Method of east Work Theor of Structures II M Shahid Mehmood epartment of ivil Engineering Swedish ollege of Engineering & Technolog, Wah antt
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.
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
Staticall Indeterminate Trusses P 1 1 +.5 S 1 S S P 1 +. 5S U S P E P S 4
Example 1 etermine the force in members of the truss shown. E is same for all members. 9 T 6 ft 6 ft 5
Solution The truss is indeterminate to the first degree. 9 T 9 T S + S 9 T 9 T 9 T 6
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
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
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/ 7 9 + S 1 7 1 11 1 + S +1 7 7 648 S 7 S 7 S 7 648 S 196 + 7 S 648 + 36S + 36S + 36S 648 + 36S 196 +11.8S 7 S 11.8S P P 14.4 + 347. 6S S 9
Solution 1 P P E S P P S 14.4 + 347.6S 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
Example nalze the truss shown. E is same for all members. 3m 4m 1 kn 11
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
Solution V 3m V 4m 1 kn M F x 1 4 + 3 + 4 + 3 133.33 4 133. 33 kn 133.33 kn 3 F V + V 1 V 1 V 13
Solution 133.33 kn 1 V 3m 133.33 kn V 4m 1 kn M F x 1 4 + 3 + 4 + 3 133.33 4 133. 33 kn 133.33 kn 3 F V + V 1 V 1 V 14
Solution Joint V F 133.33 kn 1 V F 4 5 + F 3 5 4 4 V 5 3 133.33 kn V 1 3 1 kn F x 4 133.33 F3 + F4 5 5 F3 133.33 + V 3 F 133.33 + 1.33V 3 4 5 F 4 3 4 + 133.33 kn F 3 V 15
Solution Joint 1 133.33 kn 1 V F 3 4 5 1 V F5 5 3 133.33 kn F 5 1 V 3 5 1 kn 5 5 V F5 1 V 3 3 F F F F 1 1 1 1 F x 4 133.33 + F5 133.33 kn 5 F 1 3 4 4 5 5 4 133.33 F5 133.33 1 V 5 5 3 3 1 V 133.33 133.33+ 1.333V F 5 + 1.333V 16
Solution Joint 133.33 kn 1 V 1 4 5 F 3 F + F 4 5 133.33 kn V 3 1 kn F F 3 F + 4 V F +V 5 5 3 3 5 F 1 1.333V 4 3 F 4 V.5/3 4 F 17
Solution The strain energ of a truss is U F E U F V V F E See Table on next slide 18
Universit of Engineering & Technolog, Taxila Solution Member F F/ V (m) F. F/ V Final Forces 1 +1.33V +1.333 4 7.175V +6.3 kn +V +1 3 3V +46.679 kn 3 133.33 +1.33V +1.333 4 79.315+7.75V 71.46 kn 4 VV x5/3 1.667 5 +13.894V 77.798 kn 5 5/3(1 V ) 1.667 5 1389.44+13.894V +88.868 kn F V F ( 7.175 + 3 + 7.75 + 13.894 + 13.894) V ( 79.315 + 1389.44) V 46.679 kn 19
Solution 133.33 kn 53.31 kn 3m 133.33 kn 46.679 kn 4m 1 kn
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
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.
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
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
Two inged Parabolic rch x V V For an rch the bending moment is M V. x. if V.x µ M µ. 5
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
Two inged Parabolic rch x V Equation (1) becomes, U (. ) M µ ds EI If then (. ) U M µ EI ds V 7
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
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
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
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
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
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
Example 3 Find the horizontal thrust in the arch shown. 8 T 8 T 6f ft ft 1 ft 1 ft ft 36
Solution 8 T 8 T 6 ft 1 ft 1 ft 1 ft ft V V onsidering oad 1 3 1 6 k ( ) 4 3 8 5 k k k w c + 37 + 4 3 3 1 3 1 3 1 6 6 8 8 5
Solution 8 T 8 T 1 7.16 T 6 ft 7.16 T ft 1 ft V 1 ft ft V 1 1 5 + 3 7 81 13.58 T s the load is smmetric, so total reaction will be 13.58 7.16 T 38
Solution 8 T 8 T 1 7.16 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 6 8 4 8 V 8T F V + V 8 8+ V V 8T 8 8 8 39
Example 3 Find the horizontal thrust in the arch shown. 8 T 8 T 6f ft ft 1 ft 1 ft ft 4
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 6 6 15 ( x). x. ( 6 x) ( x) 41
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
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 15 3188. 7 T-ft µ dx 8x. 15 3 6 x dx 15 3 ( 6 x ) dx 115 ft 43
