Deflection in beams

Mechanics of Solid Deflection in Beams

Introduction:

The rigidity of a flexural member depends upon the length of the beam, types of load and their magnitude. If

the deflection in a beam is beyond the permissible limit, there will be a loss of rigidity causing undesired

deflections and slopes, and also the smooth operation of a flexural member becomes impossible.

Relation between Bending Moment and Curvature:

Basic relationship between curvature (d2y/dx2) and bending moment, M, provides the starting equation for

the determination of slope and deflection at any section of the beam. Under the action of transverse loads, a

beam bends over span length L as shown in Fig. 11.1(a). Consider the length of BC = δL along the curved

beam with horizontal projection δx and vertical projection δy. The radius of curvature of small

length δL = BC is R and O is the centre of curvature. BG is the tangent to the curve at point B, and FCF is the

tangent to the curve at point C.

Slope at B = ϕ

Slope at C = ϕ + δϕ

The angle subtended by length δL at the centre of curvature = δϕ

or, Rδϕ = δL

Differentiating the above equation on both sides,

But, slope dy/dx, is a very small quantity in any beam.

This differential equation gives the relationship between the moment of resistance and curvature (in the

Cartesian co-ordinates of a point on the beam).

Figure 11.1 Bending in beam


Sign Conventions:

1. On the left side of a section, upward shear force is positive or the shear force tending to rotate the body in

clockwise direction is positive.

2. On the left side of the section, clockwise bending moment is positive, the bending moment which produces

concavity upwards is a positive bending moment.

3. In x–y Cartesian co-ordinate system, x is positive towards right and y is positive in upward direction.

Figure 11.2 shows a beam AB in bent shape showing concavity upwards, the bending moment from A to B is

positive. The deflection at A, yA, and the deflection at B, yB, are positive, while the deflection at C, yC, is

negative (below x axis). Similarly the slope at E, iE, is positive, while the slope at D, iD, is negative.

Figure 11.2 Slopes and deflections in beam

Simply Supported Beam with a Central Point Load:

A beam of length L, hinged at end A and roller supported at end B, carries a central point load W at centre C.

Flexure curve of the beam is ACB, Fig 11.3. The slope at A is –i′A, the slope at B is –i′B, and the slope at C is zero

due to symmetrical loading. Similarly, the deflection at ends A and B is zero but the deflection at centre C is yC,

the maximum. Due to symmetry, reactions at A and B will be equal, that is,

Consider a section YY at a distance of x from end A.

Then, bending moment,

Using the equation 11.1 of bending moment and curvature,

Integrating Eq. (11.2), we get

where C1 is a constant of integration.

At the centre, x = L/2, dy/dx = 0, and the slope is zero

From the above equation, we cannot find the slope at B because we have made equation of bending moment

only in portion AC of the beam.

Integrating Eq. (11.3) again,

Figure 11.3


where C2 is another constant of integration.

The deflection y = 0 at x = 0, at end A,

0 = 0 − 0 + C2

Hence, constant, C2 = 0

Finally,

Deflection at C, x = L/2, by substituting the value in Eq. (11.4),

A Beam Carrying Udl with Simply Supported Ends:

A beam AB, simply supported at ends over a span L, carries a udl (uniformly distributed load) of

intensity w per unit length. Total load on beam = wL, due to symmetrical loading about the centre of the

beam.

Reactions, RA = RB = wL/2, Fig. 11.4.

Considering a section at a distance of x from end A, then the bending moment at this section is

In this case, Eq. (11.5) is sufficient to determine the slope and deflection at any section of the beam as there is

only one portion, AB.

Integrating Eq. (11.5),


Slope dy/dx = 0, at x = L/2, by substituting this value,

Integrating Eq. (11.6),

where C2 is another constant of integration at end A, x = 0, deflection, y = 0.

So, 0 = 0 − 0 + C2

or constant, C2 = 0.

Finally,

Figure 11.4


This shows that expression (11.7) is valid from one end to other end of beam.

At the centre, x = L/2, the deflection is maximum, therefore,

Slope is maximum at ends, when x = 0, from Eq. (11.6),

At end B, x = L, by substituting this value in Eq. (11.6),

A Cantilever with the Point Load at Free End:

Figure11.5 shows a cantilever AB of length L, free at end A and fixed at end B, carrying a point load W at A.

