Lecture Strength of Materials I: Chapter 3 - PhD. Tran Minh Tu

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Chapter 3 - Axially loaded members. The following will be discussed in this chapter: Normal stress and normal strain, tension and compression test, poisson’s ratio, shearing strain, allowable stress – factor of safety, statically indeterminate problem. STRENGTH OF MATERIALS 1/10/2013 TRAN MINH TU - University of Civil Engineering, Giai Phong Str 55, Hai Ba Trung Dist Hanoi, Vietnam CHAPTER Axially loaded members 1/10/2013 Contents 3.1 Introduction 3.2.Normal Stress and Normal Strain 3.3 Tension and Compression Test 3.4 Poisson’s ratio 3.5 Shearing Strain 3.6 Allowable Stress – Factor of Safety 3.7 Statically Indeterminate Problem 1/10/2013 3.1 Introduction • 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 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 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 3.1 Introduction • Prismatic bar: Straight structural member with the section throughout its length same cross- • Axial force: Load directed along the axis of the member • Axial force can be tensile or compressive Axially loaded members are structural components subjected only to axial force (tension or compression) 1/10/2013 3.1 Introduction 1/10/2013 3.1 Introduction  Axial force diagram Using the method of section , the internal axial force is obtained from the equilibrium as a function of coordinate z Z   N  Kinematic assumptions z  Before deformation After deformation 1/10/2013 3.2 Normal stress and normal strain  Kinematic assumptions The axis of the member remains straight 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 not rotate about the axis  Normal stress Nz z  A  z  const  – normal stress at any point on the cross-sectional area Nz – internal resultant normal force A – cross-sectional area 1/10/2013 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 L N ( z )dz L     EA( z ) 1/10/2013 3.2 Normal stress and normal strain  Elongation of the bar – constant load and cross-sectional area: Nz L L    EA  Normal Strain – elongation per unit length  Nz   L EA 1/10/2013 EA – stiffness of axially loaded bar 10 Sample Problem 3.1 Normal stress  AB  BC  CD N AB 15    15(kN / cm2 ) AAB N BC    3.5(kN / cm2 ) ABC  max  15(kN / cm2 ) NCD 9    4.5(kN / cm2 ) ACD Displacement  A  LAD   AB   BC   CD N ABl AB N BC lBC NCDlCD     0.0127(cm) EAAB EABC EACD Since the result is positive, the bar elongates and so the displacement at A is upward 1/10/2013 29 Sample Problem 3.2 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 The rigid bar BDE is supported by two of B and D links AB and CD Link AB is made of aluminum (E = 70 • Work out the geometry to find the deflection at E given the GPa) and has a cross-sectional area deflections at B and D of 500 mm2 Link CD is made of steel (E = 200 GPa) and has a crosssectional area of (600 mm2) For the 30-kN force shown, determine the deflection a) of B, b) of D, and c) of E 1/10/2013 30 Sample Problem 3.2 SOLUTION: Free body: Bar BDE Displacement of B: B  PL AE   60 103 N 0.3 m   50010-6 m2 70 109 Pa   514 10 m MB  0  30 kN  0.6 m   FCD  0.2 m FCD  90 kN tension  B  0.514 mm  Displacement of D: D  PL AE  30 kN  0.4 m   FAB  0.2 m  90 103 N 0.4 m   60010-6 m2 200109 Pa  FAB  60 kN compression  300 10 m  MD   D  0.300 mm  1/10/2013 31 Sample Problem 3.2 Displacement of D: BB BH  DD HD 0.514 mm 200 mm   x  0.300 mm x x  73.7 mm EE  HE  DD HD E 0.300 mm  400  73.7 mm 73.7 mm  E  1.928 mm  E  1.928 mm  1/10/2013 32 Sample Problem 3.3 Consider the bar made from two segments having the cross-sectional areas of A1 and A2 Draw the axial force diagram Determine the max normal stress Determine the displacement of end D A2 A1 F1 F2 B C b D a with F1=10kN; F2=25kN; A1=5cm2; A2=8cm2 a=b=1m; E=2.104kN/cm2 F1 NCD SOLUTION Using method of section, the internal axial force in each segments are: NCD  F1  10kN NBC z1 D F1 F2 C z2 D a N BC  F1  F2  15kN 1/10/2013 33 Sample Problem 3.3 A2 Axial force diagram: Determine the maximum normal stress  CD A1 F1 F2 B NCD 10    2(kN / cm2 ) A1 C b N BC 15  BC     1,875(kN / cm2 ) A2   max  2(kN / cm2 ) The displacement of point D D a 10 N kN N BC b NCD a wD  LBD  lBC  lCD   EA2 EA1  15.102 10.102  2 wD     0, 0625.10 (cm)   2.10   =>displacement toward right 34 Sample Problem 3.4 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 D E  SOLUTION: Determine axial forces: Using method of join FBD of join C  X    N sin   N  N1  N 2 P (1)  N1cos  P (2) (1)  (2)  N1  N  Y   N1 N2 C P 2cos  h  C sin    Y   N1cos  N2co  P  EA EA P X 35 Sample Problem 3.4 Displacement of join C: From geometry: with: D E L1 yC  cos  EA  N1 L1 L1  EA EA h  C And: P N1  cos  h L1  cos  Ph  L1  EAcos  1/10/2013  L1 yC C’ Ph  yC  EAcos  36 Sample Problem 3.5 Bài 2.4: Three steel rods, each have a crosssectional area of A=5cm2, jointly support the load P= 50kN Determine the axial forces in these rods and the displacement of join C Using H E = 2.104kN/cm2, H=4m SOLUTION: Axial forces: FBD of join C  X    N sin 30 o  N1  N3 30 30 A o C P 30o L2 2 N1H N H   N1  N (3) EA 3EA A A N2 (2) Compatibility L1  L3  L2cos30o  o  N3 sin 30o  (1)  Y   ( N1  N3 ).cos30o  N2  P   3N1  N  P A A o 30 30 H 30o N3 N1 C P C P o A L2 L1 37 Sample Problem 3.6 1/10/2013 38 Sample Problem 3.6 1/10/2013 39 Sample Problem 3.7 1/10/2013 40 Sample Problem 3.7  B  l AC P.l AC   EA 20.103.0,04 200.109.  0,0002 0,01 => Appear reaction FB at fixed end B • Compatibility condition: l AB   FB  0.8   FB  P   0.4    0.0002 EA EA 0.12 FB  0.04P  0,0002  200  10    0.012  FB  4.05(kN ) 1/10/2013 41 Sample Problem 3.7 1/10/2013 42 THANK YOU FOR YOUR ATTENTION ! 1/10/2013 43 ... is made of steel (E = 200 GPa) and has a crosssectional area of (600 mm2) For the 30 -kN force shown, determine the deflection a) of B, b) of D, and c) of E 1/10/20 13 30 Sample Problem 3. 2 SOLUTION:.. .CHAPTER Axially loaded members 1/10/20 13 Contents 3. 1 Introduction 3. 2.Normal Stress and Normal Strain 3. 3 Tension and Compression Test 3. 4 Poisson’s ratio 3. 5 Shearing Strain 3. 6 Allowable... does not return to zero after the stress is removed, the material is said to behave plastically 1/10/20 13 23 3 .3 Tension and Compresion Test 1/10/20 13 24 3. 5 Allowable Stress – Factor of Safety
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