ARCE 302-Structural Analysis Lecture Notes, Fall 2009 Ansgar Neuenhofer Associate Professor, Department of Architectural Engineering California Polytechnic State University, San Luis Obispo Copyright © 2009 Ansgar Neuenhofer (photo by author) TABLE OF CONTENTS Review of Statics 1.1 Goal of structural analysis 1.2 Structural idealizations 1.3 Summary of properties of moment and shear force diagrams 1.4 Example 1.1 1.5 Frames 1.6 Example 1.2 1.7 Statical determinacy-Instability-Degree of indetermincay Problems 1 10 14 Principal of Virtual Forces 2.1 Introductory remarks 2.2 Virtual work-Principle of virtual forces 2.3 Procedure for analysis 2.4 Principle of virtual forces for trusses 2.5 Principle of virtual forces for beams 2.6 Principle of virtual forces for frames 2.7 Example 2.1 2.8 Example 2.2 2.9 Integration tables 2.10 Summary 2.11 Example 2.3 2.12 Example 2.4 2.13 Example 2.5 2.14 Example 2.6 2.15 Shear deformation Problems 18 18 18 19 19 19 19 20 22 23 25 27 28 20 32 35 38 The Force Method 3.1 Introduction 3.2 Discussion 3.3 Example 3.1 3.4 Example 3.2 3.5 Example 3.2 (Alternative) 3.6 Example 3.3 3.7 Example 3.4 3.8 Force method for arbitrary degree of statical indeterminacy 3.9 Maxwell's law 3.10 Summary of force method 3.11 Example 3.5 3.12 Example 3.6 3.13 Example 3.7 3.14 The force method for space frames Problems 49 49 49 50 53 55 57 60 64 64 65 65 70 73 75 78 The Slope-Deflection Method 4.1 General remarks 4.2 End moments for prismatic members 4.3 Example 4.1 4.4 Example 4.2 4.5 Example 4.3 4.6 Example 4.4 4.7 Summary of procedure 4.8 Comparison between slope-deflection and force methods 4.9 Fixed-end moments Problems 88 88 89 93 95 97 99 102 103 104 106 The Moment-Distribution Method 5.1 Introduction 5.2 General description 5.3 Example 5.1 5.4 Example 5.2 5.5 Summary of steps 5.6 More discussion and illustration 5.7 Example 5.3 5.8 Summary Problems 113 113 113 114 116 118 119 120 122 123 Approximate analysis of building frames under lateral load 6.1 Introduction 6.2 Discussion 6.3 The Portal Method 6.4 The Cantilever Method Problems 126 126 126 128 133 143 Influence Lines 7.1 Introduction 7.2 Müller-Breslau principle 7.3 Use of influence lines 7.4 Example 7.5 Properties of influence lines of statically determinate structures 7.6 Influence lines of statically indeterminate structures Problems 145 145 148 149 150 151 152 156 Review of Statics 1.1 Goal of structural analysis The objective of a structural analysis is to determine the force (stress) and displacement (strain) demand of structures using a mechanical model The analysis must be both as economical as possible and as accurate as necessary Since the exact mechanical relations are extremely complicated, we rely on many approximations that are more or less accurate With these approximations and simplifications we map the real structure onto our mechanical model It is important that we are aware of these simplifications in order to judge whether or not certain models are appropriate After selecting an appropriate structural model, we analyze the model under the most unfavorable combination of loads The results of structural analysis are the internal forces and deformations that form the basis for, say, reinforced concrete, steel or timber design We have to always remember that the results of any structural analysis can never be better than the underlying model A crude model remains a crude model and yields crude results, no matter how many digits we include in our analysis, in short garbage in-garbage out In this class we talk little about modeling but assume that we already have an appropriate mechanical representation of the real structure Thus we are omitting a very important step of engineering work, if not the most important, a fact we should always be aware of 1.2 Structural idealizations All structures are three-dimensional In structural analysis we usually work with one- or two-dimensional idealizations of the real structure (a) 3-dimensional structural elements (rarely used in structural engineering) (b) 2-dimensional structural elements (plate, shells) Ly requirement H H Lx H Ly Lx (c) 1-dimensional structural elements Structural elements whose two dimensions (width and height) are small compared to their length are commonly called truss (axial force response) or beam (bending moment response) elements In this class, we often use the term frame member or frame element to denote a combination of truss and beam elements We represent a frame element