Institute of Mechanics and Fluid Dynamics (IMFD) Chair of Applied Mechanics - Solid Mechanics Prof. Dipl.-Ing. Björn Kiefer, Ph.D. Prof. Dr. rer. nat. habil. Meinhard Kuna, i.R.

Exercises Mechanics of Materials

Winter term 2016/2017

Freiberg, October 19, 2016

The present exercise book is published as supplementary material for the lecture "Mechanics of Materials" taught at the Institute of Mechanics and Fluid Dynamics at TU Bergakademie Freiberg. All rights of this publication are reserved. Duplication and distribution are only allowed with the prior written permission of the publisher. Responsible Professors: Editors:

Prof. Dipl.-Ing. Björn Kiefer, Ph.D., Prof. Dr. rer. nat. habil. Meinhard Kuna, i.R. Dr. Matthias Scherzer, Dr. Thomas Linse, Dr. Geralf Hütter, Dr. Stephan Roth, Marcel Selent

1. edition entitled "Aufgabensammlung Werkstoffmechanik" 2. revised and extended edition September 2012 3. english edition September 2012 4. edition with new corporate design September 2013 5. revised edition October 2016

Contents

0 Repetition Mathematics and Mechanics

4

1 Basic Rheological Models

7

2 Rheological Models – Elastic-Plastic

10

3 Rheological Models – Relaxation and Creep Functions

12

4 Multiaxial Stress and Deformation States

19

5 Elastic Material Laws

22

6 Elastic-Plastic Material Laws

26

7 Failure Criteria

30

3

0 Repetition Mathematics and Mechanics 1. Metal sheet with inclined weld seam given: σ0 , α = 60◦ , see Figure 0.1 find: σα , τα τα σα

Figure 0.1 2. Strain determination with strain gauges given: measured strains see Figure 0.2 εα1 = 0, 6 · 10−3 , εα2 = 0, 75 · 10−3 , εα3 = −0, 4 · 10−3 variant A: α2 = 60◦ , α3 = 120◦ variant B: α2 = 45◦ , α3 = 90◦ E = 2, 0 · 105 MPa, ν = 0, 3 find: a) εxx , εyy , γxy b) σxx , σyy , τxy c) principal strains d) principal stresses (value, direction) y

α1 x

Figure 0.2

4

3. First order ordinary differential equation + a v(t) = bet given: dv(t) dt find: v(t) method: variation of constants 4. Statically determinable framework given: F , EA, H, α, see Figure 0.3 find: rod forces S1 , S2 , ∆H (assumption: ∆H/H  1)

1

2 H

α

ΔH F

Figure 0.3 5. Statically indeterminable framework given: F , EA, H, α, see Figure 0.4, yield stress σF , (assumption: ∆H/H  1) find: a) rod forces S1 , S2 , S3 method: kinematic constraint at node A b) At which force F does rod 3 begin to plastify? 3

1

2

α

A

H

ΔH F

Figure 0.4

5

0 Repetition Mathematics and Mechanics 6. Beam with distributed loads given: E, l, b, h, q(x) = ax, see Figure 0.5, yield stress σF find: location where plastification initiates l

Figure 0.5

6

1 Basic Rheological Models 1. Develop a one dimensional ε-σ-material law ( ε-total strain, σ-total stress) for the sketched rheological models with stiffness E, E1 , E2 and E3 as well as viscosities η and η1 , see Figure 1.1.

Figure 1.1

7

1 Basic Rheological Models 2. A material model consists of the parallel connection of K ELVIN- and M AXWELLsolids (see Figure 1.2). Determine the one-dimensional constitutive equation (ε-σ) for such a model.

Figure 1.2 3. A bar with constant cross sectional area A and length l is fixed at both ends, see Figure 1.3. A force F is applied at point C. Determine the reaction forces FA and FB assuming one dimensional linear elastic material behavior (σ = E ε).

