Method of Least Work. Theory of Structures II M Shahid Mehmood Department of Civil Engineering Swedish College of Engineering & Technology, Wah Cantt

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1 Method of east Work Theor of Structures II M Shahid Mehmood epartment of ivil Engineering Swedish ollege of Engineering & Technolog, Wah antt

2 Method of east Work / astigliano s Second Theorem Staticall Indeterminate Trusses Strain energ stored or work done in a member of a truss is P U E In the entire truss, the strain energ is U P E where P is the axial force in the member.

3 Staticall Indeterminate Trusses ifferential co efficient of U with respect to a force S in the member is U P P S E S ccording to the Theorem of east Work U S U P P S E S 3

4 Staticall Indeterminate Trusses P S 1 S S P S U S P E P S 4

5 Example 1 etermine the force in members of the truss shown. E is same for all members. 9 T 6 ft 6 ft 5

6 Solution The truss is indeterminate to the first degree. 9 T 9 T S + S 9 T 9 T 9 T 6

7 Solution 9 T S S 9 T 9 T 9 T Member Force due to Force due to Redundant P pplied oad Member Force 9 S/ 9 S/ S/ S/ S/ S/ 9 S/ 9 S/ 9 +S 9 +S +S S 7

8 Solution 9 T S 9 S 9 + S + S S 9 S Member Force due to Force due to Redundant P pplied oad Member Force 9 S/ 9 S/ S/ S/ S/ S/ 9 S/ 9 S/ 9 +S 9 + S +S S 8

9 Solution 9 T S 9 S 9 + S + S S 9 S Member P P/ S (in) (in.) P P P/ S 9 S/ 7 S/ 1 7 S/ 7 9 S/ S S S 7 S 7 S S S S + 36S + 36S S S 7 S 11.8S P P S S 9

10 Solution 1 P P E S P P S S Substituting value of S in nd olumn P 4.49T P 4.51T S 6.38T 9 T P 4.51T P 4.49T P 6.35T P 6.38T 1

11 Example nalze the truss shown. E is same for all members. 3m 4m 1 kn 11

12 Solution The truss is externall indeterminate to the first degree. et V be the redundant. Find the other reactions in terms of redundant V. V V 1 kn 1

13 Solution V 3m V 4m 1 kn M F x kn kn 3 F V + V 1 V 1 V 13

14 Solution kn 1 V 3m kn V 4m 1 kn M F x kn kn 3 F V + V 1 V 1 V 14

15 Solution Joint V F kn 1 V F F V kn V kn F x F3 + F4 5 5 F V 3 F V F kn F 3 V 15

16 Solution Joint kn 1 V F V F kn F 5 1 V kn 5 5 V F5 1 V 3 3 F F F F F x F kn 5 F F V V V F V 16

17 Solution Joint kn 1 V F 3 F + F kn V 3 1 kn F F 3 F + 4 V F +V F V 4 3 F 4 V.5/3 4 F 17

18 Solution The strain energ of a truss is U F E U F V V F E See Table on next slide 18

19 Universit of Engineering & Technolog, Taxila Solution Member F F/ V (m) F. F/ V Final Forces V V +6.3 kn +V V kn V V kn 4 VV x5/ V kn 5 5/3(1 V ) V kn F V F ( ) V ( ) V kn 19

20 Solution kn kn 3m kn kn 4m 1 kn

21 Method of east Work Staticall Indeterminate rches Fixed rches Two inged rches 6 Reactions 4 Reactions i 3 rd degree i 1 st degree 1

22 Two inged rch 4 Reactions V i 1 st degree V We have 4 unknowns, V,, V, and but onl three equations of equilibrium.,, when no external horizontal force is acting on the arch. If the supports are unielding U/, which gives the fourth equation for solving it.

