# Structural Analysis Hibbeler 8th Edition Solutions Manual

Structural Analysis Hibbeler 8th Edition Solutions Manual – Important Notice Scheduled Server Maintenance (GMT) on Sunday, June 26th from 2:00 AM to 8:00 AM. The site is not available during the specified hours.

330 CHAPTER 8 DEFLECTIONS EXAMPLE 8.14 Determine the maximum deflection of 8 kN for the steel beam shown in Fig. 8-26a. Answers were calculated. E = 200 GPa, I = 60 (106) mm4. _1_8 EI 2kN 9m B A¿ B¿ real beam 9m 3m 3m 6kN (a) coupled beam Fig. A coupled beam loaded with diagram M/EI 2_7_ is shown in Fig. 8-26b. Since the M/EI diagram is positive, the EI line load acts upward (away from the beam). Balance. First, the external reactions on the 6m 4m 2m coupled beam are determined and given in the free-body diagram of Figure 8–26c. The maximum true beam deflection occurs at 4__5 6_3_ where the beam tilt is zero. This corresponds to the same EI EI point in the coupled beam where the shear is zero. Assuming that this external reaction point operates within 0 … x … 9 m of AAA, we can isolate part (c) shown in Figure 8–26d. Note that the peak distributed stress was determined from the proportional triangles, ie. H. w > x = (18 > EI) > 9. We require V¿ = 0 so that 8 _18_ ( 9_x ) ϭ 2_x_ + c ©Fy = 0; – 45 + 1 a 2x bx = 0 EI EI EI 2 EI A¿ _4_5 x = 6.71 m 10 … x … 9 m2 OK EI M¿ For this value of x, the maximum deflection in the real beam corresponds to the moment M. Thus, x V¿ ϭ 0 d+ ©M = 0; 45 16.712 – c 1 a 216.712 b 6.71 d 1 16.712 + M¿ = 0 EI 2 EI 3 201.2 kN # m3 internal reactions ¢max = M¿ = – (d) EI #= – 201.2 k>Nm [6011062 mm411 m4> 110324 mm42] = – 0.0168 m = – 16.8 mm ans. A negative sign indicates that the deviation is downward. www.TechnicalBooksPdf.com

## Structural Analysis Hibbeler 8th Edition Solutions Manual

8.5 MID-BEAM METHOD 331 EXAMPLE 8.15 The beam in Fig. 8-27a consists of a continuous beam and is braced by cover plates in the middle, where its moment of inertia is greatest. The 12-ft end sections have a moment of inertia of I = 450 in4, and the center section A CB has a moment of inertia of I¿ = 900 in4. Determine the deflection I = 450 in4 I¿ = 900 in4 I = 450 in4 at center C. Take E = 29(103) ksi. 12 ft 6 ft 6 ft 12 ft Answers calculated. 10 k 10 k SOLUTION true beam bent beam. First, the moment diagram of the beam (a), Figure 8-27b, is determined. Since IА = 2I, for convenience we can express the coupled beam loading as a constant EI, as shown in Figure 8–27c. Figure 8-27 Equilibrium. Composite beam responses can be calculated from loading symmetry or using equilibrium equations. The results are shown in Figure 8-27d. Since the deflection at C must be determined, the internal moment at C¿ must be calculated. In the section method, section A¿C¿ is isolated and the resultant line loads and their positions are determined, Figure 8–27e. So 1116 720 360 36 + d + ©MC¿ = 0; 1182 – 1102 – 132 – 122 MC¿ = 0 EI EI EI EI #MC¿ 11 736 k ft3 M (kift) EI = – 120 144 120 If we add the numerical data of EI and convert the units, we have 1241. moment diagram #11 736k ft311728 in3>ft32 ans. x (ft) (b) 36 ¢C = MC¿ = – 2911032 k>in21450 in42 = – 1.55 in. 8 A negative sign indicates that the deviation is downward. 1__20_ _1_2_0_0_ _7_2_ 7__2_0_ _36_0_ Ei ei ei ei ei _3_6_0_ Ei _6_0_0 Vcst 10 foot 8 foot 12 foot 6 foot 12 foot 1__1_1_1_1_1_1_1_1_1_1_1_1