Considering a section YY at a distance of x from A.

Bending moment, Mx = −Wx (bending moment producing convexity upwards)

Integrating Eq. (11.8)


at fixed end x = L, slope dy/dx, so

Integrating Eq. (11.9), we get


at x = L, fixed end, deflection, y = 0

Finally,

Figure 11.5 Cantilever with point load at free end


Note that both slope and deflection are maximum at free end A, where x = 0.

;

A Cantilever with a UDL:

A cantilever AB of length L, free at end A and fixed at end B carrying a uniformly distributed load of

intensity w per unit length, is shown in Fig. 11.6. The cantilever bends as shown, point A is shifted to A′. Slope

and deflection are maximum at free ends.

Total load on the cantilever = wL′

Reaction at B, RB = wL′

Fixing moment at B, MB = wL2/2 as shown.

Consider a section YY at a distance x from end A.

Bending moment, Mx = Mx = –wx2/2 (bending moment tending to produce convexity is negative)

Integrating


At end B, fixed end, slope dy/dx = 0

Integrating again, we set


Deflection y = 0, at fixed end B, where x = L

Finally,

At end A, x = 0.

Slope and deflection at any section of the cantilever from A to B can be determined by using Eqs

(11.11) and (11.12).

Example: A steel cantilever of I section with Ixx = 1,200 × 104 mm4, and length 4 m carries a udl of

intensity w kN/m. If the maximum deflection in cantilever is not to exceed 1 mm, E = 208 GPa. Determine the

value of w.

Solution: E = 208 × 106 kN/m2 ; I = 1,200 × 10−8 m4 ; EI = 2,496 kN m2

Say, rate of loading = w kN/m

Length, L = 4 m

Figure 11.6 A cantilever with a udl

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Maximum deflection, ymax = wL4/8EI = 1 mm = 1 × 10−3 m

or

Macaulay’s Method:

This is the most versatile technique to determine slope and deflection at any section of a beam/cantilever

carrying any type of loading or a combination of loading such as point loads, udl, moment or a variable load.

In this method following steps are taken:

1. Determine the reactions at supports using the equations of equilibrium.

2. Select any one end as origin and go to the last portion of the beam (separated by loads) and take x from

origin to a section YY in the last portion.

3. Make the equation of bending moment for the section under consideration.

As example shown in fig. Last portion is DE, BM at section YY

4. Integrate this equation two times with constants of integration, C1 and C2.

5. Use boundary conditions of slope and deflection at ends and by carefully using the terms of the equation

which are valid for that end, determine constants C1 and C2.

6. These constants will be valid for all portions of the beam.

7. Once the equations for deflection and slope with known constants C1 and C2 are made, then slope and

deflection can be calculated at any section of the beam.

Example: A beam AB, 10 m long, carries point loads of 6 and 3 kN at C and D as shown in Fig.. Determine

support reactions, deflection at C and D, and slope at ends A and B, if EI is the flexural rigidity of the beam.

Solution:

Reactions

Taking moments about A,

6 × 4 + 7 × 3 = 10RB

Reaction, RB = (24 + 21)/10 = 4.5 kN

Total load on beam = 6 + 3 = 9 kN

Reaction at A, RA = 9 − RB = 9 − 4.5 = 4.5 kN.

There are three portions, that is, AC, CD and DB in the beam and if A is the origin then DB is the last portion.

Consider a section at a distance x from A in the portion DB,

Bending moment, Mx = 4.5x − 6(x − 4) − 3(x − 7)

or

Integrating

Since the beam is not symmetrically loaded about its centre, so we do not know where slope is zero.

By integrating


Boundary conditions x = 0 at end A, deflection y = 0

Moreover in portion AC, only the first term is valid and the other two terms of equation are not valid.

So, 0 = 0.75 × 0, neglected term – neglected term + C1 × 0 + C2

or, constant C2 = 0

At end B, x = 10 m, deflection y = 0, by substituting the value

0 = 0.75 × 103 – 63 – 0.5 × 33 + 10C1

0 = 750 – 216 –13.5 + 10C1

Constant, C1 = –52.05

Finally, the equations are

EIy = 0.75x3 − (x − 4)3 − 0.5(x − 7)3 − 52.05x (11.16)

Slope at end A, x = 0

EIiA = 2.25 × 0 – neglected terms –52.05

or

at end B, x = 10, so all the terms are valid

Deflection

At point C, x = 4 m, hence, the third term in the equation is invalid.

