Let's say that, based off of Jackie's and Veronica's heights, that this I-beam:

- is 6 inches wide (i.e. the left-to-right width of each of the flat parts on both sides if the I-beam were resting on the flat parts)
- is 8 inches high (i.e. the top-to-bottom length/height of the entire I-beam; parallel to the middle part if the beam were resting on the flat parts)
- has flanges (i.e. the top-to-bottom thickness of the flat parts on both sides if the I-beam were resting on the flat parts) that are each 1 inch thick
- has a web (i.e. the left-to-right thickness of the flat part in the middle if the I-beam were resting on the flat parts) that's 1 inch thick
- is 8 feet long
- is bent/deformed by about 2.5 feet/30 inches

Additionally, let's assume this I-beam is made out of mild steel, which has a Young's modulus of 200 gigapascals, or 29,010,000 pounds per square inch, since that's usually what I-beams used for construction are made of.

If we plug those values into this calculator (cross-sectional area of the beam) and this calculator (beam deflection/deformation under stress), that means she's putting 35,000,000 pounds - about 15,875 metric tons - of force on that I-beam to deform it to within the ballpark of 30 inches. For reference, an American WW2 New Orleans-class cruiser weighs somewhere between 10,110 and 12,663 metric tons. Moreover, I deliberately underestimated all those values above except for the Young's modulus and the weight of the New Orleans-class, meaning that the I-beam is probably more durable than I actually did the math for, meaning that she's exerting even more force than that.

Either way, that means that her thighs can exert more force, as a point load, than the weight of a warship.

I'm unsure as to what I should do with all this, or what purpose it serves, but I had to do it.

stuckuplatinwords said:
Let's say that, based off of Jackie's and Veronica's heights, that this beam:

- is 6 inches wide (i.e. the left-to-right width of each of the flat parts on both sides if the beam were resting on the flat parts)
- is 8 inches high (i.e. the top-to-bottom length/height of the entire beam; parallel to the middle part if the beam were resting on the flat parts)
- has flanges (i.e. the top-to-bottom thickness of the flat parts on both sides if the beam were resting on the flat parts) that are each 1 inch thick
- has a web (i.e. the left-to-right thickness of the flat part in the middle if the beam were resting on the flat parts) that's 1 inch thick
- is 8 feet long
- is bent/deformed by about 2.5 feet/30 inches

Additionally, let's assume this beam is made out of mild steel, which has a Young's modulus of 200 gigapascals, or 29010000 pounds per square inch.

If we plug those values into this calculator (cross-sectional area of the beam) and this calculator (beam deflection/deformation under stress), that means she's putting 35,000,000 pounds - about 15,875 metric tons - of force on that beam to bend it 30 inches. For reference, an American WW2 New Orleans-class cruiser weighs somewhere between 10,110 and 12,663 metric tons. Moreover, I deliberately underestimated all those values above except for the Young's modulus, meaning that the beam is probably more durable than I actually did the math for.

In otherwords, her thighs can exert more force, as a point load, than the weight of a warship.

I'm unsure as to what I should do with all this, or what purpose it serves, but I had to do it.

holy SHIT this is insanely cool, thank you for doing the maths

securipun said:
holy SHIT this is insanely cool, thank you for doing the maths

holy SHIT seeing a story-tall wolf lady bend a goddamned i-beam with her thighs is insanely cool, thank you for doing the arts

stuckuplatinwords said:
Let's say that, based off of Jackie's and Veronica's heights, that this I-beam:

- is 6 inches wide (i.e. the left-to-right width of each of the flat parts on both sides if the I-beam were resting on the flat parts)
- is 8 inches high (i.e. the top-to-bottom length/height of the entire I-beam; parallel to the middle part if the beam were resting on the flat parts)
- has flanges (i.e. the top-to-bottom thickness of the flat parts on both sides if the I-beam were resting on the flat parts) that are each 1 inch thick
- has a web (i.e. the left-to-right thickness of the flat part in the middle if the I-beam were resting on the flat parts) that's 1 inch thick
- is 8 feet long
- is bent/deformed by about 2.5 feet/30 inches

Additionally, let's assume this I-beam is made out of mild steel, which has a Young's modulus of 200 gigapascals, or 29,010,000 pounds per square inch, since that's usually what I-beams used for construction are made of.

