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ME 338 Lecture 7

# ME 338 Lecture 7 - ME 338 Lecture#7 Beam Deflections...

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ME 338 Lecture #7 – Beam Deflections

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Outline Beam Deflections By straight integration By superposition By Singularity functions
Beam Curvature and Moment Curvature of beam subjected to bending moment M Curvature of plane curve Slope of beam at any point x along the length If the slope is very small, the denominator approaches unity. Combining equations for beams with small slopes,

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Beam Deflection Equation Derivation Recall that Therefore and
Beam Deflections

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Determining Beam Deflections Determining the deflection of a beam is an exercise in integration ° = ±²? + ? 1 ³ = °²? + ? 1 ? + ? 2 ´ = µ ¶· + ? 1 ? ¸ + ? 2 ? + ? 3 ¹ = ´²? + ? 1 ? º + ? 2 ? ¸ + ? 3 ? + ? 4 Constants C 1 and C 2 can be found from boundary conditions on shear and moment functions Constants C 3 and C 4 can be found from boundary conditions on slope and deflection functions
Simple Example Pinned-Pinned beam with uniform load ° = 2 − ?? ² = °³? = 2 ? − ? 2 ? ´

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Simple Example Pinned-Pinned beam with uniform load, Determine slope and deflection equations as well as the maximum slopes and deflections ° = 2 ? − ? 2 ? ² °³? = ´µ ³? ³? = 4 ? ² ? 6 ? 3 + ¶ 1 · °³? = ´µ? = 12 ? 3 ? 24 ? ¸ + ¶ 1 ? + ¶ 2
Simple Example Pinned-Pinned beam with uniform load, Determine slope and deflection equations as well as the maximum slopes and deflections 0 = 12 (0) 3 ? 24 0 ² + ? 1 0 + ? 2 → ? 2 = 0 Boundary conditions: y = 0 at x=0, x = l 0 = 12 (±) 3 ? 24 ± ² + ? 1 ± → ? 1 = − 3 24 ³ = ?? 24´° 2±? µ − ? 3 − ± 3 Therefore, and ¶ = ·³ ·? = ? 24´° 6±? µ − 4? 3 − ± 3

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Simple Example Pinned-Pinned beam with uniform load, Determine slope and deflection equations as well as the maximum slopes and deflections Maximum Slope occurs at ends of beam ?
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