Strength of Materials WORK-BOOK

Strength of Materials W ORK -B OOK RM Nkgoeng Copyright c 2013 Mashilo Nkgoeng S ELF - PUBLISHED W W W.M A S H I L O N K G O E N G.C O.Z A Licensed under the Creative Commons Attribution-Non-commercial 3.0 Unported License (the “License”). You may not use this file except in compliance with the License. You may obtain a copy of the License at http://creativecommons.org/licenses/by-nc/3.0. Unless required by applicable law or agreed to in writing, software distributed under the License is distributed on an “AS IS” BASIS, WITHOUT WARRANTIES OR CONDITIONS OF ANY KIND, either express or implied. See the License for the specific language governing permissions and limitations under the License. First printing, December 2013 Contents 1 Deflection of Beams . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5 1.1 What is a Beam? 1.1.1 1.1.2 1.1.3 1.1.4 1.1.5 1.1.6 Beam terminology . . . . . . . . . . . . . . . . Mathematical Models . . . . . . . . . . . . . . Assumption of Classical Beam Theory . Beam Loading . . . . . . . . . . . . . . . . . . . Support Conditions . . . . . . . . . . . . . . . Stresses, strains and bending moments 1.2 Notation 7 1.3 Second Order Method for Beam Deflections 7 1.4 Double Integration Using Bracket Functions 7 1.5 Examples 9 1.6 Exercises 16 2 Continuous Beams . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 19 2.1 Introduction 2.1.1 2.1.2 Point Load . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 19 Uniformly Distributed Load . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 21 2.2 Exercise 22 2.3 Examples 23 2.4 Exercises 23 3 Energy Methods . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 25 3.1 Introduction 25 3.2 Strain Energy of Bars 26 3.3 Castigliano’s Theorem 26 3.4 Structures 27 3.5 Castigliano’s theorem applied to Curved Beams 27 3.6 Castigliano’s theorem applied to Beams 27 3.6.1 3.6.2 Cantilever beam with a Point Load at the free end . . . . . . . . . . . . . . . . . . . . . . . 27 S/S beam with a Point Load at mid-point . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 28 3.7 Examples 29 3.8 Exercises 34 5 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5 5 6 6 6 6 19 4 Unsymmetrical Bending of Beams . . . . . . . . . . . . . . . . . . . . . . . . . . . 39 4.1 Symmetric Member in Pure Bending 39 4.2 Unsymmetrical Bending 40 4.3 Alternative procedure for stress determination 42 4.4 Deflection 44 4.5 Notation 44 4.6 xxx 44 4.6.1 4.6.2 Point Load . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 44 Uniformly Distributed Load . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 44 4.7 Examples 44 4.8 Exercises 47 5 Inelastic Bending . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 49 5.1 Plastic Bending of Rectangular Beams 49 5.2 Plastic Bending of Symmetrical (I-Section) Beam 52 5.3 Partially plastic Bending of Unsymmetrical Sections 53 5.4 Limit Analysis-Bending 56 5.4.1 5.4.2 Bending . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 56 The principle of Virtual Work . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 58 5.5 Solid Shaft 65 5.6 Hollow shaft 67 5.7 Exercises 69 What is a Beam? Beam terminology Mathematical Models Assumption of Classical Beam Theory Beam Loading Support Conditions Stresses, strains and bending moments Notation Second Order Method for Beam Deflections Double Integration Using Bracket Functions Examples Exercises 1 — Deflection of Beams Introduction This lecture notes starts the presentation of methods for computed lateral deflections of plane beams undergoing symmetric bending. The workbook just summarises the topic, but the textbook1 covers in detail this topic in chapter 9 and 10. We assume that as a student you are familiar with the following: 1. integration of Ordinary Differential Equations 2. Statics of plane beams under symmetric bending. This is also covered in Chapter 5 and 6 of the prescribed textbook. 1.1 What is a Beam? Beams are the most common type of structural component, particularly in Civil and Mechanical Engineering. A beam is a bar-like structural member whose primary function is to support transverse loading and carry it to the supports. By bar-like, we mean that one of the dimensions is considered larger than the other two. This dimension is called the longitudinal dimension or beam axis. The intersection of planes normal to the longitudinal dimension with the beam member are called cross sections. A longitudinal plane is one that passes through the beam axis. A beam resist transverse loads mainly through bending action. Bending produces compressive longitudinal stresses in one side of the beam and tensile stresses in the other. the two regions are separated by a neutral surface of zero stress. The combination of tensile and compressive stresses produces an internal bending moment. This moment is the primary mechanism that transports loads to the supports. This is illustrated by Fig. # 1.1.1 Beam terminology General Beam –it is a bar like member designed to resist a combination of loading actions such as biaxial bending, transverse shears, axial stretching or compression, possibly torsion. If the internal axial force is compressive, the beam has also to be designed to resist buckling. If the beam is subject primarily to bending and axial forces, it is called a beam-column. If it is subjected primarily to bending forces, it is called simply a beam. A beam is straight if its longitudinal axis is straight. It is prismatic if its cross section is constant. Spatial Beam –it supports transverse loads that can act on arbitrary directions along the cross section. Plane Beam –it resists primarily transverse loading on a preferred longitudinal plane. 1.1.2 Mathematical Models One-dimensional mathematical models of structural beams are constructed on the basis of beam theories. Since beams are actually three-dimensional bodies, all models necessarily involve some form of approximation to the underlying physics. The simplest and best known model for straight, prismatic beams are based on the Bernoulli-Euler theory as well as the Timoshenko 1 Mechanics of Materials by Gere and Goodno Deflection of Beams 6 beam theory. The B-E theory is the one that is taught in SOM3602 and is the only one that we will be dealing with in SOM401M. The Timoshenko model incorporates a first order kinematic correction for transverse shear effects. This model assumes additional importance in dynamics and vibration. 1.1.3 Assumption of Classical Beam Theory The classical beam theory for plane beams rests on the following assumptions: 1. Planar symmetry: The longitudinal axis is straight and the cross section of the beam has a longitudinal plane of symmetry. The resultant of the transverse loads acting on each section lies on that plane. The support conditions are also symmetric about this plane. 2. Cross section variation: The cross section is either constant or varies smoothly. 3. Normality: Plane sections originally normal to the longitudinal axis of the beam remain plane and normal to the deformed longitudinal axis upon bending. 4. Strain energy: The internal strain energy of the member accounts only for bending moment deformations. All other contributions, notably transverse shear and axial force, are ignored. 5. Linearisation: Transverse deflections, rotations and deformations are considered so small that the assumptions of infinitesimal deformations apply. 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 excluded. 1.1.4 Beam Loading The transverse force per unit length that acts on the beam in the y+ direction is denoted by f (x) as shown in Figure #Point loads and moments acting on isolated beam sections can be represented with Discontinuity Functions (DF’s). 1.1.5 Support Conditions Support conditions for beams exhibit far more variety than for bar members. Two cases are often encountered in engineering practice: simple support and cantilever support. These are shown in Figs. # and # respectively. Beams often appear as components of skeletal structures called frameworks, in which case the support conditions are of more complex type. Easily solved using finite element methods. 1.1.6 Stresses, strains and bending moments The Bernoulli-Euler model assumes that the internal energy of beam member is entirely due to bending strains and stresses. Bending produces axial stresses σx which is just abbreviated σ and axial strains εx which is just ε. The strains can be linked to the displacements by differentiating the axial displacement u(x): ε= ∂u d2v = −y 2 = −yv′′ = −yκ ∂x dx (1.1) where κ denotes the deformed beam axis curvature. The bending stress is linked to e through one-dimensional Hooke’s law σ = Ee = −Ey d2v = −Eyκ dx2 (1.2) 1.2 Notation 7 The most important stress resultant in classical beam theory is the bending moment Mx , which is defined as the cross sectional integral Mx = Z A −yσ dA = E d2v dx2 Z A y2 dA = EIxx κ (1.3) The bending moment is considered positive if it compresses the upper portion: y > 0. the product EIxx is called the bending rigidity of the beam with respect to flexure about the x axis. 1.2 Notation Quantity Generic load for ODE Work Transverse Shear Force Bending Moment Slope of deflection curve Deflection Curve Symbol f (x) V (x) M(x) dv(x) ′ dx = v (x) v(x) Table 1.1: Notation 1.3 Second Order Method for Beam Deflections The second-order method to find beam deflections gets its name from the order of the ODE to be integrated: EIxx v′′ (x) = M(x) is a second order ODE. The procedure can be broken down into the following steps: 2 1. Find the bending moment M(x) = d dxv(x) directly. i.e. by cutting the beam at distance x 2 and taking moments about x. 2. Integrate M(x) once to get the slope v′ (x) = dv(x) dx 3. Integrate the slope v′ (x) once to get the deflection v(x) 4. If there are no continuity conditions, the above steps will produce two integration constants C1 and C2 . Apply kinematics boundary conditions to determine the values of the integration constants. If there are continuity conditions, more than two constants of integration may appear and are solved the same way, i.e. by using boundary conditions. 