Fundamentals of Structural Analysis Episode 2 Part 5 pdf

Influence Lines by S. T. Mau 255 Influence line solutions. Example 2. Construct the influence lines for R a , R d , M d , M b , V d , V cR and V cL of the beam shown. Beam with an internal hinge. a b c d 1 R a a b c d 1 R b a b cd 1 V c a b c d 1 M c α h α h a b cd 1 V d a b c d 1 M d a b c d Influence Lines by S. T. Mau 256 Solution. Applications of the Müller-Breslau Principle yield the following solutions. Influence line solutions. ac d 1 R a a c d 1 R d a b c d 1 M d a b c d 1 M b a b c d 1 V b a b c d V cR 1 a b c d V cL 1 Influence Lines by S. T. Mau 257 Influence lines for statically indeterminate beam and frames. The Müller-Breslau Principle is especially useful in sketching influence lines for a statically indeterminate beam or frame. The process is the same as that for a statically determinate structure but the precise shape cannot be obtained without further computation, which is very involved. We shall demonstrate only the qualitative solution process without any computations. Example 3. Sketch the influence lines for R a , R c , V d , and M d of the beam shown. Two-span beam example. Solution. The influence lines are curved because the virtual displacements must be curved to accommodate the support constraints. Influence line solutions. Example 4. Sketch the influence lines for M a of the frame shown. a b c d a b c d 1 R a R c a b c d 1 a b c d V d 1 a b c d M d 1 Influence Lines by S. T. Mau 258 Frame example. Solution. According to the Müller-Breslau Principle, we need to make a cut at section a and impose a unit relative rotation at the cut. Try-and error leads to the following sketch that satisfies all constraints of the principle. Sketch of influence line for M a of section a. Example 5. Place uniformly distributed loads anywhere on the second floor of the frame shown in Example 4 to maximize M a . Using the influence line of M a as the guide, we place the load at locations as shown in the following figure for maximum positive and negative moments at section a. Loading pattern for maximum positive M a (left) and negative M a (right). Applications of influence lines. The following exmaples illustrate the use of influence lines to find the maximum of a desired design parameter. 1 a a a a Influence Lines by S. T. Mau 259 Example 6. Find the maximum moment at c for (1) a single load of 10 kN and (2) a pair of 10- kN loads 1 m apart. A simply supported beam. Solution. The influence line for M c has been obtained earlier and is reproduced below. Influence line for M c . For a single load of 10 kN, we place it at the location of the peak of the influence line and we compute (M c ) max = 10 kN (2.5 kN-m/kN)=25 kN-m. For the pair of loads, we place them as shown below. (M c ) max = 10 kN (2.5 kN-m/kN)+ 10 kN (2.0 kN-m/kN)=45 kN-m. For this case, it turns out that the pair of loads can be placed anywhere within 1 m of the center point of the beam and the resulting maximum M c would be the same. Example 7. Find the maximum shear at c for uniformly distributed loads of intensity 10 kN/m and unlimited length of coverage. 5 m 5 m ac b 2.5 kN-m/kN M c 2.5 M c 1 m 2.0 10 kN 10 kN Influence Lines by S. T. Mau 260 Beam with an overhang. Solution. The influence line as constructed earlier is reproduced below. Influence line for V c . In beam design, the sign of shear force is often not important. Thus, we want to find the maximum shear regardless of its sign. From the influence line, the following load application produces the maximum shear force. Loads to maximize V c . The maximum value of V c is computed using the influence line and the area below the influence line of the loaded portion: (V c ) max = (−)10 [( 2 1 )(5)( 2 1 )] + (−)10 [( 2 1 )(5)( 2 1 )] = −25 kN. Deflection Influence Lines. In design we need to answer the question: what is the maximum deflection of any given point on the center line of a beam? The answer is in the influence line for deflections. Surprisingly, the deflection influence line is identical to the deflection curve under a unit load applied at the point of interest. Consider the beam and unit load configuration shown below. a b c d 5 m 5 m 5 m a b cd V c 5 m 5 m 5 m 1/2 1/2 1/2 a b c d 5 m 5 m 5 m 10 