MAHALAKSHMI ENGINEERING COLLEGE

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1 AHAAKSHI ENGINEERING COEGE TIRUCHIRAPAI QUESTION WITH ANSWERS DEPARTENT : CIVI SEESTER: V SU.CODE/ NAE: CE 5 / Strength of aterials UNIT INDETERINATE EAS 1. Define statically indeterminate beams. If the numbers of reaction components are more than the conditions equations, the structure is defined as statically indeterminate beams. E = R r E = Degree of external redundancy R = Total number of reaction components r = Total number of condition equations available. A continuous beam is a typical example of externally indeterminate structure.. State the degree of indeterminacy in propped cantilever. For a general loading, the total reaction components (R) are equal to (+) =5, While the total number of condition equations (r) are equal to. The beam is statically indeterminate, externally to second degree. For vertical loading, the beam is statically determinate to single degree. E = R r = 5 =. State the degree of indeterminacy in a fixed beam. For a general system of loading, a fixed beam is statically indeterminate to third degree. For vertical loading, a fixed beam is statically indeterminate to second degree. E = R r For general system of loading: R = + and r = E = 6- = For vertical loading: R = + and r = E = 4 = UTHUKUAR.P/Asst.Prof./CIVI-III Page 1

2 4. State the degree of indeterminacy in the given beam. The beam is statically indeterminate to third degree of general system of loading. R = = 6 E = R-r = 6- = 5. State the degree of indeterminacy in the given beam. The beam is statically determinate. The total numbers of condition equations are equal to + = 5. Since, there is a link at. The two additional condition equations are at link. E = R-r = = 5-5 E = 0 6. State the methods available for analyzing statically indeterminate structures. i. Compatibility method ii. Equilibrium method 7. Write the expression fixed end moments and deflection for a fixed beam carrying point load at centre. W A 8 W ymax 19EI 8. Write the expression fixed end moments and deflection for a fixed beam carrying eccentric point load. Wab A Wa b Wa b ymax ( under the load) EI 9. Write the expression fixed end moments for a fixed due to sinking of support. 6EI A 10. State the Theorem of three moments. (AUC Nov/Dec 01) (AUC Apr/ay 011) Theorem of three moments: It states that If C and CD are only two consecutive span of a continuous beam subjected to an external loading, then the moments, C and D at the supports, C and D are given by UTHUKUAR.P/Asst.Prof./CIVI-III Page

3 1 C ( 1 ) D. 1 _ 6a1 x Where = ending oment at due to external loading C = ending oment at C due to external loading D = ending oment at D due to external loading 1 = length of span A = length of span C a 1 = area of..d due to vertical loads on span C = area of..d due to vertical loads on span CD 1 _ 6a x a _ x 1 = Distance of C.G of the..d due to vertical loads on C from _ x = Distance of C.G of the..d due to vertical loads on CD from D. 11. What are the fixed end moments for a fixed beam of length subjected to a concentrated load w at a distance a from left end? (AUC Nov/Dec 004) (AUC Apr/ay 010) Fixed End oment: Wab A Wab 1. Explain the effect of settlement of supports in a continuous beam. (Nov/Dec 00) Due to the settlement of supports in a continuous beam, the bending stresses will alters appreciably. The maximum bending moment in case of continuous beam is less when compare to the simply supported beam. 1. What are the advantages of Continuous beams over Simply Supported beams? (i)the maximum bending moment in case of a continuous beam is much less than in case of a simply supported beam of same span carrying same loads. (ii) In case of a continuous beam, the average. is lesser and hence lighter materials of construction can be used it resist the bending moment. 14. A fixed beam of length 5m carries a uniformly distributed load of 9 kn/m run over the entire span. If I = 4.5x10-4 m 4 and E = 1x10 7 kn/m, find the fixing moments at the ends and deflection at the centre. Solution: Given: = 5m W = 9 kn/m, I = 4.5x10-4 m 4 and E = 1x10 7 kn/m (i) The fixed end moment for the beam carrying udl: UTHUKUAR.P/Asst.Prof./CIVI-III Page