by its axis, which is the connection of the centroids of the cross-sections In this class, we will be dealing with one-dimensional structural elements (truss, beam and frame members) only We use those to model plane (two-dimensional) structures and, to a smaller extent, space (three-dimensional) structures 1.3 Summary of Properties of Moment and Shear Force Diagrams In the following, we summarize key properties of shear force and bending moment diagrams • • • • • In beam segments without distributed loading, the shear force is constant and the bending moment is linearly varying In regions with a uniformly distributed load the shear force varies linearly and the bending moment is a quadratic parabola In general, if the distributed load is of order n , the functions for the shear force and bending moment are of order n+1 and n+2 , respectively At points where a concentrated force (a reaction force or an externally applied force) is applied the shear force is discontinuous It “jumps“ upward or downward according to the direction of the force The moment function has a change in slope at that point but is continuous An external moment causes a jump in the bending moment It does not change the slope of the moment function, nor does it affect the shear force at that location The shear force is the derivative of the bending moment Hence the moment function is one degree higher than the shear force function When the shear force is zero, the bending moment takes on its maximum w ML P A B L Ay wL + V =0 V By − P M linear no change in slope ML + linear quadratic M max linear change in slope no change in slope Fig 1.1 Properties of shear force and bending moment diagrams linear 1.4 Example 1.1: Beam with internal hinges 50 kN 10 kN/m [m] 5.00 2.00 6.00 2.00 5.00 Fig 1.2 Beam with internal hinges Problem: Draw the bending moment and shear force diagrams for the beam in Fig 1.2 Shown in Figure 1.2 is a beam with two internal hinges The first important observation is that this structure is not a single rigid body If we remove the beam from its supports, it ceases to be rigid We can easily see, that we can rotate against because of the hinge The consequence for calculating the support reactions is that we have to break the structure apart and look at free-body diagram of individual parts Differently put, we cannot calculate the support forces by looking at the structure as a whole because there are more unknown forces than there are equilibrium equations We have four unknown support reaction but only two equilibrium equations 50 10 A B FA F E D C FB FE FF Fig 1.3 Free-body diagram of whole structure 50 10 FA FB FC FC FD FD FE FF Fig 1.4 Free-body diagram of individual parts We cut the beam at the internal hinges resulting in the three free-body diagrams in Fig 1.4 In applying the equilibrium equations to the free-body diagrams it is important to find the right starting point The center portion CD of the beam contains only two unknown forces while the exterior portions contain three unknowns Hence ∑ MC = → FD = − ⋅ 10 ⋅ ⋅ = −30 (1.1) ∑F y =0 → FC = 30 − 50 − 60 = −80 We have applied the 50-kN concentrated force to the center free-body diagram Applying the force to the left-hand portion of the beam is an alternative that would affect the results for FC but not the final results We now apply forces FC and FD to the exterior portions of the beam and solve for the support reactions FA , FB and FE , FF , respectively Caution: Since the numerical results for FC and FD are negative, we change the direction of the arrows representing these forces (see Fig 1.5) FA 80 FB 30 FE FF Fig 1.5 Free-body diagrams of exterior parts after calculating FC and FD Applying the equilibrium equations to the free-body diagrams in Fig 1.5 yields ∑M A =0 ∑ MF = → FB = 112 ∑F =0 → FA = −32 → FE = 42 ∑ Fy = → FF = −12 y (1.2) We have now calculated all support reactions Before drawing the shear force and bending moment diagrams, it is good practice to redraw the free-body diagrams and label all forces by their magnitude 50 10 32 112 80 80 30 42 30 12 Fig 1.6 Free-body diagrams with all forces known 80 V [N] + 30 12 − 30 32 160 M [Nm] 60 − C A B + D E F 45 Fig 1.7 Shear force and bending moment diagrams Unlike the previous examples, in which we wrote explicit equations for the shear force and bending moment variation, we now use our experience gained in previous examples and try to draw the internal force moment diagrams directly from the free-body diagrams in Fig.1.6 