Strength of materials - Bending

torsion. Here we introduce the third fundamental loading: bending. When deriving the relationship between the bending moment and the stresses causes, we find it again necessary to make certain simplifying assumptions. We use the same steps in the analysis of bending that we used for torsion in chapter 6. 1/10/2013 5 7.1. Introduction Classification of Beam Supports 1/10/2013 6 7.1. Introduction  Limitation 1/10/2013 7 7.1. Introduction  Segment BC: Mx≠0, Qy=0 => Pure Bending  Segments AB,CD: Mx≠0, Qy≠0 => Nonuniform Bending 1/10/2013 8 7.1. Introduction Pure Bending: Prismatic members subjected to equal and opposite couples acting in the same longitudinal plane 1/10/2013 9 7.2. Bending stress  Simplifying assumptions 1/10/2013 10 7.2. Bending stress The positive bending moment causes the material within the bottom portion of the beam to stretch and the material within the top portion to compress. Consequently, between these two regions there must be a surface, called the neutral surface, in which longitudinal fibers of the material will not undergo a change in length. Neutral axis 1/10/2013 11 7.2. Bending stress Neutral fiber c d a b c d  d dz 1 2 1 2 1 2 1 2 y y a b Due to bending moment Mx caused by the applied loading, the cross- section rotate relatively to each other by the amount of d.  ' ' z y d ddz c d cd y dz cd d                The Normal strain of the longitudinal fiber cd that lies distance y below the neutral surface.  – radius of curvature of the neutral fiber. z y    Compatibility Consider a segment of the beam bounded by two cross-sections that are separated by the infinitesimal distance dz. 1/10/2013 12 7.2. Bending stress  Equilibrium z y E  Following Hooke’s law, we have. 1 ????  y z x dA  x y z K MxBecause of the loads applied in the plane yOz, thus: Nz=My=0 and Mx≠0. 0z z A A E N dA ydA     0x A ydA S  0y z A A E M x dA xydA     0xy A xydA I  x – neutral axis (the neutral axis passes through the centroid C of the cross-section). y - axis – the axis of symmetry of the cross-section 1/10/2013 13 7.2. Bending stress Mx>0: stretch top portion Mx<0: compress top portion => For convenient: y z x dA  x y z K Mx 2 x z x A A E E M y dA y dA I      1 x x M EI   EIx – stiffness of beam Mx – internal bending moment  – radius of neutral longitudinal fiber x z x M y I  y – coordinate of point x z x M y I   Belong to tensile zone Belong to compressive zone  Flexure formula – section modulus 7.2. Bending stress • Stress distribution - Stresses vary linearly with the distance y from neutral axis • Maximum stresses at a cross-section max max x t x M y I   min max x c x M y I   ytmax – the distance from N.A to a point farthest away from N.A in the tensile portion ycmax – the distance from N.A to a point farthest away from N.A in the compressive portion 1/10/2013 15 7.2. Bending stress x y min max h/2 h/2 z Mx max 2 x x x x M Mh I W     max min  / 2 x x I W h  max max 2 t c hy y  min 2 x x x x M Mh I W     with called the section modulus of the beam 1/10/2013 16 7.2. Bending stress 1/10/2013 17 7.2. Bending stress Properties of American Standard Shapes 1/10/2013 18 7.2. Bending stress 1/10/2013 19 7.2. Bending stress 1/10/2013 20 7.2. Bending stress 1/10/2013 21 7.3. Shearing stress in Bending In general, a beam will support both shear and moment – non-uniform bending. The shear force Qy is the results of a transverse shear-stress distribution that acts over the beam’s cross-section. As a results of the shearing stress, the shear strain will be developed and (these will) tend to distort the cross-section. This non-uniform shear-strain will cause the cross-section warp 1/10/2013 22 7.3. Shearing stress in Bending  The Shear formula When a beam is subjected to loads that produce both bending moment and shear force, then both normal and shear stresses are developed in the beam. Consider a beam of rectangular cross section subjected to a positive shear force. We assume • The Shear stresses acting on the cross- section are paralell to the shear force, that is, parallel to the vertical sides of the cross-section • The Shear stresses are uniformly distributed across the width of the beam, although they may vary along the height 1/10/2013 23 7.3. Shearing stress in Bending Let’s consider the horizontal force equilibrium of a portion (of the element) taken from the beam in fig. (a). A FBD of this element is shown in fig. (b). This distribution is caused by the bending moment M an M+dM. We have excluded the effect of V and V+dV and the transverse loading on the FBD, because they are not involved in a horizontal force summation. 