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### ME 205- Chapter 6 - Pure Bending of Beams.pdf

• 1. Stress Analysis I ME 205 – Fall 2023 Chapter (6) - Pure Bending of Beams Beams Members that are slender and support loadings applied perpendicular to their longitudinal axis are called beams (see the figure). Objectives ➢ Determine the stress in members caused by bending ➢ Derive and understand the bending theory, its limitations and its applications for strength based design and analysis of symmetric bending of beams ➢ Determine the normal stress developed in the symmetric bending of beams. 1
• 2. 2 ✓ It depends on the beam cross-section ✓ It depends on the properties of the cross section (i.e., location of centroid) How to calculate the bending stress ? Does the stress caused by bending is normal or shear stress ? Cross-section What the Bending Moment does to the Beam? ✓ It causes compression on one face and tension on the other ✓ It causes the beam to deflect
• 3. 3 Double symmetric cross-sections (Two axes of symmetry) Single symmetric cross-sections (One axis of symmetry) Asymmetric cross-sections (No axis of symmetry) Various Beam Cross-sections Axis of symmetry Axis of symmetry Axis of symmetry
• 4. 4 At least One axis of symmetry
• 6. Beam sign convention The sign convention for bending moment (M) and shear force (V) in beams are shown: 6
• 7. 7 Theory of Simple Bending of Straight Beams When a bending moment is applied to a straight prismatic beam, the longitudinal lines become curved and vertical transverse lines remain straight and undergo a rotation. Basic Assumptions 1. Beam material is homogenous and isotropic and obeys Hook’s law, 2. Longitudinal axis x (within the neutral surface) does not experience any change in the length, 3. All cross sections remain plane and perpendicular to longitudinal axis during the deformation.
• 8. 8 ➢ Member remains symmetric, i.e., y-axis is the axis of symmetry, ➢ Bending moment (M) causes beam to bends uniformly to form a circular arc, ➢ Length of top decreases and length of bottom increases, ➢ A neutral surface must exist that is parallel to the upper and lower surfaces and for which the length does not undergo any change, ➢ In this case. stresses and strains are negative (compressive) above the neutral plane and positive (tension) below it for the given M direction. Beam with a plane of symmetry in pure bending
• 9. 9 ( ) ( ) max max max max max max (strain varies linearly) or x x L y L L y y y y L y y ρ y y                  = −  = − = − − =− = =− =− = = =− Strain due to Bending of a Straight Beam ➢ A contraction will occur (-ε) in the fibers located above the neutral axis (+y). ➢ The elongation will occur (+ε) in the fibers located below the axis (-y). Consider a beam segment of length L. After deformation, the length of the neutral surface remains L’ (=JK). At other sections, where ρ is the radius of curvature. L ymax ymax
• 10. 10 Stress Due to Bending • For a linearly elastic material, max max max max (stress varies linearly) x x y y E E y y     = =− =− • For static equilibrium, max max max max 0 0 x x y F dA dA ydA y y dA y    = = = − =− → =     First moment with respect to neutral plane is zero. Therefore, the neutral surface must pass through the section centroid. • For static equilibrium, where S= I/ymax is the section modulus (‫المقطع‬ ‫)معامل‬ of the beam. ( ) ( ) 2 max max max max max max max max x I y M y dA y dA y dA y y y M y M I S        = − = − − = =     = =    x NA M y I  =− The maximum stress σmax due to bending is: max max max x NA M y I  = x dF dA  = ymax
• 11. Hollow circular section 11 Moment of Inertia for Regular Shapes 4 4 2 4 64 R d I A R    = = = N. A. C Circular cross section N. A. C ( ) ( ) 4 4 2 2 4 o i o i R R I A R R   − = = − Rectangular cross section b h N. A. h/2 3 12 bh I A bh = = C N. A. 3 3 12 12 o i i i o o i i b h b h I A b h b h = − = − bi bo hi ho h/2 Hollow Rectangular section C
• 12. 12 Section Moduli for Regular Shapes 3 max 3 2 max , 12 2 12 6 2 bh h I y bh I bh S h y = = = = = For example: The section modulus (S) for rectangular cross section of height h and width b can be obtained as:
• 13. 13 Deformations in a Transverse Cross Section • Deformation due to bending moment M is quantified by the curvature of the neutral surface max max max max max max 1 1 1 M y M y E y E y I EI M EI     = = = = = where ρ is the radius of curvature (‫االنحناء‬ ‫قطر‬ ‫)نصف‬.
• 14. 14 • Stress due to eccentric loading found by superposing the uniform stress due to a centric load and linear stress distribution due a pure bending moment ( ) ( ) centric bending x x x P My A I    = + =  Eccentric Axial Loading in a Plane of Symmetry • Eccentric loading F P and M P d = =  • Validity requires stresses below proportional limit, deformations have negligible effect on geometry, and stresses not evaluated near points of load application.
