Explanation Given information: The moment of inertia of the cross-section as follows: I y = 1.894 × 10 6 mm 4 I z = 0.614 × 10 6 mm 4 I y z = 0.800 × 10 6 mm 4 Consider the coordinates of the points Y , Z , and E are ( I y , I y z ) , ( I z , I y z ) , and ( I y + I z 2 , I y z ) . Show the points Y, Z, and E as follows: Y ( 1.894 × 10 6 mm 4 , 0.800 × 10 6 mm 4 ) Z ( 1.894 × 10 6 mm 4 , 0.800 × 10 6 mm 4 ) E ( 1.254 × 10 6 mm 4 , 0 ) Plot the points Y, Z, and E and construct the Mohr circle as shown in Figure 1. Refer to Figure 1. Calculate the radius R of the Mohr circle using the relation: R = E F ¯ 2 + F Z ¯ 2 (1) Substitute 0.640 for E F ¯ , and 0.800 for F Z ¯ in Equation (1). R = ( 0.640 × 10 6 ) 2 + ( 0.800 × 10 6 ) 2 = 1.0496 × 10 12 = 1.0245 × 10 6 mm 4 Calculate the moment of inertia about the principal axis u and v as follows: I v = E − R = 1.254 × 10 6 − 1.0245 × 10 6 mm 4 = 0.2295 × 10 6 mm 4 I u = E + R = 1.254 × 10 6 + 1.0245 × 10 6 mm 4 = 2.2785 × 10 6 mm 4 Find the orientation of the principal axis as follows: tan 2 θ m = F Z F E = 0.800 × 10 6 0.640 × 10 6 2 θ m = tan − 1 ( 1.25 ) θ m = tan − 1 ( 1.25 ) 2 θ m = 25.67 ° Calculate the moments about the principal axis v as follows: M v = M cos θ m Substitute 1.2 kN ⋅ m for M and 25.67 ° for θ m . M v = 1.2 kN ⋅ m × ( 10 6 N ⋅ mm 1 kN ⋅ m ) cos ( 25.67 ° ) = 1.2 × 10 6 ( 0.9013 ) = 1.0816 × 10 6 N ⋅ mm Calculate the moments about the principal axis u as follows: M u = − M sin θ m Substitute 1

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ISBN: 9780073398235

Author: Ferdinand P. Beer, E. Russell Johnston Jr., John T. DeWolf, David F. Mazurek

Publisher: McGraw-Hill Education

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Chapter 4.9, Problem 141P

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Problem (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/s

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 4

Chapter 4 Solutions

Mechanics of Materials, 7th Edition

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Solution Summary: The author analyzes the moment of inertia of the cross-section as follows: lY(I_y

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