Mechanics of Materials
11th Edition
ISBN: 9780137605460
Author: Russell C. Hibbeler
Publisher: Pearson Education (US)
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Chapter 11.2, Problem 32P
The tapered beam supports a uniform distributed load w. If it is made from a plate and has a constant width b, determine the absolute maximum bending stress in the beam.
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Chapter 11 Solutions
Mechanics of Materials
Ch. 11.2 - Determine the minimum dimension a to the nearest...Ch. 11.2 - of the rod to safely support the load. The rod is...Ch. 11.2 - The wood has an allowable normal stress of allow =...Ch. 11.2 - of the beam's cross section to safely support the...Ch. 11.2 - Determine the minimum dimension b to the nearest...Ch. 11.2 - The beam is made of steel having an allowable...Ch. 11.2 - Determine the minimum width of the beam to the...Ch. 11.2 - if P=10 kip.Ch. 11.2 - The beam has an allowable normal stress of allow...Ch. 11.2 - If each beam is to be designed to carry 90 lb/ft...
Ch. 11.2 - Determine its height h so that it simultaneously...Ch. 11.2 - to safely support the load. The wood has an...Ch. 11.2 - and the wood has an allowable normal stress of...Ch. 11.2 - if allow = 30 ksi and allow = 15 ksi. The journal...Ch. 11.2 - if allow = 30 ksi and allow = 15 ksi. The journal...Ch. 11.2 - The tapered beam supports a uniform distributed...Ch. 11.2 - Determine the variation in the depth d of a...Ch. 11.2 - Prob. 41PCh. 11.2 - The pulleys fixed to the shaft are loaded as...Ch. 11.2 - Prob. 43PCh. 11.2 - Prob. 44PCh. 11 - The cantilevered beam has a circular cross...Ch. 11 - Select the lightest-weight wide-flange overhanging...Ch. 11 - Prob. 3RPCh. 11 - Determine the shaft's diameter to the nearest...Ch. 11 - Select the lightest-weight wide-flange beam from...Ch. 11 - The simply supported joist is used in the...Ch. 11 - The simply supported joist is used in the...Ch. 11 - by 4-in. pieces of wood braced as shown. If the...
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- The evaporator of a vapor compression refrigeration cycle utilizing R-123 as the refrigerant isbeing used to chill water. The evaporator is a shell and tube heat exchanger with the water flowingthrough the tubes. The water enters the heat exchanger at a temperature of 54°F. The approachtemperature difference of the evaporator is 3°R. The evaporating pressure of the refrigeration cycleis 4.8 psia and the condensing pressure is 75 psia. The refrigerant is flowing through the cycle witha flow rate of 18,000 lbm/hr. The R-123 leaves the evaporator as a saturated vapor and leaves thecondenser as a saturated liquid. Determine the following:a. The outlet temperature of the chilled waterb. The volumetric flow rate of the chilled water (gpm)c. The UA product of the evaporator (Btu/h-°F)d. The heat transfer rate between the refrigerant and the water (tons)arrow_forward(Read image) (Answer given)arrow_forwardProblem (17): water flowing in an open channel of a rectangular cross-section with width (b) transitions from a mild slope to a steep slope (i.e., from subcritical to supercritical flow) with normal water depths of (y₁) and (y2), respectively. Given the values of y₁ [m], y₂ [m], and b [m], calculate the discharge in the channel (Q) in [Lit/s]. Givens: y1 = 4.112 m y2 = 0.387 m b = 0.942 m Answers: ( 1 ) 1880.186 lit/s ( 2 ) 4042.945 lit/s ( 3 ) 2553.11 lit/s ( 4 ) 3130.448 lit/sarrow_forward
- Problem (14): A pump is being used to lift water from an underground tank through a pipe of diameter (d) at discharge (Q). The total head loss until the pump entrance can be calculated as (h₁ = K[V²/2g]), h where (V) is the flow velocity in the pipe. The elevation difference between the pump and tank surface is (h). Given the values of h [cm], d [cm], and K [-], calculate the maximum discharge Q [Lit/s] beyond which cavitation would take place at the pump entrance. Assume Turbulent flow conditions. Givens: h = 120.31 cm d = 14.455 cm K = 8.976 Q Answers: (1) 94.917 lit/s (2) 49.048 lit/s ( 3 ) 80.722 lit/s 68.588 lit/s 4arrow_forwardProblem (13): A pump is being used to lift water from the bottom tank to the top tank in a galvanized iron pipe at a discharge (Q). The length and diameter of the pipe section from the bottom tank to the pump are (L₁) and (d₁), respectively. The length and diameter of the pipe section from the pump to the top tank are (L2) and (d2), respectively. Given the values of Q [L/s], L₁ [m], d₁ [m], L₂ [m], d₂ [m], calculate total head loss due to friction (i.e., major loss) in the pipe (hmajor-loss) in [cm]. Givens: L₁,d₁ Pump L₂,d2 오 0.533 lit/s L1 = 6920.729 m d1 = 1.065 m L2 = 70.946 m d2 0.072 m Answers: (1) 3.069 cm (2) 3.914 cm ( 3 ) 2.519 cm ( 4 ) 1.855 cm TABLE 8.1 Equivalent Roughness for New Pipes Pipe Riveted steel Concrete Wood stave Cast iron Galvanized iron Equivalent Roughness, & Feet Millimeters 0.003-0.03 0.9-9.0 0.001-0.01 0.3-3.0 0.0006-0.003 0.18-0.9 0.00085 0.26 0.0005 0.15 0.045 0.000005 0.0015 0.0 (smooth) 0.0 (smooth) Commercial steel or wrought iron 0.00015 Drawn…arrow_forwardThe flow rate is 12.275 Liters/s and the diameter is 6.266 cm.arrow_forward
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