Is it 8.3 for this one because of the reaction at the right hand support which is 8.3? Or why are you saying to use 8333? I used 8.4 which was the bending moment for the equation that gave the result 12.5x10^6
Here's how I got 8333 N-m: The reaction forces of the supports at the ends are 6666N at one end and 8333 at the other (did you get these same values?). One meter from the support with 8333N is the point of maximum bending moment, which is 8333N times 1 meter = 8333 N-m.
I got 6.7 kN and 8.3 kN the maximum bending moment was 8.4 kNm 2m from the left hand end is what I have. Therefore the maximum stress in the beam 100x200mm was 12.5x10^6. The new dimensions are 100mmx240mm and the maximum stress I got was 8750.....
At least we agree on where the max moment is (I said 1 m from the right end, and you said 2 m from the left end, which is the same thing). But you apparently did some rounding in your calculation of the reaction forces. To find the reaction force at the right end take sum of moments about the left end and set it equal to zero:
Sum M = 0 = -5kN x 1m - 10 kN x 2m + R x 3m
R = (5+20)Kn-m/3m = 8333 n.
You can use the same technique to find the force at the left end, or use sum of forces = 0. Either way you get R at the left end is 6666 n.
I was taught to have the reaction forces in kN not n though hence the 8.3kN. So what do you have for all the equations so far on the first page and see if wee have any others which match up plus then I know the end figures Iam trying to get for all the questions then I can work from there with formulas............
If you want to summarize your results, starting with the moment of 8.3kN-m instead of the 8.4 you were using previously, let's see what you have.
from the top..... reaction force 1 is 6.7kN reaction force 2 is 8.3kN
Maximum bending moment is 8.4kNm at 2metres from the left hand end
Maximum stress is 12.5x10^6 for 100mmx200mm
still don't have the tensile and compressive forces. What do you have there and I can work out why you got it?....
new dimensions are 100mmx240mm
New maximum stress for 100mmx240mm is 8750 don't know or not sure how to convert this to a power of ten figure. Help required.
Then just have to get the percentage.....
so almost there I hope.....
Looking good. Two comments:
1. The powers of ten for the max stress given the new dimensions is the same as for the original. So 8.75 x 10^6 Pa. I;m not sure why you are struggling with this - it's a straightforward calculation of Mc/I = 8.4x10^3 N x 0.12m/1.15x10^-4 m^4
2. You have asked several times about tensile and compressive forces, but I don't understand (a) why you're asking about it (it's not asked for in the original problem, and in any event (b) it doesn't make sense. I think what you mean is the graph of stress (not force) across the depth of the beam. I've tried to explain that to you several times. You know the max stress at the top and bottom edges of the beam, and you know it's 0 at the midpoint, so draw the straight line graph.
You have already calculated the max stress, which occurs at the section of the beam 2 meters from the left end across the top edge of the beam as a compressive stress and the bottom edge as a tensile stress. At any point in the beam the stress is My/I, where M is function of the x-coordinate of the beam (and we already know that at x=2 meters M is a max), and y is the y-coordinate measured from the center line of the beam. My suggestion is draw the profile where you already have the max value for stress. It should look something similar to the right hand side of example: