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31 Pages

### ch1-1

Course: ECEN 215, Spring 2008
School: Texas A&M
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Word Count: 6999

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Compression, 1 Tension, and Shear Normal Stress and Strain Problem 1.2-1 A solid circular post ABC (see figure) supports a load P1 2500 lb acting at the top. A second load P2 is uniformly distributed around the shelf at B. The diameters of the upper and lower parts of the post are dAB 1.25 in. and dBC 2.25 in., respectively. (a) Calculate the normal stress AB in the upper part of the post. (b) If it is desired...

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Texas A&M - ECEN - 215
32CHAPTER 1Tension, Compression, and ShearProblem 1.6-10 A flexible connection consisting of rubber pads (thickness t 9 mm) bonded to steel plates is shown in the figure. The pads are 160 mm long and 80 mm wide. (a) Find the average shear strai
Texas A&M - CVEN - 205
2Axially Loaded MembersChanges in Lengths of Axially Loaded MembersProblem 2.2-1 The T-shaped arm ABC shown in the figure lies in a vertical plane and pivots about a horizontal pin at A. The arm has constant cross-sectional area and total weight
Texas A&M - CVEN - 205
80CHAPTER 2Axially Loaded MembersProblem 2.3-8 A bar ABC of length L consists of two parts of equal lengths but different diameters (see figure). Segment AB has diameter d1 100 mm and segment BC has diameter d2 60 mm. Both segments have length
Texas A&M - CVEN - 205
106CHAPTER 2Axially Loaded MembersProblem 2.5-3 A rigid bar of weight W 750 lb hangs from three equally spaced wires, two of steel and one of aluminum (see figure). The diameter of the wires is 1/8 in. Before they were loaded, all three wires h
Texas A&M - CVEN - 205
122CHAPTER 2Axially Loaded MembersStresses on Inclined SectionsProblem 2.6-1 A steel bar of rectangular cross section (1.5 in. 2.0 in.) carries a tensile load P (see figure). The allowable stresses in tension and shear are 15,000 psi and 7,000
Texas A&M - CVEN - 205
134CHAPTER 2Axially Loaded MembersProblem 2.6-16 A prismatic bar is subjected to an axial force that produces a tensile stress 63 MPa and a shear stress 21 MPa on a certain inclined plane (see figure). Determine the stresses acting on all faces
Texas A&M - CVEN - 205
144CHAPTER 2Axially Loaded MembersProblem 2.7-9 A slightly tapered bar AB of rectangular cross section and length L is acted upon by a force P (see figure). The width of the bar varies uniformly from b2 at end A to b1 at end B. The thickness t
Texas A&M - CVEN - 205
160CHAPTER 2Axially Loaded MembersStress ConcentrationsThe problems for Section 2.10 are to be solved by considering the stress-concentration factors and assuming linearly elastic behavior. Problem 2.10-1 The flat bars shown in parts (a) and (
Penn State - CHE - 210
CHAPTER ELEVEN11.1 a. The peroxide mass fraction in the effluent liquid equals that in the tank contents, which is:xp =Mp MTherefore, the leakage rate of hydrogen peroxide is m1 M p / M b. Balance on mass: Accumulation = input outputEdM =
Penn State - CHE - 210
CHAPTER TWO2.1 (a)= 18144 10 9 ms . 1 wk 1 d 1 h 1 s 38.1 ft / s 0.0006214 mi 3600 s (b) = 25.98 mi / h 26.0 mi / h 3.2808 ft 1 h 3 wk 7d 24 h 3600 s 1000 ms(c)554 m 4 1d 1h d kg 24 h 60 min1 kg 108 cm 4 = 3.85 10 4 cm 4 / min g 1000 g 1
Penn State - CHE - 210
CHAPTER THREE3.1 (a) m =16 6 2 m3 1000 kg 2 10 5 2 103 2 105 kg 3 mbgb gb gd i4 106(b) m =8 oz 2s1 qt106 cm31g332 oz 1056.68 qt cmb3 10gd10 i3 1 102 g / s(c) Weight of a boxer 220 lb m 12 220 lb m 1 stone Wmax
Penn State - CHE - 210
CHAPTER FOUR4.1 a. b. Continuous, Transient Input Output = Accumulation No reactions Generation = 0, Consumption = 06.00c.dn kg kg dn kg - 3.00 = = 3.00 dt dt s s st=100 m3 1000 kg 1 s . = 333 s 1 m3 3.00 kg4.2a. b. c.Continuous, S
Penn State - CHE - 210
CHAPTER FIVE5.1Assume volume additivity Av. density (Eq. 5.1-1):1=0.400 0.600 + = 0.719 kg L 0.703 kg L 0.730 kg LAAOa.Dmmass of tank at time tA= mt + m0 m =mass of empty tankAb250 - 150gkg = 14.28 kg min bm = mass f
