MECHANICS OF MATERIALS

Transcription

1 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

2 Contents Stress & Strain: xial oading Normal Strain Stress-Strain Test Stress-Strain Diagram: Ductile Materials Stress-Strain Diagram: Brittle Materials Hooke s aw: Modulus of lasticity lastic vs. lastic Behavior Fatigue Deformations Under xial oading xample.01 Sample roblem.1 Static Indeterminacy xample.04 Thermal Stresses oisson s Ratio Generalized Hooke s aw Dilatation: Bulk Modulus Shearing Strain xample.10 Relation mong, n, and G Sample roblem.5 Composite Materials Saint-Venant s rinciple Stress Concentration: Hole Stress Concentration: Fillet xample.1 lastoplastic Materials lastic Deformations Residual Stresses xample.14,.15,.16 -

3 Stress & Strain: xial oading Suitability of a structure or machine may depend on the deformations in the structure as well as the stresses induced under loading. Statics analyses alone are not sufficient. Considering structures as deformable allows determination of member forces and reactions which are statically indeterminate. Determination of the stress distribution within a member also requires consideration of deformations in the member. Chapter is concerned with deformation of a structural member under axial loading. ater chapters will deal with torsional and pure bending loads. -

4 Normal Strain stress normal strain - 4

5 Stress-Strain Test - 5

6 Stress-Strain Diagram: Ductile Materials - 6

7 Stress-Strain Diagram: Brittle Materials - 7

8 Hooke s aw: Modulus of lasticity Below the yield stress Youngs Modulus or Modulus of lasticity Strength is affected by alloying, heat treating, and manufacturing process but stiffness (Modulus of lasticity) is not. - 8

9 lastic vs. lastic Behavior If the strain disappears when the stress is removed, the material is said to behave elastically. The largest stress for which this occurs is called the elastic limit. When the strain does not return to zero after the stress is removed, the material is said to behave plastically. - 9

10 Deformations Under xial oading From Hooke s aw: From the definition of strain: quating and solving for the deformation, With variations in loading, cross-section or material properties, i i i i i - 10

11 xample psi D 1.07in. d 0.618in. SOUTION: Divide the rod into components at the load application points. pply a free-body analysis on each component to determine the internal force Determine the deformation of the steel rod shown under the given loads. valuate the total of the component deflections. - 11

12 SOUTION: Divide the rod into three components: pply free-body analysis to each component to determine internal forces, lb lb lb valuate total deflection, i i i i i in in. 0.9 in 16in. 0.in in. - 1

13 Sample roblem.1 SOUTION: The rigid bar BD is supported by two links B and CD. ink B is made of aluminum ( = 70 Ga) and has a cross-sectional area of 500 mm. ink CD is made of steel ( = 00 Ga) and has a cross-sectional area of (600 mm ). For the 0-kN force shown, determine the deflection a) of B, b) of D, and c) of. pply a free-body analysis to the bar BD to find the forces exerted by links B and DC. valuate the deformation of links B and DC or the displacements of B and D. Work out the geometry to find the deflection at given the deflections at B and D. - 1

14 Sample roblem.1 SOUTION: Free body: Bar BD Displacement of B: B 6010 N 0.m m 7010 a m M F F B 0 M CD D 0 B 0 0kN 0.6m 0 0kN 0.4m F 90kN tension F CD B 60kN compression 0.m 0.m Displacement of D: D B 0.514mm 9010 N 0.4m m 0010 a m D 0.00mm - 14

15 Sample roblem.1 Displacement of D: BB DD BH HD 0.514mm 0.00 mm x 7.7 mm 00mm x x DD 0.00mm H HD 1.98mm mm 7.7 mm 1.98mm - 15

16 Static Indeterminacy Structures for which internal forces and reactions cannot be determined from statics alone are said to be statically indeterminate. structure will be statically indeterminate whenever it is held by more supports than are required to maintain its equilibrium. Redundant reactions are replaced with unknown loads which along with the other loads must produce compatible deformations. Deformations due to actual loads and redundant reactions are determined separately and then added or superposed. R 0-16