If we plug those values into this calculator (cross-sectional area of the beam) and this calculator (beam deflection/deformation under stress), that means she's putting 35,000,000 pounds - about 15,875 metric tons - of force on that I-beam to deform it to within the ballpark of 30 inches. For reference, an American WW2 New Orleans-class cruiser weighs somewhere between 10,110 and 12,663 metric tons. Moreover, I deliberately underestimated all those values above except for the Young's modulus and the weight of the New Orleans-class, meaning that the I-beam is probably more durable than I actually did the math for, meaning that she's exerting even more force than that.

Either way, that means that her thighs can exert more force, as a point load, than the weight of a warship.

I'm unsure as to what I should do with all this, or what purpose it serves, but I had to do it.

I have learned so many cool facts from porn on this site

stuckuplatinwords said:
Let's say that, based off of Jackie's and Veronica's heights, that this I-beam:

- is 6 inches wide (i.e. the left-to-right width of each of the flat parts on both sides if the I-beam were resting on the flat parts)
- is 8 inches high (i.e. the top-to-bottom length/height of the entire I-beam; parallel to the middle part if the beam were resting on the flat parts)
- has flanges (i.e. the top-to-bottom thickness of the flat parts on both sides if the I-beam were resting on the flat parts) that are each 1 inch thick
- has a web (i.e. the left-to-right thickness of the flat part in the middle if the I-beam were resting on the flat parts) that's 1 inch thick
- is 8 feet long
- is bent/deformed by about 2.5 feet/30 inches

Additionally, let's assume this I-beam is made out of mild steel, which has a Young's modulus of 200 gigapascals, or 29,010,000 pounds per square inch, since that's usually what I-beams used for construction are made of.

If we plug those values into this calculator (cross-sectional area of the beam) and this calculator (beam deflection/deformation under stress), that means she's putting 35,000,000 pounds - about 15,875 metric tons - of force on that I-beam to deform it to within the ballpark of 30 inches. For reference, an American WW2 New Orleans-class cruiser weighs somewhere between 10,110 and 12,663 metric tons. Moreover, I deliberately underestimated all those values above except for the Young's modulus and the weight of the New Orleans-class, meaning that the I-beam is probably more durable than I actually did the math for, meaning that she's exerting even more force than that.

Either way, that means that her thighs can exert more force, as a point load, than the weight of a warship.

I'm unsure as to what I should do with all this, or what purpose it serves, but I had to do it.

... so she could lift - or potentially crush - a New Orleans class with her legs?

Sorry if not, I'm terrible at math XD

onyx_lupine said:
... so she could lift - or potentially crush - a New Orleans class with her legs?

Sorry if not, I'm terrible at math XD

Naw, she just exerts force equivalent to one. In order for her to lift a New Orleans-class, she'd need to convert that compressive force to force in a single direction, which would likely involve hydraulics, since they're good at turning compressive force into push force. Basically, hydraulics works via turning a small amount of force across a very wide area into a massive amount of force in a small area, or vice versa; for instance, if you apply 100 pounds of force across a piston head with 10 square inches of space, the other piston in the system can translate it into another 100 pounds of force across another 10 square inches of space, 1 pound of force across 10,000 square inches of space, or 10,000 pounds of force across 1 square inch of space.

The problem with that is that the conversion of squeeze force to lift force isn't going to be 100% efficient; when you compress hydraulic fluid, it drags on the walls of the hydraulic cylinder. However, the efficiency of a hydraulic cylinder approaches 100% as more force is applied to it, mitigating that. However again, there's another problem: even the largest and strongest hydraulic cylinders can only handle a few tens of metric tons of force, and "only" thousands of PSI of pressure (which she's exceeding by a factor of 10, at least) - they're (a) not enough to lift a ship, and (b) much too large for her to actually apply force to. Moreover, those cylinders I linked to are the absolute top-of-the-line in regards to hydraulics; you can't just use a smaller piston for her to apply force to so as to move a larger piston, like with my example above, because these ones - which can take thousands of PSI and tens of metric tons of force - are basically the strongest there are, and they're nowhere close to enough. If she could somehow apply force to them, she'd destroy them; smaller ones she could actually wrap herself around would just get destroyed a few fractions of a second faster.

