Friday, 9 September 2011

RELATION BETWEEN SHEAR FORCE AND BENDING MOMENT

Shear Force and Bending Moment

An introduction to Shear Force and Bending Moments in Beams

Contents:

Shearing Force
Bending Moments
Types Of Load
Types Of Support
The Relationship Between W; F; M.
Apendix

1.Shearing Force:

The shearing force (SF) at any section of a beam represents the tendency for the portion of the beam on one side of the section to slide or shear laterally relative to the other portion.

The diagram shows a beam carrying loads $\displaystyle W_1\;;\;W_2\;\;and\;\;W_3$ . It is simply supported at two points where the reactions are:-

$\displaystyle R_1\;\;\;and\;\;\;R_2$

Assume that the beam is divided into two parts by a section XX The resultant of the loads and reaction acting on the left of AA is F vertically upwards and since the whole beam is in equilibrium, the resultant force to the right of AA must be F downwards. F is called the Shearing Force at the section AA. It may be defined as follows:- The shearing force at any section of a beam is the algebraic sum of the lateral components of the forces acting on either side of the section. Where forces are neither in the lateral or axial direction they must be resolved in the usual way and only the lateral components use to calculate the shear force.

2.Bending Moments:

In a similar manner it can that if the Bending moments (BM) of the forces to the left of AA are clockwise then the bending moment of the forces to the right of AA must be anticlockwise. Bending Moment at AA is defined as the algebraic sum of the moments about the section of all forces acting on either side of the sectionBending moments are considered positive when the moment on the left portion is clockwise and on the right anticlockwise. This is referred to as a sagging bending moment as it tends to make the beam concave upwards at AA. A negative bending moment is termed hogging.

3.Types Of Load:

A beam is normally horizontal and the loads vertical. Other cases which occur are considered to be exceptions. A Concentrated load is one which can be considered to act at a point although of course in practice it must be distributed over a small area. A Distributed load is one which is spread in some manner over the length or a significant length of the beam. It is usually quoted at a weight per unit length of beam. It may either be uniform or vary from point to point.

4.Types Of Support:

A Simple or free support is one on which the beam is rested and which exerts a reaction on the beam. It is normal to assume that the reaction acts at a point although it may in fact act act over a short length of beam. A Built-in or encastre' support is frequently met . The effect is to fix the direction of the beam at the support. In order to do this the support must exert a "fixing" moment M and a reaction R on the beam. A beam which is fixed at one end in this way is called a Cantilever. If both ends are fixed in this way the reactions are not statically determinate. In practice it is not usually possible to obtain perfect fixing and the fixing moment applied will be related to the angular movement of the support. When in doubt about the rigidity it is safer to assume that the beam is freely supported.

5.The Relationship Between W; F; M.

In the following diagram $\displaystyle \delta x$ is the length of a small slice of a loaded beam at a distance x from the origin O

Let the shearing force at the section x is F and at $\displaystyle x = \delta x\;\;be\;\;F + \delta F$ . Similarly the bending moment is M at x and $\displaystyle M + \delta M\;\;\;at\;\;\;x + \delta x$ . If w if the mean rate of loading of the length $\displaystyle \delta x$ , the total load is $\displaystyle w\delta x$ , acting approximately ( exactly if uniformly distributed) through the centre C. The element must be in equilibrium under the action of these forces and couples and the following equations can be obtained:-

Taking Moments about C

$M + F.\frac{\delta x}{2} + (F + \delta F)\;\frac{\delta x}{2} = M + \delta M$

.................(1)

Neglecting the product $\displaystyle \delta F.\delta x$ in the limit:-

$F = \frac{dM}{dx}$

.................(2)

Resolving vertically:-

$w\delta x + F + \delta F = F$

.................(3)

$Or\;\;\;\;\;\;\;w = -\frac{dF}{dx}$

.................(4)

$= -\frac{d^2M}{dx^2}\;\;\;\;\;from\;equation\;(2)$

..................(5)

From equation (2) it can be seen that if M is varying continuously zero shearing force corresponds to either maximum or minimum bending moment. It ca be seen from the examples that "peaks" in the bending moment diagram frequently occur at concentrated loads or reactions and these are not given by $\displaystyle F = \frac{dM}{dx} = 0$ although they may in fact represent the greatest bending moment on the beam. Consequently it is not always sufficient to investigate the points of zero shearing force when determining the maximum bending moment. At a point on the beam where the type of bending is changing from sagging to hogging, the bending moment must be zero and this is called a point of inflection or contra -flexure. By integrating equation (2) between the x = a and x = b then:-

