Problem 1 replace the distributed loading with an equivalent...

Name: problem 1 replace the distributed loading with an equivalent resultant force and specify its location on the beam measured from point a then find the reactions at the supports 22 text knm 15 86886
Uploaded: 2024-07-31T17:26:22-08:00
Duration: 8 min 49 s
Channel: Bcrypt_Sha256$$2B$12$We1Wwocamog01O5I.V2Tkouxdh4Ofnmgpwkor7Leaonfpu0Ubfpua Bcrypt_Sha256$$2B$12$We1Wwocamog01O5I.V2Tkokttmmj7Lscvwvlptp4Rlhbswcdg9.Wy
Description: problem 1 replace the distributed loading with an equivalent resultant force and specify its location on the beam measured from point a then find the reactions at the supports 22 text knm 15 86886

Problem #1 Replace the distributed loading with an equivalent resultant force, and specify its location on the beam measured from point $A$. Then find the reactions at the supports. $22 \text{ kN/m}$ $15 \text{ kN/m}$ $A$ $B$ $3 \text{ m}$ $3 \text{ m}$ $3 \text{ m}$

Transcript

00:01 We are given the following beam.

00:05 We have a pin support here, a.

00:09 We have a roller support here.

00:13 B.

00:14 We have this distributed load that goes like here we have this line here, another line here, and it goes to zero again.

00:30 So the peak point here is 15 kilo -neutons per meter and we have 10 kiloons per meter here and the dimensions are this is three meters three meters and three meters again okay with that the question is how do we replace this distributed load with an equivalent resultant force and where do we put this okay to do that let us assume some reference frame okay we have x here and suppose this is my load axis so i start measuring my x from here and this is my load function now the task is to express this load as a function of x so we have w equal to w of x and what is that w function we are going to do it piece piecewise so w of x is equal to something when x is between zero and three and something else when x is between three and six and yet something else when x is between six and nine okay now for the first part we had this straight line that goes from the origin and it has this slope okay we have 15 over three so it is five so the first part is just 5x next we have this line now we need to do some work for it suppose this is x2 y2 and suppose this is x1 y1 let's actually replace y by w now when we know the coordinates of two points on a line we can write down the equation as follows.

02:45 Okay, we have w minus w1 over w2 minus w1 equal to x minus x1 or x2 minus x1.

02:58 For this line, we have w1 equal to 15 and w2 equal to 10.

03:10 And we have x1.

03:11 1 equal to this coordinate which is 3 and x2 is this one which is 6 so if we solve this equation for w now we obtain w equal to 20 minus 5 over 3 x and this is our second function okay now for the last part we are going to to repeat this calculation but for different points.

03:55 So this time this will be our x2, w2 and this will be x1 w1.

04:01 And we will use the same formula over here...

Question

Problem #1 Replace the distributed loading with an equivalent resultant force, and specify its location on the beam measured from point $A$. Then find the reactions at the supports. $22 \text{ kN/m}$ $15 \text{ kN/m}$ $A$ $B$ $3 \text{ m}$ $3 \text{ m}$ $3 \text{ m}$

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