FIXED BEAMS IN BENDING

INTRODUCTION

Fixed or built-in beams are commonly used in building construction because they possess high rigidity in comparison to simply supported beams. When a simply supported beam of length, L, and flexural rigidity, EI, is subjected to a central concentrated load W, the maximum bending moment (BM) in beam is (WL/4), maximum deflection in beam is WI3/48EI, and maximum slope in beam is ±WL2/16EI but if the ends of the same beam are fixed (i.e.

built in walls), maximum bending moment is reduced to (WL/8), maximum deflection is reduced to WL3/192EI, and maximum slope is reduced to ±WL2/32EI. If the allowable bending stress for the fixed beam is taken to be

the same as for simply supported beam, then the load carrying capacity of the fixed beam is greatly improved. The overall reduction in bending moment, deflection, and slope in the fixed beam is due to the effect of fixing couples provided by the wall in keeping the slope and deflection at the ends to be zero. During building construction, the beams are cast along with the columns using reinforced cement concrete.

W

W/2 W/2

EIWL

y

WLWLWLLWLLWEIy

Lyat

Cyx

CxWLxW

deflection

EIWL

dx

dyWLC

CLW

slopeL

x

CxW

dx

dyEIslope

xW

dx

ydEIxM

48

48963

96216862

2

0000,0

1662

1616

24200,

2

22

2)(

3

33323

2

2

23

22

1

1

2

1

2

2

2

W/2

W/2

WL/4

x

x

y

the maximum bending moment (BM) in beam is (WL/4), maximum deflection in beam is WI3/48EI, and maximum slope in beam is ±WL2/16EI

simply supported beam

FIXED BEAM−BENDING MOMENT DIAGRAM

A fixed beam can be considered as equivalent to a simply supported beam plus a beam of the same length, same material, same section subjected to end moments as shown in Figure 1 (a), (b) and (c).

Figure 1 Fixed beam

MD

A beam ABCD, of length L, fixed at ends A and D, carries loads W1 and W2 at points B and C, respectively, is equivalent to a simply supported beam of length L, simply supported at ends A and D, carrying loads W1 and W2 at B and C, respectively, plus a beam AD of length L, subjected to end moments MA and MD. Considering the beam as simply supported (SS) beam, bending moment diagram can be drawn as shown in Figure 2 (a). Figure 2 (b) shows the bending moment diagram of beam due to end moments. Due to end couples, there is convexity in beam, producing negative bending moment throughout.

Figure 2 (a) BM diagram of SS beam (b) BM diagram of end couples (c) BM diagram of fixed beam P1, P2 are point of contraflexure

MD

Due to loads and simple supports at ends, there is concavity in the beam producing positive bending moment throughout in the beam. Superimposing the two bending moment diagrams, we get the bending moment diagram for fixed beam as shown in Figure 2 (c), in which P1 and P2 are points of contraflexure. In the combined bending moment diagram there are positive bending moments and negative bending moments. At P1 and P2 bending moment changes sign from negative to positive and from positive to negative as shown.

Figure 2 (a) BM diagram of SS beam (b) BM diagram of end couples (c) BM diagram of fixed beam P1, P2 are point of contraflexure

MB MD

Let us consider a section XX at a distance of x from end A of the beam. M, resulting bending moment at section

= Mx + M ′x

MB

where Mx = Bending moment at section when beam is considered as simply

supported, and

M′x = Bending moment at section due to end moments MA and MD.

xM

AM

DMx L-x

xM

DAAx MM

L

xMMM

MD

X

X

FIXED BEAM−SUPPORT MOMENTS

If we refer to Figures 2 (a), (b), (c), resultant bending moment at any section, M = Mx + M′x

or

= Bending moment at section when beam is SS + Bending moment at section due to support moments. Multiplying both sides of by dx, we get

xx

xx

MMdx

ydEI

MMM

2

2

dxMdxMdxdx

ydEI xx

2

2

Integrating this equation over the length of the beam

or

= Area of the BM diagram of SS beam + area of BM diagram due to

support moments

where i′D = Slope at end D, length is L, and

i′A = Slope at end A, length is zero

But beam is fixed for both the ends, so slope at ends

The equation becomes

EI × 0= a +a′ where

aadx

dyEI

dxMdxMdxdx

ydEI

L

L

x

L

x

L

0

000

2

2

L

MMa DA

2

This equation shows that the area of BM diagram due to support moments is

numerically equal to the area of the BM diagram when beam is SS.

