# Constitutive Equations (Linear Elasticity)

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1 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 the strains and viceversa. Poisson s ratio Poisson s ratio: Hooke s Law x Δl l x Nominal lateral strain (transverse strain) lateral tensile When strains are small, most of materials are linear elastic. x z strain x strain z Δl l z z Tensile: Shear: Ε τ G γ

2 Relationships between Stress and Strain An isotropic material has a stress-strain relationships that are independent of the orientation of the coordinate system at a point. A material is said to be homogenous if the material properties are the same at all points in the body Uniaxial Stresses Strains

3 ( ) ( ) Stresses Strains + Biaxial x y z YY ( ) ( ) ( ) Stresses Strains + + Triaxial x y z YY ZZ

4 For strains [] γ yz γ zx γ xy OR [ ] For elastic isotropic materials there is a relationship between stresses and strains [ ] [ S ][ ] τ τ τ yz zx xy

5 [] G G G υ υ υ υ υ υ Uniaxial Stresses

6 + + [] z y x z y x G G G υ υ υ υ υ υ [ ] Triaxial Stresses

7 + + Isotropic Materials xy xy zx zx yz yz G G G τ γ τ γ τ γ An isotropic material has stress-strain relationships that are independent of the orientation of the coordinate system at a point. The isotropic material requires only two independent material constants, namely the lastic Modulus and the Poisson s Ratio. xy zx yz xy zx yz G G G τ τ τ γ γ γ

8 xample kn forces are applied to the top & bottom of a cube ( mm edges), 6 GPa,.3. Find (i) the force exerted by the walls, (ii) z kn y x (i), and - 3 N/( -3 m) 3 7 Pa ( - v - v ) / [ -.3 (- 3 7 )]/ Pa (compressive) Force A ( -3 m) (-9 6 Pa) N (ii) ( - v - v ) / [.3 (- 3 7 ).3 (- 9 6 )]/

9 xample 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 ksi and ksi. For x 6 psi and /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. Apply the generalized Hooke s Law to find the three components of normal strain. + y psi ( ksi) ( ksi) 3 in./in. in./in in./in.

10 valuate the deformation components. δ B d A δ C d D t δ t 3 ( in./in. )( 9in. ) 3 ( +.6 in./in. )( 9in. ) 3 (.67 in./in. )(.75in. ) δ B A δ C D δ t in. in. in. Find the change in volume e x + y + z.67 3 in 3 /in 3 ΔV ev.67 3 ( ) in ΔV in

11 Mechanical Properties Axially Loaded Members Stress and Strain Linear lasticity Axially Loaded Members Torsion Shear Force and Bending Moment Diagrams

12 Typical cross sections of structural members.

13 xample Cross Sectional Area Bar BD mm Cross Sectional Area Bar C 5mm Bars are made of steel 5GPa Find maximum force P if the deflection at A is limited to mm. Assume ABC is rigid. FBD The displacements in B and C will be given by the equation: δ BD F L BD BD δc ABD FCL A C C From equilibrium: F F C BD P 3P

14 or C BD C A C B B B C A A A δ δ δ δ (P in Newtons) mm P mm GPa mm P A L F mm P mm GPa mm P A L F C C C C BD BD BD BD δ δ A mm δ P Max KN 3.

15 Bars with Intermediate Axial Loads FBD

16 Bars Consisting of Prismatic Segments ach Having Different Axial Forces, Dimensions, and Materials FBD

17 Bars with Continuously Varying Loads and/or Dimensions

18 xample Change in length of a tapered bar.

19 xample Find the total extension of the bar (square section). o.6m 5mm W δx Width of a cross-sectional element at x: Stress in this element : X.m 5mm 3 N.88 ( x/) m x x 3 W (5 m).6m 7 Pa kn x ( m) Strain of this element: 7.88 / x x 4 The extension of this element : The total extension of the whole bar is : de dx.9 x e de 4 dx.8.9 x.6 4 dx.3 x -4 m

20 Statically Indeterminate Structures Statically Determinate Structures Reactions and internal forces can be determined solely from free-body diagrams and equations of equilibrium without knowing the properties of the materials. Statically Indeterminate Structures In addition to the equilibrium equations, the relations between forces and displacements are usually needed to determine the reactions and the internal forces.

21 Analysis of a Statically Indeterminate Bar R A δ δ AC BC N AC a...( lengthen) A N BC b...( shorten) A N AC OR N BC R B

22 xample A steel cylinder encased in a copper tube, both are compressed by a force P.

23 xample

24 Thermal ffects Changes in temperature produce expansion or contraction of structural materials, resulting in thermal strains and thermal stresses. xample Strain produced in a steel bar by a change of temperature of o F. 3x 6 psi and α9.6x -6 / o F

25

26 xample Statically indetermine bar subjected to a uniform temperature increase.

27 xample Sleeve and bolt assembly with uniform temperature increase ΔT.

28 Misfits (departures from the theoretical configuration dimensions of the structure If a structure is statically determinate, small misfits in one or more members will not produce strain or stresses. If a structure is statically indeterminate, small misfits in one or more members will produce strain or stresses. The structure is not free to adjust to the misfits.