In other words, yes, she is strong enough to lift a New Orleans-class with her thighs, but she'd need to do it through a piece of technology serving as an intermediary, and no modern technology that can fit between her legs can survive the forces her body exerts long enough for her to do so.

As for crushing one, no:
- the structure of such a ship is probably made out of something tougher than mild steel
- the durability of all the materials in the ship add up; it's not just one I-beam, but a hundred of them, sort of like how it's easy to bend one straw but hard to bend 100 of them in a bundle
- there's no way she could get her thighs around one of them, which is probably the actual limiting factor here

Basically, if it fits between those human torso-sized thighs, she can crush it, unless it's made out of something like graphene, armor steel, or rock...and that last one depends on the type of rock.

I'm trying to think of a funny, non-offensive way to make a "steel beams" 9/11 joke but i just now realized that's an oxymoronic thing to do

stuckuplatinwords said:
Let's say that, based off of Jackie's and Veronica's heights, that this I-beam:

- is 6 inches wide (i.e. the left-to-right width of each of the flat parts on both sides if the I-beam were resting on the flat parts)
- is 8 inches high (i.e. the top-to-bottom length/height of the entire I-beam; parallel to the middle part if the beam were resting on the flat parts)
- has flanges (i.e. the top-to-bottom thickness of the flat parts on both sides if the I-beam were resting on the flat parts) that are each 1 inch thick
- has a web (i.e. the left-to-right thickness of the flat part in the middle if the I-beam were resting on the flat parts) that's 1 inch thick
- is 8 feet long
- is bent/deformed by about 2.5 feet/30 inches

Additionally, let's assume this I-beam is made out of mild steel, which has a Young's modulus of 200 gigapascals, or 29,010,000 pounds per square inch, since that's usually what I-beams used for construction are made of.

If we plug those values into this calculator (cross-sectional area of the beam) and this calculator (beam deflection/deformation under stress), that means she's putting 35,000,000 pounds - about 15,875 metric tons - of force on that I-beam to deform it to within the ballpark of 30 inches. For reference, an American WW2 New Orleans-class cruiser weighs somewhere between 10,110 and 12,663 metric tons. Moreover, I deliberately underestimated all those values above except for the Young's modulus and the weight of the New Orleans-class, meaning that the I-beam is probably more durable than I actually did the math for, meaning that she's exerting even more force than that.

Either way, that means that her thighs can exert more force, as a point load, than the weight of a warship.

I'm unsure as to what I should do with all this, or what purpose it serves, but I had to do it.

You're doing gods work, please continue to bless us in the future!

stuckuplatinwords said:
Let's say that, based off of Jackie's and Veronica's heights, that this I-beam:

- is 6 inches wide (i.e. the left-to-right width of each of the flat parts on both sides if the I-beam were resting on the flat parts)
- is 8 inches high (i.e. the top-to-bottom length/height of the entire I-beam; parallel to the middle part if the beam were resting on the flat parts)
- has flanges (i.e. the top-to-bottom thickness of the flat parts on both sides if the I-beam were resting on the flat parts) that are each 1 inch thick
- has a web (i.e. the left-to-right thickness of the flat part in the middle if the I-beam were resting on the flat parts) that's 1 inch thick
- is 8 feet long
- is bent/deformed by about 2.5 feet/30 inches

Additionally, let's assume this I-beam is made out of mild steel, which has a Young's modulus of 200 gigapascals, or 29,010,000 pounds per square inch, since that's usually what I-beams used for construction are made of.

If we plug those values into this calculator (cross-sectional area of the beam) and this calculator (beam deflection/deformation under stress), that means she's putting 35,000,000 pounds - about 15,875 metric tons - of force on that I-beam to deform it to within the ballpark of 30 inches. For reference, an American WW2 New Orleans-class cruiser weighs somewhere between 10,110 and 12,663 metric tons. Moreover, I deliberately underestimated all those values above except for the Young's modulus and the weight of the New Orleans-class, meaning that the I-beam is probably more durable than I actually did the math for, meaning that she's exerting even more force than that.