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 3188.7 µ 7. 16 T dx 115 44
Solution +14.9 +14.9 1.67 1.67 ending Moment iagram 45
Example 4 nalze the arch shown. 4T/ft 15 ft ft 1 ft ft 46
-- 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) 4 5 5 1 x 5 5 ( x) ft 1 ft ft μ bending moment while considering simpl supported beam equation of parabola V 47
- 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 1 3 48x 4(x 1) / 3x ft 1 ft ft 48
Solution dx 1 x 5 ( x) 4 T/ft µ 15 ft dx 5 Portion Origin imits μ 1 48x 1 3 48x 4(x 1) / 3x V 1 ft ft ft V 1 ( ) 1. x( 5 x) dx 5 1 3 µ dx ( 48x) x( 5 x) dx + 48x ( x 1) 5 1 ( 3x) x( 5 x) dx 1 + 5 163 + 197 + 7168 787 T-ft 3 49
Solution dx 1 x 5 ( x) 4 T/ft µ 15 ft dx 5 Portion Origin imits μ 1 48x 1 3 48x 4(x 1) / 3x V 1 ft ft ft V 1 5 5 dx x ( 5 x ) 6 ft 3 dx 5
Solution put these values into the following equation 4 T/ft µ dx 46.45 T dx 46.45 T 787 46. 45T 6 787 48 T 3 T 51
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
Example 4 nalze the arch shown b the method of summation. 4T/ft 15 ft ft 1 ft ft 53
Solution µ δ x δ x ivide the arch into equal segments. δ s 1 5 1 5 ft 1 x 1 4 T/ft 5 ft x 1.5 ft 4c 4 15 1 x 5 6.5 5.5 5 ( x ) x ( x ) ( ).85 ft x 1 V 48 T 5 ft V 3 T µ 1 V x 1 48.5 1 T ft 54
Solution Portion x (ft) (ft) μ (T ft) μ 1.5.85 1 8.1 344 4 T/ft 1 x 1 5 ft x 1 V 48 T V 3 T 5 ft 55
Solution Portion x (ft) (ft) μ (T ft) μ 1.5.85 1 8.1 344 7.5 7.65 36 58.55 75 4 T/ft x 5 ft x V 48 T V 3 T 5 ft 56
Solution Portion x (ft) (ft) μ (T ft) μ 1.5.85 1 8.1 344 4 T/ft 7.5 7.65 36 58.55 75 3 1.5 11.5 587 16.7 66 3 x 3 5 ft x 3 V 48 T V 3 T 5 ft 57
Solution Portion x (ft) (ft) μ (T ft) μ 1.5.85 1 8.1 344 4 T/ft 7.5 7.65 36 58.55 75 3 1.5 11.5 587 16.7 66 4 17.5 13.65 77 186.3 993 x 4 4 5 ft V 48 T V 3 T 5 ft 58
Solution Portion x (ft) (ft) μ (T ft) μ 1.5.85 1 8.1 344 4 T/ft 7.5 7.65 36 58.55 75 3 1.5 11.5 587 16.7 66 4 17.5 13.65 77 186.3 993 5.5 14.85 767.6 114 x 5 5 5 ft V 48 T V 3 T 5 ft 59
Solution Portion x (ft) (ft) μ (T ft) μ 1.5.85 1 8.1 344 4 T/ft 7.5 7.65 36 58.55 75 3 1.5 11.5 587 16.7 66 4 17.5 13.65 77 186.3 993 5.5 14.85 767.6 114 x 6 6 6 7.5 14.85 77.6 15 5 ft V 48 T V 3 T 5 ft 6
Solution Portion x (ft) (ft) μ (T ft) μ 1.5.85 1 8.1 344 4 T/ft 7.5 7.65 36 58.55 75 3 1.5 11.5 587 16.7 66 4 17.5 13.65 77 186.3 993 5.5 14.85 767.6 114 7 x 7 6 7.5 14.85 77.6 15 7 17.5 13.65 56 186.3 765 V 48 T 5 ft V 3 T 5 ft 61
Solution Portion x (ft) (ft) μ (T ft) μ 1.5.85 1 8.1 344 4 T/ft 7.5 7.65 36 58.55 75 3 1.5 11.5 587 16.7 66 4 17.5 13.65 77 186.3 993 5.5 14.85 767.6 114 8 x 8 6 7.5 14.85 77.6 15 7 17.5 13.65 56 186.3 765 8 1.5 11.5 4 16.7 45 V 48 T 5 ft V 3 T 5 ft 6
Solution \ Portion x (ft) (ft) μ (T ft) μ 1.5.85 1 8.1 344 4 T/ft 7.5 7.65 36 58.55 75 3 1.5 11.5 587 16.7 66 4 17.5 13.65 77 186.3 993 5.5 14.85 767.6 114 9 x 9 6 7.5 14.85 77.6 15 7 17.5 13.65 56 186.3 765 8 1.5 11.5 4 16.7 45 9 7.5 7.65 4 58.5 1835 V 48 T 5 ft V 3 T 5 ft 63
Solution \ Portion x (ft) (ft) μ (T ft) μ 1.5.85 1 8.1 344 4 T/ft 7.5 7.65 36 58.55 75 3 1.5 11.5 587 16.7 66 4 17.5 13.65 77 186.3 993 5.5 14.85 767.6 114 1 x 1 6 7.5 14.85 77.6 15 7 17.5 13.65 56 186.3 765 8 1.5 11.5 4 16.7 45 9 7.5 7.65 4 58.5 1835 1 5.5 85.85 8 81 8.1 8 V 48 T 5 ft V 3 T 1.4 55737 5 ft µ δ x 55,737 δ 1.4 x 46.3T 64
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
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
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
Solution onsider triangle O R ( R ) + ( 3) c R 5 m 3 sin α 5 o α 36.87.6435 rad S ength of arch R α 64. 35 m 5 kn R 3 m 3 m α R R O c 1 m µ ds ds 68
- 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
- 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 5 3 3 ( 3 Rsinθ )( R cosθ 4) Rdθ 194.5 1 kn m 7
- 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 3397.5 m 3 3 194.5 1 57. kn 3397.5 71
Thank You 7