The 10-mm-diameter shank of the steel bolt has a bronze...

The 10-mm-diameter shank of the steel bolt has a bronze sleeve bonded to it. The outer diameter of this sleeve is $20 \mathrm{mm}$. If the yield stress for the steel is $\left(\sigma_{Y}\right)_{\mathrm{st}}=640 \mathrm{MPa}$ and for the bronze $\left(\sigma_{Y}\right)_{\mathrm{br}}=520 \quad \mathrm{MPa},$ determine the magnitude of the largest elastic load $P$ that can be applied to the assembly. $E_{\mathrm{st}}=200 \mathrm{GPa}, E_{\mathrm{br}}=100 \mathrm{GPa}$.

Key Concepts

Elastic Behavior and Yield Criterion

Elastic behavior refers to the linear response of materials under load until the yield point is reached. The yield criterion is used to determine the maximum load a composite system can support without any of its constituents exceeding their yield stress, thereby ensuring that all parts of the structure remain within the elastic regime.

Composite Structures

Composite structures involve combining two or more materials into one member so that they act together under load. The overall behavior is affected by the individual material properties and the way they are bonded, leading to a distribution of stress that depends on each material’s stiffness and cross-sectional area.

Axial Load and Stress Distribution

When a member is loaded in the axial direction, the load is shared among the different materials based on their respective stiffnesses and cross-sectional areas. This requires establishing compatibility of strains across the different materials and applying force equilibrium to find the stress in each component.

Modulus of Elasticity

The modulus of elasticity, or Young's modulus, is a measure of a material's stiffness, dictating how much it will deform under a given load. In composite members, differences in modulus of elasticity between materials cause variations in stress distribution and strain compatibility under an applied load.

Transcript

00:01 Consider the diagram i've written right here.

00:03 One of my nicer.

00:04 We're told we have a 10 millimeter diameter shake of a steel bolt that has a bronze sleeve bonded to it.

00:12 The outer diameter to the sleeve is 20 millimeters.

00:15 And the yield stress for the steel is 640 megapascals, which i've converted to kiladoons per meter squared right here.

00:25 And for the steel, it's, for the bronze, it's 520 megapascals, which i have written in kilonutons per meter squared right here.

00:34 We're asked to determine the magnitude of the largest plastic load p that can be applied to assembly.

00:41 And we're given for both steel and bronze, 200 and 100 gigapascals.

00:53 Okay, let me see here.

01:00 And what else? there we go.

01:03 So recall from a formula, the formula for calculating displacement, we're going to be using delta.

01:30 Or in the case of segments, we, in the case that segments are not infinite, we use.

01:47 Okay.

01:48 So the tension on the bolt result the same compression on the bolt in the sleeve.

01:54 So we also know that for the bronze will equal for the steel.

02:00 Okay.

02:03 So let's go ahead and draw a free body diagram and i think i'm just going to go to red and do that right here.

02:12 So we're going to have p going up.

02:16 Then we'll have our p for our steel and we'll add into that a p for our bronze.

02:24 Okay.

02:30 So we know that our x, all we need is the y component.

02:38 So we know that our total forces has to equal zero.

02:44 And all we need is because there wasn't any horizontal.

02:48 Force here.

02:49 We've got f y as to equal zero, some of our forces.

02:56 Okay, so let's figure out.

03:01 This will be p minus p for a bronze minus p of her steel equals zero.

03:10 So p will equal p for a bronze plus p for a steel.

03:19 Then let's look at our first equation and go ahead and find our area of y.

03:32 Our area, let's make that a b.

03:37 That'll equal pi over four times our diameter 0 .01 meters squared.

03:48 That'll equal 7 .854 times 10 to the minus fifth meters squared.

03:58 And for e, our elastic modulus of elasticity, 200 gigapascals will be equal to 2 times 10 to the eighth kilonutons per meter squared.

04:19 And our internal force in will equal force b.

04:28 And then l.

04:36 Okay, let's go back to this page right here.

04:39 So look, well, i'll just go ahead and write it down.

04:42 I'm going to go to the next page.

04:59 And then for this one, we can set this up as f -b -l over a -b -e -a, and this will equal f -b times l over 7 .85 or times 10 to the minus 5th times 2 times 10 to the 8th.

05:32 And this will be give us this is equal to fb times l times one five seven oh eight then our sleeve displacement i should do an r there will be let's do this same thing here um that was for our bronze let's go to the next page for our steel so again let's do the area of the steel will be equal to pi over four times d minus d or b and that will equal pi over four times 0 .0 .0 .2 squared minus 0 .010 h squared.

06:42 And this will equal 2 .356 times to the minus 4 meters square...

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