Either way, that means that her thighs can exert more force, as a point load, than the weight of a warship.

I'm unsure as to what I should do with all this, or what purpose it serves, but I had to do it.

Now that's dedication.

THIGHS OF STEEEEEL

also i notice that this has a lot of facts in it so ima keep this favourited

Another favourite for me, suprisingly informative comments too. And to Securipun if you read this, I read through some of the bs comments on other posts you've had to back n forth with and I just gotta say I'm sorry you gotta put up with people like that. You got a fan in me though and if I ever have anything negative to say (unlikely with your quality) I will be constructive about it. (I saw that brain dead guy with the '0/10 too fat' comment.)

stuckuplatinwords said:
Naw, she just exerts force equivalent to one. In order for her to lift a New Orleans-class, she'd need to convert that compressive force to force in a single direction, which would likely involve hydraulics, since they're good at turning compressive force into push force. Basically, hydraulics works via turning a small amount of force across a very wide area into a massive amount of force in a small area, or vice versa; for instance, if you apply 100 pounds of force across a piston head with 10 square inches of space, the other piston in the system can translate it into another 100 pounds of force across another 10 square inches of space, 1 pound of force across 10,000 square inches of space, or 10,000 pounds of force across 1 square inch of space.

The problem with that is that the conversion of squeeze force to lift force isn't going to be 100% efficient; when you compress hydraulic fluid, it drags on the walls of the hydraulic cylinder. However, the efficiency of a hydraulic cylinder approaches 100% as more force is applied to it, mitigating that. However again, there's another problem: even the largest and strongest hydraulic cylinders can only handle a few tens of metric tons of force, and "only" thousands of PSI of pressure (which she's exceeding by a factor of 10, at least) - they're (a) not enough to lift a ship, and (b) much too large for her to actually apply force to. Moreover, those cylinders I linked to are the absolute top-of-the-line in regards to hydraulics; you can't just use a smaller piston for her to apply force to so as to move a larger piston, like with my example above, because these ones - which can take thousands of PSI and tens of metric tons of force - are basically the strongest there are, and they're nowhere close to enough. If she could somehow apply force to them, she'd destroy them; smaller ones she could actually wrap herself around would just get destroyed a few fractions of a second faster.

In other words, yes, she is strong enough to lift a New Orleans-class with her thighs, but she'd need to do it through a piece of technology serving as an intermediary, and no modern technology that can fit between her legs can survive the forces her body exerts long enough for her to do so.

As for crushing one, no:
- the structure of such a ship is probably made out of something tougher than mild steel
- the durability of all the materials in the ship add up; it's not just one I-beam, but a hundred of them, sort of like how it's easy to bend one straw but hard to bend 100 of them in a bundle
- there's no way she could get her thighs around one of them, which is probably the actual limiting factor here

Basically, if it fits between those human torso-sized thighs, she can crush it, unless it's made out of something like graphene, armor steel, or rock...and that last one depends on the type of rock.

So, as a thought experiment with only a layperson's understanding of the forces, materials, and mechanics involved: let's bring carbon nanotubes into this mess!

My understanding is that multi-tiered carbon nanotubes have one of (if not THE,) highest tensile strengths of known materials. What are the largest possible configurations of carbon nanotubes Mommy Milfy could destroy with certainty, either by volume or density of the crushed object, ASSUMING that there is a mechanical way to translate the forces she produces with 100% efficiency?

spexguy said:
So, as a thought experiment with only a layperson's understanding of the forces, materials, and mechanics involved: let's bring carbon nanotubes into this mess!

My understanding is that multi-tiered carbon nanotubes have one of (if not THE,) highest tensile strengths of known materials. What are the largest possible configurations of carbon nanotubes Mommy Milfy could destroy with certainty, either by volume or density of the crushed object, ASSUMING that there is a mechanical way to translate the forces she produces with 100% efficiency?

Multi-walled carbon nanotubes have a Young's modulus in the 34-gigapascal range, as opposed to mild steel's ~200-gigapascals.

Squish.