Figure 1.3 4. A composite structure of isotropic viscous (ηi ) and isotropic elastic (Ei ) layers under stress load is given, see Figure 1.4. Determine the effective σ-ε material law. Note that all rectangular cuboid layers have the same geometrical dimensions, i.e. base areas (A) and thicknesses (h) are identical for all layers. All layers have a unit dimension in z-direction of 1.

z ⊗

x

y h

Figure 1.4

8

5. A composite structure is given with modified arrangement compared to Exercise 4, see Figure 1.5. Determine the effective σ-ε material law for this modified structure. All layers have the same thicknesses (h) and a unit dimension in z-direction of 1. h z ⊗

x

y

Figure 1.5

6. Determine the effective σ-ε material law for the composite structure shown in Figure 1.6. The layers are made of isotropic viscous (η) and isotropic elastic (E) material. Note the different thicknesses h1 and h2 (cf. Exercises 4 and 5). All layers have a unit dimension in z-direction of 1.

z ⊗

x

y

Figure 1.6

7. A composite structure with modified arrangement compared to Exercise 6 is given, see Figure 1.7. Determine the effective σ-ε material law for this modified structure. All layers have a unit dimension in z-direction of 1.

z ⊗

x

y

Figure 1.7

9

2 Rheological Models – Elastic-Plastic 1. Measured data for a piecewise linear hardening plastic material are given: ε [%] σ [MPa]

0 0

0.2 500

0.5 600

1.0 620

Determine the material parameters E, E1 and E2 according to the measured data for the rheological model sketched in Figure 2.1 (σF1 = 500 MPa, σF2 = 600 MPa).

σ

σ

ε Figure 2.1

2. A beam, made of an elastic-ideal plastic material with material parameters E and σF , is subjected to simple bending. a) Which bending moment Mb has to be applied in order to plastify the upper and lower quarter of the cross section in case of a rectangular cross sectional area (h · b)? b) What is the value of the ultimate bending moment (the bending moment for a totally plastified cross section) in the case of a circular cross sectional area?

10

3. Determine the forces in the three bars of the sketched truss in Figure 2.2 that is loaded by the force F . The bars have identical cross sectional areas. The deformation behavior is described by σ = C εn . Assume small deformations ∆β  β.

Figure 2.2

4. Solve the problem of Exercise 3 for elastic-ideal plastic material behavior. Use as material parameters: σF - yield stress for tension E - YOUNG’s modulus Assume small deformations ∆β  β. a) Determine the state at which the middle bar begins to yield. b) Determine the condition at which the truss loses its ultimate load carrying capacity. c) Determine the plastic deformation between the states a) and b) as function of F.

11

3 Rheological Models – Relaxation and Creep Functions

1. A straight bar with a constant cross sectional area A made of viscoelastic material (standard linear solid: σ + T σ˙ = L ε + M T ε) ˙ is given, see Figure 3.1. The bar is fixed at the left end. At the right end (x = l) the constant force P is applied at time t = 0. a) Determine the elongation ∆l of the bar as function of time t. Solve the problem by using the ε-form of the material law (cf. Exercise 9 b). b) How does the solution for the elongation ∆l change if a second constant force F acts at point B at time t = 0?

Figure 3.1

2. The material is a viscoelastic standard linear solid (σ + T σ˙ = L ε + M T ε). ˙ The temporal course of applied stress is given in the diagrams in Figure 3.2. Determine the strain ε over time for the cases a) to d) and L add the diagrams for the strain into the σ, t-diagrams. (T = 4s , M = 14 ).

12

Figure 3.2

3. Develop the ε-σ-relation for the composite rheological model sketched in Figure 3.3. E1

E2

E0

σ

σ

η2

η1

Figure 3.3

4. The model of Exercise 3 is stretched by a constant deformation rate ε˙ = ε0 /t1 , see Figure 3.4. Determine the stress response σ(t).

Figure 3.4

13

3 Rheological Models – Relaxation and Creep Functions 5. The rheological model given in Figure 3.5 is characterised by the material properties Young’s modulus E, yield stress σF0 and viscosity η. Find the evolution of stress σ(t), when the evolution of the strain ε(t) is given as ( 0 für t < 0 σF0 = const. und H(t) = ε(t) = ε0 H(t) with ε0 = α . E 1 für t ≥ 0 In order to obtain the exact solution proceed as follows: a) Compile the constitutive equations consisting of elastic material law, yield condition and flow rule. b) Discuss the model’s response in principle for the cases 0 < α < 1 and α > 1. c) Write down the stress-strain relations σ ↔ ε for both cases. (Hint: In order to solve the differential equation occurring in the case α > 1, use the variation of constants method.) d) Find the asymptotic response of the model (t → ∞) in the case α > 1.

