The beam is supported by a fixed support at A and a roller...

Name: The beam is supported by a fixed support at A and a roller support at B. A 10,000-lb load is applied at point C. Find the reaction forces at A and B, draw the shear and moment diagrams, and find the maximum magnitude of normal stress in the beam. [The beam is a W8X31 (I = 110 in?, c = 4 in) section. Both the beam and rods are A36 steel (E = 29 X 10? psi).]
Uploaded: 2023-08-09T05:28:19-08:00
Duration: 8 min 9 s
Channel: Ethan Deweese
Description: The beam is supported by a fixed support at A and a roller support at B. A 10,000-lb load is applied at point C. Find the reaction forces at A and B, draw the shear and moment diagrams, and find the maximum magnitude of normal stress in the beam. [The beam is a W8X31 (I = 110 in?, c = 4 in) section. Both the beam and rods are A36 steel (E = 29 X 10? psi).]

The beam is supported by a fixed support at A and a roller support at B. A 10,000-lb load is applied at point C. Find the reaction forces at A and B, draw the shear and moment diagrams, and find the maximum magnitude of normal stress in the beam. [The beam is a W8X31 (I = 110 in?, c = 4 in) section. Both the beam and rods are A36 steel (E = 29 X 10? psi).]

Transcript

00:01 So in this problem we have our beam, so we should start with a free body diagram.

00:04 At a we have a cantilever, so we have some force in the a in the y direction, some force of a in the x direction.

00:13 Now we know this is immediately going to be zero.

00:17 A moment reaction, ma, which we could potentially have.

00:23 Force b from the roller, we know we only have the vertical component because it's a roller, and the 10 ,000 pounds of force downwards, and everything is 48 inches apart.

00:38 Now this problem looks almost symmetrical, and it's not quite because we have the extra reactions, although we're going to see that it is actually a little bit of symmetry in disguise.

00:49 So let's look at sum of moments is equal to zero.

00:54 This beam is in equilibrium.

00:55 Look at a point a, and counterclockwise positive.

01:00 So ma is there, so we need to include that.

01:03 The forces at a have no distance to the axis, so they make no moments.

01:08 10 ,000 pounds is a negative moment, and the definition is the force times the distance, so 48 inches.

01:17 Fb tends to rotate the other way, so it's a positive moment.

01:22 And 2 times 48, 96 inches away.

01:25 Now we have one equation, two unknowns, so not really helpful.

01:30 Let's look at sum of moments b equal to zero, same rotation direction, and we can do this as an equal, valid equation to take.

01:41 So from our second axis we have this being a positive moment.

01:46 Ma is still a positive moment because it doesn't really matter where it's applied.

01:50 Now fa is a negative moment, and so taking the moments at b, we have ma plus 10 ,000 times 48 minus fa 96 equal to zero.

02:13 Let's subtract these equations together.

02:16 So this is equation one, equation two.

02:19 Let's do equation one minus equation two.

02:22 We see the ma's are going to cancel, so that's our key here that it's actually equal to zero.

02:28 It's going to be symmetrical, and we have negative 20 ,000 times 48 plus fb 96 plus fa 96 equal to zero.

02:50 So one equation, two unknowns, still not good enough, and so let's look at sum of forces in the y direction is equal to zero.

02:59 Now sum of forces in the x directly gives us that x force, x reaction is zero, so we don't really need to worry about that.

03:07 So the force of a and the force of b is equal to, or not equal to, minus the 10 ,000 pounds is equal to zero.

03:21 So now we can solve, this is equation, i guess equation three and equation four.

03:27 Now we can solve these two equations, two unknowns...

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