Single-walled carbon nanotubes, on the other hand (download the PDF) are within the >1,000-gigapascal range - i.e. ~5x that of mild steel - and potentially closer to 5 terapascals - but I doubt we've seen this lady's upper limit yet, so...who knows? Probably.

Thing is, carbon nanotubes are specifically bad at surviving compressive loads, so they'd probably be slightly more vulnerable to the thighs of doom than these values suggest, but I'll stick with a resounding "maybe" for single-walled tubes and a resounding "yes" for multi-walled tubes.

stuckuplatinwords said:
Multi-walled carbon nanotubes have a Young's modulus in the 34-gigapascal range, as opposed to mild steel's ~200-gigapascals.

Squish.

Single-walled carbon nanotubes, on the other hand (download the PDF) are within the >1,000-gigapascal range - i.e. ~5x that of mild steel - and potentially closer to 5 terapascals - but I doubt we've seen this lady's upper limit yet, so...who knows? Probably.

Thing is, carbon nanotubes are specifically bad at surviving compressive loads, so they'd probably be slightly more vulnerable to the thighs of doom than these values suggest, but I'll stick with a resounding "maybe" for single-walled tubes and a resounding "yes" for multi-walled tubes.

So what I'm led to understand is that we should just start making things out of MILFs and DILFs given their superior load bearing capacities.

spexguy said:
So what I'm led to understand is that we should just start making things out of MILFs and DILFs given their superior load bearing capacities.

Of course. What else would one use?

stuckuplatinwords said:
Let's say that, based off of Jackie's and Veronica's heights, that this I-beam:

- is 6 inches wide (i.e. the left-to-right width of each of the flat parts on both sides if the I-beam were resting on the flat parts)
- is 8 inches high (i.e. the top-to-bottom length/height of the entire I-beam; parallel to the middle part if the beam were resting on the flat parts)
- has flanges (i.e. the top-to-bottom thickness of the flat parts on both sides if the I-beam were resting on the flat parts) that are each 1 inch thick
- has a web (i.e. the left-to-right thickness of the flat part in the middle if the I-beam were resting on the flat parts) that's 1 inch thick
- is 8 feet long
- is bent/deformed by about 2.5 feet/30 inches

Additionally, let's assume this I-beam is made out of mild steel, which has a Young's modulus of 200 gigapascals, or 29,010,000 pounds per square inch, since that's usually what I-beams used for construction are made of.

If we plug those values into this calculator (cross-sectional area of the beam) and this calculator (beam deflection/deformation under stress), that means she's putting 35,000,000 pounds - about 15,875 metric tons - of force on that I-beam to deform it to within the ballpark of 30 inches. For reference, an American WW2 New Orleans-class cruiser weighs somewhere between 10,110 and 12,663 metric tons. Moreover, I deliberately underestimated all those values above except for the Young's modulus and the weight of the New Orleans-class, meaning that the I-beam is probably more durable than I actually did the math for, meaning that she's exerting even more force than that.

Either way, that means that her thighs can exert more force, as a point load, than the weight of a warship.

I'm unsure as to what I should do with all this, or what purpose it serves, but I had to do it.

Yes but is it enough to crush my skull

kenui said:
Yes but is it enough to crush my skull

I don't know, is your skull made out of WW2 battleship plating?

stuckuplatinwords said:
Let's say that, based off of Jackie's and Veronica's heights, that this I-beam:

- is 6 inches wide (i.e. the left-to-right width of each of the flat parts on both sides if the I-beam were resting on the flat parts)
- is 8 inches high (i.e. the top-to-bottom length/height of the entire I-beam; parallel to the middle part if the beam were resting on the flat parts)
- has flanges (i.e. the top-to-bottom thickness of the flat parts on both sides if the I-beam were resting on the flat parts) that are each 1 inch thick
- has a web (i.e. the left-to-right thickness of the flat part in the middle if the I-beam were resting on the flat parts) that's 1 inch thick
- is 8 feet long
- is bent/deformed by about 2.5 feet/30 inches

Additionally, let's assume this I-beam is made out of mild steel, which has a Young's modulus of 200 gigapascals, or 29,010,000 pounds per square inch, since that's usually what I-beams used for construction are made of.

