Problem 443 - Analysis of Frame by Method of Members | Engineering Mechanics Review at MATHalino

Problem 443
The frame shown in Figure P-443 is hinged to rigid supports at A and E. Find the components of the hinge forces A and E and the forces in members BC and BD.

Solution 443

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From FBD for Support Reactions

$\Sigma M_E = 0$

$2A_H = 2(12)$

$A_H = 12 ~ \text{kN}$ answer

$\Sigma F_H = 0$

$E_H = A_H$

$E_H = 12 ~ \text{kN}$ answer

$\Sigma F_V = 0$

$E_V + A_V = 12$ †

From the FBD of Member AB

$\Sigma F_V = 0$

By Symmetry
$A_V = B_V = \frac{1}{2}(12)$

$A_V = B_V = 6 ~ \text{kN}$ answer

Substitute AV = 6 kN to †
$E_V + 6 = 12$

$E_V = 6 ~ \text{kN}$ answer

From FBD for Section Below M-M

$\Sigma M_D = 0$

$1.5F_{BC} = 2.5(6)$

$F_{BC} = 10 ~ \text{kN tension}$ answer

$\Sigma F_H = 0$

$\frac{3}{5}F_{BD} = 12$

$F_{BD} = 20 ~ \text{kN compression}$ answer

Problem 443 - Analysis of Frame by Method of Members

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