• 15. 15 Example (1): A member having the dimensions shown is used to resist an internal bending moment of M=90kN.m. Find the maximum stress in the member if the moment is applied about the z-axis as shown. Sketch the stress distribution over the cross-section. Solution The maximum normal bending stress is given by: The moment of inertia about the z-axis can be obtained by: max max max ( / 2) 2 M y M h where y h I I  = = = ( ) 3 3 6 4 0.2 0.15 56.25 10 12 12 bh I m −  = = =  3 max 6 90 10 0.15 120 2 56.25 10 MPa  −   = =  
• 16. 16 Example (2): An open-link chain is obtained by bending low-carbon steel rods into the shape shown. For 160lb load, find: (a) the maximum tensile and compressive stresses, and (b) distance between section centroid and neutral axis. Solution: • Equivalent centric load and bending moment 160 160 0.6=104 P lb M P d lb×in = =  =  ( )2 2 2 0.25 0.1963in 160 815 0.1963 axial A R P psi A    = = = = = = • Normal stress due to a centric load ( ) ( ) 4 4 3 4 1 1 4 4 max max 3 0.25 3.068 10 in 104 0.25 8475 0.068 10 I R M y psi I    − − = = =   = = =  • Normal stress due to bending moment
• 17. 17 (1) Maximum tensile and compressive stresses ( ) ( ) max max max max 815 8475 815 8475 axial t axial c       = + = + = − = − ( ) max 9260 t psi  = ( ) max 7660 c psi  =− (2) Neutral axis location 3 0 815 3.068 10 0.0240 105 o o M y P A I P I y in A M − = −   =  = = 0.0240 o y in = Using superposition method: ----------------------------------------------------------------------------------------------------------------------------------------------------------
• 18. 18 Section properties: The moment of inertia about the neutral axis (which passed from the centroid of the cross section) is: The maximum normal bending stress is given by: max max M y I  = ( )( ) ( )( ) ( )( ) ( )( ) ( )( ) ( )( ) 3 2 3 2 3 2 3 1 1 4 0.5 4 0.5 3.40 0.25 2 0.5 3 0.5 3 3.40 2 12 12 1 0.5 10 0.5 10 5.5 3.4 91.73 . 12 I in   = + − + + −     + + − = ( )( ) ( )( ) ( )( ) ( ) ( )( ) ( ) 1 1 0.25 4 0.5 2 2 3 0.5 5.5 10 0.5 3.40 . 4 0.5 2 3 0.5 10 0.5 i i i i i y A y in A = = + +     = = = + +   ( ) ( ) 3 t max 4 10 10.5 3.40 12 3715.12 3.715 91.73 psi ksi    −  = = = ( ) 3 max 4 10 3.40 12 1779.1 1.779 91.73 c psi ksi     = = = Example (3): Determine the maximum tensile and compressive bending stress in the beam if it is subjected to a moment of M=4 kip.ft. Solution
• 19. 19 Example (4): A cast-iron machine part is acted upon by a 3kN.m couple. Knowing E=165GPa and neglecting the effects of fillets, find (a) the maximum tensile and compressive stresses, and (b) the radius of curvature. Solution Calculate the location of section centroid (ӯ): 3 114 10 38 3000 i i i y A Y mm A  = = =   ( ) ( ) ( ) 3 2 2 3 3 2 2 9 4 12 90 20 30 40 1800 12 1200 18 868 10 12 12 x bh I I Ad Ad m  −   = + = +               = +  + +  =                       
• 20. The maximum tensile and compressive stresses: The radius of curvature (ρ): ---------------------------------------------------------------------------------------------------------------------------------------------- 20 ( ) ( ) ( ) max max 3 max 9 3 max 9 3 10 0.022 76.0 868 10 3 10 0.038 131.3 868 10 A tensile B compressive M y I M y MPa I M y MPa I    − − =   = = =    − = = =−  9 9 3 1 165 10 868 10 47.7 3 10 M EI m EI M   −    = → = = = 
• 21. 21 Example (5): The largest allowable stresses for the cast iron link are 30MPa in tension and 120MPa in compression. Determine the largest force P which can be applied to the link. Solution From Sample Example 4: 3 2 9 4 3 10 m 0.038m 868 10 m A Y I − − =  = =  • Determine equivalent centric and bending loads. 0.038 0.010 0.028m = 0.028 d P centric load M bending moment P d P = − = = =  =
• 22. 22 • Superpose stresses due to centric and bending loads ( )( ) ( )( ) P P P I Mc A P P P P I Mc A P A B A A 1559 10 868 022 . 0 028 . 0 10 3 377 10 868 022 . 0 028 . 0 10 3 9 3 9 3 − =  −  − = − − = + =  +  − = + − = − − − −   • Evaluate critical loads for allowable stresses. kN 0 . 77 MPa 120 1559 kN 6 . 79 MPa 30 377 = − = − = = = + = P P P P B A   max 77.0 P kN = • The largest allowable load: -------------------------------------------------------------------------------------------------------------------------------------------------------------
• 23. 23 Solve the problems in Sheet (6) Try to solve Problem (1): An eccentric force P is applied as shown to a steel bar of 25x90 mm cross-section. The strains at A and B have been measured and found to be: εA =+350μ and εB=-70μ. Knowing that E=200GPa, determine (a) the distance d, (b) the magnitude of the force P. (Answer : (a) 5.00mm, (b) 243kN) Problem (2): A vertical force P of magnitude 20kips is applied at point C located on the axis of symmetry of the cross-section of a short column. Knowing that y=5 in, determine (a) the stress at point A, (b) the stress at point B, (c) the location of the neutral axis. (Answer: (a) 0.576ksi, (b) -1.498ksi , (c) 2.411in)
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