FIXED BEAMS IN BENDING

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1 FIXED BEAMS IN BENDING

2 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 WI 3 /48EI, and maximum slope in beam is ±WL /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 WL 3 /19EI, and maximum slope is reduced to ±WL /3EI. 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 y x W simply supported beam x d y M ( x) EI W dx x W/ W/ W/ WL/4 the maximum bending moment (BM) in beam is (WL/4), maximum deflection in beam is WI 3 /48EI, and maximum slope in beam is ±WL /16EI W/ dy slope EI W x C1 dx x L, slope 0 0 W L C 4 WL dy C WL 1 16 dx 16EI 3 deflection W x WL xc 6 16 x 0, y C at y L 3 3 EIy W L WL L WL 3WL y WL 48EI 3 1 WL 48 3

4 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). M D Figure 1 Fixed beam

5 A beam ABCD, of length L, fixed at ends A and D, carries loads W 1 and W 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 W 1 and W 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 (a). Figure (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. M D Figure (a) BM diagram of SS beam (b) BM diagram of end couples (c) BM diagram of fixed beam P 1, P are point of contraflexure

6 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 (c), in which P 1 and P are points of contraflexure. In the combined bending moment diagram there are positive bending moments and negative bending moments. At P 1 and P bending moment changes sign from negative to positive and from positive to negative as shown. M B M D Figure (a) BM diagram of SS beam (b) BM diagram of end couples (c) BM diagram of fixed beam P 1, P are point of contraflexure

7 Let us consider a section XX at a distance of x from end A of the beam. M, resulting bending moment at section = M x + M x X X M D B M x where M x = Bending moment at section when beam is considered as simply supported, and M x = Bending moment at section due to end moments M A and M D. M M x M A x L M A M D M A M D x M x L-x

8 FIXED BEAM SUPPORT MOMENTS If we refer to Figures (a), (b), (c), resultant bending moment at any section, M = M x + M x or M M x M x d y EI M x M x dx = Bending moment at section when beam is SS + Bending moment at section due to support moments. Multiplying both sides of by dx, we get EI d y dx dx M x dx M x dx

9 Integrating this equation over the length of the beam or L 0 EI d y dx EI dx dy dx L 0 L 0 M x dx aa L 0 M x dx = 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 where EI 0= a +a a M A M D L

10 or M L A M D a 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. d y Taking EI M x M x again and multiplying both the sides by x dx, and integrating dx both the sides over length of beam: L 0 EI d y dx EI x dx dy x y dx x dx ax ax Putting the values of slope and deflection at ends EI L 0 M L 0 x x dx But in a fixed beam, at the ends, slope and deflection both are zero, therefore, L 0 M Li y 0i y ax ax D D 0ax ax A A x

11 or But So, ax ax a = a 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 am A M D L ax M L L A M 3 L M A M D 6 D (origin at A), L L 3 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 (Figure 3). With the help of equations and, support moments M A, M D can be worked out. Figure 3 a diagram

12 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 y dq dx dx dq dq dx 1 The fiber length on a radius y can be found similarly. If the difference between the two fiber lenghts is du: du ydq dq ydq By dividing both sides by dx, the last term becomes K. The normal strain will be equal to: du dx dq 1 y y y dx y y From Hooke s law, x E x Ey In this equation, the variable y can assume both positive and negative values. Since, x M I y Ey M EI

13 Curvature In Cartesian coordinates the curvature of a line is defined as 1 d y dx dy 1 dx 3/ 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. dy Since the deflections tolerated in the majority of engineering structures are very small, slope curve is also very small. Therefore, the square of the slope can be neglected compared to one. dx of the elastic So, the curvature can be taken as 1 d y dx This simplification eliminates the geometric nonlinearity from the problem. Now, the governing differential equation for small deflections of elastic beams becomes: d y dx M EI

14 There are three governing equations for determining the deflection of a loaded beam: M ( x) EI d y dx V ( x) EI 3 d y 3 dx q( x) EI 4 d y 4 dx 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.

