8
3 + ν∗ 2 2 1 + 3ν∗ 2 2 𝟏 + 𝛎∗ 𝟐 𝟐 𝜌𝑟 𝜔 ) + (𝐴 − 𝜌𝑟 𝜔 )] 𝝈𝐳 = 𝛎 [𝟐𝑨 − 𝝆𝒓 𝝎 ] 8 8 𝟐
𝜎r = 0 𝜎r = 𝐴 −
𝐵 3 + ν∗ 2 2 𝟑 + 𝛎∗ 𝟐 𝟐 − 𝜌𝑟 𝜔 𝑨 = 𝝆𝒓 𝝎 𝑟2 8 𝟖
3 + ν∗ 2 1 + ν2 2 ν𝜌𝜔2 [(3 − 2ν)𝑟o2 − 2π2 ] 𝜎z = ν [ 𝑟o − ( ) π ] 𝜌𝜔2 4 2 4(1 − ν)
14
𝜎=
2π π 𝐹 πν𝜌𝜔2 𝑟o4 𝐹 = 𝜎𝐴 𝐹 = ∫ 𝜎 d𝐴 ∫ ∫ 𝜎z 𝑟 d𝑟d𝜃 𝐴 2 0 0
π(0.3) (7 850
kg rev 2πrad min 2 ) (300 min × rev × 60 s) (0.2 m)4 3 m 𝑭 = 𝟓. 𝟖𝟒 𝐤𝐍 2
2.4 Thermal Stresses in Thick-Walled Cylinder and Disks In linear-elastic analysis, the effects of mechanical and thermal loads can be treated separately and then added together after the components are subjected to both types of loading superposition). -
consider thermal loading ONLY
Fourier’ law of heat conduction: 𝑄 = −2π𝑟𝑘
d𝑇 d𝑟
where: 𝑄 = heat flow per unit axial length in the radial direction k = thermal conductivity coefficient
𝑟
d𝑇 𝜃 =− C1 d𝑟 2π𝑘
∫ d𝑇 = ∫
d𝑟 C 𝑇 = C1 ln 𝑟 + C2 𝑟 1
where C1 and C2 are constants B. Cs. 𝑇o 𝑟
𝑇 = 𝑇i +
𝑟i 𝑇i
@ 𝑟 = 𝑟i 𝑇 = 𝑇i @ 𝑟 = 𝑟o 𝑇 = 𝑇o
𝑟o
𝑇o − 𝑇i 𝑟 𝑟 ln 𝑟 i ln 𝑟o i
15
2.4.1 Plane stress -
thin disks (externally cooled and heated) d𝑢
𝑢
d𝑢
𝑢
𝜀r = ; 𝜀θ = E E d𝑢 𝑢 d𝑟 𝑟 [𝜀 𝜎r = + ν𝜀 (1 + ν)α𝑇] ⇒ 𝜎r = [ + ν − (1 + ν)α𝑇] r θ 2 2 1−ν 1 − ν d𝑟 𝑟 𝜀r = ; 𝜀θ = E E 𝑢 d𝑢 d𝑟 𝑟 [𝜀 𝜎θ = + ν𝜀 (1 + ν)α𝑇] ⇒ 𝜎θ = − (1 + ν)α𝑇] [ +ν θ r 2 2 1−ν 1−ν 𝑟 d𝑟
Using equilibrium (Fr = 0): d𝜎r 𝜎r − 𝜎θ + =0 d𝑟 𝑟 1 1 d d𝑇 (𝑢𝑟)] = (1 + ν)α [ d𝑟 𝑟 d𝑟 d𝑟
Double integral: 𝑢 = C1 𝑟 +
𝑟 C2 1 + ν + α ∫ 𝑇𝑟 d𝑟 𝑟 𝑟 𝑟i
C1 and C2 are constants of integration
Substitute u in 𝜎r and 𝜎θ : 𝐵 E 𝑟 − α ∫ 𝑇𝑟 d𝑟 𝑟2 𝑟 2 𝑟i 𝐵 E 𝑟 𝜎θ = 𝐴 + 2 + α 2 ∫ 𝑇𝑟 d𝑟 − 𝐸α𝑇 𝑟 𝑟 𝑟i
where A and B are from boundary conditions
𝜎r = 𝐴 −
Free-free edge
𝑇o 𝑟
B. Cs.
For solid disk:
@ 𝑟 = 𝑟i 𝜎r = 0 @ 𝑟 = 𝑟o 𝜎r = 0
𝐵=0
𝑟o
𝑟i 𝑇i
αE ∫ 𝑇𝑟 d𝑟 − 𝑟i2 𝑟i 𝑟o αE𝑟i2 𝐵= 2 ∫ 𝑇𝑟 d𝑟 𝑟o − 𝑟i2 𝑟i 𝐴=
𝑟o
𝑟o2
𝐴=
αE 𝑟o ∫ 𝑇𝑟 d𝑟 𝑟o2 0
16 Steady-state temperature disk: 𝑟
∫ 𝑇𝑟 d𝑟 = 𝑇1 ( 𝑟i
𝑟 2 − 𝑟i2 𝑇o − 𝑇i 𝑟 2 𝑟 𝑟 2 − 𝑟i2 ) [ ln ( ) − )+( ( )] 𝑟 2 𝑟i 4 ln ( 𝑟o ) 2 i
2.4.2 Plane strain -
long cylinder with ends restrained 𝜎z = ν(𝜎r − 𝜎θ ) − Eα𝑇
let: E∗ =
E ν ; ν∗ = ; α∗ = α(1 + ν) 2 1−ν 1−ν
so: 𝜀r =
1 [𝜎 − ν∗ 𝜎θ ] + α∗ 𝑇 E∗ r
𝜀θ =
1 [𝜎 − ν∗ 𝜎r ] + α∗ 𝑇 E∗ θ
Stresses due to thermal loading in a cylinder with steady state temperature distribution 𝑟0 ln 𝑘 𝑟o2 𝜎r = 𝐶0 [− ln ( ) − 2 (1 − 2 )] 𝑟 𝑘 −1 𝑟 𝑟0 ln 𝑘 𝑟o2 𝜎θ = 𝐶0 [1 − ln ( ) − 2 (1 + 2 )] 𝑟 𝑘 −1 𝑟
𝜎z plane strain: 𝜎z =
where: 𝑟
k is the radius ratio, 𝑟0 i
𝐶0 =
E ′ α′ (𝑇i − 𝑇o ) ′ ′ Eα plane stress ;Eα { ∗ ∗ E α plane strain 2 ln 𝑘
αE∆𝑇 𝑟o ln 𝑘 [ν − 2 ln ( ) − 2ν 2 ] 2(1 − ν) ln 𝑘 𝑟 𝑘 −1
17
𝜎θ
𝑟 = 𝑟o
𝑟 = 𝑟i
𝜎r 𝜎z
-
use the largest and smallest 𝜎 for failure criterion
Example: A long, thick-walled steel tube with radius 𝑟i = 25 mm and 𝑟o = 50 mm is used to transport a hot, pressurized fluid @ 𝑃 = 10 MPa and 𝑇i = 250 ℃. The outer circumference is maintained @ 𝑇o = 50 ℃. Find stresses at the inner and outer radius of the cylinder. Check if yielding has occurred using Tresca’s criterion. 1 ℃
E=200 GPa, ν = 0.3, α = 11 × 10−6 , 𝜎y = 450 MPa Pressure Loads 1 = 1600 𝑟i2
𝜎θ
positive
𝜎r
𝑎
1 = 400 𝑟o2
1 𝑟o2
𝑏 1 𝑟o2
1 𝑟i2
negative
Using similar triangles: 𝑎 1600+400
𝑏
𝑃
= 400+400 = 1600−400
𝑎 (𝝈𝛉 )𝒓𝐢 = 𝟏𝟔. 𝟔𝟕 𝐌𝐏𝐚 𝑏 (𝝈𝛉 )𝒓𝐨 = 𝟔. 𝟔𝟕 𝐌𝐏𝐚
𝜎z = ν(𝜎r + 𝜎θ ) 0.3(0 MPa + 6.67 MPa) 𝝈𝐳 = 𝟐 𝐌𝐏𝐚
1 𝑟i2 𝑃i
18 Thermal Loads E ( ) (α(1 + ν))(𝑇o − 𝑇i ) E ∗ α∗ (𝑇o − 𝑇i ) 1 − ν2 𝐶0 = 2 ln 𝑘 2 ln 𝑘 200 × 109 Pa 1 ) ((11 × 10−6 ℃) (1 + 0.3)) (200 ℃) 1 − 0.32 𝑪𝟎 = 𝟒𝟓𝟑. 𝟒𝟐 𝐌𝐏𝐚 50 2 ln ( ) 25 (
@ 𝒓 = 𝒓𝐢 :
(𝝈𝐫 )ri = 𝟎 𝑟0 ln 𝑘 𝑟o2 ln 2 (1 + 4)] 𝜎θ = 𝐶0 [1 − ln ( ) − 2 (1 + 2 )] (453.42 MPa) [1 − ln 2 − 𝑟 𝑘 −1 𝑟 4−1 (𝝈𝛉 )𝒓𝐢 = −𝟑𝟖𝟗. 𝟔𝟖 𝐌𝐏𝐚 𝜎z =
αE∆𝑇 𝑟o ln 𝑘 [ν − 2 ln ( ) − 2ν 2 ] 2(1 − ν) ln 𝑘 𝑟 𝑘 −1
1 (11 × 10−6 ℃) (200 × 109 Pa)(200 ℃) 50 2(1 − 0.3) ln ( ) 25
50 ln ( ) 50 25 ] [0.3 − 2 ln ( ) − 2(0.3) 25 50 2 ( ) −1 25
𝝈𝐳 = −𝟓𝟓𝟓. 𝟒 𝐌𝐏𝐚
@ 𝒓 = 𝒓𝐨 :
(𝝈𝐫 )𝐫𝐨 = 𝟎 𝑟0 ln 𝑘 𝑟o2 ln 2 (1 + 1)] 𝜎θ = 𝐶0 [1 − ln ( ) − 2 (1 + 2 )] (453.42 MPa) [1 − ln 1 − 𝑟 𝑘 −1 𝑟 4−1 (𝝈𝛉 )𝐫𝐨 = −𝟐𝟒𝟑. 𝟏 𝐌𝐏𝐚 𝝈𝐳 = 𝟕𝟑. 𝟐 𝐌𝐏𝐚
YIELDING? @ 𝒓 = 𝒓𝐢 : (𝜎z )t = −555.4 MPa + 2 MPa (𝝈𝐳 )𝐭 = −𝟓𝟓𝟑. 𝟒 𝐌𝐏𝐚
19 𝜎1 − 𝜎2 𝜎y ≤ 2 2 𝜎2 − 𝜎3 𝜎y Tresca ≤ 2 2 𝜎1 − 𝜎3 𝜎y ≤ { 2 2 (𝜎z )temp − (𝜎r )pressure
2
≤
𝜎y 553.4 − 10 450 ≤ 2 2 2
𝟐𝟕𝟏. 𝟐𝟕 𝐌𝐏𝐚 > 𝟐𝟐𝟓 𝐌𝐏𝐚 YIELDING OCCURS!
2.4.3 Other End Conditions due to Thermal Loads
OPEN
𝜎z = 0 CLOSED
-
radial and hoop stress can be found using previous equations for plane strain axial stress, 𝜎z , from the ends
a) resultant axial force is zero
𝑟o
∫ 𝜎z 2π𝑟 d𝑟 = 0 𝑟i
b) 𝜀z = const. (axi − symmetric)
𝑟o 𝑟o 1 ∫ 𝜀z 2π𝑟 d𝑟 ∫ { [𝜎z − ν(𝜎r + 𝜎θ )] + α∆𝑇} 2π𝑟 d𝑟 𝑟i 𝑟i E 2α 1 𝑟o 𝜀z = 2 ∫ 𝑇𝑟 d𝑟 𝑘 − 1 𝑟i2 𝑟i Eα 2 1 𝑟o 𝑇i − 𝑇o 𝑟o 2 𝜎z = [ 2 ∫ 𝑇𝑟 d𝑟 − 𝑇] 𝜎z = Eα ln 𝑘] [1 − 2 ln ( ) − 2 2 1 − ν 𝑘 − 1 𝑟i 𝑟i 2(1 − ν) ln 𝑘 𝑟 𝑘 −1
20
3.0 Stress Functions Direct or exact solutions to the governing equations of elasticity are not usually possible for most practical problems Airy’s Stress Function In order to solve the compatibility equations for stress, Airy suggested that the elasticity be formulated in terms of one mathematical function 𝜑(𝑥, 𝑦) called the stress function. Airy’s stress function 𝜑(𝑥, 𝑦) is related to stress as: 𝜕2𝜑 𝜕2𝜑 𝜕2𝜑 ; 𝜎 = ; 𝜏 = − 𝜕𝑦 2 y 𝜕𝑥 2 xy 𝜕𝑥𝜕𝑦 Consider zero body forces and zero temperature change. 𝜎x =
(3.1)
The compatibility equation for stress is given by: 𝜕2 𝜕2 ∇ (𝜎x + 𝜎y ) = 0 ( 2 + 2 ) (𝜎x + 𝜎y ) = 0 𝜕𝑥 𝜕𝑦
(3.2)
𝜕4𝜑 𝜕4𝜑 𝜕4𝜑 + 2 + ∇2 (∇2 𝜑) = 0 𝜕𝑥 4 𝜕𝑥 2 𝜕𝑦 2 𝜕𝑦 4
(3.3)
2
Substitute (3.1)→ (3.2)
Bi-Harmonic Function Solution (3.3) ∇4 𝜑 = 0 must satisfy the boundary condition. -
Since (3.2) holds true for both plane stress and plain strain, so doe (3.3). Solving 𝜑 → stress → strain → displacement. For many problems, the form of the stress function is specified in advance semi-inverse method) In cases were specifying a stress function in advance is less obvious, the conformal mapping technique is used. - 𝜑 using two complex analytic functions.
3.1 Solution by Polynomial A solution to (3.3) using polynomials are used for rectangular bodies.
a)
Second degree polynomial
21 𝜕2 𝜑 𝑐2 𝜕𝑦 2 𝜕2 𝜑 where: constants 𝜎y = 𝜕𝑥 2 𝑎2 𝜕2 𝜑 𝜏 = − − xy { 𝜕𝑥𝜕𝑦
𝜎x =
𝜑2 =
𝑎2 2 𝑥 2
+ 𝑏2 𝑦 +
𝑐2 2 𝑦 ; 2
𝑏2
𝜎y = 𝑎2 𝜎x = 𝑐2
𝜎x = 𝑐2
𝜏xy = −𝑏2 𝜎y = 𝑎2 if
𝑏2 = 0; biaxial tension 𝑎2 = 𝑏2 0; uniaxial tension 𝑎2 = 𝑐2 0; pure shear
b)
Third order polynomial 𝜑2 =
𝑎3 3 𝑏3 2 𝑐3 𝑑3 𝑥 + 𝑥 𝑦 + 𝑥𝑦 2 + 𝑦 3 6 2 2 6
→ Satisfies (3.3) 𝜕2𝜑 𝑐3 𝑥 + 𝑑3 𝑦 𝜕𝑦 2 𝜕2𝜑 𝜎y = 𝑐3 𝑥 + 𝑏3 𝑦 𝜕𝑥 2 𝜕2𝜑 𝜏xy = − − 𝑏3 𝑥 − 𝑐3 𝑦 𝜕𝑥𝜕𝑦 𝜎x =
if 𝑎3 = 0; 𝑏3 = 0; 𝑐3 = 0 𝜎x = 𝑑3 𝑦 𝜎y = 𝜏xy 0 (simple bending)
𝜎x
22
𝜎y +ℎ
−ℎ 𝜎y
𝐿 if 𝑎3 = 0; 𝑐3 = 0; 𝑑3 = 0 𝜎y = 𝑏3 𝑦 𝜎x = 0 𝜏xy = −𝑏3 𝑥 Method (i) Assuming: 𝜎x ∝ 𝑦 𝜎x ∝ 𝑥 (-BM) 𝜎x = 𝑐1 𝑥𝑦
𝜕2𝜑 𝜕𝜑 𝑐1 𝑥𝑦 2 = + 𝑓1 (𝑥) 𝜕𝑦 2 𝜕𝑦 2
1 𝜑 = 𝑐1 𝑥𝑦 3 + 𝑦𝑓1 (𝑥) + 𝑓2 (𝑥) 6 ∇4 𝜑 = 0 𝑦 𝜕 4 𝑓1 (𝑥) =0 𝜕𝑥 4 𝜕 4 𝑓2 (𝑥) =0 𝜕𝑥 4
𝑦
𝜕 4 𝑓1 (𝑥) 𝜕 4 𝑓2 (𝑥) + =0 𝜕𝑥 4 𝜕𝑥 4 ; 𝑓1 (𝑥) = 𝑐2 𝑥 3 + 𝑐3 𝑥 2 + 𝑐4 𝑥 + 𝑐5 ; 𝑓2 (𝑥) = 𝑐6 𝑥 3 + 𝑐7 𝑥 2 + 𝑐8 𝑥 + 𝑐9
1 𝜑 = 𝑐1 𝑥𝑦 3 + 𝑦(𝑐2 𝑥 3 + 𝑐3 𝑥 2 + 𝑐4 𝑥 + 𝑐5 ) + 𝑐6 𝑥 3 + 𝑐7 𝑥 2 + 𝑐8 𝑥 + 𝑐9 6 𝜎y = 𝜏xy = −
𝜕2𝜑 6[(2𝑦 + 𝑐6 )𝑥 + 2(𝑐3 𝑦 + 𝑐4 )] 𝜕𝑥 2 𝜕2𝜑 1 − 𝑐1 𝑦 2 − 3𝑐2 𝑥 2 − 2𝑐3 𝑥 − 𝑐4 𝜕𝑥𝜕𝑦 2
23 BCs for all 𝑥 {
𝜎y = 0 @ ± ℎ 𝜏xy = 0 @ ± ℎ
𝑐2 , 𝑐3 , 𝑐6 , and 𝑐7 are all zero 1 𝑐4 = − 𝑐1 ℎ2 2 1 𝜏xy = − 𝑐1 (𝑦 2 − ℎ2 ) 2 c)
4th order polynomial 𝜑4 =
𝑎4 𝑏4 𝑐4 𝑑4 𝑒4 𝑥4 + 𝑥3𝑦 + 𝑥2𝑦2 + 𝑥𝑦 3 + 𝑦4 (4)(5) (4)(3) (3)(2) 2 (3)(2)
(3.8)
𝑒4 = −(2𝑐4 + 𝑎4 ) satisfies equation (3.3) 𝜎x =
𝜕 2 𝜑4 𝑏4 3 𝑐4 2 𝑑4 𝑒4 𝑥 + 𝑥 𝑦 + 3 𝑦2 + 4 𝑦3 2 (4)(3) 𝜕𝑦 6 2 6 𝑐4 𝑥 2 + 𝑑4 𝑥𝑦 − (2𝑐4 + 𝑎4 )𝑦 2
𝜕 2 𝜑4 𝑎4 𝑥 2 + 𝑏4 𝑥𝑦 + 𝑐4 𝑦 2 𝜕𝑥 2 𝜕2𝜑 𝑏4 𝑑4 𝜏xy = − − 𝑥 2 − 2𝑐4 𝑥𝑦 − 𝑦 2 𝜕𝑥𝜕𝑦 2 2 Take all of the coefficients except 𝑑4 to be zero. 𝜎y =
𝜎x = 𝑑4 𝑥𝑦, 𝜎y = 0, 𝜎x varies linearly in x and y direction @ 𝑦 = ±ℎ, 𝜏xy is uniform along the transverse direction, 𝜏xy ∝ 𝑦 2
𝜏xy = −
𝑑4 2 𝑦 2
(3.9)
24 Example
𝐿
2ℎ
𝑃
𝑥
𝑦
𝑡
𝑉
𝐵𝑦 ℎ
ℎ
1 𝑐1 𝑡(𝑦 2 − ℎ2 ) d𝑦 −ℎ 2
𝑃 = ∫ 𝜏xy 𝑡 d𝑦 + ∫ −ℎ
𝑐1 = −
3𝑃 𝑃 3 2𝑡ℎ 𝐼
−𝑃 𝑀𝑦 𝜎x = 𝑐1 𝑥𝑦 ( ) 𝑥𝑦 = 𝐼 𝐼 𝜏xy = (
−𝑃 ) (ℎ2 − 𝑦 2 ) 2𝐼
Method (iii) Superposition: 𝜑2 and 𝜑4 with 𝑎2 , 𝑐2 , 𝑎4 , 𝑏4 , 𝑐4 , and 𝑒4 all equal zero 𝜑 = 𝜑2 + 𝜑4 𝜑 = 𝑏2 𝑥𝑦 +
𝑑4 3 𝑥𝑦 6
𝜕2𝜑 𝜎x = 𝑑4 𝑥𝑦 𝜕𝑦 2 𝜎y =
𝜕2𝜑 0 𝜕𝑥 2
𝜏xy = −𝑏2 −
𝑑4 2 𝑦 2
25 𝜏xy = 0 @ ± ℎ 𝑑4 = −
2𝑏2 ℎ2
𝜏xy = −𝑏2 (1 − ℎ
𝑦 2⁄ ) ℎ2
ℎ
3𝑃 𝑃ℎ2 𝑦 2⁄ 𝑃 = ∫ 𝜏xy 𝑡 d𝑦 ∫ 𝑏2 𝑡 (1 − ) d𝑦 𝑏2 = − ℎ2 4ℎ𝑡 2I −ℎ −ℎ 𝜎x = − 𝜏xy = −
𝑃𝑥𝑦 I
𝑃 2 (ℎ − 𝑦 2 ) 2I
3.2 Stress Concentrations Stress concentrations refer to stresses which have higher than normal stresses -
Yielding, fatigue, fracture, etc.
