## How to draw Bending Moment diagrams?

In drawing of bending moment diagrams, we have to follow certain guidelines in strict manner as explained in earlier article *guide lines of beam diagram*. For illustration and clear understanding of the process of drawing of beam diagrams, we have drawn following typical beam diagrams with very common beams and their loading patterns. These are –

- Drawing of beam diagrams for a cantilever beam with point load.
- Drawing of beam diagrams for a cantilever beam with uniformly distributed load.
- Drawing of beam diagrams for a cantilever beam with uniformly varying load.
- Drawing of beam diagrams for a simply supported beam loaded with a couple.

__Beam diagram for a Cantilever with Point load__

Consider about a *cantilever beam* of length ( l ) loaded with a *point load* ( W ) at its free end as shown in figure.

### Space diagram –

Space diagram for the beam is drawn in figure ( a ).

### Shear Force diagram –** **

*If there is no other load between two points of a beam, then shear force between these two points will remain constant. Therefore, ( SF ) diagram will be parallel to the base line between these two points.*

- Point load ( W ) is loaded at free end point B .
- There is no other load between the points A and B .
- Hence, shear force diagram is a rectangle of width ( W ) between these two points.
- Effect of load has a tendency to rotate the beam in clockwise direction.
- So,
*shear force is positive*. Hence, ( SF ) diagram is drawn in positive side.

Shear Force diagram for the beam is drawn in figure ( b ).

### Bending Moment diagram –

*If there is no other load between two points of a beam, then bending moment between these two points will change linearly. Therefore, BM diagram will be an inclined straight line between these two points.*

Because, **Bending Moment = Load × Distance. **( As the distance of the beam section from load increases, bending moment also increases linearly. )

*Bending moment*of load ( W ) at point B is zero.- Bending moment of load ( W ) at point A is ( W \times l = W l ) .
- No other load is present in between A and B .
- Therefore ( BM ) diagram is a straight line inclined to base line forming a triangle.
- Bending moment has a tendency to bend the beam in ∩ shape. Hence it is
*hogging moment*or negative moment.

Bending moment diagram for the beam is drawn in figure ( c ).

## Beam diagram for a Cantilever with Uniformly Distributed Load

Consider about a cantilever beam of length ( l ) loaded with a *uniformly distributed load* ( UDL ) of magnitude, ( w ) unit per unit length of beam as shown in figure.

### Space diagram –

Space diagram for the beam is drawn in figure ( a ).

### Shear Force diagram –

*If there is an ( UDL ) between two points of a beam, then shear force between these two points will change linearly between these points. *

Rate of loading of UDL = w units per unit length of beam. Thus –

- Total load at free end will be [ W = ( w \times 0 ) = 0 ]
- Total load at a distance of ( x ) unit from free end will be [ W = ( w \times x ) = w x ]

*Therefore, ( SF ) diagram between these two points will be a straight line inclined to the base line.*

- There is no other load between the points A and B .
- Total load at point B is ( 0 ) and at point A is ( w l )
- Hence, shear force diagram is a triangle of width ( w l ) .
- Effect of load ( wx ) has a tendency to rotate the beam in clockwise direction.
- So,
*positive shear force*has developed in beam. Therefore, ( SF ) diagram is drawn in positive side.

Shear Force diagram for the beam is drawn in figure ( b ).

### Bending Moment diagram –

Rate of loading of UDL = w units per unit length of beam.

We know that, **Bending Moment ( M ) = Load × Distance of centre of gravity of load.**

Therefore, magnitude of bending moment at a distance ( x ) from free end of beam will be –

M = \left ( w \ x \right ) \times \left ( \frac {x}{2} \right )

= \left ( \frac { w x^2}{2} \right )

Thus, change in bending moment is proportional to ( x^2 ) . This is a required condition of a parabola curve.

*Therefore, bending moment diagram of a beam carrying an ( UDL ) between two points will be a parabolic curve between that points.*

- Bending moment at point B is zero.
- Total load acting at point A is katex] ( W = w \times l = w l ) [/katex]. Therefore, magnitude of bending moment will be \left ( wl \times \frac {l}{2} = \frac {wl^2}{2} \right )
- No other load is present in between A and B .
- Therefore, ( BM ) diagram is a parabola curve.
- Bending moment has a tendency to bend the beam in ∩ shape. Hence it is
*hogging moment*or negative moment.

Bending moment diagram for the beam is drawn in figure ( c ).

