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SFD PRESENTATION by 10.01.03.133
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SFD PRESENTATION by 10.01.03.133

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PowerPoint presentation on SHEAR FORCE DIAGRAM by ENAMUL NASIR, I.D.: 10.01.03.133

PowerPoint presentation on SHEAR FORCE DIAGRAM by ENAMUL NASIR, I.D.: 10.01.03.133

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Transcript

  • 1. WELCOME TO MY
  • 2. ByEnamul Nasir 10.01.03.133 4th Year 2nd Semester Dept. Of Civil Engineering Ahsanullah University Of Science and Technology
  • 3. •What is “SHEAR FORCE”? Shear force: is the algebraic sum of the vertical forces acting to the left or right of a cut section along the span of the beam.
  • 4. •What is „SHEAR FORCE DIAGRAM‟?  Shear force diagram(SFD): The graphical representation of shear force in which ordinates represents shear force and the abscissa represents the position of the section is called “shear force diagram” or it can be defined as “ A shear force diagram is the one which shows the variation shear force along the length of a beam section.”
  • 5. What is the importance of SFD in case of designing structures?  Shear force diagrams are analytical tools used in conjunction with structural analysis to help perform structural design by determining the value of shear force at a given point of a structural element such as a beam. These diagrams can be used to easily determine the type, size, and material of a member in a structure so that a given set of loads can be supported without structural failure.
  • 6. Sign convention for SFD drawing?  The normal convention used in most engineering applications is to label a positive shear force one that spins an element clockwise (up on the left, and down on the right).
  • 7.  Also for simplification in solving many structural problems the shear force at a section will be considered positive when the resultant of the forces to the left section is upwards, or to the right of the section is downward.
  • 8.  IMPORTANT POINTS FOR DRAWING SHEAR FORCE DIAGRAM 1. Consider the left or right portion of the section, 2. Add the forces ( including reactions ) normal to the beam on one of the portion. [ If right portion of the section is chosen, a force on the right portion acting downwards is positive while a force acting upward is negative, if left section is chosen then sign convention will be vice versa. ] 3. The positive values of shear force are plotted above base line, and negative values below base line.
  • 9. 4. 5. The shear force diagram will increase or decrease suddenly by a vertical straight line at a section where there is a vertical point load. The shear force between any two vertical loads will be constant and hence the shear force diagram between any two vertical loads will be horizontal.  WHAT IS THE USE OF SFD? To determine the internal shear forces at any given point on a rigid body.
  • 10. PRACTICAL APPLICATION OF SHEAR FORCE DIAGRAM The figure below is referring the practical application of shear force diagram in case of designing shear reinforcement of a beam.
  • 11. Some Examples of SFD Simply Supported beam with UDL
  • 12. Simply supported beams with triangular load
  • 13.  Cantilever Beam
  • 14.  Cantilever beam carrying Three Loads
  • 15.  Simply Supported beam carrying double point load and UDL
  • 16. A detail example of drawing SHEAR FORCE DIAGRAM  A simply supported beam overhanging on one side is subjected to a U.D.L. of 1 kN/m. Sketch the shear force diagram. Solution: Consider a section (X – X’) at a distance x from end C of the beam.
  • 17. To draw the Shear Force Diagram, RA and RB has to be defined, By taking moment of all the forces about point A ∑MA =0 RB × 3 – w/2 × (4)2 = 0 RB = 1 × (4)2 / 2 × 3 = 8/3 kN For static equilibrium; ∑Fy = 0 RA + RB – 1 × 4 = 0 RA = 4 – 8/3 = 4/3 kN
  • 18. Now, To draw SFD we need S.F. at all salient points: Taking a section between C and B, SF at a distance x from end C. we have,
  • 19. Fx = + w.x kN At x = 0; FC = 0 x = 1 m; FB just right = 1 × 1 = + 1 kN Now, again taking section between B and A, at a distance x from end C, the SF is: Fx = x – 8/3 When, x = 1 m; FB = 1 – 8/3 = –5/3 kN = –1.67 kN At x = 4 m; FA = 4 – 8/3 = + 4/3 kN = + 1.33 kN
  • 20. When The S.F. becomes zero; Fx = x – 8/3 = 0 ⇒ x = 2.67 m (The sign is taken positive taken when the resultant force is in downward direction on the RHS of the section).
  • 21.  Now the Final Shear Force Diagram The SFD can be drawn by joining the ordinates at salient point by straight line, as defined by governing equations.
  • 22. NECESSITY OF SFD IN CASE OF PRESTRESSED CONCRETE
  • 23. THANK YOU ALL

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