Shear force (1) At support A the 32 force acts down Hence the shear force jumps down by 32 (2) Nothing “happens” to the beam between supports A and B Hence the shear force is constant (3) At support B a 112 force acts up Hence the shear force jumps from –32 to –32+112=80 (4) Nothing “happens” to the beam between support B and hinge C Hence the shear force is constant (5) At hinge C the applied force 50 acts down Hence the shear force jumps from 80 to 80-50=30 Note that the two 80 forces cancel when moving across hinge C (see Fig 1.6) (6) “A lot happens” between hingesC and D A total force 10 ⋅ = 60 acts down However, this force is uniformly distributed This cause the shear force “to go down gradually” from 30 to 30-60=-30 Note that the two 30 forces cancel when moving across hinge D (7) Nothing “happens” to the beam between hinge D and support E Hence the shear force is constant (8) At support E the 42 force acts up Hence the shear force jumps from –30 to –30+42=12 (9) Nothing “happens” to the beam between supports E and F Hence the shear force is constant (10) Finally, we check whether the shear force calculated between supports E and F is consistent with the support force at F This is obviously the case Bending moment (1) At support A the 32 force acts down This force causes a linearly varying bending moment increasing from at support A to 32 ⋅ 5=160 at support B Since the support force at A acts downward (preventing uplift) it causes tension at the top and compression at the bottom of the beam The bending moment is thus negative The bending moment must not jump at support B On the other hand it must be zero sat the internal hinge C Consequently, the moment has to linearly decrease from 160 at support B to zero at hinge C (2) The beam segment CD is a simply supported beam under a uniformly distributed 10-kN/m load spanning from hinge C to hinge D The moment variation is a quadratic parabola whose maximum value is L2 62 = 10 ⋅ = 45 kNm 8 This moment is positive since it produces tension at the bottom of the beam M =w (1.3) (3) We now turn to the left-hand segment of the beam At hinge D the 30 force acts down This force causes a linearly varying bending moment increasing from at hinge D to 30 ⋅ 2=60 at support E Since the force at D acts downward, it causes tension at the top and compression at the bottom of the beam The bending moment is thus negative The bending moment must not jump at support E On the other hand it must be zero at support F Consequently, the moment has to linearly decrease from 60 at support E to zero at support F As a quick check we argue along the same line starting from support F At support F a 12-kN force acts down This force causes a linearly varying bending moment increasing from at support F to 12 ⋅ 5=60 at support E Since the support force at A acts downward (preventing uplift) it causes tension at the top of the beam and the moment is negative Whenever experience and problem complexity permit, we should try to distance ourselves from blind mathematics and adopt the approach followed in this example Remarks: The bending moment diagram changes its slope at C but does not change the slope at D Can you explain? The shear force jumps at C but is continuous at D Can you explain? Fig 1.8 Sketch of deflected shape 1.5 Frames We now apply the concepts learned before to find the internal force diagrams of frame structures w w C (a) (b) Ax A H H C B A B Ay L Bx By L Fig 1.9 Three-hinge frame (a) system and loading; (b) free-body diagram of whole structure Problem: Draw the shear force, axial force and bending moment diagrams for the frame in Fig 1.9(a) Solution: (1) As always, we first calculate the support reactions Since there are four unknown reactions but only three equilibrium equations for the structure as a whole, we have to dismember the structure to find the reactions We find the vertical reactions by looking at the structure as a whole (Fig 1.9b) Equilibrium for the whole structure also requires that the horizontal support forces Ax and Bx are equal and opposite L L ∑ M B = → Ay = w ∑ Fy = → By = w ∑ Fx = → Ax = Bx (1.4) w w Cx Cy H Cy Cx Ax B A w L L/2 Bx w L/2 L Fig 1.10 Three-hinge frame Cut at hinge to find horizontal reaction forces (2) Next, we cut the structure at hinge C and formulate equilibrium for one of the two resulting free bodies (Fig 1.10) Since we know the vertical reaction force from Eq (1.4), the two free-body diagrams in Fig 1.10 involve three unknown forces each L L L