The discussion revolves around calculating the equivalent resultant force for a distributed load, specifically focusing on the method of dividing the load into parts and summing forces and moments. The subject area includes mechanics and statics.
There is ongoing clarification regarding the calculations presented, with some participants providing guidance on interpreting the distributed load and suggesting alternative approaches. Multiple interpretations of the problem are being explored, and participants are actively questioning assumptions and definitions.
Participants note confusion regarding the problem statement and the representation of forces and moments in the calculations. There is mention of a concentrated load and the need for precise definitions of terms used in the calculations.
What do you mean by "about x" etc.? You seem to be confusing forces and moments. The 6 sin 30 is a force in the y direction, but the 3(3) looks more like a moment, though I can't match it to anything in the diagram (there is no 3kN force anywhere), and the 5(4) might be the moment of the 5kN force at D about E.goldfish9776 said:
3(3) + (5-3)(0.5)(3) + 5(4) + (7-5)(4)(0.5) , each term here represent the different region for the load distribution diagram , i already stated it in the diagramharuspex said:
You did indeed, but I still cannot see how you are getting these terms.goldfish9776 said:
You really should learn the formula for calculating the area of a trapezoid:goldfish9776 said:
SteamKing said:You really should learn the formula for calculating the area of a trapezoid:
It saves quite a bit of time. If b1 or b2 is zero, this formula reduces to that of finding the area of a triangle, A = (1/2)hb.
You should be aware that the load shown at point D is a 5 kN concentrated load. It is a separate load from the distributed load.
The value of the distributed load at point D is not 5 kN/m. The distributed loading changes by 4 kN/m over a distance of 7 m, so the value of the distributed loading at point D is calculated by linear interpolation. This step is unnecessary if you know how to calculate the area of a trapezoid, as shown above.
There is also a couple of magnitude 5 kN-m acting CCW shown at point C, which doesn't seem to appear in your calculations.
The problem statement as given is confusing, as haruspex noted. Please provide the exact wording for this problem from your text.
Yes, though I would have done it as 5+(3+4)(3+7)/2, the (3+7)/2 being the average spread load.goldfish9776 said:
