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How does the strength of a square tube depend on the side size?
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There are two square tubes of the same length, material and wall thickness, one being 40x40mm, the other 20x20mm. How does their strength against bending compare?
On the one hand, the cross-sectional area of the 40mm tube is 4 times bigger. On the other hand, I guess it is mostly the two walls that are in the same plane as the bending force that work against it. The other two walls provide very little resistance comparing to the first two. And because the surface of the "working" walls of the 40mm tube is twice as much bigger as those of the 20mm one, I am inclined to think that the 40mm tube is roughly twice as much stronger. How does this conclusion stand?
More specifically, this question is about galvanized steel tubes with 1.6mm thick walls:
I am looking at using them as two or more horizontal beams 2.4m long which will take some evenly distributed load. The vertical space is limited so that I would rather use four/eight/sixteen 20mm tubes than two 40mm ones. But how many 20mm tubes would be roughly equal to two 40mm ones?
structural-engineering materials steel strength
edited Aug 9 at 8:36
Sam Farjamirad
440211
asked Aug 9 at 8:30
Greendrake
1335
add a comment |Â
up vote
6
down vote
favorite
There are two square tubes of the same length, material and wall thickness, one being 40x40mm, the other 20x20mm. How does their strength against bending compare?
On the one hand, the cross-sectional area of the 40mm tube is 4 times bigger. On the other hand, I guess it is mostly the two walls that are in the same plane as the bending force that work against it. The other two walls provide very little resistance comparing to the first two. And because the surface of the "working" walls of the 40mm tube is twice as much bigger as those of the 20mm one, I am inclined to think that the 40mm tube is roughly twice as much stronger. How does this conclusion stand?
More specifically, this question is about galvanized steel tubes with 1.6mm thick walls:
I am looking at using them as two or more horizontal beams 2.4m long which will take some evenly distributed load. The vertical space is limited so that I would rather use four/eight/sixteen 20mm tubes than two 40mm ones. But how many 20mm tubes would be roughly equal to two 40mm ones?
structural-engineering materials steel strength
edited Aug 9 at 8:36
Sam Farjamirad
440211
asked Aug 9 at 8:30
Greendrake
1335
1
The 40x40 mm, has a bigger vertical distance to neural axis, it means more normal stress caused by bending moment. Take a look here strucalc.com/normal-stress-bending-stress-shear-stress
– Sam Farjamirad
Aug 9 at 8:39
How are the horizontal beams themselves supported... are they fixed at the ends? Resting on the ends? Supported underneath by yet more cross-members?
– elliot svensson
Aug 9 at 16:31
@elliotsvensson They rest on the ends 40mm on each side.
– Greendrake
Aug 9 at 23:36
add a comment |Â
up vote
6
down vote
favorite
up vote
6
down vote
favorite
There are two square tubes of the same length, material and wall thickness, one being 40x40mm, the other 20x20mm. How does their strength against bending compare?
On the one hand, the cross-sectional area of the 40mm tube is 4 times bigger. On the other hand, I guess it is mostly the two walls that are in the same plane as the bending force that work against it. The other two walls provide very little resistance comparing to the first two. And because the surface of the "working" walls of the 40mm tube is twice as much bigger as those of the 20mm one, I am inclined to think that the 40mm tube is roughly twice as much stronger. How does this conclusion stand?
More specifically, this question is about galvanized steel tubes with 1.6mm thick walls:
I am looking at using them as two or more horizontal beams 2.4m long which will take some evenly distributed load. The vertical space is limited so that I would rather use four/eight/sixteen 20mm tubes than two 40mm ones. But how many 20mm tubes would be roughly equal to two 40mm ones?
structural-engineering materials steel strength
edited Aug 9 at 8:36
Sam Farjamirad
440211
asked Aug 9 at 8:30
Greendrake
1335
There are two square tubes of the same length, material and wall thickness, one being 40x40mm, the other 20x20mm. How does their strength against bending compare?
On the one hand, the cross-sectional area of the 40mm tube is 4 times bigger. On the other hand, I guess it is mostly the two walls that are in the same plane as the bending force that work against it. The other two walls provide very little resistance comparing to the first two. And because the surface of the "working" walls of the 40mm tube is twice as much bigger as those of the 20mm one, I am inclined to think that the 40mm tube is roughly twice as much stronger. How does this conclusion stand?
More specifically, this question is about galvanized steel tubes with 1.6mm thick walls:
I am looking at using them as two or more horizontal beams 2.4m long which will take some evenly distributed load. The vertical space is limited so that I would rather use four/eight/sixteen 20mm tubes than two 40mm ones. But how many 20mm tubes would be roughly equal to two 40mm ones?
structural-engineering materials steel strength
edited Aug 9 at 8:36
Sam Farjamirad
440211
asked Aug 9 at 8:30
Greendrake
1335
edited Aug 9 at 8:36
Sam Farjamirad
440211
edited Aug 9 at 8:36
Sam Farjamirad
440211
edited Aug 9 at 8:36
Sam Farjamirad
440211
440211
asked Aug 9 at 8:30
Greendrake
1335
asked Aug 9 at 8:30
Greendrake
1335
asked Aug 9 at 8:30
Greendrake
1335
1335
1
The 40x40 mm, has a bigger vertical distance to neural axis, it means more normal stress caused by bending moment. Take a look here strucalc.com/normal-stress-bending-stress-shear-stress
– Sam Farjamirad
Aug 9 at 8:39
How are the horizontal beams themselves supported... are they fixed at the ends? Resting on the ends? Supported underneath by yet more cross-members?
– elliot svensson
Aug 9 at 16:31
@elliotsvensson They rest on the ends 40mm on each side.
– Greendrake
Aug 9 at 23:36
add a comment |Â
1
The 40x40 mm, has a bigger vertical distance to neural axis, it means more normal stress caused by bending moment. Take a look here strucalc.com/normal-stress-bending-stress-shear-stress
– Sam Farjamirad
Aug 9 at 8:39
How are the horizontal beams themselves supported... are they fixed at the ends? Resting on the ends? Supported underneath by yet more cross-members?
– elliot svensson
Aug 9 at 16:31
@elliotsvensson They rest on the ends 40mm on each side.
– Greendrake
Aug 9 at 23:36
1
1
The 40x40 mm, has a bigger vertical distance to neural axis, it means more normal stress caused by bending moment. Take a look here strucalc.com/normal-stress-bending-stress-shear-stress
– Sam Farjamirad
Aug 9 at 8:39
The 40x40 mm, has a bigger vertical distance to neural axis, it means more normal stress caused by bending moment. Take a look here strucalc.com/normal-stress-bending-stress-shear-stress
– Sam Farjamirad
Aug 9 at 8:39
How are the horizontal beams themselves supported... are they fixed at the ends? Resting on the ends? Supported underneath by yet more cross-members?
– elliot svensson
Aug 9 at 16:31
How are the horizontal beams themselves supported... are they fixed at the ends? Resting on the ends? Supported underneath by yet more cross-members?
– elliot svensson
Aug 9 at 16:31
@elliotsvensson They rest on the ends 40mm on each side.
– Greendrake
Aug 9 at 23:36
@elliotsvensson They rest on the ends 40mm on each side.
– Greendrake
Aug 9 at 23:36
add a comment |Â
3 Answers
3
active
oldest
votes
up vote
8
down vote
accepted
Flexural capacity is based on the stress at the extreme fiber (the point farthest vertically from the neutral axis).
$$sigma = fracMyI$$
Or, rearranging
$$M = sigma fracIy = sigma S$$
M = moment capacity
$sigma$ = allowable stress
S = I/y = section modulus
So the relative capacity of the tubes will be a function of their section moduli.
For a rectangular tube:
$$S = fracBH^26 - fracbh^36H$$
answered Aug 9 at 9:51
CableStay
1,517623
So it turns out 40mm square tube is 4.5 times stronger than 20mm one!
– Greendrake
Aug 9 at 10:10
2
+1 for your beautiful diagrams. These are always a welcome sight.
– Mark
Aug 9 at 10:34
add a comment |Â
up vote
2
down vote
The second moment area of a hollow square section, I, can be calculated by subtracting the I of hole from the I of square.
We call the outer side ,a and inner side b.
$$ I = frac (a^4-b^4)12 $$
And $$ S = frac (a^4-b^4)6a $$
So roughly, you need three of the smaller size beams for each 40x40 mm.
But since the deflection of the smaller beam will be THREE times more under the same load, we need to check on that not to become too excessive.
answered Aug 9 at 14:26
kamran
2,4971410
Given that load is distributed evenly across all beams, three small beams will deflect the same as one big one, correct?
– Greendrake
Aug 9 at 23:41
Yes the deflection is the same but the support will be springy and soft. Remember when place a load p on a beam even gently, you dynamically applying three to four times the p. It is the reason soft ceilings even though structurally sound have excessive vibration under the foot traffic. The utility of a beam is to be able to support the load while it doesn't deform or vibrate much. But if we want to design a suspension system it is desirable for it to be very soft.
– kamran
Aug 10 at 0:30
add a comment |Â
up vote
0
down vote
Contrary to popular opinion, the smaller tubes will provide greater strength under load than the larger tubes. Here's why:
Four 20x20mm tubes have a greater cross sectional surface area than a single 40×40mm tube. In fact, the 20×20 occupy the same external volume but have treble the strength according to the physical size. This issue is directly related to the square-cube law. Because the work of fracture required to cause structural failure in all 4 of the 20×20 tubes is higher, it is therefore more reliable. If the 4 20×20 tubes are spot welded together, their structural strength may be decreased but the rigidity of the whole resultant 40×40 tube will be 3 times stronger because of the internal cross of material measuring 3.2mm thick in the shape of a "+". Obviously the 40×40mm tube does not have that rigidity or ability to resist deflection as well as 4 20×20mm tubes welded together as a 40×40mm tube. Another advantage of the 20×20mm is their ability to resist torsion better when welded together as a box tube.
answered Aug 15 at 3:16
Rhodie
3049
add a comment |Â
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
8
down vote
accepted
Flexural capacity is based on the stress at the extreme fiber (the point farthest vertically from the neutral axis).
$$sigma = fracMyI$$
Or, rearranging
$$M = sigma fracIy = sigma S$$
M = moment capacity
$sigma$ = allowable stress
S = I/y = section modulus
So the relative capacity of the tubes will be a function of their section moduli.
For a rectangular tube:
$$S = fracBH^26 - fracbh^36H$$
answered Aug 9 at 9:51
CableStay
1,517623
So it turns out 40mm square tube is 4.5 times stronger than 20mm one!
– Greendrake
Aug 9 at 10:10
2
+1 for your beautiful diagrams. These are always a welcome sight.
– Mark
Aug 9 at 10:34
add a comment |Â
up vote
8
down vote
accepted
Flexural capacity is based on the stress at the extreme fiber (the point farthest vertically from the neutral axis).
$$sigma = fracMyI$$
Or, rearranging
$$M = sigma fracIy = sigma S$$
M = moment capacity
$sigma$ = allowable stress
S = I/y = section modulus
So the relative capacity of the tubes will be a function of their section moduli.
For a rectangular tube:
$$S = fracBH^26 - fracbh^36H$$
answered Aug 9 at 9:51
CableStay
1,517623
So it turns out 40mm square tube is 4.5 times stronger than 20mm one!
– Greendrake
Aug 9 at 10:10
2
+1 for your beautiful diagrams. These are always a welcome sight.
– Mark
Aug 9 at 10:34
add a comment |Â
up vote
8
down vote
accepted
up vote
8
down vote
accepted
Flexural capacity is based on the stress at the extreme fiber (the point farthest vertically from the neutral axis).
$$sigma = fracMyI$$
Or, rearranging
$$M = sigma fracIy = sigma S$$
M = moment capacity
$sigma$ = allowable stress
S = I/y = section modulus
So the relative capacity of the tubes will be a function of their section moduli.