These arise from loading discontinuities or geometrical discontinuities. Loading Discontinuity -
Loading through a knife edge, point, ball bearing or roller bearings, or gear teeth
Geometry Discontinuity -
holes, notches, change in cross-section more pronounced in brittle material than in ductile - find stress concentrations using analytical, numerical, or experimental methods often, it is more convenient to use polar coordinates
3.2.1 Use of Polar Coordinates ∇ 4 𝜑 = ∇ 2 ⁄∇ 2 𝜑 0 ∇2 = (
𝜕2 𝜕 2 rectangular + ) 𝜕𝑥 2 𝜕𝑦 2 coordinates
∇2 = (
𝜕 1 𝜕 1 𝜕2 + + ) 𝜕𝑟 2 𝑟 𝜕𝑟 𝑟 2 𝜕𝜃 2
𝜎r = (𝜎x )θ=0
𝜕2𝜑 1 𝜕 2 𝜑 1 𝜕𝜑 = + ( ) 𝜕𝑦 2 𝑟 2 𝜕𝜃 2 𝑟 𝜕𝑟
(3.12)
(3.13)
26 𝜕2𝜑 𝜕2𝜑 = θ=0 𝜕𝑥 2 𝜕𝑟 2 𝜕2𝜑 𝜕 1 𝜕𝜑 𝜏rθ = (𝜏xy ) (− =− ( ) ) θ=0 𝜕𝑥𝜕𝑦 θ=0 𝜕𝑟 𝑟 𝜕𝜃 For axis symmetry, 𝜑 is independent of angle 𝜃 and equation (3.3) reduces to: 𝜎θ = (𝜎y )
∇2 𝜑 = (
𝜕2 1 𝜕 𝜕 2 𝜑 1 𝜕𝜑 + + ) ( ) 0 Euler Equation 𝜕𝑟 2 𝑟 𝜕𝑟 𝜕𝑟 2 𝑟 𝜕𝑟
(3.14)
Equation (3.13) can be reduced to: 𝜎r = General solution of (3.14)
1 𝜕𝜑 𝜕2𝜑 ; 𝜎θ = 2 ; 𝜏rθ = 0 𝑟 𝜕𝑟 𝜕𝑟
(3.15)
𝜎r = 𝑐1 ln 𝑟 + 𝑐2 𝑟 2 ln 𝑟 + 𝑐3 𝑟 2 + 𝑐4
(3.16)
where 𝑐1 to 𝑐4 are constants 𝑐1 + 𝑐2 (1 + 2 ln 𝑟) + 2𝑐3 𝑟2 𝑐1 𝜎θ = − 2 + 𝑐2 (3 + 2 ln 𝑥) + 2𝑐3 𝑟 𝜏rθ = 0 If there are no holes, 𝑐1 and 𝑐2 are zero. 𝜎r =
(3.17)
3.2.2 Stresses due to Concentration Loads -
point load (mathematic abstraction) in practice, “point” loads are always applied at a small area analytical solutions therefore are NOT valid at the point
Example Comparison of a wedge (unit thickness) → indenter
𝜑 = 𝑐𝑃𝑟𝜃 sin 𝜃 -
𝑦
satisfies (3.3)
𝑃
𝑟
𝛼
𝜃
𝑥 𝛼
𝜎r =
1 𝜕𝜑 1 𝜕 2 𝜑 1 + 2 2 2𝑐𝑃 cos 𝜃 𝑟 𝜕𝑟 𝑟 𝜕𝜃 𝑟
27 𝜕2𝜑 0 𝜕𝑟 2
𝜎θ = 𝜏rθ =
1 𝜕𝜑 1 𝜕 2 𝜑 − =0 𝑟 2 𝜕𝜃 𝑟 𝜕𝜃𝜕𝑟
𝛼
𝛼
𝑃 − 2 ∫ 𝜎r cos 𝜃 𝑟d𝜃 = 0 𝑃 = −2 ∫ 2𝑐𝑃 0
0
cos 2𝜃 + 1 cos 𝜃 = 2 2
𝜎r =
Special Case where 𝛼 =
π 2
𝛼 cos 2𝜃 + 1 cos 2 𝜃 d𝜃 = −1 𝑟d𝜃 4𝑐 ∫ 2 𝑟 0 1 𝑐=− } 2𝛼 + sin 2𝛼
−2𝑃 cos 𝜃 2𝛼 + sin 2𝛼 𝑟
(3.19)
𝜎θ = 0, 𝜏rθ = 0
Flamant Problem -
point or line load on semi-infinite solid 𝜎r =
−2𝑃 cos 𝜃 π 𝑟
(3.20)
𝜎θ = 0, 𝜏rθ = 0
𝑃
𝜃
𝑑
𝑟 = 𝑑 cos 𝜃 𝜎r = constant for a given d at all points on the circle
(b) Bending of a wedge:
−2𝑃 π𝑑
(3.21)
28
𝑦 𝑟
𝛼
𝜃 𝐹
𝑥 𝛼
π π 𝜑 = 𝑎𝐹𝑟 ( − 𝜃) sin ( − 𝜃) 𝑎𝐹𝑟𝜃1 sin 𝜃1 2 2 𝜎r =
−2𝐹 cos 𝜃1 −2𝐹 cos 𝜃1 𝑟(2𝛼 − sin 2𝛼) 𝑟(2𝛼 − sin 2𝛼)
𝜎θ = 𝜏rθ 0 𝜎r = −
𝑦 𝑃a
2 𝑃a cos 𝜃 𝑃t sin 𝜃 + [ ] 𝑡𝑟 2𝛼 + sin 2𝛼 2𝛼 − sin 2𝛼
𝛼 𝑥
𝛽
𝛼
𝑃t
Example: An orthogonal cutting tool has a rake angle of 20°, clearance angle of 10°, and width of cut of 4 mm. The horizontal and vertical loads are 𝑃x = 1.5 kN, 𝑃y = 2.5 kN. Find the minimum tool nose radius of 𝜎r ≤ 900 MPa.
𝑃a
𝑃y
50°
𝑃b 40°
𝑃x
20°
30° 30° 10°
Mitchell’s Solution
r 60°
29 2 𝑃a cos 𝜃 𝑃b sin 𝜃 + [ ] (𝜔)𝑟 2𝛼 + sin 2𝛼 2𝛼 − sin 2𝛼 𝑃a = 𝑃x cos 50° + 𝑃y cos 40° 5.75 kN 𝑃b = 𝑃y sin 40° + 𝑃x sin 50° 0.916 kN π 𝛼 = 𝜃 30° = 6 3 2 (5.75 × 10 ) cos 30° 0.916 × 103 ) sin 30° − [ + ] π π 4𝑟 + sin 60° − sin 60° 3 3 = 900 MPa 𝜎r = −
r 30°
𝒓 = 𝟐. 𝟖𝟓 𝐦𝐦; 𝑟t = 𝑟 tan 30° 𝒓𝐭 = 𝟏. 𝟔𝟓 𝐦𝐦
3.2.3 Stress Concentrations around a Small Circular Hole (Kirsh Solution)
𝜑(𝑟, 𝜃) =
𝜎θ 𝑟 2 𝜎0 𝑏4 ( − 𝑏 2 ln 𝑟) + (2𝑏 2 − 𝑟 2 − 2 ) cos 2𝜃 2 2 4 𝑟
𝑦 𝑟
𝐴
D
𝐛
B
𝜎r =
C
π 3π , 2 2
π 2
𝜎0
1 𝜕 2 𝜑 1 𝜕𝜑 𝜎0 𝑏2 𝜎0 𝑏2 𝑏4 + − + − 4 + 3 (1 ) (1 ) cos 2𝜃 𝑟 2 𝜕𝜃 2 𝑟 𝜕𝑟 2 𝑟2 2 𝑟2 𝑟4 𝜕2𝜑 𝜎0 𝑏2 𝜎0 𝑏4 𝜎θ = 2 (1 + 2 ) + (−1 − 3 4 ) cos 2𝜃 𝜕𝑟 2 𝑟 2 𝑟 2 𝜕 1 𝜕𝜑 𝜎0 𝑏 𝑏4 𝜏rθ = − ( ) (−1 − 2 2 + 3 4 ) sin 2𝜃 𝜕𝑟 𝑟 𝜕𝜃 2 𝑟 𝑟
Points A and B have maximum stress:
Putting
𝑥
C: 0 rad D: π rad 3π and 2 into equation (3.23) for 𝜎θ at 𝑏 = 𝑟, 𝜎θ = 3𝜎0 𝜎x
(3.23)
30
Stress concentration:
3𝜎0 𝜎0
=3 @ 𝜃 = 0 and π: 𝜎θ = 𝜎y −𝜎0 (compressive) @ 𝑟 > 10𝑏 (𝜎θ )π 2
,
3π 2
𝜎0 𝑏2 𝑏4 = (2 + 2 + 3 4 ) 𝜎0 2 𝑟 𝑟
Result is valid for finite width plate 𝑤 > 10𝑏.
4.0 Shock and Impact Loading -
When forces or displacements are applied suddenly to structural members, the stress levels and deformation are often much higher than those that occur by the same forces or displacements applied gradually Rapidly moving loads: (train crossing a bridge, drop of a forge hammer, sudden loads produced during combustion in the power stroke of an internal combustion engine, collision, acceleration, etc.) Impact shock load when 1 𝑡applied ≤ 𝑇 2 } time of highest natural frequency of the structure 𝑡applied ≤ 𝑇
-
-
-
Quasi-static: 𝑡applied ≥ 3𝑇
4.1 Stress and Displacement in an Elastic Axial Bar
circular x − section 𝐴, 𝐸 𝑊(striking mass) 𝐿 ℎ 𝑦 pan
31 -
mass of the pan and the bar is small compared to the striking mass (W) 𝑦 ≠ 𝑓(𝑡) Linear elastic NO energy lost
strain energy
𝐹
𝑦 Energy balance: external work = stored strain energy of the bar 1 𝑊(ℎ + 𝑦̂) = 𝐹̂ 𝑦̂; (̂ denotes maximum) 2 𝜎̂ = 𝐸𝜀̂ 𝐸
𝑦̂ 𝜎̂𝐿 = 𝑦̂ 𝐿 𝐸
𝐹̂ = 𝜎̂𝐴 𝑊 (ℎ +
𝜎̂𝐿 1 𝜎̂𝐿 ) = 𝜎̂𝐴 𝐸 2 𝐸
𝐴𝐿 2 𝑊𝐿 𝜎̂ − ( ) 𝜎̂ − 𝑊ℎ = 0 2𝐸 𝐸
𝜎̂ =
𝑊 2ℎ𝐸𝐴 [1 + √1 + ] 𝐴 𝑊𝐿
(4.1)
(4.2)
∆ impact factor 𝑦̂ =
𝑊𝐿 2ℎ𝐸𝐴 [1 + √1 + ] 𝐴𝐸 𝑊𝐿
(4.3)
2ℎ𝐸𝐴 ] 𝑊𝐿
(4.4)
𝐹̂ = 𝑊 [1 + √1 + 𝐹̂ is the effective dynamic fore due to impact. 𝑦static =
𝑊𝐿 𝐴𝐸
32 Impact, ∆ ∆= 1 + √1 + ∆= 1 + √1 + -
2ℎ𝐸𝐴 𝑊𝐿
(4.5)
2ℎ 𝑦static
If the load is applied suddenly even at h=0; 𝜎̂ =
2𝑊 2(𝜎max )static 𝐴
𝐹̂ = 2(𝐹max )static 𝑦̂ = 2(𝑦max )static -
𝜎̂ can be reduced by increasing L or A and decreasing 𝜎static =
𝐹 𝐴
Applications: -
wood is used as railway track long bolts used to attach ends of the pneumatic cylinder of jack hammers if the bar has a weight of q per unit length, the impact factor, ∆ can be modified to:
∆∗ = [1 + √1 +
-
2ℎ𝐸𝐴 1 ( )] 𝑊𝐿 1 + 𝑞𝐿 3𝑊
Analysis is similar for cases involving impact of moving loads. (KE)moving load = 𝑢structure
Example (a) Find the max stress in a steel rod. (i) (ii)
with washer without washer
(4.7)
33 rubber washer: N mm ∅ = 15 mm 𝐸 = 200 GPa 𝑘=9
10 kg 1.5 m ℎ rubber washer 16 mm
i)
with washer 𝜎̂ =
𝑊 = (10 kg) (9.81 𝐴=
m ) 98.1 N s2
π 2 π ∅ (15 mm)2 = 177 mm2 4 4
𝑦static =
(98.1 N)(1 500 mm) (98.1 N) 𝑊𝐿 𝑊 + + 𝒚𝐬𝐭𝐚𝐭𝐢𝐜 = 𝟏𝟎. 𝟗 𝐦𝐦 N N 𝐴𝐸 𝑘 (177 mm2 ) (200 × 103 ) 9 mm mm2
without washer
with washer
𝜎̂ =
ii)
𝑊 2ℎ [1 + √1 + ] 𝐴 𝑦static
98.1 N 2(1 000 mm) ̂ = 𝟖. 𝟎𝟖 𝐌𝐏𝐚 [1 + √1 + ] 𝝈 2 177 mm 10.9 mm
without washer
𝜎̂ =
98.1 N 2(1 000 mm) ̂ = 𝟑𝟖𝟓 𝐌𝐏𝐚 [1 + √1 + ] 𝝈 2 177 mm 4.157 × 10−3 mm
Example (b) Find the max h allowed to avoid yielding.
34
𝑊 = 60 N 𝑏 = 2.5 mm
−7
𝐼 = 8.79 × 10
4
m
ℎ
𝐸 = 207 GPa 𝜎y = 207 MPa
𝑑 = 75 mm
𝑉̂
1.5 m
Equating external energy = stored strain energy: 𝑊(ℎ + 𝑉̂ ); 𝑉̂ =
𝜎̂ =
𝑃𝐿3 𝑃 = dynamic load 3𝐸𝐼
𝑃𝐿 𝑑⁄2 𝑀𝑦 𝜎̂𝐼 𝜎̂ = 𝑃 = 𝐼 𝐼 𝐿 𝑑⁄2 𝜎̂𝐼 𝐿3 ̂ 𝑉=( ) 𝐿 𝑑⁄2 3𝐸𝐼 𝜎̂𝐼 𝐿3 𝑃𝐸 = 𝑊 (ℎ + ( ) ) 𝐿 𝑑⁄2 3𝐸𝐼
Stored strain energy: 2
1 𝐿 𝑀2 1 𝐿 (𝑃𝑥)2 𝑃2 𝐿2 𝜎̂𝐼 𝐿3 𝑈= ∫ d𝑥 𝑈 = ∫ d𝑥 𝑈 =( ) 2 0 𝐸𝐼 2 0 𝐸𝐼 6𝐸𝐼 𝐿 𝑑⁄2 6𝐸𝐼 𝑃𝐸 = 𝑈 𝜎̂𝐿2 𝜎̂ 2 𝐼𝐿 𝑊 (ℎ + )= 2 3(𝑑⁄2)𝐸 6(𝑑⁄2) 𝐸 60 (ℎ +
(270 × 106 Pa)2 (8.79 × 10−7 m4 )(1.5 m) (270 × 106 Pa)(1.5 m)2 = ) 3(0.0375 m)(207 × 109 Pa) 6(0.0375 m)2 (207 × 109 Pa) 𝒉 = 𝟎. 𝟖𝟗𝟏 𝐦
35
Example (c) 𝑊 = 12 kg 𝐸 = 8 GPa (WOOD)
ℎ A
𝐼 = 1.5625 × 10−6 m4 𝐿 = 1.5 m
C′
B
C′′ 𝐿 (i) (ii)
𝐿
max h allowed if max deflection is limited to 38 mm amount of energy absorbed by each beam
𝑃𝐸 = 𝑈 𝑃𝐸 = 𝑊(ℎ + 𝛿max )
FBD
𝑃
𝑅′
A 𝑅 ′′
𝑅 B
𝑀
𝑅 ′′′ (i) Beam B (0.038 m)3(8 × 109 Pa)(1.5625 × 10−6 m4 ) 𝑅𝐿3 𝛿B 3𝐸𝐼 𝛿B = 𝑅 = 𝑅 = 𝑹 = 𝟒𝟐𝟐. 𝟐 𝐍 (1.5 m)3 3𝐸𝐼 𝐿3 Beam A
36
𝛿A = 𝑃 =
(𝑃 − 𝑅)(2𝐿)3 𝑊 ′ 𝐿3 𝛿A 6𝐸𝐼 𝑃 = +𝑅 48𝐸𝐼 48𝐸𝐼 𝐿3
(0.038 m)6(8 × 109 Pa)(1.5625 × 10−6 m4 ) + 422.2 N (1.5 m)3 𝑷 = 𝟏 𝟐𝟔𝟔. 𝟕 𝐍
∆= 1 + √1 +
2ℎ 𝑦static
∆=
𝑃 1 266.7 N m ∆= 𝟏𝟎. 𝟕𝟔 𝑊 12 kg (9.81 2 ) s
𝒉 = 𝟏𝟔𝟔. 𝟒 𝐦𝐦
𝛿static =
𝛿max 0.038 m ∆ 10.76
𝑃𝐸 = 𝑈 1 𝑊(ℎ + 𝛿max ) = 𝑃𝛿; 𝒉 = 𝟏𝟔𝟔. 𝟒 𝐦𝐦 2 (ii) 1 1 𝑈A = 𝑃′ 𝛿 ′ (𝑃 − 𝑅)𝛿 𝑼𝐀 = 𝟏𝟔. 𝟎𝟓 𝐉 2 2 1 ′ ′ 1 𝑈B = 𝑃 𝛿 𝑅𝛿 𝑼𝐀 = 𝟖. 𝟎𝟐 𝐉 2 2
37 Superposition can be used to obtain the deflection.
I.
𝑊 A
B
a
b
𝑀 = 𝑊𝑏
B
A
𝜃
𝑊 𝛿
𝑉̂ = 𝛿 + 𝜃𝑏 𝑏 tan 𝜃𝑏 = 𝜃𝑏
II.
𝑊 A
B C
A
B C
A
B C 𝑅
38
4.2 Effects of Geometry
B
A 𝐿
𝐴1
stress concentrations are neglected for beam B
𝐿
𝐴2 = 2𝐴1
3 𝐿 3
𝐴1
𝐿
𝐴2
3
1 𝑈A = 𝑃̂𝛿 2 𝑃̂ = 𝜎̂𝐴, 𝛿 = 𝜀𝐿, 𝜎̂ = 𝐸𝜀 1 𝜎̂ 2 𝐴𝐿 2𝐸 1 𝜎̂ 2 𝑈A = 𝐴 𝐿 2𝐸 1 𝑈=
𝑈B =
1 𝜎̂ 2 𝐿 1 𝜎̂ 2 𝐿 𝐴1 + 2 [ 𝐴2 ] 2𝐸 3 2𝐸 3 2 𝑈B = 𝑈𝐴 2𝐴1 3
To design or maximum resistance to impact load: -
use a low E increase length, L distribute the stress as uniformly as possible avoid stress concentrations
4.3 Stress Wave Propagation under Impact Loading The above provides a quick solution estimate of maximum stress and deflection under impact loading. Exact consideration involves the motion of stress waves traveling within the structures. Consider the 1-D problem: 𝐿 m
m′
m
m′
n
n′
n
n′
𝑃imp 𝑐
(wave velocity)
𝑑𝑥 𝑢 +
d𝑢 d𝑥 d𝑥
39 At the instant of impact of the free end, a very thin segment of the bar directly under the load is set into motion. The rest of the bar, remote from the load, 𝑃, remains undisturbed for a small, finite length of time. The deformation due to 𝑃 propagates along the bar with time in a form of an elastic deformation: stress wave. If 𝐿 is large, time required for the wave propagation becomes significant and must be considered. Two different types of stress waves:
𝑃
𝑃 𝑥
𝑥
Longitudinal Wave Particle motion in the 𝑥-direction Pressure wave (𝑃wave ) Velocity (𝑐L )
Transverse Wave Shear wave (s-wave) Particle motion in the 𝑦-direction
𝐹 = 𝑚𝑎 𝐹 = 𝜎𝐴 𝜎 = 𝐸𝜀 𝜀=
d𝑢 d𝑥
𝐹 = 𝐸𝐴
d𝑢 d𝑥 𝒄𝟐 =
𝑚 = 𝜌𝑉 𝑚 = 𝜌𝐴 𝑑𝑥 d2 𝑢 𝑎= 2 d𝑡 d𝑢 d2 𝑢 𝐸𝐴 = 𝜌𝐴 𝑑𝑥 2 d𝑥 d𝑡 𝐸 d2 𝑢 d2 𝑢 = 𝜌 d𝑥 2 d𝑡 2 𝐝𝟐 𝒖 𝐝𝟐 𝒖 + 𝐝𝒙𝟐 𝐝𝒕𝟐
1-D wave equation 𝐸 𝜌
𝑐L = √ : velocity of longitudinal wave 𝐺
𝑐S = √𝜌: velocity of transverse wave e.g. 207×109 7 850
steel: 𝑐L = √
copper: 𝑐s = √
5 135
120×109 8 900
m s
3 672
m s
40 Verify that 𝑓(𝑥 ± 𝑐𝑡) is a solution of equation (4.8) where; + is the negative 𝑥 direction − is the positive 𝑥 direction 𝑢(𝑥, 𝑡) = 𝑓1 (𝑥 − 𝑐𝑡) + 𝑓2 (𝑥 + 𝑐𝑡) 𝑢 = 𝐴 sin (
2π ) (𝑥 ∓ 𝑐𝑡) 𝐿
𝒖𝒊 = 𝑨𝒊 𝒆𝒊(𝝎𝒕−𝒌𝒊 𝒙𝒊 ) 𝑘: wave number λ
2π
wave
𝜔: angular frequency
Stress Waves at Fixed and Free Ends @ fixed end Waves are reflected and unchanged i.e. compressive stays compressive @ free end Waves are reflected with same amplitude but opposite in sign. compressive → tensile tensile → compressive 2-D (longitudinal and shear) net tensile strength exceeds rupture stress
scab
explosive internal pressure
Fracture occurs at the thickest section as compared to thin section in static loading.
