1 / 74

Fundamental of Thermal-Fluid Sciences

Fundamental of Thermal-Fluid Sciences. Fluid : chapter 9 – 12 Thermodynamic : 1 – 7 Appendix 1: Property tables and charts. Tensile Strength. Tensile strength measures the force required to pull something such as rope, wire, or a structural beam to the point where it breaks.

yhayden
Download Presentation

Fundamental of Thermal-Fluid Sciences

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  1. Fundamental of Thermal-Fluid Sciences Fluid : chapter 9 – 12 Thermodynamic : 1 – 7 Appendix 1: Property tables and charts

  2. Tensile Strength Tensile strength measures the force required to pull something such as rope, wire, or a structural beam to the point where it breaks. The tensile strength of a material is the maximum amount of tensile stress that it can be subjected to before failure. The definition of failure can vary according to material type and design methodology. This is an important concept in engineering, especially in the fields of material science, mechanical engineering and structural engineering

  3. There are three typical definitions of tensile strength: Yield strength - The stress a material can withstand without permanent deformation. This is not a sharply defined point. Yield strength is the stress which will cause a permanent deformation of 0.2% of the original dimension. Ultimate strength - The maximum stress a material can withstand. Breaking strength - The stress coordinate on the stress-strain curve at the point of rupture.

  4. CHAPTER OBJECTIVES • Review important principles of statics • Use the principles to determine internal resultant loadings in a body • Introduce concepts of normal and shear stress • Discuss applications of analysis and design of members subjected to an axial load or direct shear 4

  5. CHAPTER OUTLINE Introduction Equilibrium of a deformable body Stress Average normal stress in an axially loaded bar Average shear stress Allowable stress Design of simple connections 5

  6. 1.1 INTRODUCTION A branch of mechanics • It studies the relationship of • External loads applied to a deformable body, and • The intensity of internal forces acting within the body • Deals with the behavior of solid bodies subjected to various types of loading • Study body’s stability when external forces are applied to it. • A thorough understanding of mechanical behavior is essential for the safe design of all structures. 6

  7. 8

  8. 1.2 EQUILIBRIUM OF A DEFORMABLE BODY External loads • A body can be subjected to different types of external load; surface force, body force. • Surface forces – caused by the direct contact of one body with the surface of another. • Linear distributed force (forces distributed over the a.o.c) • Concentrated force (a.o.c smaller than total surface area) 9

  9. 1.2 EQUILIBRIUM OF A DEFORMABLE BODY External loads • FR of w(s) = area under distributed loading curve, acts through centroid C. • Centroid C (or geometric center) • Body force (e.g., weight) – one body exerts a force on another without direct physical contact eg effect of earth’s gravitation. 10

  10. 1.2 EQUILIBRIUM OF A DEFORMABLE BODY Support reactions • The surface forces that develop at the supports or points of contact between bodies 11

  11. ∑F = 0 ∑MO = 0 1.2 EQUILIBRIUM OF A DEFORMABLE BODY Equations of equilibrium • For equilibrium • balance of forces • balance of moments • Draw a free-body diagram to account for all forces acting on the body • Apply the two equations to achieve equilibrium state 12

  12. 1.2 EQUILIBRIUM OF A DEFORMABLE BODY Internal resultant loadings • Define resultant force (FR) and moment (MRo) acting within the body (3D): • Normal force, N(acts perpendicular to area) • Shear force, V(lies in the plane of area) 13

  13. 1.2 EQUILIBRIUM OF A DEFORMABLE BODY Internal resultant loadings • Torsional moment or torque, T(develop when external loads tend to twist 1 segment of the body with respect to another) • Bending moment, M (cause by external loads that tend to bend the body) 14

  14. 1.2 EQUILIBRIUM OF A DEFORMABLE BODY Internal resultant loadings • For coplanar loadings: • Normal force, N • Shear force, V • Bending moment, M 15

  15. 1.2 EQUILIBRIUM OF A DEFORMABLE BODY Internal resultant loadings • For coplanar loadings: • Apply ∑ Fx = 0 to solve forN • Apply ∑ Fy = 0 to solve for V • Apply ∑ MO = 0 to solve for M 16

  16. 1.2 EQUILIBRIUM OF A DEFORMABLE BODY Procedure for Analysis • Method of sections • Choose segment to analyze • Determine Support Reactions • Draw free-body diagram for whole body • Apply equations of equilibrium 17

