Question 7.13: Determine the internal normal force, shear force, and moment...
Determine the internal normal force, shear force, and moment acting at point C and at point D, which is located just to the right of the roller support at B
Support Reactions: From FBD (a),
\hookrightarrow +\Sigma M_A=0 ; \quad B_y(8)+800(2)-2400(4)-800(10)=0B_y=2000 \mathrm{lb}
Internal Forces: Applying the equations of equilibrium to segment E D[\mathrm{FBD}(\mathrm{b})] , we have
\overset{+}{\longrightarrow } \Sigma F_x=0 ; \quad N_D=0
+↑ \Sigma F_y=0 ; \quad V_D-800=0 \quad V_D=800 \mathrm{lb}
\hookrightarrow +\Sigma M_D=0 ; \quad-M_D-800(2)=0
M_D=-1600 \mathrm{lb} \cdot \mathrm{ft}=-1.60 \mathrm{kip} \cdot \mathrm{ft}
Applying the equations of equilibrium to segment E C[\mathrm{FBD}(\mathrm{c})] , we have
\overset{+}{\longrightarrow }\Sigma F_x=0 ; \quad N_C=0
+↑ \Sigma F_y=0 ; \quad V_C+2000-1200-800=0 \quad V_C=0
\hookrightarrow +\Sigma M_C=0 ; \quad 2000(4)-1200(2)-800(6)-M_C=0
M_C=800 \mathrm{lb} \cdot \mathrm{ft}
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