4 W A = = 1 9x(5) = KNm 1 (ii) The deflection at the centre due to udl: y c 4 W 84 EI 4 9x(5) yc x1x10 x4.5x10 Deflection is in downward direction. mm 15. A fixed beam A, 6m long is carrying a point load of 40 kn at its center. The.O.I of the beam is 78 x 10 6 mm 4 and value of E for beam material is.1x10 5 N/mm. Determine (i) Fixed end moments at A and. Solution: Fixed end moments: A W 8 50x6 A 7. 5 knm A fixed beam A of length m is having.o.i I = x 10 6 mm 4 and value of E for beam material is x10 5 N/mm. The support sinks down by mm. Determine (i) fixed end moments at A and. Solution: Given: = m = 000mm I = x 10 6 mm 4 E = x10 5 N/mm = mm A 6EI 5 6 6xx10 xx10 x = (000) =1x10 5 N mm = 1 kn m. UTHUKUAR.P/Asst.Prof./CIVI-III Page A fixed beam A, m long is carrying a point load of 45 kn at a distance of m from A. If the flexural rigidity (i.e) EI of the beam is 1x10 4 knm. Determine (i) Deflection under the oad. Solution: = m W = 45 Kn EI = 1x10 4 knm

5 Deflection under the load: In fixed beam, deflection under the load due to eccentric load y C Wa b EI y y y C C C 45x() x1x x(1) x() mm The deflection is in downward direction. m 18. A fixed beam of 5m span carries a gradually varying load from zero at end A to 10 kn/m at end. Find the fixing moment and reaction at the fixed ends. Solution: Given: = 5m W = 10 kn/m (i) Fixing oment: A W 0 and W 0 A = 10(5) knm 10(5) knm 0 0 (ii) Reaction at support: W 7 R and R W A 0 0 R R A *10*5 0 7 *10* kn 17.5 kn UTHUKUAR.P/Asst.Prof./CIVI-III Page 5

6 19. A cantilever beam A of span 6m is fixed at A and propped at. The beam carries a udl of kn/m over its whole length. Find the reaction at propped end. (ay /.June01) Solution: Given: =6m, w = kn/m Downward deflection at due to the udl neglecting prop reaction P, 4 wl y 8EI Upward deflection at due to the prop reaction P at neglecting the udl, Pl y EI Upward deflection = Downward deflection 4 Pl wl EI 8EI P = W/8 = **6/8 =4.5 Kn 19. Give the procedure for analyzing the continuous beams with fixed ends using three moment equations? The three moment equations, for the fixed end of the beam, can be modified by imagining a span of length l 0 and moment of inertia, beyond the support the and applying the theorem of three moments as usual. 0. Define Flexural Rigidity of eams. The product of young s modulus (E) and moment of inertia (I) is called Flexural Rigidity (EI) of eams. The unit is N mm. 1. What is a fixed beam? (AUC Apr/ay 011) A beam whose both ends are fixed is known as a fixed beam. Fixed beam is also called as built-in or encaster beam. Incase of fixed beam both its ends are rigidly fixed and the slope and deflection at the fixed ends are zero.. What are the advantages of fixed beams? (i) For the same loading, the maximum deflection of a fixed beam is less than that of a simply supported beam. (ii) For the same loading, the fixed beam is subjected to lesser maximum bending moment. (iii) The slope at both ends of a fixed beam is zero. (iv) The beam is more stable and stronger. 4. What are the disadvantages of a fixed beam? (i) arge stresses are set up by temperature changes. (ii) Special care has to be taken in aligning supports accurately at the same lavel. (iii) arge stresses are set if a little sinking of one support takes place. (iv) Frequent fluctuations in loadingrender the degree of fixity at the ends very uncertain. 5. Define: Continuous beam. A Continuous beam is one, which is supported on more than two supports. For usual loading on the beam hogging ( - ive ) moments causing convexity upwards at the supports and sagging ( + ve ) moments causing concavity upwards occur at mid span. UTHUKUAR.P/Asst.Prof./CIVI-III Page 6

