Beam Deflections: Second-Order Method

10 eam Deflections: Second-Order Method 10 1

Lecture 10: EM DEFLECTIONS: SECOND-ORDER METHOD TLE OF CONTENTS Page 10.1 Introduction..................... 10 3 10.2 What is a eam?................... 10 3 10.2.1 Terminolog................. 10 3 10.2.2 Mathematical Models.............. 10 4 10.2.3 ssumptions of Classical eam Theor......... 10 4 10.3 The ernoulli-euler eam Theor............. 10 4 10.3.1 eam Coordinate Sstems............. 10 4 10.3.2 eam Motion................. 10 4 10.3.3 eam Loading................. 10 5 10.3.4 Support Conditions............... 10 5 10.3.5 Strains, Stresses and ending Moments......... 10 6 10.3.6 eam Kinematics............... 10 6 10.4 Notation Summar................... 10 6 10.5Differential Equations Summar............. 10 7 10.6 Second Order Method For eam Deflections......... 10 7 10.6.1 Eample 1: Cantilever Load Under Tip Point Load.... 10 8 10.6.2 Eample 2: Cantilever eam under Triangular Distributed Load 10 8 10.6.3 Eample 3: Simpl Supported eam Under Uniform Load.. 10 9 10.6.4 ssessment of The Second Order Method........ 10 10 10.7 ddendum. oundar Condition Table........... 10 11 10 2

10.2 WHT IS EM? Note to SEN 3112 students; the material in 10.1 through 10.3 is largel classical and ma be skipped if ou remember beams in SEN 2001. The new stuff starts with the summar in 10.4. 10.1. Introduction This Lecture starts the presentation of methods for computed lateral deflections of plane beams undergoing smmetric bending. This topic is covered in great detail in Chapter 9 of the tetbook b eer, Johnston and DeWolf. The lectures posted here summarize some important points and present some eamples. Students are assumed to be familiar with: (1) integration of ODEs, and (2) statics of plane beams under smmetric bending. The latter topic is covered in SEN 2001. Chapters 4 through 6 of the Mechanics of Materials tetbook b eer, Johnston and DeWolf deal thoroughl with that subject, which is assumed to be known. 10.2. What is a eam? eams are the most common tpe of structural component, particularl in Civil and Mechanical Engineering. beam is a bar-like structural member whose primar function is to support transverse loading and carr it to the supports. See Figure 10.1. bar-like it is meant that one of the dimensions is considerabl larger than the other two. This dimension is called the longitudinal dimension or beam ais. The intersection of planes normal to the longitudinal dimension with the beam member are called cross sections. longitudinal plane is one that passes through the beam ais. Figure 10.1. beam is a structural member designed to resist transverse loads. beam resists transverse loads mainl through bending action, ending produces compressive longitudinal stresses in one side of the beam and tensile stresses in the other. The two regions are separated b a neutral surface of zero stress. The combination of tensile and compressive stresses produces an internal bending moment. This moment is the primar mechanism that transports loads to the supports. The mechanism is illustrated in Figure 10.2. 10.2.1. Terminolog Neutral surface Tensile stress Compressive stress Figure 10.2. eam transverse loads are primaril resisted b bending action. general beam is a bar-like member designed to resist a combination of loading actions such as biaial bending, transverse shears, aial stretching or compression, and possibl torsion. If the internal aial force is compressive, the beam has also to be designed to resist buckling. If the beam is subject primaril to bending and aial forces, it is called a beam-column. If it is subjected primaril to bending forces, it is called simpl a beam. beam is straight if its longitudinal ais is straight. It is prismatic if its cross section is constant. 10 3

