A solid square rod is cantilevered at one end. The rod is 0.6 m long and supports a completely reversing transverse load at the other end of + / - 2 kN. The material is AISI 1080 hot-rolled steel. If the rod must support this load for 104 cycles with a factor of safety of 1.5, what dimension should the square cross section have? Neglect any stress concentrations at the support end. Hint: Initially, assume the size endurance limit modification factor is kb=0.85; then check this assumption after the actual size is determined.

Answer / Explanation:

Given the data:

Length of the rod (L) = 0.6m

Load; F = ± 2 KN

Ultimate strength; S u t = 770 M P a = 770 / 6.89 kpsi = 112 kpsi

Yield strength; S y = 420 M P a

Design factor = 1.5

Surface factor; k a = 0.488

Size factor; k b = 0.85

Now by using mechanical engineering design data book,

We calculate for fatigue strength

So, Se¹ = 0.5 * 770

Se¹ = 385 M P a

K a = 57.7 * 770⁻⁰⁷¹⁸

Kₐ = 0.488

Assume Kb = 0.85 (check later)

Se = Kₐ Kb Se¹

Se = 0.488 * 0.85 * 385

S e = 160 M P a

By using design data book,

a = (f Sut) ² / Se

b = - 1 / 3 log f Sut / Se

S f = ₐ N ᵇ

Sf = 2553 (10⁴) ⁻ ⁰.²⁰⁰⁵

S f = 403 M P a

The expression for stress,

M max = 2000 * 0.6

M max = 1200 Newton per meter

= M (b/2) / I = M (b/2) / bb³/12

= 6M / b³ = 6 x 1200 / b³

= 7200 / b³ Pa

Now by comparing strength and stress,

S f = n * σ a

= 403 * 10 ⁶

= 1.5 * 7200 / b³

where b = 0.0299 m

b = 30 m m

So, we assume the size factor, so the expression for checking the size factor is,

By using design data book,

de = 0.808 (hb) ∧ 1/2

de = 0.808 b

de = 0.808 * 30

de = 24.2

Kb = (24.2 / 7.62) ∧-0.107

Kb = 0.88

We have assume 0.85 which is nearer to the 0.88 and it is acceptable.

So the dimension of the square is 30 mm