41 In real materials, there is a phase difference between the stress and strain due to hysteresis or internal friction damping (energy losses). Stress level decays with distance.
4.4 Changes in Material Properties under Impact Loading
rapid loading
𝜎
static loading
𝜀
𝑘1D < 𝑘1C 30% less 𝑘1C
𝑘1D
10
20
30
m s
loading speed
42
5.0 Elasto-Plastic Analysis Yield Criteria Tresca: 𝜎1 − 𝜎2 𝜎y = 2 2 𝜎2 − 𝜎3 𝜎y = 2 2 𝜎1 − 𝜎3 𝜎y = 2 2
(5.1)
Von Mises 1 √2
1
[(𝜎1 − 𝜎2 )2 + (𝜎2 − 𝜎3 )2 + (𝜎1 − 𝜎3 )2 ]2 = 𝜎y
(5.2a)
In 2D 1
(𝜎12 + 𝜎22 − 𝜎1 𝜎2 )2 = 𝜎y
(5.2b)
When the equivalent stress in parts of a component exceeds the yield stress of the material, elastoplastic deformation occurs. Elasto-plastic analysis results in an analytical solution and is only possible for relatively few types of problems with simple geometry and idealized stress-strain behaviour (i.e. bending, torsion, axis-symmetric problems.) General Stress-Strain Curve:
𝜎
×
𝜎y
𝜀
43
𝜎
𝜎
𝜎y
𝜎y
𝜀
𝜀
(a) elastic-perfectly plastic
(b) elastic-linear plastic/hardening
𝜎
𝜎p = 2(𝐴 + 𝐵𝜀 𝑛−1 )
𝜎y
𝜀 ***NOT COVERED IN THIS COURSE (c) non-linear Permanent set occurs when the component is loaded beyond its yield point. When the load is removed there are residual stress. These can be both beneficial and harmful.
𝜎 loading
𝜎y
unloading
𝜀p
𝜀
44
5.1 Beam Bending Consider (a) and (b): 𝐢𝐧𝐜𝐫𝐞𝐚𝐬𝐢𝐧𝐠 𝐥𝐨𝐚𝐝 𝑏
ℎ
𝜎y
𝜎y plastic hinge
𝑦p 𝑑 𝑦
ℎ
𝜎y
𝜎y
(i) onset of yielding
(ii) elasto-plastic deformation
𝜎=
(iii) fully plastic
𝑀𝑦 𝜎𝐼 𝑀 = 𝐼 𝑦
Elasto-plastic bending moment: 𝑀ep = 𝑀e + 𝑀p elastic
𝜎y 𝑏(𝑑 − 2ℎ)2 + 𝜎y 𝑏ℎ(𝑑 − ℎ) 6 plastic
𝜎y 𝑏𝑑2 ℎ ℎ [1 + 2 (1 − )] 6 𝑑 𝑑 𝜎y 𝑏 2 2 𝑀ep = [3𝑑 − 4𝑦p ] 12
(5.3a)
𝑀ep =
(5.3b) distance from neutral axis
Fully plastic: 𝑀ep =
𝜎y 𝑏𝑑2 1 1 𝑏𝑑2 [1 + 2 (1 − )] 𝜎y = 6 2 2 4 𝑀=
𝜎𝐼 𝑏𝑑2 𝜎y = 𝑦 6
𝑧p (shape factor) =
𝑀p 1.5 𝑀y
The shape factor five a measure of the increase in strength-carrying capacity available beyond the elastic limit.
45 Removal of loads after elasto-plastic deformation:
𝜎 𝑀ep
𝜎y
−𝑀ep
no net moment
𝜀
𝜀p
-
equivalent to superposition of a linear-elastic unloading stress distribution over the initial stress distribution unloading stress distribution must be done to a moment of equal magnitude but opposite in direction of the applied load 𝑦 A
𝑀ep
B
C
D
A B
C
ℎ
D
−𝑀ep
***not to scale
Subtract unloading from loading stress distribution to find residual stress distribution: 𝑦
Example: Rectangular section beam loaded beyond elastic limit; yielding of half the cross section. Find the residual stress distribution when the load is removed. 𝑏
𝑑 𝑦0
7.5 mm 7.5 mm
Elastic-perfectly plastic; 𝜎y = 600 MPa; 𝑏 = 10 mm; 𝑑 = 30 mm
46
𝑀ep =
𝜎y 𝑏𝑑2 (600 × 106 )(10 × 10−3 )(30 × 10−3 )2 ℎ ℎ 1 1 [1 + 2 (1 − )] [1 + 2 (1 − )] 6 𝑑 𝑑 6 4 4 𝑴𝐞𝐩 = 𝟏 𝟐𝟑𝟕. 𝟓 𝐍 ∙ 𝐦
Elastic stresses due to 𝑀ep : 𝑑 −𝑀ep 2 𝑑 𝑦 = 𝜎 = 𝝈 = −𝟖𝟐𝟓. 𝟎 𝐌𝐏𝐚 2 𝐼 𝑦 = 𝑦0 𝜎=
−𝑀ep 𝐼
𝑑 4 𝝈 = −𝟒𝟏𝟐. 𝟓 𝐌𝐏𝐚
Residual stress distribution: A: 𝜎x = −825.0 + 600 − 225 MPa B: 𝜎x = −412.5 + 600 187.5 MPa C: 𝜎x = 412.5 − 600 − 187.5 MPa D: 𝜎x = 825. − 600 225 MPa 𝑦
𝑦 A = 600
𝑀ep
A = −225
B = 600
B = 187.5
𝜎x (MPa)
C = −600 D = −600
C = −187.5
−𝑀ep
𝐷 = 225
Springback refers to the elastic recovery of the plastically deformed structure when the loads are removed. e.g. 𝛿f 𝛿ep
springback
𝛿f = 𝛿ep − 𝛿e -
𝑦
strain: 𝜀 = 𝑅; 𝑅 (radius of curvature)
𝛿e
47
𝜀=
𝑦
𝜎y 𝑦p } still elastic up to 𝑦p 𝑅ep E
A
A
𝑅ep =
𝑦p E 𝜎y
𝑦p =
𝑅ep 𝜎y E
𝑦p
Substituting 𝑦p back into equation (5.3b): 𝑀ep
𝜎y 𝑏 𝑅ep 𝜎y 2 2 = [3𝑑 − 4 ( ) ] 12 E
(5.4)
EI 𝑏𝑑 3 E 𝑅e 12 𝑅e
(5.5)
For elastic bending: 𝑀e =
𝑅 2 = 𝐿2 + (𝑅 − 𝛿)2 Since 𝛿 is small compared to R and L, neglect the 𝛿 2 term.
𝛿
𝑅−𝛿
𝐿
𝐿2 ≈ 2𝑅𝛿 𝛿 =
𝑅
𝐿2 2𝑅
𝛿f = 𝛿ep − 𝛿e 𝐿2 𝐿2 𝐿2 1 = − ; (curvature) 2𝑅f 2𝑅ep 2𝑅e 𝑅 𝑅ep 𝑅ep 𝑅ep 1 1 1 = − =1− = 0 complete springback } measure of springback; when 𝑅f 𝑅ep 𝑅e 𝑅f 𝑅e 𝑅e Since elastic loading after elastic-plastic deformation:
|𝑀e | = |𝑀ep | 𝜎 𝜎y
𝑀ep
𝑀e elastic unloading moment
𝜀
48 𝑅ep 𝜎y 2 𝑏𝑑3 E 𝜎y 𝑏 2 = [3𝑑 − 4 ( ) ] 12𝑅e 12 E 𝑅ep 𝑅ep 𝜎y 𝑅ep 𝜎y 3 = 1 − 3[ ]+4[ ] 𝑅f E𝑑 E𝑑 𝑅ep 𝑅ep 𝜎y 2𝑅ep 𝜎y 2 =( + 1) ( ) 𝑅f E𝑑 E𝑑
(5.7)
In the fully plastic case: 𝑀p =
𝑏𝑑 2 𝜎 4 y
𝑏
𝜎y ℎ=
𝑑 2
𝑑 𝑑 2
𝜎y
Elastic recovery:
|𝑀e | = |𝑀ep | EI 𝑏𝑑2 = 𝜎 𝑅e 4 y 𝑅p 1 − 𝑏𝑑2 𝜎y 𝑅p = 𝑅f 4EI Using 𝐼 =
𝑏𝑑 3 12
𝑅p 3𝜎y 𝑅p =1− 𝑅f E𝑑
(5.8)
Example: Determine the tool radius required to produce a final bend radius of 50 mm in a steel strip 20 mm wide and 2 mm thick. What is the required moment? Assume fully plastic deformation and against the assumptions.
𝜎y = 300 MPa; E = 200 GPa 3𝜎y 𝑅p 𝑅p =1− 𝑅f E𝑑 𝑅p = (50 × 10−3 m) [1 −
3(300 × 106 Pa)𝑅p ] 𝑹𝐩 = 𝟒𝟒. 𝟗 𝐦𝐦 (200 × 109 Pa)(2 × 10−3 m)
49
𝑀p =
(20 × 10−3 m)(2 × 10−3 m)2 𝑏𝑑2 (300 × 106 Pa) 𝑴𝐩 = 𝟔 𝐍 ∙ 𝐦 𝜎y 4 4
Check the elastic case:
𝜎y 𝜎y 𝑅ep 𝑦p (300 × 106 Pa)(44.9 × 10−3 m) = 𝑦p = 𝒚𝐩 = 𝟎. 𝟎𝟔𝟕 𝐦𝐦 𝑅ep E E 200 × 109 Pa Since 𝑦p is small, the assumption is correct.
Deflection To find the deflection, use moment-curvature relationship as in the elastic case. 2
𝑀 𝜎 𝐸 𝜕 𝑣 = = 𝑀 = −𝐸𝐼 2 𝐼 𝑦 𝑅 𝜕𝑥
The relationship between 𝑅ep and 𝑀ep is more complicated, integration has to be separated between the region that is plastically deformed and the region that is still elastic.
5.1.2 Elastic-Linear Strain Hardening In
𝜎 𝜎o
𝑦
tan−1 𝐸p
𝜎y
𝜎o 𝜎y
𝜎o : stresses at 𝑦p
the outer are max
𝜎
tan−1 𝐸
𝜀o
𝜀
In the elastic range: |𝑦| < |𝑦p | 𝜎e = 𝜎y
𝑦 𝑦p
In the plastic region: |𝑦| > |𝑦p |
(5.9a)
50
𝑦 − 𝑦p 𝜎p = 𝜎y + (𝜎o − 𝜎y ) (𝑦 ) − 𝑦 p 2
(5.9b)
Using the stress-strain plot:
(𝜎o − 𝜎y ) = (𝜀o −
𝜀o = 𝑦p 𝑅ep
=
𝜎y )E E p
𝑑 (2) 𝑅ep
𝜎y ; when 𝑦 = 𝑦p E 𝜀o =
𝑑 ( 2) 𝜎y E𝑦p
𝑑 ( ) Ep 𝜎 [ 2 − 1] (𝜎o − 𝜎y ) = E y 𝑦p
𝜎p = 𝜎y +
(5.10a) for |𝑦| > |𝑦p |
Ep 𝑦 − 𝑦p 𝜎y [ ] E 𝑦p
(5.10b)
Elastic-Plastic Bending Moment 𝑑 2
𝑑 2
𝑦p
𝑀ep = ∫ 𝜎 𝑏𝑦 d𝑦 2 [∫ 𝜎e 𝑏𝑦 d𝑦 + ∫ 𝜎p 𝑏𝑦 d𝑦] 0
0
𝑀ep = when
Ep E
𝜎y 𝑏𝑑2 12
[(3 − 𝑐2 ) +
𝑦p
𝑦p Ep (2 − 3𝑐 + 𝑐3 )] ; where 𝑐 = 𝑑⁄ 𝑐E 2
= 0 elastic-perfectly plastic case (5.3b)
𝜎 tan−1 𝐸p
𝜀
51 For residual stress calculations, similar to the elastic-perfectly plastic case, assume elastic unloading due to applied −𝑀ep . -
Superimpose elastic stress distribution due to −𝑀ep to that due to loading by 𝑀ep .
Springback occurs when unloading from elasto-plastic deformation due to 𝑀ep : |𝑀e | = |𝑀ep | 12𝑀ep 1 = 𝑅e 𝑏𝑑3 E Moment to initiate yielding: 𝑀y =
𝑏𝑑2 𝜎 6 y
2𝜎y 𝑀ep 1 = 𝑅e E𝑑 𝑀y
𝜎y 1 1 = ( ) 𝑅e E 𝑦p 1 1 1 = − 𝑅f 𝑅ep 𝑅e
𝜎y 2𝜎y 𝑀ep 1 = − ( ) 𝑅f E𝑦 E𝑑 𝑀y 𝑦p 1 2𝜎y 1 𝑀p = ( − ) ; where 𝑐 = 𝑑⁄ 𝑅f E𝑑 𝑐 𝑀y
(5.13)
2
Example: A beam of rectangular cross-section (𝑏 = 25 mm, 𝑑 = 40 mm) is loaded gradually so that the plastic zone spreads inwards from the outer fibers. Fine the values of the bending moment that cause: a) yielding initiation b) yielding up to a depth of 10 mm from the outer fibers c) What is the final radius of curvature of the beam upon unloading?
𝜎y = 250 MPa E = 200 GPa Ep = 10 GPa (linear strain − hardening material) 10 mm 10 mm 40 mm
𝑦p = 10 mm
𝑦p
52 a) (25 × 10−3 m)(40 × 10−3 m)2 𝑏𝑑2 (250 × 106 Pa) 𝑴𝐲 = 𝟏. 𝟔𝟔𝟕 𝐤𝐍 ∙ 𝐦 𝑀y = 𝜎 6 y 6 b) 𝑀ep =
𝜎y 𝑏𝑑2 12
[(3 − 𝑐2 ) +
𝑐=
Ep (2 − 3𝑐 + 𝑐3 )] 𝑐E
𝑦p 10 = 0.5 𝑑⁄ 20 2
(250 × 106 Pa) (25 × 10−3 m)(40 × 10−3 m)2 10 (2 − 3(0.5) + 0.53 )] [(3 − 0.52 ) + 12 0.5(250) = 𝟐. 𝟖𝟏𝟐𝟓 𝐤𝐍 ∙ 𝐦
𝑀ep = 𝑴𝐞𝐩 c)
1 2𝜎y 1 𝑀p 2(250 × 106 Pa) 1 2.8125 = − ( − ) [ ] 9 −3 (200 × 10 Pa)(40 × 10 m) 0.5 1.667 𝑅f E𝑑 𝑐 𝑀y 1 = 0.0195 m−1 𝑅f 𝑹𝐟 = 𝟓𝟏. 𝟐 𝐦
Perfectly plastic:
Ep
=0
E
𝑴𝐞𝐩 = 𝟐. 𝟐 𝐤𝐍 ∙ 𝐦 𝑹𝐟 = 𝟐𝟓. 𝟔 𝐦
5.2 Thick-Walled Cylinders under Internal Pressure Assume: i) elastic-perfectly plastic
𝑟
ii) Tresca’s maximum shear stress yield criterion
𝑟i
𝑟o
53
𝑝i 𝑟o2 𝑝i 𝑟o2 𝑟o 𝜎r = 2 (1 − 2 ) ; 𝜎θ = 2 (1 + 2 ) ; 𝑘 = 𝑘 −1 𝑟 𝑘 −1 𝑟 𝑟i In a thick-walled cylinder under internal pressure:
𝜎θ > 𝜎z > 𝜎r 𝜏̂ =
𝜎y 𝜎θ − 𝜎r 𝑝i 𝑟o2 2 ( 2) 2 2 𝑘 −1 𝑟
(5.15)
Yielding will occur first at the bore, where 𝑟 = 𝑟i
𝜎y 𝑝i − 𝑘 2 2 𝑘 −1 2
𝜏̂ =
(𝑝)elastic
=
breakdown
𝑘2 − 1 𝜎 2𝑘 2 y -
𝑟 = 𝑛𝑟i
(5.16)
consider partial yielding in the cylinder to a depth equivalent to a radius ratio if 𝑛 (i.e. yielding at 𝑟 = 𝑟i to 𝑟 = 𝑛𝑟i )
𝑟i
𝑟o elastic zone
Let the radial pressure at 𝑟 = 𝑛𝑟i to be equal to 𝑃𝑛 -
consider the elastic zone 𝑟
𝑘
substitute for 𝑘 the quantity 𝑛𝑟o = 𝑛 i
𝑃𝑛
𝜏̂ =
𝑘
𝑘
2
(𝑛) − 1 𝑃𝑛 =
2
( ) 𝑛
𝜎y (𝑘2 − 𝑛2 ) 2𝑘2
Consider the plastic zone equilibrium equation: 𝑟
d𝜎r d𝑟
+ 𝜎r − 𝜎 θ = 0
𝜎y 2
(5.17)
54
In the plastic zone: 𝜏̂ = 𝑟
d𝜎r d𝑟
𝜎y 𝜎θ − 𝜎r 2 2
+ 𝜎y = 0 d𝜎r = 𝜎y
d𝑟 𝑟
Integrating
𝜎r = 𝜎y ln 𝑟 + C Boundary condition at 𝑟 = 𝑛𝑟i 𝜎r = −𝑝𝑛 C = −𝑝𝑛 − 𝜎y ln(𝑛𝑟i ) C =
−𝜎y (𝑘2 − 𝑛2 ) 2𝑘2
(𝑘 2 − 𝑛2 ) 𝜎r = 𝜎y [ln ( ) − ] 𝑛𝑟i 2𝑘 2
− 𝜎y ln(𝑛𝑟i )
𝑟
(5.18a)
𝜎θ − 𝜎r 𝜎y = 𝜎θ = 𝜎y + 𝜎r 2 2 𝜎θ = 𝜎y [ln (
𝑟 𝑛𝑟i
)−
(𝑘 2 + 𝑛2 ) ] 2𝑘 2
(5.18b)
when 𝑟 = 𝑟i 𝜎r = −𝑝
From (5.18a) 𝑝 = 𝜎y [ln(𝑛) +
(𝑘 2 − 𝑛2 ) ] 2𝑘 2
(5.19)
𝑝 is the pressure required to cause yielding to the radial depth 𝑛𝑟i so when 𝑛 = 𝑘, complete yielding throughout the cylinder is present.