29 Bolts and Turnbuckles The pitch of the threads is the distance from one thread to the next. The distance traveled by the nut: δ np Where n is the number of revolutions, it is not necessarily an integer. Double-acting turnbuckle. (ach full turn of the turnbuckle shortens or lengthens the cable by p, where p is the pitch of the screw threads.).

30 xample Statically indeterminate assembly with a copper tube in compression and two steel cables in tension.

31 xample Two copper bars and one aluminum bar are fixed at the bottom as shown. The top ends of all three bars are supposed to be welded to a rigid steel plate. The aluminum bar is a little shorter (δ. in.) than the copper bars and it had to be heated to make it extend to the same length as the copper bars to complete the welding process. What is the temperature increase, ΔT (ºF), that is needed to bring the aluminum bar to the same length as that of copper bars? After the welding is done and the temperature returns to normal, what will the stresses be in the aluminum bar and the copper bars, respectively?

32 FBD

33 xample A Bar Subjected to ΔT and P

34 xample Determine stress and strain if ΔT C

35 xample Determine the vertical displacement (δ Β ) of joint B. Both members have the same rigidity A. Force acting in each bar: P F Cosβ Length of the bar: L H Cosβ F L U A Strain-nergy of the two bars : ( ) P H 4ACos 3 β Work of the load P : W Pδ B Conservation of energy PH ACos δ B 3 β

36 Stresses on Inclined Sections xamples of shear failure under uni-axial tension/compression Slip bands (or Lüders bands) in a polished steel specimen loaded in tension. Shear failure along a 45 plane of a wood block loaded in compression.

37 Stresses on Inclined Sections Prismatic bar in tension showing the stresses acting on cross section mn: (a) bar with axial forces P, (b) three-dimensional view of the cut bar showing the normal stresses, and (c) two-dimensional view. Stress lement: Useful way of representing the stresses in the bar, such as element label C it should show all the stresses acting on all faces of this element. The dimensions of the stress element are assumed to be very small (equal dimensions on all sides cube). Stress element at point C of the axially loaded bar; (a) three-dimensional view of the element, and (b) twodimensional view of the element. x P A

38 Stresses on Inclined Sections Prismatic bar in tension showing the stresses acting on an inclined section pg: (a) bar with axial forces P, (b) threedimensional view of the cut bar showing the stresses, and (c) twodimensional view. The normal of the plane pq is inclined an angle θ to the x-axis. The force P acting in the x- direction need to be resolved in two components: Normal force (N) perpendicular to the plane pq. Shear force (V) tangential to the plane pq N V P Cosθ P Sinθ

39 Prismatic bar in tension showing the stresses acting on an inclined section pq. The relationship between the section mn and the section pq is: A A θ Cosθ θ N A θ P Cosθ A Cosθ P Cos θ A x Cos θ θ V A θ P Sinθ A Cosθ P A Sinθ Cosθ x Sinθ Cosθ

40 Sign convention for stresses acting on an inclined section (Normal stresses are positive when in tension and shear stresses are positive when they tend to produce counterclockwise rotation.) θ x x Cos θ ( + Cosθ ) Sin θ θ Cos x θ Sin x ( θ ) Maximum Normal Stress occurs at θ Graph of normal stresses θ and shear stress τ θ versus angle θ of the inclined section. Maximum Shear Stress occurs at θ45 o

41 Maximum Normal and Shear Stresses (for a bar in tension) θ τ θ x Max θ τ θ 45 Max 45 o τ x θ 45 x

42 xample Bar under a compression force.

43 xample A square section plastic bar that was glued at section pq

44 xample A tension member is to be constructed of two pieces of plastic glued along plane pq (see figure). For purposes of cutting and gluing, the angle must be between 5 and 45. The allowable stresses on the glued joint in tension and shear are 5. MPa and 3. MPa, respectively. (a) Determine the angle so that the bar will carry the largest load P. (Assume that the strength of the glued joint controls the design.) (b) Determine the maximum allowable load Pmax if the cross-sectional area of the bar is 5 mm.

45 Strain nergy During the loading process, the load P moves slowly through the distance δ and does a certain amount of work. To find the work done, we need to know the manner in which the force and deformation change (loaddisplacement diagram). W δ Pd δ The work done by the load is equal to the area below the load-displacement curve. Strain nergy is defined as the energy absorbed by the bar during the loading process. Strain nergy (U) Work Done (W) Conservation of energy.

46 Linearly lastic Behavior lastic Strain nergy : Strain energy recovered during unloading. Inelastic Strain nergy : Strain nergy that is not recovered during unloading. nergy that is lost in the process of permanently deforming the bar. Most structures are designed with the expectation that the material will remain within the elastic range under ordinary conditions of service. U W Pδ Combining the equations: k stiffness A L Load-longation relationship P L U A δ U f k PL A Aδ L flexibility nergy on a spring Load-displacement diagram for a bar of linearly elastic material. P U k U kδ

47 Non-Uniform Bars Bar consisting of prismatic segments having different cross-sectional areas and different axial forces. The total strain energy of a bar consisting of several segments is equal to the sum of the strain energy of the individual segments. U n i U i Linearly lastic U n i Ni Li A i i Non-prismatic bar with varying axial force. U [ N( x) ] L A( x dx ) We can not obtain the strain energy of a structure supporting more than one load by combining the strain energies obtained from individual loads acting separately.