1/10/2013 24 7.3. Shearing stress in Bending Now consider the shaded top portion of the element that has been sectioned at y’ from the neutral axis. This segment has the width of bs, and the cross-sectional area of As. The Longitudinal shear  stress acts over the bottom face of the segment (constant across the width bs of the bottom face). 1/10/2013 25 7.3. Shearing stress in Bending  ' 0 s s s s s z A A F dA dA b dz          0 s s s s s x xA A M dM M ydA ydA b dz I I                    s s s x A dM ydA b dz I          Have: y x dM Q I  s s s x A ydA S then:   s y x zy s x Q S I b 1/10/2013 26   s y x zy s x Q S I b 7.3. Shearing stress in Bending 1/10/2013 27 7.3. Shearing stress in Bending x y h b= y b c max As 2 21 2 2 2 2 4 s x h h b h S y y b y                  2 2 2 2 3 3 12 6 . . 2 4 4 y y zy Q Qb h h y y bh b bh                 0 2 zy h y     ax 3 0 2 y m Q y bh    bs=b; Ix=bh 3/12; Sx s=yC.A s => • Shearing stress distributions at the rectangular cross-section: 1/10/2013 28 7.3. Shearing stress in Bending • Shearing stress distributions at the wide-flanged cross-section: 21 . 2 s x xS S d y  21 . 2 y zy x x Q S d y I d         max0 => y x x Q S y I d   2 1 1 => 2 2 2 y x x Qh h y s S d s I d                       y x h s b y d max  1  1 1/10/2013 29 7.3. Shearing stress in Bending 1/10/2013 30 7.4. Strength condition x y N K C B Mx z max min max h /2 h /2 maxmax  minmin max max    B B B B Consider a rectangular cross-section of the beam. Normal and shearing stresses distribution on the cross-section are shown in figure. K, N – uniaxial stress C- pure shearing stress B- special plane stress 1/10/2013 31 7.4. Strenght condition 1/10/2013 32 7.4. Strenght condition • Bending stress criterion Cross-section need to control: cross-section have maximum bending moment (ductile: absolute magnitude, brittle : maximum positive an negative bending momet) - Ductile materials: - Brittle materials :    max minmax , allow        max min= ; tensile compressive allow allowk n         1/10/2013 33 • Shear stress criterion Cross-section need to control : cross-section have maximum shear force - Ductile materials :  axmmax     0 n    - Experimental determine 0     2    - Failure hypothesis 3     3    - Failure hypothesis 4 7.4. Strenght condition 1/10/2013 34 Problem 7.5.1: For the timber beam and loading shown • Draw the shear and bend-moment diagrams • Determine the maximum normal stress due to bending • Check strength condition of the beam, known []=1,5 kN/cm2 7.5. Sample Problem 1/10/2013 35 SOLUTION: • Support reactions: 0 : 40kN 14kNy B B DF M R R     • Using method of section • Section 1 - 1    1 1 1 1 1 0 20 kN 0 20kN 0 20kN 0m 0 0 yF Q Q M M M             • Section 2 - 2    2 2 2 2 2 0 20 kN 0 20kN 0 20kN 2.5m 0 50kN m yF Q Q M M M               • Section 3 – 3 • Section 4 – 4 • Section 5 – 5 • Section 6 – 6 3 3 4 4 5 5 6 6 26kN 50kN m 26kN 28kN m 14kN 28kN m 14kN 0 Q M Q M Q M Q M                   7.5. Sample Problem 1/10/2013 36 • From shear force and bending moment diagram: max max26kN 50kN mBQ M M    Q kN M kNm • Maximum bending moment    221 1 6 6 6 3 3 max max 6 3 0.080m 0.250m 833.33 10 m 50 10 N m 833.33 10 m x x W b h M W             6 2 max 60.0 10 Pa=60MPa=6kN/cm   • smax > [s] => Strength condition isn’t satistify 7.5. Sample Problem 1/10/2013 37 7.5. Sample Problem A simply supported steel beam is to carry the distributed and concentrated loads shown. Knowing that the allowable normal stress for the grade of steel to be used is 160 MPa, select the wide- flange