7 6. What is mean by prop?. (AUC Nov/Dec 01) When a beam or cantilever carries some load, maximum deflection occurs at the free end. the deflection can be reduced by providing vertical support at these points or at any suitable points. (PART 16 arks) 1. A fixed beam A of length 6m carries point load of 160 kn and 10 kn at a distance of m and 4m from the left end A. Find the fixed end moments and the reactions at the supports. Draw. and S.F diagrams. (AUC Apr/ay 008)(AUC Nov/Dec006) Solution: Given: = 6m oad at C, W C = 160 kn oad at D, W C = 10 kn Distance AC = m Distance AD =4m UTHUKUAR.P/Asst.Prof./CIVI-III Page 7

8 First calculate the fixed end moments due to loads at C and D separately and then add up the moments. Fixed End oments: For the load at C, a=m and b=4m A1 A1 W C ab 160xx(4) (6) 14. knm 1 1 W C a b 160x x(4) (6) For the load at D, a = 4m and b = m A A W D a b 10x x(4) (6) W Total fixing moment at A, A D a b 160x x(4) (6) A knm knm = A1 + A = = knm knm Total fixing moment at, = 1 + = = kn m. diagram due to vertical loads: Consider the beam A as simply supported. et R A * and R * are the reactions at A and due to simply supported beam. Taking moments about A, we get R R * * R A * x6 160x 10x kn 6 = Total load - R * =( ) 1. = kn. at A = 0. at C = R A * x = x = 9.4 kn m UTHUKUAR.P/Asst.Prof./CIVI-III Page 8

9 . at D = 1. x = kn m. at = 0 S.F Diagram: et R A = Resultant reaction at A due to fixed end moments and vertical loads R = Resultant reaction at Equating the clockwise moments and anti-clockwise moments about A, R x 6 + A = 160 x + 10 x 4 + R = 10.7 kn R A = total load R = kn S.F at A = R A = kn S.F at C = = kn S.F at D = = kn S.F at = 10.7 KN. A fixed beam A of length 6m carries two point loads of 0 kn each at a distance of m from the both ends. Determine the fixed end moments. Sloution: Given: ength = 6m Point load at C = W 1 = 0 kn Point load at D = W = 0 Kn Fixed end moments: A = Fixing moment due to load at C + Fixing moment due to load at D W a b W a 0xx4 0x4x 6 6 Since the beam is symmetrical, b 40kN m A = = 40 knm UTHUKUAR.P/Asst.Prof./CIVI-III Page 9

10 . Diagram: To draw the. diagram due to vertical loads, consider the beam A as simply supported. The reactions at A and is equal to 0kN.. at A and = 0. at C =0 x = 60 knm. at D = 0 x = 60 knm. Find the fixing moments and support reactions of a fixed beam A of length 6m, carrying a uniformly distributed load of 4kN/m over the left half of the span.(ay/june 006 & 007) Solution: acaulay s method can be used and directly the fixing moments and end reactions can be calculated. This method is used where the areas of. diagrams cannot be determined conveniently. For this method it is necessary that UD should be extended up to and then compensated for upward UD for length C as shown in fig. The bending at any section at a distance x from A is given by, d y EI dx x (x ) RA x A wx +w*(x-) 4x x ) =R A x A - ( ) +4( = R A x A - x +(x-) ) UTHUKUAR.P/Asst.Prof./CIVI-III Page 10

