Section Modulus & Stress Calculation of Rail Section
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Fl e x u r e Fo Fo r m u l a Defi fine nes s be b end ndin ing g stress
Flexure Formula (derived for rectangular beam) 1 Beam with marked area 2 Bent beam deforms area into rhomboid 3 Stress block due to bending Flexure formula derivation: f y /y = f/c f y= y f/c (similar triangles) The internal resiting moment is the sum of forces F times lever arm y about NA, the Neutral Axis M=Fy F = a f y F = a y f/c M = y a y f/c M = a y2 f/c f = Mc / a y2 f = Mc / I I = a y2 I = Moment of Inertia Calculus defines the area a as differential area da and the sum sign as integration sign I = y 2 da I = Moment of Inertia Flexure Formula f = Mc / I f = bending stress at any distance c from NA. For maximum stress the flexure formula simplifies to f=M/S S = I / c = Section Modulus
Section Modulus (for rectangular beam) 1 Stress block of partial beam C = T = b (d/2) (f/2) Stress block centroids are d/3 from NA Lever arm between C and T is 2/3 d Internal resisting moment: M = C 2/3 d = T 2/3 d Substituting b (d/2) (f/2) for C and T, yields M = 2/3 d b (d/2) (f/2) M = 2/3 f bd2/4 M = f bd2/6 Solving for f (maximum stress) f = M / bd2/6 f = M/ S where S = bd 2/6 (Section Modulus) Comparing a 2”x12” joist upright and flat: 2 S=2 (12)2/6 S = 48 in3 3 S=12 (2)2/6 S = 8 in3 The upright joist is 6 times stronger !
Moment of Inertia 1 Stress block 2 Moment of Inertia as parabolic volume 3 T-beam with asymmetric stress block 4 L-bar stress blocks about X, Y, and Z-axis The Moment of Inertia formula I = y 2da reveals, resistance of areas da increases quadratic with the distance from NA (parabolic distribution). The Moment of Inertia parabolic volume is 1/3 the volume of a cube of equal dimensions: I = 1/3 bd (d/2)2 I = 1/3 bd3 / 4 I = bd 3/12 I = Moment of Inertia for rectangular beams only From previous derivation, the flexure formula f=Mc/I defines stress at any distance c from NA (needed for asymmetrical shapes, such as T or L-shapes).
Moment of Inertia Effect of shapes 1 Upright joist: 2”x12”, I = 2 (12)3/12 I = 288 in4 2 Flat joist: 12”x2”,
I = 12 (2)3/12
I = 8 in4
3 Wide flange beam: effective (flanges far from NA) 4 Cross beam: ineffective (cross bar at NA) Note: • Beams at right deform more than beams at left • Material at NA is least effective (short lever arm) • Moment of Inertia defines strength and stiffness
Area Method review • Shear at any point is: V = load area left of the point • Bending at any point is: M = shear area left of the point • Maximum bending occurs where shear goes through zero • Negative bending causes convex deflection • Positive bending causes concave deflection • Inflection point (0 bending) coincides with change of deflection curvature
Design Defines beam size for actual loads and allowable stress of selected material Analysis Checks if a given beam satisfies allowable stress of the actual material Assume: Wood Allowable bending stress Allowable shear stress (parallel to fiber) Steel Yield strength Allowable bending stress (0.6 Fy) Allowable shear stress (0.4 Fy) Note: F = allowable stress f = actual stress