# CONSTITUTIVE RELATIONS FOR LINEAR ELASTIC SOLIDS

Save this PDF as:
Size: px
Start display at page:

## Transcription

1 Chapter 9 CONSTITUTIV RLATIONS FOR LINAR LASTIC SOLIDS Figure 9.1: Hooke memorial window, St. Helen s, Bishopsgate, City of London 211

2 212 CHAPTR 9. CONSTITUTIV RLATIONS FOR LINAR LASTIC SOLIDS 9.1 Mechanical Constitutive quations Recall that in Chapters 2, 3 and 8 we briefly introduced the concept of a constitutive equation, which generally relates kinetic variables to kinematic variables in the application of interest. With respect to the application of the analysis of mechanical deformations in solids, the kinetic variable is the stress tensor, σ, whereas the kinematic variables are the displacements u x, u y, u z, and the strain tensor, ε which includes derivatives (sometimes called gradients) of the displacements. Since it is generally observed that rigid body displacements do not induce stresses, the displacement field u x, u y, u z, will not enter into a mechanical constitutive equation. Thus, the constitutive equations will in general relate stress, σ, to strains, ε, and temperature T. In 1660, Robert Hooke observed that for a broad class of solid materials called linear elastic (or Hookean), this relationship may be described by a linear relationship. Hooke originally considered the test of a uniaxial body with a force (stress) applied only in one direction and measured the corresponding elongation (strain) to obtain: y x A=cross-sectional area yy = ε yy Figure 9.2: Stress-Strain Curve for Linear lastic Material Forageneral three-dimensional state of stress, there are 6 independent stresses and 6 independent strains; therefore, the linear relationship between stress and strain can be written in matrix form as: σ =[C] ε (9.1) where [C] isa6 6 matrix of elastic constants that must be determined from experiments. expanded form, these 6 equations become: In σ xx = C 11 ε xx + C 12 + C 13 ε zz + C 14 ε yz + C 15 ε zx + C 16 ε xy = C 21 ε xx + C 22 + C 23 ε zz + C 24 ε yz + C 25 ε zx + C 26 ε xy σ zz = C 31 ε xx + C 32 + C 33 ε zz + C 34 ε yz + C 35 ε zx + C 36 ε xy σ yz = C 41 ε xx + C 42 + C 43 ε zz + C 44 ε yz + C 45 ε zx + C 46 ε xy (9.2) σ zx = C 51 ε xx + C 52 + C 53 ε zz + C 54 ε yz + C 55 ε zx + C 56 ε xy σ xy = C 61 ε xx + C 62 + C 63 ε zz + C 64 ε yz + C 65 ε zx + C 66 ε xy It is interesting to note that Robert Hooke first proposed the above law publicly in an anagram at Hampton Court (1676) given by the group of letters: ceiiinosssttuv.

3 9.1. MCHANICAL CONSTITUTIV QUATIONS 213 In 1678 he explained the anagram to be Ut tensio sic vis, which is Latin meaning as the tension so the displacement or in nglish the force is proportional to the displacement. Students may recall that during this time period, science and scientific writing was criticized and hence Hooke thought it necessary to discretely disclose his scientific finding with an anagram. Note that in the previous chapters stress and strain were represented as (3 3) matrices. It is convenient, however, here to represent them as (6 1) column vectors since they have only 6 independent components (stress due to conservation of angular momentum and strain by its definition). We then write {σ} = σ xx σ zz σ yz σ zx σ xy, {ε} = By adopting this representation for σ and ε, their linear relationship (9.2) can be easily written in matrix form: ε xx ε zz ε yz ε zx ε xy σ xx σ zz σ yz σ zx σ xy = C 11 C 12 C 13 C 14 C 15 C 16 C 21 C 22 C 23 C 24 C 25 C 26 C 31 C 32 C 33 C 34 C 35 C 36 C 41 C 42 C 43 C 44 C 45 C 46 C 51 C 52 C 53 C 54 C 55 C 56 C 61 C 62 C 63 C 64 C 65 C 66 ε xx ε zz ε yz ε zx ε xy (9.3) If a material is homogeneous then the constants (C ij, i =1,..., 6 and j =1,..., 6) are all independent of x, y, z for any time, t. If a material is isotropic, then for a given material point, C ij are independent of the orientation of the coordinate system (i.e., the material properties are the same in all directions). If a material is orthotropic, then for a given material point, C ij can be defined in terms of properties in three orthogonal coordinate directions. If a material is anisotropic, then for a given material point, C ij are different for all orientations of the coordinate system. In order to determine the material constants in equation (9.3), consider a uniaxial tensile test using a test specimen of linear elastic isotropic material with cross-sectional area A and subjected to a uniaxilly applied load F in the axial (y) direction as shown below. The cross-section may be any shape but generally a rectangular or cylindrical shape is chosen. For a rectangular specimen, assume a width W and thickness t so that the cross-sectional area is A = Wt. Assume a small gauge length of L for which the axial deformation will be measured during the load application. During the uniaxial tensile test, weobserve that the gauge length changes from L to L and the gauge width decreases from W to W. We also observe a decrease (contraction) in the z dimension. We further observe no change in angular orientation of the vertical or horizontal elements and conclude that for uniaxial loading, no shear strains are produced. This leads us to postulate the following strain state: ε xx,,ε zz 0,ε xy,ε yz,ε xz =0. The axial stress and strain in the axial (y) direction are defined to be

4 214 CHAPTR 9. CONSTITUTIV RLATIONS FOR LINAR LASTIC SOLIDS F after deformation y x L L* z A=cross-sectional area t test specimen F A=Wt W* W Figure 9.3: xperimental Measurement of Axial and Transverse Deformation = F A = L L = (L L) L The strain the in the transverse (x) direction due to the axial load is ε xx = W W = (W W ) W If we plot axial stress vs. axial strain and transverse strain vs. axial strain, weobtain the following two plots: = F A ε xx = W W 1 = ε xx = ν = L L 1 ν Figure 9.4: xperimental Results for Axial Stress vs. Axial Strain & Transverse Strain vs. Axial Strain From these two plots, we can write = and ε xx = ν for the uniaxial tension test. Consequently, we may define the following two material constants from this single uniaxial test: = slope of the uniaxial vs. curve = a material constant called Young s modulus

