NEW 375 Lecture18 - Shearing Stresses in Beams THEOREM OF...

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Unformatted text preview: Shearing Stresses in Beams THEOREM OF RECIPROCITY CONSIDER THE FOLLOWING SECTION: THE NORMAL STRESS DISTRIBUTION FOR THIS SECTION IS: WE WILL CONSIDER THE SHEARING STRESS ON A PORTION OF THE SECTION (AREA A), LOCATED A DISTANCE OF Y 1 FROM THE NEUTRAL AXIS. THE FORCE dF 1 IS THE NORMAL FORCE ACTING ON A DIFFERENTIAL AREA dA AND IS EQUAL TO 1 dA. THE RESULTANT OF THESE DIFFERENTIAL FORCES IS F 1 WHICH IS THE SUMMATION OF dF 1 OR: = = dA dF F 1 1 1 INTEGRATED OVER THE AREA NORMAL TO F 1 FROM: = dA F 1 1 WHERE 1 = FIBER STRESS AT A DISTANCE Y FROM THE NEUTRAL AXIS 1 CAN BE WRITTEN AS: I y M 1 = = = dA y I M dA I y M F 1 = dA y I M F 1 EVALUATE: NOTE: dy t dA = THEN: Q I M ) dy t ( y I M F c y 1 1 = = WHERE: dy y t Q c y 1 = SIMILARLY, THE RESULTANT FORCE ON THE RIGHT SIDE OF THE ELEMENT IS: Q I ) M M ( F 2 + = THE SUMMATION OF FORCES IN THE HORIZONTAL DIRECTION ON THE PREVIOUS FIGURE YIELDS: 1 2 H 2 1 H x F F V F F V F- = = +-- = + Q I M V Q I M Q I ) M M ( F- F V H 1 2 H =- + = = SHEAR STRESS t I Q x M Q I M t ) x ( 1 t ) x ( V A V H H H = = = = SHEAR FORCE = = t I Q dx dM t I Q x M lim x t I Q x M = TAKING THE LIMIT AS x CONSIDER: WE ALREADY DEMONSTRATED THAT: WHERE V IS THE SHEAR IN THE BEAM AND THE STRESS IS TO BE EVALUATED V dx dM = t I VQ t I Q dx dM = = EVALUATING Q ( 29 ( 29 1 1 2 1 2 c y 2 c y y c y c t 2 1 Q y 2 1 c 2 1 t y 2 1 t d y y t Q 1 1 +- = - = = = C=0.5h C=0.5h ) ( 2 1 ) 2 1 2 1 ( 2 1 2 1 ) ( 2 1 ) ( ' 1 1 1 1 1 1...
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This note was uploaded on 03/27/2008 for the course ENGIN 375 taught by Professor Miller during the Spring '08 term at University of Cincinnati.

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NEW 375 Lecture18 - Shearing Stresses in Beams THEOREM OF...

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