Taking again and multiplying both the sides by x dx, and integrating

both the sides over length of beam:

or

Putting the values of slope and deflection at ends

But in a fixed beam, at the ends, slope and deflection both are zero, therefore,

xaxaydx

dyxEI

dxxMdxxMdxxdx

ydEI

L

L

x

L

x

L

0

000

2

2

xx MMdx

ydEI

2

2

aL

MM DA 2

xaxayiyiLEI AADD 0

xaxa 0

or

But a = −a′ So,

i.e. distance of centroid of BM diagram when SS is equal to the distance of the centroid of BM diagram due to support moments from origin A, or from end of the beam. From the a’ diagram, area

(origin at A),

Dividing the diagram into two triangles, we can determine that centroid of area A′AD′ lies at

from A and centroid of triangle ADD′ lies at

With the help of equations and , support moments MA, MD can be worked out.

from A (Figure 3).

Figure 3 a′ diagram

xaxa

DA

DA

DA

MML

LLM

LLMxa

LMMa

26

32

232

2

2

GOVERNING DIFFERENTIAL EQUATION FOR DEFLECTION

In pure bending of a prismatic beam, for an element dx bent by an infinitesimal angle dq, we can write

From Hooke’s law,

dx

qd

q

q

1

dx

d

ddx

q

yy

yydx

dy

dx

du

1

yEE xx

y

The fiber length on a radius – y can be found similarly. If the difference between the two fiber lenghts is du:

qqq ydddydu

By dividing both sides by dx, the last term becomes K. The normal strain will be equal to:

In this equation, the variable y can assume both positive and negative values. Since,

yEyI

Mx

EI

M

Curvature

In Cartesian coordinates the curvature of a line is defined as

2/32

2

2

1

1

dx

dy

dx

yd

Where x locates a point on the elastic curve of a deflected beam and y gives the deflection of the same point

from its initial position.

Since the deflections tolerated in the majority of engineering structures are very small, slope of the elastic

curve is also very small. Therefore, the square of the slope can be neglected compared to one. dx

dy

So, the curvature can be taken as

2

21

dx

yd

This simplification eliminates the geometric nonlinearity from the problem. Now, the governing differential

equation for small deflections of elastic beams becomes:

EI

M

dx

yd

2

2

There are three governing equations for determining the deflection of a loaded beam:

2

2

)(dx

ydEIxM

3

3

)(dx

ydEIxV

4

4

)(dx

ydEIxq

The choice of one of these equations for determining y depends on the case with which an expression for load,

shear or moment can be formulated.

Single moment: 2

)(

axMxq O

Concentrated force: 1

)(

axFxq O

Uniformly distributed load: 0

)( axwxq O

1)( ax

dx

dwxq Linearly increasing distributed load:

2

2)( ax

cxq Quadratically increasing distributed load:

a

x

y

MO

a

x

y

FO

a

x

y

wO

a

x

y

dx

dw

a

x

y

cdx

wd

2

2

In order to keep distributed loads within

specified boundaries, same magnitude but

negative direction distributed loads must

be applied for the remaining length.