15 Macaulay s method: y Single moment: q( x) M O x a M O x a y Concentrated force: q( x) F O x a 1 F O y a x Uniformly distributed load: In order to keep distributed loads within specified boundaries, same magnitude but q( x) wo x a 0 a w O x negative direction distributed loads must be applied for the remaining length. Linearly increasing distributed load: dw q( x) x a dx 1 y dw dx x Quadratically increasing distributed load: c q( x) x a y a d w c dx x a

16 Macaulay s method:

17 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 M A and M B. Since the beam is symmetrically loaded, end moments will be equal, i.e. M A = M B Figure 4 Equivalence of a fixed beam

18 In SS beam, BM at centre area of a BM diagram, Figure 4 a BM diagram

19 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 8 (d), in which maximum positive BM maximum negative BM

20 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. (due to symmetry) End couples Figure 5 Fixed beam with central point load Take a section X X, at a distance of x from end A, in the portion CB. Bending moment, (as obtained earlier)

21 or integrating it, we get where C 1 is constant of integration. (Term in bracket is to be omitted) At end A, because end is fixed. So, integration constant, C 1 = 0 Now integrating the above equation again, we get where C is another constant of integration. At end A, x = 0, deflection, y = 0, because end is fixed. So, 0 = omitted term + C, using Macaulay s method, constant C = 0.

22 Finally, the equation for deflection is Maximum deflection will occur at centre, i.e. at Putting this value of x, or

23 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 knm for the beam, determine the fixing couple at ends and maximum deflection in the beam.

24 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 M A and M B as shown in Figure 6 (b) and (c). Since the beam is symmetrically loaded, end moments M A = M B Figure 6 (a) Fixed beam; (b) a BM diagram; (c) a BM diagram

25 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 = a a = M A L = M B L (due to end moments) Reactions at supports = (due to symmetrical loading).

26 The resultant bending moment diagram for fixed beam is shown in Figure 30 (d), in which P 1 and P are points of contraflexure. The locations of P 1 and P 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 or Integrating this, we get Figure 7 Fixed beam with uniform distributed loading

27 at x = 0, end A, slope, so constant C 1 = 0., Now Integrating this equation again, we get At end A, x = 0, deflection y = 0, so constant, C = 0 Finally, we have Maximum deflection in the beam will occur at the centre, putting the value of x =

28 Note that the deflection is reduced to only 0% 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).

29 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 = 10 GPa, I = 4,800 cm 4.

30 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. 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 M = M A + R A x W (x a) or Integrating this equation, we get C (constant of integration) At x = 0, end A, slope,, because of fixed end 0 = omitted term + C 1 Figure 8 Fixed beam with eccentric load

31 Constant C l = 0 Integrating this equation we get At end A, x = 0, deflection, y = 0, because of fixed end 0 = omitted term + C Constant C = 0. So, At the end B, x = L, both deflection and slope are zero. Putting x = L in and

32 or or From these equations, or Reaction R A

33 Reaction R B Putting the value of R A in, we get

34 Similarly we can find out that support moment at B, In Figure 3, we have taken a C Bending moment at point C, when beam is in BM diagram. Bending moment diagram for fixed beam is A A P 1 C P BB as shown in Figure 9, where P 1 and P are points of contraflexure. End moments, Figure 9 BM diagram of fixed beam

35 Example Let us take numerical values of a and b, say a = m, b = 4 m, L = 6 m, W = 8 kn. What is the equation of deflection? (SS) BM diagram of fixed beam as shown in the figure. In a fixed beam with eccentric load, we know that the reaction R A, Moment M A, Deflection equation will become

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

37 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.

38 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 = 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 knm.

39 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

40 Integrating this equation, we get (constant of integration C 1 ) (at end B, x = 0, slope, because of fixed end.) So, 0 = omitted term + C 1 Constant C 1 = 0 Integrating this equation, we get (constant of integration C ) At end B, x = 0, deflection y = 0, because end is fixed.

41 0 = omitted terms + C Constant C = 0 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 and 0 = 6M B + 18R B or 0 = 6M B + 18R B 56 0 = 18M B + 36R B 68

42 or M B + 3R B = M B + R B = From these equations, reaction R B is obtained as, R B = 5.556kN Total load on beam = = 18 kn Reaction, R A = = kn Moment, M B = R B = = = knm To know bending moment M A, let us put x = 6 in and putting values of M A, R B also M A = x 6 6 (6 ) 3(6 4) = = knm.

