INFLUENCE LINE. Structural Analysis. Reference: Third Edition (2005) By Aslam Kassimali

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1 INFLUENCE LINE Reference: Structural Analsis Third Edition (2005) B Aslam Kassimali

2 DEFINITION An influence line is a graph of a response function of a structure as a function of the position of a downward unit load moving across the structure

3 INFLUENCE LINES FOR BEAMS AND FRAMES BY EQUILIBRIUM METHOD Influence Line for: a) Reactions b) Shear at B c) Bending moment at B

4 S B x C =, 0 x < a = L x A = 1, a< x L L M B x C ( L a) = ( L a), 0 x a L = x A ( a) = 1 a, a x L L

5 Example 8.1 Draw the influence lines for the vertical reactions at supports A and C, and the shear and bending moment at point B, of the simpl supported beam shown in Fig. 8.3(a). kn Influence line for A c = 0: ( ) ( x) 120 ( x) A = 0 A M x = = Fig. 8.3(c) kn/kn Influence line for C A = 0: ( x) C ( ) 1( x) x = 0 C M = = Fig. 8.3(d)

6 Influence line for S B Place the unit load to the left of point B, determine the shear at B b using the free bod of the portion BC: S = C 0 x< 12ft B Place the unit load to the right of point B, determine the shear at B b using the free bod of the portion AB: S = A 12 ft < x 20 ft B gives S B x C =, 0 x < 12ft 20 = x A = 1, 12 ft < x 20 ft 20 Fig. 8.3(e)

7 Influence line for M B Place the unit load to the left of point B, determine the bending moment at B b using the free bod of the portion BC: M = 8C 0 x 12ft B Place the unit load to the right of point B, determine the bending moment at B b using the free bod of the portion AB: M = 12A 12 ft x 20 ft B gives M B 2x 8 C =, 0 x 12ft 5 = 3x 12A = 12, 12 ft x 20 ft 5 Fig. 8.3(f)

8 Example 8.3 Draw the influence lines for the vertical reactions at supports A, C, and E, the shear just to the right of support C, and the bending moment at point B of the beam shown in Fig. 8.5(a).

9 Influence line for E Place the unit load at a variable position x to the left of the hinge D and consider free bod diagram DE: E DE D ( ) = 0 20 = 0 E = 0 0 x 40ft M Next, the unit load is located to the right of hinge D: M DE D = 0 ( x ) E ( ) 1( x 40) x = 0 E = = 2 40ft x 60ft E 0 0 x 40ft = x 2 40ft x 60ft 20 Fig. 8.5(c)

10 Influence line for C A = 0 1( x) + C (20) + E (60) = 0 C M x = 3E 20 B substituting the expressions for E, we obtain C x x 0= 0 x 40ft = x x x 3 2 = 6 40ft x 60ft Fig. 8.5(d)

11 Influence line for A F = 0 A 1+ C + E = 0 A = 1 C E B substituting the expressions for C and E, then A x x 1 0= 1 0 x 40ft = x x x = 3 40 ft x 60 ft Fig. 8.5(e)

12 Influence line for Shear at Just to the Right of C, S C,R S CR, E 0 x 20ft = 1 E 20ft x 60ft B substituting the expressions for E, we obtain 0 0 x 20ft < SCR, = 1 0= 1 20ft < x 40ft x x 1 2 = 3 40ft x 60ft Fig. 8.5(f)

13 Influence line for M B M B ( ) 10A 1 10 x 0 x 10 ft = 10A 10 ft x 60 ft B substituting the expressions for A, we obtain x x ( 10 x) = 0 x 10 ft 20 2 x x M B = 10 1 = ft x 40 ft 20 2 x x 10 3 = ft x 60 ft 20 2 Fig. 8.5(g)

14 MULLER-BRESLAU S PRINCIPLE AND QUALITATIVE INFLUENCE LINES Developed b Heinrich Muller-Breslau in Muler-Breslau s principle: The influence line for a force (or moment) response function is given b the deflected shape of the released structure obtained b removing the restraint corresponding to the response function from the original structure and b giving the released structure a unit displacement (or rotation) at the location and in the direction of the response function, so that onl the response function and the unit load perform external work. Valid onl for influence lines for response functions involving forces and moments, e.g. reactions, shears, bending moments or forces in truss members, not valid for deflections.

