Use Hooke s Law (as it applies in the uniaxial direction),

Transcription

1 0.6 STRSS-STRAIN RLATIONSHIP Use the principle of superposition Use Poisson s ratio, v lateral longitudinal Use Hooke s Law (as it applies in the uniaxial direction), x x v y z, y y vx z, z z vx y Copyright 0 Pearson ducation South Asia Pte Ltd

2 STRSS-STRAIN RLATIONSHIP (cont) Use Hooke s Law for shear stress and shear strain xy G xy yz G yz xz G xz Copyright 0 Pearson ducation South Asia Pte Ltd

3 STRSS-STRAIN RLATIONSHIP (cont) Relationship bet. and G : Use the state of pure shear : xy G v, Using Hooke s law / ( v ) ( v)/ / ( v ) ( v)/ ( v) max in plane G G ( v) Copyright 0 Pearson ducation South Asia Pte Ltd

4 STRSS-STRAIN RLATIONSHIP (cont) V dv Dilatation (i.e. volumetric strain e ) x y z V ( ) dx( ) dy( ) dz dxdydz x y z V ( x) dx( y) dy( z) dz dxdydz e x y z dv dxdydz v ( x y z ) Copyright 0 Pearson ducation South Asia Pte Ltd

5 STRSS-STRAIN RLATIONSHIP (cont) Copyright 0 Pearson ducation South Asia Pte Ltd For special case of hydrostatic loading, Where the right-hand side is defined as bulk modulus k, i.e. v e p p z y x 3 0 ~ v k v e p k z y x v e for most metals (plastic region)

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10 XAMPL 0.0 The copper bar in Fig. 0 4 is subjected to a uniform loading along its edges as shown. If it has a length a = 300 mm, b = 500 mm, and t = 0 mm before the load is applied, determine its new length, width, and thickness after application of the load. Take cu 0 GPa, v cu Copyright 0 Pearson ducation South Asia Pte Ltd

11 XAMPL 0.0 (cont) Solutions From the loading we have 800 MPa, 500 MPa, 0, 0 x y xy z The associated normal strains are determined from the generalized Hooke s law, x x v y v z v , , y z y x z z x y The new bar length, width, and thickness are therefore a' 300 b' 50 t' mm (Ans) mm (Ans) mm (Ans) Copyright 0 Pearson ducation South Asia Pte Ltd

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14 0.7 THORIS OF FAILUR (for ductile material) Maximum-shear-stress theory (or Tresca yield criterion) Yield will occur when max. shearing stress in a body exceeds the shear strength of the material. The shear strength is determined in a uniaxial tension test. max Y max min Y Copyright 0 Pearson ducation South Asia Pte Ltd

15 THORIS OF FAILUR (for ductile material) (cont) For general stress cases: Y 3 Y 3 Y For plane-stress cases: Y Y Y Copyright 0 Pearson ducation South Asia Pte Ltd

16 THORIS OF FAILUR (for ductile material) (cont) Maximum-distortion-energy theory (or von Mises criterion): Yield will occur when the distorsion energy reaches a critical value. ( )/3 ave 3 Change in volume: u v Change in shape: u d v e x y z ( x y z ) v For the third figure: e [( ave ) ( ave ) ( 3ave )] 0 Copyright 0 Pearson ducation South Asia Pte Ltd

17 THORIS OF FAILUR (for ductile material) (cont) Copyright 0 Pearson ducation South Asia Pte Ltd v u ( ) 6 ( ) ( )( ) 6 d v u u u v v 3 ) ( 6 ) ( ) 3( 3 3 v v v u ave ave ave v ( ) ( ) ( ) 6 ( ) ( ) ( ) 6 d v u v Strain nergy Density: / u zx zx yz yz xy xy z z y y x x u ] [ ) ( zx yz xy x z z y y x z y x G v u / u

18 THORIS OF FAILUR (for ductile material) (cont) u From tension test, i.e., 3 v,, Or substituting 3 3 u u d ave ave ave v ( ) ( 3) ( 3 ) 6, 0, 0 Y 3 v v ( 0) (0 0) (0 ) 6 3 d Y Y Y v v ud ( ) ( ) ( ) 6 3 ( ) ( ) ( ) 3 3 Y 3 3 Y 3 3 For state of plane stress, i.e., 3 0 Y Copyright 0 Pearson ducation South Asia Pte Ltd

19 THORIS OF FAILUR (for ductile material) (cont) For plane or biaxial-stress cases This theory fits experimental data a little bit better than Tresca s theory. Tresca s theory is slightly more conservative than this theory. Y Copyright 0 Pearson ducation South Asia Pte Ltd

20 THORIS OF FAILUR (for brittle material) (cont) Maximum-normal-stress theory (for materials having equal strength in tension and compression) Yield will occur when the max. normal stress reaches the ultimate stress. Maximum principle stress σ in the material reaches a limiting value that is equal to the ultimate normal stress the material can sustain when it is subjected to simple tension. Copyright 0 Pearson ducation South Asia Pte Ltd

21 THORIS OF FAILUR (for brittle material) (cont) For general stress cases: 3 ult ult ult For plane stress cases: ult ult Copyright 0 Pearson ducation South Asia Pte Ltd

