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1. DEFLECTION OF BEAM x  

2. M D F L C T I O    As a beam is loaded,  D i ff e r en t r eg i on s a r e s ub j e c t ed t o V and M  T he bea m w ill de f l e c t Recall the curvature equation x   x   The slope ( q )   & deflection (y) at any spanwise location   can be derived   R

3. e Bending   e e e C T S h ea r

4. DESIGN OF BEAMS T   he cross section of a beam has to be designed in such a way that it is strong enough to limit the bending moment and shear force that are developed in the beam. This criterion is known as the STRENGTH CRITERION of design .  Another criterion for beam design is that the maximum deflection of the beam must not exceed a given permissible limit and the beam must be stiff enough to resist the deflection caused due to loading. This criterion is known as „ STIFFNESS CRITERION of design”  

5. D e f i n i t i on s : - DEFLECTION  :-The vertical distance in transverse direction   between positions of axis before and after loading at the section of the beam, is defined as the deflection of beam at that section. ELASTIC CURVE OR DEFLECTION CURVE:- The neutral axis in its deflected position after loading of the beam is known as its elastic curve or deflection curve. SLOPE:- The slope of the beam at any section is defined as the angle (in radians) of inclination of the tangent drawn at that section to the axis in its deflected position after loading, measured w. r. t. the un-deformed axis. FLEXURAL RIGIDITY(EI):- The product of modulus of elasticity and Moment of Inertia is known as Flexural rigidity.

6. ASSUMPTIONS MADE IN THE DEFLECTION:- (i)  Axis of the beam is horizontal before loading. (i   i) Deflection due to S.F. is negligible. (iii) (a)   Simple Bending equation M/I= σ /y=E/R is applicable and all the assumptions made in simple bending theory are valid. Material of the beam is homogenous, isotropic and obey H oo k ‟ s l a w .   The modulus of elasticity is same in compression as well as in tension.  Plane section remain plane before and after bending

7. D I FF E R E T I A L E Q U AT I O F O R D E F L E C T I O A D SLOPE OF BEA S    Below is shown the arc of the neutral axis of a beam subject to bending.   For small angle dy/dx = tan θ = θ   The curvature of a beam is identified as dθ /ds = 1/R   In the figure δθ is small ,and δx= ~  δs;    i.e ds /dx =1 6 x  

8.   Then dy /ds =sin θ   dx /ds= cos θ   dy/dx =tan θ   Differentiating equation (a) w.r.t. x, we get S e c 2   θ .( d θ / d x ) = d 2 y / d x 2   Therefore, d θ /dx =(d 2 y/dx 2 )/sec 2 θ   –  -(2) Other Method R = d s / d θ   = ( d s / d θ ) × ( d x / d x ) = ( d s / d x )/( d θ / d x) = s e c φ /( d θ / d x ) -----( 1 )

9.   From equation(1), R= sec θ /(d θ /dx) = sec 3 θ  /(d 2 y/dx 2 ) 1/R=(d 2 y/dx 2 )/sec 3 θ  = (d 2 y/dx 2 )/ (sec 2 θ ) 3/2 = (d 2 y/dx 2 )/ (1+tan 2 θ ) 3/2 =(d 2 y/dx 2 )/[1+(dy/dx) 2 ] 3/2 In any practical case of bending of beams, the slope (dy/dx) is very small (because curve is almost flat); hence (dy/dx) 2  can be ignored so, 1/R=d 2 y/dx 2  

10.    From the pure bending equation, we know that.  Integrating between selected limits. The deflection between limits is obtained by further integration. 9   M    I    E     R

11. SIGN CONVENTIONS : Linear horizontal distance x: positive when measured from left to right Vertical distance or deflection y is positive when measured a b xi o s v o e f t b h ea a m x . is and is negative when measured below the NOTE :