𝑝 = 𝑝c 𝜎y ln 𝑘
(5.20)
𝑝c : collapse pressure At the collapse pressure:
𝑟 𝑟o
(𝜎r )𝑝c = 𝜎y ln ( )
𝑟 𝑟o
(𝜎θ )𝑝c = 𝜎y [1 + ln ( )]
(5.21)
55 The stress distribution: a) onset of yielding at the bore 𝜎 𝑝𝑘 2 + 1 𝑘2 − 1 𝜎θ 2𝑝 −1
𝑘2 𝜎y
𝑟 𝜎r
−𝑝
b) elastic-plastic cylinder 𝜎
elastic
plastic
𝜎θ
𝜎y
𝜎r
elastic − plastic interface
𝑟
56 c) fully plastic cylinder 𝜎 𝜎θ
𝜎y
𝜎y
𝜎r
𝑟
Example: elastic-fully plastic, 𝑘 = 2, 𝑝 = −𝜎r
a) elastic breakdown pressure
𝜎y = 250 MPa
b) the pressure to cause yielding up to 50% across its thickness, and the corresponding stresses 𝜎r and 𝜎θ at the elastic-plastic interface
𝑟 𝑟i
c) the collapse pressure and the corresponding hoop stresses at the bore 𝑟o
a) 𝑝eb 𝑝eb =
𝑘2 − 1 4−1 (250 MPa) 𝒑𝐞𝐛 = 𝟗𝟑. 𝟕𝟓 𝐌𝐏𝐚 𝜎y 2 2𝑘 2(4)
b) 𝑝 (𝑘 2 − 𝑛2 ) 𝑝 = 𝜎y [ln(𝑛) + ] ; 𝑛 = 1.5 2𝑘 2 (22 − 1.52 ) 𝑝 = (250 MPa) [ln(1.5) + ] 𝒑 = 𝟏𝟓𝟔. 𝟏 𝐌𝐏𝐚 2(22 )
57 At the interface:
(𝑘 2 − 𝑛2 ) 𝑝n = 𝜎y [ ] ; 𝑛 = 1.5 2𝑘 2 (22 − 1.52 ) ( ) 𝑝n = 250 MPa [ ] 𝒑𝐧 = 𝟓𝟒. 𝟕 𝐌𝐏𝐚 2(22 ) (𝜎r )𝑛=1.5 = − 𝑝n − 𝟓𝟒. 𝟕 𝐌𝐏𝐚 (𝜎θ )𝑛=1.5 = 𝜎y + 𝜎r 250 − 54.7 𝟏𝟗𝟓. 𝟑 𝐌𝐏𝐚
c) 𝑝c 𝑝c = 𝜎y ln 𝑘 (250 MPa) ln 2 𝟏𝟕𝟑. 𝟐 𝐌𝐏𝐚 𝜎θ = 𝜎y + 𝜎r 250 − 173.2 𝟕𝟔. 𝟕 𝐌𝐏𝐚
5.2.1 Residual Stresses in Internally Pressurized Cylinders Compounding cylinders is one way to obtain favourable pre-stresses before they are put into service. This is a more effective use of material. An even more effective use of material, while achieving the same goal, is by applying enough pressure to cause yielding (residual stresses) in some or all of the material in a simple cylinder then releasing the pressure before it is put into service. The generation of favourable residual stresses in a cylinder by plastic action is called autofrettage. Consider a fully (100% overstrained) cylinder which is being unloaded after reaching the state. Assume elastic unloading. Stresses due to the applied collapse pressure, 𝑃c :
𝑟 𝑟o
(𝜎r )𝑝c = 𝜎y ln ( )
𝑟 𝑟o
(𝜎θ )𝑝c = 𝜎y [1 + ln ( )] For elastic unloading due to the 𝑃c 𝑝c 𝑟o2 − (1 ) 𝑘2 − 1 𝑟2 𝑐 𝑟o2 𝜎θ = 2 (1 + 2 ) 𝑘 −1 𝑟 𝜎r =
𝑟2o (1 − 2 ) 𝑟 𝑘2 − 1 𝑝 𝑟 𝑟2o (𝜎θ )res = 𝜎y [1 + ln ( )] − 2 c (1 + 2 ) 𝑟o 𝑟 𝑘 −1 𝑟 𝑟o
(𝜎r )res = 𝜎y ln ( ) −
𝑝c
(5.21)
58 𝑝c = 𝜎y ln(𝑘)
𝑟 𝑟o
(𝜎r )res = 𝜎y [ln ( ) −
ln 𝑘
(1 −
𝑟2o )] 𝑟2
(5.22a)
𝑘2 − 1 𝑟 ln 𝑘 𝑟2o (𝜎θ )res = 𝜎y [1 + ln ( ) − 2 (1 + 2 )] 𝑟o 𝑟 𝑘 −1 𝜏̂ res =
(5.22b)
𝜎y |(𝜎θ )res | − |(𝜎r )res | 2 ln 𝑘 𝑟2o [1 − 2 ( )] 2 2 𝑘 − 1 𝑟2
(5.23)
𝜎
(𝜎θ )res
(𝜎r )res
𝑟
Reverse Yielding: 𝜏̂ res = −𝜏̂ y −
𝜎y 2
𝜎y 𝜎y 2 ln 𝑘 𝑟o2 [1 − 2 ( 2 )] = − 2 𝑘 −1 𝑟 2 Max stresses occur at the bore: 1−
2 ln 𝑘 ln 𝑘 = −1 𝒌 = 𝟐. 𝟐𝟐 𝑘2 − 1
𝑘 = 2.22 is the limiting radius above which 100% overstrain (fully plastic condition) will give rise to reverse yield at the bore
59 100% autofrettage: the theoretical max residual stress left in the bore 50% autofrettage: 50% theoretical residual stress left in the bore
5.2.2 The “Shakedown” Condition -
reverse yielding is not desirable “shakedown” cylinder settling down to a safe working condition after initial plastic deformation Optimum (100%) autofrettage is achieved with 100% overstrain when 𝑘 ≤ 2.22 but if 𝑘 > 2.22 with less that 100% overstrain
Max allowable pressure (𝑃̂i ) for 𝑘 ≤ 2.22 𝜎y 𝜎y 𝑃̂i 𝑘 2 2 ln 𝑘 2 + − 𝑘 = [1 ] 𝑘2 − 1 2 𝑘2 − 1 2 Due to 𝑃̂i 𝑃̂i = 𝜎y ln 𝑘 𝑃c
(5.24)
𝑃̂i 𝑃eb 𝑃eb =
𝑘2 − 1 𝑃̂i 𝜎 ; for 𝑘 = 2.22 ≤2 2𝑘 2 y 𝑃eb
Load-carrying capacity increased by a factor of 2.
For 𝑘 > 2.22 𝜎y 𝜎y 𝑃̂i 𝑘 2 − = 𝑘2 − 1 2 2 𝑃̂i =
𝑘2 𝜎 2𝑃eb 𝑘2 − 1 y
Load-carrying capacity is increased by a factor of 2 for any 𝑘 > 2.22.
Example (a): Determine – for completely linear-elastic behaviour of a cylinder with radius ratio k=2.5 and 𝜎y = 400 MPa – the optimum autofrettage pressure and the resulting percentage overstain of the cylinder wall. 𝑃̂i = 2𝑃eb 2
𝑘 2 − 1 𝜎y 2.52 − 1 400 = 336 MPa 𝑘2 2 2.52
overstrain:
𝑛−1 × 100% 𝑘−1
60
(𝑘 2 − 𝑛2 ) 𝜎r = 𝜎y [ln ( ) − ] 𝑛𝑟i 2𝑘 2 𝑟
(5.18a)
@𝑟 = 𝑟i , −𝑃̂i = 𝜎r − 336 MPa (2.5 − 𝑛2 ) 1 −336 MPa = 400 [ln ( ) − ] 𝑛 2(2.5)2 (2.5 − 𝑛2 ) 1 336 ln ( ) − =− f(𝑛) 2 𝑛 2(2.5) 400 f(𝑛) 0.725 0.875 0.815
n 1.5 2 1.75 ∴ 𝑛 = 1.845 % overstrain:
1.845 − 1 × 100 % 56.3 % 2.5 − 1
Example (b): A thick-walled steel cylinder pressure vessel, optimally autofrettaged for linear-elastic behaviour in service, has inner and outer diameters of 110 mm and 160 mm, respectively. What are the radial and hoop stresses at the inner and outer radii of the cylinder when it is subjected to an internal pressure of 200 MPa. 𝜎y = 660 MPa; assume elastic-perfectly plastic 𝑘=
𝑟o 55 = 1.4545 ≤ 2.22 𝑟i 80
𝑃̂i = 𝑃c 𝜎y ln 𝑘 = 600 ln(1.4545) 247.3 MPa 𝑃̂i : optimal autofrettage pressure 𝑟 ln 𝑘 𝑟o2 (𝜎r )res = 𝜎y [ln ( ) − 2 (1 − 2 )] 𝑟o 𝑘 −1 𝑟 𝑟 ln 𝑘 𝑟o2 (𝜎θ )res = 𝜎y [1 + ln ( ) − 2 (1 + 2 )] 𝑟o 𝑘 −1 𝑟 @ 𝑟i and 𝑟o (𝜎r )res = 0 (𝜎θ )ri = 𝜎y [1 −
(2𝑘 2 ) ln 𝑘] − 278 MPa 𝑘2 − 1
𝑟 ln 𝑘 𝑟o2 (𝜎θ )ro = 𝜎y [1 + ln ( ) − 2 (1 + 2 )] 600(0.3283) = +216.7 MPa 𝑟o 𝑘 −1 𝑟
61 Elastic stresss due to 𝑃i = 200 MPa Lamé plot 1 = 330.5 𝑟i2 1 = 156.25 𝑟o2
𝜎θ
positive
𝜎r
𝑏 1 𝑟o2
𝑎 1 𝑟o2
1 𝑟i2 𝑏 (
1 1 2 + 2) 𝑟i 𝑟o
1 𝑟i2 𝑃i
negative =
𝑎 𝑃i 1 1 1 2 2 2− 2 𝑟o 𝑟i 𝑟o
(𝜎θ )𝑟o = 𝑎 858.5 MPa (𝜎θ )𝑟i = 𝑏 558.5 MPa Total stress in the autofrettage cylinder 𝑃i = 200 MPa
@ 𝑟 = 𝑟i 𝜎r = −200 MPa 𝜎θ = 558.5 − 278 280.5 MPa
@ 𝑟 = 𝑟o 𝜎r = 0 MPa (𝜎θ )𝑟o = (𝜎θ )𝑟o + [(𝜎θ )𝑟o ]
res
358.5 + 216.7
(𝜎θ )𝑟o = 575.2 MPa
5.2.3 Using von Mises 𝜎y
, instead of 𝜏y =
𝜎y
-
If von Mises criteria is used, 𝜏y =
-
All the expressions developed for Tresca can be used when 𝜎y is replaced by
√3
2
in Tresca. 2 𝜎. √3 y
1
1.1 Stress
1
1.2 Equation of Equilibrium
2
1.3 Stress-Strain Relationship
3
1.4 Vectors and Tensors Notation
4
1.5 Stress at a Point
5
2.0 Elastic Axis-symmetric Deformations
8
2.1 Basic Relation
8
2.2 Pressurized and Thick-Walled Cylinders
9
2.3 Rotating Disks and Long, Rotating Cylinders (T=0)
12
2.4 Thermal Stresses in Thick-Walled Cylinder and Disks
14
2.4.1 Plane stress
15
2.4.2 Plane strain
16
2.4.3 Other End Conditions due to Thermal Loads
19
3.0 Stress Functions
20
3.1 Solution by Polynomial
20
3.2 Stress Concentrations
25
3.2.1 Use of Polar Coordinates
25
3.2.2 Stresses due to Concentration Loads
26
3.2.3 Stress Concentrations around a Small Circular Hole (Kirsh Solution)
29
4.0 Shock and Impact Loading
30
4.1 Stress and Displacement in an Elastic Axial Bar
30
4.2 Effects of Geometry
38
4.3 Stress Wave Propagation under Impact Loading
38
4.4 Changes in Material Properties under Impact Loading
41
5.0 Elasto-Plastic Analysis 5.1 Beam Bending 5.1.2 Elastic-Linear Strain Hardening 5.2 Thick-Walled Cylinders under Internal Pressure
42 44 49 52
5.2.1 Residual Stresses in Internally Pressurized Cylinders
57
5.2.2 The “Shakedown” Condition
59
5.2.3 Using Von Mises
61
1
1.0 Elastic Theory 1.1 Stress Stress is an internal resistance of a material and is defined as an average force exerted over an area. Two types of stresses: normal stress and shear stress arise due to the force acting normal and parallel to the plane, respectively. Mathematically, stress can be written as:
𝐹N 𝐴 𝐹T 𝜏= 𝐴
(1.1)
𝜎=
where: -
(1.2)
𝜎 is normal stress 𝜏 is shear stress 𝐹N is normal force 𝐹T is tangential force 𝐴 is cross-sectional area
A double-suffix notation, 𝜏xy , is used to define the direction of shear stress acting on a plane. The first suffix defines the direction of the outward normal of the plane and the second one represents the direction of the shear stress. It can be shown that: 𝜏xy = 𝜏yx ; 𝜏yz = 𝜏zy ; 𝜏xz = 𝜏zx
(1.3)
In the 3-D case, the stress matrix [𝜎] can be written as: 𝜎xx [𝜎] = [𝜏xy 𝜏xz
𝜏xy 𝜎yy 𝜏yz
𝜏xz 𝜏yz ] 𝜎zz
(1.4)
In solid mechanics, strain is a measure of deformation due to the applied load with reference to the original length. The 3-D normal strain in a solid medium due to an applied load is given by:
𝜕𝑢 𝜕𝑥 𝜕𝑣 𝜀yy = 𝜕𝑦 𝜕𝑤 𝜀zz = 𝜕𝑧 𝜀xx =
(1.5) (1.6) (1.7)
where 𝜀xx , 𝜀yy , and 𝜀zz are normal strains in the x, y and z directions, respectively and u , v , and w are displacements in the x, y, and z directions, respectively.
2 In addition to normal strain, shear strain, 𝛾, due to shear force is given by: 𝜕𝑣 𝜕𝑢 + = 2𝜀xy 𝜕𝑥 𝜕𝑦 𝜕𝑤 𝜕𝑣 𝛾zy = 𝛾yz + = 2𝜀yz 𝜕𝑦 𝜕𝑤 𝜕𝑤 𝜕𝑢 𝛾xz = 𝛾zx + = 2𝜀xz 𝜕𝑥 𝜕𝑧 𝛾xy = 𝛾yx
(1.8) (1.9) (1.10)
Six independent stress components 𝜎xx, 𝜎yy, 𝜎zz, 𝜏xy , 𝜏yz , and 𝜏zx are needed to describe the load carried by a 3-D solid at any point. The six independent strain components 𝜀xy , 𝜀yy , 𝜀zz , 𝛾xy ,𝛾yz , 𝛾zx , which are produced by the load, are used to describe the deformation at any point.
1.2 Equation of Equilibrium Consider an infinitesimal element ∆𝑥 × ∆𝑦 × ∆𝑧 onto which non-uniform stresses are applied at different faces, as shown in Figure 1. 𝑦
𝑥
𝑧
Figure 1: An infinitesimal element showing stress distribution.
∑ 𝐹x = 0, gives (𝜎xx + ∆𝜎xx )∆𝑦∆𝑧 − 𝜎xx ∆𝑦∆𝑧 + (𝜏yx + ∆𝜏yx )∆𝑥∆𝑧 − ∆𝜏yx ∆𝑥∆𝑧 + (𝜏zx + ∆𝜏zx )∆𝑥∆𝑦 − ∆𝜏zx ∆𝑥∆𝑦 =0
(1.11)
Dividing both sides by ∆𝑥∆𝑦∆𝑧 and taking the limit as ∆𝑥, ∆𝑦, and ∆𝑧 goes to zero, (1.11) can be rewritten as:
3 𝜕𝜎xx 𝜕𝜏yx 𝜕𝜏zx + + =0 𝜕𝑥 𝜕𝑦 𝜕𝑧
(1.12)
In the presence of body forces, 𝐹x , (1.12) can be rewritten as: 𝜕𝜎xx 𝜕𝜏yx 𝜕𝜏zx + + + 𝐹x = 0 𝜕𝑥 𝜕𝑦 𝜕𝑧
(1.13)
𝜕𝜏xy 𝜕𝜎yy 𝜕𝜏yz + + + 𝐹y = 0 𝜕𝑥 𝜕𝑦 𝜕𝑧
(1.14)
𝜕𝜏xz 𝜕𝜏yz 𝜕𝜎zz + + + 𝐹z = 0 𝜕𝑥 𝜕𝑦 𝜕𝑧
(1.15)
Similarly for ∑ 𝐹y = 0:
And for ∑ 𝐹z = 0:
1.3 Stress-Strain Relationship In general for a 3-D isotropic solid, the stress-strain relationship can be written as: 1 [𝜎 − ν(𝜎yy + 𝜎zz )] E xx 1 𝜀yy = [𝜎yy − ν(𝜎xx + 𝜎zz )] E 1 𝜀zz = [𝜎zz − ν(𝜎xx + 𝜎yy )] E
(1.16)
𝜀xx =
(1.17) (1.18)
E is Young’s modulus and ν is Poisson’s ratio Shear stress is related to the shear strain by: 𝜏xy ; G where shear stress is given by: 𝛾xy =
𝛾yz =
G=
𝜏yz ; G
E 2(1 + ν)
𝛾xz =
𝜏xz ; G
(1.19)
(1.20)
4
1.4 Vectors and Tensors Notation In order to represent stresses, strains using tensor notation, it is important to set up the index notation using Einstein’s summation convention. This notation is used to simplify expressions involving vectors and tensors operations. The direction x, y, and z is referred to as 1, 2, and 3, respectively. A vector 𝑣⃗ in the index notation can be written as; 3
𝑣⃗ = 𝑣i ∑ 𝑣i ⃗⃗⃗ 𝑒i
(1.21)
i=1
where {𝑒⃗⃗⃗⃗, 𝑒2 , ⃗⃗⃗⃗⃗} 𝑒3 are the orthonormal unit vectors in 3-D space 1 ⃗⃗⃗⃗⃗ Similarly, a tensor in the index notation can be written as: 3
3
𝑇 𝑇ij = ∑ ∑ 𝑇ij ⃗⃗⃗𝑒 𝑒i ⃗⃗⃗j
(1.22)
i=1 j=1
where {𝑒⃗⃗⃗,i ⃗⃗⃗⃗} 𝑒j are unit dyads that represent an ordered pair of coordinate directions In Einstein’s summation convention, if the index appears more than once, then the summation is repeated for each index over the range of the index. The number of free indices determine whether the index notation is a scalar, a vector, or a tensor. In this method, no free index or rank 0, represents a scalar, one free index or rank 1, means it is a vector, two free indices or rank 2 represents a tensor, and so on. For the Cartesian co-ordinate system where the summation is performed in three mutually- orthogonal space, the rank of the tensor n have 3n elements. For example: 𝜕𝑢i 𝜕2 𝑓 , and are scalars (no free index, rank 0 = 1 element) 𝜕𝑥i 𝜕𝑥i 𝜕𝑥i 𝜕𝑓 ∈ijk , 𝑢i 𝑣j , , and 𝑇ii 𝑣i are vectors (one free index, rank 1 = 3 elements) 𝜕𝑥i 𝑢i 𝑣i , 𝑇ii,
+1 when any two indices i, j, k are equal ∈ijk { −1 when ijk = 1 2 3 or 3 2 1 or 2 3 1 0 otherwise
𝜕𝑢j 𝜕2 𝑓 𝑢i 𝑣j , and , and are tensors (two free indices, rank 2 = 9 element) 𝜕𝑥i 𝜕𝑥i 𝜕𝑥i 𝑐ijk is a tensor (four free indices, rank 4 = 81 elements)
Based on the aforementioned conventions, the strain tensor can be written as: 1 𝜕𝑢i 𝜕𝑢j 𝜀ij = ( + ) 2 𝜕𝑥j 𝜕𝑥i
(1.23)
5 The stress tensor is given by: 𝜎xx 𝑇ij = [𝜏xy 𝜏xz
𝜏xy 𝜎yy 𝜏yz
𝜏xz 𝜏yz ] 𝜎zz
(1.24)
The relationship of strain as a function of stress can be written as:
𝜀ij = where, 𝛿ij = {
1+ν ν 𝑇ij − (𝑇11 + 𝑇22 + 𝑇33 )𝛿ij E E
(1.25)
1 if i = j also known as Kronecker delta 0 if i ≠ j
and the stress in terms of strain is given by: 𝑇ij = λ(𝜀11 + 𝜀22 + 𝜀33 )𝛿ij + 2G𝜀ij
(1.26)
Finally, the equilibrium equations for elasticity can be written as: 𝜕𝑇ij +𝐹 =0 𝜕𝑥i
1.5 Stress at a Point
Transforming the coordinate system by angle θ and knowing 𝜏yx = 𝜏xy: 𝜎n (AC) + [𝜏yx (BC) cos 𝜃] − 𝜎x (AB) cos 𝜃 + 𝜏yx (AB) sin 𝜃 − 𝜎y (BC) sin 𝜃 = 0
Divide every term by area AC: BC AB AB BC 𝜎n + [𝜏yx ( ) cos 𝜃] − 𝜎x ( ) cos 𝜃 + 𝜏yx ( ) sin 𝜃 − 𝜎y ( ) sin 𝜃 = 0 AC AC AC AC where: BC AB = sin 𝜃 and = cos 𝜃 ; areas are constant AC AC
(1.27)
6 It can be further simplified to: 𝜎n + 𝜏yx sin 𝜃 cos 𝜃 − 𝜎x cos2 𝜃 + 𝜏yx cos 𝜃 sin 𝜃 − 𝜎y sin2 𝜃 = 0 𝜎n − 𝜎x cos 2 𝜃 − 𝜎y sin2 𝜃 + 2𝜏yx cos 𝜃 sin 𝜃
For Mohr’s circle, if the shear stress tends a clockwise (C.W.) rotation, the element is positive. Using: sin2 𝜃 =
1 − cos 2𝜃 1 + cos 2𝜃 ; cos 2 𝜃 = ; 2 sin 𝜃 + cos 𝜃 = sin 2𝜃 2 2
1 1 𝜎n = (𝜎x − 𝜎y ) + (𝜎x + 𝜎y ) cos 2𝜃 + 𝜏yx sin 𝜃 2 2 1 𝜏s = (𝜎x + 𝜎y ) sin 2𝜃 −𝜏xy cos 2𝜃 2 Squaring ① and ② and adding them: 1 2 𝜏s = (𝜎x − 𝜎y ) + 𝜏2xy 4
Comparing this to the equation of a circle: (𝑥 − 𝑎)2 + (𝑦 − 𝑏)2 = 𝑟 2
…so: 1 2 𝑟 = (𝜎x − 𝜎y ) + 𝜏2xy 4
① ②
7 Hence the Mohr’s circle:
𝜎x + 𝜎y 𝜎ave = 2 1 1 𝜎n = (𝜎x − 𝜎y ) + (𝜎x + 𝜎y ) cos 2𝜃 + 𝜏yx sin 𝜃 2 2 1 𝜏s = (𝜎x + 𝜎y ) sin 2𝜃 −𝜏xy cos 2𝜃 2 𝜎1,2 =
𝜎x + 𝜎y 𝜎x − 𝜎y 2 2 ± √( ) + 𝜏xy 2 2
tan 2𝛷 =
𝜏xy 1 2 (𝜎x − 𝜎y )
𝜎x − 𝜎y 2 2 |𝜏max | = √( ) + 𝜏xy 2 𝜎x − 𝜎y |𝜏max | = 2 (𝜎x − 𝜎y ) tan 2𝜃s = − 2𝜏xy
Principals of Solutions in Solid Mechanics: i) ii) iii) iv)
condition of state equilibrium stress-strain relationship condition for compatibility boundary conditions (BCs)
Methods of Solution a) engineering (strength of material) approach - simplifying assumptions related to the geometry of deformation, the state of distribution of stress/strain across the cross section - useful in the simple beam bending and torsion problem b) “exact” – mathematical solution to the governing equations c) numerical solution (FEM, BEM) d) experiments
8 Compatibility Equation in Terms of Stress ∂ ∂ ( 2 + 2 ) (𝜎x + 𝜎y ) = 0 ∂x ∂y ∇2 (𝜎x + 𝜎y ) = 0 True when: 𝐹x , 𝐹y , and 𝑇 are zero.