shape that should be used. SOLUTION: • Considering the entire beam as a free-body, determine the reactions at A and D. • Develop the shear diagram for the beam and load distribution. From the diagram, determine the maximum bending moment. • Determine the minimum acceptable beam section modulus. Choose the best standard section which meets this criteria. Problem 7.5.2: 38        5m 60kN 1.5m 50kN 4m 0 58.0kN 58.0kN 60kN 50kN=0 52.0kN A D D y A A M V V F V V               52.0kN 60kN 8kN A A B A q B Q V Q Q S Q          • Maximum bending moment: Q = 0 => z = 2,6 m. max ( ) 67.6kNmQM S AE  Q kN 58 66Mmax M kNm 7.5. Sample Problem Support reactions: 1/10/2013 39 • Bending stress criterion:   max min 2 6 3 3 67.6kN m 16kN/cm 422.5 10 m 422.5 m M W c             max max max x x M M W W        7.5. Sample Problem Choose the best standard section which meets this criteria. 4481.46W200 5358.44W250 5497.38W310 4749.32W360 63738.8W410 mm10 33      SShape 9.32360W 1/10/2013 40 7.6. Deflection of Beam Because the design of beams is frequently governed by rigidity rather than strength, the computation of deflection is an integral component of beam analysis. For example, building codes specify limits on deflections as well as stresses. 7.4.1. Differential equation of elastic curve • The deformed axis of the beam is called elastic curve. • Consider a cantilever beam with concentrated load acting downward at the free end. Chosen coordinates are shown in fig. 1/10/2013 41 7.6. Deflection of Beam Assumption. - Displacements and slopes are very small. - The stresses are below the elastic limit • The deflection y is the displacement in the y direction of any point on the axis of the beam. Deflection is positive when downward. m m m1 m1   y • The angle of slope  (angle of rotation) is angle between cross-section before and after deformation. • From geometry can obtain (z) = tg = y’(z) 42 7.6. Deflection of Beam • From equilibrium of bending beam – the curvature of neutral longitudinal fiber (relationship between bending moment and curvature for pure bending) ( )1 x x M z EI  3 2 2 1 "( ) "( ) (1 ' ) y z y z y      • From elementary calculus : '' ( )x x M z y EI   (small deformation) But + or - ??? 1/10/2013 43 • Prismatic beams BENDING MOMENT EQUATION SHEAR FORCE EQUATION LOAD EQUATION 7.6. Deflection of Beam z M M>0 ''( ) 0y z  z M ''( ) 0y z  M<0 ( ) "( ) x x M z y z EI    - Differential equation of elastic curve 2 2x x d y EI M dz   3 3x y d y EI Q dz   4 4 ( )x d y EI q z dz   1/10/2013 44 7.4.2. Double - integration method 7.6. Deflection of Beam       x x Mdy z dz C dz EI x x M y(z) dz C .dz D EI            • Bending moment is function of coordinate z. Then first integration give the angle of slope function • Second integration give the deflection’s function • C an D are constants of integration to be determined from the prescribed constrains (boundary conditions) 1/10/2013 45 7.6. Deflection of Beam • Boundary conditions • Continuity conditions C C y y  C C    1/10/2013 46 7.6. Deflection of Beam Problem 7.4.1: The cantilever beam shown in figure is subjected to a vertical load P at its end. Determine the deflection and angle of slope at free end. EI = const SOLUTION  M F L z   B F L-z L EIx z     '' x x x F L zM (z) y (z) EI EI                 2 x x F L z) F z z dz C Lz C EI EI 2            2 3 x F z z y z L Cz D EI 2 6 0 0 0z C     0 0 0z y D           2 B x FL z L 2EI Boundary condition     3 B x FL y y z L 3EI Bending moment function: Substitution to diff. equation of elastic curve: 1/10/2013 47 • For a beam subjected to a distributed load,     2 2 dM d M dQ Q z q z dz dz dz    • Equation for beam displacement becomes   2 4 2 4 d M d y EI q z dz dz       3 21 1 1 2 3 46 2 EI y z dz dz dz q z dz C z C z C z C          • Integrating four times yields • Constants are determined from boundary conditions.  