L L2 L2 (1.5) ∑ MC = = w ⋅ − w ⋅ − Bx ⋅ H → Bx = w 8H → Ax = w 8H due to Ay due to load Important: Since a hinge cannot transfer a moment, there is no internal moment in Fig 1.10 at point C Also note that by selecting C as the moment reference point, we avoid referencing the forces C x and C y transferred though the hinge x M P P V M I L2 8H w M V H x w V w II L2 8H L w w L L2 8H III w x P L Fig 1.11 Three cuts for internal forces (3) With reaction forces known, we are now in the position to find the internal force variation by making cuts at arbitrary locations along the girder and the two columns Figure 1.11 shows the resulting free-body diagrams Equilibrium for the three sections requires Section I ∑ Fx = → V = −w L2 8H ∑ Fy = → P = −w L2 L2 8H ∑ Fy = → V = w − wx L ∑ M = → M = −w 8H x Section II ∑ Fx = → P = −w L (1.6) ⎛ ⎞⎟ ⎜⎜ ⎟ ⎜⎜ L2 L x ⎟⎟ ∑ M = → M = w ⎜⎜⎜ − + x − ⎟⎟⎟⎟ ⎟⎟ ⎜⎜due toA due tow ⎠ due toAy x ⎝ ⎟ Section III ∑ Fx = →V = w L2 8H L2 L ∑ Fy = → P = −w ∑ M = → M = −w 8H x Note that because of the 90 degree angle at the column-girder junction, the axial force in the column turns into a shear force in the girder, and the shear force in the column turns into an axial force in the girder (4) Before drawing the internal force diagrams, it is important to emphasize that we should consider the internal force diagrams as a proper connection of values calculated at specific locations rather than plots of mathematical relations Columns: The axial force in the columns is constant and equal to the vertical reaction forces Clearly, the columns are in compression The shear force in the columns is also constant and equal to the horizontal reaction forces According to our sign convention, the shear force is negative for the left column and positive for the right column The bending moment in the columns varies linearly from zero at the pin supports to a certain value at the column-girder junction We calculate this value by substituting x =H into the moment expression for sections I or III, or x =0 into the moment expression or Section II Hence 7 7.1 Influence Lines Introduction max compression max tension Fig 7.1 Illustration of influence line and internal force diagram 145 7.1.1 General Remarks When designing structures, we have to determine the most unfavorable set of internal design forces due to all possible external design loads or load combinations In many cases, we have to consider movable loads whose most unfavorable positions cannot be readily determined In these cases, influence lines help us to determine the load positions for maximum effects An influence line is defined as a diagram that describes the variation of a load effect as a concentrated force of unit magnitude moves across the structure We can determine influence lines for all possible load effects like internal forces, stresses, displacements, strains, etc In this class, we focus on influence lines for internal forces in beams Throughout this chapter, we have to strongly differentiate between influence lines for an internal force and internal force diagrams Internal force diagrams, like moment, axial force or shear force diagrams show the value of an internal force at several, usually all locations in the structure due to a stationary loading An influence line, on the other hand, shows the variation of an internal force at one particular location say at location r as a function of the position of a unit load Figure 7.1 helps us distinguish between the concepts of influence lines and internal force diagram If we fix position of the truck in Fig 7.1 and look at all the member forces that this load position causes, we are considering an internal force diagram If, in contrast, we look at one specific member of the truss and investigate how its force varies with the position of the truck, the result of that analysis is the influence line for that member force 7.1.2 Internal Force Diagram Revisited Let us first recall the idea of an internal force diagram and look for the bending moment diagram M of the beam with cantilever shown in Fig 7.2 P w A a L Fig 7.2 Beam with cantilever (internal force diagrams revisited) We find the shear force and bending moment by cutting the beam at an arbitrary section x and considering equilibrium of the portion of the beam on one side of the section As always, we define the bending moment as positive when it causes tension at the bottom of the beam Figure 7.3 shows the two free-body diagrams w P M x1 A V M V x2 Fig 7.3 Free-body