Plane strain (𝜀z = 0): ∇2 (𝜎x + 𝜎y + Eα𝑇) = −
∂𝐹x ∂𝐹y + ) ∂y 1 − ν ∂x 1
(
Plane stress (𝜎z = 0): ∇2 (𝜎x + 𝜎y + Eα𝑇) = −(1 + ν) (
∂𝐹x ∂𝐹y + ) ∂x ∂y
2.0 Elastic Axis-symmetric Deformations -
axis-symmetric problems in stress analysis occur in numerous practical-pressure vessels, compound cylinders, turbine/compressor disks, flywheels not only geometry has to be similar, loads applied along an axis has to be similar
2.1 Basic Relation -
general transformation of Cartesian to cylindrical polar 𝑥 = 𝑟 cos 𝜃 𝑦 = 𝑟 sin 𝜃 𝑥2 + 𝑦2 = 𝑟2 tan 𝜃 =
𝑦 𝑥
2𝑟
d𝑟 d𝑟 = 2𝑥 = cos 𝜃 d𝑥 d𝑥
∂ ∂𝑟 ∂ ∂𝜃 ∂ = + ∂x ∂x ∂r ∂x ∂𝜃
2𝑟
d𝑟 d𝑟 = 2𝑦 = sin 𝜃 d𝑦 d𝑦
∂ ∂𝑟 ∂ ∂𝜃 ∂ = + ∂y ∂y ∂r ∂y ∂𝜃
d𝜃 𝑦 sin 𝜃 =− 2 − d𝑥 𝑟 𝑟
∂ ∂ sin 𝜃 ∂ = cos 𝜃 − ∂x ∂r 𝑟 ∂𝜃
d𝜃 𝑥 cos 𝜃 =− 2 d𝑦 𝑟 𝑟
∂ ∂ cos 𝜃 ∂ = sin 𝜃 + ∂y ∂r 𝑟 ∂𝜃
9
In matrix form: ∂ cos 𝜃 ∂x = [ ∂ sin 𝜃 [∂y]
∂ sin 𝜃 𝑟 ] ∙ [ ∂r ] cos 𝜃 ∂ 𝑟 ∂𝜃
−
-
for axis-symmetric, all displacements are 𝑓(𝑟) and independent of 𝜃 𝑟 = 0 (pressurized
-
cylinders and rotating disks) shear in 𝜏rθ = 0 (principle stress) 𝜎r , 𝜎θ are 𝑓(𝑟)
For equilibrium: d𝜎r 𝜎r − 𝜎θ + + 𝐹r = 0 d𝑟 𝑟
Strain-Displacement Relationship
Plane stress (𝝈𝐳 = 𝟎)
d𝑢 d𝑟 } 2.2 𝑢 𝜀θ = 𝑟
(thin rotating disk; open-ended cylinder)
𝜀r =
Hooke’s Law 1 𝜀r = [𝜎r − ν(𝜎θ + 𝜎z )] + α∆𝑇 E 1 𝜀θ = [𝜎θ − ν(𝜎r + 𝜎z )] + α∆𝑇 E 1 𝜀z = [𝜎z − ν(𝜎r + 𝜎θ )] + α∆𝑇 E
E [𝜀 + ν𝜀θ (1 + ν)α𝑇] 1 − ν2 r E [𝜀 + ν𝜀r (1 + ν)α𝑇] 𝜎θ = 1 − ν2 θ 𝜎r =
Plane strain (𝜺𝐳 = 𝟎) (long cylinders with restrained ends) 𝜎z = ν(𝜎r − 𝜎θ ) − Eα𝑇 1 − ν2 ν 𝜎 ] + (1 + ν)α𝑇 [𝜎r − E 1−ν θ 2 1−ν ν 𝜀θ = 𝜎 ] + (1 + ν)α [𝜎θ − E 1−ν r 𝜀r =
2.2 Pressurized and Thick-Walled Cylinders Under pressure loads, the stresses in the cylinder are given by Lamé’s solution: 𝜎r = 𝐴 −
𝐵 𝑟2
𝜎θ = 𝐴 +
𝐵 𝑟2
where A and B are found from the given boundary conditions.
10 Example 1: Thick-walled open-ended cylinder subject to internal pressure: 𝑝 = 100 MPa. Find the maximum stress in the cylinder given: 𝑟i = 25 mm 𝑟o = 50 mm Boundary conditions: @ 𝑟 = 𝑟i
@ 𝑟 = 𝑟o
𝐵 𝜎r = 𝐴 − 2 𝑟i
𝜎r = 𝐴 −
𝐵 ① (25 mm)2 (−𝑝 due to compression) −𝑝 = 𝐴 −
0=𝐴−
𝐵 𝑟o2
𝐵 ② (50 mm)2
②→① −100 MPa =
𝐵 𝐵 𝐵 𝐵 − − 100 MPa = − −3 2 −3 2 2 (50 × 10 m) (25 × 10 m) 0.0025 m 6.25 × 10−4 m2 (6.25 × 10−4 m2 )𝐵 − (0.0025 m2 )𝐵 1.56 × 10−6 m4 2 − 156.25 N ∙ m = (−1.88 × 10−4 m2 )𝐵 𝑩 = 𝟖. 𝟑𝟑 × 𝟏𝟎𝟒 𝐍 − 100 MPa =
𝐵 8.33 × 10−4 N 0=𝐴− 𝐴= (50 mm)2 25 × 10−3 m2 𝑨 = 𝟑. 𝟑𝟑 × 𝟏𝟎𝟕 𝐏𝐚 8.33 × 104 N 𝜎r = 3.33 × 10 Pa − 𝑟2 8.33 × 104 N 𝜎θ = 3.33 × 107 Pa + 𝑟2 7
(𝝈𝐫 )𝐦𝐚𝐱 = −𝟏𝟎𝟎 𝐌𝐏𝐚 𝝈𝛉 @ 𝒓 = 𝒓𝐢 𝟏𝟔𝟔. 𝟔 𝐌𝐏𝐚 𝝈𝛉 @ 𝒓 = 𝒓𝐨 𝟔𝟔. 𝟔 𝐌𝐏𝐚
In general, the boundary conditions are: 𝐵 𝑝i − 𝑝o 𝑘 2 𝐴 = 2 𝑟o 𝑟i 𝑘2 − 1 ; where 𝑘 = 2 𝐵 (𝑝 − 𝑝o )𝑟o 𝑟i 𝑟 = 𝑟o 𝜎ro = −𝑝o − 𝑝o = 𝐴 − 2 𝐵 = i 2 𝑟o } 𝑘 −1 𝑟 = 𝑟i 𝜎ri = −𝑝i − 𝑝i = 𝐴 −
For internal pressure ONLY (𝑝o = 0): 𝑝i 𝑟o2 𝑝i 𝑟o2 𝜎r = 2 (1 − 2 ) ; 𝜎θ = 2 (1 + 2 ) 𝑘 −1 𝑟 𝑘 −1 𝑟
For external loading ONLY (𝑝i = 0): 𝑝i 𝑘 2 𝑟i2 −𝑝o 𝑟o2 𝜎r = 2 ( 2 − 1) ; 𝜎θ = 2 (1 − 2 ) 𝑘 −1 𝑟 𝑘 −1 𝑟
11
𝜎θ > 𝜎z > 𝜎r
Example: A long, thick-walled cylinder with ends rigidly restrained is embedded in a rigid wall and subjected to an internal pressure of 𝑝i . What is the maximum p allowed to avoid yielding according to Tresca’s criteria? (Tresca:
𝜎θ + 𝜎r 𝜎y = ) 2 2 𝑟=ro
1 [𝜎 − ν(𝜎r − 𝜎θ )] E z 𝜎z = ν(𝜎θ + 𝜎r ) 𝑢 (𝜀θ )ro = 0 = 0; 𝑢 is the displacement 𝑟o 1 𝜀θ = [𝜎θ − ν(𝜎r + 𝜎z )]𝑟o 0 E
𝑟i = 20 × 10−3 m 𝑟o = 50 × 10−3 m ν = 0.33 E = 60 GPa 𝜎y = 80 MPa 1 1 = 400; 2 = 2500 2 𝑟o 𝑟i
𝜀z = 0
let: 𝑎 = (𝜎θ )ri & 𝑏 = (𝜎θ )ro @ 𝑟i : 𝑎 + 𝑃i 80 = 2 2 𝑎 + 𝑃i = 80 MPa ① @ 𝑟o : 𝜎θ = 𝑏, 𝜎r = −𝑃o 𝑢r 2 = 0
𝑎 + 𝑃i 𝑃i − 𝑃o = −2 + 400 m 2500 m−2 + 400 m−2 𝑏 + 𝑃o 400 m−2 + 400 m−2 2500 m−2
𝑎 = 0.8125𝑃1 𝑃i = 2.3125𝑃2 𝑃o = 25.6 MPa
12 𝑏 − ν[−𝑃o ν(b − 𝑃o )] = 0 ν 𝑏=− 𝑃 − 0.5𝑃o ② 1−ν o
𝑃i = 59.2 MPa 𝒂 = (𝝈𝛉 )𝐫𝟏 𝟐𝟎. 𝟖 𝐌𝐏𝐚 𝒃 = (𝝈𝛉 )𝐫𝟐 − 𝟏𝟐. 𝟖 𝐌𝐏𝐚
2.3 Rotating Disks and Long, Rotating Cylinders (T=0) -
similar to analysis as in the Lamé’s equation, for pressurized cylinders, except now due to rotation there is a body force, Fr
Plane stress case (i.e. disk): 𝐵 3+ν 2 2 − 𝜌𝑟 𝜔 𝑟2 8 𝐵 1 + 3ν 2 2 𝜎θ = 𝐴 + 2 − 𝜌𝑟 𝜔 𝑟 8
A and B are constants for a given geometry and speed (found using boundary conditions)
𝜎r = 𝐴 −
Example: A thin steel disk with 𝑟i = 80 mm, 𝑟o = 380 mm is shrink-fitted onto a rigid shaft. The radial interference (𝛿 = 0.1 mm). At what speed will the shrink fit loosen? 𝐸 = 200 GPa ν = 0.3 𝜌 = 7 850
kg m3
When the shrink fit loosens, 𝜎r = 0 @ 𝑟 = 𝑟i @ 𝑟 = 𝑟o 𝜎(ro )2 = 0 @ 𝑟 = 𝑟i 𝐵 3 + 0.3 kg 0=𝐴− − ( ) (7 850 ) (0.08 m)2 𝜔2 ① (0.08 m)2 8 m3 𝐴 = 488.3𝜔2 @ 𝑟 = 𝑟o 𝐵 = 2.933𝜔2 𝐵 3 + 0.3 kg 0=𝐴− −( ) (7 850 3 ) (0.38 m)2 𝜔2 ② 2 (0.38 m) } 8 m 𝜀θ =
𝑢 𝑟 (𝜎θ − ν𝜎r ) 𝑟 E
𝛿 = (𝑢)rmo − (𝑢)rmi 0.1 m
13 𝜎θ @ 𝑟 = 𝑟1 (0.1 × 10−3 m)(200 × 109 Pa) 𝐵 3 + 0.3 kg =𝐴+ −( ) (2 830 3 ) (0.08 m)2 𝜔2 (0.08 m)2 (0.08 m)2 8 m 𝝎 = 𝟓𝟏𝟒. 𝟔
𝐫𝐚𝐝 𝝎 = 𝟒 𝟗𝟏𝟒 𝐑𝐏𝐌 𝐬
Plane strain (long rotating cylinders) 𝐵 3 + ν∗ 2 2 − 𝜌𝑟 𝜔 𝑟2 8 𝐵 1 + 3ν∗ 2 2 𝜎θ = 𝐴 + 2 − 𝜌𝑟 𝜔 𝑟 8 𝜎r = 𝐴 −
ν∗ =
ν 1−ν
Example: A long, solid steel shaft of 200 mm radius is rotating @ 300 RPM with its ends restrained from longitudinal displacement. Find the total longitudinal force exerted at the ends. E = 200 GPa, ν = 0.3, 𝜌 = 7 850 ν∗ =
kg m3
0.3 0.42857 0.7
B. Cs. 𝜀z = 0 𝜀z =
1 [𝜎 − ν(𝜎r − 𝜎θ )] + α∆𝑇 𝝈𝐳 = 𝛎(𝝈𝐫 − 𝝈𝛉 ) E z
𝐵 = 0 for a solid steel shaft 𝜎z = ν [(𝐴 − ν [(𝐴 −
𝑟 @ 𝑟o = 200 mm,
𝐵 3 + ν∗ 2 2 𝐵 1 + 3ν∗ 2 2 − 𝜌𝑟 𝜔 ) + (𝐴 + − 𝜌𝑟 𝜔 )] 𝑟2 8 𝑟2 8
3 + ν∗ 2 2 1 + 3ν∗ 2 2 𝟏 + 𝛎∗ 𝟐 𝟐 𝜌𝑟 𝜔 ) + (𝐴 − 𝜌𝑟 𝜔 )] 𝝈𝐳 = 𝛎 [𝟐𝑨 − 𝝆𝒓 𝝎 ] 8 8 𝟐
𝜎r = 0 𝜎r = 𝐴 −
𝐵 3 + ν∗ 2 2 𝟑 + 𝛎∗ 𝟐 𝟐 − 𝜌𝑟 𝜔 𝑨 = 𝝆𝒓 𝝎 𝑟2 8 𝟖
3 + ν∗ 2 1 + ν2 2 ν𝜌𝜔2 [(3 − 2ν)𝑟o2 − 2π2 ] 𝜎z = ν [ 𝑟o − ( ) π ] 𝜌𝜔2 4 2 4(1 − ν)
14
𝜎=
2π π 𝐹 πν𝜌𝜔2 𝑟o4 𝐹 = 𝜎𝐴 𝐹 = ∫ 𝜎 d𝐴 ∫ ∫ 𝜎z 𝑟 d𝑟d𝜃 𝐴 2 0 0
π(0.3) (7 850
kg rev 2πrad min 2 ) (300 min × rev × 60 s) (0.2 m)4 3 m 𝑭 = 𝟓. 𝟖𝟒 𝐤𝐍 2
2.4 Thermal Stresses in Thick-Walled Cylinder and Disks In linear-elastic analysis, the effects of mechanical and thermal loads can be treated separately and then added together after the components are subjected to both types of loading superposition). -
consider thermal loading ONLY
Fourier’ law of heat conduction: 𝑄 = −2π𝑟𝑘
d𝑇 d𝑟
where: 𝑄 = heat flow per unit axial length in the radial direction k = thermal conductivity coefficient
𝑟
d𝑇 𝜃 =− C1 d𝑟 2π𝑘
∫ d𝑇 = ∫
d𝑟 C 𝑇 = C1 ln 𝑟 + C2 𝑟 1
where C1 and C2 are constants B. Cs. 𝑇o 𝑟
𝑇 = 𝑇i +
𝑟i 𝑇i
@ 𝑟 = 𝑟i 𝑇 = 𝑇i @ 𝑟 = 𝑟o 𝑇 = 𝑇o
𝑟o
𝑇o − 𝑇i 𝑟 𝑟 ln 𝑟 i ln 𝑟o i
15
2.4.1 Plane stress -
thin disks (externally cooled and heated) d𝑢
𝑢
d𝑢
𝑢
𝜀r = ; 𝜀θ = E E d𝑢 𝑢 d𝑟 𝑟 [𝜀 𝜎r = + ν𝜀 (1 + ν)α𝑇] ⇒ 𝜎r = [ + ν − (1 + ν)α𝑇] r θ 2 2 1−ν 1 − ν d𝑟 𝑟 𝜀r = ; 𝜀θ = E E 𝑢 d𝑢 d𝑟 𝑟 [𝜀 𝜎θ = + ν𝜀 (1 + ν)α𝑇] ⇒ 𝜎θ = − (1 + ν)α𝑇] [ +ν θ r 2 2 1−ν 1−ν 𝑟 d𝑟
Using equilibrium (Fr = 0): d𝜎r 𝜎r − 𝜎θ + =0 d𝑟 𝑟 1 1 d d𝑇 (𝑢𝑟)] = (1 + ν)α [ d𝑟 𝑟 d𝑟 d𝑟
Double integral: 𝑢 = C1 𝑟 +
𝑟 C2 1 + ν + α ∫ 𝑇𝑟 d𝑟 𝑟 𝑟 𝑟i
C1 and C2 are constants of integration
Substitute u in 𝜎r and 𝜎θ : 𝐵 E 𝑟 − α ∫ 𝑇𝑟 d𝑟 𝑟2 𝑟 2 𝑟i 𝐵 E 𝑟 𝜎θ = 𝐴 + 2 + α 2 ∫ 𝑇𝑟 d𝑟 − 𝐸α𝑇 𝑟 𝑟 𝑟i
where A and B are from boundary conditions
𝜎r = 𝐴 −
Free-free edge
𝑇o 𝑟
B. Cs.
For solid disk:
@ 𝑟 = 𝑟i 𝜎r = 0 @ 𝑟 = 𝑟o 𝜎r = 0
𝐵=0
𝑟o
𝑟i 𝑇i
αE ∫ 𝑇𝑟 d𝑟 − 𝑟i2 𝑟i 𝑟o αE𝑟i2 𝐵= 2 ∫ 𝑇𝑟 d𝑟 𝑟o − 𝑟i2 𝑟i 𝐴=
𝑟o
𝑟o2
𝐴=
αE 𝑟o ∫ 𝑇𝑟 d𝑟 𝑟o2 0
16 Steady-state temperature disk: 𝑟
∫ 𝑇𝑟 d𝑟 = 𝑇1 ( 𝑟i
𝑟 2 − 𝑟i2 𝑇o − 𝑇i 𝑟 2 𝑟 𝑟 2 − 𝑟i2 ) [ ln ( ) − )+( ( )] 𝑟 2 𝑟i 4 ln ( 𝑟o ) 2 i
2.4.2 Plane strain -
long cylinder with ends restrained 𝜎z = ν(𝜎r − 𝜎θ ) − Eα𝑇
let: E∗ =
E ν ; ν∗ = ; α∗ = α(1 + ν) 2 1−ν 1−ν
so: 𝜀r =
1 [𝜎 − ν∗ 𝜎θ ] + α∗ 𝑇 E∗ r
𝜀θ =
1 [𝜎 − ν∗ 𝜎r ] + α∗ 𝑇 E∗ θ
Stresses due to thermal loading in a cylinder with steady state temperature distribution 𝑟0 ln 𝑘 𝑟o2 𝜎r = 𝐶0 [− ln ( ) − 2 (1 − 2 )] 𝑟 𝑘 −1 𝑟 𝑟0 ln 𝑘 𝑟o2 𝜎θ = 𝐶0 [1 − ln ( ) − 2 (1 + 2 )] 𝑟 𝑘 −1 𝑟
𝜎z plane strain: 𝜎z =
where: 𝑟
k is the radius ratio, 𝑟0 i
𝐶0 =
E ′ α′ (𝑇i − 𝑇o ) ′ ′ Eα plane stress ;Eα { ∗ ∗ E α plane strain 2 ln 𝑘
αE∆𝑇 𝑟o ln 𝑘 [ν − 2 ln ( ) − 2ν 2 ] 2(1 − ν) ln 𝑘 𝑟 𝑘 −1
17
𝜎θ
𝑟 = 𝑟o
𝑟 = 𝑟i
𝜎r 𝜎z
-
use the largest and smallest 𝜎 for failure criterion
Example: A long, thick-walled steel tube with radius 𝑟i = 25 mm and 𝑟o = 50 mm is used to transport a hot, pressurized fluid @ 𝑃 = 10 MPa and 𝑇i = 250 ℃. The outer circumference is maintained @ 𝑇o = 50 ℃. Find stresses at the inner and outer radius of the cylinder. Check if yielding has occurred using Tresca’s criterion. 1 ℃
E=200 GPa, ν = 0.3, α = 11 × 10−6 , 𝜎y = 450 MPa Pressure Loads 1 = 1600 𝑟i2
𝜎θ
positive
𝜎r
𝑎
1 = 400 𝑟o2
1 𝑟o2
𝑏 1 𝑟o2
1 𝑟i2
negative
Using similar triangles: 𝑎 1600+400
𝑏
𝑃
= 400+400 = 1600−400
𝑎 (𝝈𝛉 )𝒓𝐢 = 𝟏𝟔. 𝟔𝟕 𝐌𝐏𝐚 𝑏 (𝝈𝛉 )𝒓𝐨 = 𝟔. 𝟔𝟕 𝐌𝐏𝐚
𝜎z = ν(𝜎r + 𝜎θ ) 0.3(0 MPa + 6.67 MPa) 𝝈𝐳 = 𝟐 𝐌𝐏𝐚
1 𝑟i2 𝑃i
18 Thermal Loads E ( ) (α(1 + ν))(𝑇o − 𝑇i ) E ∗ α∗ (𝑇o − 𝑇i ) 1 − ν2 𝐶0 = 2 ln 𝑘 2 ln 𝑘 200 × 109 Pa 1 ) ((11 × 10−6 ℃) (1 + 0.3)) (200 ℃) 1 − 0.32 𝑪𝟎 = 𝟒𝟓𝟑. 𝟒𝟐 𝐌𝐏𝐚 50 2 ln ( ) 25 (
@ 𝒓 = 𝒓𝐢 :
(𝝈𝐫 )ri = 𝟎 𝑟0 ln 𝑘 𝑟o2 ln 2 (1 + 4)] 𝜎θ = 𝐶0 [1 − ln ( ) − 2 (1 + 2 )] (453.42 MPa) [1 − ln 2 − 𝑟 𝑘 −1 𝑟 4−1 (𝝈𝛉 )𝒓𝐢 = −𝟑𝟖𝟗. 𝟔𝟖 𝐌𝐏𝐚 𝜎z =
αE∆𝑇 𝑟o ln 𝑘 [ν − 2 ln ( ) − 2ν 2 ] 2(1 − ν) ln 𝑘 𝑟 𝑘 −1
1 (11 × 10−6 ℃) (200 × 109 Pa)(200 ℃) 50 2(1 − 0.3) ln ( ) 25
50 ln ( ) 50 25 ] [0.3 − 2 ln ( ) − 2(0.3) 25 50 2 ( ) −1 25
𝝈𝐳 = −𝟓𝟓𝟓. 𝟒 𝐌𝐏𝐚
@ 𝒓 = 𝒓𝐨 :
(𝝈𝐫 )𝐫𝐨 = 𝟎 𝑟0 ln 𝑘 𝑟o2 ln 2 (1 + 1)] 𝜎θ = 𝐶0 [1 − ln ( ) − 2 (1 + 2 )] (453.42 MPa) [1 − ln 1 − 𝑟 𝑘 −1 𝑟 4−1 (𝝈𝛉 )𝐫𝐨 = −𝟐𝟒𝟑. 𝟏 𝐌𝐏𝐚 𝝈𝐳 = 𝟕𝟑. 𝟐 𝐌𝐏𝐚
YIELDING? @ 𝒓 = 𝒓𝐢 : (𝜎z )t = −555.4 MPa + 2 MPa (𝝈𝐳 )𝐭 = −𝟓𝟓𝟑. 𝟒 𝐌𝐏𝐚
19 𝜎1 − 𝜎2 𝜎y ≤ 2 2 𝜎2 − 𝜎3 𝜎y Tresca ≤ 2 2 𝜎1 − 𝜎3 𝜎y ≤ { 2 2 (𝜎z )temp − (𝜎r )pressure
2
≤
𝜎y 553.4 − 10 450 ≤ 2 2 2
𝟐𝟕𝟏. 𝟐𝟕 𝐌𝐏𝐚 > 𝟐𝟐𝟓 𝐌𝐏𝐚 YIELDING OCCURS!
