# ELASTICITY | Beam |Internal bending moment | Flexural rigidity |Cantilever |bending moment and shearing force

## Beam

A beam is a bar on rod of uniform rectangular or circular cross – section of a homogeneous isotropic elastic material having length much greater than its crossа -sectional area.

## Internal bending moment

A beam can be divided up into a number of straight parallel filaments having same lengths. When external torque is applied on a beam then it bents i.e. compression forces act on some layers thus those layers are contracted and extensional forces act on some other layers hence they are extended. The filament which maintain the original length is called the neutral layer .

Internal forces come into play to counteract the effect of bending. The moment of this internal forces is called internal bending moment.

Note :- At equilibrium, internal bending moment must be balanced by external bending moment or torque.

Expression –

Let us consider a portion of a bent beam whose two end faces are inclined by an angle θ. AB is the neutral layer whose radius of curvature is R.

Let us take a layer EF at a distance Z from the neutral layer AB, hence its radius of curvature is R + Z.

Original length of each layer = length of neutral layer = Rθ .

Now length of layer EF = (R+Z)θ

Hence longitudinal strain,

If α be the cross -sectional area of layer EF and F be the couple (longitudinal force), then

Young’s modulus Y

So, Couple required to bent the layer EF,

Moment of this force about the neutral Layer

Summing up all such moment of force for all the layer we get internal bending moment,

Where         ΣαZ²  = geometrical moment of inertia = AK² ;

A = cross -section of beam

K = radius of gyration .

Internal bending moment

Note :

i) S.I. unit  = Newton- meter

ii) For circular cross-section,

iii) For rectangular cross- section,

## Flexural rigidity

Bending moment per unit radius of curvature is known as flexural rigidity. i.e.

Flexural rigidity

S.I. unit – Newton – m²

## Cantilever

A cantilever is a uniform beam fixed horizontally at its one end and loaded at any other point.

There are two types of cantilever-

i) Light cantilever, ii) Heavy cantilever .

### Light cantilever : Calculate slope and depression

Let us consider a light cantilever of neglected weight. It is fixed at x =0 and loaded by w at x=l  i.e. free end.

Let us take a point C on the beam at a distance x from fixed end and C is depressed by y due to load w at free end.

The load exerts an external torque at C  = w (l-x)

At equilibrium, internal bending moment is equal to this external torque i.e.

where,

Y = Young’s modulus

A = cross -section

K = radius of gyration

Integrating,

slope at a distance x from fixed end.

Again integrating,

— Depression at a distance x from fixed end.

— Depression at a the loaded end.

NOTE-

A) For rectangular cross – section ;

B) For circular cross – section ;

### Reciprocal theorem for light cantilever

Statement: Reciprocal theorem states that the depression at any point of a light cantilever due to load applied at the second point is same as the depression at the second point when some load is applied at the first point.

Proof:

Step I

Depression at a distance x when load applied at the free end (y say)

At equilibrium, external tongue at C = internal bending moment

Step II

Depression at a distance l (i.e. free end) when a load w is applied at a distance x (y’ say)

Depression at a distance be l,

Hence y = y’ i.e. the theorem is proved.

### Heavy cantilever : Calculate slope and depression

Let us consider a heavy cantilever of weight per unit length = ρ

It is fixed at x =0 and loaded by w at x =l  i.e. free end.

Let us take a point on the cantilever at a distance x from fixed end. Let C is depressed by y due to load w at the free end and weight of the rod for the portion CB.

At equilibrium, internal bending moment = moment due to load about C+ moment due to own weight for CB                                                                                                                                 portion about C

Depression produced at the free end,

Due to an additional force F, let a further depression ẟ’ is produced, then,

Comparing equation 1 and 2, we get

; restoring force.

Hence equation of motion can be written as,

Hence the vibration is simple harmonic.

where,

Time period of oscillation,

Also,

Integrating,

Slope at a distance x,

Again integrating,

where, w’ = ρl , weight of the beam.

This expression is the depression at the loaded end.

## Define bending moment and shearing force

#### Bending moment — (B.M.)

The bending moment at any point is the algebraic sum of the moments about the point of all the vertical components of the loads acting on the any side of the point.

#### Shearing force — (S.F.)

The algebraic sum of all the vertical forces acting on the one side of any section of a loaded beam is the measure of shearing force at the section.

Relation — If F be the shearing force, then Bending moment

S.I. unit of Bending moment – Joule.

S.I. unit of Shearing force – Newton.

Remember that

i) Free end with no road =>        F = 0 and M = 0.

ii) Free end loaded by a weight =>          F = w and M = 0

iii) Clamped end =>                   F ≠ 0 and M ≠ 0

iv) Supported end =>                 F = 0 and M = 0

Relation between Bending moment and Shearing force :

Let ρ be the weight per unit length of a beam. Let us consider an element 𝛿x at a left edge at a distance x  from origin and right edge at a distance x+𝛿x from origin.

Shearing force at the left face is F and at the right face is F+𝛿F during upward displacement of the element. The corresponding internal resisting moment at the left face due to left hand portion of the beam is M (anticlockwise direction) and at the right hand portion of the beam is M+𝛿M (clockwise).

Considering vertical equilibrium of the element x,

i) force equation becomes

In the limit

ii) moment equation becomes (about A)

In the limit

From equation 1 and 2, we get