A helical spring B is placed inside the coils of a second helical spring^. havin

Question

A helical spring B is placed inside the coils of a second helical spring^. having the same number of coils and free length. The springs are made of the same material. The composite spring is compressed by an axial load of 2300 N which is shared between them. The mean diameters of the spring A and B are 100 mm and 70 mm respectively and wire diameters are 13 mm and 8 mm respectively. Find the load taken and the maximum stress in each spring. [Ans. WA = 1670 N; WB = 630 N; sigma A = 230 MPa; sigma B = 256 MPa]

Explanation / Answer

Here WA + WB = 2300

Also

WA = (RB/RA)^3 *WB

Therefore

WA =((70)^3*13^4)/(100^3*8^4)) * WB

WA = 2.4WB

Therefore

2.4WB + WB = 2300

WB   630 N

WA = 2300 - 630

= 1670 N

Stress A = 16WARA/(dA^3)

= (16*1670*50*10^-3)/(pi*(13*10^-3)^3)

230 MPa

Similarlily

B = 16WBRB/(dB^3)

= (16*630*35*10^-3)/(pi*(8*10^-3)^3)

256 MPa

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