11 Integrating, we get dy x x ( x ) EI =RA -A x - +C1 + dx (1) dy When x=0, =0. dx Substituting this value in the above equation up to dotted line, C 1 = 0 Therefore equation (1) becomes dy x x EI =RA -A x - dx Integrating we get ( x ) x Ax x ( x ) EI y RA C When x = 0, y = 0 y substituting these boundary conditions upto the dotted line, C = 0 4 RAx Ax x 1( x ) EI y y subs x =6 & y = 0 in equation (ii) 4 (ii) 0 R A 6 A R A 9 A = dy At x =6, 0 in equation (i) dx 4 1(6 ) 6 4 6R A 18 A (iii) RA x Ax6 x 6 18R x R 6 y solving (iii) & (iv) A A A A 16 6 A = 8.5 knm y substituting A in (iv) 16 = 18 R A 6 (8.5) R A = 9.75 kn R = Total load R A R =.5 kn UTHUKUAR.P/Asst.Prof./CIVI-III Page 11

12 y equating the clockwise moments and anticlockwise moments about + R A x 6 = A + 4x (4.5) =.75 knm Result: A = 8.5 knm =.75 knm R A = 9.75 kn R =.5 KN. A cantilever A of span 6m is fixed at the end and proposal at the end. it carries a point load of 50KnN at mid span. level of the prop is the same as that of the fixed end. (i) (ii) Determine The Reaction At The Prop. Draw SFD AND D. UTHUKUAR.P/Asst.Prof./CIVI-III Page 1

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14 4. Analysis the propped cantilever beam of the length 10m is subjected to point load of 10KN acting at a 6m from fixed and draw SFD and D. (AUC Nov/Dec 010 ) UTHUKUAR.P/Asst.Prof./CIVI-III Page 14

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16 5. A propped cantilever of span 6m having the prop at the is subjected to two concentrated loads of 4 KN and 48KN at one third points respective from left end (fixed support ) draw SFD and D. (AUC Nov/Dec 010 ) UTHUKUAR.P/Asst.Prof./CIVI-III Page 16

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19 6. A propped cantilever of span 6m is subjected to a UD of KN/m over a length of fixed the end. Determine the prop reaction and draw the SFD and D. (AUC ay/june 01) UTHUKUAR.P/Asst.Prof./CIVI-III Page 19

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21 7. A fixed beam of a 6m span supports two point loads 00KN each at a two meters from each end. find the fixing moments at the ends and draw the D and SFD. Find also the deflection. Take I = 9 x 10 8 mm and E = 00KN /m. (AUC Apr / AY 010) UTHUKUAR.P/Asst.Prof./CIVI-III Page 1

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23 8. For The fixed beam shown in fig. Draw the D and SFD. UTHUKUAR.P/Asst.Prof./CIVI-III Page

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25 9. A cantilever beam AC of span 6m fixed at a and propped at c is loaded with an UD of 10KN/m for the length of 4m from the fixed end. the prop reaction.the find the maximum sagging and point of concentrations. UTHUKUAR.P/Asst.Prof./CIVI-III Page 5

26 10. A continuous beam consists of three successive span of 6m and 1m and 4m and carries load of KN/m, 1KN/m and KN/ respectively on the spans. Draw D and SFD for the beam. UTHUKUAR.P/Asst.Prof./CIVI-III Page 6

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28 11. Find the support moments and reactions for the continuous beam shown in fig. Draw the D and SFD. ( AUC APR/ay 010) UTHUKUAR.P/Asst.Prof./CIVI-III Page 8

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30 1. A continuous beam AC, is loaded as shown in fig. Find the support moments three moment equation. Draw SFD and D. UTHUKUAR.P/Asst.Prof./CIVI-III Page 0

