Strength of materials - Axially loaded members

on the deformations in the structure as well as the stresses induced under loading. Statics analyses alone are not sufficient. • Considering structures as deformable ones allows us to determinate the member forces and reactions which are statically indeterminate. • Determination of the stress distribution within a member also requires the consideration of deformations in the member. • Chapter 3 is concerned with the stress and deformation of a structural member under axial loading. Later chapters will deal with torsional and pure bending loads. 1/10/2013 4 3.1. Introduction Axially loaded members are structural components subjected only to axial force (tension or compression) • Prismatic bar: Straight structural member with the same cross- section throughout its length • Axial force: Load directed along the axis of the member • Axial force can be tensile or compressive 1/10/2013 5 1/10/2013 6 3.1. Introduction 1/10/2013 7 3.1. Introduction  Axial force diagram 0 ...zZ N   Using the method of section , the internal axial force is obtained from the equilibrium as a function of coordinate z  Kinematic assumptions Before deformation After deformation 1/10/2013 8  Kinematic assumptions 3.2. Normal stress and normal strain  Normal stress z z N A   z const   – normal stress at any point on the cross-sectional area Nz – internal resultant normal force A – cross-sectional area 1. The axis of the member remains straight 2. Cross sections which are plane and are perpendicular to the axis before deformation, remain plane and remain perpendicular to the axis after deformation. And the cross sections do not rotate about the axis 1/10/2013 9 3.2. Normal stress and normal strain  Elongation of the bar: Consider the bar, which has a cross-sectional area that gradually varies along its length L. The bar is subjected to concentrated loads at its ends and variable external load distributed along its length. 0 ( ) ( ) L N z dz L EA z     1/10/2013 10 3.2. Normal stress and normal strain  Elongation of the bar – constant load and cross-sectional area: zN LL EA    zN L EA      Normal Strain – elongation per unit length EA – stiffness of axially loaded bar 1/10/2013 11   zi i N EA const   n zi i i 1 i N L L EA   3.2. Normal stress and normal strain 3.4. Poisson’s Ratio • For a slender bar subjected to axial loading: 0zz x y E       • The elongation in the x-direction is accompanied by a contraction in the other directions. Assuming that the material is isotropic (no directional dependence), 0x y   • Poisson’s ratio is defined as lateral strain axial strain yx z z          Poisson’s Ratio 1/10/2013 12 • Strength of a material can only be determined by experiment • One test used by engineers is the tension or compression test • This test is used primarily to determine the relationship between the average normal stress and the average normal strain in common engineering materials, such as metals, ceramics, polymers and composites 3.3. Tension and Compression Test Performing the tension or compression test • Specimen of material is made into “standard” shape and size • Before testing, 2 small punch marks are identified along the specimen’s length • Measurements are taken for both the specimen’s initial x-sectional area A0 and the gauge-length distance L0; between the two marks 1/10/2013 13 1Performing the tension or compression test • Seat the specimen into a testing machine as shown below - The machine will stretch the specimen at a slow constant rate until the breaking point - At frequent intervals during test, the data is recorded of the applied load P. 3.3. Tension and Compresion Test • The Elongation δ = L − L0 is measured by using either a caliper or an extensometer • δ is used to calculate the normal strain in the specimen • Sometimes, the strain can also be read directly by using an electrical-resistance strain gauge 1/10/2013 3.3. Tension and Compresion Test • A stress-strain diagram is obtained by plotting the various values of the stress and corresponding strain in the specimen Conventional stress-strain diagram • Using recorded data, we can determine the nominal or engineering stress by P A0 σ = • Likewise, the nominal or engineering strain is found directly from strain gauge reading, or by δ L0  = By plotting σ (ordinate) against  (abscissa), we