12. Deflection at support A and B are zero and maximum at the middle of Span. slope θ  is maximum at A and B and zero at middle of Span (at point of symmetry); At point of maximum deflection slope is zero θ   A θ B   A NOTE : SUPPORT or BOUNDARY CONDITIONS: (i) Simply supported beams: C  A B D Deflection at support A and B are zero but more at free end and also at the centre of span . Slope is maximum at supports B and  A

13. ( ii) Cantilever Beam: Deflection and slope both are zero at fixed support. Slope and deflections are maximum at free end θ  increases from point A towards B. B θ m ax y m ax   A

14.   Methods for finding slope and deflection of beams: Double or   Direct integration method Macaulay’s method  Area moment method Conjugate beam method (v) Unit load method

15. We have from differential equation of flexure, EI d 2 y/dx 2 =M Integrating w. r. t.. x both sides, we get E I ( dy / dx ) = ∫ M d x + C 1 Integrating again w .r .t. x both sides ,we get E I ( y ) = ∫ ∫ ( M d x ) d x + C 1 ( x ) + C 2 where C 1 and C 2 are constant of integration Thus, integrating the differential equation w .r .t. x, we get the equation for slope (dy /dx) ,and integrating it twice w. r .t. x, we get the equation for deflection ( y). DOUBLE INTEGRATION METHODS

16. The two constants of integration can be determined u an si d n g s lo th p e e k a n t o th w e n su b p o p u o n r d ts ary conditions of deflection T he m e t hod o f doub l e o r d i r e c t i n t eg r a t i on i s c on v en i en t on l y i n f e w c a s e s o f l oad i ng s . I n general case of loading, where several loads are acting the B.M. expression for each of the different regions of loading is different .Then the method of double integration will become extremely lengthy and laborious. Therefore ,it is not generally used.

17. Case--1(i): Determine the slope and deflection equation for the beam loaded as shown in fig. (ii) Find the maximum deflection and maximum slope. S o l u t i on : B.M. at section 1-1 M= - P( x) 2  2 E I d y / d x   = M = - 2 P ( x ) EI (dy /dx ) = -P (x /2) + C 1 EI y = -P (x 3 /6) + C x + C 1 2 1 1 P x   A B θ   A l y   A  At fixed end, when x = L ,(dy /dx) =0 (iii) What will be the P=6kN, L=3m, deflection and slope at free end when E=210GPa, I=16x10 4 cm 4 .

18. Therefore, - (PL 2 /2) + C = 1 2 C 1 = PL /2  At x = L, y = 0, -PL 3 /6+(PL 2 /2) L+C =0 2 3 3 or,C 2 = PL /6 - PL /2 = PL 3 /6[1-3]= - PL 3 /3 2 Therefore ,C = - PL 3 /3

19. Equation of slope; EI (dy/ dx) =-Px 2  /2 + PL 2  /2-----(1) Equation of deflection ,EI (y)=-Px 3  /6 + PL 2 x/2 - PL 3  /3-----(2) Maximum deflection : When x=0 (at free end) ,then from equation (2), EI (y)=-0+0-PL 3  /3 y = -PL 3  /3EI max Maximum Slope: Slope is maximum at free end (at x=0).hence from equation(1), EI (dy/ dx) = -0 + PL 2  /2 (dy /dx) max = PL 2  /2EI  

20. Deflection at free end (i.e; at A):= y  A = PL 3 /3EI ( iii )   A θ =PL 2 /2EI 6  10 3 ( 3  1 3 ) 2 s l op e θ   A = ( dy / dx ) a t A  2 ( 21  1 3 ) ( 1 6  10 8 )  8 . 035 7  1  5 r a d i a n 6  1 3 ( 3  1 3 ) 3 3 ( 21  1 3 ) ( 1 6  1 8 )  = 0.161mm P

