Two steel springs arranged in series as shown supports a load P. The upper spring has 12 turns of 25-mm-diameter wire on a mean radius of 100 mm. The lower spring consists of 10 turns of 20-mm diameter wire on a mean radius of 75 mm. If the maximum shearing stress in either spring must not exceed 200 MPa, compute the maximum value of P and the total elongation of the assembly. Use G = 83 GPa. Compute the equivalent spring constant by dividing the load by the total elongation. P
Two steel springs arranged in series as shown supports a load P. The upper spring has 12 turns of 25-mm-diameter wire on a mean radius of 100 mm. The lower spring consists of 10 turns of 20-mm diameter wire on a mean radius of 75 mm. If the maximum shearing stress in either spring must not exceed 200 MPa, compute the maximum value of P and the total elongation of the assembly. Use G = 83 GPa. Compute the equivalent spring constant by dividing the load by the total elongation. P
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Transcribed Image Text:Two steel springs arranged in series as shown supports a load P. The
upper spring has 12 turns of 25-mm-diameter wire on a mean radius of
100 mm. The lower spring consists of 10 turns of 20-mm diameter wire
on a mean radius of 75 mm. If the maximum shearing stress in either
spring must not exceed 200 MPa, compute the maximum value of P and
the total elongation of the assembly. Use G = 83 GPa. Compute the
equivalent spring constant by dividing the load by the total elongation.
P
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why did we use 3498.28 N in computing for the total elongation
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