LECTURE 23 and 24
Procedure for drawing shear force and bending moment diagram:
The advantage of plotting a variation of shear force F and bending moment M in a beam as a function of ‘x' measured from one end of the beam is that it becomes easier to determine the maximum absolute value of shear force and bending moment.
Further, the determination of value of M as a function of ‘x' becomes of paramount importance so as to determine the value of deflection of beam subjected to a given loading.
Construction of shear force and bending moment diagrams:
A shear force diagram can be constructed from the loading diagram of the beam. In order to draw this, first the reactions must be determined always. Then the vertical components of forces and reactions are successively summed from the left end of the beam to preserve the mathematical sign conventions adopted. The shear at a section is simply equal to the sum of all the vertical forces to the left of the section.
When the successive summation process is used, the shear force diagram should end up with the previously calculated shear (reaction at right end of the beam. No shear force acts through the beam just beyond the last vertical force or reaction. If the shear force diagram closes in this fashion, then it gives an important check on mathematical calculations.
The bending moment diagram is obtained by proceeding continuously along the length of beam from the left hand end and summing up the areas of shear force diagrams giving due regard to sign. The process of obtaining the moment diagram from the shear force diagram by summation is exactly the same as that for drawing shear force diagram from load diagram.
It may also be observed that a constant shear force produces a uniform change in the bending moment, resulting in straight line in the moment diagram. If no shear force exists along a certain portion of a beam, then it indicates that there is no change in moment takes place. It may also further observe that dm/dx= F therefore, from the fundamental theorem of calculus the maximum or minimum moment occurs where the shear is zero. In order to check the validity of the bending moment diagram, the terminal conditions for the moment must be satisfied. If the end is free or pinned, the computed sum must be equal to zero. If the end is built in, the moment computed by the summation must be equal to the one calculated initially for the reaction. These conditions must always be satisfied.
In the following sections some illustrative problems have been discussed so as to illustrate the procedure for drawing the shear force and bending moment diagrams
1. A cantilever of length carries a concentrated load ‘W' at its free end.
Draw shear force and bending moment.
At a section a distance x from free end consider the forces to the left, then F = -W (for all values of x) -ve sign means the shear force to the left of the x-section are in downward direction and therefore negative
Taking moments about the section gives (obviously to the left of the section)
M = -Wx (-ve sign means that the moment on the left hand side of the portion is in the anticlockwise direction and is therefore taken as –ve according to the sign convention)
so that the maximum bending moment occurs at the fixed end i.e. M = -W l
From equilibrium consideration, the fixing moment applied at the fixed end is Wl and the reaction is W. the shear force and bending moment are shown as,
2. Simply supported beam subjected to a central load (i.e. load acting at the mid-way)
By symmetry the reactions at the two supports would be W/2 and W/2. now consider any section X-X from the left end then, the beam is under the action of following forces.
.So the shear force at any X-section would be = W/2 [Which is constant upto x < l/2]
If we consider another section Y-Y which is beyond l/2 then
for all values greater = l/2
Hence S.F diagram can be plotted as,
.For B.M diagram:
If we just take the moments to the left of the cross-section,
Which when plotted will give a straight relation i.e.
It may be observed that at the point of application of load there is an abrupt change in the shear force, at this point the B.M is maximum.
3. A cantilever beam subjected to U.d.L, draw S.F and B.M diagram.
Here the cantilever beam is subjected to a uniformly distributed load whose intensity is given w / length.
Consider any cross-section XX which is at a distance of x from the free end. If we just take the resultant of all the forces on the left of the X-section, then
So if we just plot the equation No. (1), then it will give a straight line relation. Bending Moment at X-X is obtained by treating the load to the left of X-X as a concentrated load of the same value acting through the centre of gravity.
Therefore, the bending moment at any cross-section X-X is
The above equation is a quadratic in x, when B.M is plotted against x this will produces a parabolic variation.
The extreme values of this would be at x = 0 and x = l
Hence S.F and B.M diagram can be plotted as follows:
4. Simply supported beam subjected to a uniformly distributed load [U.D.L].
The total load carried by the span would be
By symmetry the reactions at the end supports are each wl/2
If x is the distance of the section considered from the left hand end of the beam.
S.F at any X-section X-X is
Giving a straight relation, having a slope equal to the rate of loading or intensity of the loading.
The bending moment at the section x is found by treating the distributed load as acting at its centre of gravity, which at a distance of x/2 from the section
So the equation (2) when plotted against x gives rise to a parabolic curve and the shear force and bending moment can be drawn in the following way will appear as follows:
When the beam is subjected to couple, the shear force and Bending moment diagrams may be drawn exactly in the same fashion as discussed earlier.
6. Eccentric loads.
When the beam is subjected to an eccentric loads, the eccentric load are to be changed into a couple/ force as the case may be, In the illustrative example given below, the 20 kN load acting at a distance of 0.2m may be converted to an equivalent of 20 kN force and a couple of 2 kN.m. similarly a 10 kN force which is acting at an angle of 300 may be resolved into horizontal and vertical components.The rest of the procedure for drawing the shear force and Bending moment remains the same.