5 9.1. MCHANICAL CONSTITUTIV QUATIONS 215 ν = εxx = negative ratio of the strain normal to the direction of loading over the strain in the loading direction = a material constant called Poisson s ratio If the transverse strain were measured in the z direction, we would find the same ratio for transverse to axial strain: ν = εzz. Combining these equations, we can write the two transverse strains entirely in terms of the axial stress : ε xx = ν = ν ( ) and εzz = ν = ν ( ). In order to obtain a complete description of three-dimensional constitutive behavior, consider a test where we apply normal tractions (stresses) in the x, y and z directions simultaneously and measure the strain only in the x direction. For a linear material response, we may use the principle of linear superposition and consider three separate cases as shown below: σ zz σ zz σ σ xx σ xx σ xx xx = + + σ zz σ zz Figure 9.5: xperimental Test with all Components of Normal Stresses Applied ε xx = normal strain in x direction due to σ xx + normal strain in x direction due to + normal strain in x direction due to σ zz = 1 σ xx ν ν σ zz or ε xx = 1 [σ xx ν( + σ zz )] (9.4) The stress in the x direction increases the strain in the x direction while the transverse stresses causes a contraction (decrease in ε xx ). Doing similar experiments in the y and z directions gives: = 1 [ ν(σ xx + σ zz )] (9.5) ε zz = 1 [σ zz ν(σ xx + )] xperiments with shear tractions will show that a shear stress σ xy in the x-y plane produces only shear strain ε xy in the x-y plane for a state of pure shear loading (i.e., no normal strain is observed so that the shear strain is uncoupled from the normal strain). 1 Thus, we obtain the following 1 Keep in mind that even for the case of pure shear, if one calculates shear stresses (or strains) at some angle θ from the x-axis (Mohr s circle), one may obtain non-zero normal stresses (or strains) for the off-axis planes.

6 216 CHAPTR 9. CONSTITUTIV RLATIONS FOR LINAR LASTIC SOLIDS experimental observations for the shear strains: ε xy = 1+ν σ xy ε xz = 1+ν σ xz (9.6) ε yz = 1+ν σ yz Combining equations (9.4), (9.5) and (9.6), Hooke s law for a linear elastic isotropic solid with a three-dimensional stress state becomes: Hooke s Law for a Linear lastic Isotropic Solid ε xx = 1 [σ xx ν( + σ zz )] = 1 [ ν(σ xx + σ zz )] ε zz = 1 [σ zz ν(σ xx + )] ε xy = 1+ν σ xy (9.7) ε xz = 1+ν σ xz ε yz = 1+ν σ yz where = Young s modulus and ν = Poisson s ratio. It should be noted that in materials that undergo permanent deformation, the above model is not accurate (such as metals beyond their yield point, or polymers that flow). A typical uniaxial stress-strain curve for a ductile metal is shown below: σ xx yield stress ultimate stress failure stress ε xx Figure 9.6: Typical Stress-Strain Curve for Ductile Metal An algebraic inversion of the strain-stress relationship (9.7) provides the following relationship of stress in terms of strain:

7 9.1. MCHANICAL CONSTITUTIV QUATIONS 217 {σ} = 1+ν 1 ν 1 2ν ν 1 2ν ν 1 2ν ν 1 2ν 1 ν 1 2ν ν 1 2ν 1 2ν ν 1 2ν 1 ν 1 2ν ν ε xx ε zz ε yz ε zx ε xy (9.8) or, Hooke s Law for a Linear lastic Isotropic Solid {σ} = σ xx σ zz σ xy σ xz σ yz = (1+ν)(1 2ν) [(1 ν)ε xx + ν + νε zz ] (1+ν)(1 2ν) [νε xx +(1 ν) νε zz ] (1+ν)(1 2ν) [νε xx + ν +(1 ν)ε zz ] 1+ν ε xy 1+ν ε xz 1+ν ε yz (9.9) where = Young s modulus and ν = Poisson s ratio. The term (1+ν) 2G defines a shear modulus, G, relating shear strain and shear stress (similar to Young s modulus,, for extensional strain). Thus, the shear modulus is given by: G = 2(1 + ν) (9.10) 2( 1+ν Note that the shear strain ε xy is related to engineering shear strain γ xy by γ xy = 2ε xy = )σ xy = σxy G so that σ xy = Gγ xy =2Gε xy. σ xy G = shear modulus σ xy = Gγ xy = G2ε xy G = 2(1+ν) γ xy Figure 9.7: xperimental Results for Shear Stress vs. ngineering Shear Strain Note that G is defined in terms of and ν and consequently G is not a new material property. Thus, for a homogeneous linear elastic isotropic solid, we conclude that only two material properties (Young s modulus,, and Poisson s ratio, ν) are required to completely define the three-dimensional constitutive behavior.

8 218 CHAPTR 9. CONSTITUTIV RLATIONS FOR LINAR LASTIC SOLIDS The stress-strain equations may also be written in terms of shear modulus to obtain: σ xx = = σ zz = σ yz = σ zx = σ xy = (1 + ν)(1 2ν) [(1 ν)ε xx + ν + νε zz ]= 2G 1 2ν [(1 ν) ε xx + ν + νε zz ] (1 + ν)(1 2ν) [νε xx +(1 ν) νε zz ]= 2G 1 2ν [νε xx +(1 ν) + νε zz ] (1 + ν)(1 2ν) [νε xx + ν +(1 ν)ε zz ]= 2G 1 2ν [νε xx + ν +(1 ν) ε zz ] 1+ν ε yz =2Gε yz (9.11) 1+ν ε zx =2Gε zx 1+ν ε xy =2Gε xy Side Note: Ifthe definition of ε xx and ε zz from ν = εxx = εzz is substituted into equation (9.9), we obtain for the uniaxial bar extension experiment described previously: σ xx = (1 + ν)(1 2ν) [(1 ν)( ν)+ν + ν( ν) ]=0 [ = ν 2 +(1 ν) ν 2 ] = εyy (1 + ν)(1 2ν) σ zz =0 σ xy = σ xz = σ yz =0 This result is consistent with all observations made regarding the nature of stress for the uniaxial test with an applied stress of. 9.2 Constitutive quations with Thermal Strain xperimentally, we observe for a linear isotropic metal that a temperature increase, T, produces a uniform expansion but no shear and the expansion is proportional to a material constant α (coefficient of thermal expansion). The additional strain due to heating is thus: ε xx = = ε zz = α T. Thus, the constitutive equation for a linear elastic isotropic solid (9.7) may be modified by the addition of the thermal strain to the normal strain components: ε xx = 1 [σ xx ν( + σ zz )] + α T = 1 [ ν(σ xx + σ zz )] + α T ε zz = 1 [σ zz ν(σ xx + )] + α T ε xy = ( 1+ν )σ xy (9.12) ε xz = ( 1+ν )σ xz ε yz = ( 1+ν )σ yz

