Assume the supports of the uniform cantilever shown in Fig. 12...

Question

Assume the supports of the uniform cantilever shown in Fig. 12–79 (m = 2900 kg) are made of wood. Calculate the minimum cross-sectional area required of each, assuming a safety factor of 9.0.

Diagram of a uniform cantilever with forces and dimensions labeled, illustrating support reactions and load distribution.

Accepted Answer

Hey, everyone in this problem, we're given a figure with a uniform horizontal steel H beam of mass 1500 kg that rests on two columns of aluminum. We're asked to find the minimum cross sectional area of the two columns to support the beam. If a safety factor of eight is required, we're told to assume that the ultimate strength for tension and compression for aluminum are both 2.0 multiplied by 10 to the exponent eight newtons per meter squared. So let's start with the figure we're given. And so we have our steel beam, a horizontal steel beam. It is 30 m long. The first column supporting the beam is on the far left end. And we have this force T one pointing upwards at that point 10 m from the left end, we have the second column with a force T two pointing upwards. We have our center of gravity acting on that beam as well and that's it for the diagram. OK. So we have four answer choices in this problem options A through D, each of them containing a different cross sectional area for these two columns. And all of these answer choices are in meters squared. So we're gonna come back to these answer choices once we're done working through this problem. Now, in this problem, we're given some information about the safety factor. We're given information about the ultimate strength for aluminum. We wanna find a cross sectional area. So we wanna think about the relationship between all of these things. So the first thing we can recall is that our stress, which we can write as the force divided by the cross sectional area is equal to our ultimate strength divided by the safety factor. So if we write our equation like this, we can rearrange for that cross sectional area we wanna find. And so A is going to be equal to the force F multiplied by the safety factor divided by the ultimate mystery. Yeah, we call it the ultimate strength is that maximum strength that can be applied without breaking and in this case without breaking those columns. OK? So we have this equation for our cross sectional area. A we know the safety factor, we know the ultimate strength. So what we need to figure out is the force F, OK? Now that force is gonna be different, we have the force T one and T two. So that force is different for each of those columns, which gives us those different cross sectional areas. And that's where those are gonna come into play. So let's start with the first one and for the first force we're gonna start with T one, we wanna find the force T one. And what we're gonna do is we're gonna look at our torque and in order for this column to be in equilibrium, what we're going to say is that the sum of the torques must be equal to zero, right? We're gonna take the torque about the point where the force T two acts and we're gonna do that because of the force is acting at the pivot point. It's not going to contribute to the torque. So T two is not going to be in our equation. That's gonna leave the force T one and the center of gravity which we can calculate. OK. That's one unknown we'll be able to solve for T one. So that's what we're gonna start with. And you could start the other way if you want it as well, you could take the torque about the point where T one act and find the force T two instead. Um But we'll start with T one. OK. So for T one, we are gonna look at the sum of the torques and we said that's gonna be equal to zero. Now, if we can imagine from our diagram that we look at the force T one, it's pointing upwards to the left of our pivot point. So we can imagine that's gonna cause clockwise rotation or a negative torque. So we have negative the torque from T one. Now we have that center of gravity, it's acting downwards to the right of our pivot. Again, that's gonna cause clockwise rotation, which is a negative torque. So we have minus tau the center of gravity and this is going to be equal to zero. Those are the only two forces we have that are acting away from that pivot. Now, how can we write the torque? We'll recall that the torque is going to be the force multiplied by the distance to the pivot are multiplied by sine of theta where theta is the angle between the force vector and the displacement vector pointing from the pivot to the force. So we can write our equation as negative T one multiplied by R one multiplied by sine of theta one minus the force acting at the center of gravity. FCG multiplied by RCG sign of the CG. That's all equal to zero. Hey, let's fill in what we know if we go up to our diagram. OK. If we look from the pivot to our force T one, we know that that's a distance of 10 m. The force is acting perpendicular to our support beam. So theta is gonna be 90 degrees time of theta will be