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 Determine the horizontal and vertical components of reaction at pins  \(A\)  and  \(C\)  of the two-member frame. 

 

 

 

\( \sum M_{B}=0 \)

\( 200(3)(1.5)-A _y(3)=0 \)

\( ∴ A _y=300 N \)

\( +\uparrow \sum F_{y}=0 \)

\( A_ y+(y-200(3))=0 \)

\( C _y=600-300=300 N\)

\( \sum M_{A}=0 \)

\( C_{x}(3)-(200)(3)(1.5)=0 \)

\(∴ C_x=300 N \)

\( \stackrel{+}{\longrightarrow} \sum F_{x}=0 \)

\( c_ x+A_ x=0 \longrightarrow\quad A_x=300N\leftarrow \)

 Two beams are connected together by the short  link  \(B C\) . Determine the components of reaction at the fixed  support  \(A\)  and at pin \(D\).

 

 

 

BD

\( \sum M_{p}=0 \)
\( 10(1.5)-F_{B C}(3)=0 \)

\(∴ F_{B C}=5 K N \)

\( \stackrel{+}\longrightarrow\sum F_{x}=0 \quad \longrightarrow D_{x}=0 \)

\( +\uparrow \sum F_ y=0 \quad \longrightarrow \quad D_y-10+5=0 \)

\( D_ y=5 k N \)

AC

\( \stackrel{+}\longrightarrow\sum F_{x}=0 \quad \longrightarrow A_{x}=0 \)

\( +\uparrow \sum F_ y=0 \quad \longrightarrow \quad A_y-12-5=0 \quad \longrightarrow A_y=17kN\)

\( \sum M_ A=0\quad\longrightarrow\quad M_A-(12)(1)-5(4)=0 \ \longrightarrow M_A=32kN.m \)

 

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