9 9.2. CONSTITUTIV QUATIONS WITH THRMAL STRAIN 219 These equations can be inverted to obtain stress in terms of strain: σ xx = = σ zz = σ xy = σ xz = σ yz = (1 + ν)(1 2ν) [(1 ν)ε xx + ν + νε zz ] α T (1 2ν) (1 + ν)(1 2ν) [νε xx +(1 ν) + νε zz ] α T (1 2ν) (1 + ν)(1 2ν) [νε xx + ν +(1 ν)ε zz ] α T (1 2ν) 2(1 + ν) ε xy (9.13) 2(1 + ν) ε xz 2(1 + ν) ε yz In the above, T = T ( x, y, z ) and represents the increase in temperature from a reference temperature where the thermal strain is zero. It should be noted that the first term in the extensional strain terms above (the[] term) is due to elastic behavior of the material (i.e., it has Young s modulus in it). The second part is due to thermal strain. We can separate the total strain into elastic and thermal strains: ε total xx = ε elastic xx + ε thermal ε total yy = ε elastic yy + ε thermal (9.14) ε total zz = ε elastic zz + ε thermal The elastic (also called mechanical) and thermal terms are given by: ε elastic xx = 1 [σ xx ν( + σ zz )] ε elastic yy = 1 [ ν(σ xx + σ zz )] (9.15) ε elastic zz = 1 [σ zz ν(σ xx + )] ε thermal = α T The terms ε total xx, ε total yy, ε total zz represent the total strain as measured or observed, and are thus equal to their deformation gradient definitions, i.e., for small strain, ε total xx = ε xx = u x x ε total yy = = u y y ε total zz = ε zz = u z z (9.16) We state once again that shear strains have no thermal component for an isotropic material. xamples of problems involving thermal strain will be considered in Chapter 10. Some typical values of material properties for isotropic metals are provided in the table below. Note that the values of (Young s modulus) are typically in the million psi or GPa range for engineering materials, while the values of ν are between zero and 0.5 (0 < ν < 0.5). The yield strength represents the stress level at which the metal yields (becomes inelastic). For ductile metals,

10 220 CHAPTR 9. CONSTITUTIV RLATIONS FOR LINAR LASTIC SOLIDS the ultimate tensile strength is typically 10 to 50% higher than the yield strength. It should be noted that material properties for commonly used metals must satisfy specifications established by regulatory agencies. Values in Tables 9.1 and 9.2 that are not provided are unknown for purposes of presentation herein and they should not to be interpreted as zero. The reader may with to consult other sources for the omitted values. For non-metals, properties may vary significantly depending upon many variables (for example, wood has a Young s modulus varying from psi to psi depending upon the tree species and direction of wood grain; the modulus of concrete will depend on the concrete/aggregate ratio and curing process). xamples of non-metals commonly used are concrete (ultimate compressive strength of 5 ksi but zero tensile strength; with an elastic modulus in compression of psi) and Douglas fir (parallel to grain, ultimate compressive strength of 7 ksi; with an elastic modulus of psi). Material Density ( lb in 3 ) Young s Modulus (10 6 psi) Poisson s Ratio Yield Strength (ksi) Ultimate Tensile Strength (ksi) Coefficient of Thermal xpansion Tension Shear Aluminum 2024-T (200 F) 6061-T ( F) Steel Structural (A36) AISI ( F) 5Cr-Mo-V ( F) Copper G3-Heat Treated ( F) Titanium Ti-5Al-2.5Sn ( F) Ti-6M-4V ( F) Table 9.1: Structural Material Properties for Selected Metals (US Customary Units) ( 10 6 F ) Given: xample 9-1 [σ] = y Mz I zz 0 0 where M z and I zz are constants Required : (a) Verify that the stress tensor satisfies the Conservation of Linear Momentum. (b) Determine the components of the infinitesimal strain tensor, ε. (c) Determine the components of the displacement, u x, u y, and u z. (d) Describe the displacement and physical problem described by these equations. Use as reference the figure below.

11 9.2. CONSTITUTIV QUATIONS WITH THRMAL STRAIN 221 Material Density ( Mg m 3 ) Young s Modulus (GPa) Poisson s Ratio Yield Strength (MPa) Ultimate Tensile Strength (MPa) Tension Shear Aluminum 2024-T T Steel Structural A Stainless Copper Alloy Bronze C Titanium Ti-6Al-4V , Ti-6M-4V Table 9.2: Structural Material Properties for Selected Metals (SI Units) Coefficient of Thermal xpansion ( 10 6 C ) y y z h x t L Figure 9.8: (a) x-component of linear momentum: x σ xx x + σ xy y + σ xz z + ρg x =0 ( ) y Mz I zz x =0 - Stress tensor satisfies the Conservation of Linear Momentum (b) The strains are given by: ε xx = σ xx, = ε zz = ν σ xx, ε xy = ε xz = ε yz =0 ε xx = y M z I zz, = νym z I zz, ε zz = νym z I zz

12 222 CHAPTR 9. CONSTITUTIV RLATIONS FOR LINAR LASTIC SOLIDS (c) Integrate the displacement equations and apply boundary conditions: ε xx = u x x = y M z I zz u x = y M z = u y y = νym ν z u y = I zz x + C 1 I ) zz M z ( y 2 2 I zz + C 2 ε zz = u z z = νym z I zz u z = νym z I zz z + C 3 u x x=0 =0, C 1 =0 u y y=0 =0, C 2 =0 u z z=0 =0, C 3 =0 u x = yx M z I zz u y = νv M 2 y2 z I zz u z = ν yz M z I zz (d) The displacement in the x-direction is negative (shortening) when y is positive due to the negative u x term. If y is negative the displacement in the x-direction is positive (expanding). In the y-direction and z-direction the displacement is expanding when y is greater than zero and vice versa. Displacement in the y-direction changes with respect to y 2, and in the z-direction it changes with respect to y.

13 9.2. CONSTITUTIV QUATIONS WITH THRMAL STRAIN 223 Deep Thought Ut tensio sic vis! Ut tensio sic vis!! Ut tensio sic vis!!!

14 224 CHAPTR 9. CONSTITUTIV RLATIONS FOR LINAR LASTIC SOLIDS 9.3 Questions 9.1 Which conservation laws are especially useful for describing stresses and strains? How are stress and strain related? 9.2 Write the equations that result from an inversion of the stress-strain relationship. 9.3 Describe in your own words the meanings of the state of plane stress and the state of plane strain? 9.4 Describe the two types of problems which when solved using the theory of plane elasticity provide exact solutions. 9.5 Consider small shear strain for a moment. It is often given in terms of an angle. xplain why this is done. 9.6 What is a constitutive relation? Write down the general constitutive relation in terms of Cauchy stress and strain. 9.7 For an elastic, isotropic solid material, how many constants are required to define the constitutive relations? Name these and define their meaning. 9.4 Problems 9.8 Structural steel is subjected to the deformation defined by u x ( x, y, z )=0.002x, u y ( x, y, z )= 0, u z ( x, y, z )=0(displacements in inches). Determine the following in US units: a) Infinitesimal strain tensor. b) Stress tensor. c) Draw Mohr s Circle for the given state of stress. d) Principal Stresses and Strains. 9.9 Repeat steps a and b in 9.8 for u x ( x, y, z ) = 0.002x x, u y ( x, y, z ) = 0.002xy, u z ( x, y, z )=0.001z GIVN :AHookean material with = psi and ν =0.5 experiences the following deformation: u x ( x, y, z )=0,u y ( x, y, z )=0.004x, u z ( x, y, z )=0 RQUIRD: a) Sketch u x versus x, u y versus y, and u z versus z, and calculate ux x, uy y, uz z. b) Calculate the infinitesimal strain tensor. c) Calculate the stress tensor GIVN : ν =0.25 and = Pa, and strain tensors as follows (1) , (2) RQUIRD: (1) Calculate the stress tensors; (2) How much is the relative volume change (the dilatation) for this deformation, and compare the results obtained by using both finite strain formula and the small strain formula.