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32 1. A continuous beam AC covers two consecutive span A and C of lengths 4m and 6m, carrying uniformly distributed loads of 6kN/m and 10kN/m respectively. If the ends A and C are simply supported, find the support moments at A, and C. draw also..d and S.F.D. (AUC Apr/ay 01) Solution: Given Data: ength A, 1 =4m. ength C, =6m UD on A, w 1 =6kN/m UD on C, w =10kN/m (i) Support oments: Since the ends A and C are simply supported, the support moments at A and C will be zero. y using cleyperon s equation of three moments, to find the support moments at (ie). 6a1x1 6ax A 1 + ( 1 + ) + C = 4 6 6a1x1 6a x 0 + (4+6) + 0 = 4 6 a1x1 0 = ax The..D on a simply supported beam is carrying UD is a parabola having an attitude of w 8. Area of..d = **h w = * Span * 8 The distance of C.G of this area from one end, =. a 1 =Area of..d due to UD on A, span UTHUKUAR.P/Asst.Prof./CIVI-III Page

33 6(4 ) = *4* 8 = 1 x 1 = = 4/ = m. a = Area of..d due to UD on C, 10(6 ) = *6* 8 = 180m. x = / = 6 / =m Substitute these values in equation(i). We get, ** 0 = (180*) = =1.8 knm. (ii)..d (iii) The..D due to vertical loads (UD) on span A and span C. Span A: w = * 4 = 8 =1kNm w Span C: = 8 10 * 6 = 8 =45kNm S.F.D: To calculate Reactions, For span A, taking moments about, we get (R A *4)-(6*4*) =0 4R A 48 = 1.8 ( =1.8, -ve sign is due to hogging moment. R A =4.05kN Similarly, For span C, taking moment about, UTHUKUAR.P/Asst.Prof./CIVI-III Page

34 (R c *6)-(6*10*) =0 6R C 180=-1.8 R C =4.7kN. R =Total load on AC (R A +R ) =(6*4*(10*6))-( ) =55.5kN. RESUT: A = C =0 =1.8kNm R A =4.05kN R =55.5kN R C =4.7kN 14. A continuous beam ACD of length 15m rests on four supports covering equal spans and carries a uniformly distributed load of 1.5 kn/m length.calculate the moments and reactions at the supports. Draw The S.F.D and..d. Solution: Given: ength A = 1 = 5m ength C = = 5m ength CD = = 5m u.d.l w 1 = w = w = 1.5 kn/m Since the ends A and D are simply supported, the support moments at A and D will be Zero. A =0 and D =0 For symmetry =0 (i)to calculate support moments: UTHUKUAR.P/Asst.Prof./CIVI-III Page 4

35 To find the support moments at and C, by using claperon s equations of three moments for AC and CD. For AC, 6a1x1 6ax A 1 +[ ( 1 + )]+ C = 1 6a1x1 6a x 0+[ (5+5)]+[ C (5)]= C = ( a 1x1 ax ) (i) 5 a 1 =Area of D due to UD on A when A is considered as simply supported beam. = w * A * Altitude of parabola (Altitude of parabola= 1 1 ) 8 1.5*(5) = *5* 8 =15.65 x 1 = 1 / =5/=.5m Due to symmetry.a =a 1 =15.65 x =x 1 =.5 subs these values in eqn(i) C = [(15.65*.5) (15.65*.5)] 5 =9.75 Due to symmetry = C 0 +5 =9.75 =.75kNm. = C =.75kNm. (ii) To calculate due to vertical loads: The D due to vertical loads(here UD) on span A, C and CD (considering each span as simply supported ) are shown by parabolas of altitude w * kNm (iii)to calculate support Reactions: each. et R A,R,R C and R D are the support reactions at A,,C and D. Due to symmetry R A =R D R =R C For span A, Taking moments about, We get UTHUKUAR.P/Asst.Prof./CIVI-III Page 5