Penn State - CHE - 210
CHAPTER SIX6.1 a.AB: Heat liquid - -V constantBC: Evaporate liquid - -V increases, system remains at point on vapor - liquid equilibrium curve as long as some liquid is present. T = 100 o C.CD: Heat vapor - -T increases, V increases .b. Point
Penn State - CHE - 210
CHAPTER SEVEN0.80 L 35 10 4 kJ 0.30 kJ work . 1h 1 kW = 2.33 kW 2.3 kW h L 1 kJ heat 3600 s 1 k J s7.12.33 kW 10 3 W 1.341 10 -3 hp 1 kW7.21W= 312 hp 3.1 hp .All kinetic energy dissipated by friction(a) E k =mu 2 2 5500 lbm 552 mil
Penn State - CHE - 210
CHAPTER EIGHT8.1 a.U (T ) = 25.96T + 0.02134T 2 J / molU (0 o C) = 0 J / mol U (100 o C) = 2809 J / mol Tref = 0 o C (since U(0 o C) = 0)b. We can never know the true internal energy. U (100 o C) is just the change from U (0 o C) to c.U (100 o
Penn State - CHE - 210
CHAPTER NINE9.14 NH 3 (g) + 5O 2 (g) 4NO(g) + 6H 2 O(g) H ro = -904.7 kJ / mola.When 4 g-moles of NH3(g) and 5 g-moles of O2(g) at 25C and 1 atm react to form 4 g-moles of NO(g) and 6 g-moles of water vapor at 25C and 1 atm, the change in enth
Penn State - CHE - 210
CHAPTER TEN10.1 b. Assume no combustionn 1 (mol gas),T1 (C) x 1 (mol CH4 /mol) x 2 (mol C2 H6 /mol) 1 x 1 x 2 (mol C3 H8 /mol) n 2 (mol air), T2 (C) n 3 (mol), 200C y 1 (mol CH /mol) 4 y 2 (mol C2 H6 /mol) y 3 (mol C3 H8 /mol) 1 y 1 y 2 y 3 (m
Virginia Tech - CHE - 2114
CHAPTER TWO2.1 (a)= 18144 10 9 ms . 1 wk 1 d 1 h 1 s 38.1 ft / s 0.0006214 mi 3600 s (b) = 25.98 mi / h 26.0 mi / h 3.2808 ft 1 h 3 wk 7d 24 h 3600 s 1000 ms(c)554 m 4 1d 1h d kg 24 h 60 min1 kg 108 cm 4 = 3.85 10 4 cm 4 / min g 1000 g 1
Virginia Tech - CHE - 2114
CHAPTER THREE3.1 (a) m =16 6 2 m3 1000 kg 2 10 5 2 103 2 105 kg 3 mbgb gb gd i4 106(b) m =8 oz 2s1 qt106 cm31g332 oz 1056.68 qt cmb3 10gd10 i3 1 102 g / s(c) Weight of a boxer 220 lb m 12 220 lb m 1 stone Wmax
Virginia Tech - CHE - 2114
CHAPTER FOUR4.1 a. b. Continuous, Transient Input Output = Accumulation No reactions Generation = 0, Consumption = 06.00c.dn kg kg dn kg - 3.00 = = 3.00 dt dt s s st=100 m3 1000 kg 1 s . = 333 s 1 m3 3.00 kg4.2a. b. c.Continuous, S
Virginia Tech - CHE - 2114
CHAPTER SIX6.1 a.AB: Heat liquid - -V constantBC: Evaporate liquid - -V increases, system remains at point on vapor - liquid equilibrium curve as long as some liquid is present. T = 100 o C.CD: Heat vapor - -T increases, V increases .b. Point
Virginia Tech - CHE - 2114
CHAPTER FIVE5.1Assume volume additivity Av. density (Eq. 5.1-1):1=0.400 0.600 + = 0.719 kg L 0.703 kg L 0.730 kg LAAOa.Dmmass of tank at time tA= mt + m0 m =mass of empty tankAb250 - 150gkg = 14.28 kg min bm = mass f
Virginia Tech - CHE - 2114
CHAPTER SEVEN0.80 L 35 10 4 kJ 0.30 kJ work . 1h 1 kW = 2.33 kW 2.3 kW h L 1 kJ heat 3600 s 1 k J s7.12.33 kW 10 3 W 1.341 10 -3 hp 1 kW7.21W= 312 hp 3.1 hp .All kinetic energy dissipated by friction(a) E k =mu 2 2 5500 lbm 552 mil
Virginia Tech - CHE - 2114
CHAPTER EIGHT8.1 a.U (T ) = 25.96T + 0.02134T 2 J / molU (0 o C) = 0 J / mol U (100 o C) = 2809 J / mol Tref = 0 o C (since U(0 o C) = 0)b. We can never know the true internal energy. U (100 o C) is just the change from U (0 o C) to c.U (100 o
Virginia Tech - CHE - 2114
CHAPTER NINE9.14 NH 3 (g) + 5O 2 (g) 4NO(g) + 6H 2 O(g) H ro = -904.7 kJ / mola.When 4 g-moles of NH3(g) and 5 g-moles of O2(g) at 25C and 1 atm react to form 4 g-moles of NO(g) and 6 g-moles of water vapor at 25C and 1 atm, the change in enth
Virginia Tech - CHE - 2114
CHAPTER TEN10.1 b. Assume no combustionn 1 (mol gas),T1 (C) x 1 (mol CH4 /mol) x 2 (mol C2 H6 /mol) 1 x 1 x 2 (mol C3 H8 /mol) n 2 (mol air), T2 (C) n 3 (mol), 200C y 1 (mol CH /mol) 4 y 2 (mol C2 H6 /mol) y 3 (mol C3 H8 /mol) 1 y 1 y 2 y 3 (m
Virginia Tech - CHE - 2114
CHAPTER ELEVEN11.1 a. The peroxide mass fraction in the effluent liquid equals that in the tank contents, which is:xp =Mp MTherefore, the leakage rate of hydrogen peroxide is m1 M p / M b. Balance on mass: Accumulation = input outputEdM =
Cornell - ECE - 2100
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N.C. State - MA - 241
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