15 9.4. PROBLMS GIVN : ν =0.33 and = MPa, and stress tensors as follows 10 MPa 4 MPa 0 20 MPa 50 MPa 0 (1) 4MPa 30MPa 0 (2) 50 MPa MPa RQUIRD: Calculate the strain tensors GIVN : u x = z10 4, u z = x10 4, and u y =0,and material constants = , and ν =0.3. (1) Compute infinitesimal strain tensor. (2) Compute the corresponding stress tensor. (3) What are the principal stresses and principal strains? (4) Are the principal stresses and strains acting in the same directions? 9.14 A steel plate lies flat in the x-y plane and has dimensions 20 cm 40 cm. If the plate is uniformly heated throughout at 1000 C and the thermal expansion coefficient is given by α = , calculate the new dimensions of the plate due to thermal expansion. m (m - C) 9.15 A thin rectangular sheet of linearly elastic material has an x-y coordinate system located at its lower left corner. The body extends 15 in. in the x direction and 8 in. in the y direction. The material is isotropic with an =35, 000, 000 psi and ν =0.33. A plane stress condition has been created by forces acting along the edges of the body with a displacement field of: u x = x 2 y u y = xy 2 Write expressions for the surface force normal to and for the tangential surface force along the upper 15 in. boundary as functions of x. Write expressions for the surface force normal to and for the tangential surface force along the right 8 in. boundary as functions of y. Draw the distribution of normal surface force along these two boundaries on a sketch of the body Use web resources to determine the following material properties. Provide the URL (http address) that you used. (a) Yield strength in shear of 2024-T4 and 2014-T6 aluminum. (b) Poisson s ratio and yield strength in shear for Ti-5Al-2.5Sn. (c) All of the table values as presented in Table 9.1 for 4130 heat treated alloy steel. (d) All of the table values as presented in Table 9.1 for balsa wood GIVN : The isothermal (no temperature gradient) uniaxial bar specimen of 2024-T4 Aluminum (isotropic) shown below: The axial displacements are measured to be: RQUIRD: u x = 0.02x in u y = x in x in inches!! 1. Sketch the deformed configuration of the test section boundary (using the displacements given above). 2. Calculate the infinitesimal strain tensor for the test section.

16 226 CHAPTR 9. CONSTITUTIV RLATIONS FOR LINAR LASTIC SOLIDS Problem Calculate the stress tensor for the test section GIVN :Alinear isotropic Thermolastic plate of Stainless 304 is subjected to a uniform temperature change of T and is assumed to be in a state of stress as shown below. At equilibrium, the T is known. σ = MPa RQUIRD: a) Calculate the infinitesimal strain tensor when T =0 C. (review equations 9.12 and 9.13 in the notes) b) Calculate the infinitesimal strain tensors for the two cases: when T = 100 C, and when T =25 C. c) Find the temperature change T necessary to produce zero strain GIVN : Consider the state of stress called plane stress in which non-zero stresses exist in only one plane. RQUIRD: a) For a state of plane stress in the x-y plane, show that the constitutive equations for an elastic isotropic material (isothermal case) reduce to the following. Hint: start with the constitutive equations for the general 3-D elastic, isotropic case and reduce to plane stress; see equation 10.6): SHOW ALL STPS. σ xx = = σ xy = (1 ν 2 ) [ε xx + ν ] (1 ν 2 ) [νε xx + ] (1 + ν) ε xy

17 9.4. PROBLMS 227 b) Starting with the above relations, show that the strains for plane stress in the x-y plane become those shown below (see equation 10.7). You must show all steps necessary to obtain the relations below. ε xx = 1 [σ xx ν ] = 1 [ νσ xx ] ε xy = ( 1+ν )σ xy ε zz = ν (σ xx + ) You may use Scientific Workplace to do the matrix algebra GIVN : Given that for a general orthotropic elastic material there are 12 unique coeffecients such that: [D] = 1 11 ν12 11 ν13 11 ν23 22 ν µ µ µ 12 ν ν31 33 The constitutive equation for this form would then be: where the stress have the following values RQUIRD: {σ} = σ xx =5ksi =10ksi σ zz =20ksi σ yz =0ksi σ zx =0ksi σ xy =7.5 ksi {ε} =[D] {σ} ; {ε} = a) Write the stress tensor in its more common form (i.e., as a tensor or matrix). Does this constitute generalized plane stress? Why or why not? ε xx ε zz ε yz ε zx ε xy Recall that generalized plane stress is a requirement for Mohr s Circle b) Suppose that the 12 material coefficients have the following values: 11 = 10 6 psi 22 = psi 33 = psi

18 228 CHAPTR 9. CONSTITUTIV RLATIONS FOR LINAR LASTIC SOLIDS ν 12 = 0.2 ν 13 = 0.25 ν 21 = 0.33 ν 23 = 0.43 ν 31 = 0.05 ν 32 = 0.06 µ 23 = 10 4 psi µ 31 = psi µ 12 = psi Calculate the infinitesimal strain tensor. c) Write the strain tensor in its more common form. Does this constitute generalized plane strain? Why or why not?

### Mechanical Properties of Materials

Mechanical Properties of Materials Strains Material Model Stresses Learning objectives Understand the qualitative and quantitative description of mechanical properties of materials. Learn the logic of

### 3.2 Hooke s law anisotropic elasticity Robert Hooke ( ) Most general relationship

3.2 Hooke s law anisotropic elasticity Robert Hooke (1635-1703) Most general relationship σ = C ε + C ε + C ε + C γ + C γ + C γ 11 12 yy 13 zz 14 xy 15 xz 16 yz σ = C ε + C ε + C ε + C γ + C γ + C γ yy

### MECHANICS OF MATERIALS

CHATR Stress MCHANICS OF MATRIALS and Strain Axial Loading Stress & Strain: Axial Loading Suitability of a structure or machine may depend on the deformations in the structure as well as the stresses induced

### 3D Elasticity Theory

3D lasticity Theory Many structural analysis problems are analysed using the theory of elasticity in which Hooke s law is used to enforce proportionality between stress and strain at any deformation level.