36 =(R A *5)-(1.5*5*.5) -.75=(R A *5) R A =.0kN. Due to symmetry R A =R D =.0kN R =R C R A +R +R C +R D =Total load on ACD +R +R +=1.5*15 R =8.5kN R C =8.5kN. Result: A = D = 0 = C =.75kNm. R A =R D =.0kN R =8.5kN R C =8.5kN. 15. A continuous beam ACD, simply supported at A,, C and D is loaded as shown in fig. Find the moments over the beam and draw..d and S.F.D. (Nov/ Dec 00) Solution: Given: ength A = 1 = 6m ength C = = 5m ength CD = = 4m Point load W 1 = 9kN Point load W = 8kN u.d.l on CD, w = kn/m UTHUKUAR.P/Asst.Prof./CIVI-III Page 6

37 (i)..d due to vertical loads taking each span as simply supported: W1ab 9* * 4 Consider beam A,. at point load at E = 1kNm 6 W ab 8* * Similarly. at F = 9.6kNm 6. at the centre of a simply supported beam CD, carrying U.D. w 8 * kNm (ii)..d due to support moments: Since the beam is simply supported A = D = 0 y using Clapeyron s Equation of Three oments: a) For spans A and C 6a1x1 6ax A 1 + ( 1 + ) + C = 4 6 6a1x1 6ax 0 (6 5) c (5) C a1x1 ax (i) 5 a 1 x 1 = ½*6*1*+a/ = ½*6*1*(6+)/ = 96 a x = ½*5*9.6*+b/ = ½*5*9.6*(6+4)/ = 64 Substitute the values in equation (i) + 5 C = 96+6/5* C = (ii) b) For spans C and CD + C ( + ) + D = 6a x 6 a x *5 + C (5+4) +0 = a 5 6a 4 6 x x 6ax 6ax 5 18 C (iii) 5 4 a x = ½ * 5 * 9.6 *(+a)/ =1/ * 5 * 9.6 *(5+)/ = 56 a x = / * 4*6*4/ = Substitute these values in equation (iii) UTHUKUAR.P/Asst.Prof./CIVI-III Page 7

38 5 18 C 5 18 C 6* * 4 y solving equations (ii) &(iv) = 6.84 knm and C = 4.48 knm (iii) Support Reactions: For the span A, Taking moment about, = R A * 6 9*4 = R 6 6 A R A = 4.86KN 6 For the span CD, taking moments about C 4 C RD 4 4 ( C 4.48) R D = 4.88KN For AC taking moment about C c = R R *5 8* A * 5R *11 R = 9.41 kn R C = Total load on ACD (R A +R +R D ) R C = (9+8+4*) ( ) R C = 9.85 kn Result: A = D = 0 = 6.84 knm and C = 4.48 knm R A = 4.86kN R = 9.41kN R C = 9.85 kn R D = 4.88KN UTHUKUAR.P/Asst.Prof./CIVI-III Page 8

39 16. Using the theorem of three moments draw the shear force and bending moment diagrams for the following continuous beam. (April / ay 00) Solution: Given: ength A, 1 =4m. ength C, =m. ength CD, =4m. UD on A, w=4 kn/m Point load in C, W 1 =4kN/m Point load in CD, W 1 =6kN (i) ending oment to Vertical oads: w 4* 4 Consider beam A,.= 8 8 =8kNm. Similarly for beam C, W1ab 6* *1.= =4kNm Similarly for beam CD, Wab.= =6kNm 8*1* 4 (ii) ending oment to support moments: et A,, C And D be the support moments at A,,C and D. Since the ends is simply supported, A = D =0. y using Clayperon s equation of three moments for span A and C, 6a1x1 6ax A 1 +[ ( 1 + ) ]+ C = 1 UTHUKUAR.P/Asst.Prof./CIVI-III Page 9

