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 Determine the internal normal force, shear force. and moment at point \( C\) .

 

 

 

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

\( B_ y(5)-400(1.2)=0 \)

\( \rightarrow B_ y=96 N \uparrow \)

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

\( -A_ y+B _y=0 \rightarrow A_ y=96 N \downarrow \)

\( \stackrel{+}\rightarrow \sum F_{x}=0 \longrightarrow -A _x+400=0 \rightarrow A_ x=400 N \leftarrow\)

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

\( N_{c}-400=0 \longrightarrow N_{c}=400 N \rightarrow \)

\( +\uparrow \sum F _y=0 \)

\( -V_{c}-96=0 \longrightarrow V_{c}=96 N \uparrow \)

\(\sum M_C=0\quad\longrightarrow M_c+96(1.5)=0\)

\(∴ M_{c}=144 N \cdot m \)

Determine the internal normal force, shear force,  and moment at points  \(C\) and  \(D\)  in the simply supported  beam. Point  \(D\)  is located just to the left of the  \(5 -kN\) force. 

 

 

\( \frac{W_C}{1.5}=\frac{3}{3} \)

\(∴ W_{c}=1.5 \mathrm{kN} / \mathrm{m} \)

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

\( B_ y(6)-5(3)-\frac{1}{2}(3)( 3 )(1)=0 \)

\(∴ B_ y=3.25 k N \)

Sam 

\( \stackrel{+}\rightarrow\sum F_{x}=0 \rightarrow N_{c}=0 \)

\( +\uparrow\sum F_ y=0 \rightarrow V_{c}+3.25-\frac{1}{2}(1.5) (1.5 )-5=0 \)

\(∴ V_{c}=2.875 k N \)

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

\( 3.25 (4.5)-\frac{1}{2}(1.5)(1.5)(0.5)-5(1.5)-M_c=0 \quad \longrightarrow M_c=6.56 kN.m \)

 Determine the internal normal force, shear force,  and moment at points  \(C\)  and  \(D\)  in the simply supported  beam. Point \( D\)  is located just to the left of the  \(5-kN\) force. 

 

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

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

\( V_{D}+3.25-5=0 \)

\(∴ V_{D}=1.75 k N \)

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

\( 3.25(3)-M_{D}=0 \)

\(∴ M_{D}=9.75 k N . m \)

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