Bending Moment diagram

TVP45

On the SFD, you're going in a straight line from +381 kN at reaction A to -367 kN at reaction B, so the zero crossing would be (381/(381+367))*4.30 m = 2.19 m. See if you get that also.

And, you're right that you wouldn't just drop in a straight line from the -28.9 kNm to 0 on the BMD. The parabola starts at the -28.9 kNm cause that's where the UDL starts. Zero crossing can be found by setting the summations = to 0. I always just sort of guess the parabola shape unless it's something criticial; then I plot maybe a dozen points.

The important thing to remember is that the maximum BM is wherever the SFD goes through zero.

You're doing great. Keep at it.

nemesis24

im sorry but what do you mean by setting the summations = to 0 is that just taking moments at a point and saying it is zero? also wouldnt the UDL have to cross 0 on the SFD at 2.15 as that is 4.3/2?

Nesha

Exactly that. Just calculate moments from left to right and say =0 (or from right to left). Sorry on bad english.

TVP45

Look at the SFD. From A to B, you have a left side that is 381 kN. On the right side, you have -367 kN. They are joined by a straight line. They are similar triangles. The sum of the bases is 4.30 m. The base of the left hand, by similar triangles, is (4.30 m) X (381 kN/748 kN) = 2.19 m. 748 kN is just 381 kN - (-367 kN).

About the summation. Go down to the BMD. You can see by examination that the zero crossing will be a little to the right of reaction A point. Let's call that distance s meters. Move to s and calculate the bending moments. You have (without units)
-17 X (1.70 + s ) = -28.9 - 17s
398 X s = 398s
-((174 ) X s) X (s/2) = -87s^2
Sum them to zero
-87s^2 + 381s - 28.9 = 0
s = 0.077 m and 4.30 m
You already knew 4.30 m, so the one you want is 0.077 m to the right of reaction A.