NEW 375 Lecture15(2)

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BENDING BENDING (curves)
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)
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.
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.
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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