21. Case (2) : When cantilever is subjected to an u .d. L. of intensity w unit/m run over entire span Here A is the origin. Take a section X-X at a distance x from A. x B.M at distance x from A= M  = EI d 2 y/dx 2 =-w.x.x/2=-wx 2 /2 Integrating once, EI (dy/dx) =-wx 3 /6 + C  ------------------------(1) 1 where C 1  is constant of integration B L x X w un i t / m X  A

22.  Applying boundary conditions:- from equation(1) C 1 = wL 3 /6 at x =L, dy/dx=0 0=-wL 3 /6 + C 1   therefore, EI dy/dx=-wx 3 /6+wL 3 /6---------(2) Integrating once again, EI y=-wx 4 /24 + wL 3 .x/6 +C  (3) 2 where C 2 is 2 nd  constant of integration  Applying boundary condition; at x=L, y=0

23. 0=-wL 4 /24+wL 4 /6+C 2 Therefore, 4 4 C 2 =wL /24-wL /6 4 C 2 =-wL /8. Therefore, equation (3) becomes, EI(y)=-wx 4 /24 + wL 3 .x/6  – wL 4 /8--------(4) Maximum deflection It occurs at free end where x= From (4),EIy=-0+0-wL 4 /8

24. 4 y max =-wL /8EI similarly maximum slope occurs at free end where x=0 from (2), 3 EI (dy/dx) =-0+wL /6 m ax (dy/dx ) =wL 3 /6EI

25. Case 3:- When simply supported beam is subjected to a single concentrated load at mid-span. x X M x  = (P/2) x EI d 2 y/dx 2 =(P/2)x P EI dy/dx=(P/2)x 2 /2 + C 1 Due to symmetry slope at x = L/2 is zero C 1  = -PL 2 /16 EI dy/dx=(P/2)x 2 /2 -PL 2 /16  A R  A =P/2 L/2 X P C L/2 R B = P/ 2 B Integrating again we get 3 2 EIY = (P/2)x /6  –  (PL /16) x + C 2  At x=0 , Y = C 2  =

26. Hence EIY = (Px 3 /12)  –  (PL 2 /16) x Deflection at mid span i.e. at x = L/2 is Y = -PL 3 /48EI = PL 3 /48EI (downward) Slope at support, is obtained by putting x = 0, in slope equation   E I   1 6 1 6   E I   2 2        1   WL    WL   dx     dy     x  q    A

27. C a s e4 : - R B = W L / 2   A x X W unit / run B R   A = W L / 2 X 1 C  3 2 2 2 2 w x 2 w   x 6  EI  dx   E I   2 w x 2 wL 2      x     x  d   y wL d   x 2 d y w L   x 4   M x 

28. Due to symmetry dy/dx = at x = L/2 w L 3 C 1       2 4 Integrating both side w.r.t. x, we get 2 2 4 2 4 1 2 w L x 3 w x 4 w l   3   x  C     E I   Y         At x = 0, y = C 2 =0 w L x 3 w x 4 w l   3 12 24 24    x    E I   Y    H en c e Maximum deflection y c   which occurs at centre C is obtained by substituting x = L/2 in the above equation

29. Due to symmetry dy/dx = at x = L/2 w L 3 C 1       2 4 Integrating both side w.r.t. x, we get w L x 3 w x 4 w l   3 1 2 2 4 2 4   x C   2   E I   Y          At x = 0, y = 2 C = 3 4 3  w x  w l     x 24 24 Hence   E I   Y    w L x 12 Maximum deflection y c   which occurs at centre C is obtained by substituting x = L/2 in the above equation.   A ( downward  ) 384  EI  5 w l   w L 3 24  EI  4 5 w l   4      dx     dy   Yc  Yc  384  EI    x  Slopeat end  q   

30.   MACAULAY‟S  METHOD For a general case of loading on the beam, though the expression for B.M. varies from region to region, the constants of integration remain the same for all regions. Macaulay recognized this fact and proposed a method which is known as the Macaulay‟s  method to obtain the slope and deflection equations. It is essentially modified method of double integration of the B.M. expression but following a set of rules given below:-

