Influence Line - Lecture Notes 2

EXAMPLE 9: For each beam in Fig 6.16, sketch the influence line Text book example 6-9 for the vertical reaction at A SOLUTION The support is replaced by a roller guide at A since it will resist, but not Ay. The force Ay is then applied. Again, a roller guide is placed at A and the force Ay is applied. A double-roller guide must be used at A in this case EXAMPLE 10: For each beam in Figs. 6–17a through 6–17c, sketch the influence line for the shear at B. Text book example 6-10 (a) The roller guide is introduced at B and the positive shear is applied. Notice that the right segment of the beam will not deflect since the roller at A actually constrains the beam from moving vertically, either up or down. (b) (c) EXAMPLE 11: For each beam in Figs. 6–18a through 6–18c, sketch the influence line for the Moment at B. Text book example 6-11 Solution: A hinge is introduced at B and positive moments MB are applied to the beam. The deflected shape and corresponding influence line are shown. (a) (b) Fig. 6-18 (c) 6.2 use of influence line for Beams Once the influence line for a function has been constructed, it will be possible to position live loads on the beam which will produce the maximum value of the function. Two types of loadings will be considered 1) Concentrated force 2) Uniform load Concentrated force  For any concentrated force, F acting on the beam, the value of the function can be found by multiplying the ordinate of the influence line at position x by magnitude of F. Uniform load  value of a function caused by a uniform load = the area under the influence line  intensity of the uniform load Example 7: Determine the max +ve live shear that can be developed at point C in the beam shown in Fig 6.10(a) due to: A concentrated moving load of 4kN and a uniform moving load of 2kN/m. Solution: The influence line for shear at C has been established as shown in Fig 6.10(b) Concentrated force  The max +ve positive shear at C will occur when the 4kN force is located at x = 2.5 m  The ordinate at this peak is +0.75, hence: V 0.75(4) 3 kN C Text book example 6-7 Uniform load  The uniform moving load creates the max +ve influence for VC when the loads acts on the beam between x = 2.5 m and x = 10 m  The magnitude of VC due to this loading is:  VC 1 (10m 2.5m)(0.75) 2 5.625kN  (2kN/m)   Total max shear at C due to concentrated load and uniform load : (VC )max 3 kN 5.625 kN 8.625 kN m m EXAMPLE 8: The frame structure shown in Fig. 6–11a is used to support a hoist for transferring loads for storage at points underneath it. It is anticipated that the load on the dolly is 3 kN and the beam CBhas a mass of 24 kg/m. Assume the dolly has negligible size and can travel the entire length of the beam. Also, assume A is a pin and B is a roller. Determine the maximum vertical support reactions at A and B and the maximum moment in the beam at D. Text book example 6-8 Solution: Fig 6-11a We first draw the influence lines for, reaction Ay, reaction By and moment at D. Max reaction at A: o The max value for Ay occurs when the dolly is at C o The dead load must be placed over the entire length: 1  ( Ay ) max 3000(1.33) 24(9.81)  (4)(1.33)  2  4.63 kN Max reaction at B:  Here the dolly must be at B: 1  ( By ) max 3000(1) 24(9.81)  (3)(1)  2  1  24(9.81)  (1)( 0.333) 3.31kN 2  Max moment at D: o Here the dolly must be at D: 1   (0.75)  3000 24(9.81)  (1)(  0.5) 2  1  24(9.81)  (3)(0.75) 2.46 kN 2  ( M D) max (b) (c) EXAMPLE 12: Determine the max +ve moment that can be developed at point D in the beam shown in Fig 6.19(a) due to a concentrated moving load of 16kN, a uniform moving load of 3kN/m & a beam weight of 2kN/m Solution A hinge