A curved beam with a rectangular cross section strikes a 90^∘ arc and is loaded and supported as

A curved beam with a rectangular cross section strikes a...

Key Concepts

Curved Beam Theory

Curved beam theory deals with the analysis of beams that have a non?straight geometry. Unlike rectilinear beams, curved beams experience a non-uniform stress distribution due to their geometry, and the neutral axis does not coincide with the centroid. This theory helps in understanding the complexities of bending in curved members, including how geometry, curvature, and moment distribution affect internal stresses.

Hoop Stress

Hoop stress, also called circumferential stress, is the stress component acting along the curved path of a beam. In curved beams, hoop stress varies across the section, reaching different values at the inside and outside radii. Assessing hoop stress is crucial for ensuring that the material can withstand the bending moments induced by applied loads, especially in regions where stress concentrations occur.

Neutral Axis in Curved Beams

In the context of curved beams, the neutral axis is the locus of points within the cross section where the bending stress is zero. Unlike straight beams where the neutral axis typically passes through the centroid, in curved beams it shifts toward the inside of the curvature due to geometric effects. Accurately locating this axis is essential for correctly calculating stress distributions across the section.

Influence of Cross-Sectional Geometry

The geometry of a beam's cross section, such as being rectangular, is a critical factor in stress analysis. The dimensions, including thickness and width, directly influence the moment of inertia and, consequently, the distribution of stress within the beam. In curved beams, this geometry determines how stresses vary from the inside to the outside radii and at the centroid, impacting the design and safety of the structure.

Transcript

00:01 In this problem, we're first asked to find the maximum tensile strength and the maximum compressive stress.

00:07 Stress, i should have said stress, not strength.

00:09 Due to load p acting on a simple bean as shown.

00:15 Okay, so we are going to first the sum of moments around point b.

00:23 So let's do this.

00:24 This is a.

00:29 This is b.

00:32 This is x.

00:36 This is y.

00:39 Here's p.

00:40 Okay.

00:43 So let's switch colors here.

00:45 Let's have this is h -a.

00:50 This is, oops, going this way, not that is r -a, this is r -a, this is r -b, this is d, and this is l.

01:15 Okay, so let's do, we're going to do m -b -b -e -e -e -e -e -e -e -e -e -e -e -e -e -e -e -e -e -e -e -e -e -e -e -e -e -e -e -e -e -e -e -e -e -e -e -e -e -e -e -e -e - 0 of negative r a times l plus p times d equals zero so we can take our r a will equal p times d divided by l okay so let's cut through the segment our segment where zero is less than equal and analyze the beam on the left okay so let's do this that'll be 0.

02:14 So we'll have ra minus v equals 0 or v equals ra.

02:21 We know v equals p times d over l.

02:26 So we can plug our values in here.

02:30 P, d, l, and this will be 2422 newton's, which will equal 2 .422 kilonitons.

03:05 Okay let's make a note of that then it's the moments around the cut section so i'll have m equals and m okay and then let's go back to this so right around my a to my cut section right here this is my x this is my v and this is my m on this one then let's cut through the segment and do the other forces here.

04:25 So the ra plus p minus v equals zero.

04:32 So i'll get v equals and then v will equal p times d over l minus p, which will equal p times d over l minus one.

04:51 Then let's do another sum of moments, the bending moment at this section.

05:02 Okay.

05:19 And then this one is going to be a little bit longer.

05:22 I'm going to reduce this one.

05:24 I'm going to skip some steps here.

05:39 Okay.

05:40 And now let's do this one here.

06:09 That's x.

06:11 Okay.

06:14 And then we can do the bending moments in putting values of x.

06:20 So let's do.

06:37 And then that will be equal to.

06:38 To and this will be equal to and that will be also equal to 4 .723 times 10 to the 6 newton millimeters okay and then our bending moment that will be okay and then the section modulus of our cross section will divide that cross section into two rectangles a1 and a 2 okay, so let's do this.

08:14 So let's do, and my height here will be 120, and then we're going to do a cross -section area, and then i'm going to have my first cross -section right here, and my second cross -section right here.

09:02 Okay, so this height is 90 millimeters.

09:10 Here's my b from here to here of 80 millimeters, and then my y -1 from here to here, of 80 millimeters, and then my y -1 from here to here.

09:28 Here, and that's my y2 is right to here, and my y1 is right to there.

09:44 Okay, so, so the distance to our neutral axis, c1.

10:14 So c1 will equal, and this will equal, and then the other one will equal 120 minus 75 .97 will equal 44 .03 millimeters.

10:52 Now we have c1 and c2, and then we have to find the polar moment of inertia.

11:15 Okay, so my i, this will equal i, c, equal t times h over i or over 12.

11:47 This will equal.

12:03 So this will equal 25 times 90 cubed over 12 plus 22, 250 times 75 .97 minus 45 squared.

12:29 And this will equal 3 .67 times 10 to the 6 millimeters to the fourth.

12:40 I'm going to circle that.

12:41 Uh oh.

12:46 I'm going to have to plug in here...

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