Lecture Strength of Materials I - Chapter 7: Bending. The following will be discussed in this chapter: Introduction, bending stress, shearing stress in bending, strength condition, sample problems, deflections of beam, statically indeterminate beams.
STRENGTH OF MATERIALS 1/10/2013 TRAN MINH TU - University of Civil Engineering, Giai Phong Str 55, Hai Ba Trung Dist Hanoi, Vietnam CHAPTER BENDING 1/10/2013 Contents 7.1 Introduction 7.2 Bending stress 7.3 Shearing stress in bending 7.4 Strength condition 7.5 Sample Problems 7.6 Deflections of beam 7.7 Statically indeterminate beams 1/10/2013 7.1 Introduction In previous charters, we considered the stresses in the bars caused by axial loading and 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 1/10/2013 7.1 Introduction Classification of Beam Supports 1/10/2013 7.1 Introduction Limitation 1/10/2013 7.1 Introduction Segment BC: Mx≠0, Qy=0 => Pure Bending Segments AB,CD: Mx≠0, Qy≠0 => Nonuniform Bending 1/10/2013 7.1 Introduction Pure Bending: Prismatic members subjected to equal and opposite couples acting in the same longitudinal plane 1/10/2013 7.2 Bending stress Simplifying assumptions 1/10/2013 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 1/10/2013 Neutral axis 10 7.6 Deflection of Beam 7.4.2 Double - integration method • Bending moment is function of coordinate z Then first integration give the angle of slope function Mx dy z dz C dz EI x • Second integration give the deflection’s function Mx y(z) dz C .dz D EI x • C an D are constants of integration to be determined from the prescribed constrains (boundary conditions) 1/10/2013 44 7.6 Deflection of Beam • Boundary conditions • Continuity conditions yC yC C C 1/10/2013 45 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 Bending moment function: EIx z M F L z Substitution to diff equation of elastic curve: M x (z) F L z '' y (z) EI x EI x z F z2 z3 y z L Cz D EI x z C z y D 1/10/2013 F B L-z L F L z) EI x F z2 dz C Lz C EI x 2 Boundary condition FL2 B z L 2EI x FL3 yB y z L 3EI x 46 7.6 Deflection of Beam Direct Determination of the Elastic Curve From the Load Distribution • For a beam subjected to a distributed load, dM Q z dz d M dQ qz dz dz • Equation for beam displacement becomes d 2M d4y EI q z dz dz • Integrating four times yields EI y z dz dz dz q z dz 16 C1 z 12 C2 z C3 z C4 • Constants are determined from boundary conditions 1/10/2013 47 7.7 Initial Parameters Method • 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” Fa qi F0 qi+1 q0 M0 Ma n i+1 i z y0 0 z=a y (a) i y 1/10/2013 y a y (a) i+1 (a) i (a) i+1 48 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) yi 1 ( z ) yi ( z ) ya a ( z a) EI where 1/10/2013 ( z a) ( z a )3 ( z a) ( z a ) ' M a 2! Qa 3! qa 4! q a 5! M a M a qa qi1 (a) qi (a) Qa Qa q q (a) q (a) ' a ' i 1 Note: only for EI = const ' i 49 7.7 Initial Parameters Method • Deflection of first segment: y1 ( z ) y0 0 z EI y0 ,0 , M , Q0 , q0 , q0' , z2 z3 z4 z ' M 2! Q0 3! q0 4! q0 5! - 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 ya If (i)-th (i+1)-th segments are one-piece then ya a 1/10/2013 50 7.8 Sample Problems Problem 7.8: q Use initial parameters method to determine deflection at point A C and slope at point D of the beam subjected by loading as shown in figure a P=4qa B VB a C M=qa D a VD 2a Solution: 3a Support reactions: 11 VB qa VD qa Table of initial parameters To find out yC => determine y2(z) To find out D => determine y3’z) 1/10/2013 z=0 z=a z = 2a y0 0 M0 Q0 q0 q q0, ya a M a Qa VB qa q qa, ya a M a Qa P qa qa, 510 7.8 Sample Problems Recurring formula yi 1 ( z ) yi ( z ) ya a ( z a) EI ( z a)2 ( z a )3 ( z a)4 ( z a ) ' M Q q q a a a a 2! 3! 4! 5! Consider segment 1(AB): 0≤z≤a qz y1 (z) y o o z 24EI x Consider segment (BC): a ≤ z ≤ 2a z=0 z=a z = 2a y0 0 M0 Q0 q0 q q0, ya a M a Qa VB qa q qa, ya a M a Qa P qa qa, qz q(z a)4 VB (z a)3 y (z) y o o z 24EI x 24EI x 6EI x 1/10/2013 52 7.8 Sample Problems yi 1 ( z ) yi ( z ) ya a ( z a) EI ( z a)2 ( z a )3 ( z a)4 ' ( z a) M a 2! Qa 3! qa 4! q a 5! Consider segment (CD): 2a ≤ z ≤ 3a z=0 z=a z = 2a y0 0 M0 Q0 q0 q0, ya a M a Qa VB qa q qa, ya a M a Qa P qa qa, qz q(z a)4 VB (z a)3 P(z 2a)3 y3 (z) y o o z 24EI x 24EI x 6EI x 6EI x 1/10/2013 53 7.8 Sample Problems Equations of elastic curve of every segment : qz y1 (z) y o o z 24EI x qz q(z a)4 VB (z a)3 y (z) y o o z 24EI x 24EI x 6EI x qz q(z a)4 VB (z a)3 P(z 2a)3 y3 (z) y o o z 24EI x 24EI x 6EI x 6EI x y0, 0 ??? 1/10/2013 54 7.8 Sample Problems To determine initial parameters y0 and 0 , let’s consider boundary conditions: z = a => y1(z=a) = z = 3a => y3(z=3a) = From equations of elastic curve of two segments y1(z) y3(z), using boundary condition, we have: 5qa yo 24EI x Thus 1/10/2013 qa o 6EI x 7qa y C y (z 2a) 24EI x qa D y'3 (z 3a) 6EI x 55 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 F z 0 F y 0 M A 0 The beam is statically indeterminate • Also have the beam deflection equation, z z 0 EI y dz M z dz C1 z C2 which introduces two unknowns but provides three additional equations from the boundary conditions: At z 0, y 1/10/2013 At z L, y 56 1/10/2013 57 THANK YOU FOR ATTENTION ! 1/10/2013 58 .. .CHAPTER BENDING 1/10/2013 Contents 7. 1 Introduction 7. 2 Bending stress 7. 3 Shearing stress in bending 7. 4 Strength condition 7. 5 Sample Problems 7. 6 Deflections of beam 7. 7 Statically... through the centroid C of the E M y x z dA xyd A cross-section) A A A xyd A I A 1/10/2013 xy 0 y - axis – the axis of symmetry of the cross-section 12 z 7. 2 Bending stress M x... 1/10/2013 15 7. 2 Bending stress 1/10/2013 16 7. 2 Bending stress Properties of American Standard Shapes 1/10/2013 17 7.2 Bending stress 1/10/2013 18 7. 2 Bending stress 1/10/2013 19 7. 2 Bending