|Mar2-08, 05:04 PM||#18|
Bending Moment diagram
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.
|Mar2-08, 05:07 PM||#19|
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?
|Mar2-08, 05:43 PM||#20|
|Mar2-08, 06:41 PM||#21|
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.
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