48 Strain nergy Density P L U A U Aδ L u u P A δ L P A δ L It is defined as the strain energy per unit volume of the material. combining u u The strain-energy density of the material when it is stressed to the proportional limit (pl) is called the modulus of resilience. u r pl Toughness refers to the ability of a material to absorb energy without fracturing. The Modulus of Toughness is the strain-energy density when the material is stressed to the point of failure.

49 xample Calculation of strain energy. (a) P L U a A U b n Ni Li A i (b) P ( L 5) P ( 4L 5) Ub + A ( 4A) U b P L 5A i i U 5 a n (c) Ni Li P ( L 5) P ( 4L 5) 3P L 3U U c + A A ( 4A) A i i i a

50 xample Determine the vertical displacement (δ Β ) of joint B. Both members have the same rigidity A. Force acting in each bar: P F Cosβ Length of the bar: L H Cosβ F L U A Strain-nergy of the two bars : ( ) P H 4ACos 3 β Work of the load P : W Pδ B Conservation of energy PH ACos δ B 3 β

51 Impact Loading Loads can be classified as static or dynamic depending upon whether they remain constant or vary with time. A static load is applied very slowly. A dynamic load may take many forms, some loads are applied and removed suddenly (impact loads), others persist for long periods of time and continuously vary in intensity (fluctuating loads). Consider the potential energy of the collar. Potential _ nergy Mgh The potential energy is transformed into kinetic energy. Mgh Mv During impact some the kinetic energy is transformed into strain energy and heat.

52 Assumptions: (a) The collar sticks to the flange and moves downwards with it. (b) Disregard all energy loses (heat). (c) The bar behaves always in the elastic range. (d) The stresses are uniform across the section of the bar. The total potential energy equals the strain energy of the bar. Mg( h Max ) + δ Aδ L Max Finding δ Max WL WL WL δ Max + + h A A A The first term is the elongation of the bar due to the weight of the collar under static conditions. δ δ Max Max δ δ Static Static + ( δ + hδ ) Static h + + δ Static Static δ If h>>δ Static δ Static Max hδ Static WL A MghL A

53 Maximum Stress in the Bar Max Max Max Static δ L + Max + W A W + A W + h LA + + h Static L Static Static h L W Mg Static A A Static δ L Static If h is large Max L h Static Mv AL Increase in Kinetic nergy causes an increase in stress, and an increase in the volume of the bar reduces the stress. Impact Factor The ratio of a dynamic response of a structure to the static response (for the same load) is known as the impact factor δ Impact - Factor - for - elongation δ Max Static Impact - Factor - for - stress Max Static

54 xample Round prismatic steel bar (GPa). (a) Calculate the maximum elongation and the impact factor. (b) Calculate the maximum tensile stress. δ (a) Calculate d Static and compared to the height of the mass MgL A.Kg 9.8m GPa π 4 s.m Static. 6 h δ Static 5mm.6mm 4,5 ( 5mm) mm δ Max δ 5mm.6mm.783 h Static Since the height of the fall is very large compared to the static elongation..783 Impact - Factor 68.6

55 (b) Maximum Tensile Stress δ L W A Mg A GPa.783mm mm Max Max 87. Kg 9.8m π 4 Static. (.5m) s MPa MPa 87.MPa Impact Factor Stress 68.6.MPa

57 The number of cycles to failure of the material at different stresses is known as the S-N diagram. The S-N diagram of some materials (steel) shows a horizontal asymptote known as the fatigue limit or endurance limit. ndurance curve, or S-N diagram, showing fatigue limit. Typical endurance curves for steel and aluminum in alternating (reversed) loading. Aluminum does not show an endurance limit and the fatigue limit needs to be defined, typically as the stress at 5x 8 cycles.

58 Stress Concentration P A This formula is based on the assumption that the stress distribution is uniform throughout the cross section. Saint-Venant s Principle The peak stresses occur directly under the load. However, these stresses decreases rapidly at a distance from the applied load. Uniform stresses are reached at a cross section at least a distance b away from the concentrated load (b is the largest lateral dimension of the bar). Stress distribution near the end of a bar of rectangular cross section (width b, thickness t) subjected to a concentrated Stress Concentration Factors load P acting over a small area. Bars often have holes, grooves, notches, keyways, shoulders, threads or other abrupt changes in geometry that creates disruption to a uniform stress pattern. Discontinuities in geometry, cause high stresses in very small regions of the bar. These high stresses are known as stress concentrations.

59 Stress distribution in a flat bar with a circular hole. Stress-concentration factor K for flat bars with circular holes. K Max No min al

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