2.4.3 Other End Conditions due to Thermal Loads
OPEN
𝜎z = 0 CLOSED
-
radial and hoop stress can be found using previous equations for plane strain axial stress, 𝜎z , from the ends
a) resultant axial force is zero
𝑟o
∫ 𝜎z 2π𝑟 d𝑟 = 0 𝑟i
b) 𝜀z = const. (axi − symmetric)
𝑟o 𝑟o 1 ∫ 𝜀z 2π𝑟 d𝑟 ∫ { [𝜎z − ν(𝜎r + 𝜎θ )] + α∆𝑇} 2π𝑟 d𝑟 𝑟i 𝑟i E 2α 1 𝑟o 𝜀z = 2 ∫ 𝑇𝑟 d𝑟 𝑘 − 1 𝑟i2 𝑟i Eα 2 1 𝑟o 𝑇i − 𝑇o 𝑟o 2 𝜎z = [ 2 ∫ 𝑇𝑟 d𝑟 − 𝑇] 𝜎z = Eα ln 𝑘] [1 − 2 ln ( ) − 2 2 1 − ν 𝑘 − 1 𝑟i 𝑟i 2(1 − ν) ln 𝑘 𝑟 𝑘 −1
20
3.0 Stress Functions Direct or exact solutions to the governing equations of elasticity are not usually possible for most practical problems Airy’s Stress Function In order to solve the compatibility equations for stress, Airy suggested that the elasticity be formulated in terms of one mathematical function 𝜑(𝑥, 𝑦) called the stress function. Airy’s stress function 𝜑(𝑥, 𝑦) is related to stress as: 𝜕2𝜑 𝜕2𝜑 𝜕2𝜑 ; 𝜎 = ; 𝜏 = − 𝜕𝑦 2 y 𝜕𝑥 2 xy 𝜕𝑥𝜕𝑦 Consider zero body forces and zero temperature change. 𝜎x =
(3.1)
The compatibility equation for stress is given by: 𝜕2 𝜕2 ∇ (𝜎x + 𝜎y ) = 0 ( 2 + 2 ) (𝜎x + 𝜎y ) = 0 𝜕𝑥 𝜕𝑦
(3.2)
𝜕4𝜑 𝜕4𝜑 𝜕4𝜑 + 2 + ∇2 (∇2 𝜑) = 0 𝜕𝑥 4 𝜕𝑥 2 𝜕𝑦 2 𝜕𝑦 4
(3.3)
2
Substitute (3.1)→ (3.2)
Bi-Harmonic Function Solution (3.3) ∇4 𝜑 = 0 must satisfy the boundary condition. -
Since (3.2) holds true for both plane stress and plain strain, so doe (3.3). Solving 𝜑 → stress → strain → displacement. For many problems, the form of the stress function is specified in advance semi-inverse method) In cases were specifying a stress function in advance is less obvious, the conformal mapping technique is used. - 𝜑 using two complex analytic functions.
3.1 Solution by Polynomial A solution to (3.3) using polynomials are used for rectangular bodies.
a)
Second degree polynomial
21 𝜕2 𝜑 𝑐2 𝜕𝑦 2 𝜕2 𝜑 where: constants 𝜎y = 𝜕𝑥 2 𝑎2 𝜕2 𝜑 𝜏 = − − xy { 𝜕𝑥𝜕𝑦
𝜎x =
𝜑2 =
𝑎2 2 𝑥 2
+ 𝑏2 𝑦 +
𝑐2 2 𝑦 ; 2
𝑏2
𝜎y = 𝑎2 𝜎x = 𝑐2
𝜎x = 𝑐2
𝜏xy = −𝑏2 𝜎y = 𝑎2 if
𝑏2 = 0; biaxial tension 𝑎2 = 𝑏2 0; uniaxial tension 𝑎2 = 𝑐2 0; pure shear
b)
Third order polynomial 𝜑2 =
𝑎3 3 𝑏3 2 𝑐3 𝑑3 𝑥 + 𝑥 𝑦 + 𝑥𝑦 2 + 𝑦 3 6 2 2 6
→ Satisfies (3.3) 𝜕2𝜑 𝑐3 𝑥 + 𝑑3 𝑦 𝜕𝑦 2 𝜕2𝜑 𝜎y = 𝑐3 𝑥 + 𝑏3 𝑦 𝜕𝑥 2 𝜕2𝜑 𝜏xy = − − 𝑏3 𝑥 − 𝑐3 𝑦 𝜕𝑥𝜕𝑦 𝜎x =
if 𝑎3 = 0; 𝑏3 = 0; 𝑐3 = 0 𝜎x = 𝑑3 𝑦 𝜎y = 𝜏xy 0 (simple bending)
𝜎x
22
𝜎y +ℎ
−ℎ 𝜎y
𝐿 if 𝑎3 = 0; 𝑐3 = 0; 𝑑3 = 0 𝜎y = 𝑏3 𝑦 𝜎x = 0 𝜏xy = −𝑏3 𝑥 Method (i) Assuming: 𝜎x ∝ 𝑦 𝜎x ∝ 𝑥 (-BM) 𝜎x = 𝑐1 𝑥𝑦
𝜕2𝜑 𝜕𝜑 𝑐1 𝑥𝑦 2 = + 𝑓1 (𝑥) 𝜕𝑦 2 𝜕𝑦 2
1 𝜑 = 𝑐1 𝑥𝑦 3 + 𝑦𝑓1 (𝑥) + 𝑓2 (𝑥) 6 ∇4 𝜑 = 0 𝑦 𝜕 4 𝑓1 (𝑥) =0 𝜕𝑥 4 𝜕 4 𝑓2 (𝑥) =0 𝜕𝑥 4
𝑦
𝜕 4 𝑓1 (𝑥) 𝜕 4 𝑓2 (𝑥) + =0 𝜕𝑥 4 𝜕𝑥 4 ; 𝑓1 (𝑥) = 𝑐2 𝑥 3 + 𝑐3 𝑥 2 + 𝑐4 𝑥 + 𝑐5 ; 𝑓2 (𝑥) = 𝑐6 𝑥 3 + 𝑐7 𝑥 2 + 𝑐8 𝑥 + 𝑐9
1 𝜑 = 𝑐1 𝑥𝑦 3 + 𝑦(𝑐2 𝑥 3 + 𝑐3 𝑥 2 + 𝑐4 𝑥 + 𝑐5 ) + 𝑐6 𝑥 3 + 𝑐7 𝑥 2 + 𝑐8 𝑥 + 𝑐9 6 𝜎y = 𝜏xy = −
𝜕2𝜑 6[(2𝑦 + 𝑐6 )𝑥 + 2(𝑐3 𝑦 + 𝑐4 )] 𝜕𝑥 2 𝜕2𝜑 1 − 𝑐1 𝑦 2 − 3𝑐2 𝑥 2 − 2𝑐3 𝑥 − 𝑐4 𝜕𝑥𝜕𝑦 2
23 BCs for all 𝑥 {
𝜎y = 0 @ ± ℎ 𝜏xy = 0 @ ± ℎ
𝑐2 , 𝑐3 , 𝑐6 , and 𝑐7 are all zero 1 𝑐4 = − 𝑐1 ℎ2 2 1 𝜏xy = − 𝑐1 (𝑦 2 − ℎ2 ) 2 c)
4th order polynomial 𝜑4 =
𝑎4 𝑏4 𝑐4 𝑑4 𝑒4 𝑥4 + 𝑥3𝑦 + 𝑥2𝑦2 + 𝑥𝑦 3 + 𝑦4 (4)(5) (4)(3) (3)(2) 2 (3)(2)
(3.8)
𝑒4 = −(2𝑐4 + 𝑎4 ) satisfies equation (3.3) 𝜎x =
𝜕 2 𝜑4 𝑏4 3 𝑐4 2 𝑑4 𝑒4 𝑥 + 𝑥 𝑦 + 3 𝑦2 + 4 𝑦3 2 (4)(3) 𝜕𝑦 6 2 6 𝑐4 𝑥 2 + 𝑑4 𝑥𝑦 − (2𝑐4 + 𝑎4 )𝑦 2
𝜕 2 𝜑4 𝑎4 𝑥 2 + 𝑏4 𝑥𝑦 + 𝑐4 𝑦 2 𝜕𝑥 2 𝜕2𝜑 𝑏4 𝑑4 𝜏xy = − − 𝑥 2 − 2𝑐4 𝑥𝑦 − 𝑦 2 𝜕𝑥𝜕𝑦 2 2 Take all of the coefficients except 𝑑4 to be zero. 𝜎y =
𝜎x = 𝑑4 𝑥𝑦, 𝜎y = 0, 𝜎x varies linearly in x and y direction @ 𝑦 = ±ℎ, 𝜏xy is uniform along the transverse direction, 𝜏xy ∝ 𝑦 2
𝜏xy = −
𝑑4 2 𝑦 2
(3.9)
24 Example
𝐿
2ℎ
𝑃
𝑥
𝑦
𝑡
𝑉
𝐵𝑦 ℎ
ℎ
1 𝑐1 𝑡(𝑦 2 − ℎ2 ) d𝑦 −ℎ 2
𝑃 = ∫ 𝜏xy 𝑡 d𝑦 + ∫ −ℎ
𝑐1 = −
3𝑃 𝑃 3 2𝑡ℎ 𝐼
−𝑃 𝑀𝑦 𝜎x = 𝑐1 𝑥𝑦 ( ) 𝑥𝑦 = 𝐼 𝐼 𝜏xy = (
−𝑃 ) (ℎ2 − 𝑦 2 ) 2𝐼
Method (iii) Superposition: 𝜑2 and 𝜑4 with 𝑎2 , 𝑐2 , 𝑎4 , 𝑏4 , 𝑐4 , and 𝑒4 all equal zero 𝜑 = 𝜑2 + 𝜑4 𝜑 = 𝑏2 𝑥𝑦 +
𝑑4 3 𝑥𝑦 6
𝜕2𝜑 𝜎x = 𝑑4 𝑥𝑦 𝜕𝑦 2 𝜎y =
𝜕2𝜑 0 𝜕𝑥 2
𝜏xy = −𝑏2 −
𝑑4 2 𝑦 2
25 𝜏xy = 0 @ ± ℎ 𝑑4 = −
2𝑏2 ℎ2
𝜏xy = −𝑏2 (1 − ℎ
𝑦 2⁄ ) ℎ2
ℎ
3𝑃 𝑃ℎ2 𝑦 2⁄ 𝑃 = ∫ 𝜏xy 𝑡 d𝑦 ∫ 𝑏2 𝑡 (1 − ) d𝑦 𝑏2 = − ℎ2 4ℎ𝑡 2I −ℎ −ℎ 𝜎x = − 𝜏xy = −
𝑃𝑥𝑦 I
𝑃 2 (ℎ − 𝑦 2 ) 2I
3.2 Stress Concentrations Stress concentrations refer to stresses which have higher than normal stresses -
Yielding, fatigue, fracture, etc.
These arise from loading discontinuities or geometrical discontinuities. Loading Discontinuity -
Loading through a knife edge, point, ball bearing or roller bearings, or gear teeth
Geometry Discontinuity -
holes, notches, change in cross-section more pronounced in brittle material than in ductile - find stress concentrations using analytical, numerical, or experimental methods often, it is more convenient to use polar coordinates
3.2.1 Use of Polar Coordinates ∇ 4 𝜑 = ∇ 2 ⁄∇ 2 𝜑 0 ∇2 = (
𝜕2 𝜕 2 rectangular + ) 𝜕𝑥 2 𝜕𝑦 2 coordinates
∇2 = (
𝜕 1 𝜕 1 𝜕2 + + ) 𝜕𝑟 2 𝑟 𝜕𝑟 𝑟 2 𝜕𝜃 2
𝜎r = (𝜎x )θ=0
𝜕2𝜑 1 𝜕 2 𝜑 1 𝜕𝜑 = + ( ) 𝜕𝑦 2 𝑟 2 𝜕𝜃 2 𝑟 𝜕𝑟
(3.12)
(3.13)
26 𝜕2𝜑 𝜕2𝜑 = θ=0 𝜕𝑥 2 𝜕𝑟 2 𝜕2𝜑 𝜕 1 𝜕𝜑 𝜏rθ = (𝜏xy ) (− =− ( ) ) θ=0 𝜕𝑥𝜕𝑦 θ=0 𝜕𝑟 𝑟 𝜕𝜃 For axis symmetry, 𝜑 is independent of angle 𝜃 and equation (3.3) reduces to: 𝜎θ = (𝜎y )
∇2 𝜑 = (
𝜕2 1 𝜕 𝜕 2 𝜑 1 𝜕𝜑 + + ) ( ) 0 Euler Equation 𝜕𝑟 2 𝑟 𝜕𝑟 𝜕𝑟 2 𝑟 𝜕𝑟
(3.14)
Equation (3.13) can be reduced to: 𝜎r = General solution of (3.14)
1 𝜕𝜑 𝜕2𝜑 ; 𝜎θ = 2 ; 𝜏rθ = 0 𝑟 𝜕𝑟 𝜕𝑟
(3.15)
𝜎r = 𝑐1 ln 𝑟 + 𝑐2 𝑟 2 ln 𝑟 + 𝑐3 𝑟 2 + 𝑐4
(3.16)
where 𝑐1 to 𝑐4 are constants 𝑐1 + 𝑐2 (1 + 2 ln 𝑟) + 2𝑐3 𝑟2 𝑐1 𝜎θ = − 2 + 𝑐2 (3 + 2 ln 𝑥) + 2𝑐3 𝑟 𝜏rθ = 0 If there are no holes, 𝑐1 and 𝑐2 are zero. 𝜎r =
(3.17)
3.2.2 Stresses due to Concentration Loads -
point load (mathematic abstraction) in practice, “point” loads are always applied at a small area analytical solutions therefore are NOT valid at the point
Example Comparison of a wedge (unit thickness) → indenter
𝜑 = 𝑐𝑃𝑟𝜃 sin 𝜃 -
𝑦
satisfies (3.3)
𝑃
𝑟
𝛼
𝜃
𝑥 𝛼
𝜎r =
1 𝜕𝜑 1 𝜕 2 𝜑 1 + 2 2 2𝑐𝑃 cos 𝜃 𝑟 𝜕𝑟 𝑟 𝜕𝜃 𝑟
27 𝜕2𝜑 0 𝜕𝑟 2
𝜎θ = 𝜏rθ =
1 𝜕𝜑 1 𝜕 2 𝜑 − =0 𝑟 2 𝜕𝜃 𝑟 𝜕𝜃𝜕𝑟
𝛼
𝛼
𝑃 − 2 ∫ 𝜎r cos 𝜃 𝑟d𝜃 = 0 𝑃 = −2 ∫ 2𝑐𝑃 0
0
cos 2𝜃 + 1 cos 𝜃 = 2 2
𝜎r =
Special Case where 𝛼 =
π 2
𝛼 cos 2𝜃 + 1 cos 2 𝜃 d𝜃 = −1 𝑟d𝜃 4𝑐 ∫ 2 𝑟 0 1 𝑐=− } 2𝛼 + sin 2𝛼
−2𝑃 cos 𝜃 2𝛼 + sin 2𝛼 𝑟
(3.19)
𝜎θ = 0, 𝜏rθ = 0
Flamant Problem -
point or line load on semi-infinite solid 𝜎r =
−2𝑃 cos 𝜃 π 𝑟
(3.20)
𝜎θ = 0, 𝜏rθ = 0
𝑃
𝜃
𝑑
𝑟 = 𝑑 cos 𝜃 𝜎r = constant for a given d at all points on the circle
(b) Bending of a wedge:
−2𝑃 π𝑑
(3.21)
28
𝑦 𝑟
𝛼
𝜃 𝐹
𝑥 𝛼
π π 𝜑 = 𝑎𝐹𝑟 ( − 𝜃) sin ( − 𝜃) 𝑎𝐹𝑟𝜃1 sin 𝜃1 2 2 𝜎r =
−2𝐹 cos 𝜃1 −2𝐹 cos 𝜃1 𝑟(2𝛼 − sin 2𝛼) 𝑟(2𝛼 − sin 2𝛼)
𝜎θ = 𝜏rθ 0 𝜎r = −
𝑦 𝑃a
2 𝑃a cos 𝜃 𝑃t sin 𝜃 + [ ] 𝑡𝑟 2𝛼 + sin 2𝛼 2𝛼 − sin 2𝛼
𝛼 𝑥
𝛽
𝛼
𝑃t
Example: An orthogonal cutting tool has a rake angle of 20°, clearance angle of 10°, and width of cut of 4 mm. The horizontal and vertical loads are 𝑃x = 1.5 kN, 𝑃y = 2.5 kN. Find the minimum tool nose radius of 𝜎r ≤ 900 MPa.
𝑃a
𝑃y
50°
𝑃b 40°
𝑃x
20°
30° 30° 10°
Mitchell’s Solution
r 60°
29 2 𝑃a cos 𝜃 𝑃b sin 𝜃 + [ ] (𝜔)𝑟 2𝛼 + sin 2𝛼 2𝛼 − sin 2𝛼 𝑃a = 𝑃x cos 50° + 𝑃y cos 40° 5.75 kN 𝑃b = 𝑃y sin 40° + 𝑃x sin 50° 0.916 kN π 𝛼 = 𝜃 30° = 6 3 2 (5.75 × 10 ) cos 30° 0.916 × 103 ) sin 30° − [ + ] π π 4𝑟 + sin 60° − sin 60° 3 3 = 900 MPa 𝜎r = −
r 30°
𝒓 = 𝟐. 𝟖𝟓 𝐦𝐦; 𝑟t = 𝑟 tan 30° 𝒓𝐭 = 𝟏. 𝟔𝟓 𝐦𝐦
3.2.3 Stress Concentrations around a Small Circular Hole (Kirsh Solution)
𝜑(𝑟, 𝜃) =
𝜎θ 𝑟 2 𝜎0 𝑏4 ( − 𝑏 2 ln 𝑟) + (2𝑏 2 − 𝑟 2 − 2 ) cos 2𝜃 2 2 4 𝑟
𝑦 𝑟
𝐴
D
𝐛
B
𝜎r =
C
π 3π , 2 2
π 2
𝜎0
1 𝜕 2 𝜑 1 𝜕𝜑 𝜎0 𝑏2 𝜎0 𝑏2 𝑏4 + − + − 4 + 3 (1 ) (1 ) cos 2𝜃 𝑟 2 𝜕𝜃 2 𝑟 𝜕𝑟 2 𝑟2 2 𝑟2 𝑟4 𝜕2𝜑 𝜎0 𝑏2 𝜎0 𝑏4 𝜎θ = 2 (1 + 2 ) + (−1 − 3 4 ) cos 2𝜃 𝜕𝑟 2 𝑟 2 𝑟 2 𝜕 1 𝜕𝜑 𝜎0 𝑏 𝑏4 𝜏rθ = − ( ) (−1 − 2 2 + 3 4 ) sin 2𝜃 𝜕𝑟 𝑟 𝜕𝜃 2 𝑟 𝑟
Points A and B have maximum stress:
Putting
𝑥
C: 0 rad D: π rad 3π and 2 into equation (3.23) for 𝜎θ at 𝑏 = 𝑟, 𝜎θ = 3𝜎0 𝜎x
(3.23)
30
Stress concentration:
3𝜎0 𝜎0
=3 @ 𝜃 = 0 and π: 𝜎θ = 𝜎y −𝜎0 (compressive) @ 𝑟 > 10𝑏 (𝜎θ )π 2
,
3π 2
𝜎0 𝑏2 𝑏4 = (2 + 2 + 3 4 ) 𝜎0 2 𝑟 𝑟
Result is valid for finite width plate 𝑤 > 10𝑏.
