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Lecture Strength of Materials I: Chapter 2 - PhD. Tran Minh Tu

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Chapter 2 - Axial force, shear force and bending moment. The following will be discussed in this chapter: Internal stress resultants; relationships between loads, shear forces, and bending moments; graphical method for constructing shear and moment diagrams; normal, shear force and bending moment diagram of frame.

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Nội dung Text: Lecture Strength of Materials I: Chapter 2 - PhD. Tran Minh Tu

  1. STRENGTH OF MATERIALS 1/10/2013 TRAN MINH TU - University of Civil Engineering, 1 Giai Phong Str. 55, Hai Ba Trung Dist. Hanoi, Vietnam
  2. CHAPTER 2 Axial Force, Shear Force and Bending Moment 1/10/2013 2
  3. Contents 2.1. Introduction 2.2. Internal Stress Resultants 2.3. Example 2.4. Relationships between loads, shear forces, and bending moments 2.5. Graphical Method for Constructing Shear and Moment Diagrams 2.6. Normal, Shear force and bending moment diagram of frame 1/10/2013 3
  4. 2.1. Introduction - Structural members are usually classified according to the types of loads that they support. - Planar structures: if they lie in a single plane and all loads act in that same plane. 2.1.1. Support connections. - Structural members are joined together in various ways depending on the intent of the designer. The three types of joint most often specified are the pin connection, the roller support, and the fixed joint 1/10/2013 4
  5. 2.1. Introduction - Types of supports - Pin support: prevents the translation at the end of a beam but does not prevent the rotation. A Idealized HA 1/10/2013 V 5 A
  6. 2.1. Introduction - Roller support: prevents the translation in the vertical direction but not in the horizontal direction, and does not prevent the rotation. A V A 1/10/2013 6
  7. 2.1. Introduction - Fixed (clamped) support: the bar can neither translate nor rotate. A MA HA V A 1/10/2013 7
  8. 2.1. Introduction 1/10/2013 8
  9. 2.1. Introduction 2.1.2. Types of beams 1/10/2013 9
  10. 2.2. Internal Stress Resultants In general, internal stress resultants (internal forces) consist of 6 components Mx x • Nz – Normal force Mz Qx • Qx, Qy – Shear forces NZ z • Mx, My – Bending moments My Qy • Mz – Torsional moment y  Planar structures: if they lie in a single plane and all loads act in that same plane => Only 3 internal stress x Mx resultants exert on this plane (zoy) . • Nz – axial force (N); • Qy – shear force (Q); NZ z • Mx - bending moment (M) Qy y 1/10/2013 10
  11. 2.2. Internal force Resultants  To determine the internal force resultants => Using the method of sections. N N N N Q Q N 1/10/2013 11
  12. 2.2. Internal force Resultants  Sign convention: • Axial force: positive when outward N N N N of an element, negative when inward of an element • Shear force: positive when acts clockwise against an element, negative when acts counterclockwise against an element • Bending moment: positive when compresses the upper part of the beam and negative when compresses the lower part of the beam 1/10/2013 12
  13. 2.2. Axial, Shear and Moment diagram 2.1.3. Axial, Shear and Moment diagram • Because of the applied loadings, the beams develop an internal shear force and bending moment that, in general, vary from point to point along the axis of the beam. In order to properly design a beam it therefore becomes necessary to determine the maximum shear and moment in the beam. • One way to do this is expressing N, Q and M as the functions of their arbitrary position z along the beam’s axis. These axial, shear and moment functions can then be plotted and represented by graph called the axial, shear and moment diagram 1/10/2013 13
  14. 2.2. Axial, Shear and Moment diagram Moment diagram to be plotted by side, which it is in tension N, Q z z M 1/10/2013 14
  15. 2.3. Example Example 2.1: Draw the shear and moment diagram for F the beam shown in the figure. C Solution: a b 1. Support reactions VA VB M A  VB  a  b   Fa  0 Fa  VB   a  b M B  VA  a  b   Fb  0 Fb  VA   a  b Recheck: Y  0 1/10/2013 15
  16. 2.3. Example Segment AC 1 F 2 Section 1 – 1:0  z1  a A B N 0 1 C 2 a b Fb VA Y  Q  VA  0  Q  VA   a  b M M VB N N Fbz1 z1  0 M  M  V z A 1  0  M  V z A 1   a  b  VA Q z2 Q VB Segment BC Section 2 – 2: 0  z2  b N 0 Fa Y  Q  VB  0  Q  VB    a  b Faz2 M 0  M  VB z2  0  M  VB z2   a  b 1/10/2013 16
  17. 2.3. Example Fb F AC : Q   a  b Fa C BC : Q    a  b a b VA VB Fbz1 AC : M  Fb  a  b a+b Faz2 + F BC : M   a  b Q N Fa Comment 1: a+b The section on which the concentrated force acts, the M shear force diagram has “jump” Fab 1/10/2013 a+b 17
  18. 2.3. Example 1 Example 2.2: Draw the shear and q moment diagram for the beam shown in the figure. Solution: 1 L 1. Support reactions VA VB M q.l Symmetrically:  VA  VB  q 2 N Or: Q ql 2 q.l VA z  A B 2 0 M  V .l   VB  2 qz 2 ql 2 q.l  M 0 M  VA z  2  0  M B  VA .l  2  0  VA  2 ql q  M  .z  .z 2 2 2 2. Internal force resultant’s functions: ql 1/10/2013 Segment 1-1 (0 ≤ z  L)  Y  Q  qz  VA  0 Q 2  q.z 18
  19. 2.3. Example z  0  QA  qL q ql 2 Q  q.z 2 qL L z  L  QB   2 VA VB qL/2 ql q z  0  M A  0 M  .z  .z 2 + 2 2 z  L  M B  0 Q qL L M'   qz M '  0  z  2 2 qL2 L/2 M ''  q  0  M max  M  z  L / 2  8 qL/2  Comment 2 The section on which the shear M force is equal to zero then the bending moment is maximum. qL2/8 1/10/2013 19
  20. 2.3. Example – example 2.3 1. Support reactions: 1 M 2 M A VB .(a  b)  M  0 C M  VB  ab VA a 1 b 2 VB  M B  VA.(a  b)  M  0 M M  VA  M ab 2. Stress resultants: z1 Q Q z2 VB VA AC: Section 1-1 ( 0 ≤ z1  a) M Qy  VA   ab M x  VA .z Section 2-2 ( 0 ≤ z2  b) M Qy  VA   ab 1/10/2013 M x  VB .z2 20
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