Help me to understand something. I have an 8’ famco shear rated to shear 3/16” thick steel. That is the max. I understand the shear blade coming down is only slicing a part of the metal at a time with the rake angle of the blade. Yet I have an iron worker that will shear like 14” of 1/2” thick steel or 6” of 3/4” thick steel. Why on my iron worker can I shear a thicker piece if it is not as wide? If I am using the same logic as what is on my 8 foot shear by it being able to shear a full 8’ of its maximum rated 3/16” then my ironworker would be able to shear 14 inches of three-quarter inch thick steel instead of just half inch. The iron worker blade is cutting at an angle as well. Why the difference? Thank you
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Help me understand shearing
- Thread starter bigcreek
- Start date
As you shear you have to bend the sheared piece out of the way. The heavier the material the harder it is to bend out of the way. The wider the piece being sheared the more that you have to bend out of the way requiring more energy and force.
lg
no neat sig line
lg
no neat sig line
Not an expert by any means but I think my mental picture of this is correct.
Yes, the blade is only engaging a short section of material at a time. But that length increases for thicker materials. Imagine the contact length of the blade with the material as the hypotenuse of a right triangle that changes with the angle of the blade, but also the material thickness. If the angle of the blade is invariant, the only independent variable is the material thickness.
So imagine a hypothetical blade at a 45 degree angle (because the math is easy). You are cutting a material 1/4" thick so one side of the triangle is .25", and we know one angle is 45°, so we have enough to solve for the hypotenuse which is .35".
Now imagine we cut 1" thick material on the same shear with the same 45° blade angle. If one side of the triangle is 1", we can solve for the hypotenuse which is 1.41".
So it should be clear that cutting thicker material increases the length of the blade engagement, which necessarily increases the force required to make the cut.
However, we have been ignoring the other side of our triangle (the side perpendicular to the material's thickness) . In our hypothetical 45°-45°-90° triangle, that length is equal to the material thickness. But in reality, the rake angle on the blade is much shallower which results in a much longer third side for a given material thickness. As an extreme example, a shear with a blade angle of only 5° results in a hypotenuse of 11.47" and a third leg of 11.43". In other words, for a 1" thick piece of material, the blade will be engaging the material across a width of nearly 12 inches.
In this example, it's easier to see that if you decrease the width of the material you cut (let's say to just 6"), you will reduce the blade engagement length by nearly half, significantly reducing the force required to make the cut.
So in summary, you have a choice of either reducing material thickness, or material width to reduce the hypotenuse of your triangle, whereby reducing force required to make the cut.
Yes, the blade is only engaging a short section of material at a time. But that length increases for thicker materials. Imagine the contact length of the blade with the material as the hypotenuse of a right triangle that changes with the angle of the blade, but also the material thickness. If the angle of the blade is invariant, the only independent variable is the material thickness.
So imagine a hypothetical blade at a 45 degree angle (because the math is easy). You are cutting a material 1/4" thick so one side of the triangle is .25", and we know one angle is 45°, so we have enough to solve for the hypotenuse which is .35".
Now imagine we cut 1" thick material on the same shear with the same 45° blade angle. If one side of the triangle is 1", we can solve for the hypotenuse which is 1.41".
So it should be clear that cutting thicker material increases the length of the blade engagement, which necessarily increases the force required to make the cut.
However, we have been ignoring the other side of our triangle (the side perpendicular to the material's thickness) . In our hypothetical 45°-45°-90° triangle, that length is equal to the material thickness. But in reality, the rake angle on the blade is much shallower which results in a much longer third side for a given material thickness. As an extreme example, a shear with a blade angle of only 5° results in a hypotenuse of 11.47" and a third leg of 11.43". In other words, for a 1" thick piece of material, the blade will be engaging the material across a width of nearly 12 inches.
In this example, it's easier to see that if you decrease the width of the material you cut (let's say to just 6"), you will reduce the blade engagement length by nearly half, significantly reducing the force required to make the cut.
