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Influence Lines Live Load Forces: Consider the bridge in Fig. 1. As Influence Lines for the car moves across the bridge, Determinate Structures the forces in the truss members change with the position of the Introduction car and the maximum force in each member will be at a different Previous developments have car location. The design of each been limited to structures member must be based on the subjected to fixed loads. maximum probable load each member will experience. Structures are also subjected to live loads whose position may vary on the structure. This chapter focuses on such loads for statically determinate structures. 1 2 If a structure is to be safely designed, members must be proportioned such that the maximum force produced by dead and live loads is less than the available section capacity. Figure 1. Bridge Truss Structure Subjected to a Variable Structural analysis for variable Position Load loads consists of two steps: 1.Determining the positions of Therefore, the truss analysis the loads at which the for each member would response f ti i function is involve determining the load maximum; and position that causes the 2.Computing the maximum greatest force or stress in value of the response function. each member. 3 4 1 Influence Line Once an influence line is Definitions constructed: • Determine where to place live Response Function ≡ support load on a structure to maximize reaction, axial force, shear force, or the drawn response function; bending moment. and Influence Line ≡ graph of a • Evaluate the maximum response function of a structure as magnitude of the response a function of the position of a function based on the loading. downward unit load moving across structure the structure. NOTE: Influence lines for statically determinate structures are always piecewise linear. 5 6 1 Calculating Response x MB Functions a 0<x<a (Equilibrium Method) VB Ay ∑ Fy = 0 ⇒ V B = A y − 1 ∑ Ma = 0 ⇒ M B = A y a −1(a − x) 1 ILD for Ay MB a VB a<x<L Ay 0 L ILD for Cy 1 ∑ Fy = 0 ⇒ V B = A y 0 L 7 ∑ Ma = 0 ⇒ M B = A y a 8 2 1 – a/L VB Beam Example 1 0 a L ILD for VB /L -a/L MB a (1 – a/L) 0 a L Calculate and draw the support ILD for MB reaction response functions. 9 10 Beam Example 2 Frame Example BD: Link Member Calculate and draw the response functions for RA, MA, RC and VB. Calculate and draw the response functions for Ax, Ay, AB and VB . NOTE: Unit load traverses span AC. 11 12 3 CAUTION: Principle is only valid Muller-Breslau for force response functions. Principle Releases: Muller-Breslau Principle ≡ The Support reaction - remove influence line for a response restraint. translational support restraint function is given by the deflected shape of the released structure Internal shear - introduce an due to a unit displacement (or internal glide support to allow rotation) at the location and in the differential displacement direction of the response movement. function. Bending moment - introduce an A released structure is obtained internal hinge to allow differential by removing the displacement rotation movement. constraint corresponding to the response function of interest from the original structure. 13 14 Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display. Influence Line for Shear Influence Line for Bending Moment 15 16 4 Application of Muller- Breslau Principle 17 18 y = (L – x) (a/L) θ1 + θ2 = 1 19 20 5 Qualitative Influence NOTE: An advantage of Lines constructing influence lines using the Muller-Breslau Principle is that the response function of In many practical applications, it is interest can be determined necessary to determine only the directly. It does not require general shape of the influence determining the influence lines lines but not the numerical values for other functions, as was the of the ordinates. Such an case with the equilibrium influence line diagram is known as method. a qualitative influence line dia- gram. gram An influence line diagram with numerical values of its ordinates is known as a quantitative influ- ence line diagram. 21 22 Influence Lines for Trusses In a gable-truss frame building, roof loads are usually transmitted to the top chord joints through roof purlins as shown in Fig. T.1. Similarly, highway and railway bridge truss-structures transmit floor or deck loads via stringers to floor beams to the truss joints as shown schematically in Fig. T.2. y g Fig. T.2. Bridge Truss Fig. T.1. Gable Roof Truss 23 24 6 These load paths to the truss joints Due to the load transfer provide a reasonable assurance process in truss systems, no that the primary resistance in the discontinuity will exist in the truss members is in the form of member force influence line axial force. Consequently, diagrams. Furthermore, since influence lines for axial member we are restricting our attention to forces are developed by placing a statically determinate struc- unit load on the truss and making tures, the influence line judicious use of free body diagrams will be piecewise diagrams and the equations of linear. statics. 25 26 Example Truss Structure Use of Influence Lines Point Response Due to a Single Moving Concentrated Load Each ordinate of an influence line gives the value of the response function due to a single concentrated load of Calculate and draw the response unit magnitude placed on the functions for Ax, Ay, FCI and FCD. structure at the location of that ordinate. Thus, 27 28 7 1. The value of a response P function due to any single A B C D concentrated load can be obtained by multiplying the x magnitude of the load by the ordinate of the response yB function influence line at the position of the load. D A B C 2. Maximum positive value of the response function is ILD for MB -yD obtained by multiplying the point load by the maximum positive ordinate. Similarly, the + (M B ) max ⇒ place P at B maximum negative value is obtained by multiplying the − (M B ) max ⇒ place P at D point load by the maximum 29 negative ordinate. 