NEW 375 Lecture15(2) - DIFFERENT MEMBER LOADINGS...

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DIFFERENT MEMBER LOADINGS SHEAR (translates) AXIAL (push/pull) BENDING (curves) TORSIONAL (twists)
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BENDING BENDING (curves)
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Consider a beam under a uniform load. By statics, it is clear that R H = 0. By symmetry, R 1 = R 2 . + Σ F y = 0 = -wL + R 1 + R 2 R 1 = R 2 =wL/2 If we cut the beam at “x”, there must be an internal shear and moment for equilibrium.
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What are the internal shear and the internal moment?? + Σ F y = 0 = -wx - V + R 1 V = -wx + R 1 = w(L/2-x) + Σ M V = 0 = -(wx 2 )/2 - M + R 1 x M = w(x/2)(L – x)
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SIGN CONVENTION FOR INTERNAL FORCES V V + M M N N T T
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BASIC ASSUMPTIONS OVER A SMALL LENGTH, THE BEAM BENDS INTO A CIRCULAR ARC.
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CONSIDER A STRAIGHT BEAM WHICH IS THEN BENT BY A POSITIVE MOMENT AT EACH END
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SIDES OF THE ELEMENT REMAIN STRAIGHT – PLANE SECTIONS REMAIN PLANE.
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SIDES REMAIN STRAIGHT. I δ IS LINEAR WITH RESPECT TO y. ε = δ / L THEREFORE ε IS LINEAR WITH RESPECT TO y
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SECTIONS REMAIN PLANE. CORNERS REMAIN SQUARE –
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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 Lecture15(2) - DIFFERENT MEMBER LOADINGS...

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