is placed at D & +ve moments MD are applied to the beam 2m The deflected shape & corresponding influence line are shown in Fig 6.19(b) 2m 2m 4m (a) x Fig 6-19 (b) Text book example 6-12 The concentrated moving load of 16kN creates a max +ve moment at D when it is placed at D The uniform moving load of 3kN/m must extend from C to E in order to cover the region where the area of the influence line is +ve. Finally, the uniform weight of 2 kN/m acts over the entire length of the beam The loading is shown on the beam in Fig 6.19(c) Fig. 16-9 (c) Method 1: The maximum +ve moment at D can be determined using numerical values for the influence line as in Section 6.1. By inspection of Fig 6.19(b), only the peak value h at D must be computed. This requires placing a unit load on the beam at D in Fig 6.19(a) & then solving for internal moment in the beam at D. Show that the value obtained is h = 1.333 x Fig. 16-9 (b) By proportional triangles, h '/ (4 2) 1.333 / (6 4) or h ' 1.333 Hence, with the loading on the beam as in Fig. 6–19c, using the areas and peak values of the influence line, Fig. 6–19b, we have 1 M D  5  (10 4)(1.333) 2 M D  36 kNm  1  16(1.333) 2 (4)(1.333)   2    Method 2 In Fig 6.19(d), the reactions on BE have been computed Sectioning the beam at D & using the segment DE, Fig 6.19(e), we have: (d) M D 0;  M D 20(2) 19(4)   0  M D 36 kNm (e) 6-4 Influence Lines for Floor Girders* Floor systems are constructed as shown in Fig (a) Floor loads from slabs floor beams  side girders  supporting columns Stringer (b) Idealizes model *6‐4willnotbetestedinexams EXAMPLE 13: Draw the influence line for the shear in panel CD of Text book example 6-13 the floor girder in Fig. 6–21a. Solution: The unit load is placed at each floor beam location & the shear in panel CD is calculated. Fig. 6-21 (c) 3m 3m 3m Fig. 6-21 (a) 3m Fig. 6-21 (d) The unit load is placed at each floor beam location & the shear in panel CD is calculated A table of the results is shown in Fig 6.21(b) The details for the calculations when x = 0 and x = 6 m are given in Fig 6.21(c) and 6.21(d) Finally a segment of the girder is considered & the internal panel shear VCD is calculated The tabular values are plotted and the influence line is as shown in Fig 6.21(e) Fig. 6-21 (b) 6.5 Influence Lines for Trusses In Fig 6.23, the loading on the bridge deck is transmitted to stringers which in turn transmit the loading to floor beams and then to joints along the bottom cord EXAMPLE 14: Draw the influence line for the force in member GB of the bridge truss shown in Fig. 6–24a. Text book example 6-15 Solution Tabulate Values: Each successive joint at the bottom cord is loaded with a unit load and the force in member GB is calculated using the method of sections, Fig 6.24(b). For example, placing the unit load at x = 6 m (joint B), the support reaction at E is calculated first. Then calculate force in GB. (b) The values are tabulated as shown in Fig. 6.24 (b) Plotting the tabular data and connecting the points yields the influence line for member GB, Fig 6.24(d) Since the influence line extends over the entire span of truss, member GB is referred to as primary member Fig. 6.24 (d) Use equation to draw I.L. UnitloadbeforeB If we have I.L for Ay and Ey, then use those to draw I.L for member BG Takerightsectionaftercut G Ey F y  0;  E y F GB sin 45 0  0 A B C D E  FGB  E y sin 45 0 Ay At B, FGB=0.25/sin450=0.353 UnitloadafterC Takeleftsectionaftercut F y 0;  Ay FGB sin 450 0 Ey 0.353  FGB  A y sin 450 At C, FGB=-0. 