30 Point Response Due to a dR = dP y = w dx y Uniformly Distributed Live where y is the influence line Load ordinate at x, which is the point of application of dP. Influence lines can also be employed to determine the To determine the total response values of response functions of function value at a point for a structures due to distributed distributed load between x = a to x loads. This follows directly from = b, simply integrate: point forces by treating the uniform load over a differential b b segment as a differential point force, i.e., dP = w dx. Thus, a R= ∫ w ydx = w ∫ ydx response function R at a point a a can be expressed as 31 32 8 in which the last integral expres- sion represents the area under the 2. To determine the maximum segment of the influence line, positive (or negative) value of a which corresponds to the loaded response function due to a portion of the beam. uniformly d st buted live load, u o y distributed e oad, the load must be placed over SUMMARY those portions of the structure where the ordinates of the 1. The value of a response response function influence line function due to a uniformly are positive (or negative). distributed load applied over a portion of the structure can be obtained by multiplying the load Points 1 and 2 are schematically intensity by the net area under demonstrated on the next slide for the corresponding portion of the moment MB considered in the point response function influence 33 load case. 34 line. 35 36 9 Where should a CLL (Concentrated Live Load), a ULL (Uniform Live Load) and UDL (Uniform Dead Load) be placed Typical Interior on the typical ILD’s shown below Beam Shear ILD to maximize the response functions? Typical Interior Bending Moment ILD Typical End Shear (Reaction) ILD Possible Truss Member ILD 37 38 Live Loads for To calculate the response Highway and function for a given position of Railroad Bridges the concentrated load series, simply multiply the value of each Live loads due to vehicular traffic series load Pi by the magnitude on highway and railway bridges f h i fl line diagram of the influence li di are represented by a series of ordinate yi at the position of Pi , moving concentrated loads with i.e. specified spacing between the R = ∑ Pi yi loads. In this section, we discuss i the use of influence lines to The ordinate magnitude yi can be determine: (1) the value of the ( ) calculated from the slope of the response function for a given influence line diagram (m) via position of a series of concentrated loads and (2) the maximum value yi = m x i of the response function due to a series of moving concentrated 39 40 loads. 10 where x i is the distance to point i For example, consider the ILD xi measured from the zero y-axis shown on the next slide intercept, as shown in the subjected to the given wheel schematic ILD below. loading: m Load Position 1: L dP ii 1 1 yb ya VB1 = 8( 1 20) + 10( 1 16) + 15( 1 13) +5( 1 8) 30 30 30 30 x = ( 1 )(8(20) + 10(16) + 15(13) + 5(8)) a 30 b = m∑Pi xi = 18 5k 18.5k i ya y b = ⇒ similar triangles a b y y ∴ ya = b a ; m = b 41 42 b b 2/3 10 ft. ft. 20 ft -1/3 Position 1 ILD for Internal Shear SB Position 2 Wheel Loads 43 44 11 Load Position 2: influence line ordinate before the V = 1 (−8(6) + 10(20) + 15(17) + 5(12)) B2 lighter loads in the series. In such 30 = 15.6k a case, it may not be necessary to examine all the loading positions. Thus, load position 1 results in the , p Instead the analysis can be Instead, maximum shear at point B. ended when the value of the response function begins to NOTE: If the arrangement of decrease; i.e., when the value of loads is such that all or most of the the response function is less than heavier loads are located near one the preceding load position. This of the ends of the series, then the p process is known as the analysis can be expedited by “Increase-Decrease Method”. selecting a direction of movement for the series so that the heavier loads will reach the maximum 45 46 CAUTION: This criterion is not Zero Ordinate Location valid for any general series of loads. In general, depending on Linear Influence Line the load magnitudes, spacing, b+ and shape of the influence line, 1 h l f h the value of the response m+ function, after declining for some x- loading positions, may start increasing again for subsequent m- loading positions and may attain a 1 higher maximum. x+ b- L 47 48 12 b −b Example Truss Problem: x + = −b+ ; m+ = − + Application of Loads to m+ L Maximize Response b −b x − = −b− ; m− = + − m− L NOTE: Both of these solutions are obtained from y = mx + b with y = 0. ML 49 CM 50 Place Force and Moment UDL = 1.0 k/ft; Envelopes ULL = 4.0 k/ft; A plot of the maximum CLL = 20 kips response function as a to maximize the tension and function of the location of the compression axial forces in response function is referred members CM and ML. to as the envelope of the maximum values of a Calculate the magnitudes of the response function for the tension and compression forces. particular load case being id d considered. 51 52 13 For a single concentrated For a uniformly distributed force for a simply sup- load on for a simply sup- ported beam: ported beam: (V)+ = w max ⎛ a⎞ (V) + = P ⎜1− ⎟ ( )max ( L − a )2 2L ⎝ L⎠ w a2 (V)− = − P a (V)− = − max max 2L L ⎛ a⎞ w a M max = P a ⎜1− ⎟ M ma = max (L −a ) ⎝ L⎠ 2 Plot is obtained by treating “a” Plot is obtained by treating “a” as a variable. as a variable. 53 54 55 56 14