Spring Analysis
- 1. Mechanical Engineering Technology Spring Analysis
- 2. Stephanie Ulman 1 Executive Summary There are many things that needs to done before doing an analysis on a spring. The outside diameter, inside diameter, spring length and the wire’s width needs to be measured. Also the number of coils in the spring and the ends of the wire have to under consideration. From the previous measurements the standard diameter and the allowable stress can be determined. The number of active coils, mean diameter, pitch, pitch angle; based on the pitch, number of active coils, wire diameter, and end condition free length; % difference of the free lengths, solid height, spring index, coil clearance when the operating length is 50%, spring rate and maximum shear stress were calculated from knowing the measurements and the allowable stress and standard diameter. Table 1 below shows the values that were measured and calculated. LF 1.98in OD 0.445in ID 0.371in N 12.75 coils Dw 0.032 in Dw(standard) 0.0348 in σallow 140 ksi Dm 0.4102in C 11.79 KF 1.12 Na 9.75 coils P 0.192 in θ 8.47° G 10.5 (106) psi k 2.86 lb in Ls 0.4437 in Fs 4.39 lb
- 3. Stephanie Ulman 2 τs 121.9 ksi LF (With P, Na, and Dw) 1.976 in % error (% (error in the LF) 0.2% Table 1
- 4. Stephanie Ulman 3 Table of Contents Executive Summary Introduction Objective Given Find Solution Step 1 standard diameter (Dw(standard) ) Step 2 allowable stress (σallow) Step 3 mean diameter (Dm) Step 4 spring index (c) Step 5 wahl factor (KF) Step 6 number of active coils (Na) Step 7 pitch (P) Step 8 pitch angle (θ) Step 9 shear modulus (G) Step 10 spring rate (k) Step 11 solid length (Ls) Step 12 solid force (Fs) Step 13 stress (τs)) Step 14 free length (LF) based on pitch (P), number of active coils (Na), and width of the diameter (Dw) Step 15 % error in the Free lengths (LF) Step 16 coil clearance when the operating length (Lo) is 50% of the free length (LF) Conclusion Appendix
- 5. Stephanie Ulman 4 Introduction A spring is a flexible element that can have a force on it. Meaning springs can be pushed or pulled. Most springs are made of steel. When force is placed on a spring stress is created. The maximum stress can’t be higher than the allowable stress. To calculate the allowable stress see figures 19-8 through 19-13 in the textbook. Objective A spring analysis was performed to find out if the spring is safe under an allowable stress. To do this measurements and calculations was performed. These are the measurements that were taken: Outside diameter OD Inside diameter ID Spring Length LF Wire’s width Dw Number of coils N Ends of the wire Calculations can be performed from these measurements that are above. These are the calculations that need to be done. Standard diameter (Dw(standard) ) Allowable stress (σallow) Mean diameter (Dm) Spring index (c) Wahl factor (KF) Number of active coils (Na) Pitch (P) Pitch angle (θ) Shear modulus (G) Spring rate (k) Length (Ls) Solid force (Fs) Stress (τs)) Free length (LF) based on pitch (P), number of active coils (Na), and width of thdiameter (Dw) % error in the Free lengths (LF) Coil clearance when the operating length (Lo) is 50% of the free length (LF)
- 6. Stephanie Ulman 5 Given The measurement of the spring is below. Squared Ends LF = 1.98in OD = 0.445in ID = 0.371in Dw = 0.032 in N = 12.75 coils The picture of the spring is in figure 1. Figure 1 Find Standard diameter (Dw(standard) ) Allowable stress (σallow) Mean diameter (Dm) Spring index (c) Wahl factor (KF) Number of active coils (Na) Pitch (P) Pitch angle (θ) Shear modulus (G) Spring rate (k) Length (Ls) Solid force (Fs) Stress (τs)) Free length (LF) based on pitch (P), number of active coils (Na), and width of thdiameter (Dw) % error in the Free lengths (LF) Coil clearance when the operating length (Lo) is 50% of the free length (LF)
