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Some useful techniques for solving shear force and bending moment diagrams.bilal
- 1. “Some useful techniques for solving shear force and bending moment diagrams” Muhammad Bilal 1st year- CED Uet Peshawar
- 2. Loads • Concentrated load Also called point load • Uniformly distributed load It becomes easy to solve when we convert UDL to a point load.
- 3. Things to keep in mind • Consider this figure • The area under the SFD above the x axis should equal the area between the x-axis and the SFD below the x axis. i.e. the area should sum to zero. Check this is true in our example. • Any points where the SFD cross the x-axis, will be a max or min Bending Moment • The SFD should always equal zero at both ends
- 4. How to select sections/Cuts We know when we are solving SFD and BMD we select different sections of beam and calculate shear force and moment in each section separately. each section bear different answer because there is different loading in each section. Consider this (source meriam statics)
- 5. Calculate reactions I started with a FBD and calculated the two reactions named as Ra and Rb. From categories of equilibrium It is a parallel force system so we need to find moment at any point and a force equation in the direction of forces. Such that Ra= p/3 and Rb=2p/3
- 6. Section 1 0<x<2L/3 We see that there is only one force in this first section which is actually the reaction force. We get the shear force and bending moment It should be noted that the shear force P/3 is same throughout the section but moment varies with x which means that it will be different at different point in this section
- 7. Section 2 Section 1 was before P and section 2 is after P. We can see that now in section 2 the load P is included so its shear and moment will be different from section 1.
- 8. Section 2 ( another possibility) As the total length of beam is L we can also select section 2 like this 2L/3 <X <L We can see the answer is same.in this section the load P is not included.
- 9. Shear force diagram(explanation) The diagram can be easily drawn by looking at the equations which we have derived. Now look at SFD In first section 1 ‘V was P/3’ as you see in the diagram then there is no force so it moves on horizontally when it reaches Position 2 it experience a force P opposite in direction ( P/3-P=-2P/3) again no force is their in the section 3-4 so horizontally goes no . At 4 it experience a upward force equal in magnitude opposite in direction which push it to 0.
- 10. Shear moment diagram(explanation) • And moment For section 1 it was P/3x when x=2L/3 moment becomes 2/9PL which is actually the maximum moment • For section 2 it was 2P/3(L-x) when x=L it reaches zero which can be seen in the moment diagram
- 11. Things to remember In case of uniformly distributed load we will always get a curve in moment diagram ( mostly parabola). If we take the clockwise moments positive we will get a negative parabola and vice versa.
- 12. In most of the cases • In case of horizontal line in shear force diagram we will get a linear line in bending moment diagram • And in case of linear line in shear force diagram we will get a curve in moment diagram
- 13. Still having problems Please watch https://youtu.be/iaS5VLX0NFQ?list=LLrHy- WQKbvgcLmuS5J4Z9qg https://youtu.be/IMHits0fs-g?list=LLrHy- WQKbvgcLmuS5J4Z9qg https://youtu.be/01ig6qgRj14?list=LLrHy- WQKbvgcLmuS5J4Z9qg