For a rectangular tube:
$$S = fracBH^26 - fracbh^36H$$
answered Aug 9 at 9:51
CableStay
1,517623
Flexural capacity is based on the stress at the extreme fiber (the point farthest vertically from the neutral axis).
$$sigma = fracMyI$$
Or, rearranging
$$M = sigma fracIy = sigma S$$
M = moment capacity
$sigma$ = allowable stress
S = I/y = section modulus
So the relative capacity of the tubes will be a function of their section moduli.
For a rectangular tube:
$$S = fracBH^26 - fracbh^36H$$
answered Aug 9 at 9:51
CableStay
1,517623
answered Aug 9 at 9:51
CableStay
1,517623
answered Aug 9 at 9:51
CableStay
1,517623
answered Aug 9 at 9:51
CableStay
1,517623
1,517623
So it turns out 40mm square tube is 4.5 times stronger than 20mm one!
– Greendrake
Aug 9 at 10:10
2
+1 for your beautiful diagrams. These are always a welcome sight.
– Mark
Aug 9 at 10:34
add a comment |Â
So it turns out 40mm square tube is 4.5 times stronger than 20mm one!
– Greendrake
Aug 9 at 10:10
2
+1 for your beautiful diagrams. These are always a welcome sight.
– Mark
Aug 9 at 10:34
So it turns out 40mm square tube is 4.5 times stronger than 20mm one!
– Greendrake
Aug 9 at 10:10
So it turns out 40mm square tube is 4.5 times stronger than 20mm one!
– Greendrake
Aug 9 at 10:10
2
2
+1 for your beautiful diagrams. These are always a welcome sight.
– Mark
Aug 9 at 10:34
+1 for your beautiful diagrams. These are always a welcome sight.
– Mark
Aug 9 at 10:34
add a comment |Â
up vote
2
down vote
The second moment area of a hollow square section, I, can be calculated by subtracting the I of hole from the I of square.
We call the outer side ,a and inner side b.
$$ I = frac (a^4-b^4)12 $$
And $$ S = frac (a^4-b^4)6a $$
So roughly, you need three of the smaller size beams for each 40x40 mm.
But since the deflection of the smaller beam will be THREE times more under the same load, we need to check on that not to become too excessive.
answered Aug 9 at 14:26
kamran
2,4971410
Given that load is distributed evenly across all beams, three small beams will deflect the same as one big one, correct?
– Greendrake
Aug 9 at 23:41
Yes the deflection is the same but the support will be springy and soft. Remember when place a load p on a beam even gently, you dynamically applying three to four times the p. It is the reason soft ceilings even though structurally sound have excessive vibration under the foot traffic. The utility of a beam is to be able to support the load while it doesn't deform or vibrate much. But if we want to design a suspension system it is desirable for it to be very soft.
– kamran
Aug 10 at 0:30
add a comment |Â
up vote
2
down vote
The second moment area of a hollow square section, I, can be calculated by subtracting the I of hole from the I of square.
We call the outer side ,a and inner side b.
$$ I = frac (a^4-b^4)12 $$
And $$ S = frac (a^4-b^4)6a $$
So roughly, you need three of the smaller size beams for each 40x40 mm.
But since the deflection of the smaller beam will be THREE times more under the same load, we need to check on that not to become too excessive.
answered Aug 9 at 14:26
kamran
2,4971410
Given that load is distributed evenly across all beams, three small beams will deflect the same as one big one, correct?
– Greendrake
Aug 9 at 23:41
Yes the deflection is the same but the support will be springy and soft. Remember when place a load p on a beam even gently, you dynamically applying three to four times the p. It is the reason soft ceilings even though structurally sound have excessive vibration under the foot traffic. The utility of a beam is to be able to support the load while it doesn't deform or vibrate much. But if we want to design a suspension system it is desirable for it to be very soft.
