Shear Force and Bending Moment Diagrams due to Concentrated moments

A concentrated moment has no “up and down” force, it does not cause any change in the magnitude of the shear diagram at its point of application. That does not mean that they do not influence the shear diagram, because they do. They influence of concentrated moment is in the support reactions, which in turn influences the shear diagram. Thus you will see no change in the shear diagram at the point of application of a concentrated moment. 

Concentrated moments changes the magnitude of the moment diagram to a sudden “jump” at their points of application and equal to the magnitude of moments.

 

Problem 1. 

Draw the shear force and bending moment diagrams in a simply supported beam carrying a concentrated moment M as shown in figure.

Solution : 

Step.1

Find the unknown support reactions Ax, Ay and By at supports as shown in the figure.

Along direction x, there is no imposed loads applied to the structure.
Thus the first equilibrium equation is: ∑ Fx = 0
Ax = 0
Along direction y, there is no imposed loads, we have only support reactions.
Thus the second equilibrium equation  ∑ Fy = 0

Ay + By = 0 

Ay  = - By  --- (a)
Taking moments about A and apply the third equilibrium equation ∑ Mxy = 0
M + By. L = 0 --- (b)
By = - M/L (downwards reaction)
Substitute the By value in equation (a)
Ay =  - By = - (- M/L) = M/L (upwards reaction)

Step.2 : consider segment AB

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

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 = + Ay = + M/L  ---(1)
Mx = + (Ay).x = + (M/L)x ---(2)

substituting x values in the above equations,
When x= 0,
Vx = The shear force at A, VA = + M/L
Mx = The bending moment at A, MA = 0
When x= a,
Vc = The shear force at C,
Vc = + M/L
Mx = The bending moment at C,
Mc = +(M/L).a

Step.3 : consider segment BC
Take a section x-x in between C and B points at a distance x form B

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 < b
considering sign convention (Right side), the S.F and B.M. equations are
Vx = + Ay = + M/L  ---(3)
Mx = - (Ay).x = - (M/L)x ---(4)

substituting x values in the above equations,
When x= 0,
Vx = The shear force at A, Vb = + M/L
Mx = The bending moment at A, Mb = 0
When x= b,
Vc = The shear force at C,
Vc = + M/L
Mx = The bending moment at C,
Mc = - (M/L).b

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

(The negative values are plotted to scale below a horizontal reference line and positive values are plotted above the reference line.)

Shear force and Bending moment diagrams

Note: 

1.There is no change in the shear force diagram at the point of application of a concentrated moment.

2.At a point of Concentrated moment the sudden change in the magnitude of the bending moment diagram and is equal to concentrated moment

(ie. Mc = +(M/L)a - (- M/L)b = M/L (a+b) = M )

Assignment : 

Q.1. Draw the shear force and bending moment diagrams in a simply supported overhanging beam carrying a concentrated clockwise moment 6kN/m and a point load 4kN as shown in figure.

Q.2. Draw the shear force and bending moment diagrams in a simply supported overhanging beam carrying a concentrated clockwise moment, udl and point loads as shown in figure.