one. OK. Same thing, the other direction. OK. We have this center of gravity, we're gonna assume that this beam is uniform. And so the center of gravity is gonna act directly at the middle. The total length is 30 m. And so this center of gravity will act 15 m from either end if it's 15 m from either end, that means it's gonna be 5 m from the pivot point. And we T two X if we think about the angle, once again, it is perpendicular to making a 90 degree angle with that support beam. And so sin of the is going to be one as well. OK. So with all of that, we can substitute into our equation, we get negative T one multiplied by 10 m minus 1500 kg multiplied by 9.8 m per second squared. OK. This is calculating the force of gravity. So the mass multiplied by the acceleration due to gravity multiplied by that distance, which we said was 5 m. This is all equal to zero. Now, we can add our 1500 multiplied by 9.8 multiplied by five to the right hand side. And we need to divide by 10 m to get our T one value. And we find that T one will be equal to negative 7350 newtons. OK. So we have now T one, OK. So that is that first force that we were looking for. Now that we have T one, we can calculate the cross sectional area we need for this particular column. OK. So we have that A one is gonna be equal to this force which is going to be T one multiplied by the safety factor. Divided by the ultimate history. Combining all of this A one will be equal to 7350 noons. And when we take this force, when we're looking for the cross sectional area, we want the magnitude of the force. And so we're taking the magnitude of that force. The sign just indicates the, the direction and that depends on how we define our directions. But for our area, we care about that magnitude. I multiplied by the safety factor, which is eight, all divided by the ultimate strength, which are told is two multiplied by 10 to the exponent eight Newton's per meter squared. If we work all of this out, we get an area a one of 0.000 2 9 4 m squared. And so that is the area we need for that first column. Now we're gonna repeat the process for the second column and there's a couple different ways we could go about this for the second column. We could again look at the sum of torques, but we would need to take the pivot about a different point. So that T two is included in the equation or we can look at the sum of the forces since we now know what T one is and that's what we're gonna do. That's gonna be a little bit simpler. Um So we'll take that approach. OK. So right now we're gonna find T two and we're gonna look at the sum of the forces. And if we look at the sum of the forces in the y direction, we know this is gonna be equal to 00 and equilibrium, which is what we want. We want these, this beam to be supported. We don't want these columns breaking or falling um or rotating. OK. So the sum of the forces is going to be equal to zero. If we take up to be our positive direction, then in the positive direction, we have T one and T two. And in the negative direction, we have that force of gravity or that force of the center of gravity. And so what we get is T one plus T two minus FG is equal to zero. We know T one, we can calculate FG again. So we can write that T two is going to be equal to that force of gravity minus T one. This will be the max 1500 kg multiplied by the acceleration due to gravity 9.8 m per second squared minus T one T one is negative. If we subtract a negative, we need to add. So we get plus 7350 newtons. And when we work this out, we get that this force T two is going to be equal to 22,050 newtons. We have T two. Now we can calculate that cross sectional area A two and it's going to be the exact same as a one. We're just using this new force that we found. So we're gonna have t two multiplied by the safety factor divided by the ultimate strength substituting in our values. This is going to be equal to 22,050 noons multiplied by eight, divided by two multiplied by 10 to the exponent eight newtons per meter squared. Now, when we work all of this out on our calculator, we get a cross sectional area for column 282 of 0.000882 newtons. And so that is that second area that we were looking for. Whoops and not Newton's there, that should be in meters squared meters squared. There we go, the newtons will divide up will be left with meters squared. If we go back up and compare what we found to the answer choices, we're going round to two significant digits and we can see that the correct answer is going to correspond with. Option D A one is 2.9 multiplied by 10 to the exponent negative 4 m squared and A two is 8.8 multiplied by 10 to the exponent negative 4 m squared. Thanks everyone for watching. I hope this video helped see you in the next one.

Assume the supports of the uniform cantilever shown in Fig. 12–79 (m = 2900 kg) are made of wood. Calculate the minimum cross-sectional area required of each, assuming a safety factor of 9.0.

Key Concepts

Cantilever Beam

Safety Factor

Cross-Sectional Area

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