σF

0

σ

σ

E

ε Figure 3.5

14

η

6. Determine for the rheological model shown in Figure 3.6 a) the ε-σ relation b) the relaxation function R(t) of stresses for  ε = ε0 H(t) ,

ε0 = const. ,

H(t) =

0 for t < 0 . 1 for t ≥ 0

Note: R(t) is defined by σ(t) = ε0 R(t) at a given strain jump.

ε Figure 3.6

7. Show for the linear standard solid model σ + T σ˙ = L ε + M T ε˙ that the creep function J(t) is related to the corresponding relaxation function R(t) by: R(0) J(0) = R(∞) J(∞) = 1 Zt J(τ ) R(t − τ ) dτ = t. 0

Note: J(t) is defined by means of a creep test at constant stress σ0 , where ε(t) = σ0 J(t). 8. The bar from Exercise 3 is now loaded with a) a constant force F b) a linear variable force F = p · t for a time period t ≥ 0. Determine the reaction forces FA and FB as function of time t assuming the constitutive law σ + T σ˙ = L ε + M T ε˙ for a standard linear solid.

15

3 Rheological Models – Relaxation and Creep Functions 9. The standard linear solid model of viscoelasticity in form of σ + T σ˙ = L ε + M T ε˙ for the stress σ and the strain ε is given. T , L and M are material constants. a) Determine the solution of the given equation for σ. b) Determine the solution of the given equation for ε. 10. The viscoelastic rod (with P = 0) given in Exercise 1 is loaded at the point B with the force F = p · t (linear increase over time with slope p). Calculate the resulting reaction force at the rod end, if it is fixed for all time. 11. A bar with constant cross sectional area A and length l is placed between two rigid walls. The left end of the bar is fixed at the left wall whereas a gap δ  l exists between the right end of the bar and the right wall. A force F is applied at point C for time t ≥ 0, see Figure 3.7. Determine the reaction force FB as function of time t at the right wall assuming the standard linear solid constitutive law σ + T σ˙ = L ε + M T ε, ˙ when the right end of the rod has reached the wall at point B.

Figure 3.7

16

12. The sketched two bar truss in Figure 3.8 is loaded with a constant force F at time t = 0. Both bars have the same cross sectional area A and behave according to the standard linear solid: σ + T σ˙ = L ε + M T ε. ˙ Determine the displacement of the force application point over time t. Solve the problem by using the ε-form of the material law (cf. Exercise 9 b).

Figure 3.8

13. A beam (E ULER-B ERNOULLI beam theory) made of a viscoelastic material (standard linear solid model) is loaded under simple bending. Determine the relation between the bending moment and the curvature of the deformed axis of the beam. 14. Determine the bending behavior of the beam shown in in Figure 3.9 made of a viscoelastic material (standard linear solid). The forces F1 and F2 are applied according to the depicted time behavior.

Figure 3.9

17

3 Rheological Models – Relaxation and Creep Functions 15. Determine the bending stress σ11 and the deflection w for the beam of length 2l made of a viscoelastic material according to the M AXWELL modell sketched in Figure 3.10. The beam is subjected to a permanent constant line load q applied at time t = 0.

Figure 3.10

16. Solve the problem of Exercise 15 for a linear variable line load q(t) = q0 · t.

18

4 Multiaxial Stress and Deformation States 1. A stress state is characterized by vanishing of all normal stresses σxx = σyy = σzz = 0. The shear stress component τxy vanishes, too (τxy = 0). Which stress state does it correspond to? Determine the principal stresses. 2. Determine the principal stresses for the stress state σxx = σyy = τxy = 0 ,

σzz = 2000 MPa ,

τxz = τyz = 1000 MPa.

σyy = 0 MPa , τyz = −100 MPa ,

σzz = 1100 MPa , τxz = −800 MPa .

3. A stress tensor is given by σxx = 500 MPa , τxy = 300 MPa ,

Determine the normal and shear stresses as well as the magnitude of the traction vector for a plane which is equally inclined with respect to all three Cartesian coordinate axes, see Figure 4.1.

Figure 4.1

19

4 Multiaxial Stress and Deformation States 4. A stress tensor is given by the Cartesian components σxx = 500 MPa , τxy = 500 MPa ,

σyy = 0 MPa , τyz = 700 MPa ,

σzz = −300 MPa , τxz = 800 MPa .

