Axial Loading Concept of Stress & Strain.

Axial Loading Concept of Stress & Strain mi@seu.edu.cn

Contents Introduction( 拉压变形简介 ) Diagram of Axial Forces( 轴力图 ) Concept of Stresses( 应力的概念 ) General Stress State of a Point( 点的一般应力状态 ) Stresses Acting on Cross Sections( 拉压杆横截面上的应力 ) Saint-Venant s Principle( 圣维南原理 ) Stresses Acting on Oblique Sections( 拉压杆斜截面上的应力 ) Deformation of Axially Loaded Bars( 拉压杆的变形 ) Elastic Constants of Engineering Materials( 常见工程材料的弹性常数 ) Nonuniform Tension/compression( 非均匀拉压 ) Strain Energy( 应变能 ) Strain Energy Density( 应变能密度 ) Mechanical Behavior of Materials( 材料的力学性能 ) Nominal Stress-strain Curve( 名义应力应变曲线 ) Stress and Deformation Indices of Low-carbon Steel( 低碳钢的应力和变形指标 ) 2

Contents Yield Stress, Ultimate Stress and Percent Elongation of Engineering Materials ( 常见工程材料的屈服应力强度极限和断后伸长率 ) Strain (Work) Hardening( 冷作硬化 ) Mechanical Behavior of General Ductile Materials under Tension( 塑性材料的拉伸力学性能 ) Mechanical Behavior of Brittle Materials under Tension( 脆性材料的拉伸 ) Mechanical Behavior of Low-carbon Steel under Compression( 低碳钢的压缩 ) Mechanical Behavior of Cast Iron under Compression( 铸铁的压缩力学性能 ) Mechanical Behavior of Creet( 混凝土的力学性能 ) Mechanical Behavior of Wood( 木材的力学性能 ) Mechanical Behavior of Composite Materials( 复合材料的力学性能 ) Mechanical Behavior of Viscoelastic Materials( 粘弹性材料的力学性能 ) Strength Condition( 强度条件 ) Failure of Brittle vs. Ductile Bars under Tension( 脆性和塑性杆件的拉伸失效 ) Stress Concentration( 拉压杆中的应力集中现象 ) 3

Introduction 4

5 Introduction F F F F Load: equal and opposite forces along bar axis Deformation: extension (contraction) along bar axis and contraction (extension) transversely

6 Internal Forces Illustrated by Method of Sections Internal Force The change of interaction force among various parts of a solid body, introduced by external loading. The Method of Section F 1 I m II F n F 1 I m F 1 I o m M R F 2 F 2 F 2

7 Sign Convention of Axial Forces y Coordinate: a right-handed system F 2 F 1 I z M y F Sy o m F Sz M z T F N x O: Centroid of the cross-section x: Cross-section normal (Bar axis) Forces: F N, F Sy, F Sz Moments: T, M y, M z Axial Force (F N ) F N (+) o Along bar axis F N (-) o o o Extends or contracts the bar along bar axis Positive for tension; Negative for compression Assuming positive for all unknown axial forces

Procedure of Method of Sections Sectioning the member Taking either portion Substituting the other portion with internal forces Equilibrating F m F F m F F m F N F m F N F N = F F N = -F 8

9 Diagram of Axial Forces Abscissa: position of cross sections Ordinate: axial force Positive for tension; negative for compression. F F N m m F x

Sample Problem Plot the diagram of axial force 3 kn A B 2 kn/m C D 1 kn Solution 2 m 2 m 2 m 1. Internal force in AB CD 3 kn FN AB 3 kn FN CD 1kN 2 m x 2. Internal force in BC F N x FN x 3 2x 0 x 2 3. Diagram of axial force kn 4. Maximum internal force F 3 N F N,max 3 kn 1 10

11 Concept of Stresses m m A n F t ζ Given ΔF as the force transmitted across ΔA, a stress traction vector can be defined as t lim A0 F A Units: Pa (N/m 2 ), 1 MPa = 10 6 Pa, 1 GPa = 10 9 Pa.

