10a-Bending_Stresses_in_Beams

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1 Mechanics of Wood Products Topic 10a: Bending Stresses in Beams
Introduction 2 The system of internal forces (axial, shear, and moment) which may exist at a section of a beam will cause internal stresses on that section: Axial forces N(x) will cause normal stresses Moments M(x) will also cause normal stresses Shear forces V(x) will cause shear stresses A N = σ ? = σ = ? τ
Introduction 3 Take the case where a beam is under pure bending, i.e., N(x)=V(x)=0, a segment of the beam will be in equilibrium under the action of the internal moment alone. Here, we want to study how to determine the internal stress distribution arising from the application of a bending moment M at any section of a beam. M M
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Introduction 4 Basic Assumptions: The moment at a section is assumed to be delivered statically without shock or impact. The beam is assumed to be stable under the applied moment; i.e., no lateral buckling. The beam is assumed straight and prismatic (constant cross sectional area) with at least one axis of symmetry. X-sectional view
Curvature 5 As a member under bending will deform into a curve shape, let us first introduce the concept of curvature. Consider a curve in the x-y plane as shown. And let us examine an infinitesimal element ds of the curve. x y a b o ds φ+ φ d d d φ φ φ ρ ρ c
Curvature 6 at point 'a', the slope or tangent of the curve is φ . at point 'b', the slope of the curve is φ+ d φ . From geometry, aob=d φ where the point 'o' is the intersection point of the normals to curve at points 'a' and 'b' since points 'a' and 'b' are close together; i.e., ds is small, we have oa=ob= ρ . => ρ d φ = ds
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Curvature 7 curvature of radius = where curvature = ds d 1 ρ φ ρ =
Derivation of Bending Formula Let us now consider the deformation of a straight prismatic beam with an axis of symmetry about a vertical axis, under the action of a constant bending moment as shown: 8 x y o c e b d f a
Kinematics Now consider the three neighboring sections shown on the beam: a-b, c-d, e-f. The only way these 3 sections can still fit together after the beam is bent is if the sections remain plane! This is a very important assumption in beam bending theory. 9 x y o c e b d f a
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Kinematics The fibers along the length of the beam from section a-b to section c-d will: * elongate near the bottom of the beam * shorten near the top of the beam * remain unchanged near the middle of the beam The horizontal plane which contains the fibers without any change in length is defined as the neutral plane. We will set up out coordinate system such that the x-axis passes through the neutral plane and define this as the neutral axis. The neutral axis is bend into a circular curve whose radius of curvature is ρ . 10 x y o c e b d f a
Variation of ε x in a Cross Section Consider also the deformation of an infinitesimal element abcd of length dx of the beam as shown: The normal strain ε x in any fiber located at a distance y from the neutral axis is: 11 y x d a c b d dx ρ φ ρ φ ρ φ ρ φ ρ ε y - d d - d y) - ( length initial length initial - length final x = = =
Variation of ε x in a Cross Section From , we can made another statement that: In pure bending of symmetric beams, the normal strains varies linearly along the transverse direction (y-axis). 12 y x d a c b d dx ρ φ ε ε max (compression) max (tension) x y 𝜀𝜀 𝑥𝑥 = −𝑦𝑦 𝜌𝜌
