# Problem 442 - Analysis of Frame by Method of Members

**Problem 442**

Each member of the frame shown in Figure P-348 weighs 50 lb per ft. Compute the horizontal and vertical components of the pin pressure at C, D, and F.

**Solution 442**

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$W_{AB} = 50(12) = 600 ~ \text{lb}$

$W_{CE} = 50(12) = 600 ~ \text{lb}$

$W_{DF} = 50\sqrt{8^2 + 6^2} = 500 ~ \text{lb}$

From the FBD for Member CE

$8D_V = 6(600) + 12(2000)$

$D_V = 3\,450 ~ \text{lb}$

$\Sigma F_H = 0$

$D_H = C_H$

$\Sigma M_D = 0$

$8C_V + 2(600) = 4(2000)$

$C_V = 850 ~ \text{lb}$

From the FBD for Member DF

$6D_H = 4(500) + 8(3450)$

$D_H = 4933.33 ~ \text{lb}$

$\therefore \, C_H = 4933.33 ~ \text{lb}$

$\Sigma F_H = 0$

$F_H = D_H$

$F_H = 4\,933.33 ~ \text{lb}$

$\Sigma F_V = 0$

$F_V = 500 + 3450$

$F_V = 3950 ~ \text{lb}$

Checking:

$\Sigma M_B = 0$

$12R_A = 4(500) + 6(600) + 12(2000)$

$R_A = 2466.67 ~ \text{lb}$

$\Sigma F_H = 0$

$B_H = R_A$

$B_H = 2466.67 ~ \text{lb}$

$\Sigma F_V = 0$

$B_V = 600 + 500 + 600 + 2000$

$B_V = 3700 ~ \text{lb}$

From the FBD of Member AB

$\Sigma F_H = 0$

$2\,466.67 + 4\,933.33 - 2\,466.67 - 4\,933.33 = 0$ (*okay!*)

$\Sigma F_V = 0$

$3\,700 + 850 - 600 - 3\,950 = 0$ (*okay!*)

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