3

Fixed Beams

3.1 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.

3.2 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 Figures 3.1 (a), (b), and (c).

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 M_A and M_D. Considering the beam as simply supported (SS) beam, bending moment diagram can be drawn as shown in Figure 3.2(a). Figure 3.2(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. 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 3.2(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.

Let us consider a section XX at a distance of x from end A of the beam.

M, resulting bending moment at section

 

 

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.

3.3 FIXED BEAM−SUPPORT MOMENTS

Refer to Figures 3.2(a), (b), and (c). Resultant bending moment at any section,

 

 

or

= Bending moment at section when beam is SS + Bending moment at section due to support moments.

Multiplying both sides of Eq. (3.1) by dx, we get

Integrating Eq. (3.2), over the length of the beam

or

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

Equation (3.3) becomes

 

 

where

or

This Eq. (3.5) 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 Eq. (3.1) again and multiplying both the sides by x dx, and integrating both the sides over length of beam.

Putting the values of slope and deflection at ends

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

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 from A (see Figure 3.3). With the help of Eqs (3.5) and (3.9), support moments M_A, M_D can be worked out.

3.4 FIXED BEAM WITH A CONCENTRATED LOAD AT CENTRE

Consider a fixed beam AB of length L, fixed at both ends A and C, carrying a point load W at its centre as shown in Figure 3.4(a). It is equivalent to a simply supported beam AB with central load W and a beam AB subjected to end moments M_A and M_B. Since the beam is symmetrically loaded, end moments will be equal, i.e.

 

 

In SS beam, BM at centre

area of a BM diagram,

area of a′ diagram,         a′ = M_A L = M_B L (M_A = M_B, because of symmetrical loading)

or                         a′ = −a

or end moment,

Resultant bending moment diagram is shown in Figure 3.4(a), in which

maximum + ve BM               

maximum (−ve) BM               

Reader may note that 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 3.5, reactions at ends.

End couples                (as obtained earlier)

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

Bending moment,

or

Integrating Eq. (3.10), we get

where C₁ is constant of integration. (Term in bracket is to be omitted)

At end A,

because end is fixed.

So, constant of integration,              C₁ = 0

Integrating Eq. (3.11) 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

Exercise 3.1   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 fixing couple at ends and maximum deflection in beam.

Ans. [−22.5 kNm, −4.82 mm].

3.5 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 3.6(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 M_A and M_B as shown in Figures 3.6(b) and (c). Since the beam is symmetrically loaded, end moments

 

 

When the beam is S.S, bending moment diagram is a parabola with , maximum bending moment at centre. Area of BM diagram,

area of BM diagram,                  a′ = M_AL = M_BL (due to end moments)

Reactions at supports = (due to symmetrical loading).

The resultant bending moment diagram for fixed beam is shown in Figure 3.6(d), in which P₁ and P₂ are points of contraflexure. The reader can determine the location of P₁ and P₂.

Now consider any section X−X at a distance of x from end A as shown in Figure 3.7. Bending moment at section XX is

or

Integrating Eq. (3.12), we get

at x = 0, end A, slope , so constant C₁ = 0.

Now

Integrating Eq. (3.13) 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).

Exercise 3.2   A beam 6 m span has its ends built in and carries a uniformly distributed load of 4 kN/m run. Find the (i) end moments, (ii) bending moment at the centre, and (iii) maximum deflection in beam. Given E = 210 GPa, I = 4,800 cm⁴.

Ans. [(i) −12 kNm, (ii) +6 kNm, (iii) 1.34 mm].

3.6 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 3.8). Say reactions at A and B are R_A and R_B and support moments at A and B are M_A and M_B respectively. There are four boundary conditions at two ends of fixed beam, and we can determine four unknowns R_A, R_B, M_A, and M_B.

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

 

 

or

Integrating Eq. (3.15), we get

At x = 0, end A, slope, , because of fixed end

 

Constant C_l = 0

Integrating Eq. (3.16), We get

At end A, x = 0, deflection, y = 0, because of fixed end

 

 

Constant C₂ = 0

So,

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

Putting x = L in Eqs (3.16) and (3.17)

or

or

From Eqs (3.20) and (3.21),

Reaction

Reaction

Putting the value of R_A in Eq. (3.20), we get

Similarly we can find out that support moment at B,

In Figure 3.8, we have taken a

Bending moment at point C, when beam is in BM diagram.