22 THORIS OF FAILUR (for brittle material) (cont) Mohr s failure criterion (for materials having different strength in tension and compression) Perform 3 tests on the material to obtain the failure envelope. Circle A represents compression test results: σ = σ = 0, σ 3 = (σ ult ) c Circle B represents tensile test results: σ = (σ ult ) t, σ = σ 3 = 0 Circle C represents pure torsion test results, reaching the τ ult. A C B Copyright 0 Pearson ducation South Asia Pte Ltd

23 THORIS OF FAILUR (for brittle material) (cont) For plane-stress cases: Copyright 0 Pearson ducation South Asia Pte Ltd

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26 XAMPL 0.3 The solid shaft has a radius of 0.5 cm and is made of steel having a yield stress of σ Y = 360 MPa. Determine if the loadings cause the shaft to fail according to the maximumshear-stress theory and the maximum-distortion-energy theory. Copyright 0 Pearson ducation South Asia Pte Ltd

27 XAMPL 0.3 (cont) Solutions Since maximum shear stress caused by the torque, we have P 5 x A 0.5 xy kn/cm 9 MPa Tc J kn/cm 65.5 MPa Principal stresses can also be obtained using the stress-transformation equations, x y x y, xy σ 95.6 MPa and σ 86.6 MPa Copyright 0 Pearson ducation South Asia Pte Ltd

28 XAMPL 0.3 (cont) Solutions Since the principal stresses have opposite signs, the absolute maximum shear stress will occur in the plane, Thus, shear failure of the material will occur according to this theory. Using maximum-distortion-energy theory, 86.6 Y Y Using this theory, failure will not occur. Copyright 0 Pearson ducation South Asia Pte Ltd

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33 Chapter Objectives Design a beam to resist both bending and shear loads Design a shaft to resist both bending and torsional moments Copyright 0 Pearson ducation South Asia Pte Ltd

34 . BASIC for BAM DSIGN

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36 . PRISMATIC BAM DSIGN Basis of beam design Strength concern (i.e. provide safety margin to normal/shear stress limit) Stiffness concern (i.e. deflection limit See Chapter ) Section modulus: S req d where, S allow req' d F. S M Y max allow or U F. S Copyright 0 Pearson ducation South Asia Pte Ltd

37 PRISMATIC BAM DSIGN (cont) Choices of section: Steel sections e.g. AISC standard W 460 X 68 (height = mm weight = 68 kg/m) Copyright 0 Pearson ducation South Asia Pte Ltd

38 PRISMATIC BAM DSIGN (cont) Wood sections Nominal dimensions (in multiple of 5mm) e.g. 50 (mm) x 00 (mm) actual or dressed dimensions are smaller, e.g. 50 x 00 is 38 x 89. Built-up sections Copyright 0 Pearson ducation South Asia Pte Ltd

39 PRISMATIC BAM DSIGN (cont) Procedures: Shear and Moment Diagram Determine the maximum shear and moment in the beam. Often this is done by constructing the beam s shear force and moment diagrams. For built-up beams, shear and moment diagrams are useful for identifying regions where the shear and moment are excessively large and may require additional structural reinforcement or fasteners. Copyright 0 Pearson ducation South Asia Pte Ltd

40 PRISMATIC BAM DSIGN (cont) Normal Stress If the beam is relatively long, it is designed by finding its section modulus using the flexure formula, S req d = M max /σ allow. Once S req d is determined, the cross-sectional dimensions for simple shapes can then be computed, using S req d = I/c. If rolled-steel sections are to be used, several possible values of S may be selected from the tables in Appendix B. Of these, choose the one having the smallest cross-sectional area, since this beam has the least weight and is therefore the most economical. Copyright 0 Pearson ducation South Asia Pte Ltd

41 PRISMATIC BAM DSIGN (cont) Normal Stress (cont) Make sure that the selected section modulus, S, is slightly greater than S req d, so that the additional moment created by the beam s weight is considered. Shear Stress Normally beams that are short and carry large loads, especially those made of wood, are first designed to resist shear and then later checked against the allowable-bending-stress requirements. Using the shear formula, check to see that the allowable shear stress is not exceeded; that is, use τ allow V max Q/It. Copyright 0 Pearson ducation South Asia Pte Ltd

42 PRISMATIC BAM DSIGN (cont) Shear Stress (cont) If the beam has a solid rectangular cross section, the shear formula becomes τ allow.5(v max /A), q.7-5, and if the cross section is a wide flange, it is generally appropriate to assume that the shear stress is constant over the cross-sectional area of the beam s web so that τ allow V max /A web, where A web is determined from the product of the beam s depth and the web s thickness. Copyright 0 Pearson ducation South Asia Pte Ltd

43 PRISMATIC BAM DSIGN (cont) Adequacy of Fasteners The adequacy of fasteners used on built-up beams depends upon the shear stress the fasteners can resist. Specifically, the required spacing of nails or bolts of a particular size is determined from the allowable shear flow, q allow = VQ/I, calculated at points on the cross section where the fasteners are located. Copyright 0 Pearson ducation South Asia Pte Ltd