31. (1).  Assuming origin of the beam at extreme left end, take a section in the last region of the beam at a distance x from the origin and write the expression for B.M. at that section considering all the force on the left of s n ection. Integrate the term of the form (x-a)  using the formula ∫(x -a) n  dx=(x-a) n+1  /n+1 where a=distance of load from origin. While finding slope and deflection in the form (x-a) n  ,if (x-a) becomes negative on substituting the value of x, neglect the term containing the factor (x  –  a) n   ( „a 4 ‟ )   . fr I o f m a c th o e u p o l r e ig ( i m n o o m f t e h n e t ) b o e f a m a , g th n e it n u d w e ri „ t c e ‟   i t s h e a c B ti M ng d a u t e a t o d is c t o a u n p c le in the form c (x-a) . (5). If the beam carries a U.D.L, extend it up to the extreme right end and superimpose an UDL equal and opposite to that which has been added while extending the given UDL on the beam.

32.  EXERCISE PROBLEMS : Solution:- R A   =34 KN, R B  =36 KN  A B R   A R B 5 m x Q.(1) Figure shows a simply supported beam of span 5m carrying two point loads. Find (1)the deflection at the section of the point loads. (ii) Slope at A,B,C and D, (iii) maximum deflection of the beam. Take E=200GPa, I=7332.9 cm 4   X X 1m 30KN 40KN 1.25m c D M x  = R  A x - 30(x-1) - 40(x-3.75)

33. EI d 2 y/dx 2 = 34 x -30(x-1) -40(x-3.75) Integrating once, 1   E I ( dy / dx ) = c   + 34 x 2 / 2   -30(x-1) 2 /2 - 40(x-3.75) 2 /2 ---(1)   Integrating   once again, ---------- (2) Support conditions: at x=0,y=0 at x=5m,y=0 t he r e f o r e , C 1 = - 75 . 6 ≈ - 75 . 1   6 6 6 therefore,C 2 =0    4 – 3 ( x  3 . 7 5 ) 3   (   x  1 ) 3 3 4   x 3 E I y  C    C      x  2 1 Thus the equation for slope and deflection will be

34. E I (dy / dx) = - 75.1 + 17x 2  - 15(x-1) 2 -20(x-3.75) 2 - ----------(3) E I (y)= -75.1 x  + 17x 3 /3 -5(x-1) 3   -20(x-3.75) 3 /3 ---------(4) Total deflection at section of point loads  At C x=1m , deflection y= y C (say) C EI y = -75.1 1 + 17(1) 3 /3 =-69.393 × Or, y C =-69.393/EI =-69.393   /(200 × 10 6 × 7332.9 × 10 -8 ) = - 4 . 7 3 ×   1 - 3   m = - 4 . 73 m m

35. (dy/dx) D  =[17 ×(3.75)2-75.06-15 ×(2.75)2] × 1/EI (when x=3.75m) =50.525/EI=3.445 radians. (dy/ dx) at B=1/EI[17 ×(5)2-75.06-15 ×(4)2-20 ×(1.25)2] - 3 =78.65/EI=5.363 × 10 rad =0.197 degree  Assuming the deflection to be maximum in the region CD: (at point of maximum deflection dy/dx = 0)

36. say at x=x 1 where dy / dx=0 From equation (3), 1 1 EI(0)=34/2(x ) 2 -75.06-30(x -1) 2 /2 2 2 =17 x 1  -75.06 -15x 1  +30x 1  -15 =2x 2 +30x -90.06=0 1 1 x 1 =   2.56m The assumption that the maximum deflection is within the region CD is correct. EI y m a x =34(2.56) 3 /6-75.06 × 2.56-30(2.56-1) 3 /6 -3 Y max =-116.13/EI=-7.918 × 10 m =-7.918mm