- 7. Stephanie Ulman 6 Solution Step 1: (calculate the standard diameter for the spring)) In the textbook from figure 19-2 the closest diameter to the Dw valve that was measured is located below. The standard value from the textbook is called Dw(standard). Dw = 0.032 in Dw(standard) = 0.0348 in This value above was chosen because it is the closest to the Dw valve. Step 2: (Determine the allowable stress (σallow)) From figure 19-13 located in the textbook the value Dw = 0.0348in is found. From this value the line was traced up to the bold line that said high service. After finding this out it was traced back to the stress. The allowable stress is located below. σallow =140 ksi Step 3: (Calculate the mean diameter Dm) Dm = OD − Dw Dm = 0.445in − 0.0348in = 0.4102in Dm = 0.4102in Step 4: (Calculate the spring index (c)) C = Dm Dw C = 0.4102in 0.0348in = 11.79 Step 5: (Calculate the wahl factor (KF)) From the previous step (step 4) the spring index was found. From Figure 19-4 from the textbook the spring index was traced up to the bold line then traced to the what factor (KF). The KF factor is located below. KF= 1.12
- 8. Stephanie Ulman 7 Step 6: (Calculate the number of active coils (Na)) The spring has squared ends so the equations that is used has a 3 in it. Na = N − 3 Na = 12.75 coils − 3 = 9.75 coils Step 7: (Calculate the Pitch (P)) P = LF − 3Dw Na P = 1.98in−3(0.0348in) 9.75coils = 0.192 in Step 8: (Calculate the pitch angle (θ)) θ = tan−1 P π ∙ Dm θ = tan−1 0.192in π ∙ 0.4102in = 8.47° Step 9: (Determine the shear modulus (G)) The metal of the spring was found by taking a magnet up to it. The spring was found to magenatic so therefore it is steel. The G value for steel is 10.5 (106) psi. This value is from the table 19-4 from the textbook. G =10.5 (106) psi Step 10: (Calculate the spring rate (k)) k = G ∙ Dw 8 ∙ Na ∙ C3 k = 10.5(106 )psi ∙ 0.0348in 8 ∙ 9.75 coils ∙ (11.79)3 k = 2.86 lb in
- 9. Stephanie Ulman 8 Step 11: (Calculate the solid length (Ls)) Ls = N ∙ Dw Ls = 12.75 coils ∙ 0.0348 in = 0.4437 in Step 12: (Calculate the solid force (Fs)) 𝐅𝐬 = 𝐊(𝐋 𝐅 − 𝐋 𝐒) 𝐅𝐬 = 𝟐. 𝟖𝟔 𝐥𝐛 𝐢𝐧 (𝟏. 𝟗𝟖 𝐢𝐧 − 𝟎. 𝟒𝟒𝟑𝟕 𝐢𝐧) = 𝟒. 𝟑𝟗 𝐥𝐛 Step 13: (Calculate the stress (τs)) τs = 8 ∙ KF ∙ F ∙ Dm πDw 2 τs = 8 ∙ 1.12 ∙ 4.39 lb ∙ 11.79 π ∙ (0.0348 in)2 = 121.9 ksi This is safe because it is lower than 140ksi which is the allowable stress that was found earlier. Step 14: (Free length (LF) based on Pitch (P), Number of active coils (Na), and width of the diameter (Dw)) 𝐿 𝐹 = P ∙ Na + 3𝐷 𝑤 𝐿 𝐹 = 0.192in ∙ 9.75 coils + 3(0.0348in) = 1.976 in This has 3Dw because of the squared ends on the spring. Step 15: (Calculate the % error in the Free lengths (LF)) % 𝑒𝑟𝑟𝑜𝑟 = 𝑚𝑒𝑎𝑠𝑢𝑟𝑒𝑑 − 𝑐𝑎𝑙𝑐𝑢𝑙𝑎𝑡𝑒𝑑 𝑐𝑎𝑙𝑐𝑢𝑙𝑎𝑡𝑒𝑑 × 100 % 𝑒𝑟𝑟𝑜𝑟 = 1.98 − 1.976 1.976 × 100 = 0.2% The % error is going to really small it is different because of the rounding error in the calculations.
- 10. Stephanie Ulman 9 Step 16: (Calculate the coil clearance when the operating length (Lo) is 50% of the free length (LF)) 𝐿 𝑜 = 0.5 ∙ 1.98𝑖𝑛 = 0.99𝑖𝑛 𝐶𝐶 = 𝐿 𝑜 − 𝐿 𝑠 𝑁𝑎 𝐶𝐶 = 0.99𝑖𝑛 − 0.4437𝑖𝑛 9.75 𝑐𝑜𝑖𝑙𝑠 = 0.057 𝐶𝐶 ≥ 𝐷 𝑤 10 = 0.0348 10 = 0.00348 The coil clearance is ok because 0.00348 is less than 0.057. Conclusion In conclusion the spring is safe because the stress is under the allowable stress. Appendix Mott, L. Robert. Machine Elements in Mechanical Design.4th ed. New Jersey: Pearson Hall. Print.