{SOM- I} [Unit – VII] · PDF file{SOM- I} [Unit – VII] Slope and Deflection BHCET...

{SOM- I} [Unit – VII] Slope and Deflection BHCET

Prepared by: Mohammad Amir, Lecturer, Department of Mechanical Engineering, BHCET.

Introduction: In certain situations it becomes necessary to design a machine component or a structure to minimize deflection. Thus it becomes essential to determine the deflection of the member. Also for statically indeterminate or fixed beams it becomes necessary to determine the slope and deflection at salient points.

Differential Equation of the Deflection Curve or (Relation between Bending moment, slope and Deflection.) Consider a beam AB which takes the curved shape as shown in Fig(a). Consider an elementary length CD equal to ds of the beam. Let the tangent to the elastic curve at C makes with the x-axis of the beam an angle θ. The angle at D will decrease and let it be (θ-dθ). The normals at C and D intersect at O.



If the curvature is very small then the dy/dx is also small and its square is negligible, hence

The bending moment has been taken as positive if tension is caused in the bottom fibre of the beam. The positive bending moment should cause the positive curvature. The curvature is taken as positive if the centre of curvature O, is above the bent curve of the beam as shown in Fig (b). If the curvature is positive, the angle θ decreases in going from C to D, hence with proper sign.



Methods of Solution: The following methods are generally used to determine the slope and deflection at a point of a beam.

(i) Macaulay’s method. (ii) Moment area method. (iii) Conjugate beam method.

Macaulay's Method: This method is suitable for cases of beams subjected to concentrated loads and can be extended to uniformly distributed loads also. It consists of successive integration of expression for bending moment in such a way that same constants of integration are valid for all portions of the beam even though the law of bending moment differs from portion to portion. For the beam shown in Fig., the expressions for B.M. in different parts of the beam are:

isthe generalexpressiofno rbendingm oment.T heg enerael xpressionh olds



if the convention is adopted that a bracket is disregarded when for a particular value of x the contents of the bracket becomes negative. Therefore, it follows that bracket terms must be kept in tact and not multiplied out until a numerical substitution is made for x. The form of integration for a particular terms is

The general expression for bending moment must be written by taking a section just before the right hand support or load. If there is a uniformly distributed load starting anywhere and extending upto the right end then the general equation for bending moment will hold good for the entire beam However, if the load does not extend upto the right end, the load may be extended up-to the right end, and an equal and opposite load may be added to counteract the effect due to the additional load. For example, for the beam shown in Fig(a),

Whereas for the beam shown in Fig.(b), the load is not extending the upto right end. In Fig(c), the load is extended upto the right end and an equal and opposite load is applied on the portion of the beam to the right of the load which was earlier unloaded. In this case, the bending moment expression is,

Point load on a simply supported beam. Consider a beam of length l, simply supported at the ends and carrying a concentrated load W at a distance ‘a’ from left end A. Taking a section at a distance x from left end A as shown in Fig.



Uniformly distributed load on a simply supported beam: Consider a beam AB of length ‘l’ simply supported at the ends and carrying a uniformly distributed load of intensity w per unit length as shown in the figure. Taking a section at a distance of x from the left end A, we have,



Cantilever beam carrying a concentrated load at the free end. Consider a cantilever beam AB of length ‘l’ carrying a concentrated load W at the free end as shown in Fig.



Cantilever beam carrying a uniformly distributed load over the whole span: Consider a cantilever AB of length ‘l’ and carring a uniformly distributed load of intensity w per unit length as shown in the figure. Consider a section at a distance x from the free end A, Then section x-x



Cantilever beam carrying a concentrated road not at the free end:

Consider a cantilever beam AB of length ‘/’ and carrying a concentrated load W at a distance ‘a’ from the free end A. Taking a section at a distance ‘x’ from the free end the bending moment at x-x is (Fig. 8.9)



Que.1. A horizontal girder of steel having the uniform cross-section is 14m long and is

simply supported at the its ends. It carries concentrated load of 120KN and 80KN at two points 3m and 4.5m from the two ends respectively. MOI for the section of the girder is 16x104 cm4 and E for steel is 210GPa. Calculate the deflection in the girder at points under the two loads?



Que.2. A beam AB simplys upported at the ends is 4 m long, it carries a uniformly

distributed load of intensity 20 KN/m over a length of 2 m starting at a distance of 1 m from left end support together with a concentrated 40KN, at a distance of 3 m from the left end support. Calculate the deflection at the centre, if E = 210 GPa and MOI = 9600 cm4.

Sol. Consider a section at a distance x from the left end support as shown in Fig. The

distributed load does not extend upto the right end support. So we extend this load upto the right end and apply equal and opposite load over the Portion DB.



Mohr's Moment Area Method:

Consider a part AB of the deflected beam as shown in Fig. 8.25. Consider a small element CD = ds. Draw tangent as C and D to the deflected beam. Also draw normal to these tangents to meet at point O so that <COD = dθ, In the limiting case for small deflection.



Simply supported beam carrying a point load at the centre:

Cantilever beam carrying a point load at the free end:



Cantilever beam carrying a uniformly distributed load throughout the span:

Simply supported beam carrying a uniformly distributed load:



Conjugate Beam Method:

This method is based upon the modification of moment area method. This is applicable to cases of simple beams carrying a single concentrated load but by superposition this method can be extendetd to cover any kind of loading. For a single concentrated load as shown in Fig.( a) on a simply supported beam AB, the B .M. diagram is A'C'B' as shown in Fig( b). Now A'B’ can be considered, as a beam carrying a distributed load given by the B.M. diagram A’C’B’. This imaginary beam A 'B' is called the conjugate beam.



MOHR'S THEOREMS: The results given by equation from the moment area method for deflection are known as Mohr's theorems. They are state as : I. The change of slope between any two points is equal to the net area of the B.M. diagram between these points divided by EI II. The total deflection between any two points is equal to the moment of the area of B.M. diagram between the two points about the last point (i.e., B) divided by EI. The Mohr's theorems is conveniently used for following cases: 1. Problems on cantilevers (zero slope at fixed end). 2. Simply supported beams carrying symmetrical loading (zero slope at the centre). 3. Beams fixed at both ends (zero slope at each end).

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