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Figure 12-20 Full Alternative Text 12-21. Determine the maximum deflection of the solid circular shaft. The shaft is made of steel having E= 200 GPa. It has a diameter of 100 mm. Prob 12-21

Figure 12-20 Full Alternative Text 12-21. Determine the maximum deflection of the solid circular shaft. The shaft is made of steel having E= 200 GPa. It has a diameter of 100 mm. Prob 12-21

Elements Of Electromagnetics
Elements Of Electromagnetics
7th Edition
ISBN:9780190698614
Author:Sadiku, Matthew N. O.
Publisher:Sadiku, Matthew N. O.
ChapterMA: Math Assessment
Section: Chapter Questions
Problem 1.1MA
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This bartleby answer say v = -0.0115 m, but is should be -0.0000115 m correct?

Figure 12-20 Full Alternative Text 12-21. Determine the maximum deflection of the solid circular shaft. The shaft is made of steel having E = 200 GPa. It has a diameter of 100 mm. Prob. 12-21 6 kN.m A -X 1.5 m 8 kN с 1.5 m B 6 kN.m
Transcribed Image Text:Figure 12-20 Full Alternative Text 12-21. Determine the maximum deflection of the solid circular shaft. The shaft is made of steel having E = 200 GPa. It has a diameter of 100 mm. Prob. 12-21 6 kN.m A -X 1.5 m 8 kN с 1.5 m B 6 kN.m
← Mechanics of Materials 11th Edition C₂ = 0 Substitute (-13.5 kN m²) for C₁ and 0 for C₂ in Equation (4). EI = 2x² + 6x - 13.5 du (2x² + 6x - 13.5) Calculate the Equation of the elastic curve u Substitute (-13.5 kN m²) for C₁ and 0 for C₂ in Equation (5). Elu = 2+3x²+(-13.5) x +0 Elu = 2² + 3x² - 13.5x (+3x² - 13.5x) Calculate the moment of inertia (I) using the formula: I = d+ (7) Substitute 100 mm for d in Equation (7). 4 = U = ¹001) × 2 = 1 mm x = 4.9087 x 10-6 m4 Umax = 103 m 1mm 200 GPax- (6) Calculate the maximum deflection of the solid circular shaft (Umax): Substitute 1.5 m for x, 4.9087 x 10-6 m² for I, and 200 GPa for E in Equation (6). [2015+3(1.5)²-13.5(1.5)] 106 kN/m² 1 GPa = -0.011456 mx Chapter 12.2, Problem 21P -x4.9087x10-6 10¹ mm 1m = -11.5 mm Hence, the maximum deflection of the solid circular shaft (Umax) is -11.5 mm.
Transcribed Image Text:← Mechanics of Materials 11th Edition C₂ = 0 Substitute (-13.5 kN m²) for C₁ and 0 for C₂ in Equation (4). EI = 2x² + 6x - 13.5 du (2x² + 6x - 13.5) Calculate the Equation of the elastic curve u Substitute (-13.5 kN m²) for C₁ and 0 for C₂ in Equation (5). Elu = 2+3x²+(-13.5) x +0 Elu = 2² + 3x² - 13.5x (+3x² - 13.5x) Calculate the moment of inertia (I) using the formula: I = d+ (7) Substitute 100 mm for d in Equation (7). 4 = U = ¹001) × 2 = 1 mm x = 4.9087 x 10-6 m4 Umax = 103 m 1mm 200 GPax- (6) Calculate the maximum deflection of the solid circular shaft (Umax): Substitute 1.5 m for x, 4.9087 x 10-6 m² for I, and 200 GPa for E in Equation (6). [2015+3(1.5)²-13.5(1.5)] 106 kN/m² 1 GPa = -0.011456 mx Chapter 12.2, Problem 21P -x4.9087x10-6 10¹ mm 1m = -11.5 mm Hence, the maximum deflection of the solid circular shaft (Umax) is -11.5 mm.
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