A horizontal beam ab is freely supported at a and b 8 m...

A horizontal beam AB is freely supported at A and B 8 m apart and carries a uniformly distributed load of 15 kN/m run (including its own weight). A clockwise moment of 160 kN-m is applied to the beam at a point C, 3m from the left hand support A. Calculate the slope of the beam at C, if EI = 40 x 103 kN-m

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00:01 So i imagine, i imagine this beam as of that shape.

00:07 We'll have point a here, point b and maybe point c here.

00:13 And point c is 3 meters from point a.

00:17 And the whole of this is 8 meters.

00:21 And we're having that 160 kilonewton meter.

00:29 And the whole of this is 15 kilonewton meter.

00:34 Now what i want to do is quite simple because again, we're experiencing all these downward forces on this beam.

00:45 So the first thing to do is taking the moment at a, summation of moment at a as zero.

00:52 So therefore we'll have 160 plus 15 multiplied by 8 multiplied by 4 minus vb multiplied by 8 to give you zero.

01:07 Which means from here vb will be 80 kilonewton.

01:12 So what we have here is 80 kilonewtons.

01:17 And taking those summation of f forces vertical, we'll take vb plus vb minus 15 multiplied by 8 to give you zero.

01:32 So if you substitute vb here as 80 kilonewtons, vb therefore will be equal to 40 kilonewtons.

01:39 That is what we have here as 40 kilonewtons.

01:44 So therefore we can proceed with the equations for the beam slope.

01:52 Because we are tasked with finding basically, for us to find the slope, we need to find the equation for b, where bm, where ei is normally given as d squared, y dx squared, okay, giving you 40x minus 15x squared over 2 plus 160 into x minus 3...

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