Last week, we: Defined stress and strain Defined elastic properties. Learned stress strain relations for uniaxial case

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1 Last week, we: Defined stress and strain Defined elastic properties. Learned stress strain relations for uniaxial case Gave multiaxial stress-strain relations This week, we will Solve examples Learn stress transformation and principal stresses Failure theories

2 Multiaxial Stress-Strain

3 Elastic Constants For an isotropic material two elastic constant are required and enough to define elastic behaviour Hydrostatic or mean stress Volumetric strain (D=DV/V) D = Bulk modulus

4 Example:A bar made of A-36 steel (E=200 GPa, n=0.32) has the dimensions shown in Fig If an axial force of P = 80 kn is applied to the bar, determine the change in its length and the change in the dimensions of its cross section after applying the load. What is the change in volume? The material behaves elastically. Copyright 2011 Pearson sakınca Education yoktur South Asia Pte Ltd

5 Solution The normal stress in the bar is z = A P = = Pa = 16 MPa Given that E = 200 GPa z z = E = = 10 mm/mm The axial elongation of the bar is therefore = m 120 m (Ans) = z = = z L z Copyright 2011 Pearson sakınca Education yoktur South Asia Pte Ltd

6 Solution The contraction strains in both the x and y directions are x = y = - v z = 6 6 (0.32)8010 = (Ans) The changes in the dimensions of the cross section are = L x = L y x y x y = = = 2.56m (Ans) = 1.28m (Ans) Copyright 2011 Pearson sakınca Education yoktur South Asia Pte Ltd

7 Solution To find the change in volume V = V 0 [1 + ε x + ε y + ε z] V Vo V = x y z V Vo V = ( ) = V Vo = ( ) m 6 = Copyright 2011 Pearson sakınca Education yoktur South Asia Pte Ltd

8 Example: Elastik modulusu E, Poisson oranı n, akma mukavemeti S y olan aluminyumdan yapılmış L x W x t (Uzunluk X Genişlik X Kalınlık) boyutlarında bir plaka uzunluk yönünde P yükünün etkisi altındadır, genişlik yönünde ise iki ucundan ankstredir (genişlik sabit). Bu koşullar altında uzunluğundaki değişmeyi ve kalınlığındaki azalmayı hesaplayınız. y P P x

9 Example: A circle of diameter d = 9 in. is scribed on an unstressed aluminum plate of thickness t = 3/4 in. Forces acting in the plane of the plate later cause normal stresses x = 12 ksi and z = 20 ksi. For E = 10x10 6 psi and n = 1/3, determine the change in: a) the length of diameter AB, b) the length of diameter CD, c) the thickness of the plate, and d) the volume of the plate.

10 SOLUTION: Apply the generalized Hooke s Law to find the three components of normal strain. x y z = E x 1 = psi = n = x E = n = x E n E y E y n E = n z E 12ksi 0 20ksi in./in. n z E y in./in. z E in./in. 1 3 Evaluate the deformation components. 3 B = d = in./in. 9in. A x C = d D = t t y z = = in./in. 9in. C in./in. 0.75in. t Find the change in volume D = x. y z = in /in DV = D V = in DV B A D = = 0.187in in. = = in. in.

11 AVERAGE NORMAL STRESS = P A

12 NORMAL AND SHEAR STRESS z = lim DA0 DF z DA zx zy = = lim DA0 lim DA0 DFx DA DF y DA

13 Stress Transformation Stress depends on orientation Force P has normal and shear components on the surface mm. As the orientation of the surface mm changes, so does the normal and shear stresses. Biaxial loading (Plane stress, z =0, xz =0, yz =0) Two arbitrary coordinate systems and an element subjected to arbitrary stresses Stresses on the element Forces on the element

14 Stress Transformation Simplifying da terms, we get Similarly, we can find :

15 Stress Transformation Remember Note: σ x + σ y = σ x + σ y Thus the sum of the normal stresses on two perpendicular plane is an invariant quantity and independent of orientation or angle θ.

16 Principle Stresses When there is no shear stresses acting on the planes (=0, the normal stresses acting on the planes are maximum. These planes are called the principal planes, and stresses normal to these planes are the principal stresses σ 1, σ 2 and σ 3 which in general do not coincide with the cartesian coordinate axes x, y, z. Directions of principal stresses are 1, 2 and 3. Triaxial Three principal stresses, σ 1, σ 2 and σ 3, where σ 1 > σ 2 > σ 3. Biaxial-plane stress condition Two principal stresses, σ 1 and σ 2 (σ 3 =0).

