1 / 34

Beams: Pure Bending (4.1-4.5)

Beams: Pure Bending (4.1-4.5). MAE 314 – Solid Mechanics Yun Jing. Beams in Pure Bending. Prismatic beams subject to equal and opposite couples acting in the same plane are in pure bending. Pure vs. Non-Uniform Bending. Pure bending: Shear force (V) = 0 over the section

brielle-brock
Download Presentation

Beams: Pure Bending (4.1-4.5)

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  1. Beams: Pure Bending(4.1-4.5) MAE 314 – Solid Mechanics Yun Jing Beams: Pure Bending

  2. Beams in Pure Bending • Prismatic beams subject to equal and opposite couples acting in the same plane are in pure bending. Beams: Pure Bending

  3. Pure vs. Non-Uniform Bending • Pure bending: Shear force (V) = 0 over the section • Non-uniform bending: V ≠ 0 over the section Non-uniform bending • Moment → normal stresses • Shear force → shear stresses (Ch. 6) Pure bending• Moment →normal stresses Beams: Pure Bending

  4. Pure Bending: Assumptions • Beam is symmetric about the x – y plane • All loads act in the x – y plane Beams: Pure Bending

  5. Pure Bending: Curvature • Sections originally perpendicular to the axis of the member remain plane and perpendicular: “Plane sections remain plane.” • Sign convention • Positive bending moment:beam bends towards +y direction • Negative bending moment: beam bends towards -y direction Right angle Beams: Pure Bending

  6. Pure Bending: Deformation • Since angles do not change(remain plane), there is noshear stress. • The top part of the beam contractsin the axial direction. • The bottom part of the beam expandsin the axial direction. • There exists a line in the beam thatremains the same length called theneutral line. • Set y = 0 at the neutral line. • ρ = radius of curvature • εx < 0 for y > 0 and εx > 0 for y < 0 Beams: Pure Bending

  7. Pure Bending: Axial Strain • The length of DE is LDE = ρθ • The length of JK is LJK = (ρ-y)θ • Axial strain at a distance y fromthe neutral axis (εx): • Maximum compressive strain occurs on the upper surface. • Maximum tensile strain occurs on the lower surface. LJK LDE Beams: Pure Bending

  8. Pure Bending: Axial Strain • c = maximum distance between the neutral axis and the upper or lower surface • When c is the distance to the surface in compression • When c is the distance to the surface in tension Beams: Pure Bending

  9. Pure Bending: Transverse Strain • Recall there are no transverse stresses sincethe beam is free to move in the y and z directions. • However, transverse strains (in the y and zdirections) exist due to the Poisson’s ratioof the material. • ρ' = radius of anticlastic curvature = ρ/υ Beams: Pure Bending

  10. Pure Bending: Normal Stress • Let us now assume that the beam is made of a linear-elastic material. • The normal stress varies linearly with the distance from the neutral surface. Beams: Pure Bending

  11. Pure Bending: Normal Stress • Recollecting that the applied loading is a pure moment, we calculate resultant loads on the cross-section. • The resultant axial force must be equal to zero. …which is the definition of the centroid, so the neutral axis is just the centroid of the section. Beams: Pure Bending

  12. Pure Bending: Normal Stress • The resultant moment about the z-axis must be equal to the applied moment M. Definition of the second moment of inertia, I Beams: Pure Bending

  13. Example Problem A steel bar of 0.82.5-in. rectangular cross section is subjected to two equal and opposite couples acting in the vertical plane of symmetry of the bar. Determine the value of the bending moment M that causes the bar to yield. Assume m=36 ksi. Centroids and Moments of Inertia

  14. Centroids and Moments of Inertia(A.1-A.5) MAE 314 – Solid Mechanics Yun Jing Centroids and Moments of Inertia

  15. Review From Last Time • Beams in pure bending (no shear forces) • Plane sections remain plane • +M, upper surface in compression • +M, lower surface in tension • Transverse strains exist due to υ Radius of Curvature Neutral Line Centroids and Moments of Inertia

  16. Centroids • To match with our axis convention use centroid in the y-direction. • Centroids for common shapes are located in the back cover of your textbook. • If a cross section shape has an axis of symmetry, then the centroid passes through that axis of symmetry. Centroids and Moments of Inertia

  17. Centroids for Composite Shapes • Choose a reference x-axis (normally the base of the shape). • Divide the composite shape into individual shapes. • Calculate the area for each shape. • Find the location of the centroid for each shape relative to the reference x-axis. y h1b1 0.5h1 0.5h12b1 A1 h2b2 h1+0.5h2 h2b2(h1+0.5h2) A2 A3 h3b3 h1+h2+0.5h3 h3b3(h1+h2+0.5h3) x y=0 Centroids and Moments of Inertia