get a conventional stress-strain diagram 1/10/2013 15 3.3. Tension and Compression Test Conventional stress-strain diagram • This Figure shows the characteristic stress-strain diagram for steel, a commonly used material for structural members and mechanical elements 1/10/2013 16 3.3. Tension and Compresion Test Conventional stress-strain diagram Elastic behavior. • the straight line • The stress is proportional to the strain, i.e., linearly elastic • Upper stress limit, or proportional limit; σpl • If the load is removed upon reaching the elastic limit , the specimen will return to its original shape Yielding. • The material deforms permanently; yielding; plastic deformation • Yield stress, σY • Once the yield point is reached, the specimen continues to elongate (strain) without any increase in load 17 3.3. Tension and Compresion Test Conventional stress-strain diagram Figure 3-4 Strain hardening. • Ultimate stress, σu • While the specimen is elongating, its z-sectional area will decrease • Decrease in area is fairly uniform over entire gauge length Necking. • At ultimate stress, x-sectional area begins to decrease in a localized region • As a result, a constriction or “neck” tends to form in this region as the specimen elongates further • The Specimen finally breaks at fracture stress, σf 1/10/2013 18 3.3. Tension and Compresion Test Stress – Strain Diagram: Ductile materials • Defined as any material that can be subjected to large strains before it rupture, e.g., mild steel • Such materials are used because of its capacity of absorbing shock or energy so that it, will exhibit a large deformation before failing • Ductility of material is to report its percent elongation or percent reduction in area at time of fracture 1/10/2013 19 1Ductile materials • Percent elongation is the specimen’s fracture strain expressed as a percent • Percent reduction in area is defined within necking region as • Most metals do not exhibit constant yielding behavior beyond the elastic range, e.g. aluminum • It does not have a well-defined yield point, thus it is standard practice to define its yield strength using a graphical procedure called the offset method Percent elongation = Lf − L0 L0 (100%) Percent reduction in area = A0 − Af A0 (100%) 3.3. Tension and Compression Test 1Stress – Strain Diagram: Brittle Materials • Materials that exhibit little or no yielding before failure are referred to as brittle materials, e.g., gray cast iron • Brittle materials do not have a well-defined tensile fracture stress, since the appearance of an initial cracks in a specimen is quite random 3.3. Tension and Compression Test 1/10/2013 3.3. Tension and Compresion Test • Most engineering materials exhibit a linear relationship between the stress and strain with the elastic region • Discovered by Robert Hooke in 1676 using springs, known as Hooke’s law σ = E • E represents the constant of proportionality, also called the modulus of elasticity or Young’s modulus • E has units of stress, i.e., pascals, MPa or GPa. Hooke’s Law: Modulus of Elasticity 1/10/2013 22 3.3. Tension and Compresion Test • If the strain disappears when the stress is removed, the material is said to behave elastically. • When the strain does not return to zero after the stress is removed, the material is said to behave plastically. • The largest stress for which this occurs is called the elastic limit. Elastic vs. Plastic Behavior 1/10/2013 23 1/10/2013 24 3.3. Tension and Compresion Test 3.5. Allowable Stress – Factor of Safety • When designing a structural member or mechanical element, the stress in it must be restricted to safe level • Choose an allowable load that is less than the load the member can fully support • One method used is the factor of safety (F.S.) F.S. = Ffail Fallow • If the load applied is linearly related to the stress developed within the member, then F.S. can also be expressed as: F.S. = σfail σallow F.S. = τfail τallow • In all the equations, F.S. is chosen to be greater than 1, to avoid potential for failure • Specific values will depend on the type of material used and its intended purpose 1/10/2013 25 3.5. Allowable Stress – Factor of Safety • Fail Stress: 0 – stress coresponding lossing ability to sustain a load   0 . fail allow n F S          • Allowable Stress: • Axially Loaded Members –True stress must satisfy