Shear Force and Bending Moment Diagrams in Simply supported Beams

Q.1. Draw the shear force and bending moment diagrams in a simply supported beam carrying a point load W  as shown in figure.

Solution : 

Step.1

Find the unknown support reactions HA, RA and RB at supports as shown in the figure.

HA = 0; RA = Wb/L

RB = Wa/L

Step.2 

Take a section x-x in between A and C points at a distance x form support A

Using the following sign convention for shear force and bending moments and write shear force (Fx) and bending moment (Mx) equations.

+ve Sign Convention for Shear Force and Bending Moment 

Consider left side of the section, the limits of  x is valid in between A and just left side of C only.

 ie. 0 < x < a

considering sign convention (Left side), the S.F and B.M. equations are

Vx = + RA = + Wb/L                  ---  (1)

Mx = + (RA).x =  + (Wb/L). x     ---  (2)

substituting x values in the above equations,

When x= 0,

Vx =  The shear force at A, VA = + Wb/L

Mx = The bending moment at A, MA = 0

When x= a,

Vx =  The shear force at C, Vc = + Wb/L

Mx = The bending moment at C, Mc = + Wab/L

Take another section x-x in between C and B points at a distance x form support B ( we may also take a section form support A ) 

The limits of x is  0 < x < b

considering sign convention (Right side), the S.F and B.M. equations are

Vx = + RB = - Wa/L                  --- (3)

Mx = + (RB).x =  + (Wa/L). x    --- (4)

substituting x values in the above equations,

When x= 0,

Vx =  The shear force at B, VB = - Wa/L

Mx = The bending moment at B, MB = 0

When x= b,

Vx =  The shear force at C, Vc = - Wa/L

Mx = The bending moment at C, Mc = + Wab/L

Plot the shear force and bending moment diagrams by using the above values. 

Note:

1.Bending Moment is maximum at the point where the Shear Force is zero or where it changes direction from +ve to -ve or vice versa.

2.For simply supported beam with point load at the centre (a=b=l/2), the maximum bending moment will be at the centre i.e. wl/4. 

The variation in bending moment is triangular.

Q.2. Draw the shear force and bending moment diagrams in a simply supported beam carrying a Uniformly distributed load w/m as shown in figure.

Solution : 

Step.1

Find the unknown support reactions RA and RB at supports as shown in the figure. Due to symmetrical loading,the support reactions are

RA = RB = R = wL/2 (upwords)

Step.2 

Take a section x-x in between A and B points at a distance x form support A

The limits of x is  0 < x < L

considering sign convention (Left side of the section),  the S.F and B.M. equations are

Vx = + R - wx = + wL/2 - wx                 --- (1)

Mx = + (R).x =  + (wL/2). x - wx(x/2)   --- (2)

substituting x values in the above equations,

When x= 0,

Vx = The shear force at A, VA =  +wL/2

Mx = The bending moment at B, MA = 0

When x= L/2,

Vx =  The shear force at mid point C, 

Vc = 0

Mx = The bending moment at C, Mc = + wL.L/8

When x= L,

Vx =  The shear force at point B, 

VB = - wL/2

Mx = The bending moment at B, MB = 0

Plot the shear force and bending moment diagrams by using the above values.

Note:

1.Bending Moment is maximum at the point where the Shear Force is zero or where it changes direction from +ve to -ve or vice versa.

2.For simply supported beam with uniformly distributed load w/m throughout the beam, the maximum bending moment will be at the centre i.e. (wL)L/8. 

3.The variation of shear force is linear and the variation in bending moment is a parabolic curve.

Assignment :

Q.1 Draw shear force and bending moment diagram of simply supported beam carrying point load as shown in figure below.

Q.2. Draw shear force and bending moment diagram of simply supported beam carrying point load and udl as shown in figure below.