Determine the normal and shear stresses as well as the magnitude of the traction vector for a plane that is characterised by the direction cosines of its normal vector ~n, see Figure 4.2   cos α = 12 (regarding the x-axis) ~n =  cos β = 12 (regarding the y-axis)  . cos γ = √12 (regarding the z-axis)

Figure 4.2

5. A stress tensor is given by 

 10 0 15 [σij ] =  0 20 −15  σ0 15 −15 0 Determine the hydrostatic stress, the deviator of [σij ], and the second invariant of the deviator. 6. For a stress tensor the first invariant (I1σ are known  0 D  0 [σij ] = −15

= 30 MPa) and the deviatoric stress tensor  0 −15 10 15  MPa . 15 −10

Deduce the formula for computing the principal stresses.

20

7. Determine the principal stresses and the principal directions for a stress tensor defined by the components σxx = 1000 MPa , τxy = 400 MPa ,

σyy = 500 MPa , τyz = −200 MPa ,

σzz = −100 MPa , τxz = 300 MPa .

8. A plane strain state is defined by: εxx 6= 0 ,

εxy 6= 0 ,

εyy 6= 0 ,

εzz = 0 ,

εxz = 0 ,

εyz = 0 .

Determine the invariants of the strain tensor and the principal strains. 9. Determine the principal strains and principal directions for the strain tensor characterized by the components εxx = 0.001 , εxy = 0.0001 ,

εyy = 0.0005 , εyz = −0.0001 ,

21

εzz = −0.0001 , εxz = 0.00015 .

5 Elastic Material Laws 1. Decompose the strain and the stress tensor into the hydrostatic and the deviatoric part. Insert these formulas into the strain energy density function (elastic potential) U = 12 σij εij . Split U into the volumetric strain energy density and the deviatoric strain energy density (distorsion energy). 2. Characterize the stress state given by σij = −p δij ? Determine the strains for this stress state assuming isotropic linear-elastic material behavior. 3. Decompose the isotropic linear-elastic law into a relationship between the deviatoric strain and stress tensors as well as a relationship between the first invariants of the strain and stress tensor. Why is the relationship between the deviatoric tensors alone insufficient to describe isotropic linear elastic materials completely? 4. Determine the stresses, the strains, and the displacements in a bar under tension, see Figure 5.1. a) Specify the stress tensor. b) To determine the strains use the isotropic generalised H OOKE’s law in 3Dform

ν 1+ν σij − I1σ δij εij = E E

 with

δij =

1 for i = j . 0 for i 6= j

c) Integrate the strains εij = 21 (ui,j + uj,i ) to determine the displacements ui . Use the following boundary conditions: ux (x = 0) = 0, uy (y = 0) = 0, uz (z = 0) = 0).

22

Figure 5.1

5. An isotropic linear-elastic solid is defined by the material law  1 for i = j ε σij = λ I1 δij + 2 µ εij with δij = . 0 for i = 6 j Express the strain energy density U in terms of a) the components εij , b) the components σij , c) the invariants of εij . 6. The isotropic H OOKE’s law is given by σij = λ

I1ε

 δij + 2 µ εij

with

δij =

1 for i = j . 0 for i 6= j

For the case of uniaxial tension the following presumptions are valid: σxx 6= 0 ,

τxy = τxz = τyz = σzz = σyy = 0.

Compute the components of the strain tensor. 7. For the case of uniaxial tension the YOUNG’s modulus E is defined by σxx = Eεxx . a) Write the YOUNG’s modulus E as function of λ and µ (c.f. Exercise 6). b) For the case of uniaxial tension the P OISSON’s ratio ν is defined by εyy = −ν εxx . Write ν as function of λ and µ (cf. Exercise 6).

23

5 Elastic Material Laws 8. H OOKE’s law is given by σij = λ

I1ε

 δij + 2 µ εij

with

δij =

1 for i = j . 0 for i 6= j

The load case “hydrostatic pressure” is defined as p σxx = σyy = σzz = − , 3

τxy = τxz = τyz = 0.

a) characterize the strain state, b) determine the invariants of the stress and strain tensor, c) deduce the bulk modulus K as a function of λ and µ from the relation I1σ = 3 K I1ε 9. The isotropic H OOKE’s law is given by σij = λ

I1ε

 δij + 2 µ εij

with

δij =

1 for i = j . 0 for i 6= j

A pure shear state is characterised by σxx = σyy = σzz = τxz = τyz = 0,

τxy = τ 6= 0.