EIyC = 0.75 × 43 − (4 − 4)3 − neglected term − 52.05 × 4

EIyC = 48 − 0 − 208.2 = −160.2

At point D, x = 7 m, and all the terms in the equation for deflection are valid.

EIyD = 0.75 × 73 − (7 − 4)3 − 0.5(7 − 7)3 − 52.05 × 7

257.25 − 27 − 0 − 364.35= −134 1

Example: A beam ABCD, 6 m long hinged at end A and roller supported at end D, is subjected to CCW moment

of 10 kN m at point B and a point load of 10 kN at point C as shown in Fig. Determine the deflection under

load of 10 kN and slope at point B, by taking EI as flexural rigidity of the beam.


Solution:

Taking moments at A,

10 + RD × 6 = 10 × 4

RD = 5kN ↑

Total load on the beam = 10 kN

Reaction at A, RA = 10 – 5 = 5kN ↑

There are three portions: AB, BC and CD in beam, with A as the origin portion and CD is the last. Take a

section YY in portion CD at a distance x from A as shown in the figure.

The equation of bending moment is

Note that 10 kN m is a moment and we cannot take moment of moment, but 10 kN m is applied at B at a

distance of (x – 2) m from section YY. Moreover (x – 2)°= 1, so only to locate the position of moment, the term

(x – 2)° is taken, that is, (x – 2) raised to power zero, this term locates the position of moment applied at B.

Integrating we get

where C1 is a constant of integration, we do not know where the slope is zero as the beam is not

symmetrically loaded.

Also by integrating

where C2 is another constant of integration. y = 0 at end A, x = 0, only first two terms in between are valid in

portion AB, so

or, constant, C2 = 0.

y = 0 at end D, where x = 6 m, all the terms in the equation are valid, so

Finally, the equations for slope and deflection are

The slope at B and x = 2 m, the third term is not valid

Deflection at C, x = 4 m, all the terms in equation for deflection are valid for x = 4 m


Example: A beam ABCD, 7 m long hinged at A and roller supported at D carries 7 kN load at B and 4 kN/m udl

over BC = 3 m. If EI = 14,000 kN m2 for the beam, determine the slope at A and deflection at point C.

Solution:

Reactions

Total udl on beam = 4 × 3 = 12 kN

CG of this load lies at 2 + 1.5 = 3.5 m from end A

Taking moments about A,

7 × 2 + 4 × 3(3.5) = 7 RD

RD = 8 kN ; RA = 11 kN

Last portion of the beam is CD. A section YY at a distance x from end A in the portion CD of the beam is taken

to make the equation of bending moment valid for all the three portions. The udl is extended to section YY on

both sides (upward and downward), so that its net effect becomes zero.

Bending moment at section YY

where w is the rate of loading.

Note that the first term is valid for portion AB, the first three terms are valid for portion BC, and all the four

terms are valid for portion CD of the beam.

Substituting the value of w = 4 kN/m,

Integrating


Integrating


At end A, x = 0, y = 0 (in portion AB, x = 0, their three terms are invalid)

Constant, C2 = 0.

At the end D, x = 7 m, y = 0, and all the terms in equation are valid.

Constant, C1 = –54.5

The equations will become

Slope at A, x = 0


Deflection at C, x = 5 m, and all the terms are valid

Eccentric Load on a Beam:

A beam AB of length L, hinged at end A and roller supported at end B, carries a load W at point C such that,

AC = a; CB = b; a + b = L

As shown in fig. a<b. In this case, we do not know where the slope is zero but certainly we know that the

deflection at ends A and B is zero.

Reactions

Taking moments about A, Wa = RBL

Reaction, RB = Wa/L

By taking the origin at A and x is positive towards right at section YY at a distance x from A in the portion CB.

Integrating two times, we obtain

At end A, y = 0, x = 0

EI × 0 = 0 − neglected terms + 0 × C1 + C2

Constant, C2 = 0

At end B, y = 0, x = L

The expression for slope and deflection will be

Deflection under the load x = a


If a = b = L/2 (for central load)

Example : A beam ABC, 8 m long carries an eccentric load at B, such that AB = 3 m, BC= 5 m. If EI = 5,000 kN

m2, determine (1) slope at ends A and C, and (2) maximum deflection.