40 6a1x1 6a x 0+[ (4+)] C () = C = 1.5a 1 x 1 + a x (i) a 1 x 1 = oment of area D due to UD ase = * *( ase * Altitude) 4 = * *(4*8) =4. a x = oment of area D due to point load about point 1 * = * *(*4) =5. Using these values in eqn (i), 14 + C =1.5(4.) +(*5.) 14 + C = (ii) For span C and CD, 6a x 1 +[ C ( + ) ]+ D = ()+[ C (+) ]+ D = 6 a x 6a ax 6 x +1 C = a x + a x (iii) a x = oment of area D due to point load about point C *1 =(1/)**4* =.66 a x = oment of area D due to point load about point D 1 * = *1*6* =6 Using these values in Eqn(iii), + 1 C =(.66) + (*6) + 1 C = (iv) UTHUKUAR.P/Asst.Prof./CIVI-III Page 40

41 Using eqn (ii) and (iii), = 5.69 kn m C = 0.19 kn m (iii) Support Reaction: For span A, taking moment about R A *4 4*4* = R A *4 R A *4=6.71 R A = 6.68 kn For span CD, taking moment about C * 4 8*1 C R D = R D *4-8 R D = kn Now taking moment about C for AC C RA( 7) 4* 4*5 R R 4(0) R 6 Result: C 7 A * 6* (6.68) 80 R 6 R = 1.07 kn R C = Total load (R A +R + R C ) = 4 * R C = 8.16 kn A = D = 0 = 5.69 kn m C = 0.19 kn m R A = 6.68 kn R = 1.07 kn R C = 8.16 kn R D = Kn 17. A beam A of 4m span is simply supported at the ends and is loaded as shown in fig. Determine (i) Deflection at C (ii) aximum deflection (iii) Slope at the end A. E= 00 x 10 6 kn/m and I = 0 x 10-6 m 4 Solution: Given: = 4m E= 00 x 10 6 kn/m and I = 0 x 10-6 m 4 To calculate Reaction: Taking moment about A R *4 0*1 R *4 = 0 + 0() 10*( 1 1) UTHUKUAR.P/Asst.Prof./CIVI-III Page 41

42 R = 80/4 = 0 kn R A = Total load - R = (10*+0) -0 R A = 0 kn y using acaulay s method: X Integrating we get dy EI dx Integrating we get d y EI d x 10x 0x 0( x 1) 10( x ) 5( C1 10( x 1) x 4 10x 10( x 1) 5( x ) EIy C1x C (ii) 1 When x = 0, y = 0 in equation (ii) we get C = 0 When x = 4m, y = 0 in equation (ii) (4) 4C 1 (4 1) 4 1 = 1. +4C C 1 = Hence the slope and deflection equations are Slope Equation: ) dy EI dx 10x ( x 1) 5( x ) Deflection Equation: EIy 10x 9.16x 10( x 1) 5( x ) 1 4 (i) Deflection at C, y C : Putting x = m in the deflection equation, we get 10() EIy 9.16() = = y c = 8.74 (downward) 10( 1) UTHUKUAR.P/Asst.Prof./CIVI-III Page 4

43 (ii) aximum Deflection, y max : The maximum deflection will be very near to mid-point C. et us assume that it occurs in the sections between D and C. For maximum deflection equating the slope at the section to zero, we get dy EI dx 10x ( x 1) 10x (x-1) = 0 10x (x -x+1) = 0 x = 9.16/0 =1.958 m 10(1.958) EIy 9.16(1.958) y max = -5/EI y max = 8.75 mm (downward) 10( ) (iii) Slope at the end A, θ A : Putting x = 0 in the slope equation, dy EI 9.16 dx θ A = dy/dx = -9.16/EI θ A = radians θ A = º Result: (i) Deflection at C = 8.74 mm (ii) aximum deflection = 8.75 mm (iii) Slope at the end A, θ A = º 18. A continuous beam is shown in fig. Draw the D indicating salient points. (Nov/Dec 004) Solution: Given: ength 1 = 4m ength = 8m ength = 6m Udl on C w = 10 kn/m Point load W 1 = 40 kn Point load W = 40 kn UTHUKUAR.P/Asst.Prof./CIVI-III Page 4

