Archives: Solution Manual

Instructor Resource Manual – Ethics: Theory and Contemporary Issues, 9e – MacKinnon & Fiala © 2018 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part, except for use […]

Instructor Resource Manual – Ethics: Theory and Contemporary Issues, 9e – MacKinnon & Fiala Chapter 12: Sexual Morality Learning Outcomes • Summarize disputes about same-sex marriage and other topics in sexual morality. • Explain key cases and examples, including recent […]

Instructor Resource Manual – Ethics: Theory and Contemporary Issues, 9e – MacKinnon & Fiala © 2018 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part, except for use […]

Instructor Resource Manual – Ethics: Theory and Contemporary Issues, 9e – MacKinnon & Fiala © 2018 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part, except for use […]

Instructor Resource Manual – Ethics: Theory and Contemporary Issues, 9e – MacKinnon & Fiala Chapter 9: Feminist Thought and the Ethics of Care Learning Outcomes • Describe the importance of feminist thought for ethical inquiry. • Explain some of the […]

Instructor Resource Manual – Ethics: Theory and Contemporary Issues, 9e – MacKinnon & Fiala Chapter 8: Virtue Ethics Learning Outcomes • Explain how virtue ethics differs from other approaches to ethics. • Describe some key virtues and how they are […]

Instructor Resource Manual – Ethics: Theory and Contemporary Issues, 9e – MacKinnon & Fiala Chapter 7: Natural Law and Human Rights Learning Outcomes • Describe the idea of natural law and how it relates to the idea of human rights. […]

Instructor Resource Manual – Ethics: Theory and Contemporary Issues, 9e – MacKinnon & Fiala Chapter 6: Deontological Ethics and Immanuel Kant Learning Outcomes • Explain the difference between consequentialist and non-consequentialist approaches to ethics. • Describe different deontological approaches to […]

Instructor Resource Manual – Ethics: Theory and Contemporary Issues, 9e – MacKinnon & Fiala © 2018 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part, except for use […]

Instructor Resource Manual – Ethics: Theory and Contemporary Issues, 9e – MacKinnon & Fiala © 2018 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part, except for use […]

Instructor Resource Manual – Ethics: Theory and Contemporary Issues, 9e – MacKinnon & Fiala Chapter 3: Ethical Relativism Learning Outcomes • Describe the difference between descriptive relativism and metaethical relativism. • Discuss criticisms of objectivism, subjectivism, relativism, and moral realism. […]

Instructor Resource Manual – Ethics: Theory and Contemporary Issues, 9e – MacKinnon & Fiala © 2018 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part, except for use […]

Instructor Resource Manual – Ethics: Theory and Contemporary Issues, 9e – MacKinnon & Fiala © 2018 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part, except for use […]

Maximum stiffener width bf−tw 214 −0. 5 26. 75 in. Compute the compressive strength: LcKL 0. 757757. 75 in. Lc r57. 75 3. 008 19. 20 From AISC J4.4, for compression elements with Lc/r25, the nominal strength is RnFyAg509. 0450. […]

modulus: Ix1 12 twh32Af htf 2 21 12 3/8733257. 573 2. 5 2 2 awhctw bfctfc 733/8 232. 50. 476 1 10 From AISC Equation F5-6, Rpg 1−aw 1200 300aw tw Fy hc −5. 7 E ≤1. 0 1−0. 4761 […]

CHAPTER 10 –PLATE GIRDERS 10.4–1 5. 70 29, 000 50 137. 3 Since 3. 76 E Fy h tw 5. 70 E Fy , the web is noncompact, and the provisions of AISC F4 apply. However, the slender-web provisions of […]

∑QnN1Qn4014. 6584. 0 kips 662 kips (N.G.) The beam flange will not accomodate 3 studs. Reduce the ∑Qnrequirement in order to increase the number of studs required. Do this by using a larger shape. 0. 85fc´b344 0. 8541200. 843 1 […]

LRFD Solution: Interpolate in Table 3-19. First, interpolate vertically (create an intermediate row), then horizontally. 378.400 428.488 439.135 442.424 305.000 405.000 416.000 bMn439 ft-kips ASD Solution: Interpolate in Table 3-19. First, interpolate vertically (create an intermediate row), then horizontally. Y2 […]

Min. longitudinal spacing 4d40. 52. 0 in. Min. transverse spacing 4d40. 52. 0 in. Max. longitudinal spacing 8t8432 in. (upper limit 36 in.) For one stud at each section, the required spacing will be sspan no. studs 4012 94 5. […]

CHAPTER 9 –COMPOSITE CONSTRUCTION 9.1–1 (a) Ecwc 1.5 fc´1451.5 43492 ksi, nEs 29, 000 b n81 810. 13 in. For a W18 40, As11. 8 in.2,d17. 9 in., Ix612 in.4 Take moments about top of slab: Component AyAyĪdĪAd2 Slab 40.52 […]