## Structural Analysis 5th Edition Aslam Kassimali Solutions Manual

332 CHAPTER 8 DEFLECTIONS EXAMPLE 8.16 Determine the bolt displacement at B and the slope of each beam section connected to the solid for the composite beam shown in Fig. 8-28a. E = 29 (103) ksi, I = 30 in4. 8k 8k A B 30 kèft 30 kèft 12 ft 12 ft 15 ft C A ⌬B C (uB)R B (uB)L Tensile curve of a real beam (a) (b) Figure 8-28 SOLUTION Bending beam. The elastic curve of the beam is shown in Fig. 8-28b to indicate the unknown displacement ¢B and the slopes (uB)L and (uB)R to the left and right pin. Using Table 8-2, the coupled beam is shown in Figure 8-28c. To simplify the calculation, the parts of the M/EI diagram are created according to the superposition principle, as described in Chap. 4-5 The real beam is assumed as sCuypp=or2t, kA, a.nTdhtehem3o0-mke#nftt diagrams from the left, where the reaction force load is indicated. Note that the negative areas of this graph develop the distributed load downward, while the positive areas develop the distributed load upward. 8 1__52__1 –E7–8I– EI _3_.6_ 11 ft EI 15 ft B¿ A¿ C¿ ϪE3–0I–– B¿ _2_2_8_.6_ EI 12 ft Coupled Beam _1_1_7_0 –I5 –2Eft. ft 12 ft _57__6 EI external reactions (c) (d) www.TechnicalBooksPdf.com

8.5 THE KHONYUGAT RAY METHOD 333 equilibrium. First, the external reactances of BÀ and CÀ are calculated _22_5_ EI and the results are shown in Figure 8-28d. To determine (uB)R, the bowed beam is cut immediately to the right of B¿ and the 5 ft _3_.6_ shear strength (VB¿)R is calculated, Fig. 8-28 Thus, MB¿ EI B¿ 15 ft (VB¿)R + c ©Fy = 0; 1VB¿2R + 225 – 450 – 3.6 = 0 EI EI EI 7.5 ft _4_5_0 #1uB2R = 1VB¿2R = ft2 EI 228.6 k EI (e) #= 228.6 k ft2  #= 228.6 k ft2  0, 0378 radAns. _2_25_ EI Internal moment at B¿ causes bolt displacement. Thus, 5 ft _3_.6_ MB¿ EI 15 ft d + ©MB¿ = 0; – MB¿ + 225 152 – 450 17.52 – 3.6 1152 = 0 B¿ EI EI EI (VB¿)L #¢B 2_2_8_._6 7.5 ft _45_0_ EI = MB¿ = – 2304 EI -I ( f) 2304 kft3 [291103211442 k>ft2][30>11224] ft4 = – 0.381 ft = – 4.58 inches Answer: 8 The slope (uB)L can be found from the section of the bar immediately to the left of B¿, Figure 8–28f. So +c ©Fy = 0; 1VB¿2L + 228.6 + 225 – 450 – 3.6 = 0 EI EI EI EI 1uB2L = 1VB¿2L = 0 Answer: Clearly, ¢B = MB¿ for this section is the same as previously calculated, since the moment arms in Figures 1 and 2 are only slightly different. 8-28e and 8-28f. www.TechnicalBooksPdf.com

334 CHAPTER 8 CURVES BASIC PROBLEMS F8–10. Use the moment area theorems and determine F8–16. Using the moment-area theorems, find the slope at A and the deflection at A. EI is constant. The slope at A and the displacement at C are constant. F8-11. Solve prob. F8-10 using conjugate beam F8-17. Solve prob. F8-16 using the conjugate beam method. Method: 6kN 8kN A BA B C 3m 3m 3m F8-10/8-11 F8-16/8-17 F8-12. Use the moment surface theorems and determine F8–18. Using the moment-area theorems, find the slope at B and the deflection at B. EI is constant. The slope at A and the displacement at C are constant. F8-13. Solve prob. F8-12 using conjugate beam F8-19. Solve prob. F8–18 using the conjugate beam method. Method: 4kN 4kN 2m 2m A 8kNèm A B B C 8 4m 4m 4m F8-18/8-19 F8-12/8-13 F8-14. Use the moment surface theorems and determine F8–20. Using the moment-area theorems, find the slope at A and the displacement at C. EI is constant. Slope at B and displacement at B. EI is constant. F8-15. Solve prob. F8-14 using conjugate beam F8-21. Solve prob. F8-20 using the conjugate beam method. Method: 9kN 5kNèm A B B C A 2m 1.5m 1.5m 2m F8-14/8-15 F8-20/8-21 www.TechnicalBooksPdf.com

8.5 THE CONJUGATE PATH METHOD 335 PROBLEMS 8–10. Determine the slope at B and the maximum 8-18. Determine the deflection and displacement of beam C. EI during displacement. Use the moment area theorems. is permanent. Use the moment area theorems. Take E = 29(103) ksi, I = 500 in4. 8-19 Solve Prob. 8-18 using the bending beam method. 8-11 Solve prob. 8-10 using the bending beam method. P 15 k A C B A C 6

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