Determine a) the components of the strain tensor, b) the principal stresses and principal directions, c) the principal strains and principal directions. 10. Write the three-dimensional equations for linear-orthotropic material beahavior in terms of engineering constants (elastic moduli, P OISSON ratios) in a coordinate system aligned with the axes of orthotropy. a) What are the relations between the off-diagonal elements of the stiffness matrix and the compliance matrix? b) Reduce the law to the plane strain and plane stress condition in the (x1 , x2 )plane. c) What basic differences and what similarities exist between the plane strain and the plane stress state? 11. Formulate the rule for coordinate transformation of the stress tensor in the case of two Cartesian coordinate systems, which are rotated against each other around the x3 -axis by an angle ϕ. Restrict to plane stress conditions in the (x1 , x2 )-plane.

24

12. Use the transformation rule from Exercise 11. Formulate the material law of exercise Exercise 10 for the plane stress state in a coordinate system which is rotated counterclockwise by an angle of ϕ = π2 with respect to the axes of orthotropy. a) Analyse the changes in the elasticity matrix. b) What difference exists an isotropic material? 13. Solve the problem of Exercise 12 for an angle of ϕ = eters E1 = 2 · 105 MPa,

E2 = 2 · 104 MPa,

using the material param-

G12 = 5 · 105 MPa,

What is the difference to Exercise 12 in principle?

25

π 4

ν21 = 0.3.

6 Elastic-Plastic Material Laws 1. The yield criterion is given by 1 q 2 + τ 2 + τ 2 )−σ = 0 Φ(σij ) = √ (σxx − σyy )2 + (σzz − σxx )2 + (σyy − σzz )2 + 6(τxy F xz yz 2 with σF - yield stress under tension. a) Adapt the yield criterion to the plane stress state. b) Adapt the yield criterion to the plane strain state assuming that P OISSON’s ratio is ν = 21 . 2. A stress state is given by the components σxx = 700 MPa , τxy = −100 MPa ,

σyy = 50 MPa , τyz = 200 MPa ,

σzz = −300 MPa , τxz = 600 MPa .

a) Compute the factor f , by which the given stress tensor must be multiplied, so that the yield criterion of Exercise 1 is fulfilled with σF = 2000 MPa. b) Compute the plastic strain rates (related to the plastic multiplier) that follow from the P RANTDL -R EUSS law for active plastic yielding. 3. Write the v. M ISES yield criterion expressed in terms of the principal stresses.

26

4. The yield stresses for a material are given • at uniaxial tension σFZ and • at uniaxial compression σFD . Compute the shear stress τ for a pure plane shear stress state at which the yield criteria after a) M OHR-C OULOMB: σI −

σFZ σIII = σFZ σFD

and b) D RUCKER-P RAGER: √ √ √ √ I1σ (σFZ + σFD ) σV = σFZ σFD + σFD σFZ − 3( σFD − σFZ ) 3 2



2

(σV - v. M ISES-stress) are fulfilled. 5. Assume the P RANDTL-R EUSS yield law. Compute the ratios between the plastic strain rates a) for uniaxial tension with σxx = σF , σF , σyy = b) for a 2-dimensional stress state where σxx = − √ 3 τxy = τxz = τyz = 0, σF . c) for pure shear with τxy = √ 3

σF √ 3

and σzz =

6. From the results of a tensile test the material behaviour is to be characterized using the R AMBERG-O SGOOD law σ  σ n ε= + , (σ-stress, ε-strain). E C a) Fit the constants n and C using the measured data (ε = 1 % ⇔ σ = 880 MPa) and (ε = 2.5 % ⇔ σ = 880 MPa) with the knowledge of the YOUNG’s modulus E=200 GPa. b) Determine the initial yield stress Rp0.2 based on 0.2% off-set strain from the R AMBERG-O SGOOD law with the parameters E, C and n.