  18. Moments of Inertia • Moment of inertia with respect to the x-axis • Moments of inertia for common shapes are located in the back cover of your textbook. x x x’ Ix x Ix Ix’ Ix Centroids and Moments of Inertia

  19. Moments of Inertia for Composite Shapes • Divide the composite shape into individual shapes. • Calculate I of each shape about its own centroid. • Calculate I of each shape about the centroid of the entire composite shape using the parallel axis theorem: b1h13/12 b1h1d12 b1h1(d12+h12/12) A1 x’’’ b2h23/12 b2h2d22 b2h2(d22+h22/12) A2 A3 b3h33/12 b3h3d32 b3h3(d32+h32/12) x x’’ x’ Centroids and Moments of Inertia

  20. Example Problem Find the location of the centroid and the moment of inertia of the section below (the moment is applied about the horizontal axis). 5 in 0.5 in 3 in 4 in 1.5 in 3 in Centroids and Moments of Inertia

  21. Stress in the Elastic Range • Revisiting the beam subject to pure bending • For positive bending moment • Maximum tensile stress = Mc2 / I • Maximum compressive stress = Mc1 / I • For negative bending moment • Maximum tensile stress = Mc1 / I • Maximum compressive stress = Mc2 / I • The ratio I/c1 (or I/c2) is known as S, elastic section modulus. • Larger values for S indicate a beam that is more resistant to bending. c1 c2 Centroids and Moments of Inertia

  22. Deformations in the Elastic Range • Which shapes are most resistant to bending? • S-beam • rectangular • Why? • Large portion of the cross section is located far from the neutral axis, resulting in large values of S. • The deformation of a member due to bending can be calculated by Centroids and Moments of Inertia

  23. Example Problem A beam with cross-section like the previous example is subjected to a pure bending moment of M = -4500 in·lb. Find the maximum tensile and compressive normal stresses. 5 in 0.5 in 3 in 4 in 1.5 in 3 in Centroids and Moments of Inertia

  24. Example Problem Knowing that the couple shown acts in a vertical plane, determine the stress at (a) point A, (b) point B. Centroids and Moments of Inertia

  25. Example Problem The beam shown is made of nylon with an allowable stress of 24 MPa in tension and 30 Mpa in compression. Determine the largest couple M that can be applied to the beam. Centroids and Moments of Inertia

  26. Example Problem Two vertical forces are applied to a beam of the cross section shown. Determine the maximum tensile and compressive stresses in portion BC of the beam. Centroids and Moments of Inertia

  27. Bending in Composite Materials(4.6) MAE 314 – Solid Mechanics Yun Jing Centroids and Moments of Inertia

  28. Normal Stress in Composite Materials • Strain is still the same as homogenous material because it is not dependent on material properties. • The stress distribution is no longer linear. • Use different material properties in each section. Centroids and Moments of Inertia

  29. Normal Stress in Composite Materials • Top section: • Bottom section: Original Shape New Shape Centroids and Moments of Inertia

  30. Bending in Composite Materials • The new “homogenous” beam produces the same response to bending as the composite beam. • We can now calculate the centroid and moment of inertia for this beam. • To calculate the axial stresses, use σx for material 1 and nσx for material 2. • Curvature of the composite member is Centroids and Moments of Inertia

  31. Reinforced concrete beam • Beams are reinforced by steel rods • Concrete is very weak in tension, the steel rods will carry the entire tensile load • Concrete in the beam acts effectively only in compression Centroids and Moments of Inertia

  32. Example Problem A bar having the cross section shown has been formed by securely bonding brass and aluminum stock. Determine the largest permissible bending moment when the composite bar is bent about a horizontal axis. Ea = 70 GPa, σall,a = 100 MPa, Eb = 105 GPa, σall,b = 160 MPa. Centroids and Moments of Inertia

  33. Example Problem Knowing that the bending moment in the reinforced concrete beam is +100kip.ft and that the modulus of elasticity is 3.625106 psi for the concrete and 29 106 psi for the steel, determine (a) the stress in the steel, (b) the maximum stress in the concrete. Beams: Pure Bending 33

  34. Example Problem A flitched beam consists of two 50mm×200mm wooden beams and a 12 mm×80mm steel plate. The plate is placed centrally between the wooden beams and recessed into each so that, when rigidly joined, the three units form a 100mm×200mm section as shown in the figure. Determine the bending moment the beam can withstand when the maximum bending stress in the timber is 12MN/m2. What will then be the maximum bending stress in the steel. (E for the timber is 10GPa, for steel is 120GPa) Beams: Pure Bending 34

More Related