5. Substitute the constants of integration into the deflection function to get v(x) 6. Evaluate v(x) at points of interest on the beam. 1.4 Double Integration Using Bracket Functions When the loads on a beam do not conform to standard cases, the solution for slope and deflection may be found from first principles. Macaulay developed a method for making the integration simpler. The basic equation governing the slope and deflection of beams is: EI d2y = M : Where M is a function of x dx2 (1.4) When a beam has a variety of loads it is difficult to apply this theory because some loads may be within the limits of x during the derivation but not during the solution of a particular point. Macaluay’s method makes it possible to do the integration necessary by placing all the terms containing x within a square bracket and integrating the bracket, not x. During evaluation, any Deflection of Beams 8 bracket with a negative value is ignored because a negative means that the load it refers to is not within the limit of x. The general method of solution is conducted as follows. Refer to Fig. #. In a real example, the loads and reactions would have numerical values but for the sake of demonstrating the general method we will use algebraic symbols. This has only point loads. 1. Write down the bending moment equation placing x on the extreme right hand end of the beam so that it contains all the loads. Write all terms containing x in a square bracket. EI d2y = M = RA [x] − P1 [x − a] − P2 [x − b] − P3 [x − c] dx2 2. Integrate once treating the square bracket as the variable. EI dy [x]2 [x − a]2 [x − b]2 [x − c]2 = RA − P1 − P2 − P3 +C1 dx 2 2 2 2 3. Integrate again using the same rules. EIy = RA [x − a]3 [x − b]3 [x − c]3 [x]3 − P1 − P2 − P3 +C1 x +C2 6 6 6 6 4. Use boundary conditions to solve constants C1 and C2 . 5. Solve slope and deflection by putting in appropriate value of x. IGNORE any brackets containing negative values. Evaluating the constants of integration that arise in the double-integration method can become very involved if more than two beam segments must be analysed. We can simplify the calculations by expressing the bending moment in terms of discontinuity functions, also known as Macaulay bracket functions. Discontinuity functions enable us to write a single expression for the bending moment that is valid for the entire length of the beam, even if the loading is discontinuous. By integrating a single, continuous expression for the bending moment, we obtain equations for slopes and deflections that are also continuous everywhere. As an illustration let us consider a simply supported beam with three segments as shown in Fig. 1.1. I have already determined the reactions. All we need to do is to segment it nicely and take moments about X − X as follows: Mxx = 480x − 500hx − 2i − 450 hx − 3i2 Nm 2 (1.5) Note that a bracket function is zero by definition if the expression in the brackets–namely (x − a) X 500N q0 = 450N/m B A 2m 1m X 920N 480N x 5m Figure 1.1: Macaulay’s Method 1.5 Examples 9 is negative; otherwise, it is evaluated as written. A bracket function can be integrated by the same rule as an ordinary function–namely, Z hx − ain = hx − ain+1 +C n+1 (1.6) This is called the global bending moment equation and it can be integrated to obtain the slope and the deflection equations for the entire beam. The two constants of integration as mentioned before can be computed from the boundary conditions. When you use this method ensure that you include every load on the beam and leave the last reaction if the beam is simply supported. If the beam is a cantilever, you can obtain the global bending moment equation by starting from the free end or by starting from the fixed end. If you choose to start from the fixed end, calculate the reaction first so that you can be able to include it in the global bending moment equation. At the end of the day, you have the choice to use any method that you feel comfortable with. We have not included other methods, simply because in practice we have found that we hardly use the other methods in solving deflection of beam problems. 1.5 Examples  Example 1.1 Determine the equation for the deflection curve as well as the slope in Fig. 1.2.  q N/m L Figure 1.2: Cantilever Beam with UDL Solution 1.1 Still to be done  Example 1.2 Determine the equation for the deflection curve as well as the slope in Fig. 5.25.  P L Figure 1.3: Cantilever Beam with UDL Solution 1.2 Still to be done  Example 1.3 Determine the equation for the deflection curve as well as the slope in Fig.1.4 .  Deflection of Beams 10 P L/2 B A L Figure 1.4: SS Beam with Point Load Solution 1.3 Still to be done  Example 1.4 For the cantilever beam, Fig. 1.5 under triangular distributed loading or variably distributed loading, determine the equation for the deflection curve as well as the slope. X  q0 C B A D x L X Figure 1.5: Cantilever with VDL Solution 1.4 We will use the same figure, Fig. 1.5 as the free-body diagram. Let the height CD = q and use the law of triangles to get the expression of q in terms of the maximum height or intensity q0 , i.e. q q0 = x L ∴q= q0 x L The moment about X − X is as follows: x q0 x3 1 EIv′′ (x) = − q × x × = − 2 3 6L We need to integrate the above twice to get two constants of integration q0 x 4 +C1 24L q0 x 5 +C1 x +C2 EIv(x) = − 120L EIv′ (x) = − 1.5 Examples 11 The kinematic boundary conditions are v′ (L) = 0 and v(L) = 0 q0 L 4 +C1 ; 24L q0 L 5 q0 L 4 EIv(L) = 0 = − + +C2 ; 120L 24 The equation of interest are:   q0 x5 q0 L3 x q0 L4 1 − + − v(x) = EI 120L 24 30 EIv′ (L) = 0 = − q0 L3 24 q0 L4 ∴ C2 = − 30 ∴ C1 = (1.7) and   1 q0 x4 q0 L3 v (x) = − + EI 24L 24 ′  (1.8) Example 1.5 For a given beam shown below: 1. derive the equation of the elastic curve 2. determine the point of maximum deflection 3. determine ymax Before one could jump in and provide a solution, it is advisable to do the following: • Develop an expression for M(x) and derive a differential equation for the elastic curve. • Integrate the differential equation twice and apply boundary conditions to obtain the elastic curve equation • Locate a point of zero slope or point of maximum deflection • Evaluate corresponding maximum deflection P A B C a L Figure 1.6: SS Beam with an Overhang  Solution 1.5 The reactions are found to be RA = X − X we get M(x) = − Pax L (0 < x < L) Pax d2y =− 2 dx L dy Pax2 EI = − +C1 dx 2L Pax3 +C1 x +C2 EIy = − 6L EI Pa L a L
and RB = P 1 + . Taking moments about Deflection of Beams 12 The boundary conditions are simply: x = 0 : y = 0 and x = L : y = 0. The first boundary condition leads to C2 = 0 and the second boundary condition leads to C1 = PaL 6 . The resulting equations are therefore dy Pax2 PaL EI = − + (1.9) dx 2L 6 and Pax3 PaL + x (1.10) EIy = − 6L 6 Determining the position of maximum deflection is relatively easy, set y′ (x) = 0   x 2  dy PaL m 1−3 =0= dx 6EI L L ∴ xm = √ = 0.577L 3
0.0642PaL2 PaL2 ymax = 0.577 − 0.5773 = 6EI 6EI  Example 1.6 For the uniform beam shown below, 1. Determine the reaction at A 2. Derive the equation for the elastic curve 3. Determine the slope at A (Note that the beam is statically indeterminate to the first degree) B A L Figure 1.7: LHS SS and RHS Fixed  Solution 1.6 The solution will include the following: • develop differential equation for the elastic curve. (this will be functionally dependent on reaction A) • Integrate twice and apply boundary conditions to solve for the reaction at A and to obtain the equation for the elastic curve. • Evaluate slope at A Taking moments about D, i.e. Σ MD = 0   1 w0 x2 x M = RA x − 2 L 3 3 w0 x d2y M = RA x − = EI 2 6L dx 1.5 Examples 13 2 d y We need to integrate EI dx 2 twice to get the equation for the elastic curve. d2y w0 x3 = R x − A dx2 6L 2 w0 x 4 dy RA x EI = − +C1 dx 2 24L RA x3 w0 x5 − +C1 x +C2 EIy = 6 120L EI The boundary conditions are as follows: x=0: x=L: x=L: y=0 dy =0 dx dy =0 dx C2 = 0 w0 L 3 RA L 2 − 24 2 3 w0 L RA L 2 C1 = − 120 6 C1 = From the above we find that the reaction at A is RA = w0 L 10 The resulting two equations were:
w0 −x5 + 2L2 x3 − L4 x 120EI
w0 −5x4 + 6L2 x2 − L4 θ= 120EI y=  (1.11) (1.12) Example 1.7 A cantilever beam is 4m long with a flexural stiffness of 20MNm2 . It has a point load of 1kN at the free end and a u.d.l of 300N/m along its entire length. Calculate the slope and deflection at the free end.  P q N/m L Figure 1.8: Cantilever Beam with UDL and Point Load Solution 1.7 This problem can be solved by using various method. The first one that will be used is Theory of Superposition for Combined loads and the second one is double integration method from first principles.For the point load only we will use the following equations to determine the slope as well as the deflection: 1. Superposition Method for Combined Loads y=− PL3 3EI (1.13) Deflection of Beams 14 dy PL2 = dx 2EI (1.14) 1000 × 43 = −1.06mm 3 × 20 × 106 dy 1000 × 42 = = 400 × 10−6 dx 2 × 20 × 106 y=− For the U.D.L we will use the following equations: y=− qL4 48EI (1.15) dy qL3 = dx 6EI (1.16) 300 × 44 = −0.48mm 48 × 20 × 106 dy 300 × 43 = 160 × 10−6 rad = dx 6 × 20 × 106 y=− The total deflection and slope are y = −1.54mm and 2. Double Integration Method from 1st Principles  dy dx = 560 × 10−6 rad Example 1.8 The beam shown in Fig. 1.9 is 7m long with EI = 200MNm2 . Determine the slope and deflection in the middle of the beam.  x 2m 30kN 4.5m 40kN X B A 7m X Figure 1.9: SS Beam with Point Load loads Solution 1.8 The first thing that we need to solve is the reaction on either side of interest and this is done by taking moments about either A or B. If we take moments about B then our solution will be: 0 = 7RA − 30 × 5 − 40 × 2.5 RA = 35.71kN The bending moment equation is as follows: Mxx = RA [x] − 30[x − 2] − 40[x − 4.5] (1.17) 1.5 Examples 15 d2y = 35.71[x] − 30[x − 2] − 40[x − 4.5] dx2 dy [x]2 [x − 2]2 [x − 4.5]2 EI = 35.71 − 30 − 40 +C1 dx 2 2 2 [x − 2]3 [x − 4.5]3 [x]3 − 30 − 40 +C1 x +C2 EIy = 35.71 6 6 6 The boundary conditions are at x = 0 we have y = 0 and this leads to C2 = 0. At x = L we have y = 0 and C1 needs to be worked out EI [7 − 2]3 [7 − 4.5]3 [7]3 − 30 − 40 +C1 (7) 6 6 6 C1 = −187.4 0 = 35.71 We are now able to write the required expressions as dy [x]2 [x − 2]2 [x − 4.5]2 = 35.71 − 30 − 40 − 187.4 dx 2 2 2 [x]3 [x − 2]3 [x − 4.5]3 EIy = 35.71 − 30 − 40 − 187.4x 6 6 6 From the above, it becomes easy to determine the deflection and rotation at the distance x = 3.5m. EI  Example 1.9 The beam shown in Fig. 1.10 is 6m long with EI = 300MNm2 . Determine the slope at the left hand end and the deflection at the middle of the beam.  x 2m X 30kN 2kN/m B A 7m X Figure 1.10: SS Beam with UDL and Point Load Solution 1.9 The reaction is determine by first taking moments about the right hand support. 