### [5] Stress and Strain

[5] Stress and Strain Page 1 of 34 [5] Stress and Strain [5.1] Internal Stress of Solids [5.2] Design of Simple Connections (will not be covered in class) [5.3] Deformation and Strain [5.4] Hooke s Law

### 6.4 A cylindrical specimen of a titanium alloy having an elastic modulus of 107 GPa ( psi) and

6.4 A cylindrical specimen of a titanium alloy having an elastic modulus of 107 GPa (15.5 10 6 psi) and an original diameter of 3.8 mm (0.15 in.) will experience only elastic deformation when a tensile

### Understand basic stress-strain response of engineering materials.

Module 3 Constitutive quations Learning Objectives Understand basic stress-strain response of engineering materials. Quantify the linear elastic stress-strain response in terms of tensorial quantities

### Samantha Ramirez, MSE. Stress. The intensity of the internal force acting on a specific plane (area) passing through a point. F 2

Samantha Ramirez, MSE Stress The intensity of the internal force acting on a specific plane (area) passing through a point. Δ ΔA Δ z Δ 1 2 ΔA Δ x Δ y ΔA is an infinitesimal size area with a uniform force

### INTRODUCTION TO STRAIN

SIMPLE STRAIN INTRODUCTION TO STRAIN In general terms, Strain is a geometric quantity that measures the deformation of a body. There are two types of strain: normal strain: characterizes dimensional changes,

### UNIVERSITY OF SASKATCHEWAN ME MECHANICS OF MATERIALS I FINAL EXAM DECEMBER 13, 2008 Professor A. Dolovich

UNIVERSITY OF SASKATCHEWAN ME 313.3 MECHANICS OF MATERIALS I FINAL EXAM DECEMBER 13, 2008 Professor A. Dolovich A CLOSED BOOK EXAMINATION TIME: 3 HOURS For Marker s Use Only LAST NAME (printed): FIRST

### Lecture 8. Stress Strain in Multi-dimension

Lecture 8. Stress Strain in Multi-dimension Module. General Field Equations General Field Equations [] Equilibrium Equations in Elastic bodies xx x y z yx zx f x 0, etc [2] Kinematics xx u x x,etc. [3]

### EMA 3702 Mechanics & Materials Science (Mechanics of Materials) Chapter 2 Stress & Strain - Axial Loading

MA 3702 Mechanics & Materials Science (Mechanics of Materials) Chapter 2 Stress & Strain - Axial Loading MA 3702 Mechanics & Materials Science Zhe Cheng (2018) 2 Stress & Strain - Axial Loading Statics

### Stress-Strain Behavior

Stress-Strain Behavior 6.3 A specimen of aluminum having a rectangular cross section 10 mm 1.7 mm (0.4 in. 0.5 in.) is pulled in tension with 35,500 N (8000 lb f ) force, producing only elastic deformation.

### MECHANICS OF MATERIALS

Third E CHAPTER 2 Stress MECHANICS OF MATERIALS Ferdinand P. Beer E. Russell Johnston, Jr. John T. DeWolf Lecture Notes: J. Walt Oler Texas Tech University and Strain Axial Loading Contents Stress & Strain:

### ME 243. Mechanics of Solids

ME 243 Mechanics of Solids Lecture 2: Stress and Strain Ahmad Shahedi Shakil Lecturer, Dept. of Mechanical Engg, BUET E-mail: sshakil@me.buet.ac.bd, shakil6791@gmail.com Website: teacher.buet.ac.bd/sshakil

### STANDARD SAMPLE. Reduced section " Diameter. Diameter. 2" Gauge length. Radius

MATERIAL PROPERTIES TENSILE MEASUREMENT F l l 0 A 0 F STANDARD SAMPLE Reduced section 2 " 1 4 0.505" Diameter 3 4 " Diameter 2" Gauge length 3 8 " Radius TYPICAL APPARATUS Load cell Extensometer Specimen

### Symmetric Bending of Beams

Symmetric Bending of Beams beam is any long structural member on which loads act perpendicular to the longitudinal axis. Learning objectives Understand the theory, its limitations and its applications

### NORMAL STRESS. The simplest form of stress is normal stress/direct stress, which is the stress perpendicular to the surface on which it acts.

NORMAL STRESS The simplest form of stress is normal stress/direct stress, which is the stress perpendicular to the surface on which it acts. σ = force/area = P/A where σ = the normal stress P = the centric

### MECHANICS OF MATERIALS

Third CHTR Stress MCHNICS OF MTRIS Ferdinand. Beer. Russell Johnston, Jr. John T. DeWolf ecture Notes: J. Walt Oler Texas Tech University and Strain xial oading Contents Stress & Strain: xial oading Normal

### Homework No. 1 MAE/CE 459/559 John A. Gilbert, Ph.D. Fall 2004

Homework No. 1 MAE/CE 459/559 John A. Gilbert, Ph.D. 1. A beam is loaded as shown. The dimensions of the cross section appear in the insert. the figure. Draw a complete free body diagram showing an equivalent

### MECE 3321 MECHANICS OF SOLIDS CHAPTER 3

MECE 3321 MECHANICS OF SOLIDS CHAPTER 3 Samantha Ramirez TENSION AND COMPRESSION TESTS Tension and compression tests are used primarily to determine the relationship between σ avg and ε avg in any material.

### THE GENERAL ELASTICITY PROBLEM IN SOLIDS

Chapter 10 TH GNRAL LASTICITY PROBLM IN SOLIDS In Chapters 3-5 and 8-9, we have developed equilibrium, kinematic and constitutive equations for a general three-dimensional elastic deformable solid bod.

### VYSOKÁ ŠKOLA BÁŇSKÁ TECHNICKÁ UNIVERZITA OSTRAVA

VYSOKÁ ŠKOLA BÁŇSKÁ TECHNICKÁ UNIVERZITA OSTRAVA FAKULTA METALURGIE A MATERIÁLOVÉHO INŽENÝRSTVÍ APPLIED MECHANICS Study Support Leo Václavek Ostrava 2015 Title:Applied Mechanics Code: Author: doc. Ing.