16 Qualitative Influence Lines In man practical applications, it is necessar to determine onl the general shape of the influence lines but not the numerical values of the ordinates. A diagram showing the general shape of an influence line without the numerical values of its ordinates is called a qualitative influence line. In contrast, an influence line with the numerical values of its ordinates known is referred to as a quantitative influence line.

17 Example 8.6 Draw the influence lines for the vertical reactions at supports B and D and the shear and bending moment at point C of the beam shown in the Figure 8.9(a).

19 Example 8.7 Draw the influence lines for the vertical reactions at supports A and E, the reaction moment at support A, the shear at point B, and the bending moment at point D of the beam shown in Fig. 8.10(a).

21 Example 8.8 Draw the influence lines for the vertical reactions at supports A and C of the beam shown in Fig. 8.11(a).

22 INFLUENCE LINES FOR TRUSSES Consider the Pratt bridge truss shown. A unit load moves from left to right. Suppose that we wish to draw the influence lines for the vertical reactions at supports A and E and for the axial forces in members CI, CD, DI, IJ and FL of the truss.

23 Influence Lines for Reactions E = 0 A (60) + 1(60 x) = 0 A M x = 1 60 A = 0 1( x) + E (60) = 0 E M x = 60

Influence line for force in Vertical Member CI Considering the right portion of the truss (unit load at left portion) F = 0 F + E = 0 F = E 0 x 30ft CI CI Considering the left

24 Influence line for force in Vertical Member CI Considering the right portion of the truss (unit load at left portion) F = 0 F + E = 0 F = E 0 x 30ft CI CI Considering the left portion of the truss (unit load at right portion) F = 0 A + F = 0 F = A 45ft x 90ft CI CI Unit load is located between C and D: F = 0 45 x A + FCI = x FCI = A + 30 ft x 45 ft 15

25 Influence line for force in Bottom Chord Member CD I = 0 F (20) + E (30) = 0 CD F = 1.5E 0 x 30 ft CD M I = 0 A (30) + F (20) = 0 CD F = 1.5A 30 ft x 90 ft CD M

26 Influence line for force in Diagonal Member DI 4 F = 0: FDI + E = 0 5 F = 1.25E 0 x 30 ft DI 4 F = 0: A FDI = 0 5 F = 1.25A 45 ft x 90 ft DI Influence line for force in Top Chord Member IJ F IJ D = 0: (20) + E (15) = 0 F = 0.75E 0 x 45 ft IJ D = 0: A (45) F (20) = 0 IJ F = 2.25A 45 ft x 90 ft IJ M M

27 Influence line for force in Vertical Member FL

28 Example 8.12 Draw the influence lines for the forces in members AF, CF, and CG of the Parker truss shown in Fig. 8.19(a). Live loads are transmitted to the bottom chord of the truss.

29 APPLICATION OF INFLUENCE LINES

30 Response at a particular location due to a single moving concentrated load The value of a response function due to an single concentrated load can be obtained b multipling the magnitude of the load b the ordinate of the response function influence line at the position of the load To determine the maximum positive value of a response function due to a single moving concentrated load, the load must be placed at the location of the maximum positive ordinate of the response function influence line, whereas to determine the maximum negative value of the response function, the load must be placed at the location of the maximum negative ordinate of the influence line.

31 Suppose that we wish to determine the bending moment at B when the load P is located at a distance x. M B =P Maximum Positive bending moment at B * Place the load P at point B * M B =P B Maximum Negative bending moment at B * Place the load P at point D * M B =-P D

32 Example 9.1 For the beam shown in Fig. 9.2(a), determine the maximum upward reaction at support C due to a 50 kn concentrated live load. From Example 8.8 Maximum upward reaction at C: ( ) C = =+ 70kN= 70kN

33 Response at a particular location due to a uniforml distributed live load Consider, for example, a beam subjected to a uniforml distributed live load w l. Suppose that we want to determine the bending moment at B when the load is placed on the beam, from x=a to x=b. The bending moment at B due to the load dp as B ( ) dm = dp = w dx l The total bending moment at B due to distributed load from x=a to x=b: b M = wdx = w dx B l l a b a The value of a response function due to a uniforml distributed load applied over a portion of the structure can be obtained b multipling the load intensit b the net area under the corresponding portion of the response function influence line.

