Chapter 10_Force Method

# Chapter 10_Force Method - CHAPTER 10 FORCE METHOD In this...

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1 CHAPTER 10 FORCE METHOD In this Chapter we will introduce the Castigliano’s 2 nd theorem and apply it to various structural systems of slender bodies. The Castigliano’s 2 nd theorem is convenient in handling statically indeterminate, flat and curved structures as well as truss structures. The method described in this chapter is called the Force Method. Here the term “Force” is used in a general sense to include both force and moment. The Castigliano’s second theorem Beam bending Curved arches and frames Slender body under torsional moment Slender body under axial force Truss structures

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2 Castigliano’s 2 nd theorem Consider a structure under concentrated forces and moments as shown in the sketch. N P P P " , , 2 1 : point forces or moments applied at discrete points ( N " , 2 , 1) N q q q " , , 2 1 : displacements or rotations in the direction of the applied loads The strain energy of the structure can be expressed as a function of N P P P " , , 2 1 such that ) , , ( 2 1 N P P P U U " = Then one can show that i i q P U = () N i " , 2 , 1 = The above statement is called the Castigliano’s 2 nd theorem. We will accept the theorem without a proof and apply it to various structural systems. The Castigliano’s 2 nd theorem is convenient in dealing with statically indeterminate, flat and curved structures. Note : The above theorem is a manifestation of the following principles in structural mechanics. 1) The Principle of Minimum Complementary Energy 2) The Principle of Complementary Virtual Work 1 P 1 , q 1 N P N , q N 2 P 2 , q 2
3 Beam Bending Consider bending in the x - z plane (CB/P axes, and 0, 0 z FM = = ) = = = L x x y dx x w EI U 0 2 2 2 2 1 ( 1 ) y y EI M x w C = = 2 2 2 ( 2 ) (2) (1) 2 0 1 2 xL y y y x M UE I d x EI = = ⎛⎞ = ⎜⎟ ⎝⎠ 2 0 1 2 y y x M Ud x EI = = = ( 3 ) x x =0 z , w x = L

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4 Example 1: EI y ~ constant Find the tip vertical displacement q under the tip load P . () 0 y Mx P x += y M xP x = − For constant , y EI 22 3 2 3 2 2 00 0 1 2 3 6 xL yy y y xx x PP x P L UP x d x x d x EI EI EI EI = == = =− = = = ∫∫ U q P = 3 3 y PL q EI = The above equation can be expressed as qc P = where 3 3 y L c EI = : compliance constant or Pk q = 3 3 y EI k L = : spring constant x =0 P , q x = L x P V z ( x ) M y ( x ) , Pq k
5 Example 2: Find the tip displacement q under 0 P P z = . Consider a system with force P applied at x =0 in the direction of q . P is called a “fictitious load”. 0 () ( ) () 0 x yz Mx P x x P d ξ ξξ = = ++ = = = = x z y d P x Px x M 0 ) ( ) ( ) ( with 0 ) ( P P z = 2 0 0 2 0 0 0 2 1 2 1 ) ( ) ( x P Px x P Px d x P Px x M x x y = = = = = = = 2 0 2 1 ) ( x P Px x M y = For constant y EI , P V z ( x ) M y ( x ) ξ d ξ x P, q q z, w 0 z p p =

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6 = = = = + + = = L x x y L x x y dx x P x PP x P EI dx x P Px EI U 0 4 0 3 0 2 2 0 2 2 0 4 1 2 1 2 1 2 1 U q P = () 23 0 0 44 33 00 0 1 2 2 12 4 4 xL y x yy x U Px P x dx PE I x L Px P PL P EI EI = = = = =+ ⎡⎤ ⎛⎞ ⎜⎟ ⎢⎥ ⎣⎦ ⎝⎠ + = 4 3 2 2 1 4 0 3 L P PL EI q y ( * ) By setting P =0, y EI L P q 8 4 0 =
7 Example 3:

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Chapter 10_Force Method - CHAPTER 10 FORCE METHOD In this...

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