So in summary, you have a choice of either reducing material thickness, or material width to reduce the hypotenuse of your triangle, whereby reducing force required to make the cut.
Steve W.
Well-known member
One other way to look at it, which is rather similar to GeoBruin's way of presenting it:
He presented it with shear angles, which might also determine force needed. A steeper angle will be able to shear thicker metals. 1/2" steel will require a LOT more force than 3/16" aluminum, so will require a steeper angle shear. Look at the vertical stroke capacity of the two machines, as well as the shear angle. In your iron example (14" of 1/2" steel), if the shear angle is 45°, you will need at least a 14" stroke. For your 8' shear to have the same angle, you would need to have an 8' stroke. Since that is a bit impractical, the angle is less, meaning that you have a reduced capacity. Although the capacity is reduced, it likely requires just as much force from whatever the motive power is (hydraulic, electric, whatever).
.
He presented it with shear angles, which might also determine force needed. A steeper angle will be able to shear thicker metals. 1/2" steel will require a LOT more force than 3/16" aluminum, so will require a steeper angle shear. Look at the vertical stroke capacity of the two machines, as well as the shear angle. In your iron example (14" of 1/2" steel), if the shear angle is 45°, you will need at least a 14" stroke. For your 8' shear to have the same angle, you would need to have an 8' stroke. Since that is a bit impractical, the angle is less, meaning that you have a reduced capacity. Although the capacity is reduced, it likely requires just as much force from whatever the motive power is (hydraulic, electric, whatever).
.
Not an expert by any means but I think my mental picture of this is correct. Yes, the blade is only engaging a short section of material at a time. But that length increases for thicker materials. Imagine the contact length of the blade with the material as the hypotenuse of a right triangle that changes with the angle of the blade, but also the material thickness. If the angle of the blade is invariant, the only independent variable is the material thickness. So imagine a hypothetical blade at a 45 degree angle (because the math is easy). You are cutting a material 1/4" thick so one side of the triangle is .25", and we know one angle is 45°, so we have enough to solve for the hypotenuse which is .35". Now imagine we cut 1" thick material on the same shear with the same 45° blade angle. If one side of the triangle is 1", we can solve for the hypotenuse which is 1.41". So it should be clear that cutting thicker material increases the length of the blade engagement, which necessarily increases the force required to make the cut. However, we have been ignoring the other side of our triangle (the side perpendicular to the material's thickness) . In our hypothetical 45°-45°-90° triangle, that length is equal to the material thickness. But in reality, the rake angle on the blade is much shallower which results in a much longer third side for a given material thickness. As an extreme example, a shear with a blade angle of only 5° results in a hypotenuse of 11.47" and a third leg of 11.43". In other words, for a 1" thick piece of material, the blade will be engaging the material across a width of nearly 12 inches. In this example, it's easier to see that if you decrease the width of the material you cut (let's say to just 6"), you will reduce the blade engagement length by nearly half, significantly reducing the force required to make the cut. So in summary, you have a choice of either reducing material thickness, or material width to reduce the hypotenuse of your triangle, whereby reducing force required to make the cut.