Macaulay’s method:

Macaulay’s method:

FIXED BEAM WITH A CONCENTRATED LOAD AT CENTER

Consider a fixed beam AB of length L, fixed at both ends A and B, carrying a point load W at its center as shown in Figure 4. It is equivalent to a simply supported beam AB with central load W and a beam AB subjected to end moments MA and MB. Since the beam is symmetrically loaded, end moments will be equal, i.e. MA = MB

Figure 4 Equivalence of a fixed beam

area of a BM diagram, In SS beam, BM at centre

Figure 4 a−BM diagram

area of a′ diagram, a′ = MA L = MB L

(MA = MB, because of symmetrical loading)

or a′ = −a

or end moment,

Resultant bending moment diagram is shown in Figure 28 (d), in which

maximum positive

BM

maximum negative

BM

The maximum bending moment in SS diagram is (WL/4) but when it is fixed at both the ends, maximum bending moment is reduced to 50%, i.e. (WL/8). Now let us calculate the maximum deflection in a fixed beam. Again consider a fixed beam AB of length L fixed at both the ends, load at centre is W as shown in Figure 5, reactions at ends.

Figure 5 Fixed beam with central point load

(due to symmetry)

End couples

Take a section X − X, at a distance of x from end A, in the portion CB. Bending moment,

(as obtained earlier)

or integrating it, we get

where C1 is constant of integration. (Term in bracket is to be omitted) At end A,

because end is fixed. So, integration constant, C1 = 0 Now integrating the above equation again, we get

where C2 is another constant of integration. At end A, x = 0, deflection, y = 0, because end is fixed. So, 0 = 0 + 0 − omitted term + C2, using Macaulay’s method, constant C2 = 0.

Finally, the equation for deflection is

Maximum deflection will occur at centre, i.e. at

Putting this value of x,

or

Example A beam of length L = 6 m, fixed at both the ends, carries a concentrated load of 30 kN at its centre. If EI = 7,000 kNm2 for the beam, determine the fixing couple at ends and maximum deflection in the beam.

FIXED BEAM WITH UNIFORMLY DISTRIBUTED LOAD THROUGHOUT ITS LENGTH

Consider a beam AB, of length L fixed at both the ends A and B, subjected to a uniformly distributed load of intensity w per unit length, throughout the length of the beam as shown in Figure 6 (a). It is equivalent to a SS beam of length L, subjected to udl, w throughout its length and a beam AB, subjected to end moments MA and MB as shown in Figure 6 (b) and (c). Since the beam is symmetrically loaded, end moments MA = MB

Figure 6 (a) Fixed beam; (b) a − BM diagram; (c) a′ − BM diagram

When the beam is S.S, bending moment diagram is a parabola with ,

maximum bending moment at the centre. Area of BM diagram,

area of BM diagram, a′ = MAL = MBL (due to end moments) a′ = −a

Reactions at supports = (due to symmetrical loading).

The resultant bending moment diagram for fixed beam is shown in Figure 30 (d), in which P1 and P2 are points of contraflexure. The locations of P1 and P2 can be determined.

Figure 6 (d) BM diagram for fixed beam

Now let’s consider any section X−X at a distance of x from end A as shown in Figure 7. Bending moment at section XX is

Figure 7 Fixed beam with uniform

distributed loading

or

Integrating this, we get

at x = 0, end A, slope

Now

, so constant C1 = 0.

Integrating this equation again, we get

At end A, x = 0, deflection y = 0, so constant, C2 = 0 Finally, we have

Maximum deflection in the beam will occur

at the centre, putting the value of x =

,

Note that the deflection is reduced to only 20% of the deflection when the

beam is simply supported at ends and maximum bending moment in

beam is also reduced to from (in SS beam).

Example A beam with a 6 m span has its ends built in and carries a uniformly distributed load of 4 kN/m throughout its length. Find the (i) end moments, (ii) bending moment at the centre, and (iii) maximum deflection in the beam for E = 210 GPa, I = 4,800 cm4.

