Strength of materials - Chapter 3: Axial forces - Nguyễn Sỹ Lâm

Nội dung text: Strength of materials - Chapter 3: Axial forces - Nguyễn Sỹ Lâm

1. CHAPTER 3: AXIAL FORCES 3.1 Concept, Internal Forces 3.2 Normal Strain and Normal Stress 3.3 Stress – Strain diagram 3.4 Hooke’s law: Modulus of Elasticity 3.5 Deformation under axial loading 3.6 Safety Factor, Allowable Stress 3.7 Statically Indeterminate Systems
2. 3.1 CONCEPT, INTERNAL FORCES A structural member under axial loading means: Internal forces: only axial force N (along longitudinal axis of the bar) Method of Section: the same as discussed in previous chapter Imaginary cut is made through the body in the region where the internal loading is to be determined. The two parts are separated and a free body diagram of one of the parts is drawn. Only then applying equilibrium would enable us to relate the resultant internal force and moment to the external forces.
3. 3.1 CONCEPT, INTERNAL FORCES WHAT DOES IT LOOK LIKE?
4. 3.1 CONCEPT, INTERNAL FORCES EXAMPLES: DERTEMINE INTERNAL FORCES OF STRUCTURES BELOWS? (1) (2)
5. 3.2 NORMAL STRAIN AND NORMAL STRESS Strain has no unit’s since it is a ratio of length to length. Most engineering materials do not stretch very mush before they become damages, so strain values are very small figures. It is quite normal to change small numbers in to the exponent for 10-6( micro strain). Sign convention of normal stresses: x > 0 : tensile stress x < 0 : compression stress
6. 3.2 NORMAL STRAIN AND NORMAL STRESS
7. 3.3 STRESS – STRAIN DIAGRAM Ductile Materials
8. 3.3 STRESS – STRAIN DIAGRAM Stress Strain Diagram
9. 3.4 HOOKE’S LAW: MODULUS OF ELASTICITY • Below the yield stress  E E Youngs Modulus or Modulus of Elasticity • Strength is affected by alloying, heat treating, and manufacturing process but stiffness (Modulus of Elasticity) is not.
10. 3.5 DEFORMATION UNDER AXIAL LOAD POISSON RATIO • From Hooke’s Law:  P  E  E AE • From the definition of strain:   L • Equating and solving for the deformation, PL  AE • With variations in loading, cross-section or material properties, P L   i i i Ai Ei
11. 3.5 DEFORMATION UNDER AXIAL LOAD SOLUTION: • Apply free-body analysis to each • Divide the rod into three component to determine internal forces, components: 3 P1 60 10 lb 3 P2 15 10 lb 3 P3 30 10 lb • Evaluate total deflection, PiLi 1 P1L1 P2L2 P3L3   i AiEi E A1 A2 A3 1 60 103 12 15 103 12 30 103 16 6 29 10 0.9 0.9 0.3 75.9 10 3in. L1 L2 12 in. L3 16 in. 2 2 3 A1 A2 0.9 in A3 0.3 in  75.9 10 in.
12. 3.5 DEFORMATION UNDER AXIAL LOAD SOLUTION: Displacement of B: PL Free body: Bar BDE  B AE 60 103 N 0.3m 500 10-6 m2 70 109 Pa 514 10 6 m  0.514 mm   M 0 B B Displacement of D: 0 30kN 0.6m F 0.2m CD PL  D FCD 90kN tension AE  MD 0 90 103 N 0.4m -6 2 9 0 30kN 0.4m FAB 0.2m 600 10 m 200 10 Pa FAB 60kN compression 300 10 6 m  D 0.300 mm 
13. 3.5 DEFORMATION UNDER AXIAL LOAD Example 3.03
14. 3.6 SAFETY FACTOR – ALLOWABLE STRESS SELECTION OF F.S. 1. Variations that may occur in the properties of the member under considerations 2. The number of loadings that may be expected during the life of the structure /machine 3. Types of loadings that are planned for in the design, or that may occur in the future 4. Types of failures that may occur 5. Uncertainty due to the methods of analysis 6. Deterioration that may occur in the future because of poor maintenance / because of unpreventable natural causes 7. The importance of a given member to the integrity of the whole structure
15. 3.6 SAFETY FACTOR – ALLOWABLE STRESS 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.
16. 3.7 STATICALLY INDETERMINACY Example 3.04 Determine the reactions at A and B for the steel bar and loading shown, assuming a close fit at both supports before the loads are applied. SOLUTION: • 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 A due to applied loads and the reaction found at B.
17. 3.7 STATICALLY INDETERMINACY • Require that the displacements due to the loads and due to the redundant reaction be compatible,   L  R 0 1.125 109 1.95 103 R  B 0 E E 3 RB 577 10 N 577 kN • Find the reaction at A due to the loads and the reaction at B  Fy 0 RA 300 kN 600kN 577kN RA 323kN RA 323kN RB 577kN