Not an expert by any means but I think my mental picture of this is correct. Yes, the blade is only engaging a short section of material at a time. But that length increases for thicker materials. Imagine the contact length of the blade with the material as the hypotenuse of a right triangle that changes with the angle of the blade, but also the material thickness. If the angle of the blade is invariant, the only independent variable is the material thickness. So imagine a hypothetical blade at a 45 degree angle (because the math is easy). You are cutting a material 1/4" thick so one side of the triangle is .25", and we know one angle is 45°, so we have enough to solve for the hypotenuse which is .35". Now imagine we cut 1" thick material on the same shear with the same 45° blade angle. If one side of the triangle is 1", we can solve for the hypotenuse which is 1.41". So it should be clear that cutting thicker material increases the length of the blade engagement, which necessarily increases the force required to make the cut. However, we have been ignoring the other side of our triangle (the side perpendicular to the material's thickness) . In our hypothetical 45°-45°-90° triangle, that length is equal to the material thickness. But in reality, the rake angle on the blade is much shallower which results in a much longer third side for a given material thickness. As an extreme example, a shear with a blade angle of only 5° results in a hypotenuse of 11.47" and a third leg of 11.43". In other words, for a 1" thick piece of material, the blade will be engaging the material across a width of nearly 12 inches. In this example, it's easier to see that if you decrease the width of the material you cut (let's say to just 6"), you will reduce the blade engagement length by nearly half, significantly reducing the force required to make the cut. So in summary, you have a choice of either reducing material thickness, or material width to reduce the hypotenuse of your triangle, whereby reducing force required to make the cut.
Help me to understand something. I have an 8’ famco shear rated to shear 3/16” thick steel. That is the max. I understand the shear blade coming down is only slicing a part of the metal at a time with the rake angle of the blade. Yet I have an iron worker that will shear like 14” of 1/2” thick steel or 6” of 3/4” thick steel. Why on my iron worker can I shear a thicker piece if it is not as wide? If I am using the same logic as what is on my 8 foot shear by it being able to shear a full 8’ of its maximum rated 3/16” then my ironworker would be able to shear 14 inches of three-quarter inch thick steel instead of just half inch. The iron worker blade is cutting at an angle as well. Why the difference? Thank you
makes sense thank you for the explanation.Not an expert by any means but I think my mental picture of this is correct.
Yes, the blade is only engaging a short section of material at a time. But that length increases for thicker materials. Imagine the contact length of the blade with the material as the hypotenuse of a right triangle that changes with the angle of the blade, but also the material thickness. If the angle of the blade is invariant, the only independent variable is the material thickness.
So imagine a hypothetical blade at a 45 degree angle (because the math is easy). You are cutting a material 1/4" thick so one side of the triangle is .25", and we know one angle is 45°, so we have enough to solve for the hypotenuse which is .35".
Now imagine we cut 1" thick material on the same shear with the same 45° blade angle. If one side of the triangle is 1", we can solve for the hypotenuse which is 1.41".
So it should be clear that cutting thicker material increases the length of the blade engagement, which necessarily increases the force required to make the cut.
However, we have been ignoring the other side of our triangle (the side perpendicular to the material's thickness) . In our hypothetical 45°-45°-90° triangle, that length is equal to the material thickness. But in reality, the rake angle on the blade is much shallower which results in a much longer third side for a given material thickness. As an extreme example, a shear with a blade angle of only 5° results in a hypotenuse of 11.47" and a third leg of 11.43". In other words, for a 1" thick piece of material, the blade will be engaging the material across a width of nearly 12 inches.
In this example, it's easier to see that if you decrease the width of the material you cut (let's say to just 6"), you will reduce the blade engagement length by nearly half, significantly reducing the force required to make the cut.
So in summary, you have a choice of either reducing material thickness, or material width to reduce the hypotenuse of your triangle, whereby reducing force required to make the cut.
I understand that now, makes sense, thank you.One other way to look at it, which is rather similar to GeoBruin's way of presenting it:
He presented it with shear angles, which might also determine force needed. A steeper angle will be able to shear thicker metals. 1/2" steel will require a LOT more force than 3/16" aluminum, so will require a steeper angle shear. Look at the vertical stroke capacity of the two machines, as well as the shear angle. In your iron example (14" of 1/2" steel), if the shear angle is 45°, you will need at least a 14" stroke. For your 8' shear to have the same angle, you would need to have an 8' stroke. Since that is a bit impractical, the angle is less, meaning that you have a reduced capacity. Although the capacity is reduced, it likely requires just as much force from whatever the motive power is (hydraulic, electric, whatever).
.