27

6 Elastic-Plastic Material Laws 7. Fit the stress-strain curve to the results of a tensile test according to σ = C εn . Determine the material parameters C und n from the parameters ω1 , ω2 , σ0 and ε0 shown in in Figure 6.1. ω1 - area above the curve (complementary work from σ acting through ε) ω2 - area below the curve (work from σ acting through ε)

Figure 6.1

8. The stress-strain curve of Exercise 7 is now fitted by σ = C εm (1 − β ε). Use the condition dσ = 0 as additional assumption. Determine C, m and β dε ε=ε0 from ω1 , ω2 , ε0 and σ0 . 9. Develop the yield condition and the evolution equations for the internal state variables for combined isotropic-kinematic hardening of an elastic-plastic material under uniaxial loading (σxx = σ, σij = 0 otherwise). Start from the threedimensional formulation. Derive the special cases of pure isotropic and pure kinematic hardening, respectively.

28

10. The evolution of the strain ε(t) is given as depicted in Figure 6.2 with ti = i · ∆t, i = 0 . . . 5. The material shows an elastic-plastic behavior with purely isotropic, linear hardening R(p) = Ep p. Young’s modulus amounts to E = 4σF0 and the hardening modulus is Ep = σF0 wherein σF0 is the initial yield stress. Find the evolution of the σ(t) under these conditions. a) Solve the task graphically in a σ-ε-diagram first. Now compute the exact solution. Proceed as follows: a) Compile the constitutive equations consisting of elastic material law, yield condition, flow rule and evolution equation of the accumulated plastic strain p. b) Eliminate the plastic multiplier λ˙ from this equations and simplify the flow rule. (Hint: d|x|/dx = sgn(x)). c) For each loading step i → i + 2 determine i. the initiation of the plastification (in the first step or after the unloading, respectively). ii. Integrate the evolution equation taking into account the corresponding inital conditions. With this result compute σ(t) in the loading step from the yield condition and the material law. What can be said about the influence of the strain rate ε˙ = εˆ/∆t? iii. Evaluate the plastic strain and the accumulated plastic strain at the turning point of the loading. These values form the initial conditions for the next step. d) Discuss what the stabilized stress-strain-curve σ ↔ ε looks like (i.e. between ti and ti+2 for i → ∞). Hint: Specify all stresses relative to σF0 : σ/σF0 = . . .

Figure 6.2

29

7 Failure Criteria 1. Estimate the strength of a material by applying the M OHR-C OULOMB criterion (linear limit curve) to the following principal stress state σII = −400 MPa,

σI = 150 MPa,

σIII = −600 MPa.

The critical stresses are: σcz = 300 MPa

σcd = 1000 MPa

(tension),

(compression).

2. A cube is loaded uniaxially and homogeneously by a compression stress. At a compression stress of 6300 MPa the cube failed on the octahedral plane. The normal vector of the octahedral plane is equally inclined to the three orthogonal side faces of the cube. Compute the normal and the shear stress as well as the magnitude of the traction vector at failure for the octahedral plane. 3. A plane stress state is given with σxx = 150 MPa,

σyy = −50 MPa,

τxy = 75 MPa.

Compute the factor f by which the shear stress τxy must be multiplied so that the failure criterion of maximal shear stress is fulfilled with τc = 500 MPa. 4. For the stress state of Exercise 3 the stresses σxx and σyy are to be increased proportionally by a factor f until failure is reached at σc = 1000 MPa according to the normal stress criterion. Compute the factor f . 5. A stress state is given by the components of the stress tensor σxx = 500 MPa, τxy = 90 MPa,

σyy = −100 MPa, τxz = 0 MPa,

σzz = 0 MPa, τyz = 0 MPa.

It is also known that the considered solid fails according to the M OHR-C OULOMB criterion with a linear limit curve defined by τc = τc0 − σ tan ρ where ρ = 316π and τc0 = 1500 MPa. Compute the factor f for the given stress state, by which all stress components have to be multiplied so that the material fails.

30

6. The principal stresses are given as σI = 200 MPa,

σII = 100 MPa,

σIII = −50 MPa

The strength of the material is characterised by σcz = 245 MPa

σcd = 615 MPa

(tension),

(compression).

Discuss the failure behaviour of the material according to the M OHR-C OULOMB criterion (linear limit curve). 7. Evaluate the safety against failure for the given principal stress state σI = 1300 MPa,

σII = 1200 MPa,

σIII = 1050 MPa.

Use the M OHR-C OULOMB criterion with the limit curve r σ τ = τc0 1 − σcz given in the (τ σ)-plane of M OHR’s circle (τc0 = 270 MPa, σcz = 1500 MPa).

31