Bending moment diagram for fixed beam is A ′ A P₁ c′ P₂ BB′ as shown in Figure 3.9, where P₁ and P₂ are points of contraflexure. End moments,

Example 3.1   Let us take numerical values of a and b, say a = 2 m, b = 4 m, L = 6 m, W = 8 kN.

Then

as shown is Figure 3.10

Solution: In a fixed beam with eccentric load, we know that

Reaction,

Moment,

Deflection equation will become

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

or deflection,

Exercise 3.3 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.

Ans. [2.734, 1.266 kN, −4.6875, −2.8175 kNm, 1.66 m and 2.37 m from left hand end].

Example 3.2   A fixed beam AB of length 6 m carries a udl of intensity 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.

Solution: Figure 3.11 shows a fixed beam AB of length 6 m, carrying udl and concentrated load as shown. There are unknown reactions R_A, R_B at ends and unknown moments M_A, M_B 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

Integrating Eq. (i), we get

 

 

So,

 

 

Constant C₁ = 0

Integrating Eq. (ii), we get

 

 

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

 

 

Constant

 

 

Finally, the equation of deflection is

To determine R_B, M_B, let us consider end A where x = 6 m, both slope and deflection are zero.

Putting this value of x in Eqs (ii) and (iii)

 

 

or

 

 

 

or

 

 

 

From these equations, reaction,

 

 

Total load on beam	=	6 × 2 + 6 = 18 kN
Reaction,     RA	=	18 − 5.556 = 12.444 kN
From Eq. (vi),     MB	=	9.333 − 3RB = 9.333 − 3 × 5.556
	=	9.333 −16.668 = −7.335 kNm

 

To know bending moment M_A, let us put x = 6 in Eq. (i), and putting values of M_A, R_B 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 Eq. (iii)

Deflection,

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

 

	6 R′B	=	6 × 2 × 1 + 6 × 4
		=	36
Reaction,	R′B	=	6 kN
Reaction,	R′A	=	12 + 6 − 6
		=	12 kN

 

Bending Moment

 

Figure 3.12 AC′D′BDC shows a bending moment diagram of SS beam. Let us draw a′ diagram on this, taking

 

 

Figure 3.12 shows the resultant bending moment diagram for fixed beam.

Exercise 3.4   A fixed beam AB of length L carries udl of intensity w over half the length CB as shown in Figure 3.13. Using double integration method, determine support reactions, support moments, and draw its BM diagram.

3.7 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 3.14(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,

Integrating Eq. (3.25), we get

Say at x = 0, Reaction = R_A, then

Integrating Eq. (3.26), we get

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

 

 

Constant,

 

 

Integrating Eq. (3.27), we get

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

 

 

Constant of integration, C₃ = 0

So,

Integrating Eq. (3.28), we get

At x = 0, deflection, y = 0,

 

 

Constant of integration, C₄ = 0

Finally, we have

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

Putting these values in Eqs (3.28) and (3.29)

From Eqs (3.30) and (3.31)

For equilibrium,

 

 

or

Putting the values of M_A, R_A in Eq. (3.27), bending moment at end B

Figures 3.14(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.

Example 3.3 A beam of 6 m span is fixed at both the ends. Level of right hand support sinks by 3 mm. If EI = 4,500 kNm² for the beam, determine support reactions and support moments. What is the deflection in the centre of the beam if the beam is subjected to a udl of 3 kN/m.

Solution: Figure 3.15 shows the fixed beam AB, of length 6 m, subjected to udl, w = 3 kN/m

1. There is no sinking

   Reaction, R′_A = Reaction,R′_B

   

   Support moments,

   
2. Support reactions due to sinking
   
   

   FIGURE 3.15 Example 3.3

   
3. Support moments due to sinking
   

   Resulting support reactions,

    

   R_B = R′_B + R″_B = 9 − 0.75 = 8.25 kN

    

   Resultant support moments,

   

Deflection at Centre of Beam

Refer to Figure 3.15, consider a section X−X at a distance of x from end A.

Bending moment at section

or

Putting the values of R_A, M_A, and w.

Integrating Eq. (i), we get

at x = 0, , constant, C₁ = 0

Integrating again

at x = 0, = 0, so constant C₂ = 0

Finally equation of deflection is

 

 

At the centre x = 3 m, putting this value