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**CM 197 Mechanics of Materials Chap 14: Stresses in Beams**

Professor Joe Greene CSU, CHICO Reference: Statics and Strength of Materials, 2nd ed., Fa-Hwa Cheng, Glencoe/McGraw Hill, Westerville, OH (1997) CM 197 Copyright Joseph Greene 2003 All Rights Reserved

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**Chap 14: Stresses in Beams**

Topics Introduction Normal Stresses in Beams due to Bending Flexure Formula Allowable Moment Shear Stress Formula for Beams Composite Beams Copyright Joseph Greene 2003 All Rights Reserved

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**Copyright Joseph Greene 2003 All Rights Reserved**

Introduction Introduction Stresses that are caused by bending moments are studied. Chap 13 considered loads on beams that causes shear forces and bending moments. Normal stresses along the longitudinal direction are caused by bending moments. Normal stresses due to bending are called flexural stresses. Shear stresses are caused by shear forces. Copyright Joseph Greene 2003 All Rights Reserved

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**Normal Stresses in Beams due to Bending**

Tensile forces Compressive forces Flexural stresses are in bending of a beam Fig 14-1 and 14-2 Beam with supports has a load in the middle that causes it to bend. The top of the beam has compressive forces that cause shrinking. Compressive stresses which are maximum at top of beam and minimum at center. The bottom of the beam has tensile forces that causes stretching to occur. Tensile stresses which are maximum at bottom of beam and minimum at center. The center of the beam from top to bottom has an imaginary line at the centroid that is the transition between the compressive forces in the top and tensile forces in the bottom. Neutral axis or Neutral surface where the forces and stresses are zero. M C T Neutral Axis Copyright Joseph Greene 2003 All Rights Reserved

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**Copyright Joseph Greene 2003 All Rights Reserved**

Flexure Formula Derivation of Flexure Formula Fig 14-3 and 14-4 Beam subjected to bending moment +M. Along slice of beam from top to bottom the stress varies from maximum, max at the surfaces (Top and bottom) to the minimum at the center, min (Centroid). Top half of the beam has compressive forces. Bottom half of the beam has tensile forces. Stress from to to bottom = (y/c)max where y is the location between the top and bottom surfaces, and c is the distance from the center to the edge of beam Maximum stress is the Moment times the distance c divided by moment of inertia, I max = Mc/I Then, substituting for max the stress at any point in the beam is: = My/I Where is the flexural stress at any point in the section, and y is the distance from the neutral axis to the point in question, and I is the moment of inertia for the section (Chap 8) Copyright Joseph Greene 2003 All Rights Reserved

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**Copyright Joseph Greene 2003 All Rights Reserved**

Section Modulus Section Modulus Easier to work problems if the Moment of inertia, I, and distance c are combined into section modulus, S. S= I/c Then maximum stress can be found from the ratio of Moment and Section Modulus, max = M/S Equation is widely used in engineering practice when beams are designed to handle loads. The maximum shear stress should not exceed the allowable stress of the material, which may include factor of safety. (Chap 15) Example, rectangular section of width b and height h is S = I/c = (bh3/64)/(h/2) = bh2/ Example, circular section of diameter d is S = I/c = (d4/64)/(d/2) = d3/ Values of section modulus are available in Appendix, Tables A-1 to A-5 Maximum stresses- Max tensile and compressive stresses For symmetrical parts the max stresses in beam are at the top and bottom surfaces. For non-symmetrical parts, max tensile and compressive stresses are not equal and must be calculated from equations. Fig 14-5 Example, 14-1 and and 14-3 Copyright Joseph Greene 2003 All Rights Reserved

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**Copyright Joseph Greene 2003 All Rights Reserved**

Allowable Moment Allowable Moment Maximum stress is the Moment times the distance c divided by moment of inertia, I. max = Mc/I This can be rearranged for allowable moment as Mallow = I/c max Section modulus can be defined as Mallow = S max Allowable moment is calculated mainly for the purpose of computing the allowable load that can be applied safely to the beam without causing over-stress on the beam over-stress on the beam = stress is greater than the yield strength of the material Example, 14-4 and 14-5 Copyright Joseph Greene 2003 All Rights Reserved

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**Shear Stress Formula for Beams**

Derivation of Shear Stress Formula Internal shear forces exist in beam sections Shear forces cause shear stresses in cross-sections. Shear stress exist in longitudinal sections of the beam, Fig 14-6. Simple beam subjected to concentrated load with resulting shear force and bending moment diagrams. Based upon the free body diagram in Fig 14-7, the resultant shear stress in both the longitudinal and vertical sections is: Where, is the shear stress at a point in a given section V is the shear force at a given section Q is the first moment of the Area A’ about neutral axis, Q=A’y’ A’ is the part of the area in cross section above (or below) the horizontal line where the shear stress is to be calculated. y’ is the distance from the neutral axis to the centroid I is moment of inertia of entire section with respect neutral axis t is the width of the cross section at the horizontal line where strear stress is calculated 14-10 Copyright Joseph Greene 2003 All Rights Reserved

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**Shear Stress Formula for Beams**

14-10 Equation is shear formula for beams. Used to calculate the shear stresses either on vertical section or on longitudinal planes. Shear force , V, and moment of inertia, I, are constant for a given section. Shear stresses in a section vary in accordance with the variation of Q/t. Q (first moment of shear area) is similar to moment of inertia I (second moment of entire area) in its use. Its units are in2, ft2, m2, cm2, or mm2 Maximum shear stress in rectangular section Width is constant. Max value of Q occurs at neutral axis = bh2/8 I for rectangular section is bh3/12 Maximum shear stress in circular section Width is greatest at neutral axis Max shear stress occurs at neutral axis Area A’ is a semicircle (properties of semicircle Table 7-2) Example 14-7 14-11 14-12 Copyright Joseph Greene 2003 All Rights Reserved

