Archives: Solution Manual

0 1 2 0 1 2 © 0 1 2 0 1 2  0 1 2 3 0 0.5 1 1.5 2 Euler (h=0.5) Euler (h=0.25) Midpoint Analytical RK4 © © ©  0 2 4 6 0 0.2 […]

y = -0.1618x R2 = 0.9758 -10 -8 -6 -4 -2 0 010 20 30 40 50 60 ©  ©  © © © 0 1 2 3 0 0.2 0.4 0.6 0.8 t(d) SOD (g/m2/d) 0 1 2 […]

©  © ©  © © © © © © © © © © © © © 0 0.4 0.8 1.2 1.6 010 20 30 40 50 60 70 80 0 0.01 0.02 0.03 0.04 0.05 VDerivative   […]

©  © ©   © ©     © © 01000 2000 3000 4000 5000 6000 7000 3 4 5 6 7 8 9 10 11 12 13 © © © © © © ©

            ©       © ©     ©       © © 050 100 150 200 250 300 […]

© 0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 2 -20 0 20 40 60 80 100 © © © 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 -10 -8 -6 -4 -2 0 2 4 6 […]

© © © © 020 40 60 80 100 120 140 160 180 0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 5            ©  © © 01000 […]

© © © © © © © © 0 2 4 6 8 10 4.5 5 5.5 6 6.5 7 7.5 8 8.5 ©   © ©  

0 2 4 6 8 10 0 2 4 6 © © 0 1 2 3 050 100 © © © 0 0.2 0.4 0.6 0.8 1 1.2 0500 1000 © © © © y = 0.0182×4 – 0.1365×3 + […]

© © © © y = -2.015000E-06x + 9.809420E+00 R2 = 9.999671E-01 9.5 9.6 9.7 9.8 9.9 020000 40000 60000 80000 100000 120000 © © © © © ©     ©  

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050 100 150 200 250 300 350 400 2 4 6 8 10 12 14 16 18 20 22 Plot of temperature (C) versus time (days) t (days) T (C) ©          […]

©   10-2 10-1 100101102 10-12 10-10 10-8 10-6 10-4 y = 0.4945x + 20.6 R2 = 0.8385 0 10 20 30 40 50 010 20 30 40 50 60 y = 9.9529×0.3851 R2 = 0.9553 0 10 20 […]

 -500 0 500 1000 1500 2000 020 40 60 80 100 © ©   0 500 1000 1500 2000 2500 020 40 60 80 100 ©  -2 0 2 4 6 8 0 5 10 15 © […]

0.22 0.23 0.24 0.25 0.26 0.27 0.28 0.29 0 100 200 (a) Distribution of drag cd (kg/m) 60 65 70 75 0 100 200 (b) Distribution of mass m (kg) 32 32.5 33 33.5 34 0 100 200 (c) Distribution […]

© 0100 200 300 400 0 100 200 300 x y Fit (untransformed) 10-1 100101102103 10-5 100 105 log(x) log(y) Fit (transformed)  0.0001 0.001 0.01 0.1 1 10 10 100 1000  y = 38148x-3.0134 R2 = 0.9757 0 […]

© © 0 1 2 3 4 5 6 0.8 1.2 1.6 2.0 2.4 Frequency 0 1 2 3 4 5 6 0.8 1.2 1.6 2.0 2.4 Frequency ©  © 0 1 2 3 4 5 6 7 8 […]

For four interior points (h 3/5), the resulting system of equations is Setting the determinant equal to zero and expanding it gives 4 )2 + which can be solved for the first four eigenvalues (b) The system can be normalized […]

        0 4 0 4 0 4 © © © © ©  The solution along with its second derivative can be substituted into the simultaneous ODEs. After simplification, the result is © […]

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• • • 0 20 40 60 80 0 2 4 6 8 10 1.6 2 2.4 V r © 1.2 0.20.20.7 0.30.5 1.2 0.5 0.3 0.1 1.2 0.20.20.7 0.30.5 1.2 0.5 0.3 0.1 © 50 60 70 80 0 […]

© © © © © -40 -20 0 20 40 60 0 4 8 12 © © $1,200 $1,300 $1,400 $1,500 $1,600 $1,700 M M J A S Bi-monthly Monthly 0.0% 1.0% 2.0% 0 0.5 1 1.5 2 2.5 relative […]

9.15 Using the column interaction diagrams in Appendix A ( c f′= 4000 psi and y f= 60,000 psi) with ρt approximately equal to 2%, design square columns with equal reinforcement on all sides to carry each of the following […]

8.1. A rectangular beam is 16 in. wide and 29 in. deep. For c f′= 4000 psi and y f= 60,000 psi, determine the required spacing of No. 4 (No. 13) closed stirrups for a factored shear of 80 kips […]

6.1 through 6.6 (based on the beam in Fig. 4.15, Example 4.13, fy = 60 ksi) For the No. 6 bars: ( ) 2.5 in. 1.5 0.5 0.75 2 2.375 in. 10 2 1.5 0.5 0.75 2 2 2.625 in. […]

5.1 A rectangular beam secƟon is 10 in. wide and has a structural depth of 13.5 in. Given that the concrete strength is 4000 psi and there is no shear reinforcement in the secƟon, determine the nominal shear capacity allowed […]

2/2 fs0.60fy  fs36000 psi ρ As bd  ρ0.011kρn() 22ρn ρn k 0.325j1 k 3  j 0.892 a  a 3.84 in MnAsfy da     Mn440 ft kip 0.85f’cb 2   (e) Compute the […]

7.5vf’c. f’c4000psi Es29000000psi 24“ 36″ 2‐½” Ec57000 f’cpsi3.605 106 psi b36in h24in Ig bh 3   typ a)f’c=4000psi 12 41472 in4 Mcr 7.5 f’cpsi Ig  h 2 1639 in kip Reinforcement details Calculation of transformed section n Es […]

3.1. A 16 × 20 in. column is made of the same concrete and reinforced with the same six No. 9 (No. 29) bars as the column in Examples 3.1 and 3.2, except t hat a steel with yield strength […]

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\Jo t, . l6t.o o f ftfi . – q4h ;; r– ::. , ,,*o-i , , , ,,,,= +*r*.i,p5i,r .,, %;+ W; i,. +1zv – s34 — + r?4 ftl *z– -Lffi1 +5?4 = ?q(q ft; LU %o. {, […]

21.1 Determine the required embedment, hef, for a 5/8 in. diameter hex head bolt away from edges in 4000 psi cracked concrete with no supplemental reinforcement and carrying a sustained dead load of 6000 lbs, Figure P21.1. Given: fy= 36 […]

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1/2 Problem 17.5 A column transfers a factored load of 700 kips to a 9-ft square footing shown in Figure P17.5, resulting in a factored uniform soil pressure of 8640 psf. Design the footing reinforcement using strut-and-tie methods. Material strengths […]

16.1. A cantilever retaining wall is to be designed with geometry as indicated in Fig. P16.1 . Backfill material will be well–drained gravel having unit weight w = 120 pcf, internal friction angle ϕ ‘ = 33 , and friction […]

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Notes regarding reinforcement details. In the following problems, the reinforcement details are compressed to clarify the problem solutions. The reinforcement details are noted by the line: Reinforcement details Reinforcement is shown by bar size. Thus a No. 9 (No. 29) […]