diagrams to calculate internal forcesV and M Applying the equations of equilibrium gives M (x1 ) V (x1 ) = Ax1 − w =A x12 ≤ x1 ≤ L ≤ x1 ≤ L M (x ) = −Px V (x ) = P 146 ≤ x2 ≤ a ≤ x2 ≤ a wL a with A = −P L (7.1) By plotting the preceding equations, Fig 7.4 shows the shear force and bending moment diagrams P + A + V − wL − A Pa − wL2 / + − M Fig 7.4 Moment and shear force diagrams for beam with cantilever 7.1.3 Influence Line by statics For simple structures, we can determine influence lines directly For the beam with cantilever considered above, let us determine the influence line for the bending moment M r and the shear forceVr at midspan Thus we need to determine M r andVr as a function of the position x of a unit load F = (see Fig 7.5) x F =1 r L /2 L /2 a Fig 7.5 Find influence lines for overhanging beam Obviously, we need to consider two cases, case is for the unit load acting at the left-hand side of locations r and case for the unit load acting at the right-hand side of location r The figure below shows the free-body diagrams corresponding to the two cases x F =1 r A L Case 1: x ≤ r Mr Vr L/2 A L Case 2: x > Mr Vr L/2 Fig 7.6 Free-body diagrams to find influence lines "M r " and "Vr " by statics Equilibrium for Case gives Mr Vr L L L −x L L x − 1⋅ ( − x) = ⋅ − 1⋅ ( − x) = 2 2 L x = A−1 = − L = A⋅ L L 0≤x ≤ (7.2) L < x ≤ L +a L < x ≤ L +a (7.3) 0≤x ≤ Equilibrium for Case gives Mr Vr L L −x L = ⋅ = (L − x ) 2 L L −x =A= L = A⋅ 147 Plotting the above equations gives us the two desired influence lines − + − "M r " L a + L + L /2 a L /2 "Vr " a Fig 7.7 Influence lines "M r " and "Vr " for overhanging beam It is important to note that the coordinate x in Figs 7.5 and 7.6 as well as Eqs 7.2 and 7.3 denotes the location of the load and not that of the bending moment or the shear force as in Eq (7.1) In order to clearly distinguish between force diagram and influence line it is helpful to use quotes to denote influence lines 7.2 Influence lines by kinematics-Müller-Breslau principle The direct method of determining influence lines (see previous section) is limited to simple structures A general method to determine influence lines can be derived from the principle of virtual displacements We illustrate the procedure using the same sample structure as before (see Fig 7.5) To find the influence line for the bending moment M r , we create a mechanism at the section r where the influence line is desired, i.e we remove the resistance of the beam against bending at that section We then impose a rotation at that section and set up the virtual work equation, i.e we equate internal and external work We obtain M r ⋅ ϕr′ + F ⋅ vm′ = (7.4) where M r is the bending moment at location r due to F at location m and vm′ is the deflection at m due to the imposed deformation ϕr′ If we let Fm = , ϕr′ = −1 , drop the ”prime”-symbol and consider a variable location m we can write M r = vm = v(x ) (7.5) Hence the bending moment at r due to a unit load is equal to the displacement at the location of the load due to a unit rotation at r But the moment at r due to a unit load acting at an arbitrary location x is by definition the influence line for the moment at r Hence " M r " = v(x ) (7.6) due to ϕr = −1 Figure 7.8 illustrates the procedure The result is the same as that obtained by the direct method (see Fig 7.7) 148 r m F =1 L /2 L /2 Mr a F =1 − + L a "M r " vm ϕr = −1 Fig 7.8 Influence line "M r " = deflected shape due to ϕr = −1 In the same manner we can determine influence lines for a shear force V We need to remove the resistance to the shear force and introduce a complementary, that is a vertical deformation of unit magnitude in the direction opposite to the positive sign of that force (see Fig 7.9) The desired influence line is then the deflected shape resulting from the imposed deformation The result is the same as that of "Vr " in Fig 7.7 Mr Vr r r r Δvr = −1 r Δϕr = −1 Fig 7.9 Positive internal forces, negative unit deformation The idea of finding influence lines as deflected shapes was developed by Müller-Breslau in 1886 In summary, the Müller-Breslau principle states: The influence line for an internal force at location r is the deflected shape of the structure produced by removing at r the capacity with respect to the force and then introducing a (negative) unit displacement that corresponds to the restraint removed 7.3 Use of Influence Lines 7.3.1 General Remarks Once influence lines have been determined it is straightforward to