6. Loading changes or there is an abrupt change of loading:
When there is an aabrupt change of loading or loads changes, the problem may be tackled in a systematic way.consider a cantilever beam of 3 meters length. It carries a uniformly distributed load of 2 kN/m and a concentrated loads of 2kN at the free end and 4kN at 2 meters from fixed end.The shearing force and bending moment diagrams are required to be drawn and state the maximum values of the shearing force and bending moment.
Consider any cross section x-x, at a distance x from the free end
Shear Force at x-x = -2 -2x 0 < x < 1
S.F at x = 0 i.e. at A = -2 kN
S.F at x = 1 = -2-2 = - 4kN
S.F at C (x = 1) = -2 -2x - 4 Concentrated load
= - 2 - 4 -2x1 kN
= - 8 kN
Again consider any cross-section YY, located at a distance x from the free end
S.F at Y-Y = -2 - 2x - 4 1< x < 3
This equation again gives S.F at point C equal to -8kN
S.F at x = 3 m = -2 -4 -2x3
= -12 kN
Hence the shear force diagram can be drawn as below:
For bending moment diagrams – Again write down the equations for the respective cross sections, as consider above
Bending Moment at xx = -2x - 2x.x/2 valid upto AC
B.M at x = 0 = 0
B.M at x =1m = -3 kN.m
For the portion CB, the bending moment equation can be written for the x-section at Y-Y .
B.M at YY = -2x - 2x.x/2 - 4( x -1)
This equation again gives,
B.M at point C = - 2.1 - 1 - 0 i.e. at x = 1
= -3 kN.m
B.M at point B i.e. at x = 3 m
= - 6 - 9 - 8
= - 23 kN-m
The variation of the bending moment diagrams would obviously be a parabolic curve
Hence the bending moment diagram would be
7. Illustrative Example :
In this there is an abrupt change of loading beyond a certain point thus, we shall have to be careful at the jumps and the discontinuities.
For the given problem, the values of reactions can be determined as
R2 = 3800N and R1 = 5400N
The shear force and bending moment diagrams can be drawn by considering the X-sections at the suitable locations.
8. Illustrative Problem :
The simply supported beam shown below carries a vertical load that increases uniformly from zero at the one end to the maximum value of 6kN/m of length at the other end .Draw the shearing force and bending moment diagrams.
Determination of Reactions
For the purpose of determining the reactions R1 and R2 , the entire distributed load may be replaced by its resultant which will act through the centroid of the triangular loading diagram.
So the total resultant load can be found like this-
Average intensity of loading = (0 + 6)/2
= 3 kN/m
Total Load = 3 x 12
= 36 kN
Since the centroid of the triangle is at a 2/3 distance from the one end, hence 2/3 x 3 = 8 m from the left end support.
Now taking moments or applying conditions of equilibrium
Note: however, this resultant can not be used for the purpose of drawing the shear force and bending moment diagrams. We must consider the distributed load and determine the shear and moment at a section x from the left hand end.
Consider any X-section X-X at a distance x, as the intensity of loading at this X-section, is unknown let us find out the resultant load which is acting on the L.H.S of the X-section X-X, hence
So consider the similar triangles
OAB & OCD
In order to find out the total resultant load on the left hand side of the X-section
Find the average load intensity
Now these loads will act through the centroid of the triangle OAB. i.e. at a distance 2/3 x from the left hand end. Therefore, the shear force and bending momemt equations may be written as
9. Illustrative problem :
In the same way, the shear force and bending moment diagrams may be attempted for the given problem
10. Illustrative problem :
For the uniformly varying loads, the problem may be framed in a variety of ways, observe the shear force and bending moment diagrams
11. Illustrative problem :
In the problem given below, the intensity of loading varies from q1 kN/m at one end to the q2 kN/m at the other end.This problem can be treated by considering a U.d.i of intensity q1 kN/m over the entire span and a uniformly varying load of 0 to ( q2- q1)kN/m over the entire span and then super impose teh two loadings.
Point of Contraflexure:
Consider the loaded beam a shown below along with the shear force and Bending moment diagrams for It may be observed that this case, the bending moment diagram is completely positive so that the curvature of the beam varies along its length, but it is always concave upwards or sagging.However if we consider a again a loaded beam as shown below along with the S.F and B.M diagrams, then
It may be noticed that for the beam loaded as in this case,
The bending moment diagram is partly positive and partly negative.If we plot the deflected shape of the beam just below the bending moment
This diagram shows that L.H.S of the beam ‘sags' while the R.H.S of the beam ‘hogs'
The point C on the beam where the curvature changes from sagging to hogging is a point of contraflexure.
It corresponds to a point where the bending moment changes the sign, hence in order to find the point of contraflexures obviously the B.M would change its sign when it cuts the X-axis therefore to get the points of contraflexure equate the bending moment equation equal to zero.The fibre stress is zero at such sections
Note: there can be more than one point of contraflexure.