If we plug those values into this calculator (cross-sectional area of the beam) and this calculator (beam deflection/deformation under stress), that means she's putting 35,000,000 pounds - about 15,875 metric tons - of force on that I-beam to deform it to within the ballpark of 30 inches. For reference, an American WW2 New Orleans-class cruiser weighs somewhere between 10,110 and 12,663 metric tons. Moreover, I deliberately underestimated all those values above except for the Young's modulus and the weight of the New Orleans-class, meaning that the I-beam is probably more durable than I actually did the math for, meaning that she's exerting even more force than that.

Either way, that means that her thighs can exert more force, as a point load, than the weight of a warship.

I'm unsure as to what I should do with all this, or what purpose it serves, but I had to do it.

you expaning and doing the math behind this is fucking cool

stuckuplatinwords said:
I don't know, is your skull made out of WW2 battleship plating?

Idk dude. Most people now a days have the density of one.

mollytov said:
I'm trying to think of a funny, non-offensive way to make a "steel beams" 9/11 joke but i just now realized that's an oxymoronic thing to do

you will never not offend anyone :)

mollytov said:
I'm trying to think of a funny, non-offensive way to make a "steel beams" 9/11 joke but i just now realized that's an oxymoronic thing to do

A SECOND WEREWOLF HAS HIT THE TOWERS

stuckuplatinwords said:
Let's say that, based off of Jackie's and Veronica's heights, that this I-beam:

- is 6 inches wide (i.e. the left-to-right width of each of the flat parts on both sides if the I-beam were resting on the flat parts)
- is 8 inches high (i.e. the top-to-bottom length/height of the entire I-beam; parallel to the middle part if the beam were resting on the flat parts)
- has flanges (i.e. the top-to-bottom thickness of the flat parts on both sides if the I-beam were resting on the flat parts) that are each 1 inch thick
- has a web (i.e. the left-to-right thickness of the flat part in the middle if the I-beam were resting on the flat parts) that's 1 inch thick
- is 8 feet long
- is bent/deformed by about 2.5 feet/30 inches

Additionally, let's assume this I-beam is made out of mild steel, which has a Young's modulus of 200 gigapascals, or 29,010,000 pounds per square inch, since that's usually what I-beams used for construction are made of.

If we plug those values into this calculator (cross-sectional area of the beam) and this calculator (beam deflection/deformation under stress), that means she's putting 35,000,000 pounds - about 15,875 metric tons - of force on that I-beam to deform it to within the ballpark of 30 inches. For reference, an American WW2 New Orleans-class cruiser weighs somewhere between 10,110 and 12,663 metric tons. Moreover, I deliberately underestimated all those values above except for the Young's modulus and the weight of the New Orleans-class, meaning that the I-beam is probably more durable than I actually did the math for, meaning that she's exerting even more force than that.

Either way, that means that her thighs can exert more force, as a point load, than the weight of a warship.

I'm unsure as to what I should do with all this, or what purpose it serves, but I had to do it.

and here I am thinking yea the mood of the store worker at seaing that is accurate working in retail sucks your soul out

stuckuplatinwords said:
Let's say that, based off of Jackie's and Veronica's heights, that this I-beam:

- is 6 inches wide (i.e. the left-to-right width of each of the flat parts on both sides if the I-beam were resting on the flat parts)
- is 8 inches high (i.e. the top-to-bottom length/height of the entire I-beam; parallel to the middle part if the beam were resting on the flat parts)
- has flanges (i.e. the top-to-bottom thickness of the flat parts on both sides if the I-beam were resting on the flat parts) that are each 1 inch thick
- has a web (i.e. the left-to-right thickness of the flat part in the middle if the I-beam were resting on the flat parts) that's 1 inch thick
- is 8 feet long
- is bent/deformed by about 2.5 feet/30 inches

Additionally, let's assume this I-beam is made out of mild steel, which has a Young's modulus of 200 gigapascals, or 29,010,000 pounds per square inch, since that's usually what I-beams used for construction are made of.

