DEFLECTION OF BEAM.pptx

J5109-STRENGTH OF MATERIAL 2

JJ310-STRENGTH OF MATERIALSDEFLECTION OF BEAMS

1DEFLECTION OF BEAMSIn a plane figure the neutral plane is represented by a straight line. The displacement of the central line of the neutral plane from its original position in the direction of normal at any point is termed as deflection of the beam.

Since there is almost no change in the transverse dimension of the beam cross-section, it may be understood that the top and bottom surface of the beam will deform in the same way as the neutral plane and hence for all practical purpose for calculation of deflection of the beam, the beam may be represented by the central line of the neutral plane.

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Figure 1:Figure below illustrates this definition of the deflection. 3Equation to Elastic Curve- DOUBLE INTEGRATION METHODFrom figure 2:

(i)4Where R is the radius of curvature of CD and CD being small it can be regarded as arc of a circle of radian R.Also since angle is small due to proximity of points C and D.

It can be seen that while y defines the deflection of beam, dy/dx define its slope at any section. Differentiating (ii) with respect to x.

(ii)(iii)5Now using (i) in (ii),

Bending Moment equation is;

(iv)(a)

Combining eq.(a) with (iv);

(1.1)6Eq. (1.1) is the basic equation for calculation of the beam deflection y of a beam. In this equation, M represents the bending moment at any section of the beam. E is the modulus of elasticity of the beam material (a constant) and I the moment of inertia of the section of the beam.

For prismatic beam, I will be constant for all values of x. Both E and I are positive.

The quantity EI is known as flexural rigidity and may some times be specified as a single constant for a given beam.7For calculation of y from eq. (1.1), the steps are apparently straight forward. First integration will give dy/dx (the slope) and introduce a constant of integration. The second step will result into y and introduce yet another constant of integration. These steps are illustrated below.

Bending moment The slopeThe deflection8The constants C1 and C2 are determined from boundary condition.

The boundary condition are the known values of dy/dx and y at given values of x. these values are invariably known at supports. For determining two constants at least two boundary conditions must be known.

The following boundary condition will be identified:-a. at simple supports y = 0, dy/dx 09A SIMPLY SUPPORTED BEAM- with concentrated load on the center of the beam.

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A SIMPLY SUPPORTED BEAM- with concentrated load on the center of the beamThe method will now be used for calculating the maximum deflection in case of simply supported beam loaded by a single concentrated load in the middle of the span.

ABSimply supported beam concentrated load at center of the beam11The reaction force at A and B;RA = RB = P/2

The maximum deflection will occur just under the load or at x = L/2 with origin at left hand support.

This will mean that at x = L/2, dy/dx = 0. Thus boundary conditions for the determination of constants of integration are;y = o at x = o

= dy/dx = o at x = L/2(i) First boundary condition Second boundary conditionSimply supported beam concentrated load at center of the beam12With origin at left support and x measuring positive to the right, M(x) = P/2 . x (x L/2)From eq. 1.1;

(i)(ii)Simply supported beam concentrated load at center of the beam13Integrating eq. (ii);

Using second boundary condition; dy/dx = o at x = L/2

Hence,

Integrating eq.(iii);

(iii)Simply supported beam concentrated load at center of the beam14Using first boundary condition; y= o at x = 0B = 0

Hence

The maximum deflection occurs at x = L/2,

The maximum slope, occurs at x = 0;

Simply supported beam concentrated load at center of the beam15A SIMPLY SUPPORTED BEAM- with uniformly distributed load along the beam

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A SIMPLY SUPPORTED BEAM- with uniformly distributed load along the beam

The method will now be used for calculating the maximum deflection in case of simply supported beam loaded by a single concentrated load in the middle of the span.ABSimply supported beam uniformly distributed load along the beam17The reaction force at A and B;RA = RB = wL/2

The maximum deflection will occur just under the load or at x = L/2 with origin at left hand support.

This will mean that at x = L/2, dy/dx = 0. Thus boundary conditions for the determination of constants of integration are;y = o at x = o

dy/dx = o at x = L/2(i) First boundary condition Second boundary conditionSimply supported beam uniformly distributed load along the beam18With origin at left support and x measuring positive to the right,

From eq. 1.1;

Integrating eq. (ii);

(i)(ii)

Simply supported beam uniformly distributed load along the beam19Using second boundary condition; dy/dx = o at x = L/2

Hence

Integrating eq.(iii);

(iii)

Simply supported beam uniformly distributed load along the beam20Using first boundary condition; y= o at x = 0B = 0

Hence

The maximum deflection occurs at x = L/2,

The maximum slope occurs at x = 0;x = L;

Simply supported beam uniformly distributed load along the beam21A CANTILEVER BEAM- with single concentrated load at the free end

22Figure above shows a cantilever of length L and loaded at free end by a single concentrated load P. the moment of inertia of the beam cross-section is I and modulus of elasticity of its material is E. AB

A CANTILEVER BEAM- with single concentiated load at the free end (edge of the beam)Cantilever beam single concentrated load at the free end23The origin of the coordinates is chosen at free end with x measuring positive to the right. The broken line in figure shows the elastic curve of the cantilever. To determine the equation to elastic curve, the first step is to write down the expression for BM at a section x away from the origin. M(x) = - P. x(BM has been written as M(x) to emphasise its variable character)Substituting the value of BM in Eq. (1.1):

(i)(ii)Cantilever beam single concentrated load at the free end24The double integration may now be perform to obtain y but it is worthwhile at this stage to identify the boundary conditions, which are:y = o at x = Ldy/dx = o at x = L

It may also be noted that expression eq.(i) is applicable for all values of x from x = 0 to x = L.