17 xample.04 Determine the reactions at and B for the steel bar and loading shown, assuming a close fit at both supports before the loads are applied. SOUTION: Consider the reaction at B as redundant, release the bar from that support, and solve for the displacement at B due to the applied loads. Solve for the displacement at B due to the redundant reaction at B. Require that the displacements due to the loads and due to the redundant reaction be compatible, i.e., require that their sum be zero. Solve for the reaction at due to applied loads and the reaction found at B. - 17

18 xample.04 SOUTION: Solve for the displacement at B due to the applied loads with the redundant constraint released, i i i i i m N 0.150m Solve for the displacement at B due to the redundant constraint, δ R i R ii m i i B m R B 6 m N 6 m - 18

19 xample.04 Require that the displacements due to the loads and due to the redundant reaction be compatible, R B R N 577kN R B 0 Find the reaction at due to the loads and the reaction at B F R y 0 R kn 00kN 600kN 577kN R R B kn 577kN - 19

20 Thermal Stresses temperature change results in a change in length or thermal strain. There is no stress associated with the thermal strain unless the elongation is restrained by the supports. Treat the adal support as redundant and apply the principle of superposition. T T thermal T T 0 expansion coef. 0 The thermal deformation and the deformation from the redundant support must be compatible. T 0 T T - 0

21 oisson s Ratio For a slender bar subjected to axial loading: x x y z 0 The elongation in the x-direction is accompanied by a contraction in the other directions. ssuming that the material is isotropic (no directional dependence), y z 0 oisson s ratio is defined as lateralstrain y n axial strain x z x - 1

22 Generalized Hooke s aw For an element subjected to multi-axial loading, the normal strain components resulting from the stress components may be determined from the principle of superposition. This requires: 1) strain is linearly related to stress ) deformations are small With these restrictions: x y z n x y n z n x y n z n n x y z -

23 Shearing Strain cubic element subjected to a shear stress will deform into a rhomboid. The corresponding shear strain is quantified in terms of the change in angle between the sides, xy f xy plot of shear stress vs. shear strain is similar the previous plots of normal stress vs. normal strain except that the strength values are approximately half. For small strains, xy G xy yz G yz zx G where G is the modulus of rigidity or shear modulus. zx -

24 xample.10 rectangular block of material with modulus of rigidity G = 90 ksi is bonded to two rigid horizontal plates. The lower plate is fixed, while the upper plate is subjected to a horizontal force. Knowing that the upper plate moves through 0.04 in. under the action of the force, determine a) the average shearing strain in the material, and b) the force exerted on the plate. SOUTION: Determine the average angular deformation or shearing strain of the block. pply Hooke s law for shearing stress and strain to find the corresponding shearing stress. Use the definition of shearing stress to find the force. - 4

25 Determine the average angular deformation or shearing strain of the block. xy tan xy 0.04in. in. xy 0.00rad pply Hooke s law for shearing stress and strain to find the corresponding shearing stress. xy G 9010 psi 0.00rad 1800psi xy Use the definition of shearing stress to find the force. 1800psi8in..5in. 610 lb xy 6.0kips - 5

26 Sample roblem.5 circle of diameter d = 9 in. is scribed on an unstressed aluminum plate of thickness t = /4 in. Forces acting in the plane of the plate later cause normal stresses x = 1 ksi and z = 0 ksi. For = 10x10 6 psi and n = 1/, determine the change in: a) the length of diameter B, b) the length of diameter CD, c) the thickness of the plate, and d) the volume of the plate. - 6

27 SOUTION: pply the generalized Hooke s aw to find the three components of normal strain. x y z x psi n x n x n y y n n z 1ksi 0 0ksi in./in. n z y in./in. z in./in. 1 valuate the deformation components. B d in./in. 9in. x C d D t y t z B in in./in. 9in. C D in in./in. 0.75in. t Find the change in volume e V x ev y z V in /in in in 0.187in - 7