kN-m 10 kN-m Influence Lines by S. T. Mau 261 Deflection at j due to a unit load at i. According to the Maxwell’s Reciprocal Theorem, however, δ ji = δ ij And, δ ij is defined in the figure below: Influence line for deflection at j. Thus, to find the deflection influence line of a point, we need only to find the deflection curve corresponding to a unit load applied at the point. i x 1 j δ ji i 1 j δ ij Influence Lines by S. T. Mau 262 Problem 1. (1) Construct the influence lines of V b and M d of the beam shown and find the maximum value of each for a distributed load of intensity 10 kN/m and indefinite length of coverage. Problem 1. (1) (2) Construct the influence lines of V bL and V bR of the beam shown and find the maximum value of each for a distributed load of intensity 10 kN/m and indefinite length of coverage. Problem 1. (2) (3) Construct the influence lines of V cL , V cR and M c and M e of the beam shown. Problem 1. (3). (4) Sketch the influence line of V a of the frame shown. Problem 1. (4) a b c d 5 m 5 m 5 m a b c d 5 m 5 m 5 m a b c d 2.5 m 5 m 5 m 2.5 m e a Influence Lines by S. T. Mau 263 3. Truss Influence Lines For a truss, the question relevant in design is: How does a member force change when a unit load moves along the span of the truss? The answer is again in the influence line, but the truss itself only accepts loads at the joints. Thus, we need to examine how a load, moving continuously along the truss span, transmits its force to the truss joints. As shown in the figure below, a truss has a floor system that transmits a load from the floor slab (not shown) to the stringers, then to the floor beams. The floor beams transmit force to the joints of the truss. Thus, a plane truss accepts load only at the joints. Floor system of a bridge truss. As a load is applied between the joints, the load is transmitted to the two encompassing joints by the equivalent of a simply supported beam. The resulting effect is the same as that of two forces with the magnitude as shown acting at the two joints. The magnitude of each force is a linear function of the distance from each joint. Transmission of force to truss joints. Assuming that a member force S due to a downward unit load at joint i is S i and the member force S due to a downward unit load at joint j is S j , then the member force due to a unit load applied between joints i and j and located from joint i by a distance of a is: 1 a L L L aL − L a Stringer Stringer Floor b eam Floor b eam i i j j Influence Lines by S. T. Mau 264 S = ( L aL − ) S i + ( L a ) S j We conclude that the force of any member due to a load applied between two joints can be computed by a linear interpolation of the member force due to the same load applied at each joint separately. The implication in constructing influence lines is that we need only to find the member force due to a unit load applied at the truss joints. When the member force is plotted against the location of the unit load, we can connect two adjacent points by a straight line. Example 8. Construct the influence line of member forces F IJ , F CD , and F CJ . Load is applied only at the level of lower chord members. Truss example for influence line construction. Solution. We shall use the method of section and make a cut through I-J and C-D. Two FBDs are needed: one for loads applied to the right of the section and the other for loads to the left of the section. FBD for unit load acting at x > 6m. ΣM C =0 F IJ = −1.5 R A (x > 6m) 4 m 2@3 m = 6 m B C HI R A F IJ A 1 x 4 m 6@3 m = 18 m A BCDE G HI J KL F F CJ F CD J [...]... 0.41 FCJ 1 . 25 0. 6 25 6m 1.18 9m 1. 82 Placing distributed load for maximum compression in member CJ (FCJ)max= −10[0 .5( 1. 82) (0. 6 25 )+0 .5( 9)(0. 6 25 )]= −33.8 kN (4) Distributed load of finite length 10 kN/m 6m 1 . 25 0.41 FCJ 1 . 25 0. 6 25 6m 1.18 1. 82 9m Placing finite length uniform load for maximum tension in member CJ (FCJ)max =10[0 .5( 1.18)(0.41)+0 .5( 6)(0.41)−0 .5( 1.18)(0.41)(1.18/6)] = 14. 65 kN 26 9 Influence... Influence Lines by S T Mau 10 kN/m 6m 1 . 25 0.41 FCJ 1 . 25 0. 6 25 6m 1.18 1. 82 9m Placing finite length uniform load for maximum compression in member CJ (FCJ)max = −10[0 .5( 1. 82) (0. 6 25 )+0 .5( 9)(0. 6 25 )-0 .5( 4. 82) (0. 