General Stress State of a Point ζ yy ζ yz ζ yx ζ xz ζ zx ζ zz ζ xy ζ xx ζ zy ζ zy ζ xx dy ζ xy ζ zx ζ xz y ζ zz ζ yx dx ζ yz ζ yy dz z x All stress components shown in the above figure are positive. 12

13 Stresses Acting on Cross Sections Longitudinal Transverse F F Assumption: stresses are uniformly distributed on the cross sections of axially loaded bars (Saint-Venant s Principle) F N ζ FN da da A A A F N A

Saint-Venant s Principle Loads transmitted through rigid plates result in uniform distribution of stress and strain. Concentrated loads result in large stresses in the vicinity of the load application point. Stress and strain distributions become uniform at a relatively short distance from the load application points. Saint-Venant s Principle: Stress distribution may be assumed independent of the mode of load application except in the immediate vicinity of load application points. 14

15 Sample Problem F N A A 2A 1 2 2 A2 A1

16 Sample Problem For the compound roof structure shown, determine the axial stress developed in members AE and EG. q = 20 kn/m, A AE = A EG = 23 cm 2. Solution 1. Reaction force ql 0 Fy FRA FRB 177.4 kn 2 2. Axial force in EG q 2 0 M C 4.37 4.5 2 F F EG RA 4.37 4.5 F 1.2 1 356 kn EG

17 3. Equilibrium at joint E F EG FAE cos 4.37 FAE 4.37 1 F 366 kn AE 2 2 4. Axial stresses in members AE and EG EG AE 3 F 356 10 EG 155 MPa 4 A 23 10 EG 3 F 366 10 AE 159 MPa 4 A 23 10 AE

18 Stresses Acting on Oblique Sections A p A cos F F cos cos A A 2 p cos cos p sin sin 2 2 α: defined as the angle measured from cross section normal to oblique cross section normal.

19 On cross section: maximum normal stress; zero shearing stress. On 45 oblique cross section: maximum shearing stress. 0 2 0 max 0 sin(90 ), 0 cos (45 ) 45 45 2 2 2

On 90 oblique cross section: zero normal and shearing stress. 20

b Deformation of Axially Loaded Bars Longitudinal (axial) & transverse strains b b L L+L Longitudinal strain: L L Transverse strain: b b Connection between longitudinal and transverse strain ν: Poisson s ratio / 21

22 Deformation of Axially Loaded Bars Consider a differential cube of side length a, b and c F bc L Ka Ka bc E K This linear relation is referred to as the one-dimensional Hooke s law, with E denoting the Young s modulus. The measurement of axial deformation LF F L N N L L L E AE EA EA: Tension (Compression) rigidity

Elastic Constants of Engineering Materials 23

Elastic Constants of Engineering Materials 24

Nonuniform Tension/compression Pure tension/compression formula refers to a prismatic bar subjected to axial forces acting only at the ends. Nonuniform tension/compression differs from pure tension in that the bar need not to be prismatic and the applied axial forces may act anywhere along the axis of the bar. Bars in nonuniform tension/compression can be analyzed by applying the formulas of pure tension/compression to finite segments of the bar and then adding the results, or by applying the formulas to differential elements of the bar and then integrating. L L F L Ni i i i i EA i i L L FN x dx L dl 0 0 EA x 25

Sample Problem Consider an axially loaded bar with varied cross section. Given: E = 210 GPa; Section 1 (circular): d 1 = 20 mm; Section 2 (square): side length a = 25 mm, 2 = -30 MPa; Section 3 (circular) d 3 =12mm. Find: Total change in bar length L. F 1 2 3 F 0.2 m 0.4 m 0.2 m Solution: F 2 2 2A2 30 MPa 25 mm 18.75 kn FN L 1 1 FN L 2 2 FN L 3 3 18750 0.2 0.4 0.2 L 0.272mm 9 2 2 2 EA1 EA2 EA3 21010 π0.02 0.025 π0.012 4 4 26