use them in order to determine the corresponding load effect under any loading condition We recall that the ordinate of the influence line at a point is the value of the load effect due to a unit load acting at that point Thus, all we need to is sum all the loads times their corresponding ordinate η of the influence line For an arbitrary load effect Sr we thus have Sr = ∑ Pi ηi + ∫ w(x ) η(x ) dx (7.7) i where Pi and w(x ) are concentrated forces and distributed loads, respectively Note that the above equation is based on the principle of superposition, which states that the forces in an elastic structure are proportional to the magnitude of the applied loads 149 Example: Find the value of the bending moment at midspan of the beam in Fig 7.10 (7.2) under the loading shown w P A L a − r + η a "M r " L Fig 7.10 Influence line and loading Applying Eq (7.7) gives M r = w⋅ L ⋅ 7.4 L a L2 a ⋅ − P⋅ = w − P⋅ 2 (7.8) Example 7.1 1m 100 kN 100 kN 10 kN/m r 3m 3m 4m 5m Fig 7.11 Example 7.1 Problem: (a) Draw the influence lines for the shear force and bending moment at location r of the beam above (b) Calculate the maximum and minimum values of the shear force and the bending moment due to the uniformly distributed load (dead load) and the moving loads shown The relative position of the two concentrated forces is fixed Solution: (a) We obtain the influence line for the shear force at r by first removing the capacity of the section to transmit shear (but not axial force or bending moment) and then introducing a unit displacement in the direction opposite to the positive direction of the shear force Note that we obtain all influence ordinates by simple proportions (similar triangles) We obtain the influence line for the bending moment at r by first removing the capacity of the section to transmit a bending moment (but not axial force or shear force) and then introducing a unit rotation in the direction opposite to the positive direction of the bending moment Again, we obtain all influence ordinates by simple proportions (similar triangles) 150 (0.500, 0.333) max (0.667, 0.533) 0.6667 0.5 − − "Vr " + 0.5 [−] (2.000,1.600) 2.00 max (1.000,1.500) − − "M r " + [m] 1.50 Fig 7.12 Influence lines for shear force and bending moment (b) Distributed load (dead load) The value for the shear force and moment at location r due to the distributed load is the area under the influence line multiplied by the magnitude of the load, hence Vr = −0.5 ⋅ ⋅ 0.6667 ⋅ 10 = −30 kN M r = 0.5 ⋅ (1.5 ⋅ − ⋅ 9) ⋅ 10 = −45 kNm (7.9) Moving load To calculate the maximum shear force and bending moment due to the moving load, we position the two forces such that the combined influence is maximized To calculate the minimum shear force and bending moment (or maximum negative values) due to the moving load, we position the two forces such that the combined influence is minimized Finding the controlling position for moving loads, sometimes involves some trial and error For influence lines that are piecewise linear as is the case for statically determinate structures, one of the concentrated forces is always placed at the maximum ordinates of the influence line We obtain minVr = −(0.667 + 0.533) ⋅ 100 = −120 kN maxVr = (0.5 + 0.333) ⋅ 100 = 83.33 kN M r = −(2 + 1.6) ⋅ 100 max M r = (1.5 + 1) ⋅ 100 = −360 kNm = 250 kNm (7.10) Combining dead load and live load gives minVr = −30 − 120 = −150 kN maxVr = −30 + 83.33 = 53.33 kN M r = −45 − 360 = −405 kNm max M r = −45 + 250 = 205 kNm (7.11) 7.5 Properties of influence lines of statically determinate structures Influence lines of statically determinate structures are always piecewise linear This is because the induced deformation does not cause any internal forces or curvatures in the beam Consequently, we can obtain the influence lines for statically determinate structures by simple geometry 151 7.6 Influence Lines of Statically Indeterminate Structures 7.6.1 General remarks It can be shown that Eq 7.4 holds irrespective of whether the structure is statically determinate or indeterminate That is we find influence lines for statically indeterminate structures by imposing a negative unit complementary deformation at the location where the influence line is desired The fundamental difference, however, is that statically indeterminate structures offer resistance to the imposed deformation That resistance causes deformations, i.e curvature and/or axial strains, such that the segments of the influence lines are generally curved The