Lecture 10: EM DEFLECTIONS: SECOND-ORDER METHOD spatial beam supports transverse loads that can act on arbitrar directions along the cross section. plane beam resists primaril transverse loading on a preferred longitudinal plane. This course considers onl plane beams undergoing smmetric bending. 10.2.2. Mathematical Models One-dimensional mathematical models of structural beams are constructed on the basis of beam theories. ecause beams are actuall three-dimensional bodies, all models necessaril involve some form of approimation to the underling phsics. The simplest and best known models for straight, prismatic beams are based on the ernoulli-euler beam theor (also called classical beam theor and engineering beam theor), and the Timoshenko beam theor. The ernoulli-euler theor is that taught in introductor Mechanics of Materials courses, and is the onl one dealt with in this course. The Timoshenko model incorporates a first order kinematic correction for transverse shear effects. This model assumes additional importance in dnamics and vibration. 10.2.3. ssumptions of Classical eam Theor The classical beam theor for plane beams rests on the following assumptions: 1. Planar smmetr. The longitudinal ais is straight and the cross section of the beam has a longitudinal plane of smmetr. The resultant of the transverse loads acting on each section lies on that plane. The support conditions are also smmetric about this plane. 2. Cross section variation. The cross section is either constant or varies smoothl. 3. Normalit. Plane sections originall normal to the longitudinal ais of the beam remain plane and normal to the deformed longitudinal ais upon bending. 4. Strain energ. The internal strain energ of the member accounts onl for bending moment deformations. ll other contributions, notabl transverse shear and aial force, are ignored. 5. Linearization. Transverse deflections, rotations and deformations are considered so small that the assumptions of infinitesimal deformations appl. 6. Material model. The material is assumed to be elastic and isotropic. Heterogeneous beams fabricated with several isotropic materials, such as reinforced concrete, are not ecluded. 10.3. The ernoulli-euler eam Theor 10.3.1. eam Coordinate Sstems Under transverse loading one of the top surfaces shortens while the other elongates; see Figure 10.2. Therefore a neutral surface that undergoes no aial strain eists between the top and the bottom. The intersection of this surface with each cross section defines the neutral ais of that cross section. If the beam is homogenous, the neutral ais passes through the centroid of the cross section. If the beam is fabricated of different materials for eample, a reinforced concrete beam the neutral aes passes through the centroid of an equivalent cross section. This topic is covered in Mechanics of Materials tetbooks; for eample eer-johnston-dewolf s Chapter 4. The Cartesian aes for plane beam analsis are chosen as shown in Figure 10.3. is lies along the longitudinal beam ais, at neutral ais height. is lies in the smmetr plane and points upwards. is z is directed along the neutral ais, forming a RHS sstem with and. The origin is placed at the leftmost section. The total length (or span) of the beam member is called L. 10 4

10.3 THE ERNOULLI-EULER EM THEORY pplied load p() z eam cross section Neutral ais Neutral surface L Smmetr plane Figure 10.3. Terminolog and choice of aes for ernoulli-euler model of plane beam. 10.3.2. eam Motion The motion under loading of a plane beam member in the, plane is described b the two dimensional displacement field [ ] u(, ), (10.1) v(, ) where u and v are the aial and transverse displacement components, respectivel, of an arbitrar beam material point. The motion in the z direction, which is primarit due to Poisson s ratio effects, is of no interest. The normalit assumption of the ernoulli-euler model can be represented mathematicall as u(, ) = v() = v = θ, v(, ) = v(). (10.2) Note that the slope v = v/ = dv/d of the deflection curve has been identified with the rotation smbol θ. This is permissible because θ represents to first order, according to the kinematic assumptions of this model, the rotation of a cross section about z positive CCW. 10.3.3. eam Loading The transverse force per unit length that acts on the beam in the + direction is denoted b p(), as illustrated in Figure 10.3. Point loads and moments acting on isolated beam sections can be represented with Discontinuit Functions (DFs), a topic covered in Lecture 12. ;;;;;;; ;;;;;;; ;;;;;;; Figure 10.4. simpl supported beam has end ;;;;;;; supports that preclude transverse displacements but permit end rotations. Figure 10.5. cantilever beam is clamped at one end and free at the other. irplane wings and stabilizers are eamples of this configuration. 10 5

Lecture 10: EM DEFLECTIONS: SECOND-ORDER METHOD 10.3.4. Support Conditions Support conditions for beams ehibit far more variet than for bar members. Two canonical cases are often encountered in engineering practice: simple support and cantilever support. These are illustrated in Figures 10.4 and 10.5, respectivel. eams often appear as components of skeletal structures called frameworks, in which case the support onditions are of more comple tpe. 10.3.5. Strains, Stresses and ending Moments The ernoulli-euler or classical model assumes that the internal energ of beam member is entirel due to bending strains and stresses. ending produces aial stresses σ, which will be abbreviated to σ, and aial strains ɛ, which will be abbreviated to ɛ. The strains can be linked to the displacements b differentiating the aial displacement u(): ɛ = ɛ = u = d2 v d 2 = v = κ. (10.3) Here κ denotes the deformed beam ais curvature, which to first order is κ d 2 v/d 2 = v. The bending stress σ = σ is linked to e through the one-dimensional Hooke s law σ = Ee = E d2 v = Eκ, (10.4) d2 where E is the longitudinal elastic modulus. The most important stress resultant in classical beam theor is the bending moment M z, which is defined as the cross section integral M z = σ d = E d2 v 2 d= EI d 2 zz κ. (10.5) Here I zz denotes the moment of inertia 2 top face d of the cross + face section with respect to the z (neutral) ais. The bending M moment M is considered positive if it compresses the upper z portion: > 0, of the beam cross section, as illustrated in z Figure 10.6. This convention eplains the negative sign of V in the above integral. The product EI zz is called the bending Figure 10.6. Positive sign rigidit of the beam with respect to fleure about the z ais. convention for M z and V. 10.3.6. eam Kinematics The kinematic of the classical beam model used in this course is illustrated in Figure 10.7. dditional details ma be found in Chapter 9 of eer-johnston-dewolf. Deflected cross section θ() = v'() E, Izz v() Figure 10.7. eam kinematics. 10 6