Sample Problem Determine the elongation of the bar due to the end load P. Solution: x d d d d L x A B A 2 d 2 x x A x d d d 4 4 L A B A L FN dx L 4Pdx L 0 EA 0 x x E d A db d A L x d d A db d A 4PL L L 0 2 E db d A x d A db d A L 4PL 1 B A x E d d d A db d A L 4PL 1 1 4PL E d d d d Ed d B A A B A B Cannot be assumed as a prismatic bar that has the diameter (d A +d B )/2. For the special case of a prismatic bar: L 4PL PL 2 Ed EA L 0 27 2

Sample Problem Deformation of a uniform bar due to gravity. Given: cross section area A, density ρ, Young s modulus E. Find: Maximum normal stress and axial length change. Solution: 1. Axial forces and stresses q=ρga; F N (x)= ρgax; ζ(x)= F N (x)/a= ρgx, ζ max = ρgl FN( x)d x FN( x)dx gaxdx gl 2. Deformation: d( L) L EA EA EA 2E l 0 0 l 2 x x L q dx F N (x) F N (x) F N (x) ρgal 28

Sample Problem SOLUTION: Apply a free-body analysis to the bar BDE to find the forces exerted by links AB and DC. The rigid bar BDE is supported by two links AB and CD. Link AB is made of aluminum (E = 70 GPa) and has a cross-sectional area of 500 mm 2. Link CD is made of steel (E = 200 GPa) and has a cross-sectional area of (600 mm 2 ). For the 30-kN force shown, determine the deflection a) of B, b) of D, and c) of E. Evaluate the deformation of links AB and DC or the displacements of B and D. Work out the geometry to find the deflection at E given the deflections at B and D. 29

SOLUTION: Free body: Bar BDE Displacement of B: B PL AE 3 6010 N 0.3m -6 2 9 50010 m 7010 Pa 51410 6 m B 0.514mm M 0 0 30kN0.6m F 0.2m F B CD M D 90kN tension 0 0 30kN0.4m F 0.2m F AB 60kN compression CD AB Displacement of D: D PL AE 3 9010 N 0.4m -6 2 9 60010 m 20010 Pa 30010 6 m D 0.300mm

31 Displacement of E: BB DD BH HD 0.514mm 0.300 mm x 73.7 mm 200mm x x EE DD E 0.300mm E HE HD 1.928mm 400 73.7 mm 73.7 mm E 1.928mm

Sample Problem Given: tension rigidity of bar 1, 2 and 3 = EA, AB rigid, F and L. Find: the vertical and horizontal displacement at C. x y C C FL 2EA 32

33 Sample Problem Determine the vertical and horizontal displacement of A. AB is rigid. A F E B D 0 45 C

Hint: A A y A x BB O EE O, AA O EE O B A E E O LED B E F E B L BC B A A F 0 45 BB BO LBC EB EO EB EO EE EO LED EO EO EO EO EB OB BB A A AO Ay AE EO AE EO Ay Ax AO AE EO EE EO LED EO EO D C 34

Sample Problem (Poisson s Ratio) A circular tube of inner diameter d = 60 mm and outer diameter D = 80 mm is subjected to an axial load of F = 200 kn, determine the wall thickness under the deformed state. E = 200 GPa, ν = 0.3. F Solution: F EA 3 20010 9 2 2 6 20010 80 60 10 4 6 0.345510 6 136.510 D d 6 2 6 D d 6 E S s 455 10 1 10 1136.5 10 9.998635 mm 1 251.327412 1136.5 10 251.293106 mm 1 188.495559 1136.5 10 188.469830 mm Poisson s ratio is the same for all transverse directions. (Isotropy) 6 35

36 Sample Problem (Large Deformation) Given: d = 1 mm, ε = 0.0035, E = 210 GPa. Determine the axial stress in wire, the displacement of C, and the force F. Solution: 9 1. Axial stress in the wire: E 210 10 0.0035 735 MPa 2. The displacement of C AC AC 1 0.0035 0.0035 m L L L L L L L L 2 2 2 2 C AC AC AC AC AC AC 2 2 1 0.0035 0.0035 0.0837 m 3. The force F (Equilibrium must be analyzed under the deformed state.) 2 C 6 3 0.0837 2 sin 2 2 735 10 10 4 96.3 N 1.0035 F FN BAC A LAC