analytical treatment of influence lines for statically indeterminate structures is thus beyond the scope of this class Our main objectives regarding influence lines for indeterminate structures are: (1) become familiar with the shape of influence lines for internal forces of continuous beams, (2) sketch influence lines for internal forces of continuous beams, (3) establish live load patterns to produce maximum effects in continuous beams As a first example, we consider the two-span continuous beam in Fig 7.13 and consider the influence line for the bending moment at midspan AB According to the Müller-Breslau principle, we first remove the capacity with respect to bending at that location, i.e place a hinge at midspan AB , and then introduce a relative rotation of unit magnitude at the hinge The resulting deflected shape is equal to the influence line (Fig 7.13) A B L /2 L /2 C L − "M B " + ϕ =1 Fig 7.13 Influence Line for span moment of two-span continuous beam 7.6.2 Live load patterns to maximize forces in multi-span beams (Skip Loading) Building codes require that we vary the position of the live loads to maximize a certain force at a particular section In most cases, we find the largest live loads effects by placing the live load on certain portions of the structure but not on others We may use influence lines to identify the portions of a structure that we should load to maximize the design force at critical section Shown in Fig 7.14 are examples of influence lines for a four-span continuous beam We obtain the influence line "MC " for the support moment at C by first introducing a hinge at that location and then rotating the two member ends relative to each other We obtain the influence line for the span moment M correspondingly We generate the influence line for the support force at A (equal to the shear force at A ) by first removing the vertical support at A and then introducing a vertical displacement 152 B A C D E L L L − + ϕ =1 − + − − L + "M C " "M " + ϕ =1 − "VA " + − "VCB " + + Fig 7.14 Influence lines for four-span continuous beam From Fig 7.14, we can derive the following rules regarding live load effects (1) We obtain the maximum moment in span i by loading spans , i − 2, i, i + 2, (2) We obtain the minimum moment in span i by loading spans , i − 1, i + 1, (3) We obtain the minimum support moment (maximum negative value) by loading the two spans adjacent to the support and every other span next to those This live load pattern also yields the absolute maximum values of the shear forces at this support and the maximum support force (4) We obtain the maximum support moment and the minimum support reaction force by a live load pattern that is opposite to (3) To further illustrate the above rules, we look at a five-span continuous beam in Fig 7.15 Shown are all relevant live load patterns and the corresponding design forces associated with each loading pattern Qualitative bending moment and shear force diagrams due to dead load (uniformly distributed across all spans, Case 0) and each live load pattern are plotted in Fig 7.15 Figure 7.16 shows the shear fore and bending moment envelope diagrams An envelope is a plot of the maximum and minimum forces (for beams these are shear force and bending moment) that can occur at any given section accounting for different positions of the live load Clearly, we must include the dead load forces in the envelope diagrams 153 A B C D E F max M 1, max M , max M M , M maxVAB , maxVFE , max A, max F max M , max M M 1, M , M minVAB , minVFE , A, F M B minVBA , maxVBC , max B max M B maxVBA , minVBC , B MC minVCB , maxVCD , max C max MC maxVCB , minVCD , C M D minVDC , maxVDE , max D max M D maxVDC , minVDE , D M E minVED , maxVEF , max E 10 max M E maxVED , minVEF , E Fig 7.15 Live load patterns for five-span continuous beam 154 10 M V Fig 7.16 Moment and shear force diagrams for dead load (Case 0) and different live load patterns (Cases 1-10) max M M − A B D C F E + M maxV + minV V − Fig 7.17 Moment and shear force envelope diagrams for five-span continuous beam 155 Problems 7.1 A ft 10 k 10 k B A C D ft max ft10.00 ft ft ft ft ft 3.75 (1) Find the influence line for the (a) moment at A , (b) reaction force at A , (c) reaction force at D , (d) shear force at B , (e) shear force at C , (f) moment at B , (g) moment at C of the beam above (2) Using the influence lines of (1), calculate the maximum and minimum values resulting from two moving live loads, whose relative position is fixed