$M_b - M_a = \int_{a}^{b}F\,dx$

..................(6)

Which shows that the increase in bending moment between two sections is the area under the shearing force diagram. Similarly integrating equation (4)

$F_a - F_b = \int_{a}^{b}w\,dx$

...................(7)

equals the area under the load distribution diagram. Integrating equation (5) gives:-

$M_a - M_b = \int \int_{a}^{b}w\,dx.dx$

...................(8)

These relations can be very valuable when the rate of loading cannot be expressed in an algebraic form as they provide a means of graphical solution.

6.Appendix:

Shearing Force F

$F = \frac{dM}{dx}$

Mending Moment M Rate of loading w

$w = -\,\frac{dF}{dx} = -\frac{d^2M}{dx^2}$

SFD

Shear Force and Bending Moment Diagrams for a Simply-Supported Beam Under A Uniform Load
After the support reactions are calculated, the shear force and bending moment diagrams can be drawn.
Shear force is the force in the beam acting perpendicular to its longitudinal (x) axis. For design purposes, the beam's ability to resist shear force is more important than its ability to resist an axial force. Axial force is the force in the beam acting parallel to the longitudinal axis.
The following is a drawing of a simply-supported beam of length L under a uniform load, q:

This beam has the following support reactions:

where R_l and R_r are the reactions at the left and right ends of the beam, respectively.
The shear forces at the ends of the beam are equal to the vertical forces of the support reactions. The shear force F(x) at any other point x on the beam can be found by using the following equation.

where x is the distance from the left end of the beam.
Shear force diagrams are simply plots of the shear force (on the y-axis) versus the position of various points along the beam (on the x-axis). Thus, the following is the generalized shear force diagram for the beam shown above.

The bending moment at any point along the beam is equal to the area under the shear force diagram up to that point. (Note: For a simply-supported beam, the bending moment at the ends will always be equal to zero.)
To calculate the bending moment the beam must be broken up into two sections:

(a)	one from x = 0 to x = L/2 and
(b)	the other from x = L/2 to x = L.

The bending moment M(x) at any point x along the beam can be found by using the following equations:

Bending moment diagrams are simply plots of the bending moment (on the y-axis) versus the position of various points along the beam (on the x-axis). Thus, the following is the generalized bending moment diagram for the beam shown above.

Wednesday, 31 August 2011

BENDING IN BEAM

    3. BEAMS: STRAIN, STRESS, DEFLECTIONS
        The beam, or flexural member, is frequently encountered in structures and
machines, and its elementary stress analysis constitutes one of the more interesting facets
of mechanics of materials. A beam is a member subjected to loads applied transverse to
the long dimension, causing the member to bend. For example, a simply-supported beam
loaded at its third-points will deform into the exaggerated bent shape shown in Fig. 3.1

        Cantilever beams and simple beams have two reactions (two forces or one force
and a couple) and these reactions can be obtained from a free-body diagram of the beam
by applying the equations of equilibrium.          Such beams are said to be statically
determinate since the reactions can be obtained from the equations of equilibrium.
Continuous and other beams with only transverse loads, with more than two reaction
components are called statically indeterminate since there are not enough equations of
equilibrium to determine the reactions.

Figure 3.1 Example of a bent beam (loaded at its third points)

SHEAR STRESS PROBLEM:

Problem 1: Derivation of Shear stress in rectangular crosssection

problem 1.

         Derive an expression for the shear stress distribution in a beam of solid rectangular cross-section transmitting a vertical shear V.
The cross sectional area of the beam is shown in the figure. A longitudinal cut through the beam at a distance y1, from the neutral axis, isolates area klmn. (A1).
Shear stress,
τ= VQ
      It

=   V / It ∫     y.dA
             A1


                    d/2
=     V / Ib ∫ by dy
               y1

= V / 2I{( d / 2 )*( d / 2 )− ( y1 ) *− ( y1 ) }        ---------------------- (1)

The shear stress is shown as below

Max Shear Stress occurs at the neutral axis and this can be found by putting y = 0 in the
equation 1.

τmax = Vd *d
      8I

     = 3 V
     2 bh

     = 3V
      2A

go2science