10.4. Notation Summar 10.6 SECOND ORDER METHOD FOR EM DEFLECTIONS Quantit Smbol Sign convention(s) Problem specific load varies You pick em Generic load for ODE work p() +ifup Transverse shear force V () +ifupon+ face ending moment M z () + if it produces compression on top face dv() Slope of deflection curve = v () + if positive slope, or cross-section rotates CCW d Deflection curve v() + if beam section moves upward Note 1: Some tetbooks (e.g. Vable and eer-johnston-dewolf) use V = V as alternative transverse shear smbol. This has the advantage of eliminating the sign in two of the ODEs below. V will onl be used occasionall in this course. Note 2: In our beam model the slope v = dv()/d is equal to the rotation θ() of the cross section. 10.5. Differential Equations Summar The following differential relations assume sufficient differentiabilit for the derivatives to eist. This requirement can be alleviated b using Discontinuit Functions (DFs), a topic covered in Lecture 12. Connected quantities ODEs dv From load to transverse shear force d = p or p = V = V From transverse shear to bending moment dm z d = V or M z = V = V From bending moment to deflection EI d 2 v zz d 2 = M z or v = M z EI zz From load to moment M z From load to deflection EI zz v IV = p 10.6. Second Order Method For eam Deflections The second-order method to find beam deflections gets its name from the order of the ODE to be integrated: EI zz v () = M z () (third line in above table) is a second order ODE. The procedure can be broken down into the following steps. 1. Find the bending moment M z () directl, for eample b statics. 2. Integrate M z /(EI zz ) once to get the slope v () = dv()/d. 3. Integrate once more to get v(). 4. If there are no continuit conditions, the foregoing steps will produce two integration constants. ppl kinematic Cs to find those. If there are continuit conditions, more than two integration constants ma appear. ppl kinematic Cs and continuit conditions to find those. 10 7

Lecture 10: EM DEFLECTIONS: SECOND-ORDER METHOD 5. Substitute the constants of integration into the deflection function to get v(). 6. Evaluate v() at specific sections of interest. Three eamples of this method follow. 10.6.1. Eample 1: Cantilever Load Under Tip Point Load Constant EI zz L P (a) Problem definition P X (b) FD to find M z() M () z V () Figure 10.8. eam problem for Eample 1. The problem is defined in Figure 10.8(a). Using the FD pictured in Figure 10.8(b), and stating moment equilibrium with respect to X (to eliminate ab initio the effect of the transverse shear force V at that section) gives M z () = P (10.6) For convenience we scale v() b EI zz so that the ODE linking bending moment to deflection is EI zz v () = M z () = P. Integrating twice: EI zz v () = P2 2 EI zz v() = P3 6 + C 1 + C 1 + C 2 (10.7) The kinematic Cs for the cantilever are v (L) = 0 and v(l) = 0 at the fied end. The first one gives EI zz v (L) = PL 2 /2 + C 1 = 0, whence C 1 = PL 2 /2. The second one gives EI zz v(l) = PL 3 /6 + C 1 L + C 2 = PL 3 /6 + PL 3 /2 + C 2 = 0 whence C 2 = PL 3 /3. Substituting into the epression for v() gives v() = P ( 3 3L 2 + 2L 3) = P (L ) 2 (2L + ) (10.8) 6EI zz 6EI zz Of particular interest is the tip deflection at free end, which is the largest one. Setting = 0 ields v(0) = v = PL3 3EI zz (10.9) The negative sign indicates that the beam deflects downward if P > 0. 10 8

