HW 11 - SECTION 9.9 601 Castigliano's Theorem Casligliano's...

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601 SECTION 9.9 Castigliano's Theorem Casligliano's Theorem The beams described in the problems for Section 9.9 have constant M o " .flexural rigidity El. fA B Problem 9.9-1 A simple beam AB of length L is loaded at the left-hand J£p;:=[ ==========~A end by a couple of moment M o (see figure). Determine the angle of rotation (J A at support A. (Obtain the solution by determining the strain energy of the beam and then using Castigliano' s I~, --L ------>·1 theorem.) Solution 9.9-1 Simple beam with couple M o STRAIN ENERGY M o M2dX M6 fL( X)2 M6 L : X At================;18 A U = " r -2 El = 2 El 0 1 - L dx = 6 El dU MoL CASTIGLIANO'S THEOREM (J = --= -- (clockwise) _ ~_x L -+1,1 A dM o 3El (This result agree with Case 7, Table 0-2) M o R A = - (downward) L jP Problem 9.9-2 The simple beam shown in the figure supports a concentrated load P acting at distance a from the left-hand support and A D+ B distance b from the right-hand support. Ji': A Determine the deflection i5 D at point D where the load is applied. (Obtain the solution by determining the strain energy of the beam and then using Castigliano' s theorem.) ~_~-Lh~ Solution 9.9-2 Simple beam with load P U = fM 2 dX STRAIN ENERGY 2El A a b B K D~ ~_x L --'-x'---~ b 2 2 3 _ 1 r (paX)2 _ p a b U -- - dx--- Pb Pa DB ,., El L 6 EIL 2 R B - "0 R=- =- A L L CASTIGLIANO'S THEOREM Pax M DB =RBx =-- L
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SECTION 9.9 Castigliano's Theorem 603 Problem 9.9-5 A simple beam ACB supports a uniform load of intensity q on the left-hand half of the span (see figure). Determine the angle of rotation (}B at support B. (Obtain the solution by using the modified form of Castigliano's theorem.) Solution 9.9-5 Simple beam with partial uniform load q M o = fictitious load corresponding to angle of rotation () B MODIFIED CASTIGLlANO'S THEOREM (EQ. 9-88) 3qL M o qL M o SET FICTITIOUS LOAD M o EQUAL TO ZERO R A = -8- + L R B = 8 - L U2 ? () = ~ (3 q Lx _ qr)(~)dX B £1 8 2 L J 0 BENDING MOMENT AND PARTIAL DERIVATIVE FOR SEGMENTAC 2 2 M = R x _ qx = (3 qL + Mo)x _ qx AC A 2 8 L 2 qL 3 qL 3 7qL 3 = -- + -- = -- (counterclockwise) +- 128£1 96£1 384£1 (This result agrees with Case 2, Table 0-2.) BENDING MOMENT AND PARTIAL DERIVATIVE FOR SEGMENT CB qL Mo) MCB = Rs-t + Mo = ( 8 - L x + M o (osxs~) aMCB x --= --+ 1 aM o L
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638 CHAPTER 10 Statically Indeterminate Beams Problem 10.3-7 The load on a fixed-end beam AB of length L is distributed in the fonn of a sine curve (see figure). The intensity 1TX IY . . .--~= qo sin L of the distributed load is given by the equation
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This note was uploaded on 11/14/2011 for the course ECIV 310 taught by Professor Huckelbridge during the Spring '11 term at Case Western.

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HW 11 - SECTION 9.9 601 Castigliano's Theorem Casligliano's...

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