When encountering a frame, we need to determine axial force developed in it, in addition to shear force and bending moment. For a statically determinate frame, we can use equilibrium equations to solve for the reaction forces, as always.
After that, we segmentize our frame based on the point where load pattern and orientation of members change. We then proceed with determining internal force functions using equilibrium equations again, and noting that those functions are valid within the limit of the segment.
With the derived internal force functions, we can calculate the internal forces i.e. axial force, shear force and bending moment at the points of interest. Then, these points are plotted on frame member and by joining them with a continuous line, we produce the axial force, shear force and bending moment diagrams.
The following shows the procedure to produce axial force, shear force and bending moment diagrams for frame. Watch the video above for full details.
Frame overview
4m height simply supported frame with 6m base length. 5m length inclined member slanting at 53.13 degree from horizontal.
Resolve the load acting on inclined member into transverse and longitudinal components.
Support reaction
Solve for By directly using equilibrium equation.
Support reaction
Solve for Ay using force polygon.
Support reaction
Solve for Ax using force polygon.
Segmentization of frame
The frame is divided based on the orientation of members.
Resolving the reactions into transverse and longitudinal components.
Segment 1 (valid for 0 to 5m from support A)
Segment 2 (valid for 0 to 4m from support B)
Segment 3 (valid for 0 to 3m from vertical member)
Internal forces calculated for every 0.5m interval
Axial force diagram for frame
Shear force diagram for frame
Bending moment diagram for frame
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