  17. 1.2 EQUILIBRIUM OF A DEFORMABLE BODY Procedure for analysis • Free-body diagram • Keep all external loadings in exact locations before “sectioning” • Indicate unknown resultants, N, V, M, and T at the section, normally at centroid C of sectioned area • Coplanar system of forces only include N, V, and M • Establish x, y, z coordinate axes with origin at centroid 18

  18. 1.2 EQUILIBRIUM OF A DEFORMABLE BODY Procedure for analysis • Equations of equilibrium • Sum moments at section, about each coordinate axes where resultants act • This will eliminate unknown forces N and V, with direct solution for M (and T) • Resultant force with negative value implies that assumed direction is opposite to that shown on free-body diagram 19

  19. Review of Statics • The structure is designed to support a 30 kN load • The structure consists of a beam and rod joined by pins (zero moment connections) at the junctions and supports • Perform a static analysis to determine the internal force in each structural member and the reaction forces at the supports 20

  20. Conditions for static equilibrium: Structure Free-Body Diagram • Structure is detached from supports and the loads and reaction forces are indicated • Ay and Cy can not be determined from these equations 21

  21. Consider a free-body diagram of the boom; substitute into the structure equilibrium equation • Results: Reaction forces are directed along boom and rod Component Free-Body Diagram • In addition to the complete structure, each component must satisfy the conditions for static equilibrium 22

  22. Joints must satisfy the conditions for static equilibrium which may be expressed in the form of a force triangle: Method of Joints • The boom and rod are 2-force members, i.e., the members are subjected to only two forces which are applied at member ends • For equilibrium, the forces must be parallel to an axis between the force application points, equal in magnitude, and in opposite directions 23

  23. Can the structure safely support the 30 kN load? • From the material properties for steel, the allowable stress is Stress Analysis • From a statics analysis FAB= 40 kN (compression) FBC = 50 kN (tension) • At any section through member BC, the internal force is 50 kN with a force intensity or stress of dBC = 20 mm • Conclusion: the strength of member BC is adequate 24

  24. 1.3 STRESS Concept of stress • To obtain distribution of force acting over a sectioned area • Assumptions of material: • It is continuous (uniform distribution of matter) • It is cohesive (all portions are connected together) 25

  25. 1.3 STRESS Concept of stress • Consider the sectioned area, subdivided into small areas; ΔA in figure below • Small finite force, ΔF acts on ΔA, replaced by 3 components ΔFx, ΔFy, ΔFz. • ΔA→0,The components →0. The quotient of the force & area (called stress), describes the intensity of the internal force on a specific plane. 26

  26. ΔFz ΔA lim ΔA →0 σz = 1.3 STRESS Normal stress • Intensity of force, or force per unit area • Symbol used for normal stress, is σ (sigma) • Tensile stress: normal force “pulls” or “stretches” the area element ΔA • Compressive stress: normal force “pushes” or “compresses” area element ΔA 27

  27. ΔFx ΔA lim ΔA →0 τzx = ΔFy ΔA lim ΔA →0 τzy = 1.3 STRESS Shear stress • Intensity of force, or force per unit area • Symbol used for shear stress is τ (tau) 28

  28. 29

  29. 30

  30. 1.4 AVERAGE NORMAL STRESS IN AXIALLY LOADED BAR Examples of axially loaded bar • Usually long and slender structural members • Truss members, hangers, bolts • Prismatic means all the cross sections are the same 31

  31. 1.4 AVERAGE NORMAL STRESS IN AXIALLY LOADED BAR Assumptions • Uniform deformation: Bar remains straight before and after load is applied, and cross section remains flat or plane during deformation • In order for uniform deformation, force P be applied along centroidal axis of cross section 32

  32. + FRz = ∑ Fxz ∫ dF = ∫Aσ dA P = σA P A σ = 1.4 AVERAGE NORMAL STRESS IN AXIALLY LOADED BAR Average normal stress distribution σ= average normal stress at any point on cross sectional area P = internal resultant normal force A = x-sectional area of the bar 33

  33. 1.4 AVERAGE NORMAL STRESS IN AXIALLY LOADED BAR Equilibrium • Consider vertical equilibrium of the element • When the bar is stretched by the force P, the resulting stresses are tensile stresses • If the force P cause the bar to be compressed, we obtain compressive stresses 34