Solution:

Reactions

Taking moment about A, 8 × 3 = 8 Rc

Reaction, Rc = 3 kN

Reaction, RA = 8 – 3 = 5 kN

Taking a section in portion BC,

Mx = 5x – 8(x – 3)

Integrating two times, we obtain

where C1 and C2 are constants of integration. At x = 0, y = 0,

EI × 0 = –omitted term + 0 × C1 + C2

Constant, C2 = 0

At x = 8 m, end C, deflection y = 0

Equation for slope will become

At A, x = 0

At C, x = 8 m

ymax occurs in the beam where slope dy/dx is zero. Let us find location of ymax’

x = 3.718 m

Substituting the value of x in equation of deflection,


Impact Loading of a Beam:

If a load falls from a height onto a beam, instantaneous deflection is produced in the beam, causing

instantaneous stress of high level in the beam and the beam starts vibrating, but ultimately the vibrations die

down as the amplitude of vibration goes on decreasing due to air damping. Consider a beam AB of

length L and flexural rigidity EI as shown in Fig. 11.16. A load W falls from a height h on the beam and the

deflection under the load δi is produced in the beam.

Loss of potential energy of the falling weight = W (h + δi)

Strain energy absorbed by the beam = 1/2Pδi,

where P is the equivalent gradually applied load on the beam which when applied gradually produces

deflection δi.

Say the load falls at the centre of the beam,

then

where K = stiffness constant of beam =48EI/L3

If W and h are given then δi can be calculated. The maximum instantaneous stress developed due to δi can also

be calculated. Note that once the vibrations die down. δi will approach

Example: An ISMB 150 rolled steel section is held as a cantilever of length 2 m. A weight of 200 N is dropped

at the free end of the cantilever producing an instantaneous stress of 90 N/mm2. Calculate the height from

which the weight was dropped and the maximum instantaneous deflection in the cantilever. I = 726.4 × 10–

8 m4, E = 200 GPa

Solution

Length of the beam = 2 m

Say, equivalent load = P kN

Mmax, maximum bending moment = 2P kN m

σmax, maximum stress developed = 90 MPa = 90 × 106 N/m2 = 90 × 103 kN/m2

EI = 726.4 × 10-8 × 200 × 106 = 1,452.8 kN m2

Depth = 0.075 m (Note that 150 stands for 150 mm as depth of beam)

Instantaneous deflection,

Maximum instantaneous deflection, δi = 0.008 m

W = 200 N = 0.2 kN

h = 0.087108 – 0.008 = 0.07917 m = 79.17 mm

Figure 11.16


Propped Cantilevers:

A cantilever fixed at one end and simply supported at other end is known as propped cantilever. The propped

cantilevers are of two types:

1. A known force P is applied at the free end in a direction opposite to the direction of applied load W as

shown in Fig. 11.17.

Bending moment equation is

Taking the boundary conditions at end B, that is, x = L, y = 0, dy/dx = 0, constants of integration are found

out. Hence, the slope and deflection at any section of the cantilever can be determined.

2. Free-end of the loaded cantilever is simply supported so that the deflection at free end is zero, by knowing

the boundary conditions at A and B, constants C1 and C2and reaction at A, that is, RA are determined as

shown in Fig. 11.18.

Equation of bending moment for a section YY in portion CB,

Mx = RAx – W(x – a)

Figure 11.17 Propped cantilevers Figure 11.18 Propped cantilevers

Integrating

There are three boundary conditions, that is, x = 0, y = 0, x = L, y = 0, dy/dx = 0, thus one can

determine RA, C1 and C2, (3 unknowns.)

Example: Cantilever AB, 5 m long, is simply supported at A and fixed at B. If it carries a udl of 6 kN/m

over CB = 3 m, EI of cantilever is 3,600 kN m2. Determine the reaction at Aand slope at A and also find out the

deflection at C (Fig. 11.20)?

Solution:

Say reaction at A is RA. Taking a section at distance x from A, in the portion CB,

Bending moment,

where w is rate of loading

or,

Integrating we get

where C1 is a constant of integration; dy/dx = 0 at x = 5, by substituting this value, we obtain

Figure 11.20


So,

Also by integrating

At A, x = 0, y = 0

C2 = 0

Finally,

Moreover, y = 0 at end B, x = 5 m, by substituting this value,

By substituting the value of RA, equations of slope and deflection are

At A, x = 0

Deflection at C, x = 2 m

Stepped Beam:

Consider a beam AB of length L, with moment of inertia I1 for AC and I2 for CB portions of the beam, which is

subjected to a central point load W. Let us determine the deflection at the centre of the beam.