4.0 Shock and Impact Loading -
When forces or displacements are applied suddenly to structural members, the stress levels and deformation are often much higher than those that occur by the same forces or displacements applied gradually Rapidly moving loads: (train crossing a bridge, drop of a forge hammer, sudden loads produced during combustion in the power stroke of an internal combustion engine, collision, acceleration, etc.) Impact shock load when 1 𝑡applied ≤ 𝑇 2 } time of highest natural frequency of the structure 𝑡applied ≤ 𝑇
-
-
-
Quasi-static: 𝑡applied ≥ 3𝑇
4.1 Stress and Displacement in an Elastic Axial Bar
circular x − section 𝐴, 𝐸 𝑊(striking mass) 𝐿 ℎ 𝑦 pan
31 -
mass of the pan and the bar is small compared to the striking mass (W) 𝑦 ≠ 𝑓(𝑡) Linear elastic NO energy lost
strain energy
𝐹
𝑦 Energy balance: external work = stored strain energy of the bar 1 𝑊(ℎ + 𝑦̂) = 𝐹̂ 𝑦̂; (̂ denotes maximum) 2 𝜎̂ = 𝐸𝜀̂ 𝐸
𝑦̂ 𝜎̂𝐿 = 𝑦̂ 𝐿 𝐸
𝐹̂ = 𝜎̂𝐴 𝑊 (ℎ +
𝜎̂𝐿 1 𝜎̂𝐿 ) = 𝜎̂𝐴 𝐸 2 𝐸
𝐴𝐿 2 𝑊𝐿 𝜎̂ − ( ) 𝜎̂ − 𝑊ℎ = 0 2𝐸 𝐸
𝜎̂ =
𝑊 2ℎ𝐸𝐴 [1 + √1 + ] 𝐴 𝑊𝐿
(4.1)
(4.2)
∆ impact factor 𝑦̂ =
𝑊𝐿 2ℎ𝐸𝐴 [1 + √1 + ] 𝐴𝐸 𝑊𝐿
(4.3)
2ℎ𝐸𝐴 ] 𝑊𝐿
(4.4)
𝐹̂ = 𝑊 [1 + √1 + 𝐹̂ is the effective dynamic fore due to impact. 𝑦static =
𝑊𝐿 𝐴𝐸
32 Impact, ∆ ∆= 1 + √1 + ∆= 1 + √1 + -
2ℎ𝐸𝐴 𝑊𝐿
(4.5)
2ℎ 𝑦static
If the load is applied suddenly even at h=0; 𝜎̂ =
2𝑊 2(𝜎max )static 𝐴
𝐹̂ = 2(𝐹max )static 𝑦̂ = 2(𝑦max )static -
𝜎̂ can be reduced by increasing L or A and decreasing 𝜎static =
𝐹 𝐴
Applications: -
wood is used as railway track long bolts used to attach ends of the pneumatic cylinder of jack hammers if the bar has a weight of q per unit length, the impact factor, ∆ can be modified to:
∆∗ = [1 + √1 +
-
2ℎ𝐸𝐴 1 ( )] 𝑊𝐿 1 + 𝑞𝐿 3𝑊
Analysis is similar for cases involving impact of moving loads. (KE)moving load = 𝑢structure
Example (a) Find the max stress in a steel rod. (i) (ii)
with washer without washer
(4.7)
33 rubber washer: N mm ∅ = 15 mm 𝐸 = 200 GPa 𝑘=9
10 kg 1.5 m ℎ rubber washer 16 mm
i)
with washer 𝜎̂ =
𝑊 = (10 kg) (9.81 𝐴=
m ) 98.1 N s2
π 2 π ∅ (15 mm)2 = 177 mm2 4 4
𝑦static =
(98.1 N)(1 500 mm) (98.1 N) 𝑊𝐿 𝑊 + + 𝒚𝐬𝐭𝐚𝐭𝐢𝐜 = 𝟏𝟎. 𝟗 𝐦𝐦 N N 𝐴𝐸 𝑘 (177 mm2 ) (200 × 103 ) 9 mm mm2
without washer
with washer
𝜎̂ =
ii)
𝑊 2ℎ [1 + √1 + ] 𝐴 𝑦static
98.1 N 2(1 000 mm) ̂ = 𝟖. 𝟎𝟖 𝐌𝐏𝐚 [1 + √1 + ] 𝝈 2 177 mm 10.9 mm
without washer
𝜎̂ =
98.1 N 2(1 000 mm) ̂ = 𝟑𝟖𝟓 𝐌𝐏𝐚 [1 + √1 + ] 𝝈 2 177 mm 4.157 × 10−3 mm
Example (b) Find the max h allowed to avoid yielding.
34
𝑊 = 60 N 𝑏 = 2.5 mm
−7
𝐼 = 8.79 × 10
4
m
ℎ
𝐸 = 207 GPa 𝜎y = 207 MPa
𝑑 = 75 mm
𝑉̂
1.5 m
Equating external energy = stored strain energy: 𝑊(ℎ + 𝑉̂ ); 𝑉̂ =
𝜎̂ =
𝑃𝐿3 𝑃 = dynamic load 3𝐸𝐼
𝑃𝐿 𝑑⁄2 𝑀𝑦 𝜎̂𝐼 𝜎̂ = 𝑃 = 𝐼 𝐼 𝐿 𝑑⁄2 𝜎̂𝐼 𝐿3 ̂ 𝑉=( ) 𝐿 𝑑⁄2 3𝐸𝐼 𝜎̂𝐼 𝐿3 𝑃𝐸 = 𝑊 (ℎ + ( ) ) 𝐿 𝑑⁄2 3𝐸𝐼
Stored strain energy: 2
1 𝐿 𝑀2 1 𝐿 (𝑃𝑥)2 𝑃2 𝐿2 𝜎̂𝐼 𝐿3 𝑈= ∫ d𝑥 𝑈 = ∫ d𝑥 𝑈 =( ) 2 0 𝐸𝐼 2 0 𝐸𝐼 6𝐸𝐼 𝐿 𝑑⁄2 6𝐸𝐼 𝑃𝐸 = 𝑈 𝜎̂𝐿2 𝜎̂ 2 𝐼𝐿 𝑊 (ℎ + )= 2 3(𝑑⁄2)𝐸 6(𝑑⁄2) 𝐸 60 (ℎ +
(270 × 106 Pa)2 (8.79 × 10−7 m4 )(1.5 m) (270 × 106 Pa)(1.5 m)2 = ) 3(0.0375 m)(207 × 109 Pa) 6(0.0375 m)2 (207 × 109 Pa) 𝒉 = 𝟎. 𝟖𝟗𝟏 𝐦
35
Example (c) 𝑊 = 12 kg 𝐸 = 8 GPa (WOOD)
ℎ A
𝐼 = 1.5625 × 10−6 m4 𝐿 = 1.5 m
C′
B
C′′ 𝐿 (i) (ii)
𝐿
max h allowed if max deflection is limited to 38 mm amount of energy absorbed by each beam
𝑃𝐸 = 𝑈 𝑃𝐸 = 𝑊(ℎ + 𝛿max )
FBD
𝑃
𝑅′
A 𝑅 ′′
𝑅 B
𝑀
𝑅 ′′′ (i) Beam B (0.038 m)3(8 × 109 Pa)(1.5625 × 10−6 m4 ) 𝑅𝐿3 𝛿B 3𝐸𝐼 𝛿B = 𝑅 = 𝑅 = 𝑹 = 𝟒𝟐𝟐. 𝟐 𝐍 (1.5 m)3 3𝐸𝐼 𝐿3 Beam A
36
𝛿A = 𝑃 =
(𝑃 − 𝑅)(2𝐿)3 𝑊 ′ 𝐿3 𝛿A 6𝐸𝐼 𝑃 = +𝑅 48𝐸𝐼 48𝐸𝐼 𝐿3
(0.038 m)6(8 × 109 Pa)(1.5625 × 10−6 m4 ) + 422.2 N (1.5 m)3 𝑷 = 𝟏 𝟐𝟔𝟔. 𝟕 𝐍
∆= 1 + √1 +
2ℎ 𝑦static
∆=
𝑃 1 266.7 N m ∆= 𝟏𝟎. 𝟕𝟔 𝑊 12 kg (9.81 2 ) s
𝒉 = 𝟏𝟔𝟔. 𝟒 𝐦𝐦
𝛿static =
𝛿max 0.038 m ∆ 10.76
𝑃𝐸 = 𝑈 1 𝑊(ℎ + 𝛿max ) = 𝑃𝛿; 𝒉 = 𝟏𝟔𝟔. 𝟒 𝐦𝐦 2 (ii) 1 1 𝑈A = 𝑃′ 𝛿 ′ (𝑃 − 𝑅)𝛿 𝑼𝐀 = 𝟏𝟔. 𝟎𝟓 𝐉 2 2 1 ′ ′ 1 𝑈B = 𝑃 𝛿 𝑅𝛿 𝑼𝐀 = 𝟖. 𝟎𝟐 𝐉 2 2
37 Superposition can be used to obtain the deflection.
I.
𝑊 A
B
a
b
𝑀 = 𝑊𝑏
B
A
𝜃
𝑊 𝛿
𝑉̂ = 𝛿 + 𝜃𝑏 𝑏 tan 𝜃𝑏 = 𝜃𝑏
II.
𝑊 A
B C
A
B C
A
B C 𝑅
38
4.2 Effects of Geometry
B
A 𝐿
𝐴1
stress concentrations are neglected for beam B
𝐿
𝐴2 = 2𝐴1
3 𝐿 3
𝐴1
𝐿
𝐴2
3
1 𝑈A = 𝑃̂𝛿 2 𝑃̂ = 𝜎̂𝐴, 𝛿 = 𝜀𝐿, 𝜎̂ = 𝐸𝜀 1 𝜎̂ 2 𝐴𝐿 2𝐸 1 𝜎̂ 2 𝑈A = 𝐴 𝐿 2𝐸 1 𝑈=
𝑈B =
1 𝜎̂ 2 𝐿 1 𝜎̂ 2 𝐿 𝐴1 + 2 [ 𝐴2 ] 2𝐸 3 2𝐸 3 2 𝑈B = 𝑈𝐴 2𝐴1 3
To design or maximum resistance to impact load: -
use a low E increase length, L distribute the stress as uniformly as possible avoid stress concentrations
4.3 Stress Wave Propagation under Impact Loading The above provides a quick solution estimate of maximum stress and deflection under impact loading. Exact consideration involves the motion of stress waves traveling within the structures. Consider the 1-D problem: 𝐿 m
m′
m
m′
n
n′
n
n′
𝑃imp 𝑐
(wave velocity)
𝑑𝑥 𝑢 +
d𝑢 d𝑥 d𝑥
39 At the instant of impact of the free end, a very thin segment of the bar directly under the load is set into motion. The rest of the bar, remote from the load, 𝑃, remains undisturbed for a small, finite length of time. The deformation due to 𝑃 propagates along the bar with time in a form of an elastic deformation: stress wave. If 𝐿 is large, time required for the wave propagation becomes significant and must be considered. Two different types of stress waves:
𝑃
𝑃 𝑥
𝑥
Longitudinal Wave Particle motion in the 𝑥-direction Pressure wave (𝑃wave ) Velocity (𝑐L )
Transverse Wave Shear wave (s-wave) Particle motion in the 𝑦-direction
𝐹 = 𝑚𝑎 𝐹 = 𝜎𝐴 𝜎 = 𝐸𝜀 𝜀=
d𝑢 d𝑥
𝐹 = 𝐸𝐴
d𝑢 d𝑥 𝒄𝟐 =
𝑚 = 𝜌𝑉 𝑚 = 𝜌𝐴 𝑑𝑥 d2 𝑢 𝑎= 2 d𝑡 d𝑢 d2 𝑢 𝐸𝐴 = 𝜌𝐴 𝑑𝑥 2 d𝑥 d𝑡 𝐸 d2 𝑢 d2 𝑢 = 𝜌 d𝑥 2 d𝑡 2 𝐝𝟐 𝒖 𝐝𝟐 𝒖 + 𝐝𝒙𝟐 𝐝𝒕𝟐
1-D wave equation 𝐸 𝜌
𝑐L = √ : velocity of longitudinal wave 𝐺
𝑐S = √𝜌: velocity of transverse wave e.g. 207×109 7 850
steel: 𝑐L = √
copper: 𝑐s = √
5 135
120×109 8 900
m s
3 672
m s
40 Verify that 𝑓(𝑥 ± 𝑐𝑡) is a solution of equation (4.8) where; + is the negative 𝑥 direction − is the positive 𝑥 direction 𝑢(𝑥, 𝑡) = 𝑓1 (𝑥 − 𝑐𝑡) + 𝑓2 (𝑥 + 𝑐𝑡) 𝑢 = 𝐴 sin (
2π ) (𝑥 ∓ 𝑐𝑡) 𝐿
𝒖𝒊 = 𝑨𝒊 𝒆𝒊(𝝎𝒕−𝒌𝒊 𝒙𝒊 ) 𝑘: wave number λ
2π
wave
𝜔: angular frequency
Stress Waves at Fixed and Free Ends @ fixed end Waves are reflected and unchanged i.e. compressive stays compressive @ free end Waves are reflected with same amplitude but opposite in sign. compressive → tensile tensile → compressive 2-D (longitudinal and shear) net tensile strength exceeds rupture stress
scab
explosive internal pressure
Fracture occurs at the thickest section as compared to thin section in static loading.
41 In real materials, there is a phase difference between the stress and strain due to hysteresis or internal friction damping (energy losses). Stress level decays with distance.
4.4 Changes in Material Properties under Impact Loading
rapid loading
𝜎
static loading
𝜀
𝑘1D < 𝑘1C 30% less 𝑘1C
𝑘1D
10
20
30
m s
loading speed
42
5.0 Elasto-Plastic Analysis Yield Criteria Tresca: 𝜎1 − 𝜎2 𝜎y = 2 2 𝜎2 − 𝜎3 𝜎y = 2 2 𝜎1 − 𝜎3 𝜎y = 2 2
(5.1)
Von Mises 1 √2
1
[(𝜎1 − 𝜎2 )2 + (𝜎2 − 𝜎3 )2 + (𝜎1 − 𝜎3 )2 ]2 = 𝜎y
(5.2a)
In 2D 1
(𝜎12 + 𝜎22 − 𝜎1 𝜎2 )2 = 𝜎y
(5.2b)
When the equivalent stress in parts of a component exceeds the yield stress of the material, elastoplastic deformation occurs. Elasto-plastic analysis results in an analytical solution and is only possible for relatively few types of problems with simple geometry and idealized stress-strain behaviour (i.e. bending, torsion, axis-symmetric problems.) General Stress-Strain Curve:
𝜎
×
𝜎y
𝜀
43
𝜎
𝜎
𝜎y
𝜎y
𝜀
𝜀
(a) elastic-perfectly plastic
(b) elastic-linear plastic/hardening
𝜎
𝜎p = 2(𝐴 + 𝐵𝜀 𝑛−1 )
𝜎y
𝜀 ***NOT COVERED IN THIS COURSE (c) non-linear Permanent set occurs when the component is loaded beyond its yield point. When the load is removed there are residual stress. These can be both beneficial and harmful.
𝜎 loading
𝜎y
unloading
𝜀p
𝜀
44
5.1 Beam Bending Consider (a) and (b): 𝐢𝐧𝐜𝐫𝐞𝐚𝐬𝐢𝐧𝐠 𝐥𝐨𝐚𝐝 𝑏
ℎ
𝜎y
𝜎y plastic hinge
𝑦p 𝑑 𝑦
ℎ
𝜎y
𝜎y
(i) onset of yielding
(ii) elasto-plastic deformation
𝜎=
(iii) fully plastic
𝑀𝑦 𝜎𝐼 𝑀 = 𝐼 𝑦
Elasto-plastic bending moment: 𝑀ep = 𝑀e + 𝑀p elastic
𝜎y 𝑏(𝑑 − 2ℎ)2 + 𝜎y 𝑏ℎ(𝑑 − ℎ) 6 plastic
𝜎y 𝑏𝑑2 ℎ ℎ [1 + 2 (1 − )] 6 𝑑 𝑑 𝜎y 𝑏 2 2 𝑀ep = [3𝑑 − 4𝑦p ] 12
(5.3a)
𝑀ep =
(5.3b) distance from neutral axis
Fully plastic: 𝑀ep =
𝜎y 𝑏𝑑2 1 1 𝑏𝑑2 [1 + 2 (1 − )] 𝜎y = 6 2 2 4 𝑀=
𝜎𝐼 𝑏𝑑2 𝜎y = 𝑦 6
𝑧p (shape factor) =
𝑀p 1.5 𝑀y
The shape factor five a measure of the increase in strength-carrying capacity available beyond the elastic limit.
45 Removal of loads after elasto-plastic deformation:
𝜎 𝑀ep
𝜎y
−𝑀ep
no net moment
𝜀
𝜀p
-
equivalent to superposition of a linear-elastic unloading stress distribution over the initial stress distribution unloading stress distribution must be done to a moment of equal magnitude but opposite in direction of the applied load 𝑦 A
𝑀ep
B
C
D
A B
C
ℎ
D
−𝑀ep
***not to scale
Subtract unloading from loading stress distribution to find residual stress distribution: 𝑦
Example: Rectangular section beam loaded beyond elastic limit; yielding of half the cross section. Find the residual stress distribution when the load is removed. 𝑏
𝑑 𝑦0
7.5 mm 7.5 mm
Elastic-perfectly plastic; 𝜎y = 600 MPa; 𝑏 = 10 mm; 𝑑 = 30 mm
46
𝑀ep =
𝜎y 𝑏𝑑2 (600 × 106 )(10 × 10−3 )(30 × 10−3 )2 ℎ ℎ 1 1 [1 + 2 (1 − )] [1 + 2 (1 − )] 6 𝑑 𝑑 6 4 4 𝑴𝐞𝐩 = 𝟏 𝟐𝟑𝟕. 𝟓 𝐍 ∙ 𝐦
Elastic stresses due to 𝑀ep : 𝑑 −𝑀ep 2 𝑑 𝑦 = 𝜎 = 𝝈 = −𝟖𝟐𝟓. 𝟎 𝐌𝐏𝐚 2 𝐼 𝑦 = 𝑦0 𝜎=
−𝑀ep 𝐼
𝑑 4 𝝈 = −𝟒𝟏𝟐. 𝟓 𝐌𝐏𝐚
Residual stress distribution: A: 𝜎x = −825.0 + 600 − 225 MPa B: 𝜎x = −412.5 + 600 187.5 MPa C: 𝜎x = 412.5 − 600 − 187.5 MPa D: 𝜎x = 825. − 600 225 MPa 𝑦
𝑦 A = 600
𝑀ep
A = −225
B = 600
B = 187.5
𝜎x (MPa)
C = −600 D = −600
C = −187.5
−𝑀ep
𝐷 = 225
Springback refers to the elastic recovery of the plastically deformed structure when the loads are removed. e.g. 𝛿f 𝛿ep
springback
𝛿f = 𝛿ep − 𝛿e -
𝑦
strain: 𝜀 = 𝑅; 𝑅 (radius of curvature)
𝛿e
47
𝜀=
𝑦
𝜎y 𝑦p } still elastic up to 𝑦p 𝑅ep E
A
A
𝑅ep =
𝑦p E 𝜎y
𝑦p =
𝑅ep 𝜎y E
𝑦p
Substituting 𝑦p back into equation (5.3b): 𝑀ep
𝜎y 𝑏 𝑅ep 𝜎y 2 2 = [3𝑑 − 4 ( ) ] 12 E
(5.4)
EI 𝑏𝑑 3 E 𝑅e 12 𝑅e
(5.5)
For elastic bending: 𝑀e =
𝑅 2 = 𝐿2 + (𝑅 − 𝛿)2 Since 𝛿 is small compared to R and L, neglect the 𝛿 2 term.
𝛿
𝑅−𝛿
𝐿
𝐿2 ≈ 2𝑅𝛿 𝛿 =
𝑅
𝐿2 2𝑅
𝛿f = 𝛿ep − 𝛿e 𝐿2 𝐿2 𝐿2 1 = − ; (curvature) 2𝑅f 2𝑅ep 2𝑅e 𝑅 𝑅ep 𝑅ep 𝑅ep 1 1 1 = − =1− = 0 complete springback } measure of springback; when 𝑅f 𝑅ep 𝑅e 𝑅f 𝑅e 𝑅e Since elastic loading after elastic-plastic deformation:
|𝑀e | = |𝑀ep | 𝜎 𝜎y
𝑀ep
𝑀e elastic unloading moment
𝜀
48 𝑅ep 𝜎y 2 𝑏𝑑3 E 𝜎y 𝑏 2 = [3𝑑 − 4 ( ) ] 12𝑅e 12 E 𝑅ep 𝑅ep 𝜎y 𝑅ep 𝜎y 3 = 1 − 3[ ]+4[ ] 𝑅f E𝑑 E𝑑 𝑅ep 𝑅ep 𝜎y 2𝑅ep 𝜎y 2 =( + 1) ( ) 𝑅f E𝑑 E𝑑
(5.7)
In the fully plastic case: 𝑀p =
𝑏𝑑 2 𝜎 4 y
𝑏
𝜎y ℎ=
𝑑 2
𝑑 𝑑 2
𝜎y
Elastic recovery:
|𝑀e | = |𝑀ep | EI 𝑏𝑑2 = 𝜎 𝑅e 4 y 𝑅p 1 − 𝑏𝑑2 𝜎y 𝑅p = 𝑅f 4EI Using 𝐼 =
𝑏𝑑 3 12
𝑅p 3𝜎y 𝑅p =1− 𝑅f E𝑑
(5.8)
Example: Determine the tool radius required to produce a final bend radius of 50 mm in a steel strip 20 mm wide and 2 mm thick. What is the required moment? Assume fully plastic deformation and against the assumptions.
