If a helical tension spring consisting of 12 live coils of ¼ inch steel wire and of 1 ¼ inch mean diameter is attached to the end of another helical tension spring of 18 live coils of ¼ inch steel wire and of 1 ¾ inch mean diameter, calculate the largest load (kN) that may be applied to the spring assembly without exceeding a shearing stress of 70 000 psi in the springs
To calculate the largest load that may be applied to the spring assembly, we need to use Hooke's Law and the formula for the shear stress in a helical spring.
First, let's calculate the shear stress in a helical spring using the formula:
τ = (8 * F * R) / (π * d^3 * n)
Where:
τ = Shear stress
F = Load applied on the spring
R = Spring radius (mean diameter / 2)
d = Wire diameter
n = Number of active coils
For the first spring:
Number of active coils (n1) = 12
Mean diameter (D1) = 1 ¼ inch = 1.25 inch
Wire diameter (d1) = ¼ inch = 0.25 inch
For the second spring:
Number of active coils (n2) = 18
Mean diameter (D2) = 1 ¾ inch = 1.75 inch
Wire diameter (d2) = ¼ inch = 0.25 inch
We'll start by calculating the shear stress for the first spring:
R1 = D1 / 2 = 1.25 inch / 2 = 0.625 inch
τ1 = (8 * F) / (π * (0.25 inch)^3 * 12)
τ1 = (8 * F) / (π * 0.015625 * 12)
τ1 = (8 * F) / 0.1885
Next, let's calculate the shear stress for the second spring:
R2 = D2 / 2 = 1.75 inch / 2 = 0.875 inch
τ2 = (8 * F) / (π * (0.25 inch)^3 * 18)
τ2 = (8 * F) / (π * 0.015625 * 18)
τ2 = (8 * F) / 0.2827
Now, we need to determine the largest load that can be applied without exceeding the shear stress limit of 70,000 psi. We'll set the shear stress equations equal to the limit and solve for F:
τ1 = 70,000 psi
(8 * F) / 0.1885 = 70,000
F = (70,000 * 0.1885) / 8
F1 = 1,639.5 pounds
τ2 = 70,000 psi
(8 * F) / 0.2827 = 70,000
F = (70,000 * 0.2827) / 8
F2 = 2,459.9 pounds
To get the total load applied to the spring assembly, we can simply add the loads of each spring:
Total load = F1 + F2
Total load = 1,639.5 pounds + 2,459.9 pounds
Total load ≈ 4,099.4 pounds
To convert the load from pounds to kilonewtons (kN), we can use the conversion factor:
1 pound = 0.00444822 kN
Total load in kN = 4,099.4 pounds * 0.00444822 kN
Total load ≈ 18.229 kN
Therefore, the largest load that may be applied to the spring assembly without exceeding a shearing stress of 70,000 psi is approximately 18.229 kN.
To calculate the largest load that may be applied to the spring assembly, we need to consider the shearing stress in the springs.
The shearing stress in a helical tension spring is given by the formula:
τ = (8 * W * R) / (π * d^3 * n)
Where:
τ is the shearing stress,
W is the applied load,
R is the radius of the mean coil diameter,
d is the wire diameter,
n is the number of active coils.
We can apply this formula separately to each spring and then sum the loads to find the total load that can be applied to the assembly.
For the first spring with 12 live coils:
R₁ = (1/4) + (1/2) = 3/4 = 0.75 inches (mean radius of the coil diameter)
d₁ = 1/4 = 0.25 inches (wire diameter)
n₁ = 12 (number of active coils)
For the second spring with 18 live coils:
R₂ = (1 3/4) + (1/2) = 2 1/4 = 2.25 inches (mean radius of the coil diameter)
d₂ = 1/4 = 0.25 inches (wire diameter)
n₂ = 18 (number of active coils)
Now we can calculate the shearing stress in each spring.
For the first spring:
τ₁ = (8 * W₁ * 0.75) / (π * (0.25)^3 * 12)
For the second spring:
τ₂ = (8 * W₂ * 2.25) / (π * (0.25)^3 * 18)
Since we want to find the largest load that both springs can handle without exceeding a shearing stress of 70,000 psi, we need to ensure that τ₁ and τ₂ are both less than or equal to 70,000 psi.
Once we have found the loads W₁ and W₂ that satisfy this condition, we can simply sum them to find the total load that can be applied to the assembly.