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Shear and Moment Functions

The design of a beam required knowing the shear and moment value at different locations. Shear and moment functions will allow the designer to determine the moment and shear value at any point at the member under investigation. 






Determining the function of shear and moment at a point x from an arbitrary point can be determined by the method of section. An imaginary cut will be done at x then the function of shear and moment will be defined. Imaginary section or cut should be located at x, not at a specific location. The cut at x will allow for derivate the functions of shear and moment for the whole zone, but the section to a particular place will determine the value of shear and moment at this location only. 

Moment and shear functions will be valid for the zone with the same magnitude and distribution of loads. If the loads change or removed from a part of the beam. A new function shall be made for this zone. In figure 1, from A to B, we need to define functions for moment and shear forces. From B to C distributed load removed, we need to derivate other functions. At point c there is a concentrated load and this required derivating new functions for moment and shear from point C to D.

Figure 1


Procedure for Analysis


  • Determine reactions at supports.
  • Specify ur coordination system and the origin of the coordination system like point A in figure 1. also identify the zones like A to B and B to C and C to D as in figure 1.
  • Section the beam at x distance for each zone and draw the free body diagram then determine the function of  V and M using equilibrium equations.



  • The results can be checked by noting that dM/dx=V



  • Determine the shear and moment in the beam shown in Figure 2 as a function of x

    Figure 2


    first, we should determine the support reaction as shown in figure 3
    Figure 3







    in this example, there only one distributed the load along the beam with the same slope.

    to determine the load intensity at point x, we will use triangle similarities 

    2/30=w/x
    w=x/15

    ΣFy=0

    +30-0.5*(x/15)*x-V=0
    V=30-x^2/30
    V=30-0.0333X^2

    ΣM=0
    M+0.5*(x/15)*(X)(x/3)-30*x+600=0
    M=30X-X^3/90-600
    M=30X-0.011X^3-600
    Figure 4



    Try to determine the shear and moment in the beam shown in Figure 5 as a function of x????
    Figure 5






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