𝜎y = 300 MPa; E = 200 GPa 3𝜎y 𝑅p 𝑅p =1− 𝑅f E𝑑 𝑅p = (50 × 10−3 m) [1 −
3(300 × 106 Pa)𝑅p ] 𝑹𝐩 = 𝟒𝟒. 𝟗 𝐦𝐦 (200 × 109 Pa)(2 × 10−3 m)
49
𝑀p =
(20 × 10−3 m)(2 × 10−3 m)2 𝑏𝑑2 (300 × 106 Pa) 𝑴𝐩 = 𝟔 𝐍 ∙ 𝐦 𝜎y 4 4
Check the elastic case:
𝜎y 𝜎y 𝑅ep 𝑦p (300 × 106 Pa)(44.9 × 10−3 m) = 𝑦p = 𝒚𝐩 = 𝟎. 𝟎𝟔𝟕 𝐦𝐦 𝑅ep E E 200 × 109 Pa Since 𝑦p is small, the assumption is correct.
Deflection To find the deflection, use moment-curvature relationship as in the elastic case. 2
𝑀 𝜎 𝐸 𝜕 𝑣 = = 𝑀 = −𝐸𝐼 2 𝐼 𝑦 𝑅 𝜕𝑥
The relationship between 𝑅ep and 𝑀ep is more complicated, integration has to be separated between the region that is plastically deformed and the region that is still elastic.
5.1.2 Elastic-Linear Strain Hardening In
𝜎 𝜎o
𝑦
tan−1 𝐸p
𝜎y
𝜎o 𝜎y
𝜎o : stresses at 𝑦p
the outer are max
𝜎
tan−1 𝐸
𝜀o
𝜀
In the elastic range: |𝑦| < |𝑦p | 𝜎e = 𝜎y
𝑦 𝑦p
In the plastic region: |𝑦| > |𝑦p |
(5.9a)
50
𝑦 − 𝑦p 𝜎p = 𝜎y + (𝜎o − 𝜎y ) (𝑦 ) − 𝑦 p 2
(5.9b)
Using the stress-strain plot:
(𝜎o − 𝜎y ) = (𝜀o −
𝜀o = 𝑦p 𝑅ep
=
𝜎y )E E p
𝑑 (2) 𝑅ep
𝜎y ; when 𝑦 = 𝑦p E 𝜀o =
𝑑 ( 2) 𝜎y E𝑦p
𝑑 ( ) Ep 𝜎 [ 2 − 1] (𝜎o − 𝜎y ) = E y 𝑦p
𝜎p = 𝜎y +
(5.10a) for |𝑦| > |𝑦p |
Ep 𝑦 − 𝑦p 𝜎y [ ] E 𝑦p
(5.10b)
Elastic-Plastic Bending Moment 𝑑 2
𝑑 2
𝑦p
𝑀ep = ∫ 𝜎 𝑏𝑦 d𝑦 2 [∫ 𝜎e 𝑏𝑦 d𝑦 + ∫ 𝜎p 𝑏𝑦 d𝑦] 0
0
𝑀ep = when
Ep E
𝜎y 𝑏𝑑2 12
[(3 − 𝑐2 ) +
𝑦p
𝑦p Ep (2 − 3𝑐 + 𝑐3 )] ; where 𝑐 = 𝑑⁄ 𝑐E 2
= 0 elastic-perfectly plastic case (5.3b)
𝜎 tan−1 𝐸p
𝜀
51 For residual stress calculations, similar to the elastic-perfectly plastic case, assume elastic unloading due to applied −𝑀ep . -
Superimpose elastic stress distribution due to −𝑀ep to that due to loading by 𝑀ep .
Springback occurs when unloading from elasto-plastic deformation due to 𝑀ep : |𝑀e | = |𝑀ep | 12𝑀ep 1 = 𝑅e 𝑏𝑑3 E Moment to initiate yielding: 𝑀y =
𝑏𝑑2 𝜎 6 y
2𝜎y 𝑀ep 1 = 𝑅e E𝑑 𝑀y
𝜎y 1 1 = ( ) 𝑅e E 𝑦p 1 1 1 = − 𝑅f 𝑅ep 𝑅e
𝜎y 2𝜎y 𝑀ep 1 = − ( ) 𝑅f E𝑦 E𝑑 𝑀y 𝑦p 1 2𝜎y 1 𝑀p = ( − ) ; where 𝑐 = 𝑑⁄ 𝑅f E𝑑 𝑐 𝑀y
(5.13)
2
Example: A beam of rectangular cross-section (𝑏 = 25 mm, 𝑑 = 40 mm) is loaded gradually so that the plastic zone spreads inwards from the outer fibers. Fine the values of the bending moment that cause: a) yielding initiation b) yielding up to a depth of 10 mm from the outer fibers c) What is the final radius of curvature of the beam upon unloading?
𝜎y = 250 MPa E = 200 GPa Ep = 10 GPa (linear strain − hardening material) 10 mm 10 mm 40 mm
𝑦p = 10 mm
𝑦p
52 a) (25 × 10−3 m)(40 × 10−3 m)2 𝑏𝑑2 (250 × 106 Pa) 𝑴𝐲 = 𝟏. 𝟔𝟔𝟕 𝐤𝐍 ∙ 𝐦 𝑀y = 𝜎 6 y 6 b) 𝑀ep =
𝜎y 𝑏𝑑2 12
[(3 − 𝑐2 ) +
𝑐=
Ep (2 − 3𝑐 + 𝑐3 )] 𝑐E
𝑦p 10 = 0.5 𝑑⁄ 20 2
(250 × 106 Pa) (25 × 10−3 m)(40 × 10−3 m)2 10 (2 − 3(0.5) + 0.53 )] [(3 − 0.52 ) + 12 0.5(250) = 𝟐. 𝟖𝟏𝟐𝟓 𝐤𝐍 ∙ 𝐦
𝑀ep = 𝑴𝐞𝐩 c)
1 2𝜎y 1 𝑀p 2(250 × 106 Pa) 1 2.8125 = − ( − ) [ ] 9 −3 (200 × 10 Pa)(40 × 10 m) 0.5 1.667 𝑅f E𝑑 𝑐 𝑀y 1 = 0.0195 m−1 𝑅f 𝑹𝐟 = 𝟓𝟏. 𝟐 𝐦
Perfectly plastic:
Ep
=0
E
𝑴𝐞𝐩 = 𝟐. 𝟐 𝐤𝐍 ∙ 𝐦 𝑹𝐟 = 𝟐𝟓. 𝟔 𝐦
5.2 Thick-Walled Cylinders under Internal Pressure Assume: i) elastic-perfectly plastic
𝑟
ii) Tresca’s maximum shear stress yield criterion
𝑟i
𝑟o
53
𝑝i 𝑟o2 𝑝i 𝑟o2 𝑟o 𝜎r = 2 (1 − 2 ) ; 𝜎θ = 2 (1 + 2 ) ; 𝑘 = 𝑘 −1 𝑟 𝑘 −1 𝑟 𝑟i In a thick-walled cylinder under internal pressure:
𝜎θ > 𝜎z > 𝜎r 𝜏̂ =
𝜎y 𝜎θ − 𝜎r 𝑝i 𝑟o2 2 ( 2) 2 2 𝑘 −1 𝑟
(5.15)
Yielding will occur first at the bore, where 𝑟 = 𝑟i
𝜎y 𝑝i − 𝑘 2 2 𝑘 −1 2
𝜏̂ =
(𝑝)elastic
=
breakdown
𝑘2 − 1 𝜎 2𝑘 2 y -
𝑟 = 𝑛𝑟i
(5.16)
consider partial yielding in the cylinder to a depth equivalent to a radius ratio if 𝑛 (i.e. yielding at 𝑟 = 𝑟i to 𝑟 = 𝑛𝑟i )
𝑟i
𝑟o elastic zone
Let the radial pressure at 𝑟 = 𝑛𝑟i to be equal to 𝑃𝑛 -
consider the elastic zone 𝑟
𝑘
substitute for 𝑘 the quantity 𝑛𝑟o = 𝑛 i
𝑃𝑛
𝜏̂ =
𝑘
𝑘
2
(𝑛) − 1 𝑃𝑛 =
2
( ) 𝑛
𝜎y (𝑘2 − 𝑛2 ) 2𝑘2
Consider the plastic zone equilibrium equation: 𝑟
d𝜎r d𝑟
+ 𝜎r − 𝜎 θ = 0
𝜎y 2
(5.17)
54
In the plastic zone: 𝜏̂ = 𝑟
d𝜎r d𝑟
𝜎y 𝜎θ − 𝜎r 2 2
+ 𝜎y = 0 d𝜎r = 𝜎y
d𝑟 𝑟
Integrating
𝜎r = 𝜎y ln 𝑟 + C Boundary condition at 𝑟 = 𝑛𝑟i 𝜎r = −𝑝𝑛 C = −𝑝𝑛 − 𝜎y ln(𝑛𝑟i ) C =
−𝜎y (𝑘2 − 𝑛2 ) 2𝑘2
(𝑘 2 − 𝑛2 ) 𝜎r = 𝜎y [ln ( ) − ] 𝑛𝑟i 2𝑘 2
− 𝜎y ln(𝑛𝑟i )
𝑟
(5.18a)
𝜎θ − 𝜎r 𝜎y = 𝜎θ = 𝜎y + 𝜎r 2 2 𝜎θ = 𝜎y [ln (
𝑟 𝑛𝑟i
)−
(𝑘 2 + 𝑛2 ) ] 2𝑘 2
(5.18b)
when 𝑟 = 𝑟i 𝜎r = −𝑝
From (5.18a) 𝑝 = 𝜎y [ln(𝑛) +
(𝑘 2 − 𝑛2 ) ] 2𝑘 2
(5.19)
𝑝 is the pressure required to cause yielding to the radial depth 𝑛𝑟i so when 𝑛 = 𝑘, complete yielding throughout the cylinder is present.
𝑝 = 𝑝c 𝜎y ln 𝑘
(5.20)
𝑝c : collapse pressure At the collapse pressure:
𝑟 𝑟o
(𝜎r )𝑝c = 𝜎y ln ( )
𝑟 𝑟o
(𝜎θ )𝑝c = 𝜎y [1 + ln ( )]
(5.21)
55 The stress distribution: a) onset of yielding at the bore 𝜎 𝑝𝑘 2 + 1 𝑘2 − 1 𝜎θ 2𝑝 −1
𝑘2 𝜎y
𝑟 𝜎r
−𝑝
b) elastic-plastic cylinder 𝜎
elastic
plastic
𝜎θ
𝜎y
𝜎r
elastic − plastic interface
𝑟
56 c) fully plastic cylinder 𝜎 𝜎θ
𝜎y
𝜎y
𝜎r
𝑟
Example: elastic-fully plastic, 𝑘 = 2, 𝑝 = −𝜎r
a) elastic breakdown pressure
𝜎y = 250 MPa
b) the pressure to cause yielding up to 50% across its thickness, and the corresponding stresses 𝜎r and 𝜎θ at the elastic-plastic interface
𝑟 𝑟i
c) the collapse pressure and the corresponding hoop stresses at the bore 𝑟o
a) 𝑝eb 𝑝eb =
𝑘2 − 1 4−1 (250 MPa) 𝒑𝐞𝐛 = 𝟗𝟑. 𝟕𝟓 𝐌𝐏𝐚 𝜎y 2 2𝑘 2(4)
b) 𝑝 (𝑘 2 − 𝑛2 ) 𝑝 = 𝜎y [ln(𝑛) + ] ; 𝑛 = 1.5 2𝑘 2 (22 − 1.52 ) 𝑝 = (250 MPa) [ln(1.5) + ] 𝒑 = 𝟏𝟓𝟔. 𝟏 𝐌𝐏𝐚 2(22 )
57 At the interface:
(𝑘 2 − 𝑛2 ) 𝑝n = 𝜎y [ ] ; 𝑛 = 1.5 2𝑘 2 (22 − 1.52 ) ( ) 𝑝n = 250 MPa [ ] 𝒑𝐧 = 𝟓𝟒. 𝟕 𝐌𝐏𝐚 2(22 ) (𝜎r )𝑛=1.5 = − 𝑝n − 𝟓𝟒. 𝟕 𝐌𝐏𝐚 (𝜎θ )𝑛=1.5 = 𝜎y + 𝜎r 250 − 54.7 𝟏𝟗𝟓. 𝟑 𝐌𝐏𝐚
c) 𝑝c 𝑝c = 𝜎y ln 𝑘 (250 MPa) ln 2 𝟏𝟕𝟑. 𝟐 𝐌𝐏𝐚 𝜎θ = 𝜎y + 𝜎r 250 − 173.2 𝟕𝟔. 𝟕 𝐌𝐏𝐚
5.2.1 Residual Stresses in Internally Pressurized Cylinders Compounding cylinders is one way to obtain favourable pre-stresses before they are put into service. This is a more effective use of material. An even more effective use of material, while achieving the same goal, is by applying enough pressure to cause yielding (residual stresses) in some or all of the material in a simple cylinder then releasing the pressure before it is put into service. The generation of favourable residual stresses in a cylinder by plastic action is called autofrettage. Consider a fully (100% overstrained) cylinder which is being unloaded after reaching the state. Assume elastic unloading. Stresses due to the applied collapse pressure, 𝑃c :
𝑟 𝑟o
(𝜎r )𝑝c = 𝜎y ln ( )
𝑟 𝑟o
(𝜎θ )𝑝c = 𝜎y [1 + ln ( )] For elastic unloading due to the 𝑃c 𝑝c 𝑟o2 − (1 ) 𝑘2 − 1 𝑟2 𝑐 𝑟o2 𝜎θ = 2 (1 + 2 ) 𝑘 −1 𝑟 𝜎r =
𝑟2o (1 − 2 ) 𝑟 𝑘2 − 1 𝑝 𝑟 𝑟2o (𝜎θ )res = 𝜎y [1 + ln ( )] − 2 c (1 + 2 ) 𝑟o 𝑟 𝑘 −1 𝑟 𝑟o
(𝜎r )res = 𝜎y ln ( ) −
𝑝c
(5.21)
58 𝑝c = 𝜎y ln(𝑘)
𝑟 𝑟o
(𝜎r )res = 𝜎y [ln ( ) −
ln 𝑘
(1 −
𝑟2o )] 𝑟2
(5.22a)
𝑘2 − 1 𝑟 ln 𝑘 𝑟2o (𝜎θ )res = 𝜎y [1 + ln ( ) − 2 (1 + 2 )] 𝑟o 𝑟 𝑘 −1 𝜏̂ res =
(5.22b)
𝜎y |(𝜎θ )res | − |(𝜎r )res | 2 ln 𝑘 𝑟2o [1 − 2 ( )] 2 2 𝑘 − 1 𝑟2
(5.23)
𝜎
(𝜎θ )res
(𝜎r )res
𝑟
Reverse Yielding: 𝜏̂ res = −𝜏̂ y −
𝜎y 2
𝜎y 𝜎y 2 ln 𝑘 𝑟o2 [1 − 2 ( 2 )] = − 2 𝑘 −1 𝑟 2 Max stresses occur at the bore: 1−
2 ln 𝑘 ln 𝑘 = −1 𝒌 = 𝟐. 𝟐𝟐 𝑘2 − 1
𝑘 = 2.22 is the limiting radius above which 100% overstrain (fully plastic condition) will give rise to reverse yield at the bore
59 100% autofrettage: the theoretical max residual stress left in the bore 50% autofrettage: 50% theoretical residual stress left in the bore
5.2.2 The “Shakedown” Condition -
reverse yielding is not desirable “shakedown” cylinder settling down to a safe working condition after initial plastic deformation Optimum (100%) autofrettage is achieved with 100% overstrain when 𝑘 ≤ 2.22 but if 𝑘 > 2.22 with less that 100% overstrain
Max allowable pressure (𝑃̂i ) for 𝑘 ≤ 2.22 𝜎y 𝜎y 𝑃̂i 𝑘 2 2 ln 𝑘 2 + − 𝑘 = [1 ] 𝑘2 − 1 2 𝑘2 − 1 2 Due to 𝑃̂i 𝑃̂i = 𝜎y ln 𝑘 𝑃c
(5.24)
𝑃̂i 𝑃eb 𝑃eb =
𝑘2 − 1 𝑃̂i 𝜎 ; for 𝑘 = 2.22 ≤2 2𝑘 2 y 𝑃eb
Load-carrying capacity increased by a factor of 2.
For 𝑘 > 2.22 𝜎y 𝜎y 𝑃̂i 𝑘 2 − = 𝑘2 − 1 2 2 𝑃̂i =
𝑘2 𝜎 2𝑃eb 𝑘2 − 1 y
Load-carrying capacity is increased by a factor of 2 for any 𝑘 > 2.22.
Example (a): Determine – for completely linear-elastic behaviour of a cylinder with radius ratio k=2.5 and 𝜎y = 400 MPa – the optimum autofrettage pressure and the resulting percentage overstain of the cylinder wall. 𝑃̂i = 2𝑃eb 2
𝑘 2 − 1 𝜎y 2.52 − 1 400 = 336 MPa 𝑘2 2 2.52
overstrain:
𝑛−1 × 100% 𝑘−1
60
(𝑘 2 − 𝑛2 ) 𝜎r = 𝜎y [ln ( ) − ] 𝑛𝑟i 2𝑘 2 𝑟
(5.18a)
@𝑟 = 𝑟i , −𝑃̂i = 𝜎r − 336 MPa (2.5 − 𝑛2 ) 1 −336 MPa = 400 [ln ( ) − ] 𝑛 2(2.5)2 (2.5 − 𝑛2 ) 1 336 ln ( ) − =− f(𝑛) 2 𝑛 2(2.5) 400 f(𝑛) 0.725 0.875 0.815
n 1.5 2 1.75 ∴ 𝑛 = 1.845 % overstrain:
1.845 − 1 × 100 % 56.3 % 2.5 − 1
Example (b): A thick-walled steel cylinder pressure vessel, optimally autofrettaged for linear-elastic behaviour in service, has inner and outer diameters of 110 mm and 160 mm, respectively. What are the radial and hoop stresses at the inner and outer radii of the cylinder when it is subjected to an internal pressure of 200 MPa. 𝜎y = 660 MPa; assume elastic-perfectly plastic 𝑘=
𝑟o 55 = 1.4545 ≤ 2.22 𝑟i 80
𝑃̂i = 𝑃c 𝜎y ln 𝑘 = 600 ln(1.4545) 247.3 MPa 𝑃̂i : optimal autofrettage pressure 𝑟 ln 𝑘 𝑟o2 (𝜎r )res = 𝜎y [ln ( ) − 2 (1 − 2 )] 𝑟o 𝑘 −1 𝑟 𝑟 ln 𝑘 𝑟o2 (𝜎θ )res = 𝜎y [1 + ln ( ) − 2 (1 + 2 )] 𝑟o 𝑘 −1 𝑟 @ 𝑟i and 𝑟o (𝜎r )res = 0 (𝜎θ )ri = 𝜎y [1 −
(2𝑘 2 ) ln 𝑘] − 278 MPa 𝑘2 − 1
𝑟 ln 𝑘 𝑟o2 (𝜎θ )ro = 𝜎y [1 + ln ( ) − 2 (1 + 2 )] 600(0.3283) = +216.7 MPa 𝑟o 𝑘 −1 𝑟
61 Elastic stresss due to 𝑃i = 200 MPa Lamé plot 1 = 330.5 𝑟i2 1 = 156.25 𝑟o2
𝜎θ
positive
𝜎r
𝑏 1 𝑟o2
𝑎 1 𝑟o2
1 𝑟i2 𝑏 (
1 1 2 + 2) 𝑟i 𝑟o
1 𝑟i2 𝑃i
negative =
𝑎 𝑃i 1 1 1 2 2 2− 2 𝑟o 𝑟i 𝑟o
(𝜎θ )𝑟o = 𝑎 858.5 MPa (𝜎θ )𝑟i = 𝑏 558.5 MPa Total stress in the autofrettage cylinder 𝑃i = 200 MPa
@ 𝑟 = 𝑟i 𝜎r = −200 MPa 𝜎θ = 558.5 − 278 280.5 MPa
@ 𝑟 = 𝑟o 𝜎r = 0 MPa (𝜎θ )𝑟o = (𝜎θ )𝑟o + [(𝜎θ )𝑟o ]
res
358.5 + 216.7
(𝜎θ )𝑟o = 575.2 MPa
5.2.3 Using von Mises 𝜎y
, instead of 𝜏y =
𝜎y
-
If von Mises criteria is used, 𝜏y =
-
All the expressions developed for Tresca can be used when 𝜎y is replaced by
√3
2
in Tresca. 2 𝜎. √3 y