this condition:   allow P A     • Three main problems: 1. Determine if the condition of strength is satisfied or not. 2. Determine the required cross-sectional area. 3. Determine the maximum applied load. 1/10/2013 26 3.6. Statically Indeterminate Problem 1/10/2013 27 1/10/2013 28 Sample Problem 3.1 The A-36 steel bar shown in the Figure is made from two segments having the cross-sectional areas of AAB=1cm 2; ABD=2cm 2. Draw the axial force diagram and determine the maximum normal stress and the vertical displacement of end A 1. Use the method of section 1/10/2013 29 Sample Problem 3.1 2. Normal stress 215 15( / ) 1 AB AB AB N kN cm A     27 3.5( / ) 2 BC BC BC N kN cm A     29 4.5( / ) 2 CD CD CD N kN cm A       2 max 15( / )kN cm  3. Displacement 0.0127( )BC BC CD CDAB ABA AD AB BC CD AB BC CD N l N lN l L cm EA EA EA             Since the result is positive, the bar elongates and so the displacement at A is upward 1/10/2013 30 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 mm2. Link CD is made of steel (E = 200 GPa) and has a cross- sectional area of (600 mm2). For the 30-kN force shown, determine the deflection a) of B, b) of D, and c) of E. SOLUTION: • Apply a free-body analysis to the bar BDE to find the forces exerted by links AB and DC. • 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. Sample Problem 3.2 1/10/2013 31 Displacement of B:       m10514 Pa1070m10500 m3.0N1060 6 926- 3      AE PL B  mm 514.0B Displacement of D:       m10300 Pa10200m10600 m4.0N1090 6 926- 3      AE PL D  mm 300.0D Free body: Bar BDE     ncompressioF F tensionF F M AB AB CD CD B kN60 m2.0m4.0kN300 0M kN90 m2.0m6.0kN300 0 D         SOLUTION: Sample Problem 3.2 1/10/2013 32 Displacement of D:   mm 7.73 mm 200 mm 0.300 mm 514.0       x x x HD BH DD BB  mm 928.1E   mm 928.1 mm 7.73 mm7.73400 mm 300.0       E E HD HE DD EE   Sample Problem 3.2 1/10/2013 33 b a B A2 F2 F1 A1 C D Consider the bar made from two segments having the cross-sectional areas of A1 and A2 . 1. Draw the axial force diagram. 2. Determine the max. normal stress. 3. Determine the displacement of end D. with F1=10kN; F2=25kN; A1=5cm 2; A2=8cm 2 a=b=1m; E=2.104kN/cm2 SOLUTION 1. Using method of section, the internal axial force in each segments are: z1 F1 DNCD 1 10CDN F kN  a F2 F1 C D z2 NBC 1 2 15BCN F F kN    Sample Problem 3.3 34 b a B A2 F2 F1 A1 C D 10 N kN 1 5 Axial force diagram: 2. Determine the maximum normal stress 2 1 10 2( / ) 5 CD CD N kN cm A     2 2 15 1,875( / ) 8 BC BC N kN cm A       22( / ) max kN cm  3. The displacement of point D 2 1 . .BC CD D BD BC CD N b N a w L l l EA EA        2 2 2 4 1 15.10 10.10 0,0625.10 ( ) 2.10 8 5 Dw cm         =>displacement toward right Sample Problem 3.3 35 EA EA  P D C E  h The steel bars CD an CE with Young’s modulus E, each have a cross-sectional area of A, are joined at C with a pin. Determine the axial forces in each bars and the displacement of point C cause by load P. SOLUTION: 1. Determine axial forces: Using method of join. FBD of join C. 1 2 1 2 0 sin sin 0X N N N N           1 2 1 0 os o 0 2 os Y N c N c P N c P            (1) (2) 1 2(1) (2) 2cos P N N     P C N1 N2   X Y Sample Problem 3.4 1/10/2013 36 2. Displacement of join C: EA EA  D C E  h C’ From geometry: 1L yC 1 C L y cos    1 1 1 N L L EA   1 2cos P N   And: 1 cos h L   1 22 Ph L EAcos    32 C Ph y EAcos    Sample Problem 3.4 with: 37 Bài 2.4: Three steel rods, each have a cross- sectional area of A=5cm2, jointly support the load P= 50kN . Determine the axial forces in these rods and the displacement of join C. Using E = 2.104kN/cm2, H=4m SOLUTION: 1. Axial forces: FBD of join C 1 3 1 3 0 sin30 sin30 0o oX N N N N         (1) (2) 1 3 2 1 2 0 ( ). os30 0 3 oY N N c N P N N P           N3 30o 30o C N1 N2 P Compatibility o 1 3 2 2 3 os30 2 L L L c L       1 2 1 2 2 .3 3 2 43 N H N H N N EAEA    A A 30o 30 o C A P H (3) A A 30o 30 o C A P H 2L 1L Sample Problem 3.5 1/10/2013 38 Sample Problem 3.6 1/10/2013 39 Sample Problem 3.6 1/10/2013 40 Sample Problem 3.7 1/10/2013 41 Sample Problem 3.7 3 2 9 . 20.10 .0,04 0,0002 0,01 200.10 . . 4 AC B AC P l l EA        => Appear reaction FB at fixed end B • Compatibility condition: 0ABl    0.40.8 0.0002BB F PF EA EA        2 9 0.010.12 0.04 0,0002 200 10 4 BF P         4.05( )BF kN  1/10/2013 42 Sample Problem 3.7 THANK YOU FOR YOUR ATTENTION ! 1/10/2013 43