Direct Determination of the Elastic Curve From the Load Distribution 7.6. Deflection of Beam • Consider a beam subjected non-uniform bending consist n segments, is numbered 1,2,,i, i+1,..,n in order from left to right. Bending rigidities of each segments are: E1I1, E2I2,, EnIn.. Consider two adjacent segments (i) and (i+1). Between them there is a special connection that deflection and slope have “jump”. In the cross-section between two segments there are concentrated loading and moment, and distributed loading also have “jump” 1/10/2013 48 7.7. Initial Parameters Method 0 0F 0M y0 y Fa aM q0 iq qi+1 z=a i y i+1 y (a) (a) (a) i (a) i+1   z y a 1 2 i i+1 n 1/10/2013 49 7.7. Initial Parameters Method • After mathematical manipulation (Fourier expansion deflection’s function at z=a), using relations among bending moment, shear force and transverse distributed load, we obtained recurring formula of deflection’s function (deflection (i+1)-th segment is calculated through deflection of i-th segment) 1 2 3 4 5 ' ( ) ( ) ( ) 1 ( ) ( ) ( ) ( ) ... 2! 3! 4! 5!a i i a a a a a y z y z y z a z a z a z a z a M Q q q EI                           a aM M  a aQ Q  1( ) ( )a i iq q a q a   ' ' ' 1( ) ( )a i iq q a q a   where Note: only for EI = const 1/10/2013 50 7.7. Initial Parameters Method 2 3 4 5 ' 1 0 0 0 0 0 0 1 ( ) ... 2! 3! 4! 5! z z z z y z y z M Q q q EI              • Deflection of first segment: ' 0 0 0 0 0 0, , , , , ,...y M Q q q - initial parameters: NOTES:  Positive sign of couples, concentrated load, and distributed load is shown in figure.  If connection between (i)-th and (i+1)-th segment is pinned, then  If (i)-th và (i+1)-th segments are one-piece then 0ay  0a ay     1/10/2013 51 Problem 7.8: Use initial parameters method to determine deflection at point C and slope at point D of the beam subjected by loading as shown in figure Solution: 1. Support reactions: 2. Table of initial parameters  B 11 V qa 4  D 9 V qa 4 1 2 3 DB M=qaP=4qa aaa 2q A C 2a 3a z = 0 z = a z = 2a 0 0y  0 0  0 0M  0 0Q  0q q  , 0 0q  0ay  0a  0aM  a BQ V  aq q  , 0aq  0ay  0a  0aM  aQ P   , 0aq  0aq  To find out yC => determine y2(z) VB VD 7.8. Sample Problems To find out D => determine y3’z) 1/10/2013 52 7.8. Sample Problems 1 2 3 4 5 ' ( ) ( ) ( ) 1 ( ) ( ) ( ) ( ) ... 2! 3! 4! 5!a i i a a a a a y z y z y z a z a z a z a z a M Q q q EI                          Recurring formula  Consider segment 1(AB): 0 ≤ z ≤ a     4 1 o o x qz y (z) y z 24EI z = 0 z = a z = 2a 0 0y  0 0  0 0M  0 0Q  0q q  , 0 0q  0ay  0a  0aM  a BQ V  aq q  , 0aq  0ay  0a  0aM  aQ P   , 0aq  0aq  Consider segment 2 (BC): a ≤ z ≤ 2a         4 4 3 B 2 o o x x x qz q(z a) V (z a) y (z) y z 24EI 24EI 6EI 1/10/2013 53 7.8. Sample Problems z = 0 z = a z = 2a 0 0y  0 0  0 0M  0 0Q  0 0q  , 0 0q  0ay  0a  0aM  a BQ V   aq q  , 0aq  0ay  0a  0aM  aQ P   , 0aq  0aq  1 2 3 4 5 ' ( ) ( ) ( ) 1 ( ) ( ) ( ) ( ) ... 2! 3! 4! 5!a i i a a a a a y z y z y z a z a z a z a z a M Q q q EI                           Consider segment 3 (CD): 2a ≤ z ≤ 3a           4 4 3 3 B 3 o o x x x x qz q(z a) V (z a) P(z 2a) y (z) y z 24EI 24EI 6EI 6EI 1/10/2013 54 7.8. Sample Problems Equations of elastic curve of every segment :     4 1 o o x qz y (z) y z 24EI         4 4 3 B 2 o o x x x qz q(z a) V (z a) y (z) y z 24EI 24EI 6EI           4 4 3 3 B 3 o o x x x x qz q(z a) V (z a) P(z 2a) y (z) y z 24EI 24EI 6EI 6EI y0, 0 ??? 1/10/2013 55 7.8. Sample Problems   4 o x 5qa y 24EI   3 o x qa 6EI    4 C 2 x 7qa y y (z 2a) 24EI      3 D 3 x qa y' (z 3a) 6EI  To determine 2 initial parameters y0 and 0 , let’s consider boundary conditions: z = a => y1(z=a) = 0 z = 3a => y3(z=3a) = 0  From equations of elastic curve of two segments y1(z) và y3(z), using boundary condition, we have:  Thus 1/10/2013 56 7.7. Statically Indeterminate Beams • Consider beam with fixed support at A and roller support at B. • From free-body diagram, note that there are four unknown reaction components. • Conditions for static equilibrium yield 0 0 0z y AF F M     The beam is statically indeterminate.   1 2 0 0 z z EI y dz M z dz C z C    • Also have the beam deflection equation, which introduces two unknowns but provides three additional equations from the boundary conditions: At 0, 0 0 At , 0z y z L y     1/10/2013 57 THANK YOU FOR ATTENTION ! 1/10/2013 58