0 = 6RA − 30 × 4 − 2 × 62 /2 RA = 26kN qx2 2 2 2[x] EIy′′ = 26[x] − 30[x − 2] − 2 2 2 30[x − 2] 2[x]3 26[x] − − +C1 EIy′ = 2 2 6 26[x]3 30[x − 2]3 2[x]4 EIy = − − +C1 x +C2 6 6 24 Mxx = RA [x] − 30[x − 2] − Deflection of Beams 16 1.6 Exercises Exercise 1.1 The simply supported beam ABC carries a distributed load of maximum intensity q0 over its span of length L. Determine the maximum displacement of the beam. q0 B A L/2 L  Exercise 1.2 The intensity of the distributed load on the cantilever beam varies linearly from zero to q0 . Derive the equation of the elastic curve as well as the slope. q0 L/2 L  Exercise 1.3 The intensity of the distributed load on the simply supported beam varies linearly from zero to q0 . a. Derive the equation of the elastic curve b. Find the location of the maximum deflection 1.6 Exercises 17 q0 B A L/2 L  Exercise 1.4 Determine the maximum displacement of the simply supported beam due to the distributed loading shown below. (Hint: Utilize symmetry and analyse the right of the beam only.)  Exercise 1.5 Determine the deflection at the free end for the cantilever beam below with a variably distributed loaded. EI = 10MNm2  Exercise 1.6 A 203mm × 133mm × 25kg/m I-section (parallel flange) is used as a cantilever with a span of 6m. It carries a point load of 6kN at 2m from the fixed end and a uniformly distributed load of 2kN/m from the free end to a point 1m from the fixed end. The cantilever is propped at a point 2m from the free end, such that the load in the prop is 9.7kN. Calculate the deflection at the free end and under the 6kN load. Neglect the mass of the beam.  Exercise 1.7 A cantilever 3m long carries a point load of 22kN at 1m from the fixed end as well as a uniformly distributed load of 12kN/m from the free end to a point 1m from the free end. If the deflection at the free end is limited to 16mm, calculate the minimum necessary value of I for the cross-section, neglecting the mass of the beam. A prop is now introduced 1m from the free end to reduce the deflection at the free end by half. What is the magnitude of the load in the prop.  Exercise 1.8 Calculate the deflection 2m from the left-hand end and the slope at the left-hand support of the beam shown below. EI = 10MNm2  Exercise 1.9 Calculate the slope and deflection 1m from the left-hand support of the beam shown below. EI = 10MNm2  Exercise 1.10 Calculate the slope and deflection 1m from the left-hand support of the beam shown below. EI = 10MNm2  18 Deflection of Beams Exercise 1.11 A beam AB of constant section, depth 400mm and Imax = 250 × 10−6 m4 , is hinged at A and simply supported on a non-yielding support at C. The beam is subjected to the given loading as shown in the figure. Determine: a. The vertical deflection of B b. The slope of the tangent to the bent centre line at C  E = 80GPa Exercise 1.12 #  Exercise 1.13 #  Introduction Point Load Uniformly Distributed Load Exercise Examples Exercises 2 — Continuous Beams 2.1 Introduction Definition 2.1.1 — Continuous Beams. -are beams that are supported on more than two supports. These beams are statically indeterminate. We will not dwell much on the derivation of the Three Moment Theorem also known as Clapeyron’s theorem. The following equations is what you will be using when dealing with continuous beams.  A1 x1 A2 x2 + MA L1 − 2MB (L1 + L2 ) − MC L2 = 6 L1 L2  (2.1) A highway bridge shown in Fig. 2.1 is a clear example of a continuous beam Figure 2.1: A continuous beam acting as a highway bridge 2.1.1 Point Load Let us take a look at the span AB in Fig. 2.2 and tackle the two triangles ∆ACD and ∆CDB. Continuous Beams 20 P B A a b L MAB = Pab L C • A x= a 3 • D x = a + 2b 3 Figure 2.2: Span with Point Load 6Ax L 6Ax L 6Ax L 6Ax L = = = = = =      2a b 1 6 Pab Pab 1 × + a+ ×a× ×b× L 2 L 3 2 L 3  3   2 b 6 Pa b Pab a+ + L 3L 2L 3  3  2 2 6 Pa b Pa b Pab3 + + L 3L 2L 6L Pab2 (2a + b)(a + b) BUT a + b = L 6L2 Pab (2a + b) BUT b = L − a L Pa Pa (L − a)(2a + L − a) = (L2 − a2 ) L L B 2.1 Introduction 2.1.2 21 Uniformly Distributed Load We take a look at Fig. #. The centroid of the parabola is at the centre of the semi-circle and it is x = L/2. 6Ax 6 2L qL2 L = × × × L L 3 8 2 3 6Ax qL = L 4  Example 2.1 The uniform beam shown in fig. 2.3 carries the loads as indicated. Determine the B.M at B and hence draw the S.F. and the B.M diagrams for the beam.  20kN 0.5m 30 kN/m B A C 2m 2m Figure 2.3: Solution 2.1 Still to be done  Example 2.2 A beam ABCDE is continuous over four supports and carries the loads as shown in fig 2.4. Determine the values of the fixing moment at each support and hence draw the S.F. and B.M. diagrams for the beam.  20kN 2m 10 kN/m 10 kN/m B A 5m D C 4m 5m E 2m Figure 2.4: Solution 2.2 Still to be done  Example 2.3 A beam ABCDE is continuous over four supports (A, B, C and D) and fixed at one the other end E. Span AB carries a point of 10kN at 0.5m from A, Span CD carries a point load 20kN at 0.5m from C. Span BC carries a udl of 30kN/m while span DE carries a 20kN/m udl. Determine the values of the fixing moment at each support and hence draw the S.F. and B.M. diagrams for the beam.  Solution 2.3 Still to be done Continuous Beams 22  Example 2.4 A uniform continuous beam ABC is built-in at support C and simply supported at A and B. AB is loaded with a udl of 20kN/m magnitude and BC has a point load of 10kN located at 1m from point B. Distance for AB = 2m and BC = 2m. Determine the reactions at the supports and draw the bending moment diagrams as well as the shear force diagrams.  Solution 2.4 Still to be done  Example 2.5 Still to be done  Solution 2.5 Still to be done 2.2 Exercise Exercise 2.1 A uniform continuous beam is built-in at one end. a. Determine the moments as well as reactions forces. b. Sketch the shear force and bending moment (True moment and correcting moment) diagrams. c. Determine the position of the point of contra-flexure nearest to the free end.  Exercise 2.2 For the continuous beam shown below, calculate the reactions at the supports. IxxAB = 20 × 10−6 m4 = IxxBC and IxxCD = 40 × 10−6 m4 . Also, calculate the moments and sketch shear force as well as bending moment diagrams.  Exercise 2.3 A continuous beam ABCD is simply supported over three spans AB = 1m, BC = 2m and CD = 2m. The first span carries a central load of 20kN and the third span a uniformly distributed load of 30kN/m. The central span remains unloaded. Calculate the bending moments at B and C; draw S.F. and B.M. diagrams. The supports remain at the same level when the beam is loaded.  Exercise 2.4 Calculate the magnitude of the reactions at the supports of the continuous beam shown below as well as the true bending moment and and correcting moment diagrams.  Exercise 2.5 Calculate the magnitude of the reactions at the supports of the continuous beam shown below as well as the true bending moment and and correcting moment diagrams. 20 kN 2kN/m 2m 4m 4m 1m 2m  2.3 Examples 2.3 23 Examples  Example 2.6 #  Solution 2.6 #  Example 2.7 #  Solution 2.7 #  Example 2.8 #  Solution 2.8 #  Example 2.9 #  Solution 2.9 # 2.4 Exercises Exercise 2.6 #  Exercise 2.7 A beam is continuous over four supports and are loaded as shown in Fig. ??. 1. Calculate the magnitude of the forces in the supports 2. Draw the shear force and bending moment diagrams. 20kN 1m 30kN 10 kN/m A B 4m 3m 4m 1m  Exercise 2.8 A beam is continuous over four supports and are loaded as shown in Fig. ??. 1. Calculate the magnitude of the forces in the supports 2. Draw the shear force and bending moment diagrams. 30kN 8 kN/m A 5 kN/m B 4m 10kN D C 3m 1m 4m 1m  24 Continuous Beams Exercise 2.9 #  Exercise 2.10 #  Exercise 2.11 #  Exercise 2.12 #  Introduction Strain Energy of Bars Castigliano’s Theorem Structures Castigliano’s theorem applied to Curved Beams Castigliano’s theorem applied to Beams Cantilever beam with a Point Load at the free end S/S beam with a Point Load at mid-point Examples Exercises 3 — Energy Methods 3.1 Introduction The energy stored within a material when work has been done is called the strain energy. Energy is normally defined as the capacity to do work and it may exist in many forms such as mechanical, thermal, nuclear, electrical, etc. The potential energy of a body is the form of energy which is stored by virtue of the work which has previously been done on that body. Strain energy is a particular form of potential energy (PE) which is stored within materials which has been subjected to strain, i.e. to some change in dimension. Definition 3.1.1 Strain energy is defined as the energy which is stored within a material when work has been done on the material. Strain Energy U = Work Done (3.1) When an axial force P is applied gradually to an elastic body that is rigidly fixed (no displacement, rotation permitted), the force does work as the body deforms. We can see this clearly in Fig. 3.1 P δ L Figure 3.1: Elastic Bar P B Area=U C O δ Figure 3.2: Load vs Displacement Energy Methods 26 This work can be calculated from U = of the application of the load. Rδ 0 P dδ , where δ is the work absorbing displacement 1 U = Pδ 2 (3.2) where U is the area under the force-displacement diagram. The work of several loads, i.e. P1 , P2 , P3 ,..., Pn acting on an elastic body is independent of the order in which the loads are applied. The work is thus U= 1 Pi δi 2∑ (3.3) where the same U is actually the energy (strain) stored in an elastic body. The unshaded area above the line ’OB’ is called the complimentary energy, a quantity which is utilised in some advanced energy methods of solution, [dro]. [dro] 3.2 Strain Energy of Bars Let us consider a bar of constant cross-sectional