1. Mean diameter D, inside diameter Din and spring index 2. The expected stress at the operating load of W= 70 N. 3. The deflection of the spring under the load W= 70 N.
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- given: - initial length of steel tube= 200mm - aluminum rod is 0.175 mm shorter than steel tube. - l=0.175 mm properties of materials - ALUMINUM: coefficient of linear expansion: 23x10-6 /C modulus of elasticitty= 70 GPa - STEEL Coeficient of linear expansion: 11x10-6 /C Modulus of elasticity: 200 GPa *NEGLECT SHEAR AND LATERAL DEFORMATIONS. REQUIRED: - which of the two members will have contact on the rigid switch first and at what temperature - axial stress that would develop on 1st member after the 2nd material fills in the gap safely. PLS. SHOW FULL SOLUTIONS WITH BRIEF EXPLANATIONS AND NECESSARY DIAGRAMS...A typical compound helical spring consists of two coaxial closed coil spring having the following particulars: Outer spring 1: Mean coil diameter = 100mm Wire diameter = 12.5 mm No. of turns = 15 Inner spring 2: Mean coil diameter = 75 mm No. of turns = 20 Both the springs are of equal initial length when unloaded. If the working stress for each spring is to be the same. The modulus of rigidity for the wire material is 80 GPa. Calculate the following: a. Diameter of steel wire for inner spring b. Stiffness of inner spring c. Stiffness of the compound springTwo coaxial closed coil springs make up a standard compound helical spring. When both springs are unloaded, they are not of equal length. If each spring's working stress is to be the same. The wire material has an 80 GPa modulus of rigidity. Find the diameter of steel wire for inner spring and stiffness of inner spring. Also find the stiffness of the compound spring. Outer spring a: Mean coil diameter = 100 mm No. of turns = 15 Wire diameter = 12.5 mm Inner spring b: mean coil diameter = 75 mm No. of turns = 20