Integrating eq.(ii);

First boundary condition Second boundary condition

Cantilever beam single concentrated load at the free end25Using second boundary condition; dy/dx = o at x = L

Hence

Integrating eq.(iii);Using first boundary condition; y= o at x = L

(iii)

Cantilever beam single concentrated load at the free end26Therefore the eq. become

The maximum deflection occurs at x = 0.

The maximum slope occurs at x = 0.

Cantilever beam single concentrated load at the free end27A CANTILEVER BEAM- With uniformly distributed load along the beam

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Figure above shows a cantilever of length L and loaded an uniformly distributed load along the beam. The moment of inertia of the beam cross-section is I and modulus of elasticity of its material is E. A CANTILEVER BEAM- with uniformly distributed load along the beamCantilever beam uniformly distributed load along the beam29The origin is at free end and x measure positive towards right. w is the rate of loading.To determine the equation to elastic curve, the first step is to write down the expression for BM at a section x away from the origin.

Substituting the value of BM in Eq. (1.1):

(i)(ii)

Cantilever beam uniformly distributed load along the beamCantilever beam uniformly distributed load along the beam30The double integration may now be perform to obtain y but it is worthwhile at this stage to identify the boundary conditions, which are:y = o at x = Ldy/dx = o at x = L

It may also be noted that expression eq.(i) is applicable for all values of x from x = 0 to x = L.

Integrating eq.(ii);

First boundary condition Second boundary condition

Cantilever beam uniformly distributed load along the beam31Using second boundary condition; dy/dx = o at x = L

Hence

Integrating eq.(iii);Using first boundary condition; y= o at x = L

(iii)

Cantilever beam uniformly distributed load along the beam32Therefore the eq. become

The maximum deflection occurs at x = 0.

The maximum slope occurs at x = 0.

Cantilever beam uniformly distributed load along the beam33A CANTILEVER BEAM- With a single concentrated load at a point other than free end or not at free end.

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A CANTILEVER BEAM- with a single concentrated load at a point other than free endAs known the deflection for single concentrated load at free end is

35Eq.is first used to calculate yp by

replacing L by a. Thus;

From eq., the slope at x = o, i.e

Under load P,

(i)Deflection at B

(ii)Slope at B36From triangle ABD;

whereDB = (L-a)

Hence

Deflection of free end or deflection at A:

or

37Example 1:-A beam measuring 2 m long simply supported, is the concentrated force of 10 kN at the mid-point. Determine the maximum deflection occurs. Given E = 200 GN/m2 and I = 12 x 10-6 m4.

Solution:-From the proof of the simply supported beam with concentrated load on the center of beam, has been proved that the maximum deflection is: -

Example 138

Example 139Example 2:A 4 m long simply supported beam carrying a uniform load of 10 kN/cm. Calculate the deflection at the center of the beam.Given EKeluli = 2.0 GN/cm2 and I = 2000 cm4.

Example 240Solution:-From the proof of the simply supported beam with uniformly distributed load along the beam, has been proved that the maximum deflection is: -

Example 241Example 3:A wooden beam is simply supported and carries a uniformly distributed load 4 kN/m as shown in below figure. Determine the value of b if the deflection allowed is 20 mm. Given E = 10 GN/m2.

Example 342Solution:-

Example 343Example 4:A simple beam which length 8m supports a concentrated load of 300 kN at the middle span. If the material contains a value of EI = 80 x 103 kN/m2, calculate:-a. The maximum beam deflection, ymax.b. The slope (angle) of deflection at ymax, dy/dxExample 444

Solution:-a. b.

Example 445Determine the integral of variables for the equations of the beam deflection in the diagram below.

Example 5:46Solution:-The slope and deflection of the beam equation is as follows:-

By using the conditions of a suitable limits, determine the value of A and B.

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48The position of the point in simply supported beam

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The position of the point in cantilever beam.50EXERCISE 11. Calculate:-a)the maximum deflectionb)the maximum slopefor loads of the beam as shown in each figure.Given E = 200 GN/m2.

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Answer:- i)a) 2.83 mmii)a) 1.6 mmb) 4.24 radianb) 6 x 10-4 radianiii)a) 51.2 mmiv)a) 2.025 mmb) 0.128 radianb) 3.375 x 10-4 radian

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