FIXED BEAM WITH AN ECCENTRIC LOAD For this problem it is very much time consuming to draw a and a′ BM diagrams and then to determine support reactions and moments. Let us take a fixed beam AB, of length L fixed at both the ends A and B and carrying an eccentric load W at C, at a distance of a from end A, Figure 32. Say reactions at A and B are RA and RB and support moments at A and B are MA and MB respectively. There are four boundary conditions at two ends of fixed beam, and we can determine four unknowns RA, RB, MA, and MB. Consider a section X−X at a distance of x from end A, in portion CB of the beam, bending moment at section XX is M = MA + RA x − W (x − a)

Figure 8 Fixed beam with eccentric load

At x = 0, end A, slope, ,

because of fixed end 0 = 0 + 0 − omitted term + C1

C

or

Integrating this equation, we get

(constant of integration)

Constant Cl = 0

Integrating this equation we get

At end A, x = 0, deflection, y = 0, because of fixed end 0 = 0 + 0 omitted term + C2

Constant C2 = 0. So,

At the end B, x = L, both deflection and slope are zero. Putting x = L in

and

or

or

From these equations,

or

Reaction RA

Putting the value of RA in , we get

Reaction RB

Similarly we can find out that support moment at B,

In Figure 32, we have taken a

Bending moment at point C, when beam is

Bending moment diagram for fixed beam is A′ A P1 C′ P2 BB′ as shown in Figure 9, where P1 and P2 are points of contraflexure. End moments,

in BM diagram.

Figure 9 BM diagram of fixed beam

C

Example Let us take numerical values of a and b, say a = 2 m, b = 4 m,

L = 6 m, W = 8 kN. What is the equation of deflection?

as shown in the figure.

In a fixed beam with eccentric load, we know that the reaction RA,

Moment MA,

Deflection equation will become

BM diagram of fixed beam

(SS)

Let us calculate deflection under the load W, i.e. at x = a

or deflection,

1

4

4

4

6

4

2

b

L

b

a

Example

A beam 8 m long, fixed at both the ends carries a vertical load of 4 kN at a

distance of 3 m from left hand end. Determine: (i) support reactions, (ii)

support moments, and (iii) location of points of contraflexure.

Example A fixed beam AB of length 6 m carries a uniformly distributed load (udl) of intensity w=6 kN/m over a length AC = 2 m and a point load of 6 kN at point D at a distance of 4 m from A. Using double integration method, determine support reactions, support moments, and draw BM diagram for the beam. What is the deflection at point D of the beam if EI = 1,400 kNm2.

There are unknown reactions RA, RB at ends and unknown moments MA, MB as shown. In this problem it is convenient if we take origin at B and x positive towards left, and section XX in portion CA (because udl is in portion CA).

Equation of BM,

or

(at end B, x = 0, slope

Integrating this equation, we get

, because of fixed end.)

(constant of integration C1)

So, 0 = 0 + 0 − omitted term + C1 Constant C1 = 0

Integrating this equation, we get

At end B, x = 0, deflection y = 0, because end is fixed.

(constant of integration C2)

0 = 0 + 0 − omitted terms + C2 Constant C2 = 0

Finally, the equation of deflection is

To determine RB, MB, let us consider end A where x = 6 m, both slope and

deflection are zero. Putting this value of x in

and

0 = 6MB + 18RB − 3 × 42 − 8

or

0 = 6MB + 18RB − 56 0 = 18MB + 36RB − 68

or MB + 3RB = 9.333 MB + 2RB = 3.777

From these equations, reaction RB is obtained as, RB = 5.556kN

Total load on beam = 6 × 2 + 6 = 18 kN

Reaction, RA = 18 − 5.556 = 12.444 kN

Moment, MB = 9.333 − 3RB = 9.333 − 3 × 5.556

= 9.333 −16.668 = −7.335 kNm

To know bending moment MA, let us put x = 6 in and putting values of MA, RB also

MA = −7.335 + 5.556 x 6 − 6 (6 − 2) − 3(6 − 4)2

= −7.335 + 33.336 − 24 −12

= −9.999 kNm.

