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5. Calculate the centroid of a wire rod bent into the shape of an arc defined by x(y) = 3y² from the origin to point (x,y) = (12,4). Hint: follow the procedure outline in Example 9.1 and SHOW EVERY DETAIL of your work.

5. Calculate the centroid of a wire rod bent into the shape of an arc defined by x(y) = 3y² from the origin to point (x,y) = (12,4). Hint: follow the procedure outline in Example 9.1 and SHOW EVERY DETAIL of your work.

Structural Analysis
Structural Analysis
6th Edition
ISBN:9781337630931
Author:KASSIMALI, Aslam.
Publisher:KASSIMALI, Aslam.
Chapter2: Loads On Structures
Section: Chapter Questions
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EXAMPLE 9.1 Locate the centroid of the rod bent into the shape of a parabolic arc as shown in Fig. 9-8a. SOLUTION Differential Element. The differential element is shown in Fig. 9-8a. It is located on the curve at the arbitrary point (x, y). Area and Moment Arms. The differential element of length dL can be expressed in terms of the differentials dx and dy, Fig. 9-8b, using the Pythagorean theorem. dL = V V(dx)² + (dy)² Since x = y², then dx/dy= 2y. Therefore, expressing dL in terms of y and dy, we have 302 = y = dL = √(2y)² + 1 dy As shown in Fig. 9-8a, the centroid of the element is located at x = x, y = y. Integrations. Applying Eq. 9-5 and using the integration formula to evaluate the integrals, we get 1 m 1 m [√x dL [" "xV4y² + 1 dy [ "3² √4y² + 1 dy 0 x = 1 m TAL dL 0.6063 1.479 0 dx dy 1 m V V4y² + 1 dy = 0.410 m + 1 dy 1 m fydl flyv fyV4y² + 1 dy dL 1 m dLV4y² + 1 dy √4y² √4y² + 1 dy (6 800X) 0.8484 1.479 = 0.574 m Ans Ans. NOTE: These results for C seem reasonable when they are plotted on Fig. 9-8a. on 01 y = y O (x, y) x = x 1 m- C(x, y) dL (a) = y² dL aldy dx (b) Fig. 9-8 1 m
Transcribed Image Text:EXAMPLE 9.1 Locate the centroid of the rod bent into the shape of a parabolic arc as shown in Fig. 9-8a. SOLUTION Differential Element. The differential element is shown in Fig. 9-8a. It is located on the curve at the arbitrary point (x, y). Area and Moment Arms. The differential element of length dL can be expressed in terms of the differentials dx and dy, Fig. 9-8b, using the Pythagorean theorem. dL = V V(dx)² + (dy)² Since x = y², then dx/dy= 2y. Therefore, expressing dL in terms of y and dy, we have 302 = y = dL = √(2y)² + 1 dy As shown in Fig. 9-8a, the centroid of the element is located at x = x, y = y. Integrations. Applying Eq. 9-5 and using the integration formula to evaluate the integrals, we get 1 m 1 m [√x dL [" "xV4y² + 1 dy [ "3² √4y² + 1 dy 0 x = 1 m TAL dL 0.6063 1.479 0 dx dy 1 m V V4y² + 1 dy = 0.410 m + 1 dy 1 m fydl flyv fyV4y² + 1 dy dL 1 m dLV4y² + 1 dy √4y² √4y² + 1 dy (6 800X) 0.8484 1.479 = 0.574 m Ans Ans. NOTE: These results for C seem reasonable when they are plotted on Fig. 9-8a. on 01 y = y O (x, y) x = x 1 m- C(x, y) dL (a) = y² dL aldy dx (b) Fig. 9-8 1 m
5. Calculate the centroid of a wire rod bent into the shape of an arc defined by x(y) = 3y² from the origin to point (x,y) = (12,4). Hint: follow the procedure outline in Example 9.1 and SHOW EVERY DETAIL of your work.
Transcribed Image Text:5. Calculate the centroid of a wire rod bent into the shape of an arc defined by x(y) = 3y² from the origin to point (x,y) = (12,4). Hint: follow the procedure outline in Example 9.1 and SHOW EVERY DETAIL of your work.
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