LECTURE 21 Mohr s Method for Calculation of General Displacements. 1 The Reciprocal Theorem

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1 V. DEMENKO MECHANICS OF MATERIALS 05 LECTURE Mohr s Method for Cacuaton of Genera Dspacements The Recproca Theorem The recproca theorem s one of the genera theorems of strength of materas. It foows drect from the prncpe of superposton and appes to a sstems for whch ths prncpe s vad. Defnton The work done b the frst generaed force (F ) on the dspacement of ts ponts of appcaton produced b the second generaed force ( F ) s equa to the work done b the second force on the dspacement of ts pont of appcaton produced b the frst force. Obvous Fg. ( F ) ( F ) ( F ) ( F ) = δ δ = δ δ = = () where δ k s the dspacement of the pont n the drecton of the force caused b the force k. Consder two possbe was of F and F forces appcaton. Wa We frst app the force F at the pont A. Ths force does the work (see Fg. 3 eft) W= Fδ. () Further we app the force F at the pont B. Ths force does work whch s expressed smar b Fg. 3 rght:

2 V. DEMENKO MECHANICS OF MATERIALS 05 Fg. W = Fδ. (3) At the same tme the force F does work too snce the appcaton of the force F causes a dspacement of the pont A: W = Fδ. (4) Fg. 3 B summaton we obtan the tota work done b the forces when the are apped n drect order W = W W W Wa We app frst the force F and then F. W = F δ Fδ F δ. (5) Fg. 4

3 V. DEMENKO MECHANICS OF MATERIALS 05 3 Obvous the work s expressed as B equatng the work we fnd W = Fδ Fδ F δ. (6) Fδ = F δ. (7) Sometmes the recproca theorem s nterpreted n more narrow sense. When F= F = F expresson (7) becomes δ = δ (8) and the dspacement of the pont A produced b the force apped at the pont B s equa to the dspacement of the pont B produced b the same force but apped at the pont A. Mohr's Integra. Determnaton of Generaed Dspacements Let us consder an eastc sod subjected to an arbtrar sstem of forces F and F ( F and F are generaed forces). Let us assume that we app the force F. It causes the nterna force factors Q Q M x M M n cross-sectons of the rod. Set up the expresson for the potenta energ of stran N x n = N Q x Q W U dx K dx K dx EA GA GA = = x Let us assume that we app the force F. Then M M M. (9) GI ρ EI EI n N Q Q = x W U dx K dx K dx EA GA GA = =

4 4 V. DEMENKO MECHANICS OF MATERIALS 05 x M M M. GI ρ EI EI Further we app the forces F and F smutaneous: ( N N ) ( Q Q ) ( Q Q ) n x x = EA GA GA W = U = dx K dx K dx Accordng to formua (5) W where ( Mx Mx ) ( M M ) ( M M ) (0) dx. () GIρ EI EI W = W W W W = W W W or ( ) ( ) n x x x n N N dx M M dx N dx M dx = K = K = EA EI EA EI n N x dx M dx n N x N x M M K = = K EA EI = EA EI Let us now assume that F = and denote F = F. Then N x Q δ = δ = δ n Nx N F x M M F =... = EA EI () Q M x M M are the nterna force factors caused b the unt force. Dvdng both parts of Eq. () b unt force we get n the eft part of () δ as the dspacement of pont of F = appcaton under the acton of F = F force. The ntegras obtaned are known as Mohr's ntegras for cacuaton of dspacements. Bendng dspacements are most sgnfcant n the rod under consderaton. Dspacements due to tenson and shear are as sma n reaton to bendng dspacements as the energ due to tenson and shear n reaton to energ due to bendng.

5 V. DEMENKO MECHANICS OF MATERIALS 05 5 Hence of sx Mohr s ntegras () we take the one for bendng: Exampe n = = ( ) ( ) M x M F x dx δ. (3) EI Gven: F = 0kN q = 60kN/m a = m EI = const. It s necessar to determne the vertca dspacement (defecton) of the pont A. Fg. 5. We have to determne the vertca dspacement of the pont A. For ths purpose we must app at ths pont the unt dmensoness force F = n vertca drecton.. We set up the expressons for the bendng moments for each porton takng the same portons for (a) and (b) sstems: I F M ( x) = Fx= 0x I M ( x) = x= x II qx 60x M ( x) = F( x a) = 0( x ) = 0 0x 30x F II M ( x) = ( x) = x. 3. We substtute the bendng moments n expresson (3): δ A = ( 0 )( ) ( )( ) x x dx x x x dx = EI The mnus ndcates that the dspacement of the pont A s not n the drecton of the unt force but s opposte to t. 5. Note that for determnng the anguar dspacement (ange of rotaton sope) we must app the unt dmensoness moment M = at ths pont.

6 6 V. DEMENKO MECHANICS OF MATERIALS 05 3 Vereshchagn's Method for Graphca Souton of Mohr s Integra Suppose for exampe t s necessar to take the ntegra of the product of two functons f( x) f( x ) n the porton of ength : I = f ( x) f ( xdx ). (4) 0 Note that at east one of these functons s near. Let f ( x )= b kx. Expresson (4) then becomes I = b f ( xdx ) k f ( xxdx ). (5) 0 0 The frst of the ntegras above represents the area bordered b the curve f ( x ) (the area under the f ( x ) graph on Fg. 6): 0 f ( xdx ) = ω. (6) Fg. 6

7 V. DEMENKO MECHANICS OF MATERIALS 05 7 The second ntegra represents the statc moment (frst moment) of ths area wth respect to the axs.e. 0 f ( xxdx ) = S = ω x (7) c where x c s the co-ordnate of the centrod of the frst dagram. We now obtan But Consequent ( ) I = ω b kx c. (8) b kx = f x. (9) c ( c) I = ω f xc. (0) ( ) Thus b Vereshchagn's method ntegraton s repaced b mutpcaton of the area under the frst dagram b the ordnate of the second (near) dagram drect beow the centrod of the frst one. Each of Mohr's ntegras () ncudes the product of functons. Vereshchagn's method s appcabe to an of the sx ntegras. In cases both functons f ( x ) and f ( x ) are near the operaton of mutpcaton has the commutatve propert. In such cases the area under the frst dagram shoud be mutped b the ordnate of the second dagram or the area under the second dagram shoud be mutped b the ordnate of the frst dagram.