5/sin450= - 0.707 FGB 0.707 EXAMPLE 15: Draw the influence line for the force in member CG of the bridge truss shown in Fig. 6–25a. Solution: Fig. 6-25 (a) Tabulate Values: The forces in CG for different location of unit load on the bottom chord is shown in Fig. 6–25b. These values are easily obtained by isolating joint C, Fig. 6–25c. Here it is seen that CG is a zero-force member unless the unit load is applied at joint C, in which case FCG = 1 (T). Text book example 6-16 Influence Line. Plotting the tabular data and connecting the points yields the influence line for member CG as shown in Fig. 6–25d. In particular, notice that when the unit load is at x= 9 m the force in member CG is FCG = 0.5. This situation requires the unit load to be placed on the bridge deck between the joints. The transference of this load from the deck to the truss is shown in Fig. 6–25e. From this one can see that indeed FCG = 0.5 by analyzing the equilibrium of joint C, Fig. 6–25f. x EXAMPLE 16: Determine the largest force that can be developed in member BC of the bridge truss shown in Fig. 6-26a due to a moving force of 100 kN and a moving distributed load of 7.5 kN/m. The loading is applied at the top cord. Fig. 6-26(a) Solution: Find out force in member BC using static [Fig. (c)] due to different position of unit load. The forces are tabulated in Fig. (b). (c) Text book example 6-17 Using tabulated values draw the influence line as shown in Fig (d). Concentrated Live Force. The largest force in member BC occurs when the moving force of 100 kN is placed at x= 16 m Thus FBC 1.33 (100)  133 kN Distributed Live Load. The uniform live load must be placed over the entire deck of the truss to create the largest tensile force in BC.*Thus, 1 FBC   (1.33)  7.5  160 kN 2 32   Total Maximum Force. FBC max 133 kN+160 kN = 293 kN Ans. Use equation to draw I.L. If we have I.L for Ay and Ey, then use those to draw I.L for member BG UnitloadbeforeB Takerightsectionaftercut M H Ey  0;  E y 16 F BC 6 0  FBC  16E y 6 1  2.667 E y Ay UnitloadafterC Takeleftsectionaftercut M H Ey  0;  Ay 16  FBC 6 0  FBC  16 Ay 6 1.33  2.667 Ay FBC Example-17 (using qualitative procedure) (1) (2) (3) (4) (5) (6) (7) d (8) (9) RL 1 (10) (11) L=8P 0.75 0.625 0.5 (12) (13) h (14) Constructthe influencelinesfor internalforcesin members2‐10,2‐ 9,2‐3and1‐8 RR InfluencelineforRL 0.25 0.5 1 InfluencelineforRR 0.625 Influencelinefor shearinpanel2‐10 0.25 0.625 d/h Influencelineforforce2‐10 0.25 d/h Examples-17 (cont.) (1) (2) (3) (4) (5) (6) (7) d (8) (9) (10) RL (11) (12) (13) (14) L=8P RR 0.75 0.625 0.5 1 h InfluencelineforRL 0.5 0.25 1 InfluencelineforRR 0.25 Influencelineforforce2‐9 0.625 1 + Influencelineforforce1‐8 Examples-17 (cont.) (1) (2) (3) (4) (5) (6) (7) d (8) (9) (10) RL (11) (12) L=8P (13) h (14) RR 3P(5P)/(8P)=15P/8 Influencelineforbending momentat10 (‐) (C) 15P 8h Influencelineforforce2‐3 6-6 Max Influence at a Point due to a series of concentrated loads** The max effect caused by a live concentrated force is determined by multiplying the peak ordinate of the influence line by the magnitude of the force. In some cases, e.g. wheel loadings, several concentrated loadings must be placed on structure Trial-and-error procedure can be used or a method that is based on the change in function that takes place as the load is moved ** 6‐6 will not be tested in exams Example 18: Determine maximum shear at C due to three concentrated moving loads.  Critical loading occurs when one of the loads is placed just to the right of C  By trial & error, each of the three possible cases can therefore be investigated, Fig 6.27(b) Case 1 : (VC )1 4.5(0.75) 18(0.625) 18(0.5) 23.63kN Case 2 : ( VC ) 2 4.5( 0.125)  18(0.75) 18(0.625) 24.19 kN Case 3 : ( VC )3  ( 0.125) 18(0.75) 11 .25kN 4 .5(0) 18 Case 2 yields the largest value for VC and therefore represents the critical loading