17 Principle Stresses Plane stress: To determine yield we will need maximum shear stress, max at Example:, then max =250 MPa Be aware: 1 > 2 > 3, therefore 400 >0 > -100 and

18 Mohr Circle A: x, - xy ) B: y, xy ) Normal stresses are plotted along the x axis, shear stresses along the y axis. A point on Mohr s circle gives the magnitude and direction of the normal and shear stresses on any plane in the physical element. A shear stress causing a clockwise rotation about any point in the physical element is plotted above the horizontal axis of the Mohr s circle. The stresses on the planes normal to the x and y axes are plotted as points A and B. The shear stress is zero at points D and E, representing the values of the principal stresses σ 1 and σ 2 respectively. The Angle between σ x and σ 1 on Mohr s circle is 2θ.

19 Mohr Circle Example:

20 Triaxial principle stresses where I 1, I 2, and I 3 are called stress invariants. The maximum principal shear stress, τ max is given by τ max =

21 Common Triaxial Cases (d ) Torsion (e) Hydrostatic compresion

22 Failure Criteria The Maximum Normal Stress Theory (Rankine) States that failure occurs whenever the largest principal stress at any point reaches a value equal to the strength of the specimen. 1 S yt or 3 -S yc => Failure where S yt and -S yc are the tensile and compressive yield strengths, respectively. This theory does not agree with experiments and fails to predict yield in metals. For example, metals do not yield under hydrostatic tension or compression. This theory can be used to predict failure of brittle materials. 1 S ut or 3 -S uc => Failure

23 Failure Criteria The Maximum Shear Stress Theory (Treasca) States that yielding begins whenever the maximum shear stress becomes equal to the maximum shear stress in a tension test specimen of the same material when that specimen begins to yield. => Failure

24 Failure Criteria The Distortion Energy Theory (von Mises) Observation: Hydrostatic pressure does not cause yilding. Equate the energies that produces angular distortation in the simple tension test and the given stress state. => Failure

25 Comparison of Failure Criteria Failure envolope for plane stress case. Failure occurs when stress combinations fall outside the envolope for applicable theory. MNS Maximum shear stress theory gives more conservative results. For pure shear loading DE predicts y =0.577 S y MSS predicts y =0.5S y

26 Comparison of Failure Criteria Failure envolope for plane stress case. (a) Comparison of the Rankine, von Mises, and Tresca criteria. (b) Comparison of failure criteria with test + cast iron Failure occurs when stress combinations fall outside the envolope for applicable theory. Maximum shear stress theory gives more conservative results. For pure shear loading DE predicts y =0.577 GYTE S y öğrencileri tarafından çoğaltılmasında MSS predicts y =0.5S y Mechanical Behavior of Materials, Hosford

27 Example: A region on the surface of a 6061-T4 aluminum alloy component has strain gages attached, which indicate the following stresses: σ 11 = 70MPa, σ 22 = 120MPa, σ 12 = 60MPa. Determine the yielding for both the Tresca and von Mises criteria, given that σ 0 =150 MPa (the yield stress). Solution: We first have to establish the principal stresses. This is easily accomplished by a Mohr circle construction or by its analytical expression (the equation of a circle):

28 Solution (continued) According to Tresca, max =(160 0)/2=80 MPa. The value max =80 MPa exceeds the Tresca criterion (σ 0 /2=75 MPa) and the alloy would be unsafe. The von Mises criterion gives σ e = 147 MPa σ e < σ 0 = 150 MPa therefore, the material does not yield. Plainly, the Tresca criterion is more conservative than von Mises.

29 Example: Bir uçak seyahatiniz sırasında şans! eseri uçağı tasarlayan mühendislerden birinin yanına oturuyorsunuz. Mühendis size tasarım sırasında von Mises kriterini kullandıklarını söylüyor. Tresca kriterini kullanmış olmalarını tercih eder miydiniz? Açıklayın. Answer: We prefer Tresca criterian. Tresca allows lower stresses in design (means thicker parts for the same load) therefore reducing the risc of exceeding the strength of the material.

30 Example:

31 Answer: Lower surface Upper surface max =/2 = y =Sy/2 =500/2 = x =500 Both surfaces yield at the same time.

32 Design or Safety Factors Design uncertainties mean we do not push the limit. Factor of safety, N working = y N Often N is between 1.2 and 4 Example: Calculate a diameter, d, to ensure that yield does not occur in the 1045 carbon steel rod below. Use a factor of safety of , 000N d 2 / 4 working 5 = y N d = m = 6.7 cm 1045 plain carbon steel: y = 310 MPa TS = 565 MPa F = 220,000N d L o

33 Plastic Deformation