Deflection at D: Putting x = 2 m in

Deflection,

6 R′B = 6 × 2 × 1 + 6 × 4

= 36

Reaction, R′B = 6 kN

Reaction, R′A = 12 + 6 − 6

= 12 kN

Bending Moment

BM Diagram of Fixed Beam

Let us first draw BM diagram of an SS beam with same loads. Taking moments about A, let us determine reaction R′B

MD = 6 × 2 = + 12 KNm , MB = 0

In the figure AC′D′B shows a bending moment diagram of SS beam. Let us draw a′ diagram on this, taking

MA = −9.994 KNm MB = −7.335 KN

The resultant bending moment diagram for fixed beam is obtained as seen in the figure.

EFFECT OF SINKING OF A SUPPORT IN A FIXED BEAM During the construction of a building or a structure, due to the defect in material or in workmanship, one end of a fixed beam may sink by some amount. This type of sinking is common and may cause unintentional bending moment on the beam. Let us consider that one end of the fixed beam, i.e. support B, sinks by an amount δ, as shown in Figure 10 (a), say length of beam is L. Level of support B is below the level of support A by δ. Obviously there is no rate of loading on beam, but due to vertical pressure at B, sinking has occurred, therefore,

Figure 10 (a) Sinking of a support

Integrating it, we get

Say at x = 0, Reaction = RA, then

Integrating this equation, we get

(another constant of integration)

Say at A, x = 0, bending moment is MA, so

MA = 0 + C2 and Constant, C2 = MA

Integrating this we get (constant of integration)

But slope is zero at fixed end, A, i.e. at x = 0

0 = 0 + 0 + C3 so Constant of integration, C3 = 0

So,

Integrating this we get

At x = 0, deflection, y = 0, 0 = 0 + 0 + C4 so Constant of integration, C4 = 0

(constant of integration)

Finally, we have

At end B, x = L, slope, , deflection, y = − δ

Putting these values in and

From these two equations

For equilibrium, RA = − RB or

Putting the values of MA, RA in bending moment at end B

Figures 10 (b) and (c) show the SF and BM diagrams of a fixed beam, whose

one support has sunk. If a fixed beam of length L, carrying concentrated central

load W, sinks by an amount δ at right hand support, then support moments will

be now at one end to at the other end.

Figure 10 (a) Sinking of a support, (b) Shear Force diagram; (c) Bending Moment diagram

2

6

L

EI

2

6

L

EI

2

6

L

EI

2

6

L

EI

EFFECT OF ROTATION OF A SUPPORT IN A FIXED BEAM Due to defective materials or defective construction, if one end of the fixed beam rotates (instead of keeping the slope unchanged), it causes development of support moments. Consider a beam AB, fixed at both ends but support B rotates by an angle i′B as shown in Figure 11. In this case:

Figure 11 (a) Beam in consideration

Integrating this

at end A, x = 0, shear force RA

(constant of integration)

Integrating further

(another constant of integration)

At end A, x = 0, moment = MA

Therefore, MA = 0 + C2 or constant of integration = MA

Integrating this equation, we get

(constant of integration)

At x = 0,

So, 0 = 0 + 0 + C3 or constant C3 = 0

Integrating it, we get

(constant of integration)

at x = 0, end A, deflection, y = 0. So, 0 = 0 + 0 + C4 or constant, C4 = 0 Finally we have,

Let us utilize end condition at B, where x = L,

Using these values in and

From the above equation,

Putting this value of MA in , we get

Reaction, Moment,

Figure 11 (a) Beam in consideration (b) BM diagram (c) SF diagram

To maintain equilibrium, RB = RA (but in opposite direction) Using , bending moment at B,

Figure 11 (b) shows BM diagram, and Figure 11 (c) shows SF diagram for this fixed beam.

Exercise

A fixed beam AB of length 7 m rotates at end B by 0.1° in

counterclockwise direction. If the beam carries a central load of

8 kN, determine support reactions and support moments.

Given EI = 6,500 kNm2.