37 Sample Problem (Pressure Vessel) For the thin-walled pressure vessel shown, D, δ, and p denote the vessel inner diameter, wall thickness and pressure respectively. Determine: the axial stress, circumferential stress, and the elongation of vessel circumference, diameter and length. δ l p D

F axial a a 0 2 πd p πd 4 pd 4 a Dδ a D δ p D 2 4 F circumferential 0 a D 2 c b sin pb d 0 2 2bc pbd pd c 2 b: an arbitrarily chosen axial length c p c 38

39 Circumferential stress: Axial stress: a c pd 4 pd 2 Radial stress: D p pd 2 pd 4

Elongation of the vessel circumference FL N t A D c EA EA Elongation of the vessel diameter 2 2 t pd pd D 1 2E D 2E Elongation of the vessel length FL N a A l a EA pdl EA 4E 2 2 pd D pd pd 1 2 E 2E D 2E Transverse strain t pd D pd c D L 2E L 2E t D 40

Strain Energy Strain energy is developed in solid materials due to elastic deformation induced by external loading. L L 1 F Strain energy and work in pure tension/compression L 2 2 1 FL EA N U W FN L L 2 2EA 2L F FL N EA The one-half is due to the assumption that external loading is applied gradually, starting from zero. O L 41

42 Strain Energy Nonuniform tension/compression U U 2 FN ili 2EA i i i i i U 2 L L FN ( x) dx du 2 EA( x) 0 0

Strain Energy Density (Energy per Unit Volume) Consider a differential cube of side length a, b and c 1 1 U FL bc a 2 2 U 1 u V 2 Total strain energy calculated from density 2 1 1 1 FN FN FN dx U udv dv A dx A dx V V 2 L 2 L 2 A EA L 2EA For constant E, F N, and A U FL 2 N 2EA 43

Mechanical Behavior of Materials Mechanical behavior of materials focus on the strength and deformation characteristics of solid materials under external loading. Tensile specimen For circular tensile specimen For square tensile specimen L = 10d, 5d = 11.3 A, 5.65 A L = 11.3b, 5.65b. Compressive specimen: L = 13d, 13b d b L L 44

Mechanical Behavior of Materials 45

Nominal Stress-strain Curve Find more animations at http://em2lab.yolasite.com/. 46

Stress and Deformation Indices of Low-carbon Steel ζ d b c a ζ b ζ e ζ ζ p s O p : Proportional limit e : Elastic limit Y : Yield stress u : Ultimate stress Stress-strain Diagram Percent elongation: δ = (L final - L initial ) / L initial 100% Percent reduction in area ψ = (A initial - A final ) / A initial 100% ε

Yield Stress, Ultimate Stress and Percent Elongation

Yield Stress, Ultimate Stress and Percent Elongation 49

Yield Stress, Ultimate Stress and Percent Elongation 50

Exercise Based on the three stress-strain curves shown in the figure, which one of the following regarding ultimate stress, Young s modulus and percent elongation is correct? ζ 3 2 1 A: u 1 u 2 u 3 E 1 E 2 E 3 1 2 3 B: u 2 u 1 u 3 2 1 3 1 2 3 E E E O C: D: u 3 u 1 u 2 3 1 2 3 2 1 u 1 u 2 u 3 2 1 3 2 1 3 E E E E E E ε 51

Strain (Work) Hardening Unloading at point B during hardening stage down to zero force. EB almost parallel to OA. (Unloading law) Deformation in the hardening stage is composed of elastic ( L e ) and plastic elongation ( L p ). If the test sample is reloaded at point E, the linear elastic (proportional) limit substantially increases. As a result, the overall deformation the sample can endure decreases. In engineering practice, strain hardening is often employed to increase the maximum resistance force within linear elastic scope. F O ζ O A L p E L e F B L Force-displacement Diagram A E B F p e Stress-strain Diagram ε 52