area A, length L and Young’s modulus E as shown in Fig. #. If the axial load P is applied gradually to result in displacement δ , the strain energy of the bar is 1 P2 L U = Pδ = 2 2AE (3.4) This can be better expressed as follows: U= 3.3 Z L 2 P 1 2 0 2AE dx (3.5) Castigliano’s Theorem Theorem 3.3.1 — Castigliano’s First Theorem. Castigliano’s theorem states that if an elastic body is in equilibrium under the external loads P1 , P2 , P3 ,..., Pn then δi = ∂U ∂ Pi (3.6) where δi is the displacement associated with load Pi and U is the strain energy of the body. The other way of writing Castigliano’s first theorem is as follows: Theorem 3.3.2 — Castigliano’s First Theorem. If the total strain energy expressed in terms of the external loads is partially differentiated with respect to one of the loads the result is the deflection of the point of application of that load and in the direction of that load. Deflection in direction of Pi will be ∂U ∂ Pi δPi = (3.7) In applications where bending provided practially all of the strain energy, δPi = M ∂M dA EI ∂ Pi A Z Castigliano’s theorem for angular movements states: (3.8) 3.4 Structures 27 Theorem 3.3.3 — Castigliano’s theorem for angular movements. If the total strain energy expressed in terms of the external moments be partially differentiated with respect to one of the moments, the result is the angular deflection in radians of the point of application of that moment and in its direction θ= M ∂M dA A EI ∂ Mi Z (3.9) where Mi is the actual or dummy moment at the point where θ is required. 3.4 Structures Displacement in the direction of the applied load is found using the following equation: n Fi Li ∂ Fi i=1 Ai Ei ∂ P δ=∑ (3.10) There would be plenty of examples for trusses/structures in the example section Statically Indeterminate/determinate Pin jointed structures can either be statically determinate or indeterminate. The challenge is when the structure is indeterminate. The following steps should be followed when approaching a pin jointed structure: • Count the number of joints, members and reactions. m + r − 2 j = 1. If 2 j < m + r, the structure is indeterminate. There is one redundant member on the structure. • The number of redundant members is equal to the degree of indeterminacy of the structure. Release the redundant members to render the structure statically determinate. • Calculate the forces in the statically determinate structure, subjected to any external loads plus the redundant reaction forces, using methods of joints of sections. • Use Castigliano’s method to calculate the required deflections and also slopes. 3.5 Castigliano’s theorem applied to Curved Beams The theorem can be applied to all types of beams, cantilever with point load or udl, simply supported beam with point load or udl, etc. I will leave you to derive the rest of the beams. You can confirm your answer by using first principles. 3.6 3.6.1 Castigliano’s theorem applied to Beams Cantilever beam with a Point Load at the free end When given any type of beam, follow this procedure: 1. Cut an elemental strip of width dx 2. Measure distance x from the free-end or the fixed-end. If it is from the fixed end, make sure that you calculate the reaction first 3. Take moments about X − X line, i.e. distance from free-end to where the strip starts, Mxx 4. Determine partial derivative of moment with respect to the load applied, i.e. ∂ Mxx /∂ P For a cantilever given, Fig. 3.6.1, this is what you do: Energy Methods 28 P dx x L Figure 3.3: Cantilever Beam with a Point Load ∂ Mxx =x ∂P Z L 1 ∂ Mxx = dx Mxx EI 0 ∂P Z 1 L Px · xdx = EI 0 Mxx = Px; δBP δBP δBP 3.6.2 Px3 = 3EI L = 0 PL3 3EI S/S beam with a Point Load at mid-point The simply supported beam, Fig. 3.6.2 is tackled a bit differently from the cantilever. In this example the beam has constant cross-sectional area A, length L and Young’s modulus E In this P a b B C A Pb L x dx dz z Pa L L Figure 3.4: Simply Supported Beam with a Point Load problem all we need to do is to take moments about X − X and Z − Z as follows: Pb x; L Pa MCB = z; L MAC = bx ∂ MAC = ∂P L ∂ MCB az = ∂P L 3.7 Examples 29 The deflection is thus 1 δ= EI δ= δ= Z a Pb bx 1 x × dx + L L EI 0 a 2 Pb x3 L2 EI 3 Pb2 a3 0 + Pa2 b z3 L2 EI 3 Z b Pa 0 L z× az dz L 0 Pa2 b3 + 3L2 EI 3L2 EI Pb2 a2 (a + L) δ= 3L2 EI But since a = L/2 = b then δ= 3.7 PL3 48EI Examples  Example 3.1 Calculate the vertical displacement at point B on the pin-jointed structure shown in Fig. #. The cross-sectional area of both members is 2000mm2 and E = 200GPa  Solution 3.1 We start by determining what the length of AB and BC is, i.e. LAB = 2m and LBC = 2m. The only joint that we will deal with is Joint ’B’. R The structure is statically determinate. j = 3, m = 2 and r = 4. 2 j = m + r Σ F ↑ = 0 = FAB sin 30◦ − FBC sin 30◦ − Q FBC = FBA − 2Q . . . Eq. 1 Σ F → = 0 = R − FAB cos 30 − FBC cos 30◦ FBC = 1.155R − FBA FAB = Q + 0.577R FBC = 0.577R − Q ◦ . . . Eq. 2 ∂ FBA ∂ FBA = 1; = 0.577 with ∂Q ∂R ∂ FBC ∂ FBC with = −1; = 0.577 ∂Q ∂R Member Length (m) Load ∂F ∂Q ∂F ∂R FL ∂ F AE ∂ Q FL ∂ F AE ∂ R AB BC 2m 2m Q + 0.577R −Q + 0.577R 1 -1 0.577 0.577 50 × 10−6 50 × 10−6 100µm 28 × 10−6 −28 × 10−6 0 Table 3.1: Table caption  Example 3.2 A Plate 5mm thick and 30mm wide is bent into the shape shown below.  Energy Methods 30 Solution 3.2 Taking moments about z − z ∂ Mzz = r sin θ ∂P Mzz = Fr sin θ ; 1 δ= EI Z Fr3 δ= EI 3π 2 (Fr sin θ )(r sin θ )rdθ 0  θ sin 2θ − 2 4  3π 2 0 3πFr3 (200)(0.2)3 (3π) δ= = = 60.3mm 4EI 4 × 62.5  Example 3.3 The structure shown below is made from a pipe with inner and outer diameters of 80mm and 100mm respectively. Calculate the resultant deflection at A due to bending. E = 200GPa.  Solution 3.3 #  Example 3.4 Determine the vertical deflection of point A on the bent cantilever as shown below, Fig. 3.5, when loaded at A with a vertical load of 25N. The cantilever is built-in at B and EI is constant throughout and is equal to 450Nm2 . What would be the horizontal deflection at point A?  B r= m 5m 12 A 200mm P = 25N Figure 3.5: Castigliano-Semicircular Solution 3.4 We now look at Fig. 3.6 and we let R become a dummy load. Mxx = Px , ∂ Mxx = xXX ∂P Mzz = (0.2 + r sin θ )P + Rr(1 − cos θ ), ∂ Mzz ∂ Mzz = (0.2 + r sin θ ), = r(1 − cos θ )XX ∂P ∂R 3.7 Examples 31 B r= 12 r cos θ m 5m X x A r sin θ 200mm R = 0N X P = 25N Figure 3.6: Castigliano-Semicircular The deflections are calculated as follows: 1 π 1 0.2 Px · xdx + (0.2 + r sin θ )P(0.2 + r sin θ )rdθ δP = EI 0 EI 0  3 0.2 P Px + (0.04r + 0.4r2 sin θ + r3 sin2 θ )dθ δP = 3EI 0 EI    P sin 2θ π 0.0667 2 3 θ 0.04rθ − 0.4r cos θ + r + − δP = X EI EI 2 4 0 P 0.0667 δP = + (0.0157 + 0.0125 + 0.003068) EI EI 0.8484 δP = XX EI Z δR = δR = δR = δR = δR = Z 1 π (0.2 + r sin θ )Pr(1 − cos θ )rdθ EI 0 Z Pr2 π (0.2 − 0.2 cos θ + r sin θ − r sin θ cos θ )dθ X EI 0  π r cos2 θ Pr2 0.2θ − 0.2 sin θ − r cos θ + EI 2 0   2 r Pr 0.2π − r(−2) + (0) EI 2 0.3431 XX EI Z Example 3.5 The steel truss (Fig. 3.7) supports the load P = 30kN. Determine the horizontal and vertical displacements of joint E. Use E = 200GPa. The cross-sectional area for all members is 500mm2 .   Solution 3.5 Referring to Fig. 3.8 We are given A = 500mm2 , P = 30kN and E = 200GPa. We will use a JOINT method starting with Joint E, D and C. Let us introduce a dummy load R at point E. Joint E Energy Methods 32 2m B • • A • D • E 2m • P C 2m Figure 3.7: Castigliano Truss Structure 2m B • • A • D • E 2m • P C 2m Figure 3.8: Solution Castigliano Truss Structure ΣFv = 0 = P + 0.707FEC ∴ FEC = −1.414PXX ∂ FEC = −1.414 ∂P ΣFH = FED + 0.707FEC − R ∴ FED = P + RXX ∂ FED ∂ FED =1 =1 ∂P ∂R R(Dummy) 3.7 Examples 33 Joint D ΣFV = 0 = FDC XX ΣFH = FDE − FDB + R ∴ FDB = P + RXX ∂ FDB ∂ FDB =1 =1 ∂P ∂R Joint C ΣFv = 0.707FCE + 0.707FCB ΣFv = −P + 0.707FCB ∴ FCB = 1.414PXX ∂ FCB = 1.414 ∂P ΣFH = 0 = 0.707FCE − 0.707FCB − FCA ΣFH = −P − P − FCA ∴ FCA = 2PXX ∂ FCA =2 ∂P Member AC BC BD CD DE CE L(mm) 2000 2828 2000 2000 2000 2828 Load 2P 1.414P P+R 0 P+R -1.414P ∂F ∂P ∂F ∂R FL ∂ F AE ∂ P FL ∂ F AE ∂ R 2 1.414 1 0 1 -1.414 0 0 1 0 1 0 1.2 1.696 0.6 0 0.6 -1.696 δP = 2.4mm 0 0 0.6 0 0.6 0 δR = 1.2mm Example 3.6 For the simply supported beam (Fig. 3.9) loaded with a uniformly distributed load, determine the maximum deflection in the middle of the beam. Let the udl be q N/m.   Solution 3.6 We can solve this problem by concentrating on half the length of the beam. Since we need to calculate the maximum deflection which occurs in the middle, we need to place a dummy load at the middle of the beam. We then need to cut the beam at distance x from the left hand support and take moments about X − X section as follows: qL P + = RB 2 2 qx2 qLx Px qx2 Mxx = RA x − = + − 2 2 2 2 x ∂ Mxx = ∂P 2 RA = Since deflection is determined using the following 1 δP = EI Z A M ∂M dA ∂P (3.11) Energy Methods 34 q N/m B A L/2 L Figure 3.9: SS Beam with UDL δP = = = = = = 3.8   Z 2 L/2 qLx Px qx2 x + − dx EI 0 2 2 2 2   Z 1 L/2 qLx2 Px2 qx3 + − dx EI 0 2 2 2   L/2 1 qLx3 Px3 qx4 + − EI 6 6 8 0 "