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Variation of σ x in a Cross Section The normal stress σ x in any fiber located at a distance y from the neutral axis is given by Hooke's law as: substituting into Hooke's law yields: 13 𝜀𝜀 𝑥𝑥 = −𝑦𝑦 𝜌𝜌 𝜎𝜎 𝑥𝑥 = −𝐸𝐸𝐸𝐸 𝜌𝜌 𝜎𝜎 𝑥𝑥 = 𝐸𝐸𝜖𝜖 𝑥𝑥
Variation of σ x in a Cross Section From , we have the third important statement about beam bending theory: In pure bending of symmetric beams, the normal stresses σ x (y) varies linearly along the transverse direction (y-axis). 14 𝜎𝜎 𝑥𝑥 = −𝐸𝐸𝐸𝐸 𝜌𝜌 σ σ max (compression) max (tension) x y M
Equilibrium in the x direction Now consider the equilibrium in the x-direction: Substituting into the equilibrium equation yields or 15 σ σ max (compression) max (tension) x y M Σ𝐹𝐹 𝑥𝑥 = 0 Σ𝐹𝐹 𝑥𝑥 = � 𝜎𝜎 𝑥𝑥 𝑑𝑑𝑑𝑑 = 0 𝜎𝜎 𝑥𝑥 = −𝐸𝐸𝐸𝐸 𝜌𝜌 −𝐸𝐸 𝜌𝜌 � 𝐸𝐸 𝑑𝑑𝑑𝑑 = 0 � 𝐸𝐸 𝑑𝑑𝑑𝑑 = 0
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16 � 𝐸𝐸 𝑑𝑑𝑑𝑑 = 0 Σ𝐹𝐹 𝑥𝑥 = � 𝜎𝜎 𝑥𝑥 𝑑𝑑𝑑𝑑 = 0 𝜎𝜎 𝑥𝑥 = −𝐸𝐸𝐸𝐸 𝜌𝜌 Σ𝐹𝐹 𝑥𝑥 = 0 => Neutral axis located at Centre of gravity of cross section
Rotational Equilibrium about the y- axis Where z is the distance from the y axis. Substituting into the equation above yields or 17 Σ𝑀𝑀 𝑦𝑦 = 0; � 𝜎𝜎 𝑥𝑥 𝑧𝑧𝑑𝑑𝑑𝑑 = 0 𝜎𝜎 𝑥𝑥 = −𝐸𝐸𝐸𝐸 𝜌𝜌 −𝐸𝐸 𝜌𝜌 � 𝐸𝐸 𝑧𝑧 𝑑𝑑𝑑𝑑 = 0 � 𝐸𝐸 𝑧𝑧 𝑑𝑑𝑑𝑑 = 0 σ σ max (compression) max (tension) x y M
Rotational Equilibrium about the y- axis is satisfied if and only if the y and z axes are the principle axes of the section; i.e., the product of inertia = 0. This result confirm our initially stated assumption on axes of symmetry. 18 � 𝐸𝐸 𝑧𝑧 𝑑𝑑𝑑𝑑 = 0 okay no good
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Rotational Equilibrium about the z- axis Substituting into the equation above yields or Rearranging yields Where Iz= moment of inertia about the z axis 19 Σ𝑀𝑀 𝑧𝑧 = 0; 𝑀𝑀 + � 𝜎𝜎 𝑥𝑥 𝐸𝐸 𝑑𝑑𝑑𝑑 = 0 𝜎𝜎 𝑥𝑥 = −𝐸𝐸𝐸𝐸 𝜌𝜌 𝑀𝑀 − 𝐸𝐸 𝜌𝜌 � 𝐸𝐸 2 𝑑𝑑𝑑𝑑 = 0 𝜌𝜌 = 𝐸𝐸 𝑀𝑀 � 𝐸𝐸 2 𝑑𝑑𝑑𝑑 = 𝐸𝐸 𝑀𝑀 𝐼𝐼 𝑧𝑧 σ σ max (compression) max (tension) x y M 𝐼𝐼 𝑧𝑧 = � 𝐸𝐸 2 𝑑𝑑𝑑𝑑 𝐸𝐸 𝜌𝜌 = 𝑀𝑀 𝐼𝐼 𝑧𝑧
Bending Stress Formula Combining from before and yields the bending stress formula which is an expression for the normal stress distribution at a section as: 20 σ σ max (compression) max (tension) x y M 𝜎𝜎 𝑥𝑥 = −𝐸𝐸𝐸𝐸 𝜌𝜌 𝐸𝐸 𝜌𝜌 = 𝑀𝑀 𝐼𝐼 𝑧𝑧 𝜎𝜎 𝑥𝑥 = −𝑀𝑀 𝐸𝐸 𝐼𝐼 𝑧𝑧
Bending Stress Formula The maximum compressive and tensile stresses at a section are given by: where C 1 and C 2 are the distance from the neutral axis to the outer fiber in tension and compression, respectively. 21 𝜎𝜎 max 𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡 = 𝑀𝑀 𝐶𝐶 1 𝐼𝐼 𝑧𝑧 𝜎𝜎 max 𝑐𝑐𝑡𝑡𝑐𝑐𝑐𝑐𝑐𝑐𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡 = 𝑀𝑀 𝐶𝐶 2 𝐼𝐼 𝑧𝑧 y σ σ max (compression) max (tension) x M 𝐶𝐶 1 𝐶𝐶 2
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Bending Stress Formula For an isotropic and homogeneous material such as steel, the material properties can be assumed the same in tension and compression; therefore, we can simply get the absolute maximum normal stress σ max (tension or compression) as: where C is the distance from the neutral axis to the outer most fiber and S is the elastic section modulus 22 𝜎𝜎 𝑐𝑐𝑚𝑚𝑥𝑥 = 𝑀𝑀 𝐶𝐶 𝐼𝐼 𝑧𝑧 = 𝑀𝑀 𝑆𝑆
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Bending Stress Formula Note the both S and I z are geometrical properties of the section. S and I z have units of mm 3 and mm 4 , respectively. They are usually tabulated for common cross sections (See Appendix of text book) For a rectangular cross section with width b and depth d: 𝐼𝐼 𝑧𝑧 = 𝑏𝑏𝑑𝑑 3 12 23 𝑆𝑆 = 𝑏𝑏𝑑𝑑 2 6
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