Determine the maximum shear force vmax and maximum bending...

Determine the maximum shear force $V_{max}$ and maximum bending moment $M_{max}$ in the beam shown below: R a P P a P represents a concentrated load acting at a distance of a from both ends of the beam. R $V_{max} = 2P, M_{max} = \frac{1}{8}Pl^2$ $V_{max} = \frac{P}{4}, M_{max} = 2Pa$ $V_{max} = \frac{P}{2}, M_{max} = Pl$ $V_{max} = P, M_{max} = Pa$

Transcript

00:02 So in problem 43 here we have a beam and now we have basically it's supported on a ground where we're assuming that the reaction here is uniformly distributed.

00:15 So what we need to do is figure out the total force acting on the upper part of the beam and then divide that by the total length of the beam to figure out what this distributed load is here.

00:30 And so in this problem they only gave us, they didn't give us absolute actual numerical values.

00:35 So we just know that this has a magnitude of w and that would be some kind of force per unit length, same here, and then this would have the units of force.

00:50 And we're told that p equals a times w, which would be have the units of force.

00:57 And so we know that this distributed load has a value of three, a value of three, force w so this would be three force w and then w be again a force per unit length so we can we take cuts in these regions here they where things change and we see that in in there's zero shear force forces at the end and also zero moments at the end and so in this region here we have basically two counteracting distributed loads so we wind up with a net a net slope that's negative here.

01:38 So we wind up getting down to minus 0 .25.

01:44 And what i've plotted here is sheer force divided by a times w.

01:50 So as a increases or w increases, this value would actually increase.

01:56 So here the value would be the sheer force would be minus one quarter a times w.

02:03 Now in this region we have just the reaction from the ground and so we wind up with a positive slope and we wind up back up here at 0 .5 so v over a w is 0 .5 since p is a times w then we jump back down here to minus 0 .5 so this gives us our maximum absolute value so that occurs right here in the center and then over this region we have a positive slope again so we come back up and then over this region we have a negative slope because this is bigger than this and we come back down to zero.

02:44 So again another good check on this is symmetry because we can see we have symmetry in this problem.

02:51 So that fair check confirms that.

02:55 And so then after the moment diagram again we have linear regions everywhere so everything is going to be a quadratic function.

03:04 I wrote down what these kind of are...

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