4 # 3 3 P L2 q L2 1 qL L2 + − EI 6 6 8  3  1 PL qL4 qL4 + − EI 48 48 128  3  4 1 PL 5qL + EI 48 384 Exercises Exercise 3.1 Calculate the vertical displacement as well as the horizontal displacement at point B on the pin jointed structure shown in the figure below. The cross-sectional area of both members is 2000 mm2 and E = 200GPa. 3.8 Exercises 35 • A 2m 2m 2m •B 10 kN • C  Exercise 3.2 Calculate the magnitude of the force R on the pin-jointed structure, shown in the figure below, if the vertical deflection at node E is zero. The cross-sectional area of all the members is the same. R •E •F 4m D• A• •C • B 10 kN 3m 3m  Exercise 3.3 Calculate the resultant displacement at point E on the pin-jointed structure shown in the figure below. The cross-sectional area for all members is 200mm2 and E = 200GPa. Energy Methods 36 • 20 kN • 60◦ 60◦ 2m 90◦ • 60◦ •  Exercise 3.4 Calculate the vertical displacement at point D and the horizontal displacement at C on the pin-jointed structure shown in the figure below. The cross-sectional area for all members is 1200mm2 and E = 200GPa. 3m B • A• •D •C 20 kN 4m 4m  Exercise 3.5 Calculate the resultant deflection at point A on the pin-jointed structure shown in the figure below. 2m • • 3m • • 10 kN 3.8 Exercises 37  Exercise 3.6 Calculate the vertical deflection at point B on the pin-jointed structure shown below. The cross-sectional are of the members in tension is 30mm2 and for those in compression is 200mm2 . E = 200GPa • 0 .6 m 0.5m • 0.8 m 1m • 10 kN •  Exercise 3.7 Calculate the resultant deflection at point D on the pin-jointed structure shown below. The cross-sectional area of the members in tension is 1000mm2 and for those in compression is 2000mm2 . E = 200GPa • • • 0.6m • 8 kN 1m  Symmetric Member in Pure Bending Unsymmetrical Bending Alternative procedure for stress determination Deflection Notation xxx Point Load Uniformly Distributed Load Examples Exercises 4 — Unsymmetrical Bending of Beams Introduction The most common type of structural member is a beam. I actual structures beams can be found in an infinite variety of: • sizes • shapes • orientations Definition 4.0.1 — Beam. A beam may be defined as a member whose length is relatively large in comparison with its thickness and depth, and which is loaded with transverse loads that produce significant bending effects as oppose to twisting or axial effects. Beams are generally classified according to their geometry and the manner in which they are supported. Geometrical classification includes such features as the shape of the cross section whether the beam is straight, curved, tapered or has constant cross-section. Beams can also be classified according to the manner in which they are supported. Some types that occur in ordinary practice are shown in the figures below, Fig. 4.1 and 4.2. There are many other types not shown. Figure 4.1: Cantilever Beam 4.1 Symmetric Member in Pure Bending • Internal forces in any cross section are equivalent to a couple. The moment of the couple is the section bending moment. • From statics, a couple M consist of two equal and opposite forces. • The sum of the components of the forces in any direction is zero. • The moment is the same about any axis perpendicular to the plane of the couple and zero about any axis contained in the plane. Unsymmetrical Bending of Beams 40 Figure 4.2: Continuous Beam-Bridge Support • These requirements may be applied to the sums of the components and moments of the statically indeterminate elementary internal forces 4.2 Unsymmetrical Bending Simple bending theory applies when bending takes place about an axis which is perpendicular to a plane of symmetry. If such an axis is drawn through the centroid of a section, and another mutually perpendicular to it also through the centroid, then these axes are principal axes. Thus a plane of symmetry is automatically a principal axis. Second moments of area of a cross-section about its principalRaxes are found to be maximum and minimum values, while the product second moment of area, xy dA, is found to be zero. All plane sections, whether they have an axis of symmetry or not, have two perpendicular axes about which the product second moment of area is zero. Principal axes are thus defined as the axes about which the product second moment of area is zero. Simple bending can then be taken as bending which takes place about a principal axis, moments are being applied in a plane parallel to one such axis. In general, however, moments are applied about a convenient axis in the cross-section; the plane containing the applied moment may not then be parallel to a principal axis. Such cases are termed unsymmetrical bending1 . The most simple type of unsymmetrical bending problem is that of skew loading of the sections containing at least one axis of symmetry as shown in Fig. #. This axis and the axis perpendicular to it are then principal axes and the term skew loading implies load applied at some angle to these principal axes. The method of solution in this case is to resolve the applied moment MA into Muu and Mvv How to approach an Unsymmetrical Bending problem: 1. Determine the position of the centroid (if not already known) 2. Calculate values of Ixx , Iyy and Ixy 3. Calculate angle θ p using tan 2θ = 1 Mechanics 2Ixy Ixx − Iyy of Materials, EJ Hearn (4.1) 4.2 Unsymmetrical Bending 41 4. Calculate principal second moments of area Ixx + Iyy ± I11,22 = 2 s Ixx − Iyy 2 2 2 + Ixy (4.2) 5. Calculate the moment M and resolve it into components of Muu = M cos θ and Mvv = M sin θ 6. Calculate combined bending stress σ= Muu × v Mvv × u + Iuu Ivv (4.3) Figure 4.3: Asymmetrical Bending 7. Find position of neutral axis on cross-section (through centroid): σ = 0 8. Identify points on cross-section which are greatest distance from the neutral axis(one tensile and one compressive) and determine x, y coordinates of each point. 9. Use the following two equations to determine the distances u and v which are both positive in the quadrant UGV. u = x cos θ + y sin θ (4.4) v = y cos θ − x cos θ (4.5) 10. Determine the maximum bending stress using Eqn. 4.3 Moment applied along the principal axis The conditions that one should consider when working with this type of a problem are: • Stress distribution acting over entire cross-sectional area to be a zero force resultant. • Resultant internal moment about y- axis to be zero. • Resultant internal moment about z- axis to be equal to M. Unsymmetrical Bending of Beams 42 • Express the three conditions mathematically by considering forces acting on differential element dA located at (0, y, z). Force is dF = σ dA, therefore FR = Fy ; 0= Z σ dA AZ (MR )y = σ My ; 0= (MR )z = σ Mz ; 0= ZA A zσ dA −yσ dA Figure 4.4: Asymmetrical Bending • If material has linear-elastic behaviour, Rthen we can substitute σ = − R A −yσ dA and after integrating, we get A yz dA = 0 • The resultant general normal stress at any point on the cross section is σ= Mz y My z + Izz Iyy y
c σmax into 0 = (4.6) BUT in out notation we will use x instead of z. You will not be penalized if you decide to use what the book says you must use. The direction is important, so make sure that z is pointing the right direction for a positive/ negative sense. Mz = M cos θ and My = M sin θ Orientation of neutral axis • Angle α of the neutral axis can be determined by applying the stress equation with σ = 0, since normal stress acts on neutral axis. The resulting equation is thus: tan α = Izz tan θ Iyy (4.7) • For unsymmetrical bending, the angle θ defining direction of moment M is not equal to angle α, angle defining inclination of neutral axis unless Izz = Iyy 4.3 Alternative procedure for stress determination I prefer this method over the other methods. Let us consider any unsymmetrical section as shown in Fig. 4.5. The assumption that we make 4.3 Alternative procedure for stress determination 43 Figure 4.5: Alternative Procedure as we start this is that the stress at any point on the unsymmetrical section us given by σ = Px + Qy (4.8) where P and Q are constants; in other words it is assumed that bending takes place about the X and Y axes at the same time, stresses resulting from each effect being proportional to the distance from the respective axis of bending. Let there be a tensile stress σ on the element of area dA. Then the force F acting on the element is F = σ dA. The moment of this force about the X axis is then σ dAy Mxx = = Z σ dAy Z (Px + Qy)ydA = Z PxydA + Z Qy2 dA but we know from statics that Ixx = Iyy = Ixy = Z Z Z y2 dA x2 dA xydA therefore substituting the above in the moment equation, we get Mxx = PIxy + QIxx (4.9) In a similar manner the moments about the Y axis is Myy = − Myy = − Z Z σ dAx (Px + Qy)xdA = − ∴ Myy = −PIyy − QIxy Z PxydA − Z Qy2 dA (4.10) Since the stresses resulting from bending are zero on the neutral axis, the equation of the neutral axis is derived by setting the stress to zero, i.e. 0 = Px + Qy P y = − = tan αN.A x Q Unsymmetrical Bending of Beams 44 4.4 Deflection The deflections of unsymmetrical sections in the directions of the principal axes may always be determined by application of the standard deflection formulae, i.e. δ= FL3 3EI (4.11) and this we know because it is the maximum deflection of a cantilever with a point load F at the free end. The vertical deflection is determined as follows: δv = (Fyy )L3 3EIxx (4.12) and the horizontal deflection δh = 4.5 (Fxx )L3 3EIyy (4.13) Notation Quantity Generic load for ODE Work Transverse Shear Force Bending Moment Slope of deflection curve Deflection Curve Symbol f (x) V (x) M(x) dv(x) ′ dx = v (x) v(x) Table 4.1: Notation 4.6 xxx # 4.6.1 Point Load # 4.6.2 Uniformly Distributed Load 4.7 Examples Example 4.1 A z-section shown in Fig.# is subjected to bending moment of M = 20kNm. The principal axes y and z are oriented as shown such that they represent the maximum and minimum principal moments of inertia, Iyy = 900 × 10−6 mm4 and Izz = 7540 × 10−6 mm4 respectively. Determine the normal stress at point P and orientation of the neutral axis.   Solution 4.1 #  Example 4.2 Due to load misalignment, the bending moment acting on the channel sections is inclined at an angle of 3◦ with respect to the y axis. If the allowable flexural stress for this beam is σal = 150MPa, what is the maximum moment, Mmax that may be applied.  4.7 Examples 45 Solution 4.2 We start the solution by determining the components of the moment Myy = −M sin 3◦ Mxx = M cos 3◦ The normal stresses due to the moment components are Myy × x (M sin 3◦ ) × x =− Iyy Iyy Mxx × y (M cos 3◦ ) × y = = Ixx Ixx σzz1 = σzz2 The combines stress is σzz = σzz1 + σzz2 (M sin 3◦ ) × x (M cos 3◦ ) × x =− + Iyy Ixx A1 = A3 = 500mm2 A2 = 800mm2 AT = 1800mm2     10 × 503 80 × 103 2 2 Ixx = 2 + 500(8.9) + + 800(11.1) = 0.3927 × 106 mm4 12 12   10 × 803 50 × 103 2 Iyy = 2 + 500(45) + = 2.46 × 106 mm4 12 12 (M sin 3◦ ) × (−50) (M cos 3◦ ) × (−33.9) + 2.46 × 106 0.3927 × 106 M = 1.17MNm −100 × 106 = − NB: I am not quite happy with the answer.  