310 53 Sample Problem Given: E = 200 GPa, unloading at ζ = 310 MPa during hardening stage, total strain ε e + ε p = 0.02155. Find: ε e and ε p. 0.0215

310 54 Solution: e E 310 2001000 0.00155 p 0. 02 e e 0.0215 Remark: in experimental reports, strain often takes the unit of micro-strain, i.e. 0.02 (ε)=20000 (με) 1 10 6

55 Sample Problem Low-carbon steel loading and unloading experiments. Given: E = 210 GPa, ζ P = 210 MPa. Find: 1. normal stress at the strain level ε = 0.001 (under linear elastic limit); 2. normal stress at a point in hardening stage, unloading from which results in ε = 0.08 and ε p = 0.078. Solution: 1. Under linear elastic limit E 9 210 10 0.001Pa 210MPa p 2. During hardening stage: E e 9 210 10 (0.08 0.078) Pa 420 MPa

56 Mechanical Behavior of General Ductile Materials under Tension Possess distinctive ζ Y and ζ u ζ May not have yielding and/or 1500 ζ 0.2 35CrMnSi steel localized deformation stage Relatively large extension rate after fracture (δ 5%) 1000 500 O 0.2% 45 # steel Q235 steel Al alloy brass ε If there is no obvious yielding stage: take the normal stress corresponding to 0.2% plastic strain as yield limit (ζ p0.2 )

Mechanical Behavior of Brittle Materials under Tension ζ-ε is a slightly curved line and approximately obeys the Hooke s law No yielding, hardening and localized deformation stage u tangent The ultimate stress is the only index Relatively small percent elongation after fracture (δ = 2%-5%) Tangential modulus: slope at any point of ζ-ε curve intersection secant modulus: can be defined at ε = 0.1% O 0.1% 57

Mechanical Behavior of Low-carbon Steel under Compression ζ compressive tensile ζ Y ζ p O ε E, ζ p, ζ Y take the same values as in tensile tests. Cannot obtain compressive ultimate stress due to the (constant) initial cross-sectional area used in normal stress calculation. Compressive mechanical behavior of low-carbon steel is obtained from its tensile indices. 58

Mechanical Behavior of Cast Iron under Compression ζ ζ c u Compressive O Tensile There is only very short linear ζ-ε curve (only approximately follows Hooke s law). Missing yielding stage (no obvious ζ Y ). Much larger compressive ultimate stress (typically 4-5 times that of tensile strength limit). Final damage is in the form of shearing along a surface 45 0 from axis. ζ t u ε 59

Mechanical Behavior of Concrete Compressive ultimate stress is much larger than its tensile counterpart (5-20 times) Compressive behavior depends on the frictional state at sample ends Under frictional end-condition, sample is composed of two trapezoidal cones under mirror-symmetry upon damage ζ-ε comprises a short straight line, followed by an obvious curve. Elastic modulus can be defined by the slope of secant line at ζ = 0.4ζ u Small compressive ratio after fracture. u u Frictional ends Lubricated ends 60

Mechanical Behavior of Wood Material behavior is transversely isotropic. Longitudinal tensile strength is unstable due to knots. Longitudinal compressive strength is insensitive to knots. Much larger longitudinal compressive strength than the transverse one. Most often used as compressive or support bars in engineering Transverse compression Longitudinal Compression Longitudinal tension Longitudinal compression Transverse compression 61

Mechanical Behavior of Composite Materials Mechanical properties of composite materials is mostly determined by the way how fibers are arranged (anisotropic). Influence from temperature, stress concentration (geometric factors), and strain rate (viscoelasticity) High ultimate stress and low plasticity under low temperature. Low ultimate stress and low plasticity under high stress concentration state Low ultimate stress and high plasticity under high strain rate 62

63 Mechanical Behavior of Viscoelastic Materials ζ = ε f(t) - linear visco-elasticity ζ = f(ε, t) - non-linear visco-elasticity Creep: increasing strain with time under constant stress Relaxation: decreasing stress with time under constant strain