Example A beam AB of uniform section throughout and of span L is fixed at end A, and carries a central point load P. During loading, support B sinks by δ and rotates by an angle δ/L in anticlockwise direction. Determine the fixing moments at ends if EI is the flexural rigidity of the beam. The figure shows a fixed beam AB of span length L, carrying a load P at its centre, support B is below the level of support B by δ, and support B rotates by angle

in counterclockwise direction. Let’s say moments and reaction at A and B are MA, RA; MB and RB, respectively. Now consider a section X−X at a distance x from A in the portion CB of the beam.

δ/L

Bending moment at the section

or

Integrating this, we get

At end A, x = 0, slope

So, 0 = 0 + 0 − omitted term + C1, Constant, C1 = 0

Integrating this again

At end A, x = 0, deflection, y = 0, so 0 = 0 + 0 − omitted term + C2 Constant, C2 = 0. Finally the equation is,

Putting x = L, at end B, where

In and

or

Adding the last two equations, we get

or

Substituting the value of MA

Reaction,

Now putting the value of RA in

Substituting the values of MA, RA in

at x = L, this equation gives BM at end B

Example A fixed beam of length L carries a linearly increasing distributed load of intensity zero at the left hand end to w up to a distance ‘a’ from the same end. Determine: (i) support reactions, (ii) support moments, if EI is the flexural rigidity of the beam. The figure shows a fixed beam of length L, carrying a linearly increasing distributed load of intensity zero at end A and w at a distance ‘a’ from A, at section c. Consider a section X-X, at a distance x from A, Rate of loading at section X-X,

Elementary load for small length of dx

Considering as an eccentric load,

Support moments, dMA

Support moment MA,

Similarly support moment MB,

For eccentric loading

Support Reactions:

Total load on beam

Taking moments,

or

Reaction,

Example A small beam in a bridge deck has to be propped temporarily at a particular point so that the prop can carry half the concentrated load occurring at that point. The beam is 5 m long and has both the ends built in at the same level as shown in the figure. Concentrated load occurs at 3 m from the left wall. The prop or column is a circular bar. Calculate its diameter so that as stated, the beam and column carry half the applied load. Second moment of area of beam is 36,000 cm4. Young’s modulus of the beam material is double the Young’s modulus of the column material.

At the section under consideration for the beam,

Load is distance b = 3 m, distance a = 2 m, L = 5 m

Eb = Young’s modulus of beam, then Ib = Moment of inertia of beam section Deflection in beam at point C (from page 35) (y positive downwards, a and b shift)

Putting the values,

Deflection in column, length of column = 1 m

But both deflections are the same, so

Area of cross-section of column,

Ac, area of column section

Diameter of column,

Example A horizontal steel bar 60 mm in diameter is rigidly fixed at each end where the fixings are 1.2 m (L) apart. A rigid bracket is welded in the middle of the bar at right angles to the axis and in the same horizontal plane. Determine the radius of the arm of the bracket (R) at which a vertical load of 1200 N can be suspended if the deflection of the load is not to exceed 0.5 mm. Given E = 200 GPa, shear modulus, G = 80 GPa.

Bar diameter, d = 60 mm

Moment of inertia,

Polar moment of inertia, J = 2I = 127.24 × 104 mm4

E = 200 × 106 kN/m2

EI = (200 × 106) × (63.62 × 104) = 127.24 kNm2

Deflection in the centre of the bar as fixed beam (from page 21), (y positive downwards)

f = 60 mm

δ′ deflection in the bar due to twisting = 0.5 − 0.085

δ′ = 0.415 mm

Twisting moment, T = WR = 1.2R kNm Torque will be equally divided in two portions AC and CB

Torque on AC,

Using Torsion formula

0.5 0.085

T

(L'=L/2)

Putting these values

or

Example A beam of span L is fixed at both the ends. A couple M is applied to the beam at its centre about a horizontal axis at right angle to the beam. Determine the fixing couples at each support and slope at the centre of the beam. EI is the flexural rigidity of the beam.