44 (i). due to vertical loads: Consider beam A,. = For beam C, w. = 8 For beam CD, W. = 4 (ii). due to support moments: 10(8) 8 W1ab 1 80 knm 40 * 6 60 knm 4 40**1 4 0 knm et A,, C, D be the support moments at A,, C, D. Since the end A and D are simply supported A = D = 0 y using Clapeyron s Equation of Three moments. For Span A and C: A 1 ( 1 ) C 6a1x 1 1 6ax 0 (4 8 ) C (8) 6a x 4 6a x (1) +8 C = -1.5a 1 x a x 4 +8 C = -1.5a 1 x a x (i) a 1 x 1 = oment of area of..d due to point load = ½*4*0*/* = 10 a x = oment of area of..d due to udl = / (ase x Altitude) x ase/ = / (8*80)*8/ = Using these values in equation (i) 4 +8 C = -1.5(10) 0.75 ( ) 4 +8 C = (ii) For Span C and CD: C ( ) D 6ax 6ax 6a x 6ax (8) C (8 6) C = a x - a x (iii) a x = oment of area of..d due to udl UTHUKUAR.P/Asst.Prof./CIVI-III Page 44

45 = / (ase x Altitude) x ase/ = / (8*80)*8/ = a x = oment of area of..d due to point load = ½ * b*h*/ = ½ * 6*60*6/ = 60 Using these values in equation (iii) C = ( ) C = (iv) From (ii) & (iv) C = knm = knm Result: A = D = 0 C = knm = knm 19. For the fixed beam shown in fig. draw D and SFD. (AUC Nov / Dec 004) Solution: (i)..d due to vertical loads taking each span as simply supported: Consider beam A as simply supported. The. at the centre of A w 8 1 *() 8.5 knm (ii)..d due to support moments: As beam is fixed at A and, therefore introduce an imaginary zero span AA 1 and 1 to the left of A and to the right of. The support moments at A 1 and 1 are zero. et 0 = Support moment at A 1 and 1 and it is zero. A = Fixing moment at A = Fixing moment at C = Support moment at C To find A, and C, Theorem of three moments is used. (a) For the span A 1 A and AC, 0 *0 A (0 A ) 1 () C C 1 () 6a0x a1x 1 6a1x 1 1 UTHUKUAR.P/Asst.Prof./CIVI-III Page 45

46 6 A + C = - a 1 x (i) a 1 x 1 = moment of area of..d due to udl on A when it is considered as simply supported beam about = / * ase * Altitude * 1 / = / * *.5 * / a 1 x 1 = 6.75 subs this values in equation (i) we get 6 A + C = (ii) (b) For the span AC and C: ( ) 6a x 6a x 1 1 A 1 C 1 1 6a1x 1 6ax A ( ) C ( ) () A + 1 C + = a 1 x 1 + a x a 1 x 1 = moment of area of..d due to udl on A when it is considered as simply supported beam about = / * ase * Altitude * 1 / = / * *.5 * / a 1 x 1 = 6.75 a x = 0 A + 1 C + = (ii) ( c ) For the span C and 1 C ( ) 6a () x C 0 0 *0 6ax 6a0x 0 0 C + 6 = a x a x = 0 C + 6 = 0 y solving (iii), (iv), (ii) C = 1.15 knm A = knm = knm UTHUKUAR.P/Asst.Prof./CIVI-III Page 46

47 (iii) Support Reactions: et R A, R, and R C are the support reactions at A, and C. For the span AC, taking moment about C, we get R A x x x A = C R A x = 1.15 R A =.1875 kn For the span C, taking moment about C, we get R x + C = R x = R = kn R C = Total load (R A + R ) = **1.5 ( ) R C = 5.65 kn Result: C = 1.15 knm A = knm = knm R A =.1875 kn R = kn R C = 5.65 kn UTHUKUAR.P/Asst.Prof./CIVI-III Page 47