43 Deflection at D: Putting x = m in Deflection,

44 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 = = 36 Reaction, R B = 6 kn Reaction, R A = = 1 kn Bending Moment

45 M D = 6 = + 1 KNm, M B = 0 In the figure AC D B shows a bending moment diagram of SS beam. Let us draw a diagram on this, taking M A = KNm M B = KN The resultant bending moment diagram for fixed beam is obtained as seen in the figure.

46 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, Integrating it, we get Figure 10 (a) Sinking of a support

47 Say at x = 0, Reaction = R A, then Integrating this equation, we get (another constant of integration) Say at A, x = 0, bending moment is M A, so M A = 0 + C and Constant, C = M A Integrating this we get (constant of integration) But slope is zero at fixed end, A, i.e. at x = 0 0 = C 3 so Constant of integration, C 3 = 0

48 So, Integrating this we get (constant of integration) At x = 0, deflection, y = 0, 0 = C 4 so Constant of integration, C 4 = 0 Finally, we have At end B, x = L, slope,, deflection, y = δ

49 Putting these values in and From these two equations For equilibrium, R A = R B or Putting the values of M A, R A in bending moment at end B

50 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 6EI L 6EI L 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 6EI L 6EI L

51 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: Integrating this (constant of integration) Figure 11 (a) Beam in consideration at end A, x = 0, shear force R A

52 Integrating further (another constant of integration) At end A, x = 0, moment = M A Therefore, M A = 0 + C or constant of integration = M A Integrating this equation, we get (constant of integration) At x = 0, So, 0 = C 3 or constant C 3 = 0

53 Integrating it, we get (constant of integration) at x = 0, end A, deflection, y = 0. So, 0 = C 4 or constant, C 4 = 0 Finally we have,

54 Let us utilize end condition at B, where x = L, Using these values in and From the above equation, Putting this value of M A in, we get Reaction, Moment,

55 To maintain equilibrium, R B = R A (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. Figure 11 (a) Beam in consideration (b) BM diagram (c) SF diagram

56 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 knm.

57 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 M A, R A ; M B and R B, respectively. Now consider a section X X at a distance x from A in the portion CB of the beam. δ/l

58 Bending moment at the section or Integrating this, we get At end A, x = 0, slope So, 0 = omitted term + C 1, Constant, C 1 = 0

59 Integrating this again At end A, x = 0, deflection, y = 0, so 0 = omitted term + C Constant, C = 0. Finally the equation is, Putting x = L, at end B, where In and or

60 Adding the last two equations, we get or Substituting the value of M A Reaction,

61 Now putting the value of R A in Substituting the values of M A, R A in at x = L, this equation gives BM at end B

62 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 d x

63 Considering as an eccentric load, Support moments, dm A For eccentric loading Support moment M A, Similarly support moment M B,

64 Support Reactions: Total load on beam Taking moments, or Reaction,

65 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 cm 4. Young s modulus of the beam material is double the Young s modulus of the column material.

66 At the section under consideration for the beam, Load is distance b = 3 m, distance a = m, L = 5 m E b = Young s modulus of beam, then I b = 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,

67 Deflection in column, length of column = 1 m But both deflections are the same, so Area of cross-section of column, A c, area of column section Diameter of column,

68 Example A horizontal steel bar 60 mm in diameter is rigidly fixed at each end where the fixings are 1. 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 100 N can be suspended if the deflection of the load is not to exceed 0.5 mm. Given E = 00 GPa, shear modulus, G = 80 GPa.

69 Bar diameter, d = 60 mm f = 60 mm Moment of inertia, Polar moment of inertia, J = I = mm 4 E = kn/m EI = ( ) ( ) = 17.4 knm Deflection in the centre of the bar as fixed beam (from page 1), (y positive downwards)

70 δ deflection in the bar due to twisting = δ = mm T Twisting moment, T = WR = 1.R knm Torque will be equally divided in two portions AC and CB Torque on AC, Using Torsion formula (L'=L/)

71 Putting these values or

72 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.

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