If we plug those values into this calculator (cross-sectional area of the beam) and this calculator (beam deflection/deformation under stress), that means she's putting 35,000,000 pounds - about 15,875 metric tons - of force on that I-beam to deform it to within the ballpark of 30 inches. For reference, an American WW2 New Orleans-class cruiser weighs somewhere between 10,110 and 12,663 metric tons. Moreover, I deliberately underestimated all those values above except for the Young's modulus and the weight of the New Orleans-class, meaning that the I-beam is probably more durable than I actually did the math for, meaning that she's exerting even more force than that.

Either way, that means that her thighs can exert more force, as a point load, than the weight of a warship.

I'm unsure as to what I should do with all this, or what purpose it serves, but I had to do it.

thats one hell of a watermellon then~

stuckuplatinwords said:
Let's say that, based off of Jackie's and Veronica's heights, that this I-beam:

- is 6 inches wide (i.e. the left-to-right width of each of the flat parts on both sides if the I-beam were resting on the flat parts)
- is 8 inches high (i.e. the top-to-bottom length/height of the entire I-beam; parallel to the middle part if the beam were resting on the flat parts)
- has flanges (i.e. the top-to-bottom thickness of the flat parts on both sides if the I-beam were resting on the flat parts) that are each 1 inch thick
- has a web (i.e. the left-to-right thickness of the flat part in the middle if the I-beam were resting on the flat parts) that's 1 inch thick
- is 8 feet long
- is bent/deformed by about 2.5 feet/30 inches

Additionally, let's assume this I-beam is made out of mild steel, which has a Young's modulus of 200 gigapascals, or 29,010,000 pounds per square inch, since that's usually what I-beams used for construction are made of.

If we plug those values into this calculator (cross-sectional area of the beam) and this calculator (beam deflection/deformation under stress), that means she's putting 35,000,000 pounds - about 15,875 metric tons - of force on that I-beam to deform it to within the ballpark of 30 inches. For reference, an American WW2 New Orleans-class cruiser weighs somewhere between 10,110 and 12,663 metric tons. Moreover, I deliberately underestimated all those values above except for the Young's modulus and the weight of the New Orleans-class, meaning that the I-beam is probably more durable than I actually did the math for, meaning that she's exerting even more force than that.

Either way, that means that her thighs can exert more force, as a point load, than the weight of a warship.

I'm unsure as to what I should do with all this, or what purpose it serves, but I had to do it.

I call this bs
She is no way to do that
Unless she is demi God or superhero
Are you sure you did math right?

stuckuplatinwords said:
Let's say that, based off of Jackie's and Veronica's heights, that this I-beam:

- is 6 inches wide (i.e. the left-to-right width of each of the flat parts on both sides if the I-beam were resting on the flat parts)
- is 8 inches high (i.e. the top-to-bottom length/height of the entire I-beam; parallel to the middle part if the beam were resting on the flat parts)
- has flanges (i.e. the top-to-bottom thickness of the flat parts on both sides if the I-beam were resting on the flat parts) that are each 1 inch thick
- has a web (i.e. the left-to-right thickness of the flat part in the middle if the I-beam were resting on the flat parts) that's 1 inch thick
- is 8 feet long
- is bent/deformed by about 2.5 feet/30 inches

Additionally, let's assume this I-beam is made out of mild steel, which has a Young's modulus of 200 gigapascals, or 29,010,000 pounds per square inch, since that's usually what I-beams used for construction are made of.

If we plug those values into this calculator (cross-sectional area of the beam) and this calculator (beam deflection/deformation under stress), that means she's putting 35,000,000 pounds - about 15,875 metric tons - of force on that I-beam to deform it to within the ballpark of 30 inches. For reference, an American WW2 New Orleans-class cruiser weighs somewhere between 10,110 and 12,663 metric tons. Moreover, I deliberately underestimated all those values above except for the Young's modulus and the weight of the New Orleans-class, meaning that the I-beam is probably more durable than I actually did the math for, meaning that she's exerting even more force than that.

Either way, that means that her thighs can exert more force, as a point load, than the weight of a warship.

I'm unsure as to what I should do with all this, or what purpose it serves, but I had to do it.

Alright I checked little bid I all i can say is you make mistake
I beam sure is strong to bending but not from sides
It is actually more weaker to bend this way like femur in human bone
Not to mention it is not i beam but H beam