A 40 mm diameter bar has been machined from an AISI 1050 cold-drawn bar. This part is to withstand a fluctuating tensile load varying from 0 to 70 kN. Because of the ends, and the fillet radius, a fatigue stress-concentration factor $K_f$ is 1.85 for $10^6$ or larger life. Find $sigma_a$ and $sigma_m$ and the factor of safety guarding against fatigue and first-cycle yielding, using (a) the Gerber fatigue line and (b) the ASME-elliptic fatigue line.
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Given that the fluctuating tensile load varies from 0 to 70 kN (0 to 70,000 N) and the diameter of the bar is 40 mm (0.04 m), we first need to calculate the stress in the bar. Stress (σ) is given by the formula σ = F/A, where F is the force and A is the Show more…
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(a) A 25 mm diameter bar is subjected to an axial tensile load of 100 kN. Under the action of this load, a 200 mm gauge length is found to extend 0.19 x 10^-3 mm. Determine the modulus of elasticity for the bar material. (b) If, in order to reduce weight whilst keeping the external diameter constant, the bar is bored axially to produce a cylinder of uniform thickness, what is the maximum diameter of bore possible given that the maximum allowable stress is 240 MN/m^2? The load can be assumed to remain constant at 100 kN. (c) What will be the change in the outside diameter of the bar under the limiting stress quoted in (b)? (E = 210 GN/m^2 and ν = 0.3).
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'A bar ABC of length L consists of two parts of equal lengths but different diameters. Segment AB has diameter d1= 100 mm while segment BC has diameter d2 60 mm: Both segments have length L/2 = 0.6 m. A longitudinal hole of diameter d is drilled through segment AB for one-half of its length (distance L/4= 0.3 m) The bar is made of plastic having modulus of elasticity E 4.0 GPa: Compressive loads P= 110 kN act at the ends of the bar: If the shortening of the bar is limited to 8.0 mm; what is the maximum allowable diameter dmax (mm) of the hole? dmax'
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