– kamran
Aug 10 at 0:30
add a comment |Â
up vote
2
down vote
up vote
2
down vote
The second moment area of a hollow square section, I, can be calculated by subtracting the I of hole from the I of square.
We call the outer side ,a and inner side b.
$$ I = frac (a^4-b^4)12 $$
And $$ S = frac (a^4-b^4)6a $$
So roughly, you need three of the smaller size beams for each 40x40 mm.
But since the deflection of the smaller beam will be THREE times more under the same load, we need to check on that not to become too excessive.
answered Aug 9 at 14:26
kamran
2,4971410
The second moment area of a hollow square section, I, can be calculated by subtracting the I of hole from the I of square.
We call the outer side ,a and inner side b.
$$ I = frac (a^4-b^4)12 $$
And $$ S = frac (a^4-b^4)6a $$
So roughly, you need three of the smaller size beams for each 40x40 mm.
But since the deflection of the smaller beam will be THREE times more under the same load, we need to check on that not to become too excessive.
answered Aug 9 at 14:26
kamran
2,4971410
edited Aug 9 at 14:32
answered Aug 9 at 14:26
kamran
2,4971410
answered Aug 9 at 14:26
kamran
2,4971410
answered Aug 9 at 14:26
kamran
2,4971410
2,4971410
Given that load is distributed evenly across all beams, three small beams will deflect the same as one big one, correct?
– Greendrake
Aug 9 at 23:41
Yes the deflection is the same but the support will be springy and soft. Remember when place a load p on a beam even gently, you dynamically applying three to four times the p. It is the reason soft ceilings even though structurally sound have excessive vibration under the foot traffic. The utility of a beam is to be able to support the load while it doesn't deform or vibrate much. But if we want to design a suspension system it is desirable for it to be very soft.
– kamran
Aug 10 at 0:30
add a comment |Â
Given that load is distributed evenly across all beams, three small beams will deflect the same as one big one, correct?
– Greendrake
Aug 9 at 23:41
Yes the deflection is the same but the support will be springy and soft. Remember when place a load p on a beam even gently, you dynamically applying three to four times the p. It is the reason soft ceilings even though structurally sound have excessive vibration under the foot traffic. The utility of a beam is to be able to support the load while it doesn't deform or vibrate much. But if we want to design a suspension system it is desirable for it to be very soft.
– kamran
Aug 10 at 0:30
Given that load is distributed evenly across all beams, three small beams will deflect the same as one big one, correct?
– Greendrake
Aug 9 at 23:41
Given that load is distributed evenly across all beams, three small beams will deflect the same as one big one, correct?
– Greendrake
Aug 9 at 23:41
Yes the deflection is the same but the support will be springy and soft. Remember when place a load p on a beam even gently, you dynamically applying three to four times the p. It is the reason soft ceilings even though structurally sound have excessive vibration under the foot traffic. The utility of a beam is to be able to support the load while it doesn't deform or vibrate much. But if we want to design a suspension system it is desirable for it to be very soft.
– kamran
Aug 10 at 0:30
Yes the deflection is the same but the support will be springy and soft. Remember when place a load p on a beam even gently, you dynamically applying three to four times the p. It is the reason soft ceilings even though structurally sound have excessive vibration under the foot traffic. The utility of a beam is to be able to support the load while it doesn't deform or vibrate much. But if we want to design a suspension system it is desirable for it to be very soft.
– kamran
Aug 10 at 0:30
add a comment |Â
up vote
0
down vote
Contrary to popular opinion, the smaller tubes will provide greater strength under load than the larger tubes. Here's why:
Four 20x20mm tubes have a greater cross sectional surface area than a single 40×40mm tube. In fact, the 20×20 occupy the same external volume but have treble the strength according to the physical size. This issue is directly related to the square-cube law. Because the work of fracture required to cause structural failure in all 4 of the 20×20 tubes is higher, it is therefore more reliable. If the 4 20×20 tubes are spot welded together, their structural strength may be decreased but the rigidity of the whole resultant 40×40 tube will be 3 times stronger because of the internal cross of material measuring 3.2mm thick in the shape of a "+". Obviously the 40×40mm tube does not have that rigidity or ability to resist deflection as well as 4 20×20mm tubes welded together as a 40×40mm tube. Another advantage of the 20×20mm is their ability to resist torsion better when welded together as a box tube.