Base metal shear strength of column web: Rupture: Rn0. 45Futw20. 45650. 3702 21. 7 kips/in. 3.00 kips/in. Base metal shear strength of stiffener: Yielding: Rn0. 6Fytst 20. 6363/8216. 2 kips/in. 3.00 kips/in. Rupture: Rn0. 45Futst 20. 45583/8219. 6 kips/in. 3.00 […]

Shear areas: Agv 1 41. 5 32. 51 492. 25 in.2 Anv 1 0. 6FyAgv UbsFuAnt 0. 6362. 251. 0580. 281364. 92 kips The nominal block shear strength is therefore 64.92 kips. The design block shear strength is Rn0. 7564. […]

8.4–19 For D 3(w 3/16 in.), Cmin  C P 1 u Dℓ 0. 751. 55. 0 239  2. 7 Try a C-shaped weld. Use Table 8-8, Angle 30°. 60º kl l = 9″ 8″ xl ex8−9/2 tan 60 […]

f2xMy J68. 991. 5 28. 67 3. 610 kips/in.  8.4–8 If eccentricity is not considered, f1x150 2677. 895 kips/in.  6″ 2.50″ 7″ 3.5″ 3.5″ x Load when eccentricity is considered: Locate centroid with respect to right side. x […]

Let 0. 062 5Ru29. 82, Ru477 kips The reaction capacity is therefore based on the slip-critical strength. Ru111 kips (b) ASD solution 0. 1Ra7. 380, Ra73. 8 kips Tension: ∑MNA 2rt36236rt MaPae2. 25Ra From 36rt2. 25Ra,rt0. 062 5Ra Interaction of […]

CHAPTER 8 –ECCENTRIC CONNECTIONS 8.2–1 Direct shear components: Px3 5159kips, Py4 51512 kips 8.2–2 Eccentricity: ex33410 in., ey5. 5 22. 75 in. ∑x2y2262323262102. 752255. 6 in.2 (a) Direct shear components: PxPucos 30 ° 0. 866 Pu,PyPusin 30 ° 0. 5Pu […]

will be used. To determine the required length of the longitudinal welds, investigate 3. 712 28. 29 in. length of longitudinal welds 28. 29 −5 211. 65 in. For the second option, the strength of the longitudinal welds is 0. […]

For the hole nearest the edge, ℓcℓ e−h 21. 5 −0. 9375 21. 031 in. Rn 1 1. 2ℓctFu1 2. 00 1. 21. 0313/858 Summary: Use 6 bolts for tension member-to-gusset plate connection; use 6 bolts for gusset-plate-to connection angles […]

Ant 3 83−11. 438 0. 585 8 in.2 0. 6FyAgv UbsFuAnt 0. 63610. 881. 0580. 5858269. 0 kips Rn0. 75269. 0201. 8 kips 220 kips (N.G.) Try ℓe3 in. and s41 2in. Agv 3 834. 53212. 38 in.2 Anv 3 […]

CHAPTER 7 –SIMPLE CONNECTIONS 7.3–1 (a) Minimum spacing 22 3d2. 6677/82. 33 in. 2.75 in. (OK) Minimum edge distance from AISC Table J3.4 1.125 in. actual ℓe(OK) 2. 4dtFu2. 47/83/85845. 68 kips 26.91 kips ∴use Rn26. 91 kips Rn26. 9 […]

br  2Pr Lb 18 125. 776 kips/in. 2. 00 2311. 9 The length of the brace is L18 12/sin21. 80 °581. 6 in. Let AE Lcos2 2Pr Lb 5. 776 A5. 776L Ecos25. 776581. 6 29000 cos221. 8 °0. […]

Pa 2Pn/c Max Mnx/b May Mny/b 92 2520214 297 31 135 1. 03 1. 0 (N.G.) Try a W14 82. Calculate B1for each axis: Pe1x2EIx Lcx2229000881 0. 8 16 1221. 069 104kips B1xCmx 0. 3009 0. 305 1 1.0 1−1. […]

Pe1y2EIy K1L22EIy KyL2229000116 12 1221601 kips B1yCmy 1−Pr/Pe1yCmy 1−1. 00Pu/Pe1y1. 0 1−28. 0/16011. 018 Muy B1Mnty B2Mℓty 1. 018100. 80102. 6 ft-kips Compressive strength: From Manual Table 6-2, for a W10 60 with KL 12 ft, cPn633 kips Pu cPn […]

CHAPTER 6 –BEAM–COLUMNS 6.2–1 (a) LRFD solution: From Manual Table 6-2, the compressive design strength of a W12 106 with Fy50 ksi and LcKyL1. 0 14 14 feet is cPn1130 kips From Table 6-2, for Lb14 ft and Cb1. 0, […]

5.14–4 (a) The factored load is 0. 65 0. 853A1 1212 A1 ≥104, A127. 34 in.2 Check upper limit: c1. 7fc´A10. 651. 7327. 3490. 6 kips 104 kips (N.G.) Let c1. 7fc´A1Pu,0.651. 73A1104 A131. 37 in.2 The plate must be […]