Example 4.3 A rectangular-section beam 80 mm × 50 mm is arranged as a cantilever 1.3m long and loaded at its free end with a point load of 5kN inclined at an angle of 60◦ to the horizontal axis as shown in Fig. #. Determine the position and magnitude of the greatest tensile stress in the section. What would be the vertical deflection at the end? E = 210GPa  Solution 4.3 The moments of inertia are simple to calculate and are as follows: 50 × 803 = 2.133 × 106 mm4 12 80 × 503 Iyy = = 0.833 × 106 mm4 12 Mxx = 5000 × 1300 cos 30◦ = 5629 × 103 Nmm Ixx = Myy = −5000 × 1300 sin 30◦ = −3250 × 103 Nmm Using the general method of determining stress at any point, i.e. σ= Mxx y Myy x ± Ixx Iyy Unsymmetrical Bending of Beams 46 Y P A 80mm D • X B C 50mm Figure 4.6: Rectangular section We will determine the stresses at points A(25, 40), B(25, −40), C(−25, −40) and D(−25, 40) Mxx y Myy x (5629)(40)(1000) (−3250)(25)(1000) − = 203.1MPa ± = Ixx Iyy 2.133 × 106 0.833 × 106 Mxx y Myy x (5629)(−40)(1000) (−3250)(25)(1000) σB = − = −8.021MPa ± = Ixx Iyy 2.133 × 106 0.833 × 106 σA = Using the alternative method, we must determine the constants P and Q. Since the section is symmetric about both axes, we know that Ixy = 0. Mxx = 5629 × 106 = 2.133Q ∴ Q = 2639 × 106 Myy = −3250 × 106 = −0.833Q ∴ P = 3901.56 × 106 × 106 The stresses at various points are σA = 3901.56(25) + 2639(40) = 203.1MPa σB = 3901.56(25) + 2639(−40) = 8.021MPa σC = 3901.56(−25) + 2639(−40) = −203.1MPa σD = 3901.56(−25) + 2639(40) = −8.021MPa  Example 4.4 A cantilever if length 1.2m and of the cross-section shown in Fig. 4.7 carries a vertical load of 10kN at its outer end, the line of action being parallel with the longer leg and arranged to pass through the shear centre of the section (i.e. there is no twisting of the section). Working from first principles, find the stress set up in the section at points A, B and C, given that the centroid is located as shown. Determine also the angle of inclination of the neutral axis αNA . Given: Ixx = 4 × 10−6 m4 and Iyy = 1.08 × 10−6 m4  Solution 4.4 #  Example 4.5 #  Solution 4.5 #  Example 4.6  4.8 Exercises 47 Figure 4.7: Unequal Leg subjected to vertical load Solution 4.6 #  Example 4.7 #  Solution 4.7 #  Example 4.8 #  Solution 4.8 # 4.8 Exercises Exercise 4.1 #  Exercise 4.2 #  Exercise 4.3 #  Exercise 4.4 A T-section shown below has two loads acting on it. It is supported as a cantilever of length 3m and P = 500N is acting at 2m from the fixed end and P2 = 707N.  Exercise 4.5 A beam of 3m length has a cross-section as shown below and is subjected to a constamt bending moment of 600Nm about the X-axis. Calculate: 1. the maximum stress induced in the section 2. the magnitude and the direction of the maximum deflection of the beam. The following is given Ixx = 363.05 × 10−9 m4 , Iyy = 49.72 × 10−9 m4 , x = 9.45mm, y = 27.22mm and E = 200GPa.  Exercise 4.6 #  Exercise 4.7 #  Exercise 4.8 #  48 Unsymmetrical Bending of Beams Exercise 4.9 #  Exercise 4.10 #  Exercise 4.11 #  Exercise 4.12 #  Exercise 4.13 #  Exercise 4.14 #  Exercise 4.15 #  Exercise 4.16 #  Plastic Bending of Rectangular Beams Plastic Bending of Symmetrical (I-Section) Beam Partially plastic Bending of Unsymmetrical Sections Limit Analysis-Bending Bending The principle of Virtual Work Solid Shaft Hollow shaft Exercises 5 — Inelastic Bending Introduction When the design of components is based upon the elastic theory, i.e. the simple bending or torsion theory, the dimensions of the components are arranged in such a way that the maximum stresses which are likely to result do not exceed the allowable working stress. This is obtained by taking the yield stress and dividing it by the applicable safety factor. Under normal service conditions, we want to present yielding because the resulting permanent deformation is generally undesirable. However, permanent deformation does not necessarily lead to catastrophic failure; it may only make the structure or component undesirable and considered unsafe or unfit for further use. At the outer fibres yield stress may have been exceeded but some portion of the component may be found to be still elastic and capable of carrying the load. The strength of a component will normally be much greater than that assumed on the basis of initial yielding at any position. To take advantage of the inherent additional strength, a different design procedure is used which is often referred to as plastic limit design. Definition 5.0.1 — Inelastic Bending. Inelastic materials are materials which follow Hooke’s law up to the yield stress σY and then yield plastically under constant stress (see Fig. 5.1). The figure below, Fig. 5.1, assumes material behaviour which: 1. Ignores the presence of upper and lower yields and suggests only a single yield point 2. takes the yield stress in tension and compression to be equal 3. When a plastic hinge has developed at one point, the moment of resistance at that point remains constant until collapse of the whole structure takes place due to the formation of the required number of plastic hinges at other points. 4. transverse sections of beams in bending remain in plane throughout the loading process, i.e. strain is proportional to distance from the neutral axis. It is now possible on the basis of assumption (4) to determine the moment which must be applied to produce: • maximum or limiting elastic condition in the beam material with yielding just initiated at the outer fibres. • yielding to a specific depth. • yielding across the complete section, i.e. fully plastic state or plastic hinge. Depending on the support and loading conditions, one or more plastic hinges may be required before complete collapse of the beam or structure occurs and the load required to produce this situation is called the collapse load. 5.1 Plastic Bending of Rectangular Beams Let us consider a cantilever beam loaded at the tip with a point load P which is large enough to cause yielding in the shaded area. At section a-a, the stresses on the outer fibres have just reached yield stress, but the distribution is elastic as shown in Fig. 5.3. Applying the flexure Inelastic Bending 50 σ σY εY O εY ε σY Figure 5.1: Idealized Stress-Strain Diagram Figure 5.2: Cantilever Beam subjected to load P formula Mmax = σmax S bh2 6 we find that the magnitude of the bending moment at this section is = σmax bh2 σY (5.1) 6 This moment is called yield moment because it is the moment responsible for yielding. At section b-b, the cross section is elastic over the depth of 2yi but plastic outside this depth as shown in Fig. 5.4 The stress is constant at σY over the plastic portion and varies linearly over the elastic region. The bending moment carried by this elastic region is given by the following formula Ii Mpp = (5.2) yi MY = 5.1 Plastic Bending of Rectangular Beams 51 Figure 5.3: Rectangular section-Elastic Figure 5.4: Rectangular section Partially Plastic where Ii is the moment of inertia of the elastic region of the cross section about the neutral axis. For the plastic region, which is symmetrical about the neutral axis, the bending moment is Mpp = moment of elastic portion + total moment of the plastic region       h 1 h b(2yi )2 σY + 2 σY b − yi − yi + yi = 6 2 2 2 2σY by2i σY bh2 = + − σY by2i 2 4
σY b 3h2 − 4y2i Mpp = 12 This moment is referred to as partial plastic moment. Instead of the stress at the outside increasing due to an increase in loading, more and more of the section reaches the yield stress. At section c-c, the beam is fully plastic. The stress is constant at σY over the tensile and compressive portion of the cross section. The bending moment that causes this stress distribution is called the fully plastic moment M f p . When the loading has been continues until the stress distribution is as shown in Fig. 5.5, the beam will collapse. We note that yi = 0, all that remains is h/2. The fully plastic moment is determined as follows: M f p = 2σY yA   bh h = 2σY 2 4 2 bh = 4 Inelastic Bending 52 Figure 5.5: Rectangular section Fully plastic It is worth noticing that M f p = 2/3MY and this is valid for beams of rectangular cross-section. This ratio is also called the shape factor. Other shape factors for various cross sections are shown in the table below, Tab. 5.1 For a rectangular section, this shape factor means that the beam can Cross Section Solid Rectangle Solid Circle Thin-walled Circular Tube Thin-walled Wide-flange Beam M f p /MY 1.5 1.7 1.27 1.1 Table 5.1: Shape factors of different sections carry 50% additional moment to that which is required to produce initial yielding at the edge of the beam section before a fully plastic hinge is formed. 5.2 Plastic Bending of Symmetrical (I-Section) Beam Figure 5.6: I-section Elastic and Fully Plastic 5.3 Partially plastic Bending of Unsymmetrical Sections 53 The elastic moment or yield moment MY is determined as follows I MY = σY y I= BH 3 2 b2 h3 − 12 12 ! = BH 3 bh3 − 12 12 H 2   BH 3 bh3 2 − MY = σY 12 12 H
σY = BH 3 − bh3 6H y= The fully plastic moment is found to be:
σY BH 2 − bh2 Mf p = 4 (5.3) The value of the shape factor is 1.18 for the I-beam indicating that only an 18% increase in strength capacity using plastic design procedures. 5.3 Partially plastic Bending of Unsymmetrical Sections Let us consider a T-section beam shown below in Fig. 5.7 and 5.8. Whilst stresses remain within the elastic limit the position of the neutral axis can be obtained by taking moment of are about the neutral axis. Figure 5.7: T section-Elastic It does not matter what state this section is in, i.e. elastic, partially plastic or fully plastic, equilibrium of forces must always be maintained. At any section the tensile forces on one side of the neutral axis must equal the compressive forces on the other side of the neutral axis. Summation of stresses × area above the N.A = Summation of stresses × area below the N.A In the fully plastic condition, the stresses σY will be equal throughout the section, the equation then becomes: ΣAabove = ΣAbelow = Atotal 2 For partially plastic condition we will have F1 + F2 = F3 + F4 Inelastic Bending 54 Figure 5.8: T section Fully Plastic The sum of the moments of these forces about the N.A yields that partially plastic moment Mpp . This is best explained with an example, Eg. 5.1.  Example 5.1 Determine the shape factor of a T-section of dimensions 100mm × 170mm × 20mm as shown in Fig. 5.9  Figure 5.9: T Section Elastic and Fully Plastic Solution 5.1 With reference to Fig. 5.9, ΣAi yi 100 × 20 × 160 + 20 × 150 × 75 = = 109mm ΣAi 100 × 20 + 20 × 150     100 × 203 20 × 1503 2 2 INA = + 2000(160 − 109) + + 3000(75 − 109) = 14.362 × 106 mm4 12 12 y= Yielding will start at the bottom of the cross-section when bending moment reaches MY , i.e. σY I 14.362 × 106 σY = y 109 = 131.761 × 103 σY mm3 MY = When the section becomes fully plastic the N.A is positioned such that the area below NA=half 5.3 Partially plastic Bending of Unsymmetrical Sections 55 the total area. We now must locate the plastic neutral axis, i.e. y p above the base 20 × y p = 100 × 20 + 20(150 − y p ) 20 × y p = 2000 + 3000 − 20 × y p 40 × y p = 5000 y p = 125mm The fully plastic moment is then obtained by considering the moments of forces on convenient rectangular parts of the section, each being subjected to a uniform stress σY 125 1 M f p = σY (100 × 20)(45 − 10) + σY (45 − 20)(20) × (45 − 20) + σ (125 × 20)( ) 2 2 = 70000σY + 6250σY + 156250σY = 232.5 × 103 σY mm3 232.5 × 103 131.761 × 103 = 1.765 ∴f=  Example 5.2 A cantilever is to be constructed from a T-section beam of Example 5.1 and is designed to carry a udl over its entire length of 2m. Determine the maximum udl that the cantilever beam can carry if yielding is permitted over the lower part of the web to a depth of  25mm. The yield stress of the material is σY = 225MPa. Figure 5.10: T-section Fully Plastic Inelastic Bending 56 Solution 5.2 σY σYi σY = ∴ σYi = (145 − y) y 145 − y y F1 = σY (20 × 25) = 500σY σY i.e. Average stress for the triangle F2 = (20 × y) = 10yσY 2   10σY σY 125 − y × 20(125 − y) = (125625 − 250y + y2 ) F3 = 2 y y       σY 270 − 2y 125 − y 145 − y F4 = + (100 × 20) = 1000σY 2 y y y F1 + F2 = F3 + F4   10σY 270 − 2y (125625 − 250y + y2 ) + 1000σY 500σY + 10yσY = y y y = 85.25mm We are now able to calculate the values for the forces and we find them to be F1 = 112.5kN, F2 = 191.8kN, F3 = 41.7kN and F4 = 262.6kN. The moment of resistance MR that the beam carry can now be obtained by taking moments about the neutral axis. 2 2 MR = F1 (y + 12.5) + F2 × y + F3 (125 − y) + F4 [(125 − y) + 10] 3 3 191.8 × 2 × 85.25 41.7 × 2 × (125 − 85.25) + + 262.6 [(125 − 85.25 + 10)] = 112.5(85.25 + 12.5) + 3 3 = 10996.875 + 10901.34 + 1105.05 + 12065 = 36068.265Nm The maximum bending moment present on the beam will occur at the fixed end and it will be calculated using the following formula qL2 2 = 18.034kN/m Mmax = 5.4 Limit Analysis-Bending Limit analysis is a method of determining the loading that causes a statically indeterminate structure to collapse. This method applies only to ductile materials, which in this simplified discussion are assumed to be elastic, perfectly plastic. The method is straightforward, consisting of two steps. The first step is a kinematic study of the structure to determine which parts must become fully plastic to permit the structure as a whole to undergo large deformations. The second step is an equilibrium analysis to determine the external loading that creates these fully plastic parts. We will be presenting only bending in the form of examples. 5.4.1 Bending Revisiting Fig. 5.2, as the load P is increased, section c-c at the fixed end goes through elastic and partially plastic states until it becomes fully plastic, whereas the rest of the beam remains elastic. The fully plastic section is called a plastic hinge because it allows the beam to rotate about the support without an increase in the bending moment. The bending moment at the plastic hinge is called the limiting moment MP . Once the plastic hinge has formed the beam will collapse. 5.4 Limit Analysis-Bending 57 The collapse mechanism of a beam depends on the supports. Each extra support constrain requires an additional plastic hinge in the collapse mechanism. A plastic hinge on the beam is shown by a solid circle. A simply supported beam requires only one plastic hinge whereas a beam with built-in ends will require three plastic hinges, that is if there is only one point load acting on the beam. See Figs. 5.11, 5.12 and 5.13. W 2L/3 L/3 B y A • Figure 5.11: Simply Supported Beam W 2L/3 L/3 y • • Figure 5.12: Simply Supported and Fixed W 2L/3 L/3 • y • • Figure 5.13: Fixed on both ends In general, plastic hinges form where the bending moment is a maximum, which excludes built-in supports and sections with zero shear force. The location is usually obvious for beams subjected to concentrated loads. With statically indeterminate beams carrying distributed loads, the task tends to become difficult. Sometimes there is more than one collapse mechanism in Inelastic Bending 58 which case we must compute the collapse load for each mechanism and choose the smallest collapse load as the actual limiting load. The principle of Virtual Work Virtual Displacement Definition 5.4.1 Virtual Work: states that for a structure that is in equilibrium and that is given a small virtual displacement, the sum of the work done by the internal forces is equal to the work done by the external force. Virtual displacement either linear or rotational is an imaginary or hypothetical displacement given to a mechanism or a structure and has no relation to the actual displacements produced by the real loads. The work done by the real loads acting through a virtual displacement is called virtual work. If a structure or mechanism is to remain in equilibrium the work-done by the actual loads acting through a virtual displacement must be zero. See more explanation in Examples, Eg. 5.3 and 5.4. Virtual work method on Plastic Hinges Let us consider the following simply supported loaded beam in Fig. 5.14 Let AB have a virtual W B A L/2 L/2 θ θ y 5.4.2 • 2θ Figure 5.14: Simply Supported Beam-Collapse Mechanism rotation of θ radians producing a virtual linear displacement of θ L/2 at B through which it acts. The bending moment at B MP , just before a plastic hinge is formed is considered to be negative since it opposes the work done by Wc which is the collapse load. This energy is then dissipated through till a plastic hinge is formed. So in simple terms, Work done by the load = Energy dissipated 5.4 Limit Analysis-Bending 59 θL = 2MP θ 2 4MP 4 f MY Wc = = OR L L Wc L MP = 4 Wc ×  Example 5.3 The simply supported beam ABC shown in Fig. 5.15 has a cantilever overhang and supports two loads 4W and W . Determine the value of W at collapse in terms of the plastic moment, MP , of the beam.  W 4W x B D A L/2 X L/2 C L/2 Figure 5.15: Simply Supported Beam with an overhang Solution 5.3 Referring to Fig. 5.16 We start by taking moments about B, and we don this as follows: 0 = RA L − 4W × L WL + 2 2 3W 2   L 3W x − 4W x − Mxx = 2 2 RA = At x = L/2 we have MD = 3W L/4 and at x = L we have MB = −W L/2. This means that the moment that will cause collapse is MP = MD = 3Wu L/4 because it is lowest of the two moments. The bending moment diagram for the beam has been constructed. Clearly as W is increased, a plastic hinge will form first at D, the point of application of the load 4W . Thus at collapse 3Wu L 4 4MP Wu = 3L MP = where Wu is the value of W that will cause collapse. The formation of a plastic hinge in a statically determinate beam produces large, increasing deformation which ultimate result in failure with no increase in load. In this condition the beam behaves as a mechanism with different lengths of beam rotating relative to each other about the plastic hinge. In a statically indeterminate system the formation of a single plastic hinge doesn’t necessarily mean collapse. We can demonstrate this by an example, in which a propped cantilever beam is shown in example 5.4. The bending moment diagram may be drawn after the reaction at C has been determined. Inelastic Bending 60 WL 2 3W L 4 Figure 5.16: Bending Moment Diagram  Example 5.4 A propped cantilever beam shown in Figure 5.17 is loaded at mid-point with a point load W . Determine the load that will cause the collapse.  W L/2 L/2 Figure 5.17: Propped Cantilever with W Solution 5.4 Using the virtual work method, let us refer to Fig. 5.18, Wu y = MP (θ ) + MP (2θ ) Wu Lθ = 3MP θ 2 6MP Wu = L As the value of W is increased a plastic hinge will form first at A where the bending moment is greatest. This doesn’t mean that the beam will collapse. Instead, it behaves as a statically determinate beam with a point load at B and a moment MP at A. Further increases in W will eventually result in the formation of a second plastic hinge at B when the bending moment B reaches the value of MP . The beam now behaves as a mechanism and failure occurs with no further increase in the load. The elastic bending moment diagram has a maximum at point A. After the formation of the plastic hinge at A, the bending moment remains constant while the bending moment at B 5.4 Limit Analysis-Bending θ θ y • 61 • 2θ Figure 5.18: Virtual Work Method-Propped Cantilever increases until the second plastic hinge forms. This distribution of moments tends to increase the ultimate strength of statically indeterminate structures since failure at one section leads to other portions of the structure supporting additional load. It must be noted that in this example, it is unnecessary to determine the elastic bending moment diagram to solve for the ultimate load which we obtained by using virtual work method. Plastic hinges forms at beam sections where the bending moment diagram attains a peak value. It then follows that for beams carrying a series of point loads, plastic hinges are located at the load positions. This will be shown by a figure and subsequent figures showing mechanisms. A propped cantilever that supports two point loads is shown in Fig. 5.19 below. Three possible mechanisms (see Figs. 5.20, 5.21 and 5.22) are possible and each possible mechanism should be analysed and the lowest ultimate load gets selected. W1 L1 W2 L3 L2 Figure 5.19: Propped Cantilever with a series of loads y1 y2 • • Figure 5.20: Possible Mechanism No. 1 Inelastic Bending 62 y2 y1 • • Figure 5.21: Possible Mechanism No. 2 y1 • • Figure 5.22: Possible Mechanism No. 3 Example 5.5 A propped cantilever beam of Fig. 5.4 is 10m long and is required to carry a load of 100kN at mid-span. If the yield stress of mild steel is 300MPa, suggest a suitable section using a load factor against failure of 1.5.   