23 Two inged rch V V n arch is subjected to M, SF and xial Thrust N. Total strain energ for the arch is given b U M V F EI G s GE ds + ds + s ds 3

24 Two inged rch V V If we consider onl bending deformation for simplified analsis, then M U ds EI ccording to Theorem of east Work U M M EI ds (1) 4

25 Two inged Parabolic rch x V V For an rch the bending moment is M V. x. if V.x µ M µ. 5

26 Two inged Parabolic rch x V For an rch the shear force, S is V S V cos θ sinθ For an rch the xial thrust, N is N V sinθ + cosθ 6

27 Two inged Parabolic rch x V Equation (1) becomes, U (. ) M µ ds EI If then (. ) U M µ EI ds V 7

28 Two inged Parabolic rch x V Partial derivative of M with respect to is M Equation (1) becomes U U 1 EI ( µ. )( ) ds V 8

29 Two inged Parabolic rch U 1 EI µ ds + EI ds 1 µ ds EI EI ds µ EI EI ds ds µ ds ds 9

30 Parabolic rch Subjected to a Single oncentrated oad µ w dx dx x c Equation of Parabola V V k k 4c x ( x ) w( k) V w wk V k k V V 3

31 Parabolic rch Subjected to a Single oncentrated oad For k µ w ( k) x ( k ) k w 4 cx µ dx x ( x )dx k For k µ ( k) w x w ( x k) V x k w c k w V k µ dx ( k) w 4 c ( x k) x( x) dx x w k V k k V 33

32 Parabolic rch Subjected to a Single oncentrated oad For k ( k) x w µ x ( k) w 4 x k k c µ dx For k µ k ( k) w x w ( x k) ( k) x ( x )dx w 4 µ dx x w k c ( x k) x( x) dx 34

33 Parabolic rch Subjected to a Single oncentrated oad dx µdx dx ( k ) 4 c. x ( x) dx ( k ) k w 4 c w 4 x x ( x ) dx + x w k 4c x ( x ) dx c ( x k ) x ( x ) dx 5 w + 8 c ( 3 4 k k + k ) 35

34 Example 3 Find the horizontal thrust in the arch shown. 8 T 8 T 6f ft ft 1 ft 1 ft ft 36

35 Solution 8 T 8 T 6 ft 1 ft 1 ft 1 ft ft V V onsidering oad k ( ) k k k w c

36 Solution 8 T 8 T T 6 ft 7.16 T ft 1 ft V 1 ft ft V T s the load is smmetric, so total reaction will be T 38

37 Solution 8 T 8 T T 6 ft 7.16 T ft 1 ft V 8 T 1 ft ft V 8 T To find V and V, appl equations of equilibrium M V V 8T F V + V 8 8+ V V 8T

38 Example 3 Find the horizontal thrust in the arch shown. 8 T 8 T 6f ft ft 1 ft 1 ft ft 4

39 Solution 8 T 8 T 6 ft ft 1 ft V 1 ft ft 8 T 8 T V µ dx dx 4c 4 6 x x ( x). x. ( 6 x) ( x) 41

40 Solution 8 T 8 T 6 ft ft 1 ft V 1 ft ft 8 T 8 T V Portion Origin imits μ 8x 4 8x 8(x ) 8x 4

41 Solution 8 T 8 T µ dx dx ft 1 ft 6 ft V 1 ft ft 8 T 8 T V x x. ( 6 x) dx + ( 8x 8( x ) ). ( 6 x) dx T-ft µ dx 8x x dx 15 3 ( 6 x ) dx 115 ft 43

42 Solution 8 T 8 T 7.16 T 6 ft ft 1 ft V 8 T 1 ft ft 8 T 8 T 7.16 T V 8 T dx µ T dx

43 Solution ending Moment iagram 45

44 Example 4 nalze the arch shown. 4T/ft 15 ft ft 1 ft ft 46

45 -- Solution 4 T/ft µ 15 ft dx dx 4 c x 4 15 x 5 ( x ) V ft 1 ft ft ( x ) V 48T V 3T 5 3 x( x) x 5 5 ( x) ft 1 ft ft μ bending moment while considering simpl supported beam equation of parabola V 47

46 - Solution 4 T/ft µ 15 ft dx dx ft μ bending moment while considering 1 ft simpl supported beam V equation of parabola ft V V 48T V 3T Portion Origin imits μ 1 48x x 4(x 1) / 3x ft 1 ft ft 48