### A short review of continuum mechanics

A short review of continuum mechanics Professor Anette M. Karlsson, Department of Mechanical ngineering, UD September, 006 This is a short and arbitrary review of continuum mechanics. Most of this material

### MECH 5312 Solid Mechanics II. Dr. Calvin M. Stewart Department of Mechanical Engineering The University of Texas at El Paso

MECH 5312 Solid Mechanics II Dr. Calvin M. Stewart Department of Mechanical Engineering The University of Texas at El Paso Table of Contents Thermodynamics Derivation Hooke s Law: Anisotropic Elasticity

### Basic Equations of Elasticity

A Basic Equations of Elasticity A.1 STRESS The state of stress at any point in a loaded bo is defined completely in terms of the nine components of stress: σ xx,σ yy,σ zz,σ xy,σ yx,σ yz,σ zy,σ zx,andσ

### Mechanics of Earthquakes and Faulting

Mechanics of Earthquakes and Faulting www.geosc.psu.edu/courses/geosc508 Overview Milestones in continuum mechanics Concepts of modulus and stiffness. Stress-strain relations Elasticity Surface and body

### Constitutive Equations (Linear Elasticity)

Constitutive quations (Linear lasticity) quations that characterize the physical properties of the material of a system are called constitutive equations. It is possible to find the applied stresses knowing

### Module III - Macro-mechanics of Lamina. Lecture 23. Macro-Mechanics of Lamina

Module III - Macro-mechanics of Lamina Lecture 23 Macro-Mechanics of Lamina For better understanding of the macromechanics of lamina, the knowledge of the material properties in essential. Therefore, the

### Constitutive Equations

Constitutive quations David Roylance Department of Materials Science and ngineering Massachusetts Institute of Technology Cambridge, MA 0239 October 4, 2000 Introduction The modules on kinematics (Module

### MECHANICS OF MATERIALS

Third CHTR Stress MCHNICS OF MTRIS Ferdinand. Beer. Russell Johnston, Jr. John T. DeWolf ecture Notes: J. Walt Oler Texas Tech University and Strain xial oading Contents Stress & Strain: xial oading Normal

### Exercise: concepts from chapter 8

Reading: Fundamentals of Structural Geology, Ch 8 1) The following exercises explore elementary concepts associated with a linear elastic material that is isotropic and homogeneous with respect to elastic

### Solid Mechanics Chapter 1: Tension, Compression and Shear

Solid Mechanics Chapter 1: Tension, Compression and Shear Dr. Imran Latif Department of Civil and Environmental Engineering College of Engineering University of Nizwa (UoN) 1 Why do we study Mechanics

### Chapter 3. Load and Stress Analysis

Chapter 3 Load and Stress Analysis 2 Shear Force and Bending Moments in Beams Internal shear force V & bending moment M must ensure equilibrium Fig. 3 2 Sign Conventions for Bending and Shear Fig. 3 3

### Mechanics of Materials Primer

Mechanics of Materials rimer Notation: A = area (net = with holes, bearing = in contact, etc...) b = total width of material at a horizontal section d = diameter of a hole D = symbol for diameter E = modulus

### Fundamentals of Linear Elasticity

Fundamentals of Linear Elasticity Introductory Course on Multiphysics Modelling TOMASZ G. ZIELIŃSKI bluebox.ippt.pan.pl/ tzielins/ Institute of Fundamental Technological Research of the Polish Academy

### PDDC 1 st Semester Civil Engineering Department Assignments of Mechanics of Solids [ ] Introduction, Fundamentals of Statics

Page1 PDDC 1 st Semester Civil Engineering Department Assignments of Mechanics of Solids [2910601] Introduction, Fundamentals of Statics 1. Differentiate between Scalar and Vector quantity. Write S.I.

### ME 2570 MECHANICS OF MATERIALS

ME 2570 MECHANICS OF MATERIALS Chapter III. Mechanical Properties of Materials 1 Tension and Compression Test The strength of a material depends on its ability to sustain a load without undue deformation

### The University of Melbourne Engineering Mechanics

The University of Melbourne 436-291 Engineering Mechanics Tutorial Four Poisson s Ratio and Axial Loading Part A (Introductory) 1. (Problem 9-22 from Hibbeler - Statics and Mechanics of Materials) A short

### 4.MECHANICAL PROPERTIES OF MATERIALS

4.MECHANICAL PROPERTIES OF MATERIALS The diagram representing the relation between stress and strain in a given material is an important characteristic of the material. To obtain the stress-strain diagram

### Mechanics PhD Preliminary Spring 2017

Mechanics PhD Preliminary Spring 2017 1. (10 points) Consider a body Ω that is assembled by gluing together two separate bodies along a flat interface. The normal vector to the interface is given by n

### five Mechanics of Materials 1 ARCHITECTURAL STRUCTURES: FORM, BEHAVIOR, AND DESIGN DR. ANNE NICHOLS SUMMER 2017 lecture

ARCHITECTURAL STRUCTURES: FORM, BEHAVIOR, AND DESIGN DR. ANNE NICHOLS SUMMER 2017 lecture five mechanics www.carttalk.com of materials Mechanics of Materials 1 Mechanics of Materials MECHANICS MATERIALS

### 20. Rheology & Linear Elasticity

I Main Topics A Rheology: Macroscopic deformation behavior B Linear elasticity for homogeneous isotropic materials 10/29/18 GG303 1 Viscous (fluid) Behavior http://manoa.hawaii.edu/graduate/content/slide-lava

### Strength of Material. Shear Strain. Dr. Attaullah Shah

Strength of Material Shear Strain Dr. Attaullah Shah Shear Strain TRIAXIAL DEFORMATION Poisson's Ratio Relationship Between E, G, and ν BIAXIAL DEFORMATION Bulk Modulus of Elasticity or Modulus of Volume

### STRENGTH OF MATERIALS-I. Unit-1. Simple stresses and strains

STRENGTH OF MATERIALS-I Unit-1 Simple stresses and strains 1. What is the Principle of surveying 2. Define Magnetic, True & Arbitrary Meridians. 3. Mention different types of chains 4. Differentiate between

### 2. Rigid bar ABC supports a weight of W = 50 kn. Bar ABC is pinned at A and supported at B by rod (1). What is the axial force in rod (1)?

IDE 110 S08 Test 1 Name: 1. Determine the internal axial forces in segments (1), (2) and (3). (a) N 1 = kn (b) N 2 = kn (c) N 3 = kn 2. Rigid bar ABC supports a weight of W = 50 kn. Bar ABC is pinned at

### ELASTICITY (MDM 10203)

ELASTICITY () Lecture Module 3: Fundamental Stress and Strain University Tun Hussein Onn Malaysia Normal Stress inconstant stress distribution σ= dp da P = da A dimensional Area of σ and A σ A 3 dimensional

### STRESS, STRAIN AND DEFORMATION OF SOLIDS

VELAMMAL COLLEGE OF ENGINEERING AND TECHNOLOGY, MADURAI 625009 DEPARTMENT OF CIVIL ENGINEERING CE8301 STRENGTH OF MATERIALS I -------------------------------------------------------------------------------------------------------------------------------

### Finite Element Method in Geotechnical Engineering

Finite Element Method in Geotechnical Engineering Short Course on + Dynamics Boulder, Colorado January 5-8, 2004 Stein Sture Professor of Civil Engineering University of Colorado at Boulder Contents Steps

### N = Shear stress / Shear strain

UNIT - I 1. What is meant by factor of safety? [A/M-15] It is the ratio between ultimate stress to the working stress. Factor of safety = Ultimate stress Permissible stress 2. Define Resilience. [A/M-15]

### SEMM Mechanics PhD Preliminary Exam Spring Consider a two-dimensional rigid motion, whose displacement field is given by