Solution 5.5 The required ultimate load of the beam is 1.5 × 100 = 150kN. The required plastic moment, MP is calculated as follows MP = Wu L = 250kN 6 The minimum plastic modulus of the beam section is ZP = MP = 833, 300 mm3 σY A universal beam 406 × 140 × 46kg/m is found to have a plastic modulus of 886.3cm3 > This section therefore possesses the required ultimate strength and includes a margin to allow for its self-weight. Note that unless some allowance has been made for self-weight in the estimate of the working loads, the design should be rechecked to include this effect.  Example 5.6 Determine the force P required to maintain equilibrium on the following lever.  Solution 5.6 Let θ be a small virtual rotational displacement. Then −θ will be a linear displacement of A and 3θ a linear displacement of B. The work done by real loads through these displacements must be zero 0 = 400 × 3θ − Pθ ∴ P = 1200N  Example 5.7 The beam shown in the figure below, Fig. # is pin-jointed at point C. Determine the value of P for equilibrium.  5.4 Limit Analysis-Bending 63 400N P • 1m 3m Figure 5.23: Equilibrium-Lever Solution 5.7 Let y be a virtual displacement at C. Then 0.5 1 and yE = y × 4 0.8 0 = PyB − 60yE y 0.5y 0 = P × − 60 × 4 0.8 P = 150N yB = y ×  Example 5.8 A double symmetric hollow box shown below in Fig. 5.24 has a yield stress σY = 220MPa. It is subjected to a bending moment M of such magnitude that the flanges yield but the webs remain linearly elastic. Determine the magnitude of the moment M if the dimensions of the cross section are B = 150mm, b = 130mm, H = 200mm and h = 160mm.  B h H b Figure 5.24: Box Section Solution 5.8 #  Example 5.9 A rectangular steel beam 120mm by 60mm is simply supported at its ends 2m apart, and supports a central point load. If the 120mm side is vertical and the yield stress for the material is 260MPa, determine: Inelastic Bending 64 1. 2. 3. 4. the load when yielding first occurs the load which will cause yielding to a depth of 30mm the length of the beam over which yielding had occurred the load required to produce a plastic hinge P L Figure 5.25: Cantilever Beam with a point load  Solution 5.9 The solution for this problem can actually be summed up by a diagram. MP θ = Py = PLθ MP Pu = L L P y θ Figure 5.26: Cantilever Beam with a point load-Solution  Example 5.10 A cantilever of a rectangular section 100mm by 50mm is mounted with the 100mm side vertical and supports a U.D.L of 12kN/m over its entire length. Determine the maximum length for this cantilever before collapse occurs, if the yield stress for the material is 270MPa. For what length would yielding have occurred?  Solution 5.10 #  Example 5.11 #  Solution 5.11 # 5.5 Solid Shaft  Example 5.12 # 65  Solution 5.12 #  Example 5.13  Solution 5.13 #  Example 5.14 #  Solution 5.14 # 5.5 Solid Shaft Let us consider a solid shaft shown in Fig. 5.27. Let us recap on the equations that are useful in this section: J= πd 4 32 (5.4) τJ πd 3 τ = (5.5) r 16 When the torque is applied in the elastic core, the shear stress reaches τY , then we will have T= TE = πdE3 τY 16 (5.6) When the torque is beyond TE then plasticity will result and we will use the following equation to calculate the plastic torque TP =
πτY 3 d − dE3 12 (5.7) The above was obtained as follows: TP = Z r RE πr2 τY dr
2πτY 3 r − rE3 but note that: r = d/2 3   2πτY d 3 dE3 = − 3 8 8
πτY 3 d − dE3 = 12 = The total torque is obtained by adding the elastic and plastic torques together
πdE3 πτY 3 τY + d − dE3 16 12 3 τY πd τY πdE3 − = 12 48  τY π d3 = d3 − E 12 4 T= For a fully plastic condition, dE = 0 and the equation simplifies to TFP = τY πd 3 12 (5.8) Inelastic Bending 66 τY τ dE d Figure 5.27: Solid Shaft The shape factor is obtained the same way we did with other cross-sections, which is simply a ratio of the two torques TFP 16πτY d 3 = TE 12πτY d 3 4 = 3 f=  Example 5.15 A Steel shaft of φ = 100mm, 1m long is under 30kNm torque in the elastic core. Determine the diameter of the elastic core of the shaft and the angle of twis. What is the value of the fully plastic torque of the shaft. Given: τY = 120MPa and G = 80GPa  Solution 5.15 Let us start by determining the easiest part which is the diameter of the elastic core   dE3 τY π 3 d − T= 12 4   30, 000 × 12 3 3 −dE = 4 − 0.1 π(120 × 106 ) dE3 = 180.281 × 10−4 dE = 56.49mm 5.6 Hollow shaft 67 The polar moment of intertia πd 4 32 0.1 × π = 9.817 × 10−6 m4 = 32 J= The elastic torque is therefore πτY dE3 16 π(0.05649)3 (120 × 106 ) = 16 = 4, 247Nm TE = The plastic torque is determined as follows:
πτY 3 d − dE3 TP = 12 π(120 × 106 )(0.13 − 0.056493 ) = 12 = 25, 753Nm The total torque can be confirmed as follows T = TE + TP = 4, 247 + 25, 753 = 30, 000Nm The fully plastic torque can be determined as follows: πτY d 3 12 π(120 × 106 )(0.1)3 = 12 = 31, 416Nm TFP = How do we then determine the angle of twist? It’s very easy τ G TL φ= JG γ= 30, 000 × 1 (9.817 × 10−6 )(80 × 109 ) = 0.0382 rad = 5.6 Hollow shaft Let us consider a hollow shaft with an internal radius R1 and external radius R subjected to a torque large enough to produce yielding to a radius R2 , see Fig. 5.28. Now if we ignore the central hole
πτY T= 4R3 − R32 (5.9) 6 Inelastic Bending 68 τ τY τ R1 R2 R Figure 5.28: Hollow Shaft Consider now an imaginary shaft of radius R1 . Now, the torque on it will be T= πR31 τ 2 (5.10) From the law of triangles τ τY = R1 R2 R1 τ = τY R2 Therefore the torque on the imaginary shaft is given by πR31 R1 × τY 2 R2 πR41 = τY 2R2 T= T= πR41 τY 2R2 (5.11) 5.7 Exercises 69 Now, the partially plastic torque can be determined as follows
πR4 πτY 4R3 − R32 − 1 τY 6 2R2
πτY 3 4 = 4R R2 − R2 − 3R41 6R2 TPP = TPP =
πτY 4R3 R2 − R42 − 3R41 6R2 (5.12) The fully plastic torque occurs when R2 = R1 , i.e.,
πτY 4R3 R1 − 4R41 6
πτY = 4R3 − 4R31 6
2πτY 3 R − R31 = 3 TFP = TFP =
2πτY 3 R − R31 3 (5.13) Tips when Working with Shafts Description Total Torque Torque in Elastic core Torque in Plastic condition Slope of deflection curve Deflection Curve Formula to Use T= πτY d 3 − 3 dE 4 12 πτY dE3 16 πτY (d 3 −dE3 ) TP = 12 dv(x) ′ (x) = v dx TE = v(x) Table 5.2: Tips when Working with Shafts 5.7 Exercises Exercise 5.1 Determine the plastic moment and shape factor of a beam of solid circular cross-section having a radius r and yield stress σY .  Exercise 5.2 Determine the plastic moment and shape factor for a thin-walled box girder whose cross-section has a breadth b and depth d and a constant wall thickness t. Calculate the shape factor f for b = 200mm and h = 300mm  Exercise 5.3 A beam having the cross section shown below is fabricated from mild steel which has a yield stress of 300MPa. Determine the plastic moment of the section and its shape factor. The cross-section has a common thickness of 15mm. Inelastic Bending 70 300mm 75mm 250mm  Exercise 5.4 A cantilever beam of length 6m has an additional support at a distance of 2m from its free end as shown in the figure below. Determine the minimum value of W at which collapse occurs if the section of the beam is identical to the section above. State clearly the form of the collapse mechanism corresponding to this ultimate load. 2W W B A 2m D C 2m 2m  Exercise 5.5 A beam of length L is rigidly built-in t each end and carries a uniformly distributed load of intensity w along its complete span. Determine the ultimate strength of the beam in terms of the plastic moment, MP , of its cross- section.  Exercise 5.6 A simply supported beam has a cantilever overhang and supports loads as shown below. Determine the collapse load of the beam, stating the position of the corresponding plastic hinge. 5.7 Exercises 71 2W W W L/3 L/3 L/3 L/2  Exercise 5.7 Determine the ultimate strength of the propped cantilever shown below and specify the corresponding collapse mechanism. W L/3 W L/3 L/3  Exercise 5.8 A beam shown below has a length L = 3m, a load factor, L.F = 2, shape factor, f = 1.16, elastic section modulus, ZE = 270.3 × 10−6 m3 and the yield stress of the material is σY = 310MPa. N.B: MP = ZP σY . 1. Calculate the maximum value of the ultimate load responsible for collapse of the beam. 2. Calculate the maximum bending stress induced in the beam. W L/3 2L/3  Exercise 5.9 A T-section shown below is constructed as a cantilever beam of length 2m. The yield stress, σY = 300MPa. Determine: a) The position of the neutral axis if yielding is permitted over the lower part of the web to a depth of 25mm. b) The moment of resistance of the section if the plastic penetration is 25mm. Inelastic Bending 72 c) The uniformly distributed load over the whole span that will cause a plastic penetration of 25mm.  Exercise 5.10 For the beam shown in the figure below, determine the mode of collapse and the value of the collapse load. 1m W • 3m 3m  Exercise 5.11 The two beams shown below both have a shape factor, f = 1.5, yield stress, σY = 200MPa but different cross sections. The Beam AC has a rectangular cross-section 75mm × 40mm with 75mm lying vertically while beam CD has a rectangular cross-section 100mm × 60mm with 75mm lying vertically. Hint: MP = f ME = f σY ZP = f σY bd 2 6 Determine the value of the load Wu that will cause collapse of the beam W 1m A B • D C 2m 3m  Exercise 5.12 A mild steel coupling in a metal working process has a diameter of 40mm and length L = 250mm. It is subjected to a torque of 1800Nm which is known to have caused shear yielding in the shaft. τY = 120MPa and G = 80GPa  Exercise 5.13 A beam is fixed at both ends and loaded with a point load at B. The shape factor is 1.18 and the load factor is 2.2. If the yield stress is 300MPa and ZE = 305 × 10−6 m4 , determine the permissible load as well as the maximum stress.  5.7 Exercises 73 Exercise 5.14 #  Exercise 5.15 #  Exercise 5.16 #  Bibliography [1] D.L. Logan, "A First Course in Finite Element Method, 4th Ed.” [2] P. P. Chandrupatla and R. J. Belegundu, "Introduction to Finite Elements in engineering”, 4th , Pearson-Prentice Hall." [3] M.J. Fagan, "Finite Element Analysis, Theory and practice,Pearson-Prentice Hall." [4] J. Drotsky, "Strength of Materials for Technologists, Self-published." [5] P.P. Benham, et.al, "Mechanics of Engineering Materials, Prentice Hall."