10.6 SECOND ORDER METHOD FOR EM DEFLECTIONS 10.6.2. Eample 2: Cantilever eam under Triangular Distributed Load The problem is defined in Figure 10.9(a). Using the FD pictured in Figure 10.9(b), again doing moment equilibrium with respect to X gives M z () = 1 2 w() ( 1 3 ) = w 3 6L (10.10) Constant EI zz w() = w /L L w w() = w /L X M () z V () (a) Problem definition (b) FD to find M z () Figure 10.9. eam problem for Eample 2. Integrating EI zz v () = M z () twice ields EI zz v () = w 4 24L + C 1, EI zz v() = w 5 120L + C 1 + C 2. (10.11) s in Eample 1, the kinematic Cs for the cantilever are v (L) = 0 and v(l) = 0 at the fied end. The first one gives EI zz v (L) = wl 3 /24 + C 1 = 0, whence C 1 = w L 3 /24. The second one gives EI zz v(l) = w L 4 /120 + C 1 L + C 2 = 0 whence C 2 = w L 4 /30. Substituting into the epression for v() we obtain the deflection curve as v() = 1 ( w 5 EI zz L 120 + w L 4 w L 5 ) = w 24 30 120EI zz L ( 5 5L 4 + 4L 5 ) (10.12) Of particular interest is the tip deflection at, which is the largest one. Setting = 0 ields v(0) = v = w L 4 (10.13) 30EI zz The negative sign indicates that the beam deflects downward if w > 0. 10 9

Lecture 10: EM DEFLECTIONS: SECOND-ORDER METHOD Constant EI zz w uniform along beam span C w X M () z V () L/2 L/2 L (a) Problem definition R = wl/2 (b) FD to find M z () Figure 10.10. eam problem for Eample 3. 10.6.3. Eample 3: Simpl Supported eam Under Uniform Load The problem is defined in Figure 10.10(a). Using the FD pictured in Figure 10.10(b), and writing down moment equilibrium with respect to X gives the bending moment M z () = R w 2 2 = wl 2 w 2 2 = w (L ) (10.14) 2 Integrating EI zz v () = M z () twice ields EI zz v () = wl2 4 EI zz v() = wl3 12 w 3 6 + C 1 = w 2 12 (3L 2) + C 1, w 4 24 + C 1 + C 2 = w 3 24 (2L ) + C 1 + C 2. (10.15) The kinematic Cs for a SS beam are v = v(0) = 0 and v = v(l) = 0. The first one gives C 2 = 0 and the second one C 1 = wl 3 /24. Substituting into the epression for v() gives, after some algebraic simplifications, v() = w (L )(L 2 + L 2 ) (10.16) 24 EI zz Note that since v() = v(l ) the deflection curve is smmetric about the midspan C. The midspan deflection is the largest one: 5w L4 v C = v(l/2) = (10.17) 384 EI zz The negative sign indicates that the beam deflects downward if w>0. 10.6.4. ssessment of The Second Order Method Good Points. The governing ODE v () = M z ()/EI zz is of second order, so it onl needs to be integrated twice, and onl two constants of integration appear. Finding M z () directl from statics is particularl straightforward for cantilever beams since the reactions at the free end are zero. 10 10

10.7 DDENDUM. OUNDRYCONDITION TLE eam oundar Conditions for Shear, Moment, Slope & Deflection Condition Shear force ending moment Slope (= rotation) Deflection V () M z() v'()= θ() v() & Simple support 0 0 Fied end 0 0 Free end 0 * # 0 Smmetr 0 0 ntismmetr 0 0 * Unless a point force is applied at the free end & Unless a point moment is applied at the simple support # Unless a point moment is applied at the free end lank entr means that value is unknown and has to be determined b solving problem Figure 10.11. eam boundar conditions for some common support configurations. ad Points. Clums if continuit conditions appear within the beam span, unless Discontinuit Functions (DFs) are used to represent M z () as a single epression. If DFs are not used, the method is prone to errors in human calculations. This disadvantage is shared b the fourth-order method covered in the net Lecture, but the representation of p() in terms of DFs (to be covered in the lecture of Tu October 4* ) is simpler. 10.7. ddendum. oundar Condition Table The C table provided in Figure 10.11 for some common support configurations is part of a Supplementar Crib Sheet allowed for Midterm Eams 2 and 3, in addition to the student-prepared two-sided crib sheet. * This topic (DFs) will not be in midterm eam #2 but in #3. 10 11