Dua kayu persegi panjang memiliki ukuran 75 x 125 mm akan disambung seperti gambar diatas. Jika diketahui tegangan tarik maksimum yang diijinkan dalam sambungan tersebut adalah 500 kPa, maka tentukan besar beban P dan tegangan geser tersebut
\begin{aligned} A_o&=&\left(0,075\right)\left(0,125\right)=9,{375x10}^{-3}m^2\\ \theta\ \ &=& 90°-70°=20°\\\ \sigma&=&\frac{P}{A_o}{cos}^2\theta\\ P&=&\frac{A_o\sigma}{{cos}^2\theta}\\ P&=&\frac{\left(9,{375x10}^{-3}\right)\left(500{x10}^3\right)}{{cos}^2{20}^o}\\ P&=&5,3085\ x{10}^3N\\ P&=&5,31kN\\\\\\ \tau&=&\frac{Psin2\theta}{2A_o}\\ \tau&=&\frac{\left(5,3085\ x{10}^3\right)sin40}{2\left(9,375\ x\ {10}^{-3}\right)}\\ \tau&=&181,99\ x\ {10}^3Pa\\ \tau&=&182\ kPA \end{aligned}
\begin{aligned} A_o&=&\left(0,075\right)\left(0,125\right)=9,{375x10}^{-3}m^2\\ \theta\ \ &=& 90°-70°=20°\\\ \sigma&=&\frac{P}{A_o}{cos}^2\theta\\ P&=&\frac{A_o\sigma}{{cos}^2\theta}\\ P&=&\frac{\left(9,{375x10}^{-3}\right)\left(500{x10}^3\right)}{{cos}^2{20}^o}\\ P&=&5,3085\ x{10}^3N\\ P&=&5,31kN\\\\\\ \tau&=&\frac{Psin2\theta}{2A_o}\\ \tau&=&\frac{\left(5,3085\ x{10}^3\right)sin40}{2\left(9,375\ x\ {10}^{-3}\right)}\\ \tau&=&181,99\ x\ {10}^3Pa\\ \tau&=&182\ kPA \end{aligned}
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