47 Solution dx 1 x 5 ( x) 4 T/ft µ 15 ft dx 5 Portion Origin imits μ 1 48x x 4(x 1) / 3x V 1 ft ft ft V 1 ( ) 1. x( 5 x) dx µ dx ( 48x) x( 5 x) dx + 48x ( x 1) 5 1 ( 3x) x( 5 x) dx T-ft 3 49

48 Solution dx 1 x 5 ( x) 4 T/ft µ 15 ft dx 5 Portion Origin imits μ 1 48x x 4(x 1) / 3x V 1 ft ft ft V dx x ( 5 x ) 6 ft 3 dx 5

49 Solution put these values into the following equation 4 T/ft µ dx T dx T T T 3 T 51

50 Method of Summation The arch rib is divided into a number of equal parts of length δ s. The values of M and are found at mid points of each of these lengths and the evaluation of numerator and denominator of expression for is carried out b addition µ δ δ s s 5

51 Example 4 nalze the arch shown b the method of summation. 4T/ft 15 ft ft 1 ft ft 53

52 Solution µ δ x δ x ivide the arch into equal segments. δ s ft 1 x 1 4 T/ft 5 ft x 1.5 ft 4c x ( x ) x ( x ) ( ).85 ft x 1 V 48 T 5 ft V 3 T µ 1 V x T ft 54

53 Solution Portion x (ft) (ft) μ (T ft) μ T/ft 1 x 1 5 ft x 1 V 48 T V 3 T 5 ft 55

54 Solution Portion x (ft) (ft) μ (T ft) μ T/ft x 5 ft x V 48 T V 3 T 5 ft 56

55 Solution Portion x (ft) (ft) μ (T ft) μ T/ft x 3 5 ft x 3 V 48 T V 3 T 5 ft 57

56 Solution Portion x (ft) (ft) μ (T ft) μ T/ft x ft V 48 T V 3 T 5 ft 58

57 Solution Portion x (ft) (ft) μ (T ft) μ T/ft x ft V 48 T V 3 T 5 ft 59

58 Solution Portion x (ft) (ft) μ (T ft) μ T/ft x ft V 48 T V 3 T 5 ft 6

59 Solution Portion x (ft) (ft) μ (T ft) μ T/ft x V 48 T 5 ft V 3 T 5 ft 61

60 Solution Portion x (ft) (ft) μ (T ft) μ T/ft x V 48 T 5 ft V 3 T 5 ft 6

61 Solution \ Portion x (ft) (ft) μ (T ft) μ T/ft x V 48 T 5 ft V 3 T 5 ft 63

62 Solution \ Portion x (ft) (ft) μ (T ft) μ T/ft x V 48 T 5 ft V 3 T ft µ δ x 55,737 δ 1.4 x 46.3T 64

63 ircular rch The circular arches are a portion of circle. Usuall span & central rise are given. onsider the arch shown in Figure below x c R x ( R ) c R R R α R 65

64 Semi ircular rch n.r w d s R θ V V U M dx EI R 4wn Rsinθ ( 1 n ) π µ ds ds ds Rd θ For U 4wR 3π 66

65 Example 5 Two hinged circular arch carries a concentrated load of 5 kn at the crown. The span and rise are 6m and 1m respectivel. el Find the horizontalontal thrust at the abutment. 5 kn R 3 m 3 m c 1 m R O α R 67

66 Solution onsider triangle O R ( R ) + ( 3) c R 5 m 3 sin α 5 o α rad S ength of arch R α m 5 kn R 3 m 3 m α R R O c 1 m µ ds ds 68

67 - Solution µ ds ds ( ) µ 5 3 x OE R cosθ x Rsinθ OE O O 4 m R cosθ 4 ds Rdθ 5 kn x E c 1 m R R 3 m 3 m θ R α R O x 5 kn 5 kn 69

68 - Solution 5 kn µ ds ds R 3 m x θ E R 3 m c 1 m R α R O.6437 µ ds 5 ( 3 x )( R cos θ 4 ).6437 Rd θ ( 3 Rsinθ )( R cosθ 4) Rdθ kn m 7

69 - Solution 5 kn µ ds ds R 3 m x θ E R 3 m c 1 m ds.6437 ( R cos θ 4 ) Rd θ R O α R m kn

70 Thank You 7

Theory of Structures

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