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Algebra & Trigonometry with Analytic Geometry
Algebra & Trigonometry with Analytic Geometry
13th Edition
ISBN:9781133382119
Author:Swokowski
Publisher:Swokowski
Chapter11: Topics From Analytic Geometry
Section11.5: Polar Coordinates
Problem 98E
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A composite section is detailed in Fig. 3. (a) Determine the location of the centroid "C" with respect to point O; (b) Compute the moments of inertia with respect to the X and Y axes. (c) Determine the polar moment of inertia. (d) Determine the maximum deflection of the beam shown in figure 4 under given loading condition. 250-250- 150, 200 Fig. 3 150 100 300mm
Transcribed Image Text:A composite section is detailed in Fig. 3. (a) Determine the location of the centroid "C" with respect to point O; (b) Compute the moments of inertia with respect to the X and Y axes. (c) Determine the polar moment of inertia. (d) Determine the maximum deflection of the beam shown in figure 4 under given loading condition. 250-250- 150, 200 Fig. 3 150 100 300mm
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the answer is given can you explain the working outs and how did they got the centroid of middle 

A composite section is detailed in Fig. 3. (a) Determine the location of the centroid "C" with respect to point O; (b) Compute the moments of inertia with respect to the X and Y axes. (c) Determine the polar moment of inertia. (d) Determine the maximum deflection of the beam shown in figure 4 under given loading condition. E=250 GPa Solution 3: a) By virtue of symmetry, x = 0 A₁ A₂ Therefore: y = = Σ Aivi Σ Ai A1 150. A₁ (mm²) 100 x 500 50,000 = 250-250- = 300 x 200 60,000 ΣΑ, = 110,000 = -6,500,000 110,000 A2 200 = -59.1 mm 150. K 100 300mm +50 -150 A₁y₁ (mm³) +2,500,000 -9,000,000 Σαν. = -6,500,000
Transcribed Image Text:A composite section is detailed in Fig. 3. (a) Determine the location of the centroid "C" with respect to point O; (b) Compute the moments of inertia with respect to the X and Y axes. (c) Determine the polar moment of inertia. (d) Determine the maximum deflection of the beam shown in figure 4 under given loading condition. E=250 GPa Solution 3: a) By virtue of symmetry, x = 0 A₁ A₂ Therefore: y = = Σ Aivi Σ Ai A1 150. A₁ (mm²) 100 x 500 50,000 = 250-250- = 300 x 200 60,000 ΣΑ, = 110,000 = -6,500,000 110,000 A2 200 = -59.1 mm 150. K 100 300mm +50 -150 A₁y₁ (mm³) +2,500,000 -9,000,000 Σαν. = -6,500,000
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