Solution: (2) max M A = 137.5 k-ft max D = 33.75 k maxVC = k max MC = 20 k-ft M A D minVC MC = −150 k-ft = −7.5 k = −13.75 k = −55 k-ft max A = 20 k maxVB = 20 k max M B = 68.75 k-ft A = −13.75 k minVB = −13.75 k M B = −68.75 k-ft 7.2 1m 50 kN A 50 kN 6m kN/m B 4m C 2m (1) Find the influence line for (a) the moment at A , (b) the shear force at A , (c) the shear force at B , (d) the reaction force at C of the beam above (2) Using the influence lines of (1), calculate the maximum and minimum values resulting from a uniformly distributed load of magnitude kN/m (dead load across the entire beam) and two moving live loads, whose relative position is fixed (3) Repeat (2) considering the distributed load a live load of variable length Solution: (2) due to w = kN/m M A = −135 kNm A = 37.5 kN VB = 7.5 kN due to P = 50 kN max M A = 225 kNm M A = −550 kNm maxVB = 87.5 kN minVB = −37.5 kN C = 22.5 kN maxVA = 100 kN max C = 137.5 kN 156 minVA = −37.5 kN C = 7.3 P w A C B 5.00 m 5.00 m − − " M B " [m] x x The influence line " M B " for the support moment at B of a two-span continuous beam is given by the function 1 x " MB " = − x + 0≤x ≤5 100 (1) Using the above expression for the influence line, calculate by integration the support moment M B for a uniformly distributed load applied to both spans Check your result by using the “fixed-end moment table” (2) At what location must a concentrated force P be placed to produce minimum M B ? What is the value for M B ? (3) Place P at the location found in (2) and verify the value for M B using the force method of structural analysis Solution: (1) M B = 3.125w (ans ) (2) x = 100 = 2.8867 m (ans ) 12 M B = −0.4811P (ans ) 7.4 A B C D E F G 15 ft 20 ft 25 ft 25 ft 20 ft 15 ft A dead load of wD =1k/ft and a live load of wL =1.5 k/ft are applied to the six-span continuous beam shown Use skiploading for the live load and find (electronically) the maximum and minimum design forces (dead load plus live load) for the span moments, support moments, support reactions and the shear forces at the supports Use symmetry to minimize the number of live load patterns you need to analyze For simplicity, assume that the maximum span moments occur at midspan The beam has constant flexural stiffness 157 kN/m 7.5 A C B 5.00 m 5.00 m − − x " MB " x x0 [m] 1.00 m P P 5.00 m 5.00 m The influence line " M B " for the support moment at B of a two-span continuous beam is given by the function 1 x " MB " = − x + 100 0≤x ≤5 (1) Using the above expression for the influence line, calculate by integration the support moment M B due to the triangular load shown (2) At what location x must two concentrated forces P be placed to produce minimum M B ? What is the value for M B ? w live = k/ft 7.6 B C A D 20 ft 20 ft 0.342 0.576 0.714 0.768 0.75 0.672 0.546 0.384 0.198 0.198 0.384 0.546 0.672 0.75 0.768 0.714 0.576 0.342 20 ft − − "M " [ft] 0.46 1.04 1.74 2.56 3.5 2.56 1.74 1.04 0.46 + A live load of intensity w live = k/ft acts on a three-span continuous beam whose influence line for the bending moment at mid-span is given (ordinates every ft) Use the influence line and the principle of skip loading to find the minimum and maximum moments at mid-span of span You may assume a linear variation of the influence line between given ordinates 158 7.7 w live = k/ft B C A D − 20 ft 20 ft 0.78 1.28 1.54 1.6 1.5 1.28 0.98 0.64 0.3 0.528 1.02 1.46 1.79 2.0 2.05 1.9 1.54 0.912 20 ft − "M B " [ft] 0.228 0.384 0.476 0.512 0.5 0.448 0.364 0.256 0.132 + A live load of intensity w live = k/ft acts on a three-span continuous beam whose influence line for the bending moment at support B is given (ordinates every ft) Use the influence line and the principle of skip loading to find the minimum and maximum moments at support B You may assume a linear variation of the influence line between given ordinates 159 ... 4.000 13 .33 0 .33 3 0.250 4.000 13 .33 DE 10 .0 - 13 .3 -0.667 0.500 4.000 35 .47 -26.6 AD -18 .0 -0.6 01 0.4 51 3. 606 39 . 01 -29 .3 DB 6. 01 0.6 01 -0.4 51 3. 606 13 .02 -9.77 BE 6. 01 0.6 01 0.4 51 3. 606 13 .02... illustration 5.7 Example 5 .3 5.8 Summary Problems 11 3 11 3 11 3 11 4 11 6 11 8 11 9 12 0 12 2 12 3 Approximate analysis of building frames under lateral load 6 .1 Introduction 6.2 Discussion 6 .3 The Portal Method... kN/m 1. 1 C D F G 6m E A B 5m 3m 3m 4m Solution: 15 3 18 8 80 11 .25 − − 35 M 40.8 20 40.8 + 35 .2 − − + [kNm] − 91. 3 N [kN] 30 .6 40.0 + − 11 .7 30 .6 41. 7 + V [kN] 28.0 sketch of deflected shape 1. 2 10