  34. 1.4 AVERAGE NORMAL STRESS IN AXIALLY LOADED BAR • Sign convention for normal stresses is: • (+) for tensile stresses and • (-) for compressive stresses • Because the normal stress σ is obtained by dividing the axial force by the cross–sectional area, it has units of force per unit of area. • In SI units: • Force is expressed in newtons (N) and area in square meters (m2). A N/m2 is a pascals (Pa). 35

  35. 1.4 AVERAGE NORMAL STRESS IN AXIALLY LOADED BAR Maximum average normal stress • For problems where internal force P and x-sectional A were constant along the longitudinal axis of the bar, normal stress σ = P/A is also constant • If the bar is subjected to several external loads along its axis, change in x-sectional area may occur • Thus, it is important to find the maximum average normal stress • To determine that, we need to find the location where ratio P/A is a maximum 36

  36. 1.4 AVERAGE NORMAL STRESS IN AXIALLY LOADED BAR Maximum average normal stress • Draw an axial or normal force diagram (plot of P vs. its position x along bar’s length) • Sign convention: • P is positive (+) if it causes tension in the member • P is negative (−) if it causes compression • Identify the maximum average normal stress from the plot 37

  37. 1.4 AVERAGE NORMAL STRESS IN AXIALLY LOADED BAR Procedure for Analysis Average normal stress • Use equation of σ = P/A for x-sectional area of a member when section subjected to internal resultant force P 38

  38. 1.4 AVERAGE NORMAL STRESS IN AXIALLY LOADED BAR Procedure for Analysis Axially loaded members • Internal Loading: • Section member perpendicular to its longitudinal axis at pt where normal stress is to be determined • Draw free-body diagram • Use equation of force equilibrium to obtain internal axial force P at the section 39

  39. 1.4 AVERAGE NORMAL STRESS IN AXIALLY LOADED BAR Procedure for Analysis Axially loaded members • Average Normal Stress: • Determine member’s x-sectional area at the section • Compute average normal stress σ = P/A 40

  40. EXAMPLE 1.6 Bar width = 35 mm, thickness = 10 mm Determine max. average normal stress in bar when subjected to loading shown. 41

  41. EXAMPLE 1.6 (SOLN) Internal loading Normal force diagram By inspection, largest loading area is BC, where PBC = 30 kN 42

  42. PBC A 30(103) N (0.035 m)(0.010 m) = 85.7 MPa σBC = = EXAMPLE 1.6 (SOLN) Average normal stress 43

  43. 1.5 AVERAGE SHEAR STRESS • Shear stress is the stress component that act in the plane of the sectioned area. • Consider a force F acting to the bar • For rigid supports, and F is large enough, bar will deform and fail along the planes identified by AB and CD • Free-body diagram indicates that shear force, V = F/2 be applied at both sections to ensure equilibrium 44

  44. P A τavg= 1.5 AVERAGE SHEAR STRESS Average shear stress over each section is: τavg = average shear stress at section, assumed to be same at each pt on the section V = internal resultant shear force at section determined from equations of equilibrium A = area of section 45

  45. 1.5 AVERAGE SHEAR STRESS • Case discussed above is example of simple or direct shear • Caused by the direct action of applied load F • Occurs in various types of simple connections, e.g., bolts, pins, welded material 46

  46. 1.5 AVERAGE SHEAR STRESS Single shear • Steel and wood joints shown below are examples of single-shear connections, also known as lap joints. • Since we assume members are thin, there are no moments caused by F 47

  47. 1.5 AVERAGE SHEAR STRESS Single shear • For equilibrium, x-sectional area of bolt and bonding surface between the two members are subjected to single shear force, V = F • The average shear stress equation can be applied to determine average shear stress acting on colored section in (d). 48

  48. 1.5 AVERAGE SHEAR STRESS Double shear • The joints shown below are examples of double-shear connections, often called double lap joints. • For equilibrium, x-sectional area of bolt and bonding surface between two members subjected to double shear force, V = F/2 • Apply average shear stress equation to determine average shear stress acting on colored section in (d). 49

  49. Single Shear Double Shear Shearing Stress Examples 50

  50. 1.5 AVERAGE SHEAR STRESS Procedure for analysis Internal shear • Section member at the pt where the τavg is to be determined • Draw free-body diagram • Calculate the internal shear force V Average shear stress • Determine sectioned area A • Compute average shear stress τavg = V/A 51

More Related