34 b M = w dx = w dx B l l a b a This equation also indicates that the bending moment at B will be maximum positive if the uniforml distributed load is placed over all those portions of the beam where the influence-line ordinates are positive and vice versa. Maximum positive bending moment at B M = w ( areaunder the inf luenceline A C) B l 1 = wl ( 0.75L)( B) = 0.375wlBL 2 Maximum negative bending moment at B M = w ( areaunder the inf luencelinec D) B l 1 = wl ( 0.25L)( D) = 0.125wlDL 2 To determine the maximum positive (or negative) value of a response function due to a uniforml distributed live load, the load must be placed over those portions of the structure where the ordinates of the response function influence line are positive (or negative).

35 Example 9.2 For the beam shown in Fig. 9.4(a), determine the maximum upward reaction at support C due to a 15 kn/m uniforml distributed live load (udl). To obtain the maximum positive value of C, we place the 15 kn/m udl over the portion AD of the beam, Fig. 9.4(c). From Example 8.8 Maximum upward reaction at C: C 1 = ( )( ) =+ 189kN = 189kN

36 Response at a particular location due to a series of moving concentrated loads Suppose we wish to determine the shear at B of the beam due to the wheel loads of a truck when the truck is located as in figure S B = 4(0.16) 16(0.3) + 16(0.4) = 0.96k

37 Influence lines can also be used for determining the maximum values of response functions at particular locations of structures due to a series of concentrated loads. Suppose that our objective is to determine the maximum positive shear at B due to the series of four concentrated loads. During the movement of the series of loads across the entire length of the beam, the (absolute) maximum shear at B occurs when one of the loads of the series is at the location of the maximum positive ordinate of the influence line for S B. We use a trial-and-error procedure

38 Let the loads move from right to left, the 8k load placed just to the right of B: S B = 8(20) + 10(16) + 15(13) + 5(8) = 18.5k S B S B = 8(6) + 10(20) + 15(17) + 5(12) = k = 8(3) 10(7) + 15(20) + 5(15) = 9.367k S B = 10(2) 15(5) + 5(20) = 0.167k Maximum positive S B =18.5k Fig.(c)

39 Example 9.4 Determine the maximum axial force in member BC of the Warren truss due to the series of four moving concentrated loads shown in Fig. 9.8(a). We move the load series from right to left, successivel placing each load of the series at point B, where the maximum ordinate of the influence line for F BC is located (see Fig. 9.8(c)-(f)).

41 For loading position 1 (Fig. 9.8(c)): 1 FBC = [ 16(60) + 32(50) + 8(35) + 32(15) ] = 41.5 kt ( ) 80 For loading position 2 (Fig. 9.8(d)): FBC 3 1 = 16(10) + [ 32(60) + 8(45) + 32(25) ] = 44.5 kt ( ) For loading position 3 (Fig. 9.8(e)): FBC 3 1 = 32(5) + [ 8(60) + 32(40) ] = 28.0 kt ( ) For loading position 4 (Fig. 9.8(f)): FBC 1 = 32(60) = 24.0 kt ( ) 80 Maximum F BC = 44.5 k (T)

42 Absolute maximum response Thus far, we have considered the maximum response that ma occur at a particular location in a structure. In this section, we discuss how to determine the absolute maximum value of a response function that ma occur at an location throughout a structure. Although onl simpl supported beams are considered in this section, the concepts presented herein can be used to develop procedures for the analsis of absolute maximum responses of other tpes of structures.

43 Single Concentrated Load

UNIT II 1. Sketch qualitatively the influence line for shear at D for the beam [M/J-15]

UNIT II 1. Sketch qualitatively the influence line for shear at D for the beam [M/J-15] 2. Draw the influence line for shear to the left of B for the overhanging beam shown in Fig. Q. No. 4 [M/J-15] 3.