Reactions are RA = RB = W/2 (as shown).

Taking portion CB, the equation of bending moment becomes

The above equation shows the variable moment of inertia. The equation is modified because the section of the

beam is not uniform.

Integrating two times, we obtain,

at x = 0, y = 0

0 = 0 – omitted term + 0 + C2

Constant,C2 = 0 Figure 11.23


at x = L, y = 0, other end B

By substituting the value of RA = W/2,

Therefore,

At the centre x = L/2,

Say I1 = I2 = I

Let us take I1 = 2I2

Slope and Deflection by Area Moment Method:

Slope and deflection at any section of a beam can be obtained by (a) area of bending moment diagram, and (b)

first moment of area of BM diagram.

Equations of curvature and moment are

multiplying both order by dx

Mdx is the area of BM diagram over a small length dx.

Integrating both sides

area of BM diagram between sections Y2Y2 and Y1Y1

or between distances x2 and x1

or i2 − i1 = area of BM diagram between Y2Y1.


Multiply by xdx on both the sides, and then integrating, we get

a = area of diagram abcdef.

= distance of CG of this diagram from A

Example: A simply supported beam of span length L carries a udl of intensity w throughout its length.

Determine the slope at A and deflection at C by moment area method.

Solution: BM diagram is parabolic for this case with maximum bending moment at centre = + wL2/8

Area of the bending moment diagram from A to C that is, area AC′C

Because of symmetrical loading, ic = 0, and slope at centre is zero.

Now area,

CG of area AC’C lies at a distance of (5/8 × L/2) from A.

Now,

But xc = L/2, ic = 0

Therefore, xA = 0, yA = 0

Finally,

Conjugate Beam Method:

In this method, bending moment diagram for beam due to transverse loads on it is considered as loading

diagram (but in term of variation of bending moment). Taking this bending moment diagram as loading

diagram on the beam, reactions are calculated at supports (in term of bending moment). Thus,


Figure 11.27 Figure 11.28

Deflection

Moment at any section due to variable bending moment can be determined, say M′c is moment at C due to

bending moment diagram AC′ D′B, and M′D is the moment at D due to variable moment AC′ D′B, then,

A simple example will help in understanding the concepts of conjugate beam method. A beam AB of length L is

simply supported at ends, which carries a concentrated load at the centre as shown in Fig. 11.28. EI is the

flexural rigidity of the beam.

Figure 11.28(b) shows the bending moment diagram of the beam with maximum bending moment, WL/4 at

the centre supported over length L, at A and B.

Reactions at A and B are given as

At the point C, moment

Slope and Deflection of Stepped Beams:

For stepped cantilevers/beams, conjugate beam method is very conveniently applied. In such cases, bending

moment diagram is plotted for the beam/cantilever with bending moment diagram as a load diagram and the

reactions at ends are obtained. The ratio of reaction/EI gives slope at the end, ratio of moment (of bending

moment) divided by EI gives deflection at any section. In case of cantilever, maximum slope and deflection

occur at free end, while both slope and deflection are zero at fixed end. Therefore in conjugate beam method,

cantilever free end becomes the fixed one and the fixed end becomes the free one, so that the reaction and

moment can be obtained at this fixed end (which is initially free end) of the cantilever.

Example: A beam of length L carries a central load W as shown in Fig. 11.32(a). Moment of inertia for quarter

length from ends is I1 and for the middle half length moment of inertia is I2, such that I2 = 2I1, now draw the

conjugate beam diagram.

Solution:


If E is Young’s modulus of the material, diagram AE′B is the bending moment diagram such that EE’ = WL/4

Conjugate beam diagram gives,

Beam is symmetrically loaded, therefore reactions, RA′ = RB′.

RA′ = area of conjugate beam diagram up to centre.

iE = slope at centre is zero

iE – iA = area of conjugate beam diagram up to centre.

Moment at centre, M′E

Figure 11.32

Deflection in beams

Engineering

Transcript of Deflection in beams