Bending Moment diagram


by nemesis24
Tags: bending, diagram, moment
nemesis24
nemesis24 is offline
#19
Mar2-08, 05:07 PM
P: 21
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
Nesha is offline
#20
Mar2-08, 05:43 PM
P: 14
Quote Quote by nemesis24 View Post
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?
Exactly that. Just calculate moments from left to right and say =0 (or from right to left). Sorry on bad english.
TVP45
TVP45 is offline
#21
Mar2-08, 06:41 PM
P: 1,130
Quote Quote by nemesis24 View Post
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?
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.


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