SEMM Mechanics PhD Preliminary Exam Spring 2014 1. Consider a two-dimensional rigid motion, whose displacement field is given by u(x) = [cos(β)x 1 + sin(β)x 2 X 1 ]e 1 + [ sin(β)x 1 + cos(β)x 2 X 2 ]e

### Name (Print) ME Mechanics of Materials Exam # 1 Date: October 5, 2016 Time: 8:00 10:00 PM

Name (Print) (Last) (First) Instructions: ME 323 - Mechanics of Materials Exam # 1 Date: October 5, 2016 Time: 8:00 10:00 PM Circle your lecturer s name and your class meeting time. Gonzalez Krousgrill

### Stress/Strain. Outline. Lecture 1. Stress. Strain. Plane Stress and Plane Strain. Materials. ME EN 372 Andrew Ning

Stress/Strain Lecture 1 ME EN 372 Andrew Ning aning@byu.edu Outline Stress Strain Plane Stress and Plane Strain Materials otes and News [I had leftover time and so was also able to go through Section 3.1

### 3.22 Mechanical Properties of Materials Spring 2008

MIT OpenCourseWare http://ocw.mit.edu 3.22 Mechanical Properties of Materials Spring 2008 For information about citing these materials or our Terms of Use, visit: http://ocw.mit.edu/terms. Quiz #1 Example

Name: Date: Solid Mechanics Homework nswers Please show all of your work, including which equations you are using, and circle your final answer. Be sure to include the units in your answers. 1. The yield

### Mechanical properties 1 Elastic behaviour of materials

MME131: Lecture 13 Mechanical properties 1 Elastic behaviour of materials A. K. M. B. Rashid Professor, Department of MME BUET, Dhaka Today s Topics Deformation of material under the action of a mechanical

### Lab Exercise #5: Tension and Bending with Strain Gages

Lab Exercise #5: Tension and Bending with Strain Gages Pre-lab assignment: Yes No Goals: 1. To evaluate tension and bending stress models and Hooke s Law. a. σ = Mc/I and σ = P/A 2. To determine material

### 2. Mechanics of Materials: Strain. 3. Hookes's Law

Mechanics of Materials Course: WB3413, Dredging Processes 1 Fundamental Theory Required for Sand, Clay and Rock Cutting 1. Mechanics of Materials: Stress 1. Introduction 2. Plane Stress and Coordinate

### ME Final Exam. PROBLEM NO. 4 Part A (2 points max.) M (x) y. z (neutral axis) beam cross-sec+on. 20 kip ft. 0.2 ft. 10 ft. 0.1 ft.

ME 323 - Final Exam Name December 15, 2015 Instructor (circle) PROEM NO. 4 Part A (2 points max.) Krousgrill 11:30AM-12:20PM Ghosh 2:30-3:20PM Gonzalez 12:30-1:20PM Zhao 4:30-5:20PM M (x) y 20 kip ft 0.2

### Module 2 Stresses in machine elements

Module 2 Stresses in machine elements Lesson 3 Strain analysis Instructional Objectives At the end of this lesson, the student should learn Normal and shear strains. 3-D strain matri. Constitutive equation;

### Continuum Mechanics. Continuum Mechanics and Constitutive Equations

Continuum Mechanics Continuum Mechanics and Constitutive Equations Continuum mechanics pertains to the description of mechanical behavior of materials under the assumption that the material is a uniform

### PEAT SEISMOLOGY Lecture 2: Continuum mechanics

PEAT8002 - SEISMOLOGY Lecture 2: Continuum mechanics Nick Rawlinson Research School of Earth Sciences Australian National University Strain Strain is the formal description of the change in shape of a

### Rock Rheology GEOL 5700 Physics and Chemistry of the Solid Earth

Rock Rheology GEOL 5700 Physics and Chemistry of the Solid Earth References: Turcotte and Schubert, Geodynamics, Sections 2.1,-2.4, 2.7, 3.1-3.8, 6.1, 6.2, 6.8, 7.1-7.4. Jaeger and Cook, Fundamentals of

### Stress, Strain, Mohr s Circle

Stress, Strain, Mohr s Circle The fundamental quantities in solid mechanics are stresses and strains. In accordance with the continuum mechanics assumption, the molecular structure of materials is neglected

### The science of elasticity

The science of elasticity In 1676 Hooke realized that 1.Every kind of solid changes shape when a mechanical force acts on it. 2.It is this change of shape which enables the solid to supply the reaction

### ELASTICITY (MDM 10203)

LASTICITY (MDM 10203) Lecture Module 5: 3D Constitutive Relations Dr. Waluyo Adi Siswanto University Tun Hussein Onn Malaysia Generalised Hooke's Law In one dimensional system: = (basic Hooke's law) Considering

### UNIT I SIMPLE STRESSES AND STRAINS

Subject with Code : SM-1(15A01303) Year & Sem: II-B.Tech & I-Sem SIDDHARTH GROUP OF INSTITUTIONS :: PUTTUR Siddharth Nagar, Narayanavanam Road 517583 QUESTION BANK (DESCRIPTIVE) UNIT I SIMPLE STRESSES

### 16.21 Techniques of Structural Analysis and Design Spring 2003 Unit #5 - Constitutive Equations

6.2 Techniques of Structural Analysis and Design Spring 2003 Unit #5 - Constitutive quations Constitutive quations For elastic materials: If the relation is linear: Û σ ij = σ ij (ɛ) = ρ () ɛ ij σ ij =

### Donald P. Shiley School of Engineering ME 328 Machine Design, Spring 2019 Assignment 1 Review Questions

Donald P. Shiley School of Engineering ME 328 Machine Design, Spring 2019 Assignment 1 Review Questions Name: This is assignment is in workbook format, meaning you may fill in the blanks (you do not need

### Macroscopic theory Rock as 'elastic continuum'

Elasticity and Seismic Waves Macroscopic theory Rock as 'elastic continuum' Elastic body is deformed in response to stress Two types of deformation: Change in volume and shape Equations of motion Wave

### Tuesday, February 11, Chapter 3. Load and Stress Analysis. Dr. Mohammad Suliman Abuhaiba, PE

1 Chapter 3 Load and Stress Analysis 2 Chapter Outline Equilibrium & Free-Body Diagrams Shear Force and Bending Moments in Beams Singularity Functions Stress Cartesian Stress Components Mohr s Circle for

### Chapter Two: Mechanical Properties of materials

Chapter Two: Mechanical Properties of materials Time : 16 Hours An important consideration in the choice of a material is the way it behave when subjected to force. The mechanical properties of a material

### Purpose of this Guide: To thoroughly prepare students for the exact types of problems that will be on Exam 3.

ES230 STRENGTH OF MTERILS Exam 3 Study Guide Exam 3: Wednesday, March 8 th in-class Updated 3/3/17 Purpose of this Guide: To thoroughly prepare students for the exact types of problems that will be on

### High Tech High Top Hat Technicians. An Introduction to Solid Mechanics. Is that supposed to bend there?

High Tech High Top Hat Technicians An Introduction to Solid Mechanics Or Is that supposed to bend there? Why don't we fall through the floor? The power of any Spring is in the same proportion with the

### CHAPTER THREE SYMMETRIC BENDING OF CIRCLE PLATES

CHAPTER THREE SYMMETRIC BENDING OF CIRCLE PLATES * Governing equations in beam and plate bending ** Solution by superposition 1.1 From Beam Bending to Plate Bending 1.2 Governing Equations For Symmetric

### Strength of Materials (15CV 32)

Strength of Materials (15CV 32) Module 1 : Simple Stresses and Strains Dr. H. Ananthan, Professor, VVIET,MYSURU 8/21/2017 Introduction, Definition and concept and of stress and strain. Hooke s law, Stress-Strain

### Free Body Diagram: Solution: The maximum load which can be safely supported by EACH of the support members is: ANS: A =0.217 in 2

Problem 10.9 The angle β of the system in Problem 10.8 is 60. The bars are made of a material that will safely support a tensile normal stress of 8 ksi. Based on this criterion, if you want to design the

### CHAPTER 3 THE EFFECTS OF FORCES ON MATERIALS

CHAPTER THE EFFECTS OF FORCES ON MATERIALS EXERCISE 1, Page 50 1. A rectangular bar having a cross-sectional area of 80 mm has a tensile force of 0 kn applied to it. Determine the stress in the bar. Stress

### QUESTION BANK SEMESTER: III SUBJECT NAME: MECHANICS OF SOLIDS

QUESTION BANK SEMESTER: III SUBJECT NAME: MECHANICS OF SOLIDS UNIT 1- STRESS AND STRAIN PART A (2 Marks) 1. Define longitudinal strain and lateral strain. 2. State Hooke s law. 3. Define modular ratio,

### MATERIALS FOR CIVIL AND CONSTRUCTION ENGINEERS

MATERIALS FOR CIVIL AND CONSTRUCTION ENGINEERS 3 rd Edition Michael S. Mamlouk Arizona State University John P. Zaniewski West Virginia University Solution Manual FOREWORD This solution manual includes

### EE C247B ME C218 Introduction to MEMS Design Spring 2017

247B/M 28: Introduction to MMS Design Lecture 0m2: Mechanics of Materials CTN 2/6/7 Outline C247B M C28 Introduction to MMS Design Spring 207 Prof. Clark T.- Reading: Senturia, Chpt. 8 Lecture Topics:

### Mechanics of Earthquakes and Faulting

Mechanics of Earthquakes and Faulting www.geosc.psu.edu/courses/geosc508 Standard Solids and Fracture Fluids: Mechanical, Chemical Effects Effective Stress Dilatancy Hardening and Stability Mead, 1925

### NDT&E Methods: UT. VJ Technologies CAVITY INSPECTION. Nondestructive Testing & Evaluation TPU Lecture Course 2015/16.

CAVITY INSPECTION NDT&E Methods: UT VJ Technologies NDT&E Methods: UT 6. NDT&E: Introduction to Methods 6.1. Ultrasonic Testing: Basics of Elasto-Dynamics 6.2. Principles of Measurement 6.3. The Pulse-Echo

### 3 2 6 Solve the initial value problem u ( t) 3. a- If A has eigenvalues λ =, λ = 1 and corresponding eigenvectors 1

Math Problem a- If A has eigenvalues λ =, λ = 1 and corresponding eigenvectors 1 3 6 Solve the initial value problem u ( t) = Au( t) with u (0) =. 3 1 u 1 =, u 1 3 = b- True or false and why 1. if A is

### Mechanics of Materials

Mechanics of Materials Notation: a = acceleration = area (net = with holes, bearing = in contact, etc...) SD = allowable stress design d = diameter of a hole = calculus symbol for differentiation e = change

### GATE SOLUTIONS E N G I N E E R I N G

GATE SOLUTIONS C I V I L E N G I N E E R I N G From (1987-018) Office : F-16, (Lower Basement), Katwaria Sarai, New Delhi-110016 Phone : 011-65064 Mobile : 81309090, 9711853908 E-mail: info@iesmasterpublications.com,

### Bone Tissue Mechanics

Bone Tissue Mechanics João Folgado Paulo R. Fernandes Instituto Superior Técnico, 2016 PART 1 and 2 Introduction The objective of this course is to study basic concepts on hard tissue mechanics. Hard tissue

### STRAIN. Normal Strain: The elongation or contractions of a line segment per unit length is referred to as normal strain denoted by Greek symbol.

STRAIN In engineering the deformation of a body is specified using the concept of normal strain and shear strain whenever a force is applied to a body, it will tend to change the body s shape and size.

### CHAPTER 4: BENDING OF BEAMS

(74) CHAPTER 4: BENDING OF BEAMS This chapter will be devoted to the analysis of prismatic members subjected to equal and opposite couples M and M' acting in the same longitudinal plane. Such members are

### CHAPTER 6 MECHANICAL PROPERTIES OF METALS PROBLEM SOLUTIONS

CHAPTER 6 MECHANICAL PROPERTIES OF METALS PROBLEM SOLUTIONS Concepts of Stress and Strain 6.1 Using mechanics of materials principles (i.e., equations of mechanical equilibrium applied to a free-body diagram),

### Chapter 2: Elasticity

OHP 1 Mechanical Properties of Materials Chapter 2: lasticity Prof. Wenjea J. Tseng ( 曾文甲 ) Department of Materials ngineering National Chung Hsing University wenjea@dragon.nchu.edu.tw Reference: W.F.

Chapter 5 CENTRIC TENSION OR COMPRESSION ( AXIAL LOADING ) 5.1 DEFINITION A construction member is subjected to centric (axial) tension or compression if in any cross section the single distinct stress

### Sub. Code:

Important Instructions to examiners: ) The answers should be examined by key words and not as word-to-word as given in the model answer scheme. ) The model answer and the answer written by candidate may

### Chapter 2. Rubber Elasticity:

Chapter. Rubber Elasticity: The mechanical behavior of a rubber band, at first glance, might appear to be Hookean in that strain is close to 100% recoverable. However, the stress strain curve for a rubber

### Composites Design and Analysis. Stress Strain Relationship

Composites Design and Analysis Stress Strain Relationship Composite design and analysis Laminate Theory Manufacturing Methods Materials Composite Materials Design / Analysis Engineer Design Guidelines

### Theory at a Glance (for IES, GATE, PSU)

1. Stress and Strain Theory at a Glance (for IES, GATE, PSU) 1.1 Stress () When a material is subjected to an external force, a resisting force is set up within the component. The internal resistance force

### Chapter 7. Highlights:

Chapter 7 Highlights: 1. Understand the basic concepts of engineering stress and strain, yield strength, tensile strength, Young's(elastic) modulus, ductility, toughness, resilience, true stress and true