6 25 )(4. 82/ 9)] = −33.0 kN 27 0 Influence Lines by S T Mau Problem 2 (1) Construct the influence line of member forces FHI, FHC, and FCI Load is applied only at the level of the upper chord members H A I... 1m 1kN 2m 2 kN 1kN 1 . 25 0.41 FCJ 1 . 25 0. 6 25 6m Placing group load for maximum tension in member CJ (FCJ)max= [2( 0.41)+1(0.41)(4/6)+1(0.41)(3/6)]= 1.30 kN (3) Distributed load of indefinite length 10 kN/m 1 . 25 0.41 FCJ 1 . 25 0. 6 25 6m 1.18 1. 82 9m Placing distributed load for maximum tension in member CJ (FCJ)max=10[0 .5( 1.18)(0.41)+0 .5( 6)(0.41)] = 14.7 kN 26 8 Influence Lines by S T Mau 10 kN/m 1 . 25 0.41... the influence line (FCJ)max = 2 (0. 6 25 )= 1 . 25 kN (2) Group load: The group load can be applied in any orientation Try-and-error leads to the following location of the group load 2m 2 kN 1kN 1m 1kN 1 . 25 0.41 FCJ 1 . 25 0. 6 25 7m 9m Placing the group load to maximize FCJ 26 7 Influence Lines by S T Mau (FCJ)max= − [2( 0. 6 25 )+1(0. 6 25 )(7/9)+1(0. 6 25 )(6/9)]= 2. 15 kN This is a compression force maximum To find... right of the panel Example 9 For the truss in Example 8 find the maximum force in member CJ for the four kinds of loads shown in the figure below 1m 2m 1kN 1kN 2 kN 10 kN/m 10 kN/m x 2 kN 6m A single load, a group load, and uniform loads with indefinite and finite length (1) Single concentrated load 2 kN 0.41 1 . 25 FCJ 1 . 25 0. 6 25 Placing load at peak point on the influence line (FCJ)max = 2 (0. 6 25 )= 1 . 25 ... L G 4m C B F D E 6@3 m = 18 m Problem 2. (1) (2) Construct the influence line of member forces FHI, FBI, and FCI Load is applied only at the level of the upper chord members H A I J K L G 4m B C D E 6@3 m = 18 m Problem 2. (2) 27 1 F Influence Lines by S T Mau 27 2 Other Topics 1 Introduction The present text covers mainly the two major methods of linear structural analysis, the force method and the displacement... factors to the reaction influence lines 1 RA 6m 1 RG 9m FIJ 1 1 .5 3 3 3 1 .5 FCD 0.41 1 . 25 FCJ 1 . 25 0. 6 25 Constructing member force influence lines using support reaction influence lines From the above three influence lines, we observe that the upper chord-member IJ is always in compression, the lower chord-member CD is always in tension and web266 Influence Lines by S T Mau member CJ can be in tension or... of linear static analysis or beyond, that are fundamental to structural analysis We will briefly touch on these topics and outline the relevant issues and encourage readers to study in more depth in another course of structural engineering or through self-study 2 Non-Prismatic Beam and Frame Members In actual structural design, especially in reinforced concrete or prestressed concrete design, the structural. .. 3.0 RG (x > 9m) ΣFy =0 FCJ = −1 . 25 RG (x > 6m) I FIJ J K L 4m FCJ 1 A C FCD D E 4@3 m = 12 m x G F RG FBD for unit load acting at x < 6m ΣMC =0 FIJ= − 3.0 RG (x < 6m) ΣMJ =0 FCD = 3.0 RG (x < 9m) ΣFy =0 FCJ = 1 . 25 RG (x < 6m) We need to find influence lines of RA and RG first before we can construct the influence lines of the three members IJ, CD and CJ Use the FBD of the whole truss as shown below,... 4m A B C RA 1 D E 6@3 m = 18 m x FBD of the whole truss to find reactions 26 5 G RG Influence Lines by S T Mau ΣMA =0 RG = x 18 ΣMG =0 RA = 18 − x 18 The influence lines of the two support reactions are identical in shape to those of a simply supported beam and are shown below together with the influence lines of FIJ, FCD, and FCJ which are obtained by cut-and-paste of and applying the proper factors . 1kN 2 kN 1m 2m 2 kN 10 kN/m 10 kN/m x 6 m 1 . 25 1 . 25 0. 6 25 0.41 F CJ 2 kN 1 . 25 1 . 25 0. 6 25 0.41 F CJ 2 kN 1kN 1kN 1m2m 7 m 9 m Influence Lines by S. T. Mau 26 8 (F CJ ) max = − [2( 0. 6 25 )+1(0. 6 25 )(7/9)+1(0. 6 25 )(6/9)]=. CJ. (F CJ ) max = −10[0 .5( 1. 82) (0. 6 25 )+0 .5( 9)(0. 6 25 )-0 .5( 4. 82) (0. 6 25 )(4. 82/ 9)] = −33.0 kN 10 kN/m 6 m 1 . 25 1 . 25 0. 6 25 0.41 F CJ 6 m 9 m 1.18 1. 82 Influence Lines by S. T. Mau 27 1 Problem 2. (1) Construct. =10[0 .5( 1.18)(0.41)+0 .5( 6)(0.41)−0 .5( 1.18)(0.41)(1.18/6)] = 14. 65 kN 10 kN/m 6 m 1 . 25 1 . 25 0. 6 25 0.41 F CJ 10 kN/m 6 m 9 m 1.18 1. 82 1 . 25 1 . 25 0. 6 25 0.41 F CJ 6 m 9 m 1.18 1. 82 Influence Lines by S. T. Mau 27 0 Placing