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1 CE840-STRENGTH OF TERIS - II PGE 1 HKSHI ENGINEERING COEGE TIRUCHIRPI QUESTION WITH NSWERS DEPRTENT : CIVI SEESTER: IV SU.CODE/ NE: CE 840 / Strength of aterials -II UNIT INDETERINTE ES 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. 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 =
2 CE840-STRENGTH OF TERIS - II PGE 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 8 W ymax 19EI 8. Write the expression fixed end moments and deflection for a fixed beam carrying eccentric point load. Wab 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 10. State the Theorem of three moments. (UC Nov/Dec 01) (UC pr/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
3 CE840-STRENGTH OF TERIS - II PGE 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 = 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? (UC Nov/Dec 004) (UC pr/ay 010) Fixed End oment: Wab 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. 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:
4 CE840-STRENGTH OF TERIS - II PGE 4 W = = 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. fixed beam, 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 and. Solution: Fixed end moments: W 8 50x knm fixed beam 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 and. Solution: Given: = m = 000mm I = x 10 6 mm 4 E = x10 5 N/mm = mm 6EI 5 6 6xx10 xx10 x = (000) =1x10 5 N mm = 1 kn m. 17. fixed beam, m long is carrying a point load of 45 kn at a distance of m from. 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 CE840-STRENGTH OF TERIS - II PGE 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. fixed beam of 5m span carries a gradually varying load from zero at end 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: W 0 and W 0 = 10(5) knm 10(5) knm 0 0 (ii) Reaction at support: W 7 R and R W 0 0 R R *10*5 0 7 *10* kn 17.5 kn
6 CE840-STRENGTH OF TERIS - II PGE cantilever beam of span 6m is fixed at 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? (UC pr/ay 011) 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. 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.
7 CE840-STRENGTH OF TERIS - II PGE 7 6. What is mean by prop?. (UC 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. (PRT 16 arks) 1. fixed beam of length 6m carries point load of 160 kn and 10 kn at a distance of m and 4m from the left end. Find the fixed end moments and the reactions at the supports. Draw. and S.F diagrams. (UC pr/ay 008)(UC Nov/Dec006) Solution: Given: = 6m oad at C, W C = 160 kn oad at D, W C = 10 kn Distance C = m Distance D =4m
8 CE840-STRENGTH OF TERIS - II PGE 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 1 1 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 W D a b 10x x(4) (6) W Total fixing moment at, D a b 160x x(4) (6) knm knm = 1 + = = knm knm Total fixing moment at, = 1 + = = kn m. diagram due to vertical loads: Consider the beam as simply supported. et R * and R * are the reactions at and due to simply supported beam. Taking moments about, we get R R * * R * x6 160x 10x kn 6 = Total load - R * =( ) 1. = kn. at = 0. at C = R * x = x = 9.4 kn m
9 CE840-STRENGTH OF TERIS - II PGE 9. at D = 1. x = kn m. at = 0 S.F Diagram: et R = Resultant reaction at due to fixed end moments and vertical loads R = Resultant reaction at Equating the clockwise moments and anti-clockwise moments about, R x 6 + = 160 x + 10 x 4 + R = 10.7 kn R = total load R = kn S.F at = R = kn S.F at C = = kn S.F at D = = kn S.F at = 10.7 KN. fixed beam 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: = 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 = = 40 knm
10 CE840-STRENGTH OF TERIS - II PGE 10. Diagram: To draw the. diagram due to vertical loads, consider the beam as simply supported. The reactions at and is equal to 0kN.. at 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 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 is given by, d y EI dx x (x ) R x wx +w*(x-) 4x x ) =R x - ( ) +4( = R x - x +(x-) )
11 CE840-STRENGTH OF TERIS - II PGE 11 Integrating, we get dy x x ( x ) EI =R - 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 =R - x - dx Integrating we get ( x ) x x x ( x ) EI y R C When x = 0, y = 0 y substituting these boundary conditions upto the dotted line, C = 0 4 Rx x x 1( x ) EI y y subs x =6 & y = 0 in equation (ii) 4 (ii) 0 R R 9 = dy t x =6, 0 in equation (i) dx 4 1(6 ) 6 4 6R (iii) R x x6 x 6 18R x R 6 y solving (iii) & (iv) 16 6 = 8.5 knm y substituting in (iv) 16 = 18 R 6 (8.5) R = 9.75 kn R = Total load R R =.5 kn
12 CE840-STRENGTH OF TERIS - II PGE 1 y equating the clockwise moments and anticlockwise moments about + R x 6 = + 4x (4.5) =.75 knm Result: = 8.5 knm =.75 knm R = 9.75 kn R =.5 KN. cantilever 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 t The Prop. Draw SFD ND D.
13 CE840-STRENGTH OF TERIS - II PGE 1
14 CE840-STRENGTH OF TERIS - II PGE nalysis 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. (UC Nov/Dec 010 )
15 CE840-STRENGTH OF TERIS - II PGE 15
16 CE840-STRENGTH OF TERIS - II PGE 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. (UC Nov/Dec 010 )
17 CE840-STRENGTH OF TERIS - II PGE 17
18 CE840-STRENGTH OF TERIS - II PGE 18
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