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Lecture 12 deflection in beams

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Lecture 12 deflection in beams

1. 1. Unit 2- Stresses in BeamsTopics Covered  Lecture -1 – Review of shear force and bending moment diagram  Lecture -2 – Bending stresses in beams  Lecture -3 – Shear stresses in beams  Lecture -4- Deflection in beams  Lecture -5 – Torsion in solid and hollow shafts.
2. 2. Beam DeflectionRecall: THE ENGINEERING BEAM THEORY σ M E = = y I RMoment-Curvature Equation v (Deflection) y € NA A B x A’ B’If deformation is small (i.e. slope is “flat”):
3. 3. 1 dθ ∴ ≈ R dx R δy and δθ ≈ (slope is “flat”) δx B’ δy 1 d2y € A’ ⇒ ≈ 2 R dx € €Alternatively: from Newton’s Curvature Equation y € ⎛ d 2 y ⎞ ⎛ dy ⎞ 2 R ⎜ 2 ⎟ if ⎜ ⎟ <<<< 1 1 ⎝ dx ⎠ ⎝ dx ⎠ = 3 y = f (x) R ⎛ 2 ⎞ 2 ⎛ dy ⎞ ⎜1+ ⎜ ⎟ ⎟ 1 d2y x ⎝ ⎝ dx ⎠ ⎠ ⇒ ≈ 2 € R dx € €
4. 4. From the Engineering Beam Theory: M E 1 M d2y = = = 2 I R R EI dx d2y ⇒ ( EI ) 2 = M dx €Flexural € € Bending Stiffness Moment Curvature €
5. 5. RelationshipA C B Deflection = y dy Slope = dx d2yA C B Bending moment = EI 2 y dx d3y Shearing force = EI 3 dx d4 y Rate of loading = EI 4 dx €
6. 6. Methods to find slope and deflection  Double integration method  Moment area method  Macaulay’s method
7. 7. Double integration method d 2 y ⎛ 1 ⎞ ⎟ M Curvature Since, 2 = ⎜ dx ⎝ EI ⎠ dy ⎛ 1 ⎞ ⇒ = ⎜ ⎟ ∫ M ⋅ dx + C1 Slope dx ⎝ EI ⎠ € ⎛ 1 ⎞ ⇒ y = ⎜ ⎟ ∫ ∫ M ⋅ dx⋅ dx + ∫ C ⋅ dx + C 1 2 Deflection ⎝ EI ⎠€ Where C1 and C2 are found using the boundary conditions.€ Curvature Slope Deflection y R dy dx
8. 8. Double integration methodSimple supported W Slope Deflection L/2 L/2 dyA C B Slope = Deflection = y c dx yc 2 WL3 WL =− = θA = θB = − 48EI 16EI LUniform distributed load x € Slope € Deflection w/Unit lengthA C dy B Slope = Deflection = y c dx yc 2 5 WL3 WL =− = θA = θB = − 384 EI 24 EI L € €
9. 9. Macaulay’s method  The procedure of finding slope and deflection for simply supported beam with an eccentric load is very laborious.  Macaulay’s method helps to simplify the calculations to find the deflection of beams subjected to point loads.
10. 10. Moment-Area Theorems •  Consider a beam subjected to arbitrary loading, dθ d 2 y M = = dx dx 2 EI θD xD M ∫ dθ = ∫ EI dx θC xC xD M θ D − θC = ∫ EI dx xC dx CD = Rdθ = dx € € R dθ •  First Moment-Area Theorem: € area under BM diagram between € C and D.9 - 11
11. 11. Moment-Area Theorems •  Tangents to the elastic curve at P and P’ intercept a segment of length dt on the vertical through C. M dt = xdθ = x dx EI xD x M 1 D 1 − tC D = ∫ x EI dx = ∫ xMdx = EI A x EI xC xC − A= total area of BM diagram between C & D € x = Distance of CG of BM diagram from C •  Second Moment-Area Theorem: The tangential deviation of C with respect to € D is equal to the first moment with respect to a vertical axis through C of the area under the BM diagram between C and D.9 - 12
12. 12. Moment Area Method
13. 13. Moment Area Method
14. 14. An Exercise- Moment of Inertia – Comparison1 Load Maximum distance of 4 inch to the centroid I2 2 x 8 beam Load 2 I1 Maximum distance of 1 inch to 2 x 8 beam the centroid I2 > I1 , orientation 2 deflects less Ken Youssefi Engineering 10, SJSU 15