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SHEAR FORCE AND BENDING MOMENT DIAGRAM FOR SIMPLY SUPPORTED BEAMS

2.1 Shear force and Bending moment diagram for simply supported beam with a point Load at the centre of the span:

                 
                            (fig. 2.1)

Consider a simply supported beam  of span L carrying a point load w at a distance of L/2 from both the ends. 

The reactions at both the supports will be equal to W/2 as the load is acting at mid-span. 

Consider a section X-X between A & C at a distance of x from end A. The S.F at section X-X will be equal to the resultant forces acting on left portion of this section since we have considered the left portion of the section. But the resultant force acting on left portion of the section is W/2 acting upwards. 

But the resultant force acting upwards in left portion is considered positive as per sign conventions. 

Let Fx  = S.F at X-X section
Fx  = +W/2

Hence S.F between A & C is constant & equal to +W/2

Now consider a section between C &  B at a distance of x from end A.
The resultant force acting on left portion of this section will be equal to  (W/2 -  W) = - W/2

This force remains constant between C & B,

Hence S.F between C & B is equal to 
- W/2

    Hence shear force at C changes from +W/2 to - W/2

     SFD is shown below, 

                              (fig. 2.2)

    The bending moment at any section between A & C at a distance of x from end A is given by,

    Mx  = RA  (x)  = +W/2 (x)___________(1)

    This B.M will be positive as the moment of all the forces in left portion about this section is clockwise,  i.e convexity will be at the top. 

    At A,  x = 0
    B.M at A = 0
    At C,  x=L/2
    Mc  = W/2(L/2) = WL /4
    • From eq (1) it is evident that the Bending moment varies according to straight line law between A & C. 
    B.M at any section between C & B at a distance of x from end A is given by, 

    Mx  = RA (x) - W(x- L/2)
          = W/2(x)-W(x-L/2)  
           =Wx/2 -  Wx -  WL/2
           = (Wx -  2Wx)/2 - WL/2
           = (-Wx -  WL)/2 = - W(L-x)/2

    At B,  x = L
    MB  = - W(L-L)/2 =0

    • Eq(2) shows that B.M between C &  B varies by a straight line law. 
    BMD is shown below, 

                                    (fig. 2.3)

    2.2 Shear force and bending moment diagram for simply supported beam with an eccentric point load:

                         
                             (fig. 2.4)

    Consider a simply supported beam  of span L carrying a point load w at a distance of 'a'  from end A & 'b' from end B. 

    Using equilibrium equations for finding the reactions RA  &  RB  at the supports A &  B respectively. 

    Taking moments of all the forces about point A we get, 

    ΣMA  = 0
    RB x L -  W x a = 0
    RB = Wa/L

    ΣVy = 0

    RA + RB  -  W = 0
    RA + Wa/L -  W = 0

    RA  = W -  Wa/L = W(1 -  a/L) 
         = W(L-a)/L = Wb/L               
     = W -  Wa/L = W(1 -  a/L) 
         = W(L-a)/L = Wb/L          (  •   L-a = b)


    Consider a section X-X between A & C at a distance of x from end A. The S.F at section X-X will be equal to the resultant forces acting on left portion of this section since we have considered the left portion of the section. But the resultant force acting on left portion of the section is 
    +Wb/L  acting upwards. 

    But the resultant force acting upwards in left portion is considered positive as per sign conventions. 

    Let Fx  = S.F at X-X section
    Fx = +Wb/L

    Hence S.F between A & C is constant & equal to +Wb/L

    Now consider a section between C &  B at a distance of x from end A.
    The resultant force acting on left portion of this section will be equal to  (Wb/L- W) 
    = W(b-L) /L = - W(L-b)/L = - Wa/L  
                                                    (• L-a=b)

    This force remains constant between C & B,

    Hence S.F between C & B is equal to - Wa/L 


    SFD is shown below,

                               (fig. 2.5)

    The bending moment at any section between A & C at a distance of x from end A is given by,

    Mx  = RA (x)  = +Wb/L(x)___________(1) 
    This B.M will be positive as the moment of all the forces in left portion about this section is clockwise,  i.e convexity will be at the top. 

    At A,  x = 0
    B.M at A = 0
    At C,  x=a
    Mc  = Wb/L (a)  = Wab/L

    • From eq (1) it is evident that the Bending moment varies according to straight line law between A & C. 

    B.M at any section between C & B at a distance of x from end A is given by, 
    Mâ‚“ = RA  (x) - W(x-a) 
     = Wbx/L- Wx - Wa_________________(2) 

    At B,  x =L
    MB  = WbL/L - WL -  Wa
         = W(b - L- a) = W(b-b)  =0

    • Eq(2) shows that B.M between C &  B varies by a straight line law. 

    BMD is shown below, 

                                (fig. 2.6)

    PROBLEM:

    Draw SFD & BMD for the simply supported beam shown below with given loading. 

                             (fig. 2.7)

    Solution:

    Using equilibrium equations for calculating the reactions,

    Taking moments of all the forces about point A we get, 

    ΣMA  = 0
    RD x 10 - 200 x 10 - 100 x 6 -  50 x 2 = 0
    10RD = 2700
    RD = 270 N

    Now adding all the vertical forces we get, 

    ΣVy = 0
    RA + RD  - 200-100-50 = 0
    RA + 270 -  350 = 0
    RA = 350-270 = 80 N 

    Shear force diagram:

    S.F at A = RA = +80N
    S.F just to the left of B = +80N
    S.F just to the right of B = 80 - 50
     = +30N
    S.F just to the left of C = +30
    S.F just to the right of C = 30 -100
     = -70N
    S.F just to the left of D = - 70
    S.F just to the right of D = - 70 - 200 + 270=0 


    SFD is shown below, 
                           (fig. 2.8)

    Bending Moment diagram:

    MA = 0

    MB  = 80 x 2 = +160 Nm

    MC = 80 x 6 - 50 x 4 = +280 Nm

    MD = 80 x 10 - 50 x 8 - 100 x 4 = 0

    BMD is shown below, 

                              (fig. 2.9)

    2.3 Shear force and bending moment diagram for simply supported beam carrying a uniformly distributed load:

                         
                                 (fig. 2.10)

    Consider a simply supported beam carrying a uniformly distributed load "w/unit length " over it's entire span 

    The support reactions will be equal and their magnitude will be equal to half the total load acting on the entire span. 

    RA  = RB = Total load/2 = wL/2

    Shear force diagram:


    Consider a section X-X at a distance of x from left end A. The shear force at this section will be given by, 

    Fx  = +RA  -  wx = +wL/2 -  wx
    (varies by a straight line law) 

    At A,  x = 0
    FA  = +wL/2

    At C,  x = L/2
    Fc = wL/2 -  wL/2 = 0

    At B, x = L
    FB  = wL/2 -  wL = - wL/2

    SFD is shown below, 

                                  (fig. 2.11)

     Bending Moment diagram:

    The bending moment at the section X-X will be given by, 

    Mâ‚“  = +RA (x)  -  w(x) (x/2) 
           = wL/2 (x) - wx² /2  

    At A, x=0
    MA = 0
    At C, x=L/2
    Mc  = WL/2 (L/2) -  WL² /8
        = +WL² /8
    At B, x=L
     MB  = WL/2 (L)  -  W(L)² /2
           = WL²/2-WL²/2 = 0


    BMD is shown below, 

                                (fig. 2.12)

    Comments

    1. SmithNovember 16, 2021 at 2:12 PM

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      ReplyDelete

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