37. ( Q - 2 ) Obtain the equation for slope and elastic curve for the beam loaded as shown in figure and find the deflection and slope at mid-point of beam. 2 T ake E I = 1 5 ×   1 3   k N m Find the slope at A,C and D X Solution:- Reactions R A  =1/4[80 × 3+120] =90KN( ) R B  =80-90=-10kN=10( ) × 1 m 120 k N m 80 K N 2 m c x X 1 m   A D B

38.  Alternatively, R B  =1/4[80 1-120] =-10KN =10 KN( ) M X = 90x - 80(x-1) -120(x-3) 2 2 EI (dy /dx )= C 1  + 90x /2 - 80(x-1) /2  – 120 (x-3) -------(1) 3 3 2 EI(y)= C 2 +C 1 (x)  + 90 x /6 -80(x-1) /6 -120(x-3) /2 ------(2) Support reactions at x = ,y = , C 2 = M x  = 90 x  –  80 (x-1)  –  120 (x-3)

39.  At x=4,y=0 ,C 1 =-135 Equation for slope (dy/dx):- EI (dy /dx)= - 135  + 90 x 2 /2 -40(x-1) 2  -120(x-3) Equation for deflection (y):- EIy = - 135x   + 90 x 3 /6 -80(x-1) 3 /6 -120(x-3) 2 /2 To find de 3 flection at centre (i.e. 3 x=2m, at mid span ):- EIy=90(2) /6 -135(2) -80/6(2-1) /6 =-163.33 y=-163.33/(15 × 10 3 )=-10.89 × 10 -3  m =-10.89 mm To find slope at centre (i.e. x=2m, at mid span ):- EI (dy/dx) =90 × (2) 2 /2-135-80(2-1) 2 /2 =+5

40. . 01 9 °   dy/dx =5/(15 × 10 3 )=3.33 × 10 -4  radians~ =19.1 × 10 -3  degree ~ 0.02 °   (b) θ  A (at x=0) =-135/EI=-135/(15 × 10 3 ) θ 2 3 C   ( a t x =   1 m ) = [ 45 ×   ( 1 ) - 13 5 × 1 ) / 1 5 × 1 = - 90 / 1 5 × 10 3 radian θ 2 2 3 D  (at x=3m )= [45 × (3) -135-40(2) ] =110/15 × 10 radian

41. (Q.3)   Find maximum slope and maximum deflection of the beam loaded as shown in fig. Take E=200KN/mm 2 ,   I=60 × 10 6 mm4 2 m 1 m 25 k N m 15k N x X X   A B c EI(d 2 y/dx 2 )=-15x -25(x-2)   2 EI(dy/dx)= C 1   -15x /2 -25(x- 2) …………(1)   3 2 EI(y)= C 2  + C 1 x -(15/2) x /3 -25(x-2) /2 …(2)  

42. Support conditions: slope and deflection are zero at fixed support at x=3m,dy/dx=0 from equation (1), EI(0)=-15(3) 2 /2 + C - 25(3-2)=0 1 C =25+67.5=92.5 1  At C, y=0, x=3 from (2) EI(0)=-15 (3) 3 /6+92.5(3) +C  -25(3-2) 2 /2 2 C 2 =12.5-277.5+67.5=-197.5 Now equation (1) EI(dy/dx)=-15x 2 /2+92.5 -25(x-2)

43. ---------(1) Now equation (1) EI(dy/dx)=-15x 2 /2 +92.5 -25(x-2) and   A Maximum slope at A when x=0 from (1) EI(dy/dx)  A =92.5 (dy/dx) =92.5/EI=7.708 10 -3 radian × =0.441degree EI(y)=-15/2*x 3 /3+92.5x-197.5 -25(x-2) 2 /2 -----(2) Maximum Deflection at free end when x=0 EI(y)  A =-197.5, yA=-197.5/EI=-16.458mm.