5.11–4 Dead loads: Slab weight 4. 5 12 15056. 25 psf wD56. 25 55. 5336. 9 lb/ft (neglect beam wt. and check it later.) Ix510 in.4588 in.4required (N.G.) Try a W18 40, Ix612 in.4,MnbMp294 ft-kips Check beam weight: Mu214 1 […]

From the Zxtable, bMnbMp203 ft-kips 192 ft-kips (OK) Vu144 kips 131 kips (N.G.) Beam not adequate (vVncan also be found in the Zxtable.) (b) Pa60 kips 6′ 90 k 30 k 1′ 1′ 60 k 60 k 30 k -30 […]

Cb1.67 5.5–11 =15′ 18.86′ 26.41 k 29.29 k 26.41 k – 29.29 k – 8.59 k – 15.29 k Mmax 26. 4118. 86−1. 418. 862/2 249. 1 ft-kips MA26. 4113. 75−1. 413. 752/2 230. 8 ft-kips MB26. 4117. 5−1. 417. […]

CHAPTER 5 –BEAMS 5.2–1 (a) Flange area 0. 57. 53. 75 in.2 3. 75 3. 188 6. 682 in. ZA 2a3. 75 3. 18826. 68292. 72 in.3 MpFyZ5092. 724636 in-kips 386 ft-kips Z92. 7 in.3,Mp386 ft-kips (b) Moment of inertia: […]

UseanHSS10x8x1/4 (b) PaDL33 82 115 kips 4.7–8 (a) Column AB:GA10, GB∑Icol/Lcol ∑Ig/Lg 2510/20 0. 991 2303/12 From the alignment chart, Kx≈1. 90 Kx1. 90 (b) Column BC:GCGB0. 991 and Kx1. 31 (c) Column AB: Lcx rx KxL 1. 9012 12 […]

CHAPTER 4 –COMPRESSION MEMBERS 4.3–1 2. 65 45. 28 Fe2E Lc/r2229000 45. 282139. 6 ksi 4. 71 E Fy 4. 71 29000 50 113. 4 Since Lc/r45. 28 113. 4, use AISC Eq. E3-2. Fcr 0. 658Fy/FeFy0. 65850/139.65043. 04 ksi […]

with an upper limit of 0. 6FyAgv UbsFuAnt 0. 6365. 6251. 0580. 7969167. 7 kips The nominal block shear strength of the gusset plate is therefore 167.7 kips. The gusset plate controls, and the nominal block shear strength of the […]

CHAPTER 3 –TENSION MEMBERS 3.2–1 For yielding of the gross section, For rupture of the net section, Ae3/87−13 16 2. 180 in.2 PnFuAe582. 108122. 3 kips a) The design strength based on yielding is tPn0. 9094. 585. 05 kips The […]

CHAPTER 2 – CONCEPTS IN STRUCTURAL STEEL DESIGN 2-1 D = 9 kips, Lr = 5 kips, S = 6 kips, R = 7 kips, W = 8 kips (a) 1: 1.4D = 1.4(9) = 12.6 kips 3: 1.2D + […]

1.5-3 A Pf P A    (. ) / . . 0510 4 02043 250 250 02043 1224 2 in. For lb, psi 2 CHAPTER 1 – INTRODUCTION 1.5-1 (a) A  (. ) / .0550 4 02376 […]

1 Chapter 13: Intercultural Communication and Health Care Martin, Experiencing Intercultural Communication, 6e Chapter 13 Intercultural Communication and Health Care Study Objectives After studying the material in this chapter, students should be able to accomplish the following objectives. 1. Understand […]

175 Chapter 17 17.1 Tube 1: 56.4924.18.50 =−=−= tDD oi mm Eq. (17.6): 10%5.06% ≤=× − = − =100 )56.49( )56.49()8.50( )100((%) 2 22 2 22 i io R D DD A Tube 2: mm 55.7465.12.76 =−=−= tDD oi Eq. […]

159 Chapter 16 16.1 .28 ° = ′ φ From Table 16.1, = c N31.61; = q N 17.81; = γ N 13.70 )7.13)(2)(19( )81.17)(19)(5.1()61.31)(31( 2 kN/m 499.4=      ++=      ++ […]

Chapter 15 15.1 Eq. (15.15): β ββ tan tancos 2 + =Hγ F s φ tan ′ ′ c m 6.28= += += H H H 886.1 634.7 1.3 19tan 33tan )19)(tan19)(cos)(20( 47 1.3 2 15.2 Eq. (15.16): ft 41.94= […]

Chapter 14 14.1 From Figure 14.4, for ° = ′ 40 φ and ° = ′ 20δ , K p ≈ 9. So, kN/m 2,997== 2 )6)(5.18)(9( 1 P 2 p 14.2 From Table 14.1, for ° = ′ 40 […]