answered Aug 15 at 3:16
Rhodie
3049
add a comment |Â
up vote
0
down vote
Contrary to popular opinion, the smaller tubes will provide greater strength under load than the larger tubes. Here's why:
Four 20x20mm tubes have a greater cross sectional surface area than a single 40×40mm tube. In fact, the 20×20 occupy the same external volume but have treble the strength according to the physical size. This issue is directly related to the square-cube law. Because the work of fracture required to cause structural failure in all 4 of the 20×20 tubes is higher, it is therefore more reliable. If the 4 20×20 tubes are spot welded together, their structural strength may be decreased but the rigidity of the whole resultant 40×40 tube will be 3 times stronger because of the internal cross of material measuring 3.2mm thick in the shape of a "+". Obviously the 40×40mm tube does not have that rigidity or ability to resist deflection as well as 4 20×20mm tubes welded together as a 40×40mm tube. Another advantage of the 20×20mm is their ability to resist torsion better when welded together as a box tube.
answered Aug 15 at 3:16
Rhodie
3049
add a comment |Â
up vote
0
down vote
up vote
0
down vote
Contrary to popular opinion, the smaller tubes will provide greater strength under load than the larger tubes. Here's why:
Four 20x20mm tubes have a greater cross sectional surface area than a single 40×40mm tube. In fact, the 20×20 occupy the same external volume but have treble the strength according to the physical size. This issue is directly related to the square-cube law. Because the work of fracture required to cause structural failure in all 4 of the 20×20 tubes is higher, it is therefore more reliable. If the 4 20×20 tubes are spot welded together, their structural strength may be decreased but the rigidity of the whole resultant 40×40 tube will be 3 times stronger because of the internal cross of material measuring 3.2mm thick in the shape of a "+". Obviously the 40×40mm tube does not have that rigidity or ability to resist deflection as well as 4 20×20mm tubes welded together as a 40×40mm tube. Another advantage of the 20×20mm is their ability to resist torsion better when welded together as a box tube.
answered Aug 15 at 3:16
Rhodie
3049
Contrary to popular opinion, the smaller tubes will provide greater strength under load than the larger tubes. Here's why:
Four 20x20mm tubes have a greater cross sectional surface area than a single 40×40mm tube. In fact, the 20×20 occupy the same external volume but have treble the strength according to the physical size. This issue is directly related to the square-cube law. Because the work of fracture required to cause structural failure in all 4 of the 20×20 tubes is higher, it is therefore more reliable. If the 4 20×20 tubes are spot welded together, their structural strength may be decreased but the rigidity of the whole resultant 40×40 tube will be 3 times stronger because of the internal cross of material measuring 3.2mm thick in the shape of a "+". Obviously the 40×40mm tube does not have that rigidity or ability to resist deflection as well as 4 20×20mm tubes welded together as a 40×40mm tube. Another advantage of the 20×20mm is their ability to resist torsion better when welded together as a box tube.
answered Aug 15 at 3:16
Rhodie
3049
answered Aug 15 at 3:16
Rhodie
3049
answered Aug 15 at 3:16
Rhodie
3049
answered Aug 15 at 3:16
Rhodie
3049
3049
add a comment |Â
add a comment |Â
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1
The 40x40 mm, has a bigger vertical distance to neural axis, it means more normal stress caused by bending moment. Take a look here strucalc.com/normal-stress-bending-stress-shear-stress
– Sam Farjamirad
Aug 9 at 8:39
How are the horizontal beams themselves supported... are they fixed at the ends? Resting on the ends? Supported underneath by yet more cross-members?
– elliot svensson
Aug 9 at 16:31
@elliotsvensson They rest on the ends 40mm on each side.
– Greendrake
Aug 9 at 23:36