44. Q. 3 Determine the equation for the elastic curve for the beam loaded as shown in figure. Find the maximum deflection. 1 m 3 m   A C B 2kN/m D 1 m Solution: R  A =R B  = 2x3/2 = 3KN   D   A 1 m 3 m C 1m 2k N / m x M =3(x) -2(x-1) 2 /2 +2 (x-4) 2 /2

45. M =3(x) -2(x-1) 2 /2 +2 (x-4) 2 /2 EI (dy/dx)=3/2(x) 2  +C    – (x-1) 3 /3 +(x-4) 3 /3 -------(1) 1 EIy=3 x 3 /6 +C x +C     –  (x-1) 4 /12 +(x-4) 4 /12 ------(2) 1 2

46. Support reactions: at x=0,y=0, at x=5m,y=0 , C 2 =0 C 1 =-8.25 Equation for slope:- EI (dy/dx)=3x 2 /2-8.25  – (x-1) 3 /3 +(x-4) 3 /3 ------(3) Equation for the elastic curve : EIy=x 3 /2 -8.25x  – (x-1) 4 /12 +(x-4) 4 /12 ---------(4) Due to symmetry, deflection is maximum at centre at x=2.5m, 3 4 EI y max =(2.5) /2 -8.25 x (2.5)-(2.5-1) /12=-13.23 y max = -13.23/EI

47. Practice problems :-   (Q-1 ) intensity A cantilever beam of span L carries a udl of w/unit length for half of its span as shown in figure.If E is the modulus of elasticity and I is moment of inertia,determine the following in terms of w,L,E and I. (i)A expression for slope(dy/dx)at free end (ii)An expression for deflection( y ) at free end (iii)The magnitude of upward vertical force to be applied at free end in order to resume this end to the same horizontal level as built in end.

48. [Ans (i)θ A =WL 3  /48EI (ii)   ∂  A   =7wL 4  /384EI( ) (iii)P=7wL/128] L/2 L / 2 w / m  A B

49. Q- (2 )  Determine the values of deflections at points C,D and E in the beam as shown in figure.Take E=2*10 5 MPa ; I= 60 *10 8  mm 4 1m 2 m 10k N / m 1 m 1 m 20 k N  A C D 30 k N E B  Answers: [ ᵟ C =0.0603mm(downward), ᵟ D =0.0953mm(downward) ᵟ E =0.0606mm(downward)]

50. . Q-(3)   Find the position and magnitude of maximum deflection for the beam loaded as shown in fig. Take E=200GPa ,I=7500cm 4  . 20 KN   A D B X 4m 4m 4 m C X 3 K N / m X [Answers: y max  at 3.7 m from A=-118/EI=7.99mm y c =-32/EI=-2.13mm]

51. DOUBLE INTEGRATION METHOD 5 1 .

52. 51

53. EXAMPLE SIMPLY SUPPORTED BEAM    Consider a simply supported uniform section beam with a single load F at the centre. The beam will be deflect symmetrically about the centre line with a slope (dy/dx) at the centre line. It is convenient to select the origin at the centre line. 52

54. 53

55. 54

56. THE DOUBLE INTEGRATION METHOD  T he c on s t an t s o f i n t eg r a t i on a r e de t e r m i ned b y evaluating the functions for slope or displacement at a particular point on the beam where the value of the function is known.  T he s e v a l ue s a r e c a ll ed bounda r y c ond i t i on s . 55

57. 2. THE DOUBLE INTEGRATION METHOD Consider the beam shown in Fig. Here x 1 and x 2  coordinates are valid within the regions  AB & BC. 56

58. 3-3 THE DOUBLE INTEGRATION METHOD  O n c e t he f un c t i on s f o r t he s l ope an d deflections are obtained, they must give the same values for slope & deflection at point B.  This is so as for the beam to be physically continuous. 57