## Beam diagram for Cantilever with Varying Load

Consider about a cantilever beam of length ( l ) loaded with a gradually varying load from zero at free end B to ( w ) per unit length at the fixed end A as shown in figure.

### Space diagram –

Space diagram for the beam is drawn in figure ( a ).

### Shear Force diagram –** **

We know that, **Magnitude of load on beam ( W ) = Total area of the triangular load diagram.**

Therefore, \quad W = \left ( \frac {1}{2} \right ) \times \text {base} \times {\text {height}}

Also, height of load triangle at a distance ( x ) from free end of beam will be \left ( \frac {wx}{l} \right )

Therefore, load \quad W = \left ( \frac{1}{2} \right ) \times x \times \left ( \frac {w x}{l} \right )

= \left ( \frac {w x^2}{2l} \right )

Thus, change in load i.e. shear force is proportional to ( x^2 ) . This is a required condition of a parabola curve.

*Hence, shear force diagram between two points of a beam carrying an uniformly varying load will be a parabola curve.*

Effect of load \left ( \frac {w x^2}{2l} \right ) has a tendency to rotate the beam in clockwise direction. Hence ( SF ) diagram is positive.

Shear force diagram for the beam is drawn in figure ( b ).

### Bending Moment diagram –

We know that, **Bending Moment ( M ) = Load × Distance of centre of gravity of load.**

And, *centre of gravity* of a triangle is \left ( \frac {\text {height}}{3} \right ) from base.

Therefore, magnitude of bending moment at a distance ( x ) from free end of beam will be –

M = \left ( \frac {w x^2}{2l} \right ) \left ( \frac {x}{3} \right )

= \left ( \frac {wx^3}{6l} \right )

Thus, change in bending moment is proportional to ( x^3 ) . This is a required condition of a cubic curve.

*Hence, bending moment between the two points of a beam carrying an uniformly varying load will be a cubic curve between these points.*

Bending moment diagram for the beam is drawn in figure ( c ).

__Loading of a Couple__

Consider about a simply supported beam AB of length ( l ) which is loaded with an anti-clockwise *couple* ( \mu ) at point C

Tendency of the couple ( \mu ) is to lift the beam at support B and press the beam at support A . Therefore, reaction ( R_A ) will be upward and ( R_B ) will be downward.

**STEP – 1 ( Find end reactions )**

Taking the moments of reactions ( R_A ) and ( R_B ) at point A we get –

R_B \times l = \mu ( Reaction R_B is downward. )

So, \quad R_B = \left ( \frac {\mu}{l} \right ) ( Downward direction ↓ )

Also, ( R_A + R_B ) = \mu

So, \quad R_A = \left ( \frac {\mu}{l} \right ) ( Upward direction ↑ )

**STEP – 2 ( Draw SF diagram )**

- Reaction ( R_A ) is upward and at ( R_B ) is downward.
- There is no other force between points A and B
- The effect of reactions has a tendency to rotate the beam in clockwise direction.
- Therefore, shear force is positive. SF diagram is a rectangle in positive direction and its side is \left ( + \frac {\mu}{l} \right )

**STEP – 3 ( Draw BM diagram )**

Taking the moments about point A we have –

- Bending moment at point A will be –

M_A = R_A \times 0 = 0 ( Because distance is zero )

- Bending moment at just left of point C will be –

M_{C ( left )} = R_A \times a

= \left ( \frac {\mu}{l} \right ) \times a

= \left ( \frac {\mu a}{l} \right )

- Bending moment at just right of point C will be –

M_{C ( right )} = R_A \times a - \mu

= \left ( \frac {\mu}{l} \right ) \times a - \mu

= \mu \frac {\left ( a - l \right )}{l}

Because, \quad \left ( l = a + b \right )

Therefore, M_{C ( right )} = - \frac {\mu b}{l}

- Bending moment at point B will be –

M_B = \left ( R_A \times l \right ) - \mu

= \left ( \frac {\mu}{l} \times l \right ) - \mu = 0

( SF ) and ( BM ) diagrams are shown in figure.

__Loading on a Bracket__

Sometimes a load is applied on a ( L ) shaped bracket which is mounted on a beam.

Consider about a beam which has a point load ( W ) loaded on a bracket of length ( l ) . In this type of loading the load has twin effects –

- It provides an effect of point load of magnitude ( W ) at the point of attachment of bracket with beam.
- It produces a clockwise couple ( positive moment ) of magnitude ( W \times l )
*.*

These two effects of point load and couple are then treated separately in addition to other similar loads on beam.

See numerical problems based on this article.