Menurunkan rumus titik berat penampang persegi dan segitiga
Setelah mempelajari Teori Varignon serta mendapatkan persamaan, maka teori tersebut serta dan persamaan akan kita gunakan untuk mencari titik berat dari beberapa penampang pada tulisan ini pada penampang persegi dan segitiga :
1. Penampang Persegi
a. Terhadap sumbu X
\(Y \times A = \int_{0}^{h} y \times dA\)
\(Y \times A = \int_{0}^{h} y \times b \times dy\)
\(Y \times A = b \times \left [ y \times dy \right ]_{0}^{h}\)
\(Y \times A = b \times [\frac{1}{2} y^{2}]_{0}^{h}\)
\(Y \times A = \times \frac{1}{2}h^{2}\)
\(Y \times\left [ b \times h \right ] = b \times \frac{1}{2}h^{2}\)
\(Y = \frac{\frac{1}{2}b \times h^{2}}{b \times h}\)
\(Y = \frac{1}{2}h\)
b. Terhadap sumbu Y
\(X \times A = \int_{0}^{b} x \times dA\)
\(X \times A = \int_{0}^{b} x \times h \times dx\)
\(X \times A = h \times \left [ x \times dx \right ]_{0}^{b}\)
\(X \times A = h \times [\frac{1}{2}x^{2}]_{0}^{b}\)
\(X \times A = h \times \frac{1}{2}b^{2}\)
\(X \times \left [h \times b \right ] = \frac{1}{2}h.b^{2}\)
\(X = \frac{\frac{1}{2}h.b^{2}}{h.b}\)
\(X = \frac{1}{2}b\)
2. Penampang Segitiga
a. Terhadap sumbu X
\(Y \times A = \int_{0}^{h} y \times dA\)
\(Y \times A = \int_{0}^{h} y \times b^{'} \times dy\)
\(Y \times A = \int_{0}^{h} \frac{b.y.\left ( h-y \right )}{h} \times dy\)
\(Y \times A = \int_{0}^{h} b.y - \frac{b.y^{2}}{h} \times dy\)
\(Y \times A = \left [ \frac{1}{2}b.y^{2}-\frac{1}{3}\frac{b.y^{3}}{h} \right ]_{0}^{h}\)
\(Y \times A = [ \frac{1}{2}b.h^{2}-\frac{1}{3}b.h^{2}]_{0}^{h}\)
\(Y \times \frac{1}{2}b.h = \frac{1}{6}b.h^{2}\)
\(Y = \frac{1}{3}h\)
b. Terhadap sumbu Y
\(Y \times A = \int_{0}^{b} x \times dA\)
\(Y \times A = \int_{0}^{b} x \times h^{'} \times dx\)
\(Y \times A = \int_{0}^{b} \frac{h.x.\left ( b-x \right )}{b} \times dx\)
\(Y \times A = \int_{0}^{b} h.x - \frac{h.x^{2}}{b} \times dx\)
\(Y \times A = \left [ \frac{1}{2}h.x^{2}-\frac{1}{3}\frac{h.x^{3}}{b} \right ]_{0}^{b}\)
\(Y \times A = [ \frac{1}{2}h.b^{2}-\frac{1}{3}h.b^{2}]_{0}^{b}\)
\(Y \times \frac{1}{2}h.b = \frac{1}{6}h.b^{2}\)
\(Y = \frac{1}{3}b\)
Ditulis oleh ; Civil Engineering Library