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- 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. 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. 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. 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. 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. Methods to find slope and deflection Double integration method Moment area method Macaulay’s method
- 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. 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. 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. 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. 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. Moment Area Method
- 13. Moment Area Method
- 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

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