59. EXAMPLE 3  o T su h b e j e c c a t n e t d il e to ve a r e c d o u b p e l a e m m s o h m o e w n n t i M n F  a ig t . i t 8 s . 1 e 1 n ( d a . ) is  D e t e r m i ne t he equa t i on o f t he e l a s t i c c u r v e  EI is constant. 58

60. E X A M P L E 3 . 3 - S O L U T I O N  T sho e w lo n a i d n t F e i n g d . s 8 . t 9 o ( a d ) e . flect the beam as  B y i n s pe c t i on , t he i n t e r na l m o m en t c an b e represented throughout the beam using a single x coordinate.  From the free-body diagram, with M acting in +ve direction as shown in Fig. 8.11(b), we ha v e : o   M        M   59

61. E X A M P L E 3 . 3 - S O L U T I O N 60   Applying equation 8.4 & integrating twice yields: 2 d x d   2 v dx 2   E I   o 1 o   E I       y    M   o   x  C     x  C   2 1 2  Using boundary conditions, dy/dx = at x = & y = at x = then C 1  = C 2  =0   E I   d v    M     x  C      M  

62. E X A M P L E 3 . 3 - S O L U T I O N  Substituting these values into earlier equations, we get: w i t h q    d y / d x o o   M     x   M     x 2 q   ;   y    A  EI  2  EI   M a x s l ope & d i s p l a c e m en t o cc u r a t A ( x = L ) for which 2 q      M   o   L ;     y   A    M   o   L  EI  2  EI  61

63. E X A M P L E  The +ve result for q  indicates counterclockwise rotation & the +ve r   e A sult for y  A  indicates that it is upwards.  This agrees with results sketched in Fig. 8.11(a).  I o n f t o h r e d e s r l o to p e obtain some idea to the actual magnitude  Consider the beam in Fig. 8.11(a) to:  Have a length of 3.6m  Support a couple moment of 20kNm  Be made of steel having E st  = 200GPa 62

64. MOMENT-AREA THEOREMS  To develop the theorems, reference is made to Fig. 8.13(a).  I b f y w t e h e d r f l e a w x t u h r a e l r i g m o i d m i t e y n , t E d I , i a t h g e r a “ m M / f E o r I t d h i a e g b r a m a e m ” s & h o t h w e n n i n i d v F i i d g e . i t 8.13(b) results.  By equation    M    d  q      dx    EI    63

65. 3-4 MOMENT-AREA THEOREMS  d q  on either side of the element dx = the lighter shade area under the M/EI diagram.  Integrating from point A on the elastic curve to point B, Fig 8.13(c), we have   This equation forms the basis for the first moment- area theorem. d x  B  M   A  EI    B /   A       q   64

66. MOMENT-AREA THEOREMS  a T n h y e o tw re o m p o 1 in : t T s h o e n c t h h a e n e g l e a s in ti c s l c o u p r e v e b e t q w u e a e ls n the area of the M/EI diagram between the two points.  The second moment-area theorem is based on the relative derivation of tangents to the elastic curve.  S vi h e o w w o n f i t n h e F i v g e 8 rt . i 1 ca 2 l ( c d ) e i v s i a a t i g o r n e a dt l y o f e t x h a e ggerated tangents on each side of the differential element, dx. 65

67. MOMENT-AREA THEOREMS  F i g 8 . 12 ( c ) 66

68.  Since slope of elastic curve & its deflection are assumed to be very small, it is satisfactory to approximate the length of ea c h t angen t li ne b y x & t he a r c d s ‟ b y d t .    U s i ng q s = r    d t = x d q    U s i ng equa t i on 8 . 2 , d q   = ( M / E I ) d x  T he v e r t i c a l de v i a t i on o f t he t angen t a t A i w n i t t e h g t r h a e ti o ta n n . gent at B can be found by d x e q n 8 . 6  M    E I   t     B   A   A /   B         x MOMENT-AREA THEOREMS 67