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**Shear Stress Formula for Beams**

14-10 Features of Example 14-7: plate girder To calculate shear stresses in a structure and at particular point or sections Use equation for shear stress and or for max shear stress Calculate the moment of inertia for the section (Chap 8) or look it up in Table A in the Appendix. Moment of inertia Create the shear diagrams for force and moment along beam. Determine location of max shear stress. To Calculate the max shear stress form Equation Calculate the First moment Q from shear area, A’, and distance, y. Calculate the max shear stress form Equation 14-10 Pay attention to the direction of shear stresses Note: the shear stresses are in the top half and bottom half of the web. Note: the maximum shear stress is at the neutral axis! (Opposite of normal stress where the max stress is at the surface and 0 at center) Copyright Joseph Greene 2003 All Rights Reserved

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**Shear Stress Formula for Beams**

14-10 Features of Example 14-7: plate girder To calculate the shear stress at a junction (say web and flange) Use equation but notice The thickness, t, is 5in for flange OR 0.5 in for web shear stress. The area, A’, is the areas of the flange First calculate Q from Q=A’y’ Shear stress in flange (t = 5 in) is found from Eqn Shear stress is web (t = ½ in) is found from Eqn Shear stress distribution Shear stress in flange is very small since the t is big (5 in). Shear stress is bigger in the flange where t is small (½ in). Shear stress is maximum at the neutral axis from Eqn Shear stress is symmetric above and below the neutral axis because the section is symmetric. Copyright Joseph Greene 2003 All Rights Reserved

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**Shear Stress Formula for Beams**

Features of Example 14-7: plate girder Average Web shear. Most of the shear force is resisted by the web in a wide-flange section. Design codes from AISC specs allow use of an average web shear to calculate the maximum shear stress in a fabricated of hot-rolled side flange or I-beam section. Average shear stress is approximately equal to: Where, avg is the average shear stress in the web Vmax is the max shear force along beam D is the full depth of the beam Tw is the web thickness of the beam section For plate girder example of 14-7, average web stress is equal to: This is 16% less than The max shear stress of 9.91ksi Used in practice because in most beams Maximum shear stress is well within the allowable shear stress 14-13 Copyright Joseph Greene 2003 All Rights Reserved

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**Shear Stress Formula for Beams**

Features of Example 14-7 Average Web shear. Most of the shear force is resisted by the web in a wide-flange section. Design codes from AISC specs allow use of an average web shear to calculate the maximum shear stress in a fabricated of hot-rolled side flange or I-beam section. Average shear stress is approximately equal to: Where, avg is the average shear stress in the web Vmax is the max shear force along beam D is the full depth of the beam Tw is the web thickness of the beam section For plate girder example of 14-7, average web stress is equal to: This is 16% less than The max shear stress of 9.91ksi Used in practice because in most beams Maximum shear stress is well within the allowable shear stress 14-13 Copyright Joseph Greene 2003 All Rights Reserved

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**Copyright Joseph Greene 2003 All Rights Reserved**

Composite Beams Beams made from different materials are called composite beams. Example, Timber is strengthened with metal plates Concrete is reinforced with steel rebar. Assumption: steel plate is fastened properly to the wood so that there is no sliding during bending. Basic bending deformation assumes that plane sections at right angles to the beam axis remain planes. Strains of longitudinal fibers vary linearly from the neutral axis. For elastic deformations, stress is proportional to strain, Fig When section is subjected to a positive moment, the flexural stress is distributed to all members. Since E steel > E wood, the flexural stress in steel is greater than that in the wood. Stress = modulus times strain. Copyright Joseph Greene 2003 All Rights Reserved

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**Copyright Joseph Greene 2003 All Rights Reserved**

Composite Beams Stresses in composite beams Then, the ratio of the two is Note: the strains are equal If the ratio of the moduli is n, then Transformation technique for composite beams If have a beam of more than one material then you can modify the thickness of one of the materials (usually less stiff) and transform the beam from two materials into one material with a change in width. The flexural stress can be determined by constructing a section of one material and different width. The section properties of the section will be the same. Example, to calculate the flexural stress of sections with wood and steel, the wood section is transformed into an equivalent steel section. The dimensions of the original steel section remains unchanged. The vertical dimensions remained unchanged. The width of the transformed steel section becomes much thinner (a factor of b/n where b is the original width. Figure 14-11 Copyright Joseph Greene 2003 All Rights Reserved

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**Copyright Joseph Greene 2003 All Rights Reserved**

Composite Beams Transformation technique for composite beams Maximum normal stresses in steel and in wood are Where, Isteel is the moment of inertia of the transformed section in steel with respect to the neutral axis. Where, c2 is the distance from steel section to the neutral axis and Where c1 is the distance from wood section to the neutral axis Alternatively, the steel section could be transformed to an equivalent wood section with the following Fig 14-11 Where, Iwood is the moment of inertia of the transformed section in steel with respect to the neutral axis. Example 14-8 14-14 14-15 14-16 14-17 Copyright Joseph Greene 2003 All Rights Reserved

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**Copyright Joseph Greene 2003 All Rights Reserved**

Composite Beams Example Solution for Composite Beam Step 1: Calculate ratio of moduli, n. Step 2: Convert equivalent geometry of one material to transformed section. Step 3: Calculate neutral axis location y. Then c1 and c2 Step 4: Calculate moment of inertia,I, of transformed section about neutral axis. Step 5: Calculate maximum bending stress from equations and Copyright Joseph Greene 2003 All Rights Reserved

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