64 Strength Condition Limit stress ( lim ): the stress under which mechanical components get damaged. Damage Criteria: yielding (ζ Y ) for ductile materials; fracture (ζ u ) for brittle materials Allowable Stress []: the maximum stress allowed in engineering practice. It is typically taken as one n th (Safety Factor) of the limit stress Strength Condition: Strength Analysis: (1) Strength check: (2) Cross-section design: (3) Find allowable load: max max n lim A F [ ] N max FN max A [ ]

65 Sample Problem Given: d = 14 mm, [ζ] = 170 MPa, uniaxial tension load F = 2.5 kn. Check the strength condition of the circular bar. Solution max F 2.510 π 14 10 4 3 N, max 2 6 A Pa 162 MPa<[ ] The strength condition is satisfied.

Sample Problem Given: A AC = 450 mm 2, A BC = 250 mm 2, E AC = E BC, [ζ] = 100 MPa. Find: the maximum allowable load F of the truss. Solution: 1. Find F NAC and F NBC Method of joint at C: F x = 0, F NBC sin45 0 - F NAC sin30 0 = 0 F y = 0, F NBC cos45 0 - F NAC cos30 0 - F = 0 A F NAC 30 0 45 0 C F F NBC B We get: F NAC = 0.732F, F NBC = 0.517F 66

67 2. Allowable axial force in bar AC and BC so: [F] 61.48 kn So: [F] 48.36 kn 6 2 6 m FN AC 0.732F A 450 10 100 10 6 2 6 m FN BC 0.517F A 250 10 100 10 Pa Pa 3. Allowable load [F] Take the smaller value: [F] 48.36 kn.

Sample Problem Given: F = 75 kn, [ζ] = 160 MPa. Find: the minimum A AB and A BC. Solution: A 1. Find F NAB and F NBC By the Method of joint at B: F y = 0, F NBC cos45 0 + F = 0 F x = 0, F NBC cos45 0 + F NAC = 0 45 We get: F NAC = F,F NBC = -1.414F F B 45 2. Minimum cross-sectional area A A AB BC F 7510 [ ] 16010 F 10610 [ ] 16010 3 NAB 4 2 2 4.68710 m 4.687 cm 6 3 NBC 4 2 2 6.629 10 m 6.629 cm 6 C 68

69 Failure of Brittle vs. Ductile Bars under Tension Brittle Ductile Under uniaxial tension, brittle bars break along crosssections while ductile ones glide along 45 0 sections during yielding?

70 Stress Concentration Stress concentration: rapid stress increase at specific locations where geometric defects and/or abrupt cross-sectional area change occurs

Stress-concentration factor for flat bars with circular holes 71

Stress-concentration factor for flat bars with shoulder fillets. The dashed line is for a full quarter-circular fillet. 72

Stress-concentration factor for round bars with shoulder fillets. The dashed line is for a full quarter-circular fillet. 73

74 Sample Problem SOLUTION: Determine the geometric ratios and the stress concentration factor. Determine the largest axial load P that can be safely supported by a flat steel bar consisting of two portions, both 10 mm thick, and respectively c = 40 and b = 60 mm wide, connected by fillets of radius R = 8 mm. Assume an allowable normal stress of 165 MPa. Find the allowable average normal stress using the material allowable normal stress and the stress concentration factor. Apply the definition of normal stress to find the allowable load.

75 Determine the geometric ratios and the stress concentration factor. b 60 mm R 8mm 1.50 0.20 K 1.82 c 40mm c 40mm

76 Determine the geometric ratios and the stress concentration factor. b c 60 mm R 8mm 1.50 0.20 40mm c 40mm Find the allowable average normal stress using the material allowable normal stress and the stress concentration factor. ave K max 165MPa 1.82 90.7MPa 1.82 Apply the definition of normal stress to find the allowable load. 3 P A ave 40 mm 10 mm 90.7 MPa 36.3 10 N K