69.  68   x  d i s t a n c e f r o m t h e v e rt i c a l a x i s   t h r o u g h A t o t h e c e n tr o i d o f   t h e a r e a   b e t w ee n A & B .    E I    C en t r o i d o f an a r ea   x  d A     x d A  B   x   M   d x e q n 8 . 7   A t     A /  B MOMENT-AREA THEOREMS

70.  T heo r e m 1  The vertical deviation of the tangent at a point (A) on the elastic curve with the tangent extended from another point (B) equals the “moment” of the area under the M/EI diagram between the 2 points (A & B).  This moment is computed about point A where the derivation is to be determined. MOMENT-AREA THEOREMS 69

71.  P r o v i ded t he m o m en t o f a + v e M / E I a r ea f r o m A to B is computed as in Fig. 8.14(b), it indicates that the tangent at point A is above the tangent to the curve extended from point B as shown in Fig. 8.14(c).  - v e a r ea s i nd i c a t e t ha t t he t angen t a t A i s be l o w the tangent extended from B. MOMENT-AREA THEOREMS 70

72.   71 MOMENT-AREA THEOREMS

73.  It i s i m po r t an t t o r ea li z e t ha t t he m o m en t - a r e a theorems can only be used to determine the angles and deviations between 2 tangents on 72 t he bea m ‟ s e l a s t i c c u r v e .    I n gene r a l , t he y do no t g i v e a d i r e c t s o l u t i on f o r the slope or displacement at a point. MOMENT-AREA THEOREMS

74. 3-5 CONJUGATE-BEAM METHOD  Here the shear V compares with the slop q e   , the moment M compares with the displacement Y & the external load w compares with the M/EI diagram.  T o m a k e u s e o f t h i s c o m pa r i s on w e w ill no w consider a beam having the same length as the r b e e a a l m be ” a m s s b h u o t w re n f e in r r F e i d g . t o 8 . a 2 s 2 . t   he “conjugate 73

75. 3-5 CONJUGATE-BEAM METHOD 74

76. 3-5 CONJUGATE-BEAM METHOD  T o s ho w t h i s s i m il a r i t y , w e c an w r i t e t he s e equation as shown d V   dx  w    w d   2   M   dx  d q      M  d x   E I   dx 2  EI  2 d   2 v    M   75

77. 3-5 CONJUGATE-BEAM METHOD  O r i n t e r g r a t i n g V       w d x     M        w d x  d x d x   M        EI          q       M       dx  dx       EI          v     76

78. 3-5 CONJUGATE-BEAM METHOD  T he c on j uga t e bea m i s l oaded w i t h t h e M/EI diagram derived from the load w on the real beam.  From the above comparisons, we can state 2 theorems related to the conjugate beam.  T heo r e m 1  The slope at a point in the real beam is numerically equal to the shear at the corresponding point in the conjugate beam. 77

79. 3-5 CONJUGATE-BEAM METHOD  T heo r e m 2  The displacement of a point in the real beam is numerically equal to the moment at the corresponding point in the conjugate beam.  When drawing the conjugate beam, it is important that the shear & moment developed at the supports of the conjugate beam account for the corresponding slope and displacement of the real beam at its supports. 78

80. 3-5 CONJUGATE-BEAM METHOD 79  For example, as shown in Table 8.2, a pin or roller support at the end of the real beam provides zero displacement but the beam has a non-zero slope.  Consequently from Theorem 1 & 2, the conjugate beam must be supported by a pin or roller since this support has zero moment but has a shear or end reaction.  When the real beam is fixed supported, both beam has a free end since at this end there is zero shear & moment.

81. 8/10/2019 Deflection of Beam- IsM http://slidepdf.com/reader/full/deflection-of-beam-ism 81 / 8 1 3-5 CONJUGATE-BEAM METHOD  s C u o p rr p e o s r p ts o n fo d r i n o g th r e e r a c l a & s e co s n a ju re g a lis te te b d e i a n m the table