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NEW 375 Lecture18

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

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Shearing Stresses in Beams THEOREM OF RECIPROCITY

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CONSIDER THE FOLLOWING SECTION:
THE NORMAL STRESS DISTRIBUTION FOR THIS SECTION IS:

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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 σ 1 dA. THE RESULTANT OF THESE DIFFERENTIAL FORCES IS F 1 WHICH IS THE SUMMATION OF dF 1 OR: σ = = dA dF F 1 1 1 INTEGRATED OVER THE AREA NORMAL TO F 1

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FROM: σ = dA F 1 1 WHERE σ 1 = FIBER STRESS AT A DISTANCE Y FROM THE NEUTRAL AXIS σ 1 CAN BE WRITTEN AS: I y M 1 = σ = = 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 =

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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 0 F F V F - = = + - - = +

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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 = τ = = = τ SHEAR FORCE
= τ = τ t I Q dx dM t I Q x M lim 0 x t I Q x M = τ TAKING THE LIMIT AS 0 x CONSIDER:

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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 = τ = τ
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 dy y t Q 1 1 + - = - = = =

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C=0.5h C=0.5h ) ( 2 1 ) 2 1 2 1 ( 2 1 2 1 ) ( 2 1 ) ( ' 1 1 1 1 1 1 1 ` c y y c y y y c y y y c y y y c t A + = + = - + = - + = - = TAKE THE FIRST MOMENT OF THE AREA HIGHLIGHTED BELOW ABOUT THE NEUTRAL AXIS ( 29 ( 29 + - = c y 2 1 y c t y '
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NEW 375 Lecture18 - Shearing Stresses in Beams THEOREM OF...

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