part-15/#Difference between sfd and bmd for point load and udl.(#with Pdf note 📚)

 Difference between sfd and bmd for point load and udl

◼️ Bending moment and shering force diagrams:

 In drawing the bending moment diagram (B.M.D) and shearing force diagram ( S.F.D), the following sign conventions as stated in the previous article are followed (fig. 6.2). The shear force is positive if the net resultant external force to the left of a section is upward. Positive bending moment produce compression on the top fibers of the beam and negative bending moment produces tension on the top fibres of the beam.

◼️ Simple supported beam carrying a concentrated load at mid - span.

      Taking moment about R2, we have (fig. 6.3 a)

                 RA × l = w× (l/2)

                 RA = RB = (w/2)

      Taking a section at a distance x from R1, for values of x laying between RA and W, the shear force is W/2 and it is positive in nature. For values of x laying between W RB, the shear force is (W– W/2) = W/2, and it is negative in nature. Hence the S.F. diagram is shown in Fig. 6.3 (b).

    Similarly for values of x laying between RA and W, the bending moment at any point is (W/2)x, and is positive in nature. This bending moment is maximum at x = l/2, i.e. at the mid-span and is given by,

                  Mmax = (WL/4)

     For values of x laying between W and RB the bending moment is 

    (W/2)x – W(x–l/2) 

     = –(Wx/2) + (WL/2) = (W/2)l–x

     This bending moment is also positive. Hence the B.M. diagram is as shown in Fig. 6.3(c).

◼️Simply supported beam carrying a concentrated load:

      Consider a beam simply supported at the ends and carrying a concentrated load W at a distance a from left hand support as shown in fig. 6.4(a). Taking moment about RB, we get

                   RA(a+b) = Wb

                   RA = (Wb/a+b)

                   RB = (Wa/a+b)

      Take a section at a distance x from RA. When x lies between RA and W and W, the shear force is RA and is positive in nature. When x lies between W and RB, the shear force is 

                  W – RA = Wa/(a+b) = Wa/l

and is negative in nature. Hence the S.F. diagram is as shown in Fig. 6.4(b).

       Similarly for x laying between RA and W bending moment is,

                  M = RAx

 or             M = Wb/(a+b)x = Wbx/l

       This is the equation of a straight line. When x = a, the bending moment is maximum and becomes,

                  Mmax = Wab/(a+b)

                            = Wab/l

    This is a positive bending moment. For x laying between W and RB, the bending moment is,

                 M = RAx – W(x–a)

                      = Wb/(a+b).x – W(x–a)

                     = Wa – Wa/(a+b).x

                     = Wa(1–x/l)

     This also represents a straight line. When x = a, M = Wab/(a+b) and when x= a+b, M = 0. This bending moment is also positive. Hence the B.M. diagram is shown in Fig. 6.4(c).

◼️Simply supported beam carrying many concentrated load:

      Consider a beam simply supported at the ends and carrying many concentrated loads as shown in fig. 6.5(a). Taking moment about R2, we get

           R1 × l = W1(l – a)+ W2(l – b)+ W3(l –c)

           R1 = W1(1– a/l)+ W2(1– b/l)+ W3(1– c/l)

           R2 = (w1+ W2 + W3) – R1

[To be continued in next part]

****************************************

Stay connected for more updates

Thanks for watching

Please like, comments and share

Official YouTube channel : Click to subscribe