Successfully reported this slideshow.
We use your LinkedIn profile and activity data to personalize ads and to show you more relevant ads. You can change your ad preferences anytime.

Forces acting on the beam with shear force & bending moment


Published on

Forces acting on the beam with shear force & bending moment

Published in: Engineering
  • Sex in your area is here: ❤❤❤ ❤❤❤
    Are you sure you want to  Yes  No
    Your message goes here
  • Dating direct: ❤❤❤ ❤❤❤
    Are you sure you want to  Yes  No
    Your message goes here

Forces acting on the beam with shear force & bending moment

  1. 1. NAME ROLL NO. ENROLLMENT NO. •Taral Soliya 45 130030119109
  2. 2. Introduction Structural Members are usually classified according to the types of loads that they support (axially loaded bar, etc.) Now we will begin to look at beams, which are structural members subjected to lateral loads. The first beams we will investigate are called planar structures, because they lie in a single plane.
  3. 3. Introduction If all loads act in that same plane, and if all deflections act in that plane, then the plane is called the Plane of Bending. First we will look at shear forces and bending moments in beams. Once we know these, we can find the stresses, strains and deflections of a beam
  4. 4. Types of Beams Beams are usually described by the way they are supported. Simply supported beam – pin support at one end and a roller support at the other Cantilever beam – fixed at one end and free at the other. Beam with an Overhang – simply supported beam that projects beyond the support (similar to a cantilever).
  5. 5. Types of Beams Simply Supported Beam Overhanging Beam Cantilever Beam Continuous Beam Propped Cantilever Beam Fixed Beam
  6. 6. Types of Loads Concentrated Loads Couple Distributed Loads – Uniformly distributed – Linearly varying
  7. 7. Reactions Simple Beam FBD Cantilever Beam FBD Beam with Overhang FBD
  8. 8. Shear & Bending Moment in Beams If we have a beam that is loaded by a system of forces all in the y direction. The beam is classified as a simple beam (pin support at one end and a roller support at the other) The beam is in equilibrium with the application of these forces and its reactions.
  9. 9. Shear & Bending Moments To determine the internal effects of the applied loads we imagine a cutting plane to isolate either the left side or right side of the beam. In order for the isolated section of the beam to be in equilibrium a force (V) and couple (M) are required at the cut. – V is termed Shear force – M is termed Bending Moment
  10. 10. Shear & Bending Moments The shear V and Moment M are the force and couple applied to the left part of the beam by the right side of the beam to maintain equilibrium. Equilibrium of both sides is required because the entire beam is in equilibrium. The section can be made anywhere along the length of the beam. Shear and Moment are a function of the distance from the origin.
  11. 11. Shear & Moment Sign Convention The signs associated with the shear force and bending moment are defined in a different manner than the signs associated with forces and moments in static equilibrium. •The Shear Force is positive if it tends to rotate the beam section clockwise with respect to a point inside the beam section. •The Bending Moment is positive if it tends to bend the beam section concave facing upward. (Or if it tends to put the top of the beam into compression and the bottom of the beam into tension.) + shear + moment
  12. 12. Shear & Moment Sign Convention The figure shows the shear force V and Bending Moment M acting in positive directions. Notice there is a possibility for confusion with sign notation. When summing forces, the direction of V is in the negative y-direction, but is positive shear. We will deal with this confusion by always selecting the V and M directions as shown in the diagram. This will simplify the sign conventions.
  13. 13. Simple Beam A simply supported loaded beam An exaggerated view of the way the bending caused by the load is shown
  14. 14. Simple Beam If the beam were cut and we looked at the left side section For the sake of clarity, left out the vertical shear force which develops, but have shown horizontal forces (-Fx and + Fx). These forces develop since, as the beam bends, the top region of the beam is put into compression and the bottom region of the beam is put into tension.
  15. 15. Simple Beam As a result there are internal horizontal (x- direction) forces acting in the beam; however for every positive x-force, there is an equal and opposite negative x-force. Thus the net horizontal (x-direction) internal force in the beam section is zero. However, even though the actual x-forces cancel each other, the torque produced by these x-forces is not zero.
  16. 16. Simple Beam Looking at figure c and mentally summing torque about the center of the beam, we see that the horizontal x-forces cause a net toque - which we call the internal bending moment, M. This is the cause of the internal bending moment (torque) inside a loaded beam.
  17. 17. Bending deformation of a straight member Observation: - bottom line : longer - top line: shorter - Middle line: remain the same but rotate (neutral line)
  18. 18. Shear & Moment at Specified Sections To understand the concept of shear and moment at any section along the beam, it is best to calculate the values at arbitrarily sections along the beam. We will look at a simple example.
  19. 19. Example Here is a simply supported 20 ft. beam with a load of 10,000 lb. acting downward at the center of the beam. Due to symmetry, the two support forces will be equal, with a value of 5000 lb. Each This is the static equilibrium condition for the whole beam.
  20. 20. Section of the beam We will cut the beam a arbitrary distance (x) between 0 and 10 feet, and apply static equilibrium conditions to the left end section as shown We can do this since as the entire beam is in static equilibrium, then a section of the beam must also be in equilibrium. Here is the left section of the beam, x feet, long - where x is an arbitrary distance greater than 0 ft. and less than 10 ft. Notice if we just include the 5000 lb. external support force, the section of the beam is clearly not in equilibrium. Neither the sum of forces (translational equilibrium), nor the sum of torque (rotational equilibrium) will sum to zero - as required for equilibrium. Therefore, since we know the beam section is in equilibrium, there must be some forces and/or torque not accounted for.
  21. 21. Sectioning Continued In 2b, we have shown the missing force and torque When we cut the beam, the internal shear force and bending moment at that point then become an external force and moment (torque) acting on the section. These are shown 2b, and labeled V (shear force) and M (bending moment).
  22. 22. Equilibrium Conditions Sum of Forces in y-direction: + 5000 lb. -10,000 lb. - V = 0 , solving V2 = -5000 lb. Sum of Toque about left end: -10,000 lb * x (ft) -V * x (ft) + M = 0 Next substitute the value of V from the force equation into the torque equation : -10,000 lb * 10 ft. - (-5000 lb) * x (ft) + M = 0 , solving for M2 = [5000x (ft-lb.) - 100,000] ft-lb
  23. 23. What do I do with this? A useful way to visualize this information is to make Shear Force and Bending Moment Diagrams - which are really the graphs of the shear force and bending moment expressions over the length of the beam.
  24. 24. Shear & Moment Diagrams Axial load diagram Torque diagram
  25. 25. Shear & Moment Diagrams These are a quite useful way of visualizing how the shear force and bending moments vary through out the beam. We have completed our first Shear Force/Bending Moment Problem. We have determined the expressions for the shear forces and bending moments in the beam, and have made accompanying shear force and bending moment diagrams
  26. 26. Shear & Moment Diagrams Of special importance are the maximum and minimum values of these quantities. The Shear & Moment diagrams are useful ways of understanding how the shear forces and bending moments vary throughout the length of the beam. To provide a clear understanding of these diagrams, a detailed explanation of how to construct and interpret is given.
  27. 27. Concentrated Load If we start with a simple beam with a concentrated load. Find reactions Cut through beam to the left of the load P (a distance x from the left end), FBD – Equilibrium yields V and M for the left side of the beam Cut through the beam to the right of P, FBD – Equilibrium yields V and M for the right side of the beam.
  28. 28. Concentrated Load Shear Diagram The Shear Force diagram – Shear force at the end is equal to the reaction – Remains constant until the point of load P – Shear force decreases abruptly by an amount equal to P – In the right hand part, the shear force is constant but equal to the reaction at B.
  29. 29. Concentrated Load Moment Diagram The equations for moment are plotted below the Shear diagram of the beam The Bending Moment diagram – The bending moment in the left side increases linearly from zero at the support to P(ab/L) at the concentrated load x=a – In the right side, the bending moment is again a linear function of x, varying from P(ab/L) at x=a to zero at the support x=L. – The maximum bending moment is therefore P(ab/L), which occurs at the concentrated load.
  30. 30. Uniform Load Next look at a simple beam with a uniformly distributed load of constant intensity w. Find reactions, because the beam and its loading is symmetric, the reactions are equal to wL/2 The slope of the shear diagram at each point equals the negative distributed load intensity at each point  xw dx dV 
  31. 31. Uniform Load Therefore, the shear force and bending moment at a distance x from the left end are: These equations are valid through the length of the beam and can be plotted as shear and bending moment diagrams. – The maximum value occurs at the midpoint where dM/dx=V = 0. Mmax=wL2/8 The slope of the moment diagram at each point equals the shear at each point.   V dx dM xw dx dV  ,
  32. 32. Procedure 1. Draw the free-body-diagram of the beam with sufficient room under it for the shear and moment diagrams – (if needed, solve for support reactions first). 2. Draw the shear diagram under the free-body- diagram. – The change in shear V equals the negative area under the distributed loading. – Label all the loads on the shear diagram  dxxwV 
  33. 33. Procedure 3. Draw the moment diagram below the shear diagram. – The shear load is the slope of the moment and point moments result in jumps in the moment diagram. – The area under the shear diagram equals the change in moment over the segment considered (up to any jumps due to point moments). – Label the value of the moment at all important points on the